Classical physics in Ashtekar's variables
Nathan Urban
and geometric quantities in classical GR in terms of Ashtekar's variables.
For example: if/how parallel transport (or Fermi-Walker transport)
of the metric-derived Levi-Civita connection can be written in terms
of the parallel transport of the SU(2) generalized Ashtekar connection
(or the parallel transport of the "spin connection associated with the
triad" in terms of which I've seen the generalized Ashtekar connection
written), the geodesic or Raychaudhuri equations in Ashtekar variables,
whether 4-vectors such as 4-velocity or tensors such as stress-energy
can/should be reformulated in some other form (e.g., spinorial) when
using the Ashtekar variables in order to most simplify the equations
which involve them, etc. (By "generalized Ashtekar connection" I mean
A^i_a = \Gamma^i_a + \gamma k^i_a, of which the complex Ashtekar and real
Barbero connections are special cases; \Gamma^i_a is the spin connection
I mentioned.)
I've looked through Bruegmann's bibliography in gr-qc/9303015 and I
don't see anything that sounds like it specifically addresses what I'm
looking for; I see many papers with ambiguous titles that _might_ have
something relevant, but before I search through a gazillion papers I
wanted to know if anyone could save me the trouble.
By the way, I read a remark of Smolin's: "Sen found [his connection
...] by following the hint from twistor theory and asking how a neutrino
would see the geometry of spacetime. This then inspired Abhay Ashtekar
to invent a complete reformulation of Einstein's theory in terms
of new variables adapted to describe the propagation of a neutrino
through spacetime." Is it literally true that the connections describe
neutrino physics, or is it just an over-simplified way of saying that
the connection is chiral? The fact that Sen published a paper entitled
"On the existence of neutrino `zero-modes' in vacuum spacetimes" suggests
that it might be literally true. In that case, what is the physical
connection (pun not intended) between the Ashtekar connection and the
propagation of a neutrino through spacetime?
john baez
Nathan Urban <nur...@vt.edu> wrote:
>By the way, I read a remark of Smolin's: "Sen found [his connection
>...] by following the hint from twistor theory and asking how a neutrino
>would see the geometry of spacetime. This then inspired Abhay Ashtekar
>to invent a complete reformulation of Einstein's theory in terms
>of new variables adapted to describe the propagation of a neutrino
>through spacetime." Is it literally true that the connections describe
>neutrino physics, or is it just an over-simplified way of saying that
>the connection is chiral?
I'll answer this easy question now and perhaps the other, harder
ones later. (I'm afraid my replies to those other ones may not
be very satisfactory.)
In the Standard Model neutrinos are massless and described by
a chiral spinor field, also known as a "Weyl spinor". Recent
experiments suggests that neutrinos may actually be massive,
hence not described by a Weyl spinor field. But let me ignore
that possibility and talk about Standard Model neutrinos, since
that's what Smolin is doing here.
Saying that the Ashtekar connection describes the propagation of
a neutrino is simply a way of saying that the connection is chiral,
but it's by no means an oversimplification, since it's actually *true*
that true that the propagation of a massless chiral spinor particle
through spacetime depends only on one chiral part of complexified
Lorentz connection.
Remember, the Lorentz Lie algebra is isomorphic to sl(2,C), so
when we complexify it, it "splits", and we get a Lie algebra
isomorphic to sl(2,C) + sl(2,C). These are called the left-handed
and right-handed parts. A neutrino, being left-handed, transforms
according to the obvious 2-dimensional representation of the left-
handed sl(2,C). This is basically what we mean by saying that it's
a chiral spinor field, or more precisely, a left-handed spinor field.
You may worry about why I'm complexifying the Lorentz Lie algebra here.
Perhaps I can allay some of your fears by noting that any representation
of the complexification of a Lie algebra gives by restriction a
representation of the original Lie algebra. So you may think of the
complexification as a cute trick to help describe representations of
the original Lie algebra.
Now: whenever we move any sort of particle along a path in a curved
spacetime, we have to apply the holonomy of the Levi-Civita
connection - which we may think of as an element of the Lorentz
group - to its wavefunction. If we know how to do this, we can
figure out everything about how the particle propagates - at
least if it's uncharged. But to do this, we need to know what
representation of the Lorentz group the particle transforms under.
For neutrinos, we use the left-handed spinor representation.
Now, if we take the holonomy of the Levi-Civita connection, but
then apply the left-handed spinor representation to that, we are
just taking the holonomy of the Ashtekar connection! After all,
the Ashtekar connection is *defined* to be the left-handed part
of the Levi-Civita connection (viewing the latter as a connection
taking values in the *complexified* Lorentz Lie algebra).
So in short neutrinos propagate in a way which is completely
described by the Ashtekar connection.
The less obvious thing, of course, is that general relativity
becomes simpler in some ways if you formulate it in terms of
the Ashtekar connection.
It's worth noting that all this was dimly foreseen by Wheeler
quite some time ago. He wrote:
"It is impossible to accept any description of elementary particles
that does not have a place for spin 1/2. What, then, has any purely
geometric description to offer in explanation of spin 1/2 in general?
More particularly and more importantly, what place is there in quantum
geometrodynamics for the neutrino --- the only entity of half-integral
spin that is a pure field in its own right, in the sense that it has
zero rest mass and moves at the speed of light? No clear or satisfactory
answer is known to this question today. Unless and until an answer
is forthcoming, *pure geometrodynamics must be judged deficient as a
basis of elementary particle physics*."
For a bit of how Penrose, Newman, Rindler and Ashtekar dealt with
this problem by developing spinorial formulations of general relativity,
see:
http://math.ucr.edu/home/baez/week109.html
Nathan Urban
> In article <6tfj92$bn7$1...@crib.corepower.com>,
> Nathan Urban <nur...@vt.edu> wrote:
> >Is it literally true that the connections describe
> >neutrino physics, or is it just an over-simplified way of saying that
> >the connection is chiral?
> I'll answer this easy question now and perhaps the other, harder
> ones later. (I'm afraid my replies to those other ones may not
> be very satisfactory.)
Okay. I don't really need direct answers to the other ones, I just
wanted to know if these things have been studied and where I could go
for more information.
> You may worry about why I'm complexifying the Lorentz Lie algebra here.
> Perhaps I can allay some of your fears by noting that any representation
> of the complexification of a Lie algebra gives by restriction a
> representation of the original Lie algebra. So you may think of the
> complexification as a cute trick to help describe representations of
> the original Lie algebra.
So is the left-handed sl(2,C) representation a subrepresentation of the
original sl(2,C)?
> Now: whenever we move any sort of particle along a path in a curved
> spacetime, we have to apply the holonomy of the Levi-Civita
> connection - which we may think of as an element of the Lorentz
> group -
That's because the Levi-Civita connection is compatible with the metric,
hence parallel transport preserves the inner product, therefore must be
an element of the Lorentz group, correct?
> to its wavefunction. If we know how to do this, we can
> figure out everything about how the particle propagates - at
> least if it's uncharged. But to do this, we need to know what
> representation of the Lorentz group the particle transforms under.
> For neutrinos, we use the left-handed spinor representation.
Okay, I get this.
> Now, if we take the holonomy of the Levi-Civita connection, but
> then apply the left-handed spinor representation to that, we are
> just taking the holonomy of the Ashtekar connection! After all,
> the Ashtekar connection is *defined* to be the left-handed part
> of the Levi-Civita connection (viewing the latter as a connection
> taking values in the *complexified* Lorentz Lie algebra).
