The Immirzi parameter
John Baez
Something really cool just happened.
First for some historical background. Starting in 1997, Abhay
Ashtekar, Kirill Krasnov and I spent about 3 years working on
some papers in which we tried to calculate the entropy of
black holes using loop quantum gravity:
http://www.arXiv.org/abs/gr-qc/9710007
http://www.arXiv.org/abs/gr-qc/0005126
Much earlier, Hawking showed by semiclassical calculations that
the entropy of a black should equal 1/4 the area of its event
horizon, measured in Planck units. Our calculation also said
that the entropy was proportional to the area, at least for black
holes much bigger than the Planck mass. However, we only got
the constant of proportionality to be 1/4 if we set the Immirzi
parameter equal to ln(2) / pi sqrt(3).
The Immirzi parameter??
Yes! This is a strange new constant that's built into loop
quantum gravity. You can think of it as determining the
fundamental quantum of area, i.e. the smallest possible unit
of area. We don't yet know how to derive this parameter from
first principles, so we are free to set it to whatever we want
to get this black hole entropy calculation to work.
Of course, that's a bit unsatisfying. One would like loop
quantum gravity to predict the Immirzi parameter all on
its own, without relying on Hawking's calculation. But it was
hard to see how this would work, especially since the number
ln 2 / pi sqrt(3) seems so weird.
But on Sunday, a postdoc named Olaf Dreyer at the Perimeter
Insitute, who had been a student of Ashtekar, came out with
an amazing paper that changes everything:
http://www.arXiv.org/abs/gr-qc/0211076
In this paper, he calculates the Immirzi parameter in a clever
new way, using numerical results on the vibrational modes of
*classical* black holes. And his answer agrees with the value
obtained by Ashtekar, Krasnov and myself... but only if we
switch to using SO(3) instead of SU(2) as the gauge group
for loop quantum gravity.
If we use SO(3), our calculation changes only slightly, but
the end result is that the Immirzi parameter needs to be
ln 3 / 2 pi sqrt(2). This number looks just as bizarre
as the other one... but Dreyer has done a calculation
which derives this value in a *completely different way*.
Dreyer's new method only uses a tiny bit of information about
loop quantum gravity - and it doesn't use Hawking's work at
all. It's actually very similar to Bohr's early calculation of
the spectrum of hydrogen.
But the tooth-gnashingly nerve-wracking exciting thing about it is
that it relies heavily on computer calculations of the frequencies
of the vibrational modes of a black hole. These calculations,
done by a fellow named Nollert, say that if we consider highly
damped modes, the frequency of these modes approaches 0.04371235
divided by black hole's mass (in Planck units) in the limit where
the damping approaches infinity. Of course, since these are
numerical calculations, the number here is not exact.
The cool part is, Dreyer's calculation of the Immirzi parameter
gives the same answer as the SO(3) loop quantum gravity calculation
if the *exact* number is:
ln(3) / 8 pi = 0.043712394070757472250...
This agrees with 0.04371235 to 6 significant figures!
But is the agreement for real, or just a coincidence?
I hope we'll find out soon. If the numbers really match,
we'll have two completely different calculations that both give
the same formula for the Immirzi parameter, and thus the
fundamental quantum of area: a calculation using loop quantum
gravity, and a semiclassical calculation a la Bohr based on the
properties of vibrating black holes.
There are a lot of possible problems, and a lot of things
here I don't understand at all. But, all of a sudden that
weird Immirzi parameter is starting to smell a lot sweeter.
So, I wish everyone a happy Thanksgiving!
Squark
news:<as4bit$a9n$1...@glue.ucr.edu>...
> The Immirzi parameter??
>
> Yes! This is a strange new constant that's built into loop
> quantum gravity.
Btw, am I right thinking it arises through the change of
quantization variables from a complex connection to a real one,
which was originally needed solely because of the lack of
appropriate functional analysis theory for the space of complex
connection? Does it mean that if anyone eventually comes up
with the right complex theory he might be able to compute the
Immirzi parameter? That assumes, I suppose, the Immirzi
parameter really makes sense, i.e., it might in principle turn
out the "correct" complex theory is inequivalent to the real
one for all values of the Immirzi parameter.
> In this paper, he calculates the Immirzi parameter in a clever
> new way, using numerical results on the vibrational modes of
> *classical* black holes. And his answer agrees with the value
> obtained by Ashtekar, Krasnov and myself... but only if we
> switch to using SO(3) instead of SU(2) as the gauge group
> for loop quantum gravity.
What is the meaning of this for the 4D formulation? That is,
the SU(2) group appears when you pass from the manifestly
invariant connection formulation of general relativity to an
ADM style "3 + 1" formulation. What should you change in the
original "unpolarized" formulation to make that one SO(3)
instead? Does it involve using the actual Lorentz group as the
gauge group rather than its double cover? If so, doesn't it
pose a problem incorporating fermion matter into such a
theory?
> Dreyer has done a calculation
> which derives this value in a *completely different way*.
Both ways are essentially using a semiclassical approximation.
I suppose, though, that the reason this calculation is so exciting
is that it strengthens the claim LQG for the "correct" Immirzi
parameter has classical general relativity as its classical limit.
Btw, while we're on it, is the LQG black-hole entropy calculation
still valid only for non-rotating black-holes? If so, is anyone
currently conducting research of what happens there with the
rotating ones?
> So, I wish everyone a happy Thanksgiving!
Happy Channuka :-)
Best regards,
Squark
------------------------------------------------------------------
Write to me using the following e-mail:
Skvark_N...@excite.exe
(just spell the particle name correctly and use "com" rather than
"exe")
Mike Mowbray
John Baez wrote:
> > [...] his answer agrees with the value obtained by
> > Ashtekar, Krasnov and myself... but only if we switch
> > to using SO(3) instead of SU(2) as the gauge group
> > for loop quantum gravity.
Is this then a prediction that spin-half particles are not
relevant to (physical) quantum gravity at all, or just
a prediction that they don't contribute to area?
- MikeM.
John Baez
In article <939044f.02120...@posting.google.com>,
Squark <fii...@yahoo.com> wrote:
>ba...@galaxy.ucr.edu (John Baez) wrote in message
>news:<as4bit$a9n$1...@glue.ucr.edu>...
>> The Immirzi parameter??
>>
>> Yes! This is a strange new constant that's built into loop
>> quantum gravity.
>Btw, am I right thinking it arises through the change of
>quantization variables from a complex connection to a real one,
>which was originally needed solely because of the lack of
>appropriate functional analysis theory for the space of complex
>connection?
Exactly.
>Does it mean that if anyone eventually comes up
>with the right complex theory he might be able to compute the
>Immirzi parameter?
Perhaps! People are working on this. Freidel and Livine
have written a paper which constructs a theory of SL(2,C)
spin networks:
hep-th/0205268
Title: Spin Networks for Non-Compact Groups
Authors: Laurent Freidel (PI, ENS-Lyon), Etera R. Livine (CPT Marseille)
On the other hand, Ashtekar recently told me that there are problems
with applying this theory to quantum gravity - and that Lewandowski
has shown there's no easy fix. I still don't understand what these
problems are, or what range of "easy fixes" have been considered and
ruled out. So, it's premature for me to draw any grand conclusions.
However, Ashtekar suggested that this might mean switching to SU(2)
or SO(3) spin networks is really necessary, not just a technical trick.
Even if this is the case, it might ultimately be possible to determine
the correct value of the Immirzi parameter from within loop quantum
gravity. It's hard to tell.
In short: your question is fascinating, and people are hard at
work trying to answer it!
>That assumes, I suppose, the Immirzi
>parameter really makes sense, i.e., it might in principle turn
>out the "correct" complex theory is inequivalent to the real
>one for all values of the Immirzi parameter.
Right. Or if you're feeling even more pessimistic, you could
argue that loop quantum gravity can never possibly work, and that
the Immirzi parameter is so weird that it *proves* this.
That's what some string theorists like Lubos Motl have said -
and sometimes I've been on the verge of agreeing with them, when
I'm in a bad mood. That's why I'm so excited about Dreyer's
paper: it gives a second very *different* way of calculating
the Immirzi parameter, which gives the same answer in the SO(3)
theory. In fact, if we believe Dreyer's calculation, we can
use it to *predict* the next few decimal places of the frequency
of the highly damped vibrations of a Schwarzschild black hole -
and then we can go out and test this numerically! More importantly,
we might be able to use these ideas to fight our way to a better
understanding of the classical limit of loop quantum gravity.
>> In this paper, he calculates the Immirzi parameter in a clever
>> new way, using numerical results on the vibrational modes of
>> *classical* black holes. And his answer agrees with the value
>> obtained by Ashtekar, Krasnov and myself... but only if we
>> switch to using SO(3) instead of SU(2) as the gauge group
>> for loop quantum gravity.
>What is the meaning of this for the 4D formulation? That is,
>the SU(2) group appears when you pass from the manifestly
>invariant connection formulation of general relativity to an
>ADM style "3 + 1" formulation. What should you change in the
>original "unpolarized" formulation to make that one SO(3)
>instead? Does it involve using the actual Lorentz group as the
>gauge group rather than its double cover?
Yes, exactly.
>If so, doesn't it pose a problem incorporating fermion matter
>into such a theory?
Yes - and I mentioned in "week189" that this is why Ashtekar,
Krasnov and I did not include the SO(3) calculation in our
detailed paper on black hole entropy. So, this is another
big mystery.
There is never any shortage of problems in quantum gravity!
>> Dreyer has done a calculation
>> which derives this value in a *completely different way*.
>Both ways are essentially using a semiclassical approximation.
Right; more precisely, both use a mixture of loop quantum gravity
and semiclassical information. However, it's interesting to
compare the two. The ABCK calculation uses a *lot* of loop
quantum gravity, and also Hawking's entropy calculation, which
in turn relies on quantum field theory on a Schwarzschild background.
The Hod-Dreyer calculation uses a *little* loop quantum gravity,
a semiclassical argument a la Bohr, and also numerical results on
vibrational modes of a purely classical black hole.
If we could get all these things to fit together in a consistent
way, that would be very nice!
>I suppose, though, that the reason this calculation is so exciting
>is that it strengthens the claim LQG for the "correct" Immirzi
>parameter has classical general relativity as its classical limit.
Right! But also, it suggests lots of avenues for research,
and provides a specific thing to check: does the vibrational
frequency of highly damped modes of a vibrating nonrotating
black hole approach
ln 3 / 8 pi M,
where M is the black hole's mass?
>Btw, while we're on it, is the LQG black-hole entropy calculation
>still valid only for non-rotating black-holes?
Yes. People have extended the calculation to black holes
interacting with various forms of matter: dilatons, scalar fields,
and even non-minimally-coupled scalar fields, where Wald has noted
that the entropy should *not* be proportional to the area.
In every case the same Immirzi parameter works! However, the
rotating case is much harder. Ashtekar and collaborators have
carried out a detailed study of the "isolated horizon" boundary
conditions for a rotating black hole. The problem is to quantize
this theory.
>If so, is anyone currently conducting research of what happens
>there with the rotating ones?
Yes! - but so far it's not been easy, so we really need
your help. If we could do it, we could compare the answer to
calculations using the vibrational modes of a rotating black hole!
John Baez
Mike Mowbray <mi...@despammed.com> wrote:
>John Baez wrote:
Good question. If I knew anything about this, I'd be
too busy writing a paper on it to post anything to s.p.r.!
It's worth noting that the theory being studied in all
these calculation is pure gravity without spin-1/2 particles.
So, it's just a total mystery how the calculations should
change when you include spin-1/2 particles. A very irksome
mystery, I admit.
You can see some remarks on this in Dreyer's paper.
Steve Carlip
> But on Sunday, a postdoc named Olaf Dreyer at the Perimeter
> Insitute, who had been a student of Ashtekar, came out with
> an amazing paper that changes everything:
> http://www.arXiv.org/abs/gr-qc/0211076
> In this paper, he calculates the Immirzi parameter in a clever
> new way, using numerical results on the vibrational modes of
> *classical* black holes. And his answer agrees with the value
> obtained by Ashtekar, Krasnov and myself... but only if we
> switch to using SO(3) instead of SU(2) as the gauge group
> for loop quantum gravity.
I agree that this is a very exciting result. But it's also worth
emphasizing just how weird it is. It now shows the same very
bizarre number arising in two very different places, one purely
classical (the quasinormal modes) and one quite quantum
mechanical. But it doesn't explain, in any deep way, why this
number should arise in either place. This is reminiscent of
many of the ``coincidences'' in string theory, that give one a
good reason to believe there's something deep going on but
don't explain what.
Steve Carlip
John Baez
Steve Carlip <sjca...@ucdavis.edu> wrote:
>I agree that this is a very exciting result. But it's also worth
>emphasizing just how weird it is.
Yes: IT'S REALLY WEIRD!
It could even be a meaningless coincidence.
>It now shows the same very
>bizarre number arising in two very different places, one purely
>classical (the quasinormal modes) and one quite quantum
>mechanical. But it doesn't explain, in any deep way, why this
>number should arise in either place. This is reminiscent of
>many of the ``coincidences'' in string theory, that give one a
>good reason to believe there's something deep going on but
>don't explain what.
This is what keeps theoretical physicists going when they can't
predict the answers to experiments. :-) I have some ideas about
how to proceed. Here are some comments and questions that I
mailed to a bunch of loop quantum gravity people; maybe you
have some comments and answers!
.......................................................................
By the way, if there really is a link between the old calculation
of the Barbero-Immirzi parameter using Hawking's entropy formula and the
new calculation using quasinormal modes of black holes, it probably lies
in the fact that one can compute both the *gravitational Hawking
radiation* and the *frequency of quasinormal modes* by linearizing
Einstein's equations about the Schwarzschild solution!
And this raises an obvious question: has somebody already tried to
compute the gravitational Hawking radiation by quantizing the
quasinormal modes of a black hole? This might be the quickest
way to prove the two calculations of the Barbero-Immirzi parameter
really give the same answer!
It would be even more fun to predict "correction terms" in the
asymptotic formula for quasinormal mode frequencies coming from
spin network edges labelled by spins > 1. Wouldn't it be cool
if we could start making predictions about classical GR using
loop quantum gravity?
Jay Olson
> Btw, am I right thinking it arises through the change of
> quantization variables from a complex connection to a real one,
> which was originally needed solely because of the lack of
> appropriate functional analysis theory for the space of complex
> connection? Does it mean that if anyone eventually comes up
> with the right complex theory he might be able to compute the
> Immirzi parameter? That assumes, I suppose, the Immirzi
> parameter really makes sense, i.e., it might in principle turn
> out the "correct" complex theory is inequivalent to the real
> one for all values of the Immirzi parameter.
I'm sticking my neck out here (I don't know for sure), but I thought
that the original Ashtekar variables just correspond to "Immirzi = i"
and that this made the Hamiltonian constraint look nice, but it made
the reality conditions very non-trivial because you want to restrict
yourself to a real metric. So the way I understand it, simply
promising to use only the Ashtekar connection doesn't get you out of
the problem -- it's just a particular choice of the parameter.
> What is the meaning of this for the 4D formulation? That is,
> the SU(2) group appears when you pass from the manifestly
> invariant connection formulation of general relativity to an
> ADM style "3 + 1" formulation. What should you change in the
> original "unpolarized" formulation to make that one SO(3)
> instead? Does it involve using the actual Lorentz group as the
> gauge group rather than its double cover? If so, doesn't it
> pose a problem incorporating fermion matter into such a
> theory?
Yeah, I'm confused by this too. I seem to remember people saying
things like "spin networks tell you how to parallel transport a spin
__ particle, the spin depending on the labelling of the edge you want
to parallel transport along." Maybe this statement is too simplistic,
but it seems to make sense. So the question is immediately now "what
if you want to parallel transport a fermion?"
I wonder if this result could be due to the specifics of the computer
simulation, which uses the lorentz group as the gauge group of GR. It
makes one wonder if the result would differ exactly by the relevant
factor if the simulation allowed full SL(2,C) gauge symmetry. Of
course I have no idea how these simulations are done, so I don't even
know if this suggestion makes sense...
Also I wonder how this paper will effect the great A = j(j+1) vs. A =
j+1/2 debate... It looks like the j(j+1) side may have some
additional room to argue now, since the j+1/2 side can't reconcile
their value of the Immirzi parameter with this calculation so easily.
zirkus
> I agree that this is a very exciting result. But it's also worth
> emphasizing just how weird it is. It now shows the same very
> bizarre number arising in two very different places, one purely
> classical (the quasinormal modes) and one quite quantum
> mechanical. But it doesn't explain, in any deep way, why this
> number should arise in either place.
Also, the result might be in contradiction with this paper, but I
don't know how legitimate this paper is:
John Baez
Jay Olson <sjay...@yahoo.com> wrote:
>I'm sticking my neck out here (I don't know for sure), but I thought
>that the original Ashtekar variables just correspond to "Immirzi = i"
>and that this made the Hamiltonian constraint look nice, but it made
>the reality conditions very non-trivial because you want to restrict
>yourself to a real metric.
This is exactly right.
>So the way I understand it, simply
>promising to use only the Ashtekar connection doesn't get you out of
>the problem -- it's just a particular choice of the parameter.
That's true in a way, but if your Barbero-Immirzi parameter is
imaginary, "Immirzi = +i" and "Immirzi = -i" are geometrically
favored over all other choices: they correspond to working with
only the left-handed or only the right-handed part of an
so(3,1) tensor C = sl(2,C) + sl(2,C)
connection. So, these values have a nice conceptual meaning, unlike
all other imaginary values of the Barbero-Immirzi parameter.
In short: if you're working with imaginary values of the
Barbero-Immirzi parameter, there really is something very
special about i and -i, but if you're working with real values,
no particular nonzero real value seems special...
... until you try to use loop quantum gravity to make predictions
about black holes!
This is a weird situation, so it seems unlikely to be the
final word on the matter. But this is the situation we're
in right now.
>Yeah, I'm confused by this too. I seem to remember people saying
>things like "spin networks tell you how to parallel transport a spin
>__ particle, the spin depending on the labelling of the edge you want
>to parallel transport along." Maybe this statement is too simplistic,
>but it seems to make sense. So the question is immediately now "what
>if you want to parallel transport a fermion?"
And the answer is: "good question!" This is completely mysterious to me.
>I wonder if this result could be due to the specifics of the computer
>simulation, which uses the lorentz group as the gauge group of GR. It
>makes one wonder if the result would differ exactly by the relevant
>factor if the simulation allowed full SL(2,C) gauge symmetry. Of
>course I have no idea how these simulations are done, so I don't even
>know if this suggestion makes sense...
The "computer simulation" apparently just computes a certain
clever formula for the vibrational frequencies of a nonrotating
classical black hole. This formula comes from the classical
Einstein equations. Classically there aren't two versions of
these equations, like an "SO(3,1)" and an "SL(2,C)" version.
There's just one version! So, there appears to be nothing to tweak here.
But it's tempting to say this: in classical general relativity,
the vibrational modes are described by spherical harmonics, which
can only have *integer* angular momentum. So quantizing these
modes, we'd naively expect to get a theory more closely related
to SO(3) loop quantum gravity than SU(2) loop quantum gravity.
Of course the question remains: what if we included spin-1/2 matter?