Does it have to be left-handed for it to describe GR? Or just to describe
neutrino propagation? Can you do GR with an anti-self-dual connection?
> It's worth noting that all this was dimly foreseen by Wheeler
> quite some time ago. He wrote:
> "It is impossible to accept any description of elementary particles
> that does not have a place for spin 1/2. What, then, has any purely
> geometric description to offer in explanation of spin 1/2 in general?
> More particularly and more importantly, what place is there in quantum
> geometrodynamics for the neutrino --- the only entity of half-integral
> spin that is a pure field in its own right, in the sense that it has
> zero rest mass and moves at the speed of light? No clear or satisfactory
> answer is known to this question today. Unless and until an answer
> is forthcoming, *pure geometrodynamics must be judged deficient as a
> basis of elementary particle physics*."
I've seen that quote, and have never understood it. I still don't
completely understand what he was getting at, even after reading Week 109.
I guess I don't see why wormholes can't describe a spin-1/2 particle
and thus fit in with the geometrodynamics program. And if a fermion
can't be a wormhole, then does it have some other nice description
within classical GR that can be understood in terms of the conventional
geometric formulation?
> For a bit of how Penrose, Newman, Rindler and Ashtekar dealt with
> this problem by developing spinorial formulations of general relativity,
> see:
> http://math.ucr.edu/home/baez/week109.html
Hmm, okay. That reminds me.. what's going on with twistors nowadays?
Have they made any connection with quantum gravity? Most of what I've
seen is classical, and I haven't heard anything about them recently.
I have another question about Ashtekar connections. I still don't
understand it completely, but I'm starting to get a bit of a handle on
the original complexified Ashtekar connection. But I'd like to see how
it relates to the real and generalized connections; I know that most
of the literature is on the complexified connection but I think it may
be one of those cases where I'll understand the special case better
if I also understand the general case. If I understand it correctly,
the complexified connection is defined on a bundle over the base space
MxC^4. What are the other connections defined over? And what are their
symmetries? The recent literature always speaks of the real connection
being SU(2), but I looked at Barbero's paper and all I see is SO(3).
Now, infinitesimally their parallel transport ought to behave the same
way since they share a Lie algebra, but globally they don't seem to be
the same thing. I haven't yet been able to see quite how the contents
of his paper connect with the definition in, say, Rovelli's Living Review.
How would I fill in the following table?
Ashtekar holonomy is defined over
connection an element of the bundle
-------------------------------------------------
complexified
real
generalized
A. Garrett Lisi
>
> In article <6tihif$8dc$1...@pravda.ucr.edu>, ba...@galaxy.ucr.edu (john baez) wrote:
> >
> >
> > "It is impossible to accept any description of elementary particles
> > that does not have a place for spin 1/2. What, then, has any purely
> > geometric description to offer in explanation of spin 1/2 in general?
> > More particularly and more importantly, what place is there in quantum
> > geometrodynamics for the neutrino --- the only entity of half-integral
> > spin that is a pure field in its own right, in the sense that it has
> > zero rest mass and moves at the speed of light? No clear or satisfactory
> > answer is known to this question today. Unless and until an answer
> > is forthcoming, *pure geometrodynamics must be judged deficient as a
> > basis of elementary particle physics*."
>
> I've seen that quote, and have never understood it. I still don't
> completely understand what he was getting at, even after reading Week 109.
> I guess I don't see why wormholes can't describe a spin-1/2 particle
> and thus fit in with the geometrodynamics program. And if a fermion
> can't be a wormhole, then does it have some other nice description
> within classical GR that can be understood in terms of the conventional
> geometric formulation?
I spent a year or so trying to figure out the physical basis of spinor
fields within the context of GR. I was, and still am, deeply
unsatisfied with the fundamental definition of spinors as nothing more
then a field transforming under a complex irreducible representation of
the Lorentz group. For those who may share an interest in this
question, I have arrived at a compatible alternative definition that
obtains the classical spinor degrees of freedom elegantly from those of
geometrodynamics.
The idea is to take the "frame field", "vielbein", or simply
"orthonormal basis vectors" of the manifold as fundamental physical
objects. This frame may be factored into a part that contains the
information of the metric and a separate part that encodes the
rotational orientation of the frame. The rotational part is shown most
elegantly through the use of Clifford algebra to transform as a spinor
field. All this is worked out in
http://xxx.lanl.gov/abs/gr-qc/9804033
I also argue in that paper that the dynamics of the spinor field,
unconstrained by metric geometrodyamics alone, should be constrained
such that the time vector (1-form) is integrable to a scalar time field,
a restriction that requires spinor dynamics and nicely resolves the
problem of scalar time on a manifold as posed by Rovelli. The next step
is to use Kaluza-Klein theory to get the gauge fields and couplings and
then quantize the whole batch, a task I would love to undertake if I can
manage to find post-doc funding.
Best,
Garrett Lisi
UCSD physics
g...@lisi.com
john baez
Nathan Urban <nur...@vt.edu> wrote:
>In article <6tihif$8dc$1...@pravda.ucr.edu>, ba...@galaxy.ucr.edu
(john baez) wrote:
>> I'll answer this easy question now and perhaps the other, harder
>> ones later. (I'm afraid my replies to those other ones may not
>> be very satisfactory.)
>
>Okay. I don't really need direct answers to the other ones, I just
>wanted to know if these things have been studied and where I could go
>for more information.
After thinking about it a bit more, I decided that I'd never seen
most of the things you were talking about done in terms of Ashtekar's
"new variables". So what you should really learn, if you haven't
yet, are the formulas that let you express traditional variables
(most fundamentally the metric, but also the Levi-Civita connection)
in terms of the new variables (the densitized triad field and chiral
connection). Once you know these formulas you can translate anything
you want into the new variables.
For this, it's crucial that you learn about the *spacetime* version
of the new variables, not just Ashtekar's original version, which
dealt with space at a fixed time. The spacetime version is sometimes
called the "Plebanski" formulation of general relativity, since it
was developed by the Mexican physicist Plebanski before Ashtekar did
his thing, in a paper that never got much attention until *after*
Ashtekar did his thing.
You can read a quick summary of the Plebanski formulation of general
relativity in my paper "Spin networks in nonperturbative quantum gravity" -
this is probably the easiest introduction for someone who doesn't like
indices. But for more details it's best to read chapter III of Ashtekar's
book _New Perspectives in Canonical Gravity_, and also Domenico Giulini's
paper "Ashtekar variables in classical general relativity", available as
gr-qc/9312032. (That book by Ashtekar is a crucial reference, but
unfortunately a bit hard to find. It's supposedly being distributed
by the American Physical Society. If you get it, make sure to get ahold
of the errata, which are available at the website of the Center for
Gravitational Physics and Geometry.)
>> You may worry about why I'm complexifying the Lorentz Lie algebra here.
>> Perhaps I can allay some of your fears by noting that any representation
>> of the complexification of a Lie algebra gives by restriction a
>> representation of the original Lie algebra. So you may think of the
>> complexification as a cute trick to help describe representations of
>> the original Lie algebra.
>
>So is the left-handed sl(2,C) representation a subrepresentation of the
>original sl(2,C)?
I'm not sure what you're asking, because if you speak of "a subrepresentation
of a group" you must really mean "a subrepresentation of some representation
R of that group", but you're not telling me *which* representation R!
So instead of answering your question, I'll just fire off a volley of
words in its general direction.
sl(2,C) is a sub-Lie-algebra of its complexification, which is isomorphic
to sl(2,C) + sl(2,C). It's the "diagonal" sub-Lie-algebra consisting of
elements of the form (x,x). This is how it always goes when we complexify
a Lie algebra that's already complex.
sl(2,C) + sl(2,C) has two blitheringly obvious 2-dimensional representations,
coming from the obvious rep of sl(2,C). We can restrict *any* rep
of sl(2,C) + sl(2,C) to the diagonal sub-Lie-algebra. If we do this
to the two blitheringly obvious 2-dimensional reps, we get two reps of
sl(2,C) called the left-handed and right-handed spinor reps.
So we're not forming a subrepresentation of a Lie algebra here, we're
restricting a representation to a sub-Lie-algebra.
>> Now: whenever we move any sort of particle along a path in a curved
>> spacetime, we have to apply the holonomy of the Levi-Civita
>> connection - which we may think of as an element of the Lorentz
>> group -
>
>That's because the Levi-Civita connection is compatible with the metric,
>hence parallel transport preserves the inner product, therefore must be
>an element of the Lorentz group, correct?
Correcto. Here of course we are secretly trivializing our bundle at
both endpoints in order to think of the holonomy as an element of
the structure group. Physicists do this without even mentioning it.
>> Now, if we take the holonomy of the Levi-Civita connection, but
>> then apply the left-handed spinor representation to that, we are
>> just taking the holonomy of the Ashtekar connection! After all,
>> the Ashtekar connection is *defined* to be the left-handed part
>> of the Levi-Civita connection (viewing the latter as a connection
>> taking values in the *complexified* Lorentz Lie algebra).
>
>Does it have to be left-handed for it to describe GR? Or just to describe
>neutrino propagation? Can you do GR with an anti-self-dual connection?
You can do general relativity with either a self-dual or anti-self-dual
connection, and it scarcely makes a difference. It's just a matter of
convention. In fact, half the people working on general relativity seem
to do it one way, and half the other. General relativity is invariant
under parity, so it doesn't care about left versus right.
What's *not* a matter of convention is that a neutrino, coming towards
you, will be spinning a certain way. (An antineutrino will spin the
opposite way, but it's *not* just a matter of convention which we call
neutrinos and which we call antineutrinos. More precisely, P violation
and CP violation allow one to distinguish between left versus right,
and also between particle versus antiparticle.)
>> It's worth noting that all this was dimly foreseen by Wheeler
>> quite some time ago. He wrote:
>
>> "It is impossible to accept any description of elementary particles
>> that does not have a place for spin 1/2. What, then, has any purely
>> geometric description to offer in explanation of spin 1/2 in general?
>> More particularly and more importantly, what place is there in quantum
>> geometrodynamics for the neutrino --- the only entity of half-integral
>> spin that is a pure field in its own right, in the sense that it has
>> zero rest mass and moves at the speed of light? No clear or satisfactory
>> answer is known to this question today. Unless and until an answer
>> is forthcoming, *pure geometrodynamics must be judged deficient as a
>> basis of elementary particle physics*."
>
>I've seen that quote, and have never understood it. I still don't
>completely understand what he was getting at, even after reading Week 109.
>I guess I don't see why wormholes can't describe a spin-1/2 particle
>and thus fit in with the geometrodynamics program.
Well, unless you pull some sly theoretical trick or other, when you
turn a wormhole around 360 degrees it's going to look just like it
did at first, so it'll be bosonic, not fermionic. Since Wheeler wrote
this, people have come up with a couple of interesting ways to try to
get around this sort of problem, like Sorkin's trick, or the trick Witten
used to get fermions in the Skryme model. But back in the good old days,
everybody thought that the only way you could get half-integral spin *out*
of a system was by explicitly putting it *in* right at the beginning.
And since general relativity (or for that matter the Yang-Mills equations)
didn't seem to refer to half-integral spin, Wheeler was worried for his
attempt to get "charge without charge".
>And if a fermion
>can't be a wormhole, then does it have some other nice description
>within classical GR that can be understood in terms of the conventional
>geometric formulation?
You can certainly couple general relativity to fermionic fields, but
Wheeler wasn't wanting to put fermions in "by hand", he was wanting
them to fall out.
>That reminds me.. what's going on with twistors nowadays?
I don't know - nothing I've seen yet has convinced me to study them
more, though when I went to a talk by Penrose they seemed vaguely
comprehensible and somewhat charming.
>I have another question about Ashtekar connections. I still don't
>understand it completely, but I'm starting to get a bit of a handle on
>the original complexified Ashtekar connection. But I'd like to see how
>it relates to the real and generalized connections [...]
Wait a minute. The subject of "generalized connections" goes off
in a direction largely orthogonal to what you're actually interested in
here. Given *any* sort of principal bundle on a manifold, we can talk
about connections on it, and we can also talk about generalized connections
on it, which are like a stripped-down notion of connection that *only*
tells you how to do parallel transport along smooth paths, *without
any requirement that the holonomy depend smoothly on the path*. Generalized
connections thus allow more singular behavior than old-fashioned connections.
That's why they are important in quantum field theory. None of this has
anything in particular to do with Ashtekar's study of general relativity
using self-dual connections, except for the historical accident that Ashtekar
happened to help invent this idea, too.
So pick one thing or another to think about!
>I know that most
>of the literature is on the complexified connection but I think it may
>be one of those cases where I'll understand the special case better
>if I also understand the general case. If I understand it correctly,
>the complexified connection is defined on a bundle over the base space
>MxC^4.
No, it's still defined on a bundle over the same base space M,
just a different bundle. We're not complexifying *spacetime*
here!
>What are the other connections defined over?
Well, I'm not sure which "other connections" you're talking about here,
but as long as we're doing general relativity and not doing something
funky like complexifying spacetime or working in some other dimension,
the connections we're interested in are typically connections on good
old "spacetime" - some smooth 4-manifold or other - or else on "space" -
a smooth 3-manifold.
>And what are their symmetries?
Well, you really have to tell me what "they" are, to get me to answer
this. And if you tell me what "they" are, you'll probably tell me the
answer!
>The recent literature always speaks of the real connection being SU(2),
>but I looked at Barbero's paper and all I see is SO(3). Now, infinitesimally
>their parallel transport ought to behave the same way since they share a Lie
>algebra, but globally they don't seem to be the same thing.
First of all, you should note that physicists are often sloppy about the
difference between SU(2) and SO(3) and SL(2,C). I'm not accusing Barbero
*in particular* of this - it's just something you gotta get used to dealing
with. Basically you gotta read the papers and figure out what they must
mean in order to make sense, without worrying much about what they *say*.
Or you can act like a physicist and just sort of mellow out and stop worrying
about the difference between these various groups (which I don't actually
advise you to do).
Anyway, what you probably really need to know is: under what situations can
I get from one sort of connection to another, and in what ways? To take
the simplest example: if you have a manifold M with an SO(3) bundle P over
it *equipped with a trivialization*, you can just form the double cover
P' of P and that will be an SU(2) bundle, and any connection on P will
naturally lift to a connection on P', and any connection on P' will naturally
push down to a connection on P. Since physicists often work locally and
fix a trivialization of every bundle in sight, this allows them to be quite
relaxed about the difference between SO(3) and SU(2) connections. However,
since I often work globally and don't want to fix a trivialization unless
I have to, my life is not so easy: I need to think about "spin structures"
and stuff like that.