>Also I wonder how this paper will effect the great A = sqrt(j(j+1))
>vs. A = j+1/2 debate...
Read Dreyer's paper! It's only 4 pages long, and it talks about
this issue near the end. He also talks about the A = j option,
which also has a few fans.
[Btw, I put in a square root that you left out.]
>It looks like the j(j+1) side may have some
>additional room to argue now, since the j+1/2 side can't reconcile
>their value of the Immirzi parameter with this calculation so easily.
Right! That's one thing Dreyer says. It would incredibly cool
if we could use Dreyer's calculation to determine the right area
operator *and* the right value of the Barbero-Immirzi parameter!
But I don't think we can say much for sure until we understand
what's going on a lot better than we do now. Right now, for
all we know, it's just a huge perverse coincidence that
0.04371235
is close to
ln(3) / 8 pi = 0.043712394070757472250...
At a bare minimum, we need to calculate out some more decimal
places and see if this holds up. Of course, it would be even
better to figure out what the heck is really going on!!!
Ilja Schmelzer
> But on Sunday, a postdoc named Olaf Dreyer at the Perimeter
> Insitute, who had been a student of Ashtekar, came out with
> an amazing paper that changes everything:
>
> http://www.arXiv.org/abs/gr-qc/0211076
>
> In this paper, he calculates the Immirzi parameter in a clever
> new way, using numerical results on the vibrational modes of
> *classical* black holes. And his answer agrees with the value
> obtained by Ashtekar, Krasnov and myself... but only if we
> switch to using SO(3) instead of SU(2) as the gauge group
> for loop quantum gravity.
Really nice.
> of the vibrational modes of a black hole. These calculations,
> done by a fellow named Nollert, say that if we consider highly
> damped modes, the frequency of these modes approaches 0.04371235
> divided by black hole's mass (in Planck units) in the limit where
> the damping approaches infinity. Of course, since these are
> numerical calculations, the number here is not exact.
And the physical meaning of these quasinormal modes is quite suspect,
it seems. At least this is my conclusion from Nollert, gr-qc/9602032.
If we replace the potential by a discrete one, that means if we use
another numerical discretization, we obtain something completely
different.
So, my first guess is that the mode limit 0.04371235 gives some
information about the discretization method used in the numerical
computation, not necessarily about the classical problem itself.
LQG is also in some sense a discretization.
Ilja
--
I. Schmelzer, <il...@ilja-schmelzer.net>, http://ilja-schmelzer.net
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On Mon, 2 Dec 2002 23:49:35 +0000 (UTC), ultram...@hotmail.com
(mandro) wrote:
>I think this "hopefully" is hoping for too much, I've been driven
>crazy by this
>idea and not had any resolution yet, i.e., I wanted to approximate the
>usual position operator (multiplication by x) with an arrangement like
>this,
>I.e., by a bunch of adjacent detectors all arranged in a line, but
>there's
>always the problem of more than one of them "clicking at a time"
>--by the way, whats the usual detector corresponding to the operator
>x? this is the millionth time I've asked this and no one wants to
>put their neck on the line!
Why don't you install Java on your computer and then go to
http://www.cassiopaea.org/perseus/applets/eeqt.html
and click "Start". Then go to "Simulation", add couple of detectors (at
present we offre only Gaussian shape detectors, but it is easy to add
any shape) , add a potential (Gaussian or rectangular, but later on can
be of more general shapes), if you wish, and play. You see not only
what happens to the amplitude of the wave packet, but also what
happens to its phase! A unique experience! We should have
patented it :-)
ark
--
Arkadiusz Jadczyk
http://www.cassiopaea.org/quantum_future/homepage.htm
--
Oz
>It's worth noting that the theory being studied in all
>these calculation is pure gravity without spin-1/2 particles.
>So, it's just a total mystery how the calculations should
>change when you include spin-1/2 particles. A very irksome
>mystery, I admit.
Fools go where angels fear to tread ....
Doesn't it imply that spacetime (which I guess is what you are
modelling) is 'controlled' by integer-spin particles. It would be even
nicer if it only worked for spin-2 particles (graviton?) in which case
one might wonder if spacetime isn't an artefact of spin 2 (or one, for
that matter) particle interactions.
Fermions would then be flaws in spacetime.
NB Your ucr mailhost has been bouncing valid emails.
--
Oz
This post is worth absolutely nothing and is probably fallacious.
Note: soon (maybe already) only posts via despammed.com will be accepted.
Steve Carlip
> And this raises an obvious question: has somebody already tried to
> compute the gravitational Hawking radiation by quantizing the
> quasinormal modes of a black hole?
Sort of. Look at an old paper by Jim York, Phys. Rev. D 28, 2929 (1983).
The abstract is:
An explanation of the thermal quantum radiance of black holes
discovered by Hawking is offered in terms of a black-hole metric
undergoing quantum zero-point fluctuations of zero mean in its
gravitational quasinormal modes. It is shown that such zero-
temperature fluctuations, governed by the uncertainty principle,
lead to the formation of a quantum ergosphere that enables matter
of all species to tunnel out of the hole. The results confirm that a
black hole cannot be in equilibrium at zero temperature. A dynamical
temperature is calculated by equating the mean irreducible mass
associated with the quantum ergosphere to the mean thermal energy
of a quantum oscillator with the lowest quasinormal frequency. The
result agrees with the Hawking temperature to within two per cent.
The nature of the dynamical equilibrium and the higher modes are
discussed, and it is calculated that the thermal excitations of the
resonant modes have the canonical distribution to within several
percent. A calculation of the black-hole entropy using the statistical
mechanics of the quasinormal modes yields a value
(0.27654) [h-bar] -1(16 pi M2), which is near the value usually
assumed, (0.25) [h-bar] -1(16 pi M2). Characteristic fluctuation
scales are derived. The rms energy fluctuation of the physical
("dressed") event horizon is about 1017 GeV, independent of M.
The physical metric fluctuations near the hole are of order unity
when the hole has mass ~ (0.15)(Planck mass) ~ 1.8 x 1018 GeV.
It ought to be relevant, but I haven't managed to see how...
Steve Carlip
John Baez
"It's worth noting that the theory being studied in all
these calculation is pure gravity without spin-1/2 particles.
So, it's just a total mystery how the calculations should
change when you include spin-1/2 particles. A very irksome
mystery, I admit."
Oz let out a cough and raised his hand. "Fools go where
angels fear to tread...."
The Wiz glowered at him. "Yes, but that saying was never meant
as a *suggestion*, now, was it? The idea was rather that bearing
it in mind, one might refrain from certain...."
Oz cut in before the sarcasm could get any worse. "Doesn't
this result imply that spacetime (which I guess is what you are
modelling) is 'controlled' by integer-spin particles?"
The Wiz growled, summoned up a rather hefty fireball,
raised it in his hand and... pocketed it, smiling.
"Maybe. You're making a big leap going from the spins
labelling spin network edges, which describe *area*, to
spins in the sense of *angular momentum*. Nobody has ever
made that connection precise, though of course it occurs to
everyone working on this subject. The big puzzle is, how do
we get spin-1/2 spin network edges to contribute negligibly
to the black hole entropy while still leaving a place for
half-integer spin particles in the theory? To deal with
spin-1/2 particles we need an SU(2) connection, and the
representations of SU(2) include the half-integer spin ones.
Dreyer's calculation works by switching to an SO(3) connection,
but this only knows how to parallel transport integer-spin
particles. And in these loop quantum gravity entropy
calculations, it's always spin network edges with the lowest
nonzero spin that dominate. So it's hard to see a resolution.
In a sense all these questions are premature, since the
calculations thus far concern *pure* gravity, without
spin-1/2 particles. But one wants to leave the door
open for them, as it were."
Excited by the Wizard's momentary lack of grumpiness,
Oz continued, "It would be even nicer if it only worked for
spin-2 particles (graviton?) in which case one might wonder
if spacetime isn't an artefact of spin 2 (or one, for
that matter) particle interactions."
"Hmm..." said the Wiz, beginning to frown again.
"I don't see how to get these ideas to make sense,
even though they sound appealing... and don't forget,
Dreyer's result crucially depends on that ln 3 in Hod's
magic formula, where 3 is the dimension of the spin-1
spin-1 representation of SO(3). If we went up to spin 2,
we'd get a ln 5, and we just don't see that."
Undeterred, Oz pressed on to his grand goal: "Fermions would
then be flaws in spacetime!"
The Wiz sighed. "I've been struggling to concoct theories
like that for years. But this new result seems to make it
even harder!" His hand twitched about in his pocket, feeling
for that fireball, and Oz decided to stop talking.
But then the Wiz brightened momentarily. "However, I did
receive an email from Lubos Motl today. He's excited about
Dreyer's result, and he's giving a talk at Harvard about -
guess what? - loop quantum gravity! So at least some string
theorists are starting to get interested in this stuff...."
"Lubos Motl?? Loop quantum gravity??" Oz gaped in amazement
and could say no more.
Ralph Hartley
John Baez wrote:
> Something really cool just happened.
> But on Sunday, a postdoc named Olaf Dreyer at the Perimeter
> Insitute, who had been a student of Ashtekar, came out with
> an amazing paper that changes everything:
>
> http://www.arXiv.org/abs/gr-qc/0211076
Anyone who is at all interested should definately read this paper!
It is only four pages, and is very clearly written. Even I understood it.
The basic idea is that the frequency of the strongly damped vibration modes
corespond to a change in energy, which, because the relation between mass
and area of a black hole is known, gives a change in area.
If this change is assumed to be the smallest possible unit of area, then we
know the unit of area, the Immirzi parameter. To get the correct entropy
requires that each unit of area has 3 states, not 2, so for loop quantum
gravity to work the guage group needs to be SO(3) not SU(2).
The surprising thing is that it matches at all. Since this is actually pure
mathematics (even the numerical results) it's hard to say that it's a
coincidence. Numbers are what they are, if two unrelated calculations both
gave the answer "2" you would not conclude that the two calculations were
deeply related, but ln(3)/8pi doesn't seem the same. Of course there is
still the posiblility that better numerical calculations will show that the
number isn't really ln(3)/8pi after all.
It occurs to me that the purely classical numerical calculations of the
black hole vibration modes might somehow have SO(3) instead of SU(2) built
in. After all, GR really doesn't care if you rotate something by 2pi.
Ralph Hartley
zirkus
ba...@galaxy.ucr.edu (John Baez) wrote in message news:
> Right. Or if you're feeling even more pessimistic, you could
> argue that loop quantum gravity can never possibly work, and that
> the Immirzi parameter is so weird that it *proves* this.
>
> That's what some string theorists like Lubos Motl have said -
> and sometimes I've been on the verge of agreeing with them, when
> I'm in a bad mood.
Well, string theorists have not given up even though they face some
very difficult problems. Here is an explanation that is simple enough
for any reader to understand that basically describes how string
theory has to currently deal with issues such as quantum-like
phenomena in de Sitter spacetime, the black hole information problem
etc.:
http://www.aip.org/education/sps/images/miracle.jpg
Furthermore, Master Yoda is much older and wiser than we are and he
says that one should not necessarily give up hope at least until there
is clear evidence to the contrary. He wanted me to tell you that there
is a theorist named Aleksandar Mikovic who might have some suggestions
regarding the Dreyer result. For example, Mikovic has a new paper
which shows that if you want to consider the coupling of matter in the
spin foam framework then you have to study the corresponding
discretized path integral:
Steve Carlip
zirkus <zir...@hotmail.com> wrote:
> Also, the result might be in contradiction with this paper, but I
> don't know how legitimate this paper is:
> http://arxiv.org/abs/astro-ph/0208333
This paper is fairly trivially wrong at the very beginning. The
authors claim that black hole formation violates the principle
of equivalence because ``If an event horizon were to form
during collapse, this would require the time-like world line of
the collapsing matter to become null...'' This is clearly not
the case---one can write down explicit collapsing solutions in
which a horizon forms, and the world lines of the collapsing
matter remain perfectly well-behaved and timelike. This is a
pretty weird misconception, and I don't know where it came
from, but I suspect it may have originated in a bad coordinate
choice (an attempt to use Schwarzschild coordinates all the
way through the horizon).
Steve Carlip
Oz
John Baez <ba...@galaxy.ucr.edu> writes
>The Wiz rather pompously proclaimed:
>
>"It's worth noting that the theory being studied in all
>these calculation is pure gravity without spin-1/2 particles.
>So, it's just a total mystery how the calculations should
>change when you include spin-1/2 particles. A very irksome
>mystery, I admit."
Isn't that implied when you choose SU(2)?
Umm. Er.. What's the relationship between SU(2) and SO(3)?
I presume SO(3) excludes spin1/2, but does SU(2) include integer spin
particles too?
>Oz let out a cough and raised his hand. "Fools go where
>angels fear to tread...."
[He is also a *very* slow learner ...]
>In a sense all these questions are premature, since the
>calculations thus far concern *pure* gravity, without
>spin-1/2 particles. But one wants to leave the door
>open for them, as it were."
Ah. So what you seem to be saying is that it's not impossible that it
could all work out in the wash...
>"Hmm..." said the Wiz, beginning to frown again.
>"I don't see how to get these ideas to make sense,
>even though they sound appealing... and don't forget,
>Dreyer's result crucially depends on that ln 3 in Hod's
>magic formula, where 3 is the dimension of the spin-1
>spin-1 representation of SO(3). If we went up to spin 2,
>we'd get a ln 5, and we just don't see that."
Ah. Now hang on a tick. So what would do, would be a space such that
spin1/2 is expressible and it's representation in this space is 3?
Are there not other spaces other than SU(2) where spin 1/2 particles can
exist?
>But then the Wiz brightened momentarily. "However, I did
>receive an email from Lubos Motl today. He's excited about
>Dreyer's result, and he's giving a talk at Harvard about -
>guess what? - loop quantum gravity! So at least some string
>theorists are starting to get interested in this stuff...."
>
>"Lubos Motl?? Loop quantum gravity??" Oz gaped in amazement
>and could say no more.
gasp! ..... <faints clean away> .....
.... thus avoiding the fireball ....
zirkus
> By the way, if there really is a link between the old calculation
> of the Barbero-Immirzi parameter using Hawking's entropy formula and the
> new calculation using quasinormal modes of black holes, it probably lies
> in the fact that one can compute both the *gravitational Hawking
> radiation* and the *frequency of quasinormal modes* by linearizing
> Einstein's equations about the Schwarzschild solution!
According to [1], in the kinematical case, neither Einstein equations
nor Bekenstein entropy are useful in deriving Hawking radiation (which
is not a test of QG). So what are the specific physical reasons for a
more significant consideration of Hawking radiation in the dynamical
case (I'm confused about this)?
> And this raises an obvious question: has somebody already tried to
> compute the gravitational Hawking radiation by quantizing the
> quasinormal modes of a black hole? This might be the quickest
> way to prove the two calculations of the Barbero-Immirzi parameter
> really give the same answer!
According to [2], in 4 dimensions, quasinormal frequencies of black
holes have only previously been obtained via numerical methods.
Jay Olson
standard model are bosons. I suppose that to include some gauge
fields into the spin network formalism, you just have to color the
edges additionally with reps of the gauge group in question. Is there
already a nice explanation for why all gauge fields are bosons in the
standard model? If not, this might be a really cool result! All
gauge fields are bosons because fundamental "edges" in space are
labelled by reps of SO(3) instead of SU(2).
John Baez
Oz <ozac...@despammed.com> wrote:
>Umm. Er.. What's the relationship between SU(2) and SO(3)?
SU(2) is the "double cover" of SO(3), meaning that there
are two elements of SU(2) for every element of SO(3).
Elements of SO(3) are precisely rotations in 3-dimensional space.
So, SU(2) is a funny group which distinguishes between no
rotation at all and a "360-degree rotation about any axis" -
these two elements of SU(2) that correspond to the same
element of SO(3).
>I presume SO(3) excludes spin-1/2, but does SU(2) include integer spin
>particles too?
Yes.
Every integer-spin particle is a representation of SO(3).
Every integer-spin *or* half-integer-spin particle is a representation
of SU(2).
So, for example, an electron can be described using a representation
of SU(2), but not of SO(3), because to describe what happens
when you rotate an electron, we need to distinguish between no
rotation at all and a "360-degree rotation about any axis". The
former does nothing to an electron; the latter multiplies its
wavefunction by -1.
On the other hand, a photon can be described using a representation
of SU(2) or of SO(3). The reason is that in this case we can - but
don't *need* to - distinguish between no rotation at all and a "360-degree
rotation about any axis" - since both these operations have the same
effect on a photon, i.e. no effect at all.
>>In a sense all these questions are premature, since the
>>calculations thus far concern *pure* gravity, without
>>spin-1/2 particles. But one wants to leave the door
>>open for them, as it were."
>Ah. So what you seem to be saying is that it's not impossible that it
>could all work out in the wash...
Right; everything is so confusing right now that it's premature
to feel pessimistic.
>>"Hmm..." said the Wiz, beginning to frown again.
>>"I don't see how to get these ideas to make sense,
>>even though they sound appealing... and don't forget,
>>Dreyer's result crucially depends on that ln 3 in Hod's
>>magic formula, where 3 is the dimension of the spin-1
>>spin-1 representation of SO(3). If we went up to spin 2,
>>we'd get a ln 5, and we just don't see that."
>Ah. Now hang on a tick. So what would do, would be a space such that
>spin-1/2 is expressible and it's representation in this space is 3?
Eh? If I understand your crazy hope here, what "would do" would
be if the dimension of the spin-1/2 representation were 3.
But it's not. It's 2. That's just a fact.
>Are there not other spaces other than SU(2) where spin 1/2 particles can
>exist?
Eh???
If that question made sense, the short answer would be "no".
I think we've reached the point where the best explanation is
a swift fireball....
[The Wiz reaches into his pocket.]
>>But then the Wiz brightened momentarily. "However, I did
>>receive an email from Lubos Motl today. He's excited about
>>Dreyer's result, and he's giving a talk at Harvard about -
>>guess what? - loop quantum gravity! So at least some string
>>theorists are starting to get interested in this stuff...."
>>
>>"Lubos Motl?? Loop quantum gravity??" Oz gaped in amazement
>>and could say no more.
>
>gasp! ..... <faints clean away> .....
>
>.... thus avoiding the fireball ....
Damn!
Oz
>In article <CFPfZUAa...@btopenworld.com>,
>Oz <ozac...@despammed.com> wrote:
>
>>Umm. Er.. What's the relationship between SU(2) and SO(3)?
>
>SU(2) is the "double cover" of SO(3), meaning that there
>are two elements of SU(2) for every element of SO(3).
>
>Elements of SO(3) are precisely rotations in 3-dimensional space.
Ah, hang on, do I remember some of this from way back?
Isn't the 'U' something to do with 'unitary'?
Weren't the elements some 2x2 complex matrices?
It's just I'm having trouble matching a 3-D SO(3) to a 2-D SU(2), let
alone doubled up. You wouldn't be so kind as to give example elements of
SO(3) and SU(2)?
Please, sir?
Hmmm. I seem to remember that SO(3) had three complex elements
(magnitude one) and SU(2) was a 2x2 complex matrix with det = 1(?).
Ohh sh*t, now I'm for it ....
Must learn to keep my mouth shut....
>So, SU(2) is a funny group which distinguishes between no
>rotation at all and a "360-degree rotation about any axis" -
>these two elements of SU(2) that correspond to the same
>element of SO(3).
OK, I kinda guessed that.
>>Ah. Now hang on a tick. So what would do, would be a space such that
>>spin-1/2 is expressible and it's representation in this space is 3?
>
>Eh? If I understand your crazy hope here, what "would do" would
>be if the dimension of the spin-1/2 representation were 3.
>But it's not. It's 2. That's just a fact.
Oh. Oh dear.
That means I've already been fireballed up the thread for asking a
stupid question. I rather thought SO(3) had three dimensions, and
represented (in effect) rotations about three axes. If SU(2) is a double
cover one imagined that it represented all three possible rotation axes
AND differentiated the 360 deg rotation. That is, it was 'bigger' (well,
exactly twice as big).
But there again this may be coded in the little word 'unitary', which I
can't quite remember ....
Ralph Hartley
Jay Olson wrote:
> I seem to remember people saying
> things like "spin networks tell you how to parallel transport a spin
> __ particle, the spin depending on the labelling of the edge you want
> to parallel transport along." Maybe this statement is too simplistic,
> but it seems to make sense. So the question is immediately now "what
> if you want to parallel transport a fermion?"
For the black hole calculations, is parallel transport something physical?
We are, after all, talking about edges that pass through the horizon. You
obviously can't physically transport a particle arround a loop that crosses
the horizon.
Another thing to note is that both calculations are specific to black holes
with exactly 0 rotation. Things might be very different if you relaxed that
even a little. I wouldn't guess that they would be simpler.
Ralph Hartley
zirkus
Hmm, the paper looks like it should be wrong but I can't make sense of
what the authors are trying to say, so I am going to defer to your
expertise. Also, I don't trust such strict adherence to the POE
because of considerations about possible violation of the POE from QG
and VSL cosmology.
Jay Olson
simply because connections are (lie algebra valued) 1-forms. (duh,
me)
So, the thing that's confusing me here is this. When people suggest
that particles should be "free edges" of spin networks, what kinds of
particles are we talking about? All particles or just gauge bosons or
just the leptons? Before I was thinking that these free edges
represented the leptons, but that was purely because I was thinking
about the spin (1/2 was a possibility). But now I'm confused, due to
this SO(3) business, and wondering if a "free edge" corresponds to a
"real" gauge boson, since 1) an edge (labelled appropriately) lets you
parallel transport a charged particle and 2) it now looks like we can
only have integer spin edges. Can someone help me sort this out?
eb...@lfa221051.richmond.edu
Oz <ozac...@despammed.com> wrote:
>You wouldn't be so kind as to give example elements of
>SO(3) and SU(2)?
SO(3) means the set of all orthogonal (That's the O) 3x3 (that's the 3)
matrices with determinant 1 (that's the S, which stands for "special.")
These matrices describe arbitrary rotations in three dimensions.
For instance, if you rotate 90 degrees about the y axis, you
transform the coordinates (x,y,z) into (-z,y,x). The matrix that
describes this is
0 0 -1
0 1 0
1 0 0
That's an element of SO(3). Another one is
1 0 0
0 cos(theta) sin(theta)
0 -sin(theta) cos(theta)
That describes a rotation through an angle theta about the x axis.
Elements of SU(2) are unitary (that's the U) 2x2 (that's the 2)
matrices with determinant 1 (that's the S).
I'll give examples in a minute, but first let me remind you about
unitarity. "Unitary" is the complex version of "orthogonal." A
matrix M is orthogonal if M times its own transpose is 1. It's
unitary if M times its "hermitian conjugate," which is the complex
conjugate of the transpose, is 1.
The nifty thing about orthogonal matrices is that they preserve
lengths: if M is orthogonal, and x is any real vector, then x and Mx
have the same length. For short, we can write |x| = |Mx|.
The nifty thing about unitary matrices is that they do the same thing
for complex vectors: if M is unitary and z is a complex vector, then
|z| = |Mz|. Remember that absolute values for complex numbers involve
complex conjugation; that's why unitarity involves complex
conjugation.
Here's an example of an element of SU(2):
exp(i theta/2) 0
0 exp(-i theta/2)
This one describes what happens if you rotate an electron through an
angle theta about the z-axis. If we represent the electron's spin
state by a "spinor" (a 2-dimensional vector of complex numbers
representing the spin-up and spin-down parts), then rotating it
changes the spinor by multiplying it by this matrix.
Here's another one:
0 1
-1 0
That one just rotates the electron 180 degrees about either the x or y
axis. (I'm not sure which.) Note that it turns spin-up into
spin-down: if you hit the vector (1,0) (representing spin-up) with it,
you get (0,1) (representing spin-down). And vice versa, except that
you pick up an all-important minus sign. If you apply this matrix
twice, you get minus the identity matrix. That is, if you rotate your
electron 180 degrees, then rotate it 180 degrees again about the same
axis, you get back minus what you started with. That's spin-1/2 for
you.
>That means I've already been fireballed up the thread for asking a
>stupid question. I rather thought SO(3) had three dimensions, and
>represented (in effect) rotations about three axes. If SU(2) is a double
>cover one imagined that it represented all three possible rotation axes
>AND differentiated the 360 deg rotation. That is, it was 'bigger' (well,
>exactly twice as big).
OK. Let's count dimensions. You're quite right that SO(3) is
3-dimensional. The easy way to see that is to note that elements
of SO(3) represent rotations, and to describe a rotation in three
dimensions you need to specify three numbers. (You can specify
three Euler angles, or equivalently you can use two numbers to specify
an axis and a third number to specify how far to rotate about that
axis.)
We can confirm this another way: An element of SO(3) is a 3x3 matrix,
so you have to specify nine numbers to pick one element. But any
element M has to satisfy the orthogonality equation: M times transpose
of M equals identity matrix. That's six constraints. (It's a 3x3
matrix equation, but it's automatically symmetric, so the equation
that says that this is true for element ij is the same as the equation
that says its true for element ji.) 9-6=3, so SO(3) is 3-dimensional.
It turns out that SU(2) is also three-dimensional. We can see that
with a similar counting argument, although it's a bit messier.
Each element of SU(2) is a 2x2 matrix, so you need 4 numbers to
specify it. But those are complex numbers, so you really need 8 real
numbers to specify an element of SU(2).
Now how many constraints are there on those 8 numbers? Well,
unitarity tells us that M times M-dagger is the identity matrix.
(M-dagger is the Hermitian conjugate of M. Its elements are the
complex conjugates of the elements of the transpose of M.) That gives
us four different equations for complex numbers, but the two
off-diagonal elements turn out to give the same equation, so there are
really only three. The off-diagonal one has both real and imaginary
parts, but the two diagonal elements turn out to be automatically
real, so unitarity gives us four conditions that the original 8
numbers had to satisfy. So we're down to four dimensions from the
original 8.
The "special" part of SU(2) (the determinant-1 part) gives one
more constraint, knocking the dimension down to 3. If we leave
off the S and consider U(2), we have a four-dimensional group.
The determinants of those guys all have to have magnitude
1, but they can have any phase. So picking out the ones
with determinant 1 knocks us down by one dimension.
[Puzzle: When I did the counting argument for SO(3) above, I stopped
when I had counted up the constraints required by orthogonality. I
didn't worry there about the determinant-1 condition. Why doesn't
that reduce the dimension of the group by 1? To put it another way,
O(3) and SO(3) have the same dimension. Why? What's the relation
between those two groups?]
That's a bit of a messy argument! There are no doubt dozens
of slicker arguments to show that SU(2) is three-dimensional.
The simplest one is to just use the fact that JB mentioned
earlier: SU(2) is the double cover of SO(3). That is, every
element of SO(3) corresponds to exactly two elements
of SU(2). That means that the two groups have to have the
same dimension.
-Ted
--
[E-mail me at na...@domain.edu, as opposed to na...@machine.domain.edu.]
Jeremy Henty
eb...@lfa221051.richmond.edu wrote:
> [Puzzle: When I did the counting argument for SO(3) above, I stopped
> when I had counted up the constraints required by orthogonality. I
> didn't worry there about the determinant-1 condition. Why doesn't
> that reduce the dimension of the group by 1?
Because the determinant of any element of O(3) can only be 1 or -1 .
So requiring it to be 1 does not reduce the dimensionality any
further, it just throws away one of the two disjoint parts of O(3).
In fact something similar happened in your analysis of SU(3). When
you added the constraint that the determinant be 1 you dropped one
real dimension. But in this context the determinant is
complex-valued. Why should we not drop one *complex* dimension,
ie. two real ones? Because the determinant of any element of U(3)
must have unit norm, so one of those two real dimensions had already
disappeared by the time you got to U(3).
Regards,
Jeremy Henty
--
Oh, wouldn't it be great if I *was* crazy? Then the world would be
okay!
-- Twelve Monkeys
Oz
>In article <asr2v2$usa$1...@lfa222122.richmond.edu>,
>Oz <ozac...@despammed.com> wrote:
>>You wouldn't be so kind as to give example elements of
>>SO(3) and SU(2)?
>SO(3) means the set of all orthogonal (That's the O) 3x3 (that's the 3)
>matrices with determinant 1 (that's the S, which stands for "special.")
>These matrices describe arbitrary rotations in three dimensions.
>Another one is
>
>1 0 0
>0 cos(theta) sin(theta)
>0 -sin(theta) cos(theta)
>
>That describes a rotation through an angle theta about the x axis.
OK. Let's just check if I have a few other deductions right.
It looks to me that there should be three transformations that will
generate the complete set (where s=sin(theta), c= cos(theta)) something
like this (give or take a sign)
1 0 0 0 1 0 0 0 1
0 c s c 0 s c s 0
0 -s c -s 0 c -s c 0
X Y Z
Each of which give a rotation around x, y, or z (obviously by any angle
you choose round any axis). So presumably if you have an arbitrary
transformation and you want to check if it is in SO(3) then you just:
1) Check it has determinant = 1.
2) Put it into the form XYZ, with X,Y,Z having the above form.
Where the axes are arbitrary but orthogonal.
Having read down further I presume (2) is more easily done by checking
that multiplying by the transpose gives 1.
I note that the coefficients are real.
>Elements of SU(2) are unitary (that's the U) 2x2 (that's the 2)
>matrices with determinant 1 (that's the S).
>
>I'll give examples in a minute, but first let me remind you about
>unitarity. "Unitary" is the complex version of "orthogonal."
OK.
>A
>matrix M is orthogonal if M times its own transpose is 1.
Is it?
I don't think anyone bothered to mention this previously.
It turns the use of the transpose from magic to rational.
Anyway, it's not implausible.
>It's
>unitary if M times its "hermitian conjugate," which is the complex
>conjugate of the transpose, is 1.
OK.
>The nifty thing about orthogonal matrices is that they preserve
>lengths: if M is orthogonal, and x is any real vector, then x and Mx
>have the same length. For short, we can write |x| = |Mx|.
>The nifty thing about unitary matrices is that they do the same thing
>for complex vectors: if M is unitary and z is a complex vector, then
>|z| = |Mz|. Remember that absolute values for complex numbers involve
>complex conjugation; that's why unitarity involves complex
>conjugation.
OK. But I would have thought that the length-preserving characteristic
would be det=1, and the orthogonality would be MM_t = 1.
>Here's an example of an element of SU(2):
>
>exp(i theta/2) 0
>0 exp(-i theta/2)
Ok, det =1 and the hermitian conjugate = 1. OK
>This one describes what happens if you rotate an electron through an
>angle theta about the z-axis. If we represent the electron's spin
>state by a "spinor" (a 2-dimensional vector of complex numbers
>representing the spin-up and spin-down parts), then rotating it
>changes the spinor by multiplying it by this matrix.
Ok, so a spinor is of form
(a + bi)
(c + di)
Where (a+bi) is up and ....
Hang on, there must be more criteria than that.
You can't have some arbitrary spin, it's gotta be (for e-) 1/2.
Without thought (something I'm good at) I would want a spinor to be in
something like SU(1).
>Here's another one:
>
>0 1
>-1 0
>
>That one just rotates the electron 180 degrees about either the x or y
>axis. (I'm not sure which.) Note that it turns spin-up into
>spin-down: if you hit the vector (1,0) (representing spin-up) with it,
>you get (0,1) (representing spin-down). And vice versa, except that
>you pick up an all-important minus sign. If you apply this matrix
>twice, you get minus the identity matrix. That is, if you rotate your
>electron 180 degrees, then rotate it 180 degrees again about the same
>axis, you get back minus what you started with. That's spin-1/2 for
>you.
UGH! How untidy.
You only get points if you can express an integer spin particle as a
spinor and show that the same transformation doesn't result in a sign
switch.
Of course if you do that, then I'm right in there with you.
>>That means I've already been fireballed up the thread for asking a
>>stupid question. I rather thought SO(3) had three dimensions, and
>>represented (in effect) rotations about three axes. If SU(2) is a double
>>cover one imagined that it represented all three possible rotation axes
>>AND differentiated the 360 deg rotation. That is, it was 'bigger' (well,
>>exactly twice as big).
>
>
>OK. Let's count dimensions.
<much snipping>
>We can confirm this another way: An element of SO(3) is a 3x3 matrix,
>so you have to specify nine numbers to pick one element. But any
>element M has to satisfy the orthogonality equation: M times transpose
>of M equals identity matrix. That's six constraints. (It's a 3x3
>matrix equation, but it's automatically symmetric, so the equation
>that says that this is true for element ij is the same as the equation
>that says its true for element ji.) 9-6=3, so SO(3) is 3-dimensional.
I'll believe you, it's not implausible.
>It turns out that SU(2) is also three-dimensional. We can see that
>with a similar counting argument, although it's a bit messier.
>
>Each element of SU(2) is a 2x2 matrix, so you need 4 numbers to
>specify it. But those are complex numbers, so you really need 8 real
>numbers to specify an element of SU(2).
OK
[I will try and remember that: dagger is the hemitian conjugate]
<snip counting for SU(2)>
>[Puzzle: When I did the counting argument for SO(3) above, I stopped
> when I had counted up the constraints required by orthogonality. I
> didn't worry there about the determinant-1 condition. Why doesn't
> that reduce the dimension of the group by 1? To put it another way,
> O(3) and SO(3) have the same dimension. Why? What's the relation
> between those two groups?]
Det is simply a scaling factor, does not affect the dimensions.
>That's a bit of a messy argument!
It is. Can't you do it by expressing the complex numbers in SU(2) by 2x2
matrices and then (somehow) end up with a sum of a pair of 3x3 matrices,
one of which is in SO(3) and the other also in SO(3) but 'different'? I
would expect one to be the other rotated by pi. I have, given my
splendidly elementary knowledge of matrix algebra, no hope of doing
this, but it looks plausible. Hmm, probably easier to do it the other
way round.
Anyway, I am prepared to accept this. However I won't be happy until I
see both spin 1/2 and spin 1 particles operated on by SU(2) and giving
the correct answers. I'm not very happy about having one methodology for
spin1/2 and another for spin1. Ugly.
I would also like to thank you, Ted, for a splendidly clear and well set
out expose of this small area. In particular I hope to be able to follow
up on learning about spin. I am already much clearer on why the groups
are used, and the reasons for their selection. Probably, though, my
previous efforts (and those of exasperated teachers) are slowly sinking
in.
Thank you.
Ralph Hartley
understanding the implications of Dreyer's result
(http://www.arXiv.org/abs/gr-qc/0211076).
Let's consider the total angular momentum of the horizon. I don't really
know how to calculate that from a spin net state, so let's just consider
whether or not it is an integer. If my understanding of spin nets is
correct, the spin of the BH is an integer iff the number of half integer
spin edges crossing the horizon is even.
Dreyer's calculation assumes that the most strongly damped modes of the
classical non-rotating black hole correspond to transitions that remove one
unit of area, i.e. one spin-net edge.
So the transition is, in ascii art:
| Radiation
|1
| <----->
|
______|_____ _____________
Horizon Horizon
But if the guage group is SU(2), then this cannot be. The two states of the
horizon have angular momenta differing by a half integer, so they can't
both be non-rotating. Not only that, but since this is pure gravity, the
radiation must have integer spin, so angular momentum isn't even conserved!
In order to calculate the Immirzi parameter we need to know what spin net
transitions the strongly damped BH transitions *do* correspond to. To get
that, you would need some sort of dynamics on the horizon.
The simplest allowed transitions would involve a spin one edge, or a pair
of spin 1/2 edges. It is possible that some states are completely
unreachable. If just the right number of states are, the entropy could
still come out right. This would amount to pairs of SU(2) edges acting like
"composite" SO(3) edges.
I still don't know if you can get a factor of log_2(3), which is what you need.
SO(3) still looks good. The same concern doesn't appear to apply, and
Dreyer's calculation works. What could save SU(2) is that it might come out
looking just like SO(3) when you take angular momentum constriants into
account.
Ralph Hartley
John Baez
zirkus <zir...@hotmail.com> wrote:
>ba...@galaxy.ucr.edu (John Baez) wrote in message
>news:<asguvr$r3r$1...@glue.ucr.edu>...
>> By the way, if there really is a link between the old calculation
>> of the Barbero-Immirzi parameter using Hawking's entropy formula and the
>> new calculation using quasinormal modes of black holes, it probably lies
>> in the fact that one can compute both the *gravitational Hawking
>> radiation* and the *frequency of quasinormal modes* by linearizing
>> Einstein's equations about the Schwarzschild solution!
>According to [1], in the kinematical case, neither Einstein equations
>nor Bekenstein entropy are useful in deriving Hawking radiation (which
>is not a test of QG). So what are the specific physical reasons for a
>more significant consideration of Hawking radiation in the dynamical
>case (I'm confused about this)?
People usually calculate the *electromagnetic* Hawking radiation
by quantizing *Maxwell's equations* on the spacetime describing
a Schwarzschild black hole. But you can also calculate the
*gravitational* Hawking radiation by quantizing the *Einstein
equations* linearized about the solution describing a Schwarzschild
black hole. In both cases you get radiation emitted at the
same temperature, leading to the same formula for the entropy
of a black hole.
So, my point had nothing to do with "kinematical" versus "dynamical".
Instead, I was noting that both Hawking's calculation of the
entropy of a black hole and Nollert's calculation of the
vibrational frequencies of a black hole can be thought of as
spinoffs of the same theory: Einstein's equations linearized about
the Schwarzschild solution. This might eventually make it less
mysterious that they lead to the same formula for the Immirzi
parameter in loop quantum gravity. However, right now I'm afraid
it *is* mysterious - and Steve Carlip's post about York's
calculations shows just *how* mysterious!
John Baez
In article <6f477239.02120...@posting.google.com>,
Jay Olson <sjay...@yahoo.com> wrote:
>So, the thing that's confusing me here is this. When people suggest
>that particles should be "free edges" of spin networks, what kinds of
>particles are we talking about? All particles or just gauge bosons or
>just the leptons?
The fermions (leptons and quarks) and possibly the Higgs.