Anyway, the little fact in the previous paragraph is one of a bunch of
facts you need to learn, and when you have all these facts at your disposal,
then you can figure out what physicists must mean when they say the things
they do. I can dole out more such facts if you want. (Of course, it's
even better to understand why they are true, so you can re-derive them when
you need to know them!)
>How would I fill in the following table?
>
> Ashtekar holonomy is defined over
>connection an element of the bundle
>-------------------------------------------------
>complexified
>real
>generalized
Well, as I said, the third item on your list is a red herring.
Also, there are several choices of correct answer for the first
two items. And are we talking about *space* or *spacetime* here?
Let's say we're talking about *space*. Then I would tend to say
that the *real* Ashtekar connection is a connection on the internal
spinor bundle over space. This is an SU(2) bundle, so the holonomy
is an element of SU(2) after we trivialize the bundle over the
endpoints of our path. And I would say that the *complex* Ashtekar
connection is a connection on the restriction to space of the left-handed
spinor bundle over spacetime to space. This is an SL(2,C) bundle,
so the holonomy is an element of SL(2,C) after we trivialize the bundle
over the endpoints of our path. And then there is the ordinary
Levi-Civita connection, and the internal Lorentz connection, and
complexified internal Lorentz connection... but you have a habit of
trying to learn about 43 different things at once, and I don't want
to get into all 43 simultaneously.
Nathan Urban
compose a more in-depth reply. However, I do want to correct one large
terminological error on my part that considerably changed the intent of
my questions:
In article <6u67h3$uck$1...@pravda.ucr.edu>, ba...@galaxy.ucr.edu (john baez) wrote:
> In article <6u0igp$8l0$1...@crib.corepower.com>, Nathan Urban <nur...@vt.edu> wrote:
> >I have another question about Ashtekar connections. I still don't
> >understand it completely, but I'm starting to get a bit of a handle on
> >the original complexified Ashtekar connection. But I'd like to see how
> >it relates to the real and generalized connections [...]
> Wait a minute. The subject of "generalized connections" goes off
> in a direction largely orthogonal to what you're actually interested in
> here.
Oooops!!
I guess I accidentally used a term that was already in common usage; I
didn't know that there was such a thing as a "generalized connection", and
it certainly is orthogonal to what I was interested in. In my original
post I defined what I meant by a "generalized Ashtekar connection" but
you probably missed it or got confused when I started abbreviating it
to just "the generalized connection". I said:
"(By `generalized Ashtekar connection' I mean A^i_a = \Gamma^i_a + \gamma
k^i_a, of which the complex Ashtekar and real Barbero connections are
special cases; \Gamma^i_a is the spin connection ...)". \gamma is the
Immirzi parameter and k^i_a is the extrinsic curvature.
I called it the "generalized Ashtekar connection" because you can get
the common (classically-) equivalent Ashtekar-type connections from it
by choosing the Immirzi parameter appropriately. I don't know what its
real name is.
Perhaps you could look at my questions about it again in light of this,
while I read over the rest of your post?
> >What are the other connections defined over?
> >And what are their symmetries?
> Well, you really have to tell me what "they" are, to get me to answer
> this. And if you tell me what "they" are, you'll probably tell me the
> answer!
The real Barbero and the "generalized" Ashtekar connections. I wanted
to know what their bundles and symmetry groups were compared to the
complex Ashtekar connection.
john baez
john baez <ba...@galaxy.ucr.edu> wrote:
>I would say that the *complex* Ashtekar connection is a connection
>on the restriction to space of the left-handed spinor bundle over
>spacetime to space.
I just got a phone call from the Department of Redundancy Department
congratulating me on that sentence. I meant to say:
The *complex* Ashtekar connection is a connection on the restriction
to space of the left-handed internal spinor bundle over spacetime.
What does that mouthful mean, anyway? Well, if spacetime is 4-dimensional
and equipped with a Lorentzian metric, certain topological conditions
allow us to pick a "spin structure". This is a bundle with gauge group
SL(2,C) that double covers the orthonormal frame bundle. (Remember,
SL(2,C) is the double cover of the Lorentz group.)
If we complexify this "spin structure" bundle, we get a bundle whose
gauge group is the complexification of SL(2,C), namely SL(2,C) x SL(2,C).
This "complex spin bundle" is the sum of two bundles with gauge group
SL(2,C), which are called the left-handed and right-handed spin bundles.
In the Ashtekar-Plebanski approach to general relativity we throw
in a slight extra twist. We don't want to assume spacetime has a
god-given Lorentzian metric on it. But this is a slight nuisance,
since one can't really define spinors without picking such a metric! [1]
So we pull a sneaky trick: we fix a bundle *isomorphic* to the tangent
bundle, which we call the "internal space", and fix a Lorentzian metric
on *that*. If certain topological conditions hold, the orthonormal
frame bundle of this "internal space" bundle has a double cover with
gauge group SL(2,C), which we call the "internal spin bundle". Then
we just copy what we did before and get a "left-handed" and "right-
handed" internal spin bundle.
The Ashtekar connection on spacetime is a connection on the left-handed
internal spin bundle. [2]
If we now pick a 3-dimensional submanifold of spacetime and call
it "space", we can restrict the left-handed internal spin bundle
to space and get an SL(2,C) bundle over space. If we are studying
"canonical" general relativity, i.e., working on space rather than
spacetime, then the (complex) Ashtekar connection is a connection
on this bundle. [3]
-----------------------------------------------------------------------
[1] Technically speaking, one can get away with less: one needs,
first of all, a reduction of the gauge group of the frame bundle
from GL(4) to SO(3,1) - this is a little bit less data than a metric,
I guess, but still a lot - and then a lift of that to SL(2,C).
[2] We can also form a vector bundle called the bundle of "left-handed
internal spinors", using the fundamental representation of SL(2,C).
We can also think of the Ashtekar connection as a connection on this
"left-handed internal spinor bundle".
[3] Or, equivalently, on the restriction to space of the left-handed
internal spinor bundle! That's what I was trying to say.
john baez
Nathan Urban <nur...@vt.edu> wrote:
>Oooops!!
>
>I guess I accidentally used a term that was already in common usage; I
>didn't know that there was such a thing as a "generalized connection", and
>it certainly is orthogonal to what I was interested in. In my original
>post I defined what I meant by a "generalized Ashtekar connection" but
>you probably missed it or got confused when I started abbreviating it
>to just "the generalized connection". I said:
>
>"(By `generalized Ashtekar connection' I mean A^i_a = \Gamma^i_a + \gamma
>k^i_a, of which the complex Ashtekar and real Barbero connections are
>special cases; \Gamma^i_a is the spin connection ...)". \gamma is the
>Immirzi parameter and k^i_a is the extrinsic curvature.
Oh, whew, so you're actually sane after all! Sorry for not noticing this
passage where you defined what you meant by "generalized Ashtekar
connection"! For real values of the Immirzi parameter, people usually
just call this the "real Ashtekar connection" or something like that.
Sometimes they call it the "Barbero connection", which is the historically
accurate name, since Fernando Barbero was the first to publish something
about it, *including* the parameter \gamma, whose problematic role was
later discussed by Immirzi and thus acquired his name.
Anyway, for any real value of the Immirzi parameter, this "real Ashekar
connection" is a connection on the same bundle I discussed in my previous
reply: the internal spinor bundle on space. It's therefore an SU(2)
connection.
If you see someone calling it an SO(3) connection, though, don't be freaked:
you can easily *get* an SO(3) connection from it, and in this situation it's
very easy to go back, too, since every orientable 3-manifold has a
trivializable tangent bundle, and thus admits a spin structure. (I can
flesh out the details of you go back and forth if you insist.)