I.e., everything *except* the forces themselves, which are
described by connections. See:
John Baez and Kirill Krasnov,
Quantization of diffeomorphism-invariant theories with fermions,
Jour. Math. Phys. 39 (1998), 1251-1271,
http://www.arXiv.org/abs/hep-th/9703112
and also Thiemann's papers on loop quantum gravity
and the Standard Model, which are also available on the
preprint arXiv.
Tim S
> eb...@lfa221051.richmond.edu writes
>> A matrix M is orthogonal if M times its own transpose is 1.
> Is it?
> I don't think anyone bothered to mention this previously.
> It turns the use of the transpose from magic to rational.
> Anyway, it's not implausible.
Here's a sketch of how it works. Suppose x and y are column vectors. Their
inner product (i.e. their dot product x.y) is given by x_t y, where x_t is
the row vector equal to the transposition of x, and we're doing matrix
multiplication.
Suppose we apply a matrix M to all the column vectors in our space, giving
Mx, My etc. What happens to the inner products?
Well, the inner product of the transformed vectors Mx and My is going to be
(Mx)_t (My). A theorem of matrix multiplication says that the transpose of
the product of two matrices A and B is given by (AB)_t = B_t A_t. I.e. take
the transposes and multiply them in the opposite order. So we have
(Mx)_t (My)
= (x_t M_t) (My)
= x_t (M_t M) y (since matrix multiplication is associative).
Obviously, if M_t M = I -- i.e. the matrix M is orthogonal, then this
= x_t y
That is, transformation by an orthogonal matrix preserves the inner product.
Since the inner product of two vectors is the product of their lengths and
the cosine of the angle between them, preserving inner products
automatically preserves all lengths and angles, which is the property we
want for orthogonal transformations.
The determinant gives the volume scaling induced by the matrix. Obviously if
all lengths and angles are preserved, then so are volumes. The only funny
thing that can happen to volumes is if we have a reflection, which preserves
lengths and angles but inverts the volume, giving a determinant of -1. Hence
O(3) consists of two separate 3-dimensional pieces, one containing
transformations which involve reflections, and one containing
transformations (including the identity) which don't. That latter are pure
rotations and make up SO(3).
>> It's
>> unitary if M times its "hermitian conjugate," which is the complex
>> conjugate of the transpose, is 1.
> OK.
A unitary matrix preserves the _complex_ inner product x* y, where I'm
using * to mean Hermitian conjugate. I.e. we're taking the complex
conjugate of the row vector got by transposing the complex column
vector x, then matrix-multiplying it by y.
>> The nifty thing about orthogonal matrices is that they preserve
>> lengths: if M is orthogonal, and x is any real vector, then x and Mx
>> have the same length. For short, we can write |x| = |Mx|.
>> The nifty thing about unitary matrices is that they do the same thing
>> for complex vectors: if M is unitary and z is a complex vector, then
>> |z| = |Mz|. Remember that absolute values for complex numbers involve
>> complex conjugation; that's why unitarity involves complex
>> conjugation.
> OK. But I would have thought that the length-preserving characteristic
> would be det=1, and the orthogonality would be MM_t = 1.
No. Length preservation has the side-effect that |det M| = 1. That is, the
absolute magnitude of det M is 1. If M is real, this means that det M = 1 or
det M = -1. If M can be complex, all we can conclude is that
det M = exp (i phi) for some arbitrary phase phi.
> Without thought (something I'm good at) I would want a spinor to be in
> something like SU(1).
With thought, on the other hand, you can work out that SU(1) is a very
small, boring group consisting only of the number 1. That can't be what you
mean.
Quantum mechanically, we want to be able to multiply both components of our
spinor
(a + bi)
(c + di)
by any arbitrary non-zero complex number (but the same number for both
components) and get, as a result, just another way of writing the same
state. However, when we're worrying about the effects of SU(2) on spinors,
it's easier to pick a particular complex vector with the right ratio between
the components -- e.g. a complex vector of unit norm -- and act on that.
When we are trying to calculate the component of angular momentum in a given
direction (is this what you mean by the "spin"?) we divide out the norm of
the vector anyway.
>> Here's another one:
>>
>> 0 1
>> -1 0
>>
>> That one just rotates the electron 180 degrees about either the x or y
>> axis. (I'm not sure which.) Note that it turns spin-up into
>> spin-down: if you hit the vector (1,0) (representing spin-up) with it,
>> you get (0,1) (representing spin-down). And vice versa, except that
>> you pick up an all-important minus sign. If you apply this matrix
>> twice, you get minus the identity matrix. That is, if you rotate your
>> electron 180 degrees, then rotate it 180 degrees again about the same
>> axis, you get back minus what you started with. That's spin-1/2 for
>> you.
> UGH! How untidy.
> You only get points if you can express an integer spin particle as a
> spinor and show that the same transformation doesn't result in a sign
> switch.
Er...but a spin 1 particle isn't a spinor -- it's a vector! A vector in 3D
space, to be precise. To work out how it transforms under an element of
SU(2), pick the corresponding element of SO(3) and apply it as you normally
would to a vector. Obviously, rotation by 360 degrees gives you back your
original vector with no nasty minus signs.
Tim
Ralph Hartley
Ralph Hartley wrote:
> It is possible that some states are completely
> unreachable. If just the right number of states are, the entropy could
> still come out right.
This bit is not right. The entropy calculation is over all states.
Restrictions on the dinamics only come into it trough the value of the
Immirzi parameter.
My argument changes it by a factor of sqrt(2)/sqrt(3), but what we "need"
is ln(2)/ln(3).
Ralph Hartley
zirkus
> For example, Mikovic has a new paper
> which shows that if you want to consider the coupling of matter in the
> spin foam framework then you have to study the corresponding
> discretized path integral:
>
> http://arxiv.org/abs/gr-qc/0210051
It still seems to me that Dreyer's suggestion that SO(3) could be the
correct gauge group may not run into trouble with incorporating
fermionic matter because the suggestion could still be compatible with
the above paper. See the remarks by Mikovic about the tensor category
of irreps of subgroup H = SO(3) e.g. in the Conclusion, and his
earlier work on SO(3) and coset space G/H spin networks in
(gr-qc/0202026). I don't see any reason yet why Dreyer's suggestion
should be incompatible with Mikovic's work, but perhaps I am missing
something.
Also, you might want to read a new paper by Kunstatter (gr-qc/0212014)
which builds on Dreyer's result.
Someone should look to see if Kunstatter's new paper is related to the
work of Jim York that Steve Carlip mentioned earlier. As Ralph Hartley
asked, what would happen if you looked at the older classical
numerical calculations of quasimodes with SO(3) instead of SU(2) built
in?
I asked earlier, "According to (hep-th/0106111), in the kinematical
case, neither Einstein equations nor Bekenstein entropy are useful in
deriving Hawking radiation (which is not a test of QG). So what are
the specific physical reasons for a more significant consideration of
Hawking radiation in the dynamical case (I'm confused about this) ?"
Now consider this abstract:
http://arxiv.org/abs/gr-qc/0011024
I wonder if this could be somehow related to Carlip's work, perhaps
even to the speculated connection with Carlip's work suggested at the
end of Kunstatter's new paper. I'm not really familiar with Carlip's
work so maybe one of you (hopefully, Carlip himself) could comment on
this.
I am certainly no great expert on spin foams physics (although I can
understand some of the math), so I am going to try to think of a
relevant derivation of Hawking radiation in the dynamic case within a
string theory context, and also about some comments that Lubos Motl
sent me.
Aaron Bergman
sjay...@yahoo.com (Jay Olson) wrote:
> Hey, I wonder if this could explain why all the gauge fields in the
> standard model are bosons.
Gauge fields are 1-forms. By spin-statistics, they must be bosons.
Aaron
--
Aaron Bergman
<http://www.princeton.edu/~abergman/>
<http://aleph.blogspot.com>
Matt McIrvin
Oz <aco...@btopenworld.com> wrote:
> John Baez <ba...@galaxy.ucr.edu> writes
>
> >Eh? If I understand your crazy hope here, what "would do" would
> >be if the dimension of the spin-1/2 representation were 3.
> >But it's not. It's 2. That's just a fact.
>
> Oh. Oh dear.
> That means I've already been fireballed up the thread for asking a
> stupid question. I rather thought SO(3) had three dimensions, and
> represented (in effect) rotations about three axes. If SU(2) is a
double
> cover one imagined that it represented all three possible rotation
axes
> AND differentiated the 360 deg rotation. That is, it was 'bigger'
(well,
> exactly twice as big).
The groups SO(3) and SU(2) are both three-dimensional spaces. But
SU(2) has a two-dimensional *representation*, and SO(3) does not.
SU(2) has all the same representations as SO(3) (which all have
odd-numbered dimensions), and then in addition SU(2) has other
representations with even-numbered dimension that represent
half-integer spins. That's what being twice as big gets it.
--
Matt McIrvin http://world.std.com/~mmcirvin/
Oz
>The groups SO(3) and SU(2) are both three-dimensional spaces. But
>SU(2) has a two-dimensional *representation*, and SO(3) does not.
>
>SU(2) has all the same representations as SO(3) (which all have
>odd-numbered dimensions), and then in addition SU(2) has other
>representations with even-numbered dimension that represent
>half-integer spins. That's what being twice as big gets it.
Ah. That's a characteristically pithy, direct and to the point
observation. Now 'all' I have to do is see it.
Hmmmm.
Let's look at an element of SO(3) (c=cos(t), s=sin(t))
1 0 0
0 c s
0 -s c
And SU(2)
exp(i theta/2) 0
0 exp(-i theta/2)
hmmm one has an irresistible urge to rewrite this as
(ignoring irritating 1/2's)
c + i s 0
0 c - is
hmmm now what? Lessee
c s is -s
+
-s c s -is
I have no idea what the latter matrix is telling me.
Which is really irritating, it looked like it was just going to fall
out.
Anyway, run out of time.
Toby Bartels
>Ted Bunn wrote:
>>A matrix M is orthogonal if M times its own transpose is 1.
>Is it?
>I don't think anyone bothered to mention this previously.
>It turns the use of the transpose from magic to rational.
>Anyway, it's not implausible.
It's been mentioned; you've even written it down yourself.
It's on your Lie theory crib sheet, at <http://math.ucr.edu/~toby/Oz/Lie/>.
-- Toby
Oz
>>Oz:
>> OK. But I would have thought that the length-preserving characteristic
>> would be det=1, and the orthogonality would be MM_t = 1.
>
>No. Length preservation has the side-effect that |det M| = 1. That is, the
>absolute magnitude of det M is 1. If M is real, this means that det M = 1 or
>det M = -1. If M can be complex, all we can conclude is that
>
>det M = exp (i phi) for some arbitrary phase phi.
Ahh. Good point.
>> Without thought (something I'm good at) I would want a spinor to be in
>> something like SU(1).
>
>With thought, on the other hand, you can work out that SU(1) is a very
>small, boring group consisting only of the number 1. That can't be what you
>mean.
I rather thought it wouldn't be. It was to give the idea.
>Quantum mechanically, we want to be able to multiply both components of our
>spinor
>
>(a + bi) = A (I assume this is a matrix)
>(c + di)
>
>by any arbitrary non-zero complex number (but the same number for both
>components) and get, as a result, just another way of writing the same
>state. However, when we're worrying about the effects of SU(2) on spinors,
>it's easier to pick a particular complex vector with the right ratio between
>the components -- e.g. a complex vector of unit norm -- and act on that.
I think you have lost me here.
I *think* you are saying that for some complex scalar a then
aA is the same state as A,
which is presumably a definition (more or less)
Now if B is in SU(2) I *guess* you are saying BA is different state to A
But if B can be rewritten as aC (a=scalar, C=spinor)
then the "a" is redundant so you tend to a methodology where everything
is, er, a cmoplex vector of unit norm. Ie is effectively = 1?
>> UGH! How untidy.
>> You only get points if you can express an integer spin particle as a
>> spinor and show that the same transformation doesn't result in a sign
>> switch.
>
>Er...but a spin 1 particle isn't a spinor -- it's a vector!
I would be surprised if you couldn't express a vector in spinor form.
I would expect some zeros to appear. Perhaps
(a + bi)
(0 )
Job done (although this one doesn't feel right).
>A vector in 3D
>space, to be precise. To work out how it transforms under an element of
>SU(2), pick the corresponding element of SO(3) and apply it as you normally
>would to a vector.
Que? A simple example would help.
>Obviously, rotation by 360 degrees gives you back your
>original vector with no nasty minus signs.
So far no points ....
John Baez
Oz <ozac...@despammed.com> wrote:
>John Baez <ba...@galaxy.ucr.edu> writes
>>SU(2) is the "double cover" of SO(3), meaning that there
>>are two elements of SU(2) for every element of SO(3).
>>
>>Elements of SO(3) are precisely rotations in 3-dimensional space.
>Ah, hang on, do I remember some of this from way back?
You should... but *do* you?
>Isn't the 'U' something to do with 'unitary'?
Yes, precisely, just as the 'O' means 'orthogonal'.
The 'S' means 'special' - that is, determinant = 1.
>Weren't the elements some 2x2 complex matrices?
Yes: 2x2 unitary matrices with determinant = 1. SU(2). Get it?
>It's just I'm having trouble matching a 3-D SO(3) to a 2-D SU(2), let
>alone doubled up.
Yes, it's sort of wonderful and surprising, isn't it?
You can't possibly understand spin in a deep way without
a good gut feeling for how this works. You can read about
this in enormous detail in my book "Gauge Fields, Knots
and Gravity".
>You wouldn't be so kind as to give example elements of
>SO(3) and SU(2)?
Any matrix describing a rotation is an element of SO(3).
So, for example, a rotation of angle t counterclockwise
about the z axis:
cos t -sin t 0
sin t cos t 0
0 0 1
There are two elements of SU(2) corresponding to the above
element of SO(3) One of them is
exp(it/2) 0
0 exp(-it/2)
and the other is minus this matrix.
Of course this is of limited interest until you know the
recipe whereby *any* element of SO(3) comes from two elements
of SU(2); then you can check that this is an example.
>Please, sir?
>
>Hmmm. I seem to remember that SO(3) had three complex elements
>(magnitude one) and SU(2) was a 2x2 complex matrix with det = 1(?).
What the...? I thought people were just telling you that
SO(3) consisted of 3x3 *real* matrices O which are orthogonal:
O O^t = 1 (t = transpose)
and have determinant = 1:
det(O) = 1.
Where did "three complex elements" come from???
>Ohh sh*t, now I'm for it ....
>Must learn to keep my mouth shut....
Either that or keep your ears open and remember what you're told!
[Another fireball. WHAM!]
Oz
>In article <asr2v2$usa$1...@lfa222122.richmond.edu>,
>Oz <ozac...@despammed.com> wrote:
>
>Yes, it's sort of wonderful and surprising, isn't it?
>You can't possibly understand spin in a deep way without
>a good gut feeling for how this works. You can read about
>this in enormous detail in my book "Gauge Fields, Knots
>and Gravity".
One of those books that makes everything hugely simple and enormously
reasonable whilst you read it, and then somehow it's all terribly
unclear ten minutes after you have shut it. I put this down to my
profound ignorance of the detail.
>>You wouldn't be so kind as to give example elements of
>>SO(3) and SU(2)?
>
>Any matrix describing a rotation is an element of SO(3).
>So, for example, a rotation of angle t counterclockwise
>about the z axis:
>
> cos t -sin t 0
> sin t cos t 0
> 0 0 1
>
>There are two elements of SU(2) corresponding to the above
>element of SO(3) One of them is
>
> exp(it/2) 0
> 0 exp(-it/2)
>
>and the other is minus this matrix.
>
>Of course this is of limited interest until you know the
>recipe whereby *any* element of SO(3) comes from two elements
>of SU(2); then you can check that this is an example.
Indeed. Upon some thought I conclude that I need to know a bit about
matrices and the various representations of vectors therein. In
particular moving to 3-D.
Now in 2-D I am very comfortable with a vector being written as either a
co-ordinate - that is two reals (a,b) or in complex form (a+ib). In 3-D
I am happy about three reals (a,b,c) but have never used a complex
notation. One ought to be able to arbitrarily describe one as (a, b+ic),
but this loses symmetry which isn't nice. This one could be seen as a
plane (b+ic) moved a along the x-axis.
Hmmm.
On the other hand you assert that an element of SU(2) is a rotation.
This would suggest that I should be looking at a form (a+bi, c+di), but
one of the reals must be redundant. Ah, it's two orthogonal planes.
That's a bit of a mindbender. Let's try and convert this to the (a,b,c)
representation and see if any light falls.
Ok let's have (a+bi) in the x-y plane, (c+di) in the y-z plane.
<sketches diagram>
Ugh, messy ...., one of those brick questions.
Ahh, trivial.
Take y as the complex direction, slice the brick diagonally and it falls
out with no effort.
It's (x +iny, z +i(1-n)y).
Obvious, really. Might be more fun in 4-D, but I'm not going there.
hmmmm, a problem.
How to calculate lengths?
Typically (a+ic,0) is not orthogonal to (0,b+ic).
Ah, but the condition is that b+d = y so the length must be
(which doesn't seem altogether nice)
L^2 = x^2+y^2+z^2 = a^2 +(b+d)^2 +c^2
Not as symmetrical as I would like.
Hmm that's because I've aligned the axes.
What if they aren't aligned?
Oh, god, this is turning into a mission ....
Well, I can rotate around one axis by multiplying by P= p+iq, where
PP*=1. Er, hang you on a sec,
if I could show exp(it)* = exp(-it)
I could kill several birds with one stone.
Checkout the series expansion
exp(it) = 1 +it - at^2 -bit^3 +ct^4 ...
exp(-it) = 1 -it - at^2 +bit^3 +ct^4 ...
Ooooh yes. I am thick .... (pounds head on table)
and happily
exp(it/2) exp(-it/2) = 1
Now all we have to do is show it's a rotation by t.
Hmmm.
Rotate some vector to (a+ib)exp(it/2)
Initial vector has angle cos(s) = a/L
Assume final vector has angle cos(s+t), simplify, shown.
OK, now where was I?
Ah, yes, it is asserted that
[a+ic,b+id][exp(it/2) 0] is a rotation by t
[0 exp(-it/2)]
= [a exp(it/2) + ic exp(it/2), b exp(-it/2) + id exp(-it/2)]
Ok, just a rotation around both axes.
Ahh, but we have to show we get the same rotation from -exp(it/2).
One axis will do for thickoes like me.
Hmm. Doesn't work. Are you sure you didn't mean exp(-it/2), which does?
My head hurts ....
I ought to try and find the general expression switching a paired
complex co-ordinate to a 3-D real one, but I'm all washed out. It ought
to be trivial, but somehow doesn't kinda seem to be.
Anyway, assuming I haven't blown it, I now see 2D complex co-ordinates
as expressing a 3-D vector in two co-ordinate planes, so to speak. Not
very magical really, when you think about it.
So I guess you can express a boson as a 2D complex vector but you are
going to want to distinguish between SU(2) type operations and SO(3).
This may not be a trivial thing to do on reflection. Oh, maybe not. If
the spin is double, then I guess somehow you can have the rotation as
doubled by some trick. Wouldn't that do it?
>>Please, sir?
>>
>>Hmmm. I seem to remember that SO(3) had three complex elements
>>(magnitude one) and SU(2) was a 2x2 complex matrix with det = 1(?).