Nathan Urban
> In article <6u0igp$8l0$1...@crib.corepower.com>,
> Nathan Urban <nur...@vt.edu> wrote:
> After thinking about it a bit more, I decided that I'd never seen
> most of the things you were talking about done in terms of Ashtekar's
> "new variables".
Hmm. Well, in that case, it gives me something fun to work on in a couple
months when I have more time to really get deep into the mathematics.
> [To change variables], it's crucial that you learn about the *spacetime*
> version of the new variables, not just Ashtekar's original version, which
> dealt with space at a fixed time. The spacetime version is sometimes
> called the "Plebanski" formulation of general relativity, since it
> was developed by the Mexican physicist Plebanski before Ashtekar did
> his thing, in a paper that never got much attention until *after*
> Ashtekar did his thing.
Side issue: how does this tie in to the Capovilla-Dell-Jacobson
Lagrangian? This is also a spacetime formulation of general relativity,
and is obtainable directly from the Ashtekar Hamiltonian via a Legendre
transformation. Since that's how you usually go from a Hamiltonian to a
Lagrangian, it would seem that the CDJ action would be the most natural
spacetime version of Ashtekar's formulation. To get to the Plebanski
Lagrangian, you have to do a Legendre transformation and then let a
field become independent. (I'm using Peldan's "Actions for Gravity",
gr-qc/9305011, as a reference here.) Though of course the two are still
equivalent classically.
I've seen the Plebanski formulation mentioned in recent research,
I think. I see the CDJ formulation mentioned a lot in older research
and then everyone seemed to stop talking about it. Why is that? Is it
just awkward to work with?
> You can read a quick summary of the Plebanski formulation of general
> relativity in my paper "Spin networks in nonperturbative quantum gravity" -
> this is probably the easiest introduction for someone who doesn't like
> indices.
Okay. I've looked at that, but didn't remember that it had anything
about the Plebanski formulation.
> But for more details it's best to read chapter III of Ashtekar's
> book _New Perspectives in Canonical Gravity_, and also Domenico Giulini's
> paper "Ashtekar variables in classical general relativity", available as
> gr-qc/9312032.
Oh, that looks like a really great paper -- a reference on Ashtekar's
variables that's both comprehensive _and_ humane. :)
> (That book by Ashtekar is a crucial reference, but
> unfortunately a bit hard to find.
I can get that at the local university library. I tried ordering it
once from Amazon.com but it seemed to be out of print.
> It's supposedly being distributed by the American Physical Society.
Hmm, I'll have to try again.
> >So is the left-handed sl(2,C) representation a subrepresentation of the
> >original sl(2,C)?
> I'm not sure what you're asking, because if you speak of "a subrepresentation
> of a group" you must really mean "a subrepresentation of some representation
> R of that group", but you're not telling me *which* representation R!
Yeah, that was silly of me. But you addressed the intent of my
question with:
> So we're not forming a subrepresentation of a Lie algebra here, we're
> restricting a representation to a sub-Lie-algebra.
Which is exactly what you said before, except with fewer words. :)
> >> [...] we have to apply the holonomy of the Levi-Civita
> >> connection - which we may think of as an element of the Lorentz
> >> group -
> >That's because the Levi-Civita connection is compatible with the metric,
> >hence parallel transport preserves the inner product, therefore must be
> >an element of the Lorentz group, correct?
> Correcto. Here of course we are secretly trivializing our bundle at
> both endpoints in order to think of the holonomy as an element of
> the structure group.
I don't understand what it means to trivialize a bundle "at both
endpoints". (Endpoints of what? The curve you're taking the holonomy
around?) Isn't trivialization a choice you make over the whole bundle?
Are you talking about a bundle which is locally trivializable at a few
points, but possibly not everywhere?
I also don't see why you have to do this in the first place to think of
the holonomy as an element of the structure group. Since the holonomy is
generated by the connection which lives in the Lie algebra of the group,
I guess I can see how a global holonomy might not end up being an element
of the group itself, but perhaps a cover of it sharing the same algebra.
Is that what being able to trivialize at the endpoints means? That you're
ensured of getting something within the group and not its cover? But in
a sense, getting something within the cover _is_ getting something within
the group, so maybe I have no idea what I'm talking about.
> Physicists do this without even mentioning it.
That's legal in the sense that it's like choosing a set of coordinates,
right? It doesn't affect the physics. I'm not quite sure when it's
possible to trivialize a bundle, though. Is that known to always be
possible, or is it thought that any "physically realistic" situation
will be trivializable?
> You can do general relativity with either a self-dual or anti-self-dual
> connection, and it scarcely makes a difference. It's just a matter of
> convention. In fact, half the people working on general relativity seem
> to do it one way, and half the other.
I bet that makes it even more fun to work through the literature.
> >I guess I don't see why wormholes can't describe a spin-1/2 particle
> >and thus fit in with the geometrodynamics program.
> Well, unless you pull some sly theoretical trick or other, when you
> turn a wormhole around 360 degrees it's going to look just like it
> did at first, so it'll be bosonic, not fermionic.
Okay, but how do the new spinorial formulations help with that?
They're still physically equivalent to GR. Or do spin-1/2 particles
still not fit in to Wheeler's program?
> Since Wheeler wrote
> this, people have come up with a couple of interesting ways to try to
> get around this sort of problem, like Sorkin's trick, or the trick Witten
> used to get fermions in the Skryme model.
I haven't heard of these tricks. References?
> >And if a fermion
> >can't be a wormhole, then does it have some other nice description
> >within classical GR that can be understood in terms of the conventional
> >geometric formulation?
> You can certainly couple general relativity to fermionic fields, but
> Wheeler wasn't wanting to put fermions in "by hand", he was wanting
> them to fall out.
What I'm getting at is, in the new formulations or in quantum gravity
or whatever, do fermions "fall out"? Can a fermion have a geometric
description analogous to how bosons could have a wormhole description?
> >I know that most
> >of the literature is on the complexified connection but I think it may
> >be one of those cases where I'll understand the special case better
> >if I also understand the general case. If I understand it correctly,
> >the complexified connection is defined on a bundle over the base space
> >MxC^4.
> No, it's still defined on a bundle over the same base space M,
> just a different bundle. We're not complexifying *spacetime*
> here!
Ack, so C^4 is the fiber. I got mixed up.
> >What are the other connections defined over?
> Well, I'm not sure which "other connections" you're talking about here,
> but as long as we're doing general relativity and not doing something
> funky like complexifying spacetime or working in some other dimension,
> the connections we're interested in are typically connections on good
> old "spacetime" - some smooth 4-manifold or other - or else on "space" -
> a smooth 3-manifold.
I guess I meant what are they defined _on_, not _over_ (i.e., the bundle,
not the base space), which you seem to have answered in another post:
"the internal spin bundle on space". I don't quite understand this yet
so I don't see how the various Ashtekar connections all come from the
same bundle, but I'll study your other post more.
> First of all, you should note that physicists are often sloppy about the
> difference between SU(2) and SO(3) and SL(2,C).
I know the relation between SU(2) and SO(3); SU(2) is the double cover
of SO(3) and they share the same Lie algebra. But how can you treat
SL(2,C) the same as SU(2) and SO(3)? It doesn't have the same Lie
algebra, does it? I guess what you're doing is SL(2,C) on spacetime,
which you restrict down onto space to either SU(2) or SO(3) at your whim?