>
>What the...? I thought people were just telling you that
>SO(3) consisted of 3x3 *real* matrices O which are orthogonal:
>
>O O^t = 1 (t = transpose)
>
>and have determinant = 1:
>
>det(O) = 1.
>
>Where did "three complex elements" come from???
Er, typo, m'lud, sir, y'r'onor, y'r wizardship, really, honestly ...
mind on other things and all that ....
>>Ohh sh*t, now I'm for it ....
>>Must learn to keep my mouth shut....
>
>Either that or keep your ears open and remember what you're told!
>
>[Another fireball. WHAM!]
<ouch!>
<d*mn>
Please stop doing that!
It takes AGES for my hair to regrow .....
Oz
In a post a few minutes ago (my time) I wrote:
============
So I guess you can express a boson as a 2D complex vector but you are
going to want to distinguish between SU(2) type operations and SO(3).
This may not be a trivial thing to do on reflection. Oh, maybe not. If
the spin is double, then I guess somehow you can have the rotation as
doubled by some trick. Wouldn't that do it?
============
This is unlikely to be right.
Wishing to avoid further fireballs I would like to slightly edit this:
So I guess you can express a boson as a complex bi-vector (?? form
(a+bi,c+di)) but you are going to want to distinguish between SU(2) type
rotations and SO(3). If fermions follow SU(2) operations they are
presumably normally described using (a+bi,c+di) and if bosons follow
S0(3) and presumably get rotated with a 3-D real and are expressed using
(x,y,z) forms. These are not the same. It would be nice to express both
using the same formulation.
This may not be a trivial thing to do on reflection. Oh, maybe not. If
the spin is double, then I guess somehow you can have the rotation as
doubled by some trick. Wouldn't that do it? After all S0(3) is contained
within SU(3).
Michael Weiss
I changed the subject header from "Re: The Immirzi parameter" to something
hopefully more apt.
John Baez gave you an example of an element of SO(3):
a rotation of angle t counterclockwise about the z axis:
cos t -sin t 0
sin t cos t 0
0 0 1
and the corresponding elements of SU(2):
exp(it/2) 0
0 exp(-it/2)
and minus this matrix:
-exp(it/2) 0
0 -exp(-it/2)
He then teased you with this remark:
: Of course this is of limited interest until you know the
: recipe whereby *any* element of SO(3) comes from two elements
: of SU(2); then you can check that this is an example.
You rose to the bait with many fireball-worthy remarks, but having just seen
part II of The Lord of the Rings, I'm suffering from fireball-fatigue, so
you get off easy. Also, I just happen to have the recipe lying around here
somewhere. Did a bunch of cooking last week, turned up a lot of stuff I'd
forgotten about. Hmm, let's see.... (rummages around) ah yes: "Three pounds
of oysters, two bunches coriander..." no wait, that's lamb stuffing...
OK, here we are. Let's start with the ingredients.
a) One three-dimensional vector space R^3, whose elements are 3-tuples of
real numbers, (x,y,z).
b) One group SO(3), whose elements are 3x3 real matrices subject to the
conditions:
1. A^t A = I (where A^t is the transpose of A)
2. det(A) = 1
c) One action of SO(3) on R^3. That is, if A is a matrix in SO(3) and v is
a vector in R^3, we can use A to *rotate* v:
v --> Av
Here we write v as a column vector.
d) One 1-1 correspondence between R^3 and a set of *complex* matrices:
(x,y,z) <--> [ iz y+ix ]
[ -y+ix -iz ]
This furnishes an alternate way of representing vectors in 3-space. Let's
call this set of matrices H.
(You are wondering, "Where the heck did *that* matrix come from?"
>>From my write-up on the basics of SU(2) and SO(3), hosted at JB's website,
this page to be exact: http://math.ucr.edu/home/baez/lie/node6.html )
e) One group SU(2), whose elements are complex matrices of the form:
[ a+id c+ib ]
[ -c+ib a-id ]
(This comes from the same URL.)
f) One action of SU(2) on the set H from (d), above. That is, given the
matrix V corresponding to (x,y,z):
V = [ iz y+ix ]
[ -y+ix -iz ]
and a matrix B in SU(2):
B = [ a+id c+ib ]
[ -c+ib a-id ]
we can apply B to V in the following way:
V --> B^* V B (where B^* is the hermitian conjugate of B)
Now for the cooking steps:
1) Show that this matrix B^* V B is again in H, and so corresponds to a
vector (x',y',z'). So for every matrix B in SU(2), we have a mapping
(x,y,z) --> (x',y',z').
2) Show that this mapping (x,y,z) --> (x',y',z') is in fact a rotation.
So every matrix B in SU(2) determines a rotation of 3-space, and hence
determines an element A of SO(3).
3) Show that this mapping B --> A from SU(2) to SO(3) is 2-to-1 and onto.
4) As a garnish, calculate the explicit formula for this mapping from SU(2)
to SO(3).
I don't recommend that you try to carry out this recipe as given. You'll
need to pour a lot of ink into the mixing bowl, but when the end result pops
out of the oven, it's somewhat lacking in insight.
I suggest you start instead with:
http://math.ucr.edu/home/baez/lie/lie.html
and post questions as you go along.
Michael Weiss
I changed the subject header from "Re: The Immirzi parameter" to something
hopefully more apt.
John Baez gave you an example of an element of SO(3):
a rotation of angle t counterclockwise about the z axis:
cos t -sin t 0
sin t cos t 0
0 0 1
and the corresponding elements of SU(2):
exp(it/2) 0
0 exp(-it/2)
and minus this matrix:
-exp(it/2) 0
0 -exp(-it/2)
He then teased you with this remark:
: Of course this is of limited interest until you know the
: recipe whereby *any* element of SO(3) comes from two elements
: of SU(2); then you can check that this is an example.
You rose to the bait with many fireball-worthy remarks, but having just seen
Oz
Michael Weiss <mic...@spamfree.net> writes
>a) One three-dimensional vector space R^3, whose elements are 3-tuples of
>real numbers, (x,y,z).
OK.
>b) One group SO(3), whose elements are 3x3 real matrices subject to the
>conditions:
> 1. A^t A = I (where A^t is the transpose of A)
> 2. det(A) = 1
OK.
>c) One action of SO(3) on R^3. That is, if A is a matrix in SO(3) and v is
>a vector in R^3, we can use A to *rotate* v:
> v --> Av
>Here we write v as a column vector.
OK
>d) One 1-1 correspondence between R^3 and a set of *complex* matrices:
>
>(x,y,z) <--> [ iz y+ix ]
> [ -y+ix -iz ]
Insight required. (You call this set "H" later).
Firstly there should be a whole bunch of possible alternative forms,
just switching x, y, z for a start; and I'll bet you could substitute
2iz for iz as well.
Secondly H is asymmetrical. Don't much like this but out of interest is
it more symmetrical if you chose one equivalent to R^4 (presumably a
3x3 complex matrix). Hmm, no, probably not.
Thirdly how would you show this correspondence in an elementary way?
Hmm. No idea. I don;t think 'obvious' counts.
>This furnishes an alternate way of representing vectors in 3-space. Let's
>call this set of matrices H.
>(You are wondering, "Where the heck did *that* matrix come from?"
>>>From my write-up on the basics of SU(2) and SO(3), hosted at JB's website,
>this page to be exact: http://math.ucr.edu/home/baez/lie/node6.html )
That's a really scary page ...
Can we do a trick on this though?
[ iz y+ix ] = z[i 0] + y[0 1] + x[0 i]
[ -y+ix -iz ] [0 -i] [-1 0] [i 0]
Hmm. I'm sure this ought to show that for any element in R^3 there is
only one element in H and vice versa because the coefficients are
constants. Somehow this seems too easy (tremble).
>e) One group SU(2), whose elements are complex matrices of the form:
>
>[ a+id c+ib ]
>[ -c+ib a-id ]
Hey! H is in SU(2) with d=0. That ought to imply that SU(2) is
infinitely bigger than R^3. I'll bet that it's equal to R^4 though.
>(This comes from the same URL.)
<shudder>
>f) One action of SU(2) on the set H from (d), above. That is, given the
>matrix V corresponding to (x,y,z):
>
>V = [ iz y+ix ]
> [ -y+ix -iz ]
>
>and a matrix B in SU(2):
>
>B = [ a+id c+ib ]
> [ -c+ib a-id ]
>
>we can apply B to V in the following way:
>
>V --> B^* V B (where B^* is the hermitian conjugate of B)
Why in this not-very-obvious way?
<snip>
>
>I don't recommend that you try to carry out this recipe as given. You'll
>need to pour a lot of ink into the mixing bowl, but when the end result pops
>out of the oven, it's somewhat lacking in insight.
To be honest I'm more interested in the insight than the detail other
than to do a few simple examples just to get the flavour.
Michael Weiss
Oz wrote (quoting very selectively) :
: Insight required. ...
: Hmm. No idea. I don't think 'obvious' counts. ...
: Why in this not-very-obvious way? ...
: To be honest I'm more interested in the insight than the detail
Yes, exactly why I posted those formulas! They're *not* obvious. The
double-covering of SO(3) by SU(2) is one of those amazing, stunning
math-special-effects that leave you gasping, "How did they DO that?" and
NOT, "I could've done that!"
I don't know the precise history of this discovery. I do know that
Sir William Rowan Hamilton, certified super-genius, spent many years trying
invent
quaternions. And I know his motivation: the complex numbers give us a neat
way to handle 2-dim rotations; can we extend this to 3-dim rotations? And
quaternions are closely related to this SU(2) - SO(3) jazz.
Even after we get all the way through those scary
web-pages, an irreducible kernel of magic will remain. The SU(2) - SO(3)
stuff does fit into a broad general theory, namely the theory of Lie groups
and Lie algebras. Now, broad general theories do help demystify math-magic
tricks. But in some respects, the SU(2) - SO(3) covering looks like a
coincidence. For example, it doesn't generalize to some theorem relating
SO(m) to SU(n).
OK, let's look at what else you wrote.
: Can we do a trick on this though?
:
: [ iz y+ix ] = z[i 0] + y[0 1] + x[0 i]
: [ -y+ix -iz ] [0 -i] [-1 0] [i 0]
:
: Hmm. I'm sure this ought to show that for any element in R^3 there is
: only one element in H and vice versa because the coefficients are
: constants. Somehow this seems too easy (tremble).
Easy or not, that's absolutely right.
: Firstly there should be a whole bunch of possible alternative forms
Yeah, there are lots of variations. However, it'll be easier if we just
pick one set of conventions for the first pass.
: Secondly H is asymmetrical.
If you look near the bottom of http://math.ucr.edu/home/baez/lie/node6.html,
you'll see one of the variations:
[ a+d b-ic ]
[ b+ic a-d ]
which looks a bit more symmetrical, though not *completely* symmetrical. To
be precise, the matrix I just wrote is a so-called Hermitian matrix, which
means that it satisfies the equation
B* = B
(* means Hermitian conjugate--- if you don't know what that means, just
ask!)
Physicists are extremely fond of Hermitian matrices, for various reasons. I
wrote up my notes focussing on anti-Hermitian matrices instead. These
satisfy:
A* = -A
I had my own nefarious reasons for doing it this way; we'll get into them
later. But then, B is anti-Hermitian if and only if iB is Hermitian
(exercise!!), so it's at least partly "you say maths, I say math"....
: Hey! H is in SU(2) with d=0. That ought to imply that SU(2) is
: infinitely bigger than R^3. I'll bet that it's equal to R^4 though.
By "infinitely bigger", I guess you're referring to the *dimensionality* of
SU(2). And you *would* be right, except that I inexcusably (hits self on
head--- thump! thump!; smashes fireball on himself-- !FOOM!) left an
ingredient out of my "recipe" post (but fortunately, not out of the
webpages):
e) One group SU(2), whose elements are complex matrices of the form:
[ a+id c+ib ]
[ -c+ib a-id ]
WITH THE ADDITIONAL REQUIREMENT
a^2 + b^2 + c^2 + d^2 = 1
This is just the requirement that the matrix has determinant 1
(exercise!)--- which is the "S" in SU(2). Dropping that requirement gives
us U(2), the group of unitary 2x2 complex matrices.
(S is for special. Why having det=1 makes a matrix "special", I don't
know.)
OK, let's get started on those scary pages. I'll assume, unless notified
otherwise, that you can get through the intro OK
(http://math.ucr.edu/home/baez/lie/node1.html). (But remember that the
intro is just a *taste* of what's to come. For example, when I write
"Technically, SU(2) is a double cover of SO(3)", you're not supposed to
know what that means yet.)
Things don't really get going until a couple of pages later:
http://math.ucr.edu/home/baez/lie/node3.html
Let's start with the first two paragraphs
[begin quote]
A Lie matrix group is a continuous subgroup of the group of all
non-singular n x n matrices over a field K, where K is either
R or C. "Continuous'' really is a shorthand for saying that
the Lie group is a manifold. The rough idea is that the components of a
matrix in the group can vary smoothly; thus, concepts like ``differentiable
function f: R --> G should make sense. I'll just say
"Lie group'' for Lie matrix group, though many mathematicians would groan
at this.
Example: O(n) is the group of all orthogonal n x n matrices, i.e.
all matrices A with real components such that A^t A = I. This is just
the group of all isometries of R^n which leave the origin fixed.
(Standard calculation: let x be a column vector. Then (Ax)^t (Ax)=x^t
x, i.e., the norm of x equals the norm of Ax.) Note also that the
equations A^t=A^{-1} and A A^t=I follow from A^tA=I.
[end quote]
Any questions?
Michael Weiss
ago.]
Oz wrote (quoting very selectively) :
: Insight required. ...
: Hmm. No idea. I don't think 'obvious' counts. ...
: Why in this not-very-obvious way? ...
: To be honest I'm more interested in the insight than the detail
Yes, exactly why I posted those formulas! They're *not* obvious. The
double-covering of SO(3) by SU(2) is one of those amazing, stunning
math-special-effects that leave you gasping, "How did they DO that?" and
NOT, "I could've done that!"
I don't know the precise history of this discovery. I do know that
Sir William Rowan Hamilton, certified super-genius, spent many years trying
invent quaternions. And I know his motivation: the complex numbers give us
a neat
way to handle 2-dim rotations; can we extend this to 3-dim rotations? And
quaternions are closely related to this SU(2) - SO(3) jazz.
Even after we get all the way through those scary
web-pages, an irreducible kernel of magic will remain. The SU(2) - SO(3)
stuff does fit into a broad general theory, namely the theory of Lie groups
and Lie algebras. Now, broad general theories do help demystify math-magic
tricks. But in some respects, the SU(2) - SO(3) covering looks like a
coincidence. For example, it doesn't generalize to some theorem relating
SO(m) to SU(n).
OK, let's look at what else you wrote.
: Can we do a trick on this though?
:
: [ iz y+ix ] = z[i 0] + y[0 1] + x[0 i]
: [ -y+ix -iz ] [0 -i] [-1 0] [i 0]
:
: Hmm. I'm sure this ought to show that for any element in R^3 there is
: only one element in H and vice versa because the coefficients are
: constants. Somehow this seems too easy (tremble).
Easy or not, that's absolutely right.
: Firstly there should be a whole bunch of possible alternative forms
Yeah, there are lots of variations. However, it'll be easier if we just
pick one set of conventions for the first pass.
: Secondly H is asymmetrical.
If you look near the bottom of http://math.ucr.edu/home/baez/lie/node6.html,
you'll see one of the variations:
[ a+d b-ic ]
[ b+ic a-d ]
which looks a bit more symmetrical, though not *completely* symmetrical. To
be precise, the matrix I just wrote is a so-called Hermitian matrix, which
means that it satisfies the equation
A* = A
(* means Hermitian conjugate--- if you don't know what that means, just
ask!)
Physicists are extremely fond of Hermitian matrices, for various reasons. I
wrote up my notes focussing on anti-Hermitian matrices instead. These
satisfy:
B* = -B
I had my own nefarious reasons for doing it this way; we'll get into them
later. But then, B is anti-Hermitian if and only if iB is Hermitian
(exercise!!), so it's at least partly "you say maths, I say math"....
: Hey! H is in SU(2) with d=0. That ought to imply that SU(2) is
: infinitely bigger than R^3. I'll bet that it's equal to R^4 though.
By "infinitely bigger", I guess you're referring to the *dimensionality* of
SU(2). And you *would* be right, except that I inexcusably (hits self on
head--- thump! thump!; smashes fireball on himself-- !FOOM!) left an
ingredient out of my "recipe" post (but fortunately, not out of the
webpages):
e) One group SU(2), whose elements are complex matrices of the form:
[ a+id c+ib ]
[ -c+ib a-id ]
WITH THE ADDITIONAL REQUIREMENT
mike james
"Michael Weiss" <mic...@spamfree.net> wrote in message
news:ecYR9.1469$0g6...@nwrddc01.gnilink.net...
..Trimmed to -
>But in some respects, the SU(2) - SO(3) covering looks like a
> coincidence. For example, it doesn't generalize to some theorem relating
> SO(m) to SU(n).
Really?
I've just started finding out about Lie groups etc and I was assuming that
sooner or later I'd find a theorem relating SO(m) to SU(n) or perhaps
something
even more general that the SU(2)-SO(3) covering was a special case of.
Help?!
mikej
Michael Weiss
: >But in some respects, the SU(2) - SO(3) covering looks like a
: > coincidence. For example, it doesn't generalize to some theorem
relating
: > SO(m) to SU(n).
and mike james replied
: Really?
: I've just started finding out about Lie groups etc and I was assuming that
: sooner or later I'd find a theorem relating SO(m) to SU(n) or perhaps
: something
: even more general that the SU(2)-SO(3) covering was a special case of.
Now you've made me nervous. After all, "some theorem relating SO(m) to
SU(n)" covers a lot of ground; who knows what theorems lurk in the miles of
dusty journals in library basements?
To reassure myself a little, I looked at the list of classical Lie groups in
Helgason's "Differential Geometry and Symmetric Spaces". Right off the bat
I encountered this factlet(Lemma 4.1(a), Chapter 9):
SO(2n) intersect Sp(n) is topologically isomorphic to U(n)
Well, that sorta-kinda relates SO(m) to U(n). Not that reassuring!
On the other hand, this doesn't generalize the SU(2) -- SO(3) covering. (3
isn't 2n, U(n) isn't SU(n), topological isomorphism isn't double covering,
and the symplectic group Sp(n) has disappeared, for one thing, or actually
four things.)
Looking at the dimensionalities of SU(n) and SO(m) tells us a little more:
dim SU(n) = n^2 - 1
dim SO(m) = (m^2 - m)/2
So we'd need a solution to the diophantine equation
n^2 - 1 = (m^2 - m)/2
to have even a hope of a covering map.
There's the trivial solution n=m=1. SU(1) and SO(1) are both the trivial
group, consisting of the 1x1 matrix [1]. So I guess you can say that SU(1)
covers SO(1).
Next we have n=2, m=3, and indeed SU(2) covers SO(3).
Next comes n=4, m=6. The Lie algebras su(4) and so(6) are isomorphic; this
is a famous coincidence, and if I recall correctly, it has something to do
with Penrose's twistor theory.