> I'm not accusing Barbero
> *in particular* of this - it's just something you gotta get used to dealing
> with. Basically you gotta read the papers and figure out what they must
> mean in order to make sense, without worrying much about what they *say*.
> Or you can act like a physicist and just sort of mellow out and stop worrying
> about the difference between these various groups (which I don't actually
> advise you to do).
That's good, because if you had advised that I wouldn't have listened to
you. The last thing I need is to conveniently pretending like different
things are the same when they're not, even if you can get away with it.
That's like ignoring the distinction between upstairs and downstairs
indices on tensors because you've got a Euclidean metric around; it only
serves to confuse you when you don't have that crutch around.
> Anyway, what you probably really need to know is: under what situations can
> I get from one sort of connection to another, and in what ways? To take
> the simplest example: if you have a manifold M with an SO(3) bundle P over
> it *equipped with a trivialization*, you can just form the double cover
> P' of P and that will be an SU(2) bundle, and any connection on P will
> naturally lift to a connection on P', and any connection on P' will naturally
> push down to a connection on P.
I still don't quite see how this works. This can't be a bijective
mapping between P and P', can it, since P' is a double cover? If so,
doesn't it make a real difference which you're working in, since they're
not equivalent? I guess it's like how you work with SU(2) and SO(3);
you can work with either, but in SO(3) your unitary representations
become projective unitary so you have to carry around little cocycle
terms all over the place.
It just seems to me that even though _infinitesimal_ parallel transports
will end up the same (since SU(2) and SO(3) have the same Lie algebra),
_finite_ ones won't (because they are different groups), and hence will
lead to a _physically_ different global distinction.
> Since physicists often work locally and
> fix a trivialization of every bundle in sight, this allows them to be quite
> relaxed about the difference between SO(3) and SU(2) connections.
So it doesn't make sense to say that the Ashtekar connection is "more
naturally" an SU(2) connection or an SO(3) connection or an SL(2,C)
connection, because they're interchangable? What if there isn't a
trivialization? Does it end up being one or the other? I guess maybe
that's answered in your other post on spin structures, but I haven't
digested it yet.
> Anyway, the little fact in the previous paragraph is one of a bunch of
> facts you need to learn, and when you have all these facts at your disposal,
> then you can figure out what physicists must mean when they say the things
> they do. I can dole out more such facts if you want.
Then I suspect you will end up doing so. :) But perhaps not until I
get deeper into this stuff.
> Let's say we're talking about *space*. Then I would tend to say
> that the *real* Ashtekar connection is a connection on the internal
> spinor bundle over space. This is an SU(2) bundle, so the holonomy
> is an element of SU(2) after we trivialize the bundle over the
> endpoints of our path. And I would say that the *complex* Ashtekar
> connection is a connection on the restriction to space of the left-handed
> spinor bundle over spacetime to space.
(Or rather, "The *complex* Ashtekar connection is a connection on the
restriction to space of the left-handed internal spinor bundle over
spacetime.")
The complex Ashtekar connection sounds more complicated; it looks
(from your description) that you can formulate the real connection over
space directly, while you have to formulate the complex connection over
spacetime first and then restrict down to space.
> And then there is the ordinary
> Levi-Civita connection, and the internal Lorentz connection, and
> complexified internal Lorentz connection... but you have a habit of
> trying to learn about 43 different things at once, and I don't want
> to get into all 43 simultaneously.
Okay.. well, one more question about one specific connection: we
have the Ashtekar connection A^i_a = \Gamma^i_a + \gamma k^i_a where
\Gamma^i_a is the "spin connection associated to the triad e^i_a".
What kind of connection is this; what is the bundle and structure group?
Incidentally, the definition of the spin connection,
\partial_{[a} e^i_{b]} = \Gamma^i_{[a} e_{b]j}, really seems to be begging
for a formulation in terms of an exterior derivative (since that
antisymmetrized partial derivative is in there), but I can't quite see
how to rewrite that identity in terms of differential forms over some
bundle.
I also see the words "spin connection" a lot, used in seemingly different
ways. What is the definition of a "spin connection"?
Nathan Urban
> In article <6u75ta$n6m$1...@crib.corepower.com>,
> Nathan Urban <nur...@vt.edu> wrote:
> >I guess I accidentally used a term that was already in common usage; I
> >didn't know that there was such a thing as a "generalized connection", and
> >it certainly is orthogonal to what I was interested in. In my original
> >post I defined what I meant by a "generalized Ashtekar connection" but
> >you probably missed it or got confused when I started abbreviating it
> >to just "the generalized connection".
> Oh, whew, so you're actually sane after all!
Well, I wouldn't jump to conclusions.
> Anyway, for any real value of the Immirzi parameter, this "real Ashekar
> connection" is a connection on the same bundle I discussed in my previous
> reply: the internal spinor bundle on space. It's therefore an SU(2)
> connection.
Yeah, but what if the Immirzi parameter is complex? Then you get
something closer to the complex Ashtekar connection, but probably
not identical since I think you can only pull that self-dual trick
if the parameter is 'i' (or perhaps pure imaginary). I guess I'm
looking for some sort of unified formulation, that will work whether
or not the Immirzi parameter is real or imaginary -- how the real and
complex Ashtekar connections both come from the same connection, group,
and bundle.
> If you see someone calling it an SO(3) connection, though, don't be freaked:
> you can easily *get* an SO(3) connection from it, and in this situation it's
> very easy to go back, too, since every orientable 3-manifold has a
> trivializable tangent bundle, and thus admits a spin structure. (I can
> flesh out the details of you go back and forth if you insist.)
That would be helpful. As you may have seen in my other response, I
don't understand why you're allowed to do this. And is the reason that
you can treat SU(2) and SO(3) connections the same due to the fact that
we have an orientable 3-dimensional space, and this trick may break down
for other dimensions?
Also, what is the definition of a spin structure? You defined it in the
other article (which I'm not going to respond to right now) as "a bundle
with gauge group SL(2,C) that double covers the orthonormal frame bundle."
Is this the most general definition of a spin structure? It's also not
quite clear to me whether the "double covers the orthonormal frame bundle"
is a consequence of it being a spin structure, or a necessary condition
for it to be one.
Whew. I'm just not used to working with bundles and spinors and things
yet, as you can see. This will require some acclimation to sort all
these things out.
john baez
Nathan Urban <nur...@vt.edu> wrote:
>John Baez wrote:
>> Physicists do this [picking a trivialization of a bundle] without
>> even mentioning it.
>That's legal in the sense that it's like choosing a set of coordinates,
>right? It doesn't affect the physics.
Trivializing a bundle is very much like choosing a coordinate system.
It's a way of making a situation very concrete so you can easily do
computations. Like choosing a coordinate system, you can always do it
*locally*, but not necessarily *globally*. This is how things are set
up in the first place: a manifold M is defined to be a space that *locally*
looks like R^n, and a bundle P -> M with fiber F is defined to be a map
that *locally* looks like the obvious map F x R^n -> R^n. I'm not giving
the precise definitions here, of course - just trying to point out that
differential geometry studies situations that *locally* look very simple
and manageable.
>> You can certainly couple general relativity to fermionic fields, but
>> Wheeler wasn't wanting to put fermions in "by hand", he was wanting
>> them to fall out.
>
>What I'm getting at is, in the new formulations or in quantum gravity
>or whatever, do fermions "fall out"? Can a fermion have a geometric
>description analogous to how bosons could have a wormhole description?
Perhaps so! Look at the paper by Carlo Rovelli and Hugo Morales-Tecotl
entitled "Loop representation of quantum fermions and gravity", and the
paper by Kirill Krasnov and myself entitled "Quantization of
diffeomorphism-invariant theories with fermions". They're both on
gr-qc and I think they're both in the list of papers at
http://math.ucr.edu/home/baez/twf.html
The idea is that spin-1/2 particles coupled to gravity arise naturally
as the *ends* of spin networks that have "loose ends". It's pretty neat
how it works out. I think Lee Smolin also has a paper on this subject
called something like "Fermions as minimalistic wormhole ends". Same
basic idea except he regards the "loose ends" of spin networks (or loops)
as places where the spin network edge actually ducked into a wormhole.
More later.
john baez
Nathan Urban <nur...@vt.edu> wrote:
>> If you see someone calling it an SO(3) connection, though, don't be freaked:
>> you can easily *get* an SO(3) connection from it, and in this situation it's
>> very easy to go back, too, since every orientable 3-manifold has a
>> trivializable tangent bundle, and thus admits a spin structure. (I can
>> flesh out the details of you go back and forth if you insist.)
>
>That would be helpful. As you may have seen in my other response, I
>don't understand why you're allowed to do this.
I'm not sure if you want a "cheap and dirty" reply or whether you
want to learn things in a systematic from-the-ground-up way. If the latter,
it would help to get some definitions straightened out, like this:
>Also, what is the definition of a spin structure? You defined it in the
>other article (which I'm not going to respond to right now) as "a bundle
>with gauge group SL(2,C) that double covers the orthonormal frame bundle."
>Is this the most general definition of a spin structure?
Not really, since I was just talking about spin structures for a 4d
manifold with a Lorentzian manifold in that context. Also, it wasn't
the precise definition, it was just an allusion to the definition.
Let me briefly give the precise definition. I will assume you know:
1) the definition of "principal bundle with structure group G"
2) that the group SO(n,m) has a double cover Spin(n,m).
(In particular, SO(3) has a double cover Spin(3) = SU(2), while
SO(3,1) has a double cover Spin(3,1) = SL(2,C) and SO(4) has a
double cover Spin(4) = SU(2) x SU(2). These are the three most
important cases in physics and one should know by heart how they
work.)
Okay: suppose we have an smooth oriented manifold M with a metric of signature
(n,m). Then its oriented orthonormal frame bundle P -> M is a principal
bundle with structure group SO(n,m). In particular this implies that SO(n,m)
acts on the right on the total space P. But since we have a 2-1
homomorphism Spin(n,m) -> SO(n,m), the group Spin(n,m) also acts on P.
Now: given this situation, a "spin structure" for M is a principal
bundle P' -> M with structure group Spin(n,m), together with a double
cover P' -> P with the following properties:
1) each fiber P'_x is mapped to the corresponding fiber P_x
2) the map P' -> P is compatible with the right action of Spin(n,m) on
P' and P.
That's the definition. What does a spin structure let you *do*? It
lets you define spinor fields on M. Remember, a spinor is a particular
representation of Spin(n,m). So, given our spin structure, we can use
the standard trick for getting vector bundles from principal bundles to
get a bundle over M which we call the bundle of spinors.
There may be many or no spin structures on a given oriented manifold with
metric. If you know a bunch of algebraic topology, you can often figure
out how many spin structures there are in a given situation. Physically,
different spin structures correspond to different things that might happen
to a spin-1/2 particle when you carry it around a noncontractible loop in
your manifold M. Basically there's a sign ambiguity about what it'll do,
since a spinor switches sign when rotated 360 degrees, and you gotta *decide*
whether or not it's gotten an extra 360 degree rotation when it goes around
that loop... there's no a priori way of knowing. The spin structure records
these decisions. However, it's a bit subtler than this may sound, because
as I said, there are manifolds that don't admit *any* spin structure: for
these, there's no globally consistent way of defining spinor fields.
>It's also not
>quite clear to me whether the "double covers the orthonormal frame bundle"
>is a consequence of it being a spin structure, or a necessary condition
>for it to be one.
Well, I'm having trouble telling the difference between a "consequence"
of a proposition and a "necessary condition" for it to be true, so I'll
just say that while any spin structure double covers the (oriented) orthonormal
frame bundle, there are other conditions in the definition of spin structure,
too.
>Whew. I'm just not used to working with bundles and spinors and things
>yet, as you can see. This will require some acclimation to sort all
>these things out.
Yup. But don't worry: most people *never do* sort it out, so if you do,
you'll be way ahead of the pack. Most people who mess with spinors work
locally, with coordinates, with all bundles trivialized - basically
steam-rollering all the topological subtleties out of the situation.
And this is one reason they can get away with conflating all sorts of things
that you have to keep distinct if you're gonna worry about global, topological
issues. Heck, maybe they're right: maybe spacetime is R^4. :-)
john baez
john baez <ba...@galaxy.ucr.edu> wrote:
Nathan Urban wrote:
>>Also, what is the definition of a spin structure? You defined it in the
>>other article (which I'm not going to respond to right now) as "a bundle
>>with gauge group SL(2,C) that double covers the orthonormal frame bundle."
>>Is this the most general definition of a spin structure?
>
>Not really, since I was just talking about spin structures for a 4d
>manifold with a Lorentzian manifold in that context.
^^^^^^^^
should be: metric
By the way, I have been writing lots of replies to your questions, and
given the vagaries of the multiple moderations system for sci.physics.
research, they will show up in a random order.
john baez
Nathan Urban <nur...@vt.edu> wrote:
>In article <6u67h3$uck$1...@pravda.ucr.edu>, ba...@galaxy.ucr.edu (john baez)
wrote:
>> [To change variables], it's crucial that you learn about the *spacetime*
>> version of the new variables, not just Ashtekar's original version, which
>> dealt with space at a fixed time. The spacetime version is sometimes
>> called the "Plebanski" formulation of general relativity, since it
>> was developed by the Mexican physicist Plebanski before Ashtekar did
>> his thing, in a paper that never got much attention until *after*
>> Ashtekar did his thing.
>Side issue: how does this tie in to the Capovilla-Dell-Jacobson
>Lagrangian? This is also a spacetime formulation of general relativity,
>and is obtainable directly from the Ashtekar Hamiltonian via a Legendre
>transform. Since that's how you usually go from a Hamiltonian to a
>Lagrangian, it would seem that the CDJ action would be the most natural
>spacetime version of Ashtekar's formulation. To get to the Plebanski
>Lagrangian, you have to do a Legendre transformation and then let a
>field become independent. (I'm using Peldan's "Actions for Gravity",
>gr-qc/9305011, as a reference here.)
Since you have that reference and seem to understand it, I'm not
quite sure what *else* you want me to tell you about the relation
between the Capovilla-Dell-Jabson (CDJ) Lagrangian and the Plebanski
Lagrangian. But here's one bit of wisdom I'll gladly impart: what
some people call the CDJ Lagrangian, others call the Plebanski
Lagrangian, and vice versa! I think a certain usage became fashionable,
and then someone went back and actually read the papers by Plebanski
and Capovilla, Dell and Jacobson, and realized that everyone had it
backwards. Of course it doesn't really matter what you *call* things,
but sometimes it can trip you up, like when you actually want to communicate
with someone (which even people working on quantum gravity occasionally
do).