I believe SU(4) covers SO(6), but I forget the details.
For n=11, m=16, we again have a solution: SU(11) and SO(16) both have
dimension 120. I don't know of any other relation between these two groups;
Helgason does not include
su(11) and so(16) in his list of "coincidences", so I presume the Lie
algebras are not isomorphic (and so SU(11) can't cover SO(16)).
Time for some real experts to weigh in!
John Baez
mike james <mike....@infomax.demon.co.uk> wrote:
>"Michael Weiss" <mic...@spamfree.net> wrote in message
>news:ecYR9.1469$0g6...@nwrddc01.gnilink.net...
>>But in some respects, the SU(2) - SO(3) covering looks like a
>> coincidence. For example, it doesn't generalize to some theorem
>>relating SO(m) to SU(n).
>Really?
Really. Michael never lies, and he's hardly ever wrong.
SU(2) is a double cover of SO(3), and SU(4) is a double cover of SO(6),
but there are no other SU(n)'s that double cover SO(m)'s.
The easiest way to see this is just to calculate the dimension of
SU(n) and of SO(m), and see when they agree. That almost does it.
>I've just started finding out about Lie groups etc and I was assuming
>that sooner or later I'd find a theorem relating SO(m) to SU(n) or
>perhaps something even more general that the SU(2)-SO(3) covering
>was a special case of.
Well, SO(m) has a double cover called Spin(m), and there is a beautiful
and important general theory of how to get your hands on Spin(m), which
you should learn someday. But it doesn't involve SU(n)'s: it involves
"Clifford algebras" and "spinor". There are a few magical
"coincidences" that occur in low dimensions, and the fact that
SU(2) is a double cover of SO(3) is one of these.
I put "coincidence" in quotes because nothing in math happens by
accident. However, some things don't fit into general patterns
in the way you'd at first expect. These are like the "spice"
of mathematics: they keep it from becoming bland.
Philip Charlton
> I wrote:
>
>: >But in some respects, the SU(2) - SO(3) covering looks like a
>: > coincidence. For example, it doesn't generalize to some theorem
> relating
>: > SO(m) to SU(n).
>
I think this is just a coincidence, the coincidence being that SU(2)
happens to be isomorphic to Spin(3) (which by definition, *is* the
simply-connected double cover of SO(3)). There are some other coincidences
between low-dimensional Lie groups - eg. Lawson & Michelsohn list a few
like Spin(4) = SU(2) x SU(2), Spin(6) = SU(4), but I'm not aware of
any general principles relating the special-unitary and Spin groups.
Cheers,
Philip
ab
Michael Weiss:
>Sir William Rowan Hamilton, certified super-genius, spent
>many years trying invent quaternions. And I know his motivation: the
>complex numbers give us a neat way to handle 2-dim rotations; can we
>extend this to 3-dim rotations?
This is (part of) the modern view of it, but did really Hamilton think
in such terms? That he was trying to "divide vectors" is well
documented, but rotations???
Michael Weiss
: I believe SU(4) covers SO(6), but I forget the details.
Searching the archives of this newsgroup reveals that SU(4) is indeed the
double-cover of SO(6). I quote from "Lie groups - not for grunts:
octonions?", the 2000-11-10 entry by John Baez:
[quote]
A) U(1) is the double cover of SO(2).
B) Sp(1) = SU(2) is the double cover of SO(3).
C) SU(2) x SU(2) is the double cover of SO(4).
D) Sp(2) is the double cover of SO(5).
E) SU(4) is the double cover of SO(6).
At this point these nice coincidences stop. Anyone who really
understands Lie groups can give a couple of different explanations
for each of these coincidences, and explain why they stop here.
There's a nice explanation involving Dynkin diagrams, and a nice
explanation involving Clifford algebras, and a nice geometrical
explanation too. So I leave the proof of these facts as a challenge
to all would-be Lie group experts... but not the grunts!
[end-quote]
John Baez
Michael Weiss <mic...@spamfree.net> wrote:
>Looking at the dimensionalities of SU(n) and SO(m) tells us a little more:
>
>dim SU(n) = n^2 - 1
>dim SO(m) = (m^2 - m)/2
>
>So we'd need a solution to the diophantine equation
>n^2 - 1 = (m^2 - m)/2
>to have even a hope of a covering map.
Right, I talked about this once upon a time - see the end of this post.
It's a cute equation, since it says:
FIND A TRIANGULAR NUMBER THAT'S ONE LESS THAN A PERFECT SQUARE!
That's sort of thing you could imagine number junkies studying
just to get their daily fix!
So you get the solutions
n=m=1,
n=2, m=3,
and
n=4, m=6
all of which give covers...
>I believe SU(4) covers SO(6), but I forget the details.
Yup, it's does. Even better, it's the universal cover,
which is a double cover.
And then you get *this*, you rascal you:
>For n=11, m=16, we again have a solution: SU(11) and SO(16) both have
>dimension 120.
Aaaaargh!
I don't know of any other relation between these two groups;
>Helgason does not include
>su(11) and so(16) in his list of "coincidences", so I presume the Lie
>algebras are not isomorphic (and so SU(11) can't cover SO(16)).
No, SU(11) can't cover SO(16), but I'm a bit dismayed to find
their dimensions are equal... I had been hoping the Diophantine
equation would suffice to rule out all the cases except those
where double covers actually occur! I hadn't noticed that 121
is square and 120 is triangular.
Are there any more solutions of this accursed equation?
>Time for some real experts to weigh in!
I know four proofs that SU(4) is the double cover of SO(6):
the exterior algebra proof, the Clifford algebra proof,
the projective geometry proof and the Dynkin diagram proof.
But for now, I'll try to placate you with this old post, which
merely philosophizes on the subject at hand.
.......................................................................
From: ba...@galaxy.ucr.edu (John Baez)
Newsgroups: sci.physics.research
Subject: The law of small numbers
Date: Sun, 27 Jan 2002 04:04:43 +0000 (UTC)
In article <ayq38.417834$W8.14...@bgtnsc04-news.ops.worldnet.att.net>,
Danny Ross Lunsford <antima...@yahoo.com> wrote:
>John Baez should have written:
>> Besides Spin(n), SU(n) and Sp(n), there are exactly 5 compact
>> simply-connected simple Lie groups, the so-called "exceptional
>> Lie groups"...
>Why is it that these closed families come in smallish numbers? Why for
>example are there not exactly 2,718,281 exceptional Lie groups? Why is the
>table of Clifford algebras 8-periodic? What is the origin of this "law of
>small numbers"?
I can explain why there are 5 exceptional Lie groups and why the
table of Clifford algebras has period 8 - but it sounds like you're
actually interested in the deeper question of why small numbers tend
to show up a lot.
In some ways this question underlies a lot of my studies of math
and physics. Why does spacetime look 4-dimensional instead of
123897119813282349-dimensional? Why are there 3 generations of
quarks and leptons, instead of 11? Why do certain numbers like
8 and 24 show up all over the place in seemingly disparate ways...
which then turn out to be connected in a grand, magnificent
pattern whose full extent we are still struggling to grasp?
I don't think we are ready to answer questions like this yet.
At present we can't even say for sure if questions like this have
answers! So for now, I think it's best to keep gathering clues:
keep trying to understand the fabric of mathematics and physics
and learn how it's woven. A lot of this amounts to taking facts
that are already known and seeing how they fit together. I spend
much of my time trying to do just this.
Anyway, here's one obvious thing:
Small numbers stand at the tip of a lot of patterns, so these
patterns interact more strongly there, and more cool stuff happens.
To take an elementary example: the dimensions of the spin groups
Spin(n) are the triangular numbers
0, 1, 3, 6, 10, 15, 21, 28, 36, ...
while the dimensions of the special unitary groups SU(n) are
one less than the square numbers:
0, 3, 8, 15, 24, 35, 48, 63, ....
As so often the case, these sequences are packed tight at
first and spread out as they go on, so there is more opportunity
for coincidences near the beginning.
Both Spin(1) and SU(1) are 0-dimensional, and indeed we have
an isomorphism
Spin(1) = SU(1).
But this coincidence is too simple to be exciting. We consider
it "trivial".
Both Spin(3) and SU(2) are 3-dimensional, and indeed we have
Spin(3) = SU(2).
This coincidence is like Helen of Troy: so beautiful that it has
launched a thousand papers! It's a basic part of the fabric of
physics. One might say that it explains why complex numbers
automatically show up when we consider the simplest sort of
spinning particle in 3-dimensional space.
Both Spin(6) and SU(4) are 15-dimensional, and indeed we have
Spin(6) = SU(4).
This coincidence is even more amazing, but it's subtler and less
well-known than the previous one. We can't say for sure yet that
it's important in physics - but it lies near the heart of Penrose's
theory of twistors. Actually he uses a different manifestation
of the same coincidence, namely so(2,4) = su(2,2): the Lie algebra
of the conformal group in 4 dimensions is the same as su(2,2).
And this is where the fun STOPS. There are no more isomorphisms
between spin groups and special unitary groups. (I don't know if
there any more coincidences between triangular numbers and one-less-
than-square numbers.)
In short: small numbers are where the coincidences are more likely
to occur!
Mark Horn
through n = 1024 there's additionally:
n = 23, m = 33 for 528 dimensions,
n = 64, m = 91 for 4095 dimensions,
n = 134, m = 190 for 17955 dimensions,
n = 373, m = 528 for 139128 dimensions...
could one surmise then that there are an infinite number of (useless)
solutions to the diophantine equation?
mark
Graham Jones
In article <avmk2e$2fr$1...@glue.ucr.edu>, John Baez <ba...@galaxy.ucr.edu>
writes
[...]
>Are there any more solutions of this accursed equation?
[...]
Infinitely many, I'm afraid.
Putting r = 2m-1,
n^2 - 1 = (m^2 - m)/2
can be rearranged as
8n^2 - r^2 = 7
That's a well-known kind of equation, and once upon a time I used to
know how to solve things like that properly. I can now only remember
that it is related to continued fractions and approximations by
rationals of square roots of integers.
Note that in the solutions mentioned so far, ie
n=1, m=1,r=1
n=2, m=3, r=5
n=4, m=6, r=11
n=11, m=16, r=31
r/n is an increasingly good approximation to sqrt(8).
The solutions follow a pattern: if (n,r) is a solution, then
(3n+r,8n+3r) is another. So you can make as many as you like once you
have one. There seem to be two infinite series here, one starting from
(1,1), the other from (2,5). I don't know if there might be more.
Graham
--
Graham Jones, author of SharpEye Music Reader
http://www.visiv.co.uk
21e Balnakeil, Durness, Lairg, Sutherland IV27 4PT, Scotland, UK
Michael Weiss
I wrote:
: >Sir William Rowan Hamilton, certified super-genius, spent
: >many years trying invent quaternions. And I know his motivation: the
: >complex numbers give us a neat way to handle 2-dim rotations; can we
: >extend this to 3-dim rotations?
ab replied
: This is (part of) the modern view of it, but did really Hamilton think
: in such terms? That he was trying to "divide vectors" is well
: documented, but rotations???
Good question! Of course, real history is always more complicated and messy
than any one paragraph summary. In this case, it's easy for the curious to
delve deeper. Professor David Wilkins at Trinity College, Dublin, has made
most if not all of Hamilton's original works available on the web:
http://www.maths.tcd.ie/pub/HistMath/People/Hamilton/
Hamilton made his famous engraving (of the fundamental quaternion formula)
on Broome Bridge on October 16, 1843. The following day he wrote a letter
to his friend John T. Graves, describing the train of thought that led to
his discovery:
http://www.maths.tcd.ie/pub/HistMath/People/Hamilton/QLetter/QLetter.html
The desire to extend the geometrical interpretation of complex numbers to
three dimensions was present from the beginning:
...You know that I have long wished, and I believe that you have
felt the same desire, to possess a Theory of Triplets, analogous
to my published Theory of Couplets, and also to
Mr. Warren's geometrical representation of imaginary quantities. ...
My train of thoughts was of this kind. Since sqrt(-1) is in a certain
well-known sense, a line perpendicular to the line 1, it seemed
natural that there should be some other imaginary to express a line
perpendicular to the former; and because the rotation from this to this
also being doubled conducts to -1, it ought also to be a square root
of negative unity, though not to be confounded with the former.
Calling the old root, as the Germans often do, i, and the new one j, I
inquired what laws ought to be assumed for multiplying
together a + ib + jc and x + iy + jz.
Hamilton multiplied this out, using the rules i^2=j^2=-1, but was now faced
with question, what to do with ij?
For complex numbers, the absolute value of a product is the product of the
absolute values. Hamilton tried to generalize this to his triplets:
...this might tempt us to take ij = 1 or ij = -1; but with neither
assumption shall we have the sum of the squares of the
coefficients of 1, i, and j in the product = to the product
of the corresponding sums of squares in the factors.
Hamilton then noticed that if he took ij = -ji, he would get the desired
rule. He considered first the special case of a + ib + jc times itself;
then generalized slightly to (a + ib + jc) times ( x + ib + jc) --- "keeping
still, as you see, the two factor lines in one common plane with the unit
line." And here he observes something which must have convinced him he was
on the right track: the angles of rotation add!
[the results] are easily found to be the correct coordinates
of the product-point, in the sense that the rotation from
the unit line to the radius vector of a, b, c, being added
in its own plane to the rotation from the same unit-line
to the radius vector of the other factor-point x, b, c,
conducts to the radius vector of the lately mentioned product-point;
and that this latter radius vector is in length the product of the
two former. Confirmation of ij = - ji; but no information yet of the value
of k [=ij]
Hamilton now took a look at the general case, (a+bi+cj) times (x+iy+cj).
Comparing absolute values again, he discovered that (a^2 + b^2 + c^2) times
(x^2 + y^2 + z^2) is *not* equal to the sum of three squares, but *is* in
fact equal to the sum of *four* squares. And now it's lightbulb time:
And here there dawned on me the notion that we must admit,
in some sense, a *fourth dimension* of space for the purpose
of calculating with triplets; or transferring the paradox to algebra,
must admit a third distinct imaginary symbol k, not to be
confounded with either i or j...
The emphasis on "fourth dimension" is in the original. Nowadays of course
we hear the words "fourth dimension", or even "n-dimensional" and we yawn
ho-hum, but things were different in 1843. Arthur Cayley published his
first paper on n-dimensional analytic geometry around the same time
(actually a few months after Hamilton's discovery, but before its
publication).
eb...@lfa221051.richmond.edu
>FIND A TRIANGULAR NUMBER THAT'S ONE LESS THAN A PERFECT SQUARE!
>
>That's sort of thing you could imagine number junkies studying
>just to get their daily fix!
>
>So you get the solutions
>
>n=m=1,
>
>n=2, m=3,
>
>and
>
>n=4, m=6
>
>all of which give covers...
>In article <4ksS9.8185$xb....@nwrddc02.gnilink.net>,
>Michael Weiss <mic...@spamfree.net> wrote:
>>I believe SU(4) covers SO(6), but I forget the details.
>Yup, it's does. Even better, it's the universal cover,
>which is a double cover.
>
>And then you get *this*, you rascal you:
>>For n=11, m=16, we again have a solution: SU(11) and SO(16) both have
>>dimension 120.
>Aaaaargh!
>I don't know of any other relation between these two groups;
>>Helgason does not include
>>su(11) and so(16) in his list of "coincidences", so I presume the Lie
>>algebras are not isomorphic (and so SU(11) can't cover SO(16)).
>No, SU(11) can't cover SO(16), but I'm a bit dismayed to find
>their dimensions are equal... I had been hoping the Diophantine
>equation would suffice to rule out all the cases except those
>where double covers actually occur! I hadn't noticed that 121
>is square and 120 is triangular.
>
>Are there any more solutions of this accursed equation?
A question I can answer by writing a one-line program. I like those.
SO(m) and SU(n) have the same dimension when (m,n) is
0, 1
1, 1
3, 2
6, 4
16, 11
33, 23
91, 64
190, 134
528, 373
1105, 781
3075, 2174
6438, 4552
17920, 12671
37521, 26531
104443, 73852
218686, 154634
608736, 430441
Those are all the ones up to m = 1000000.
These look to me like they're growing exponentially. In fact,
if you take ratios of successive values of m, you get an interesting
sequence:
3., 2., 2.666666667, 2.062500000, 2.757575758, 2.087912088, 2.778947368, 2.092803030, 2.782805430, 2.093658537, 2.783473128, 2.093805804, 2.783587857, 2.093831085, 2.783607547
That alternating pattern -- something around 2.78 alternating with
something around 2.09 -- sure looks too good to be a coincidence!
Before trying it out, I'll predict that the next two solutions
will be approximately
m = (2.09)(608736) = 1.27e6
m = (2.78)(1.27e6) = 3.54e6
And in fact, they are
1274593, 901273
3547971, 2508794
Being a big fan of what Hume called "the scandal of induction," I'll
conjecture that the pattern goes on forever, and that this
equation has infinitely many solutions.
-Ted
(Actually, Google seems to associate the phrase "scandal of
induction" mostly with W.V.O. Quine, but I was pretty
sure it was Hume's first.)
--
[E-mail me at na...@domain.edu, as opposed to na...@machine.domain.edu.]
Patrick Schaaf
ba...@galaxy.ucr.edu (John Baez) writes:
>>n^2 - 1 = (m^2 - m)/2
>FIND A TRIANGULAR NUMBER THAT'S ONE LESS THAN A PERFECT SQUARE!
>Are there any more solutions of this accursed equation?
For what it's worth, here's a brute force scan up to n == 10^9.
(trivial) C source available by mail request.
SU(1) ~ SO(1) dim 0
SU(2) ~ SO(3) dim 3
SU(4) ~ SO(6) dim 15
SU(11) ~ SO(16) dim 120
SU(23) ~ SO(33) dim 528
SU(64) ~ SO(91) dim 4095
SU(134) ~ SO(190) dim 17955
SU(373) ~ SO(528) dim 139128
SU(781) ~ SO(1105) dim 609960
SU(2174) ~ SO(3075) dim 4726275
SU(4552) ~ SO(6438) dim 20720703
SU(12671) ~ SO(17920) dim 160554240
SU(26531) ~ SO(37521) dim 703893960
SU(73852) ~ SO(104443) dim 5454117903
SU(154634) ~ SO(218686) dim 23911673955
SU(430441) ~ SO(608736) dim 185279454480
SU(901273) ~ SO(1274593) dim 812293020528
SU(2508794) ~ SO(3547971) dim 6294047334435
SU(5253004) ~ SO(7428870) dim 27594051024015
SU(14622323) ~ SO(20679088) dim 213812329916328
SU(30616751) ~ SO(43298625) dim 937385441796000
SU(85225144) ~ SO(120526555) dim 7263325169820735
SU(178447502) ~ SO(252362878) dim 31843510970040003
SU(496728541) ~ SO(702480240) dim 246739243443988680
best regards
Patrick
Arvind Murugan
map, I was wondering if this helps, since it is the easiest way I can
see to understand the above map:
SU(2) = The 3 dimensional sphere and SO(3) is RP3.
Are these again coincidences? Are the higher maps also simple maps of
manifolds (well they ( SU(n) and SO(m) ) are all manifolds obviously
but is one the universal covering space of the other? Do Spin groups
try to sneak in if there is space in the homotopy group of SO(m)? )
Just my 2 homotopy-cents..
Arvind
Arvind Murugan
On reading more carefully(!), I see that people have been talking
about covering groups. Sorry about that!
But there was something else I found interesting (and hopefully wasn't
covered already) - motivation for Hamilton's quaternions.
I think that this giving a multiplicative structure on S3 is cool!
The complex number algebra gives the structure for U(1) or S1.
Since the only group of size 3 is Z3 and the group ring over this
(with the reals) is not good enough (is it because of 1 + x + x^2 =
0?), there is no multiplicative structure on S2. I wonder if
multiplicative structures on other manifolds have to do with any other
group rings..
In the case of spheres, if one has a multiplicative structure on S^n,
by extending it on each ray from the origin, we can get a n+1 dim.
field extension over R (but the word is that we have only C and H over
R?).
Do other multiplicative manifolds (Lie Groups?) also create
fields/algebras?
Arvind
David Einstein
It can be written as
8 n^2 - (2 m -1)^2 = 7
which is equvalent to finding integers of norm 7 in Q(sqrt(2)).
Any of the numbers should have a nice generating function, but I'm too
lazy to find it.
> mark
David Einstein
http://www.research.att.com/cgi-bin/access.cgi/as/njas/sequences/eisA.cgi?Anum=A006451
for the entry in the On-Line Encyclopedia of Integer Sequences.
mike james
news:avdg3u$pb6$1...@glue.ucr.edu...
> Really. Michael never lies, and he's hardly ever wrong.
I wasn't doubting him - it was an expression of suprise.
> SU(2) is a double cover of SO(3), and SU(4) is a double cover of SO(6),
> but there are no other SU(n)'s that double cover SO(m)'s.
So the structure of a rotation in a 2D complex vector space is "the same"
as a rotation in 3D real space and a 4D complex vector space
and a 6D real space but after that the "gears" don't mesh...
> Well, SO(m) has a double cover called Spin(m), and there is a beautiful
> and important general theory of how to get your hands on Spin(m), which
> you should learn someday.
Just getting to it.
>But it doesn't involve SU(n)'s: it involves
> "Clifford algebras" and "spinor". There are a few magical
> "coincidences" that occur in low dimensions, and the fact that
> SU(2) is a double cover of SO(3) is one of these.
>
> I put "coincidence" in quotes because nothing in math happens by
> accident. However, some things don't fit into general patterns
> in the way you'd at first expect. These are like the "spice"
> of mathematics: they keep it from becoming bland.
Yes I agree and hence the "Really" exclaimation.
Often its what is not said that's important.
For example when I did a first course in group theory I
came away that we knew everything there was to know about
the classification of finite simple groups. I was very shocked
when the monster was discovered... It was then I realised that
I'd missed the point of the course almost completely.
Its imporant to say what isn't know as well as what is...
Many books just introduce the double cover of SO(3) by SU(2) as
if it was a common place occurence and say nothing more about
its status as a "coincident".
mikej
mike james
[Snip]
> To take an elementary example: the dimensions of the spin groups
> Spin(n) are the triangular numbers
>
> 0, 1, 3, 6, 10, 15, 21, 28, 36, ...
>
> while the dimensions of the special unitary groups SU(n) are
> one less than the square numbers:
>
> 0, 3, 8, 15, 24, 35, 48, 63, ....
>
> As so often the case, these sequences are packed tight at
> first and spread out as they go on, so there is more opportunity
> for coincidences near the beginning.
[snip]
>
> In short: small numbers are where the coincidences are more likely
> to occur!
So in other words
All "small" numbers are the same - or almost so - and
all "large" number are different - very much so.
It certainly explains the reason for small number "accidentals" in a lot of
subjects...
mikej
Robert Coquereaux
It just happens that, in small dimensions, there are a few coincidences
between members of different series of Lie groups. For instance Spin(3)
coincides with SU(2). There are a few others. Here is an almost complete
list in the compact case:
Spin(3)=SU(2)= Usp(2), Spin(4)=SU(2)xSU(2), Spin(5)= U(2,H)=Usp(4),
Spin(6)=SU(4).
For n>6, Spin(n) is just... Spin(n), defined as the simply connected
covering (which is always two-fold) of SO(n).
The above H was denoting the field of quaternions ("H" standing for
Hamilton); remember that unitary groups over H are unitary symplectic
over the complex (and that real symplectic groups are particular non
compact real forms of these U(n,H)).
In small dimensions we have also several identifications for a few non
compact cases: call Spin^(p,q) the connected component of the identity
in Spin(p,q), then we have: Spin^(2,1)=Sl(2,R), Spin^(3,1)=Sl(2,C) -- as
everybody space-time physicist knows! -- , Spin^(4,1)=U(1,1,H),
Spin^(5,1)= Sl(2,H), Spin^(3,2)=Sp(4,R), Spin^(4,2)=SU(2,2).
I think that this list was complete.
One way to prove these results is to work out the structure of the real
Clifford algebra associated with a given signature (p,q) and represent
Spin(p,q) as a subspace of the its even subalgebra.
mike james a Ècrit:
Patrick Schaaf
>>n^2 - 1 = (m^2 - m)/2
>(I don't know if there any more coincidences between triangular numbers
>and one-less-than-square numbers.)
Given the 24 integer coincidences for n up to 10^9 (see below),
it occured to me that asymptotically, m approaches (sqrt(2)*n)+(1/2).
I have no idea what I'm writing about. Fun fun fun!
best regards
Patrick
SU(1) = SO(1) dim 0
SU(2) = SO(3) dim 3
SU(4) = SO(6) dim 15
SU(11) = SO(16) dim 120
SU(23) = SO(33) dim 528
SU(64) = SO(91) dim 4095
SU(134) = SO(190) dim 17955
SU(373) = SO(528) dim 139128
SU(781) = SO(1105) dim 609960
SU(2174) = SO(3075) dim 4726275
SU(4552) = SO(6438) dim 20720703
SU(12671) = SO(17920) dim 160554240
SU(26531) = SO(37521) dim 703893960
SU(73852) = SO(104443) dim 5454117903
SU(154634) = SO(218686) dim 23911673955
SU(430441) = SO(608736) dim 185279454480
SU(901273) = SO(1274593) dim 812293020528
SU(2508794) = SO(3547971) dim 6294047334435
SU(5253004) = SO(7428870) dim 27594051024015
SU(14622323) = SO(20679088) dim 213812329916328
SU(30616751) = SO(43298625) dim 937385441796000
SU(85225144) = SO(120526555) dim 7263325169820735
SU(178447502) = SO(252362878) dim 31843510970040003
SU(496728541) = SO(702480240) dim 246739243443988680
Michael K Murray
get SU(2) x SU(2) -> SO(4) as well for free.
Let H be the quaternions. H is R^4 where we write
q = t + xi + yj + z k
where i j k satisfying the usual famous relations you can find
scrawled on any bridge in Ireland. ij = k, i^2 = j^2 = k^2 = -1 ,
ijk = -1 etc . Define the conjugate of a quarternion by
c(q) = t - xi - yj - zk (usually written just q with an
overbar like complex conjugate)
Check that c(pq) = c(q)c(p).
Then q c(q) = ||q||^2 = t^2 + x^2 + y^2 + z^2
In particular if q \neq 0 then q is invertible with
q^{-1}= c(q)/||q|^2.
We call the quaternions that satisfy c(q) = -q the imaginary
quaternions. These are the ones that satisfy t=0 so are a copy
of R^3.
Let Sp(1) be the group of all quaternions with ||q||^2 = 1.
Clearly Sp(1) is the 3-sphere.
If q \in Sp(1) and x is an imaginary quaternion then c(qxq^{-1})
= c(q^{-1}) c(x) c(q) = - q x q^{-1} as c(q) = q^{-1}.
So the function x |-> qxq^{-1} is a linear map preserving
imaginary quaternions hence an element of GL(3).
Now ||qxq^{-1}||^2 = || x||^2 hence this function is
norm so inner product preserving. Sp(1) is connected and
its image under the continuous function x |-> qxq^{-1}
is connected. It contains the identity so have have a
homomorphism
Sp(1) -> SO(3)
Its easy to check the kernel is +/- 1 . To see its
onto you have to show the image contains all reflections and
use the fact that reflections generate SO(n).
You can check that if x is ANY quaternion and p and q in Sp(1)
then x |-> pxq^{-1} is in SO(4) and hence defines a homomorphism
Sp(1) x Sp(1) -> SO(4)
which is onto and has kernel +/- 1.
Finally identify H with C^2 by (v, w) |-> (z + wj)
for z and w complex and you can show that Sp(1) = SU(2).
Maybe its jw instead.
I'll bet this is better done in one of John Baez's articles !
Michael
John Baez
In article <88e613c0.03011...@posting.google.com>,
Arvind Murugan <twist_u...@yahoo.com> wrote:
>In the case of spheres, if one has a multiplicative structure on S^n,
>by extending it on each ray from the origin, we can get a n+1 dim.
>field extension over R
Yes, if by "field" you include so-called "skew fields", where
multiplication doesn't need to commute. These are better known
as "division algebras" over R, or "normed division algebras"
if they have a norm with
|xy| = |x| |y|
>(but the word is that we have only C and H over R?).
Right: as we often like to emphasize around here, the only
spheres that are Lie groups are S^0, S^1 and S^3. By the trick
you mention, these give the normed division algebras over R,
namely the reals, complexes and quaternions.
There's just one more sphere that can be made into a Lie "loop" -
which is like a group, but nonassociative. This is S^7.
Applying your trick to this we get the octonions. This is
the only *nonassociative* normed division algebra.
>Do other multiplicative manifolds (Lie Groups?) also create
>fields/algebras?
Well, not in the same way, but any group gives rise to an
algebra called its "group algebra", which is quite a noble
entity. You might enjoy studying that.
Oz
>Hamilton now took a look at the general case, (a+bi+cj) times (x+iy+cj).
>Comparing absolute values again, he discovered that (a^2 + b^2 + c^2) times
>(x^2 + y^2 + z^2) is *not* equal to the sum of three squares, but *is* in
>fact equal to the sum of *four* squares. And now it's lightbulb time:
>
> And here there dawned on me the notion that we must admit,
> in some sense, a *fourth dimension* of space for the purpose
> of calculating with triplets; or transferring the paradox to algebra,
> must admit a third distinct imaginary symbol k, not to be
> confounded with either i or j...
Oh.
Whilst I peek at quaternionic stuff I haven't really considered it
seriously, believing it was 'just' an extension to 3-D.
Obviously I was in error.
So I'd better just check that I have the basic concept right, because I
suspect that I haven't.
In 2-D we can express any point/position vector conveniently as a+ib
where rotations about the origin can be expressed as multiplication by
p+iq where p^2+q^2 =1. Obviously we can scale and reflect and so on,
too.
Now to extend that to 4-D then we need a structure of form a+bi+cj+dk.
and then what? What is the form for a 3-D rotation and how is this
conveniently expressed graphically, well presumably using a 4D space.
Hmmm, I just can't resist speculating and in this case, for once, back
will come the 'right' answer in a flash.
In 1-D there isn't such a thing as rotation. Just a position, really.
In 2-D we can have a position and a rotation that are simply related.
In 3-D position is 3-D but rotation really needs 4-D.
If that is right then a generalised transformation in 3-D, requires 4
independent variables.
So (waxing lyrical far from his realms of knowledge) we can expect
really fun things in 4-D. One might hope for another 'invariant'-type
property over and above 'position' and 'rotation'. This could be a
'different' type of spin or something completely strange.
I think I ought to stop there for reasons of self-preservation.
John Baez
>Michael Weiss <mic...@spamfree.net> writes
>>Hamilton now took a look at the general case, (a+bi+cj) times (x+iy+cj).
>>Comparing absolute values again, he discovered that (a^2 + b^2 + c^2) times
>>(x^2 + y^2 + z^2) is *not* equal to the sum of three squares, but *is* in
>>fact equal to the sum of *four* squares. And now it's lightbulb time:
>>
>> And here there dawned on me the notion that we must admit,
>> in some sense, a *fourth dimension* of space for the purpose
>> of calculating with triplets; or transferring the paradox to algebra,
>> must admit a third distinct imaginary symbol k, not to be
>> confounded with either i or j...
>Oh.
>
>Whilst I peek at quaternionic stuff I haven't really considered it
>seriously, believing it was 'just' an extension to 3-D.
>
>Obviously I was in error.
The quaternions are 4-dimensional numbers like this:
a+bi+cj+dk
You add them in the obvious way. You multiply them in a way
that's also obvious if you can just remember these rules:
i^2 = j^2 = k^2 = -1 (they're all square roots of -1)
(just like minus the dot product of vectors!)
ij = k and cyclic permutations: -
jk = i |
ki = j | just like the vector
| cross product!
ji = -k and cyclic permutations: |
kj = -i |
ik = -j -
They are very good for studying rotations in 3 dimensions AND
in 4 dimensions! But the reason *you* should learn them is that
they make it a lot easier to understand SU(2), the double cover
of the rotation group. SU(2) is just the unit quaternions:
a+bi+cj+dk with a^2 + b^2 + c^2 + d^2 = 1 !!!
Now, remember how Michael showed you some scary formulas involving SU(2)
and the 3d rotation group SO(3)?
You probably said to yourself "Ack! From which province of hell
did *those* formulas come from?"
Well, the reason the formulas were so nasty was that Michael
was using a fiendish trick invented by Pauli to write quaternions
as 2x2 complex matrices! This trick involves a bunch of arbitrary
choices, so it makes things look arbitrary and complicated. Most
physicists don't know any better! But if you define SU(2)
using quaternions, the formulas become much easier to understand!
This is yet another way in which Hamilton was too far ahead of
his time. By the time quaternions would have been really useful,
for understanding spin in quantum mechanics, almost everyone had
forgotten about them!
Michael Weiss
: [Quaternions] are very good for studying rotations in 3 dimensions AND
: in 4 dimensions! But the reason *you* should learn them is that
: they make it a lot easier to understand SU(2), the double cover
: of the rotation group.
.... as was explained in Michael Murray's excellent post in this thread. I
order you to read it, or if necessary, re-read it.
M. Murray did make one small error:
: ...so [we] have have a
: homomorphism
:
: Sp(1) -> SO(3) [Sp(1) stands for the unit quaternions]
:
: Its easy to check the kernel is +/- 1 . To see it is
: onto you have to show the image contains all reflections and
: use the fact that reflections generate SO(n).
Actually, O(3) contains reflections, but SO(3) contains only proper
rotations.
About the dimensionalities: H, the space of quaternions, is indeed
4-dimensional. The unit quaternions are a 3-dimensional subspace:
Unit quaternions = {t + xi + yj + zk | t^2 + x^2 + y^2 + z^2 = 1}
This is just like the 2-dimensional sphere sitting inside ordinary
3-dimensional space --- except pumped up one dimension.
Exercise: convince yourself that it takes 3 real numbers to specify a
rotation in 3-space. There are *lots* of ways to choose these numbers. For
example, there are the three Euler angles. Mathworld has a pretty good
graphic:
http://mathworld.wolfram.com/EulerAngles.html
complete with links to articles on quaternions, Cayley-Klein parameters,
Pauli matrices, and other stuff belonging to this circle of ideas.
By the way, how are you making out with
http://math.ucr.edu/home/baez/lie/lie.html ?
Oz
>About the dimensionalities: H, the space of quaternions, is indeed
>4-dimensional. The unit quaternions are a 3-dimensional subspace:
Ah, yes, of course. One is redundant.
> Unit quaternions = {t + xi + yj + zk | t^2 + x^2 + y^2 + z^2 = 1}
>
>This is just like the 2-dimensional sphere sitting inside ordinary
>3-dimensional space --- except pumped up one dimension.
OK. Sounds good.
>Exercise: convince yourself that it takes 3 real numbers to specify a
>rotation in 3-space.
I don't need to convince myself. That 4 were needed didn't gell, hence
the astonishment. Now I have to see why rotations in 4-D need the non-
unit quarternions. Currently it's not remotely clear to me other than
it's 4-D. OTOH rotations in 2-D are covered by the unit complex, not any
old complex number.
>By the way, how are you making out with
>http://math.ucr.edu/home/baez/lie/lie.html ?
Er, um, ohhh, weeelll, a bit busy right now .... honest ....
Oz
>>Whilst I peek at quaternionic stuff I haven't really considered it
>>seriously, believing it was 'just' an extension to 3-D.
>>
>>Obviously I was in error.
>
>The quaternions are 4-dimensional numbers like this:
>
>a+bi+cj+dk
>
>You add them in the obvious way. You multiply them in a way
>that's also obvious if you can just remember these rules:
>
>i^2 = j^2 = k^2 = -1 (they're all square roots of -1)
> (just like minus the dot product of vectors!)
>
>ij = k and cyclic permutations: -
>jk = i |
>ki = j | just like the vector
> | cross product!
>ji = -k and cyclic permutations: |
>kj = -i |
>ik = -j -
Ok. I had gleaned this at least from reading here.
>They are very good for studying rotations in 3 dimensions AND
>in 4 dimensions!
Both? Well that's unexpected. I would have thought something more
complex would be needed for 4-D.
>But the reason *you* should learn them is that
>they make it a lot easier to understand SU(2), the double cover
>of the rotation group. SU(2) is just the unit quaternions:
>
>a+bi+cj+dk with a^2 + b^2 + c^2 + d^2 = 1 !!!
I have to say that is more elegant.
>Now, remember how Michael showed you some scary formulas involving SU(2)
>and the 3d rotation group SO(3)?
>
>You probably said to yourself "Ack! From which province of hell
>did *those* formulas come from?"
Not really, but they seemed contrived, somehow.
>Well, the reason the formulas were so nasty was that Michael
>was using a fiendish trick invented by Pauli to write quaternions
>as 2x2 complex matrices! This trick involves a bunch of arbitrary
>choices, so it makes things look arbitrary and complicated. Most
>physicists don't know any better! But if you define SU(2)
>using quaternions, the formulas become much easier to understand!
So how about 4-D? How might things change in a (i,1,1,1) vs (1,1,1,1)?
There almost seems to be a direct relationship with (1,i,j,k).
>This is yet another way in which Hamilton was too far ahead of
>his time. By the time quaternions would have been really useful,
>for understanding spin in quantum mechanics, almost everyone had
>forgotten about them!
Ah, it's the attack of the mutant academic old fogeys again.
Having spent 30 years learning one way of doing it they can't face
another 30 years doing it properly, so inelegant ways persist.
Forever <shudder>.
John Baez
Oz <ozac...@despammed.com> wrote:
>John Baez <ba...@galaxy.ucr.edu> described how to multiply
>quaternions:
>>i^2 = j^2 = k^2 = -1 (they're all square roots of -1)
>> (just like minus the dot product of vectors!)
>>
>>ij = k and cyclic permutations: -
>>jk = i |
>>ki = j | just like the vector
>> | cross product!
>>ji = -k and cyclic permutations: |
>>kj = -i |
>>ik = -j -
>Ok. I had gleaned this at least from reading here.
Good! LEARN THIS STUFF! QUATERNIONS ARE COOL!
>>They are very good for studying rotations in 3 dimensions AND
>>in 4 dimensions!
>Both? Well that's unexpected. I would have thought something more
>complex would be needed for 4-D.
"More complex" - ha ha, that's a good one! Complex numbers...
quaternions... I get it! You can never resist a pun, can you?
But I said *rotations*, so I was talking about 4d SPACE, not
4d SPACETIME! You can describe a rotation in 4d space by a pair
of unit quaternions, just as you can describe a rotation in 3d
space by a single unit quaternion.
Surely you've heard me say it before: SU(2) is the double cover of
SO(3), SU(2) x SU(2) is the double cover of SO(4). And just as
surely, your eyes glazed over before any useful information got in.
But it's just a more precise way of saying what's in the previous paragraph!
On the other hand, *Lorentz transformations* are all about 4d SPACETIME.
For these we do need something more complex: namely the *complexified*
quaternions, also known as "biquaternions", also known as 2x2 complex
matrices.
As wizards are known to whisper: SL(2,C) is the double cover of the
Lorentz group.
>>But the reason *you* should learn them is that
>>they make it a lot easier to understand SU(2), the double cover
>>of the rotation group. SU(2) is just the unit quaternions:
>>
>>a+bi+cj+dk with a^2 + b^2 + c^2 + d^2 = 1 !!!
>I have to say that is more elegant.
Indeed: and more elementary too, since every kid with a little
vector algebra under his belt knows about i, j, and k.
>So how about 4-D? How might things change in a (i,1,1,1) vs (1,1,1,1)?
Huh? I presume these bizarre lists of numbers are your attempt
to talk about 4d SPACETIME versus 4d SPACE, and allude to Minkowski's
idea of working with "it" instead of "t". If so, the answer to your
question is that you need to complexify your quaternions to handle
4d spacetime. This amounts to throwing in a new extra "i" on top
of your quaternion 1,i,j,k. Probably best to call it by some other
name, to keep from going insane.
>>This is yet another way in which Hamilton was too far ahead of
>>his time. By the time quaternions would have been really useful,
>>for understanding spin in quantum mechanics, almost everyone had
>>forgotten about them!
>Ah, it's the attack of the mutant academic old fogeys again.
>Having spent 30 years learning one way of doing it they can't face
>another 30 years doing it properly, so inelegant ways persist.
>
>Forever <shudder>.
Not necessarily: old fogeys eventually become dead fogeys.
But, unless we get the word out, the *young* fogeys will never
learn that there's a better way of doing things.
So, all you kids out there: LEARN ABOUT CLIFFORD ALGEBRAS! That's
the systematic approach to this business. The real numbers, the
complex numbers, the quaternions, the biquaternions: they're
all just Clifford algebras. I'll repeat those references:
H. Blaine Lawson, Jr. and Marie-Louise Michelson, Spin Geometry,
Princeton U. Press, Princeton, 1989.
P. Budinich and A. Trautman, The Spinorial Chessboard,
Springer-Verlag, Berlin, 1988.
F. Reese Harvey, Spinors and Calibrations, Academic Press,
Boston, 1990.
Ian R. Porteous, Clifford Algebras and the Classical Groups,
Cambridge University Press, Cambridge, 1995.
Pertti Lounesto, Clifford Algebras and Spinors, Cambridge U.
Press, Cambridge, 1997.
http://math.ucr.edu/home/baez/week82.html
http://math.ucr.edu/home/baez/week93.html
http://math.ucr.edu/home/baez/week105.html
Oz
>In article <u2PkV0Df...@btopenworld.com>,
>Oz <ozac...@despammed.com> wrote:
>
>>John Baez <ba...@galaxy.ucr.edu> described how to multiply
>>quaternions:
>
>
>Good! LEARN THIS STUFF! QUATERNIONS ARE COOL!
<shock> ..... what's this new word "learn"?
>But I said *rotations*, so I was talking about 4d SPACE, not
>4d SPACETIME! You can describe a rotation in 4d space by a pair
>of unit quaternions, just as you can describe a rotation in 3d
>space by a single unit quaternion.
Ugh! That's cheating.
You can probably do it with four complex numbers, too.
>Surely you've heard me say it before: SU(2) is the double cover of
>SO(3), SU(2) x SU(2) is the double cover of SO(4). And just as
>surely, your eyes glazed over before any useful information got in.
Ah. Oh, I see. SU(2) is the unit quarternions so SU(2) x SU(2) is a pair
of unit quarternions. Duh!
>But it's just a more precise way of saying what's in the previous paragraph!
Ahh, precision, yes.
>On the other hand, *Lorentz transformations* are all about 4d SPACETIME.
>For these we do need something more complex: namely the *complexified*
>quaternions, also known as "biquaternions", also known as 2x2 complex
>matrices.
Pah! How about the octonians?
Now I seem to remember, in dark and distant times, that you can't just
keep adding complexity forever. In fact I seem to remember you saying
that the octonionas are as 'complex' as you can get.
So one might infer that rotations don;t get any trickier than octonian-
like, no matter how many dimensions you have.
Now that is a mathematical fact, so if we live in a n-dimensional
universe and my inference above is valid, then there will be a limited
number of possible variants of rotation. One presumes each will carry
some conserved quantity.
>As wizards are known to whisper: SL(2,C) is the double cover of the
>Lorentz group.
Ohhh! I don't remember what this is, presumably it's the biquarternions.
>>>But the reason *you* should learn them is that
>>>they make it a lot easier to understand SU(2), the double cover
>>>of the rotation group. SU(2) is just the unit quaternions:
>>>
>>>a+bi+cj+dk with a^2 + b^2 + c^2 + d^2 = 1 !!!
>
>>I have to say that is more elegant.
>
>Indeed: and more elementary too, since every kid with a little
>vector algebra under his belt knows about i, j, and k.
Eh? The same i,j,k? Gor blimey!
>>So how about 4-D? How might things change in a (i,1,1,1) vs (1,1,1,1)?
>
>Huh? I presume these bizarre lists of numbers are your attempt
>to talk about 4d SPACETIME versus 4d SPACE, and allude to Minkowski's
>idea of working with "it" instead of "t".
You have it in one.
>If so, the answer to your
>question is that you need to complexify your quaternions to handle
>4d spacetime. This amounts to throwing in a new extra "i" on top
>of your quaternion 1,i,j,k. Probably best to call it by some other
>name, to keep from going insane.
This would be wise, but does it work using 1,i,j,k,l?
Because if so it looks a nice relationship.
Mind you, it may be that a self-consistent mathematics cannot be put
together with this combination (due to symmetry for example) and you may
have to add a few more. One might consider these as some sort of 'hidden
dimensions'. Only joking, there's no need to ..... <ouch>.
>Not necessarily: old fogeys eventually become dead fogeys.
Unfortunately by then the young fogeys have become old fogeys, so
there's never any change.
<sigh>
Oh, and they don;t become much more old fogey than me, stuck in the
middle of the last century!
Michael Weiss
Oz:
: So one might infer that rotations don;t get any trickier than octonian-
: like, no matter how many dimensions you have.
ZAAAPPP!!!!!
Remember that we said that the double-covering of SO(3) by SU(2) (i.e., by
the unit quaternions) is in a sense a coincidence?
You still have the rotation groups SO(n), which do indeed get more
complicated, in some sense, as n increases. It's just that you can't
*represent* the rotations using division algebras (or hypercomplex number
systems, as they used to say back in the good old days --- i.e., late
nineteenth and early twentieth century).
Dave Rusin has a web page giving the whole inside scoop:
http://www.math.niu.edu/~rusin/known-math/index/products.html
Oz
Michael Weiss <mic...@spamfree.net> writes
>Oz:
>
>: So one might infer that rotations don;t get any trickier than octonian-
>: like, no matter how many dimensions you have.
>
>ZAAAPPP!!!!!
Yowch!
>Remember that we said that the double-covering of SO(3) by SU(2) (i.e., by
>the unit quaternions) is in a sense a coincidence?
Ah, yes, so you did ...
>You still have the rotation groups SO(n), which do indeed get more
>complicated, in some sense, as n increases. It's just that you can't
>*represent* the rotations using division algebras (or hypercomplex number
>systems, as they used to say back in the good old days --- i.e., late
>nineteenth and early twentieth century).
Oh dang. Just when it was all looking so, er, well organised.
So really you are back to stuff for experts only. What a pain.
Michael Weiss
Oz gripes:
: So really you are back to stuff for experts only. What a pain.
I'm not really sure what you mean by this. Of course, some succulent fruit
is always too far over one's head, even for giraffes.
However, I though you had made your peace with the basic idea of SO(n) ---
we have rotations in n-dimensions, for any n. The product of rotations is a
rotations, and the inverse of a rotation is a rotation, a fortunate state of
affairs that allows the set of all rotations (SO(n) herewith to adorn itself
with the semi-exalted title of "group". No big deal!
It just so happens that in 3 and 4 dimensions, there are sneaky ways to
express rotations using quaternions (or unitary matrices, if one prefers).
The deep significance of this is not known to me, nor apparently to The
Wizard, so we shrug it off as a coincidence.
But the shallow significance --- i.e., how x |--> q* x q determines an
element of SO(3) --- is quite within your grasp. Have you chomped through
this tasty morsel yet?
Squark
"Michael Weiss" <mic...@spamfree.net> wrote in message news:<8hZZ9.25018$uR.2...@nwrddc01.gnilink.net>...
> It just so happens that in 3 and 4 dimensions, there are sneaky ways to
> express rotations using quaternions (or unitary matrices, if one prefers).
> The deep significance of this is not known to me, nor apparently to The
> Wizard, so we shrug it off as a coincidence.
In fact, as the Wizard mentioned, the way to generalize this stuff is
Clifford algebras. You are right the relation to normed division algebras
is "coincidental".
Best regards,
Squark
------------------------------------------------------------------
Write to me using the following e-mail:
Skvark_N...@excite.exe
(just spell the particle name correctly and change the
extension in the obvious way)
John Baez
Michael Weiss <mic...@spamfree.net> wrote:
>Oz:
>: So one might infer that rotations don't get any trickier than octonian-
>: like, no matter how many dimensions you have.
>ZAAAPPP!!!!!
Hmm... sounds like the apprentice wizard is using a bug-zapper instead
of a full-fledged thunderbolt.
>Remember that we said that the double-covering of SO(3) by SU(2) (i.e., by
>the unit quaternions) is in a sense a coincidence?
Yes.
>You still have the rotation groups SO(n), which do indeed get more
>complicated, in some sense, as n increases. It's just that you can't
>*represent* the rotations using division algebras (or hypercomplex number
>systems, as they used to say back in the good old days --- i.e., late
>nineteenth and early twentieth century).
Hmm: since Oz probably doesn't know what a "division algebra" is,
I'm not sure how much this comment will help. Maybe we should be
a bit more explicit:
There's a nice relation between rotations in 2 dimensions and
the unit complex numbers:
U(1) is the same as SO(2)
^ ^
unit complex rotations in
numbers 2 dimensions
Inspired by this, Hamilton found a nice relation between rotations
in 3 dimensions and the unit quaternions:
SU(2) is a double cover of SO(3)
^ ^
unit quaternions rotations in
3 dimensions
But don't be fooled into thinking there's a simple pattern at work here!
There's nothing quite like these wonderful facts in higher dimensions.
The pattern fizzles out. There's one last feeble flicker of life in
7 dimensions: unit octonions give rotations in 7 dimensions, but they
don't give all of them, and they don't form a group, since the
octonions aren't associative. These two problems cancel each
other to a certain extent, but it's a bit subtle... and it only works
in 7 dimensions.
So what's going on?
I'll just say the all-important buzzword: Clifford algebras.
These are what you need to think of the rotation group SO(n),
or actually its double cover Spin(n), as sitting inside an
associative algebra of some sort. They work in all dimensions.
William Clifford was born in 1845, two years after Hamilton
discovered the quaternions. He died at the age of 37, and
he didn't write much, but his generalization of the quaternions
is incredibly important for physics - especially for the theory
of spinors.
John Baez
Oz <ozac...@despammed.com> wrote:
>John Baez <ba...@galaxy.ucr.edu> write:
>>You can describe a rotation in 4d space by a pair
>>of unit quaternions, just as you can describe a rotation in 3d
>>space by a single unit quaternion.
>Ugh! That's cheating.
No, it's not. It's incredibly perfect. Quaternions form
a 4-dimensional space. You can multiply a quaternion X
on the LEFT by a unit quaternion U, and you can multiply it
on the RIGHT by a unit quaternion V, and if you do both:
X |----> UXV
you get an arbitrary rotation in 4-dimensional space.
Note something important here: quaternion multiplication
doesn't commute, so multiplying on the left is different
from multiplying on the right. That's why we need two
quaternions U and V.
The complex numbers do commute, so we can get all rotations
in 2-dimensional space just by multiplying on the LEFT by
a unit complex number:
X |----> UX
You don't get extra mileage from multiplying on the right
in this case.
From this viewpoint, quaternions are naturally connected
with 4 dimensions just as complex numbers are connected with 2
dimensions. From this viewpoint, it's sort of an extra miracle that
we can also use quaternions to describe rotations in *3* dimensions!
I won't write down the formula for this one... you'll have fun
racking your brain trying to guess it, and I'll have even more
fun laughing at your guesses. It's a slight but sneaky variation of
the formulas I just wrote down.
>>Surely you've heard me say it before: SU(2) is the double cover of
>>SO(3), SU(2) x SU(2) is the double cover of SO(4). And just as
>>surely, your eyes glazed over before any useful information got in.
>Ah. Oh, I see. SU(2) is the unit quarternions so SU(2) x SU(2) is a pair
>of unit quarternions. Duh!
Yeah. But please, don't write quaRternions - there's no "R"
right there... a quaternion is not a "quarter" of "nion".
(I assume you know the passage in the Bible which mentions
quaternions; this is where Hamiltonian got the word.)
>>On the other hand, *Lorentz transformations* are all about 4d SPACETIME.
>>For these we do need something more complex: namely the *complexified*
>>quaternions, also known as "biquaternions", also known as 2x2 complex
>>matrices.
>Pah! How about the octonians?
Grrr! You're begging for a thunderbolt. It's octoniOns, not octoniAns.
Anyway, just as the quaternions are very related to rotations in
4 and also 3 dimensions, octonions are very related to rotations
in 8 and also 7 dimensions.
>Now I seem to remember, in dark and distant times, that you can't just
>keep adding complexity forever. In fact I seem to remember you saying
>that the octonions are as 'complex' as you can get.
Yes, while keeping certain nice properties, particularly the
ability to divide by anything except zero.
>So one might infer that rotations don't get any trickier than octonian-
>like, no matter how many dimensions you have.
Well, as I've said elsewhere, we are currently chatting about
special coincidences that happen in low dimensions - I'm afraid
even 8 dimensions counts as "low" to us math wizards, who can
count considerably further than that. To systematically understand
rotations in ALL dimensions using algebras, we need to give up this
"real, complex, quaternion, octonion" game and switch to Clifford
algebras, which go "real, complex, quaternion, biquaternion, [...]".
>>As wizards are known to whisper: SL(2,C) is the double cover of the
>>Lorentz group.
>Ohhh! I don't remember what this is, presumably it's the biquaternions.
I don't know which "this" you're referring to, but neither
SL(2,C) nor the Lorentz group is the biquaternions. However,
you are close.
As you should know by know, SL(2,C) is the 2x2 complex matrices
with determinant 1: the "2" tells you they're 2x2, the "C"
means "complex", the "S" means "special" i.e. determinant 1,
and the "L" just reassures you that they are linear transformations.
The biquaternions are just a fun different way of talking about
2x2 complex matrices.
So, SL(2,C) does indeed sit inside the biquaternions, very much
as SU(2) sits inside the quaternions.
It's all part of a wonderful pattern... but I don't want to put
too much information in your brain at once, or other important
stuff will fall out.
>>>>But the reason *you* should learn them is that
>>>>they make it a lot easier to understand SU(2), the double cover
>>>>of the rotation group. SU(2) is just the unit quaternions:
>>>>
>>>>a+bi+cj+dk with a^2 + b^2 + c^2 + d^2 = 1 !!!
>>>I have to say that is more elegant.
>>
>>Indeed: and more elementary too, since every kid with a little
>>vector algebra under his belt knows about i, j, and k.
>Eh? The same i,j,k? Gor blimey!
Yes! Hamilton invented them when he made up the quaternions a+bi+cj+dk.
Gibbs took the quaternions and chopped them into their "scalar" part "a"
and their "vector" part "bi+cj+dk". The quaternionists thought this
was a heinous crime, but they lost the great battle that followed.
Now schoolkids learn their scalars and vector separately... and only
when you become a math wizard do you learn it can be handy to work
with the whole quaternion a+bi+cj+dk at once.
If only more people knew this, SU(2) wouldn't seem so scary!
>>>So how about 4-D? How might things change in a (i,1,1,1) vs (1,1,1,1)?
>>Huh? I presume these bizarre lists of numbers are your attempt
>>to talk about 4d SPACETIME versus 4d SPACE, and allude to Minkowski's
>>idea of working with "it" instead of "t".
>You have it in one.
Okay. You should have said (-1,1,1,1) vs (1,1,1,1).
>>If so, the answer to your
>>question is that you need to complexify your quaternions to handle
>>4d spacetime. This amounts to throwing in a new extra "i" on top
>>of your quaternion 1,i,j,k. Probably best to call it by some other
>>name, to keep from going insane.
>This would be wise, but does it work using 1,i,j,k,l?
Well, this is the biquaternion business, but I'm getting too
tired to explain it, and I know if I do you'll forget something
more important that I said earlier on in this post.
>Because if so it looks a nice relationship.
It's great, beautiful, wondrous... it's the sort of thing
that keeps me happy when the skies are grey! It's the sort
of thing that makes me glad I didn't become a rich businessman!
>Mind you, it may be that a self-consistent mathematics cannot be put
>together with this combination (due to symmetry for example) and you may
>have to add a few more. One might consider these as some sort of 'hidden
>dimensions'. Only joking, there's no need to ..... <ouch>.
Heh. I'm not as opposed to extra dimensions as you may think.
They give pretty math, but the question is whether they give a
theory of physics that really explains anything, instead of just
predicting wads of particles that seem not to exist. The Standard
Model is brimming with mathematical mysteries, waiting for someone
to come along and with a truly good idea.
>>Not necessarily: old fogeys eventually become dead fogeys.
>Unfortunately by then the young fogeys have become old fogeys, so
>there's never any change.
><sigh>
It's not really as bad as that.
>Oh, and they don't become much more old fogey than me, stuck in the
>middle of the last century!
You said that last century, too.
Jeffery
> As you should know by know, SL(2,C) is the 2x2 complex matrices
> with determinant 1: the "2" tells you they're 2x2, the "C"
> means "complex", the "S" means "special" i.e. determinant 1,
> and the "L" just reassures you that they are linear transformations.
What does "Z" stand for in SL(2,Z)?
Jeffery
[Moderator's note: here and elsewhere in math, "Z" stands for
the integers. - jb]
Michael Weiss
: the integers. - jb]
Thanks to the German word Zahlen (numbers). Also, Q is the rationals
(quotient), and H is the quaternions (Hamilton). N, R and C are obvious.
I've occasional seen B used for {0,1} (Boolean).