If I recall correctly, the Lagrangian actually invented by Plebanski
contained the fields F and E, where F was a chiral spin connection
and E was an sl(2,C)-valued 2-form. It also contained a Lagrange
multiplier field that forces E to be of the form
(e ^ e)_+
(the self-dual part of e ^ e, where e is a cotetrad field). In short,
this Lagrangian expresses general relativity as a constrained BF theory
with extra constraints. By this I mean that the Lagrangian looks like
tr(E ^ F)
plus a term involving the Lagrange multiplier field.
And if I recall correctly, Capovilla, Dell and Jacobson consider a
number of Lagrangians, including the one previously discussed by
Plebanski, but the most interesting *new* one is the one that
looks like this:
tr(e ^ e ^ F)
where F is a chiral spin connection and e is the cotetrad field.
>From what you write, it sounds like Peldan calls the one first invented
by Plebanski the "CDJ Lagrangian", and the one first invented by CDJ,
the "Plebanski Lagrangian". No problemo - I used to do that too, you'll
see it that way in some of my papers.
>Though of course the two are still equivalent classically.
Actually they're only equivalent when the metric is nondegenerate!
Things become much subtler when the metric is degenerate.
>I've seen the Plebanski formulation mentioned in recent research,
>I think. I see the CDJ formulation mentioned a lot in older research
>and then everyone seemed to stop talking about it. Why is that? Is it
>just awkward to work with?
Well, don't forget that what many people used to call the CDJ formulation,
they now call the Plebanski formulation!
Which formulation is better depends completely on what you are trying
to do.
>> >> [...] we have to apply the holonomy of the Levi-Civita
>> >> connection - which we may think of as an element of the Lorentz
>> >> group -
>
>> >That's because the Levi-Civita connection is compatible with the metric,
>> >hence parallel transport preserves the inner product, therefore must be
>> >an element of the Lorentz group, correct?
Yeah, it must "be" an element of the Lorentz group. I use quotes because...
>> Here of course we are secretly trivializing our bundle at
>> both endpoints in order to think of the holonomy as an element of
>> the structure group.
>
>I don't understand what it means to trivialize a bundle "at both
>endpoints". (Endpoints of what? The curve you're taking the holonomy
>around?)
Both endpoints of the curve we're taking the holonomy *along*. If
the curve is a loop, of course, these endpoints are the same, and we
speak of taking the holonomy "around" the loop.
>Isn't trivialization a choice you make over the whole bundle?
If you're an crude all-or-nothing sort of guy, sure, but more refined
sorts only trivialize the bundle where they actually *need to*. To
think of the holonomy along a curve as an element of the structure
group, we only need to trivialize the bundle at both endpoints of the
curve.
>Are you talking about a bundle which is locally trivializable at a few
>points, but possibly not everywhere?
Right. And this is why it pays to be a more refined sort! You can
always trivialize a bundle over a finite set of points in your manifold,
but you can't always trivialize it over the whole manifold.
>I also don't see why you have to do this in the first place to think of
>the holonomy as an element of the structure group.
Oh-oh. Think about it. In general relativity the structure group is SO(3,1),
a particular group of 4x4 matrices. Say I do parallel transport along a
curve from one point in spacetime to another. Which 4x4 matrix describes
the resulting holonomy?
This question *doesn't make sense* unless we make some extra choices. For
example, if we pick an orthonormal basis of tangent vectors at the starting
point of the curve and at the endpoint of the curve, *then* we can describe
the holonomy as a 4x4 matrix. But what we're doing here is choosing a
trivialization at both endpoints of the curve!
And like I said...
>> Physicists do this without even mentioning it.
More later... I'm being called for dinner!
john baez
>I know the relation between SU(2) and SO(3); SU(2) is the double cover
>of SO(3) and they share the same Lie algebra. But how can you treat
>SL(2,C) as the same as SU(2) and SO(3)? It doesn't have the same Lie
>algebra, does it?
The Lie algebra of SL(2,C) is the "complexification" of the Lie algebra
of SU(2). Remember, su(2) is the traceless hermitian 2x2 complex matrices,
while sl(2,C) is *all* the traceless 2x2 complex matrices. Any matrix in
sl(2,C) can thus, with a little head-scratching, be uniquely written as
A+iB with A and B lying in su(2). That's basically what we mean by saying
that sl(2,C) is the complexification of su(2).
(There's nothing special about the number 2 here, by the way. sl(n,C)
is the complexification of su(n). Similarly gl(n,C) is the complexification
of u(n). Everything in life comes in two flavors: real and complex.
Complexification is how we go from real to complex: we can take any real
vector space and tensor it with C to get a complex vector space. Conversely
we can go from complex to real in two ways: forgetting that our complex
vector space is complex and thinking of it as a real vector space of twice
the dimension, or choosing a "real part" of it, which is not really a
"process", since it involves a *choice*. For example, gl(n,C) has u(n)
as a "real part", but it also has gl(n,R) as another "real part". (In
this context the elite say "real form", so don't say "real part" - they'll
snicker at you behind your back.))
Physicists are often relaxed about whether they are working with a Lie
algebra or its complexification, so sometimes they'll talk about an SU(2)
connection when they mean an SL(2,C) connection. And sometimes, in a half-
hearted bow to accuracy, they will call an SL(2,C) connection a "complex
SU(2) connection".
The reason they're so relaxed is that you have to really stare at the
formulas sometimes to see if the numbers floating around are real or
complex.
But this issue gets woven in with another issue, as you note:
>I guess what you're doing is SL(2,C) on spacetime,
>which you restrict down onto space to either SU(2) or SO(3) at your whim?
Sorta. More precisely, the Lorentz group is SO(3,1), any choice of
timelike vector gives a subgroup isomorphic to SO(3) which fixes that
vector - or in physics lingo, it lets us split "spacetime" into "space"
and "time" and define rotations of "space". The double cover of SO(3,1)
is SL(2,C), the double cover of SO(3) is SU(2), so any choice of timelike
vector gives a subgroup of SL(2,C) that's isomorphic to SU(2).
So you see there are two interesting relations between SU(2) and SL(2,C):
complexification, and this business of splitting spacetime into space
and time. Whenever you have *two* interesting relations between mathematical
objects, you can do all sorts of cool stuff by playing them off against
each other. And that's what this Penrose/Newman/Plebanski/Ashtekar stuff
is exploiting!
john baez
Nathan Urban <nur...@vt.edu> wrote:
>> Since Wheeler wrote
>> this, people have come up with a couple of interesting ways to try to
>> get around this sort of problem, like Sorkin's trick, or the trick Witten
>> used to get fermions in the Skryme model.
>
>I haven't heard of these tricks. References?
I don't have a reference at hand for Sorkin's trick; the basic idea
was to find manifolds that look like R^3 at spatial infinity, but
nontrivial near the middle, such that twisting around the middle
part 360 degrees gives you something different, but twisting it around
twice gets you back where you started. (If you don't understand exactly
what I mean here, don't worry: it was an incredibly vague explanation.)
Witten's trick is in Nucl. Phys. B223 (1983) 422. The basic idea was
to note that in Skyrme's explanation of baryons as topological solitons
in an SU(3) nonlinear sigma model, you could explain the fermionic
nature of the baryons as a topological phase arising from a Wess-Zumino-
Witten term in the action. (If you don't understand exactly what I
mean here, don't worry: you might need to study some stuff first.)
I just happen to have a book at hand that seems to give a nice exposition
of Witten's trick:
Rajat K. Bhaduri, Models of the Nucleon: From Quarks to Soliton,
Addison-Wesley, 1998.
I'm hoping I'll learn some fun stuff about nucleons from reading this
book.
>> Let's say we're talking about *space*. Then I would tend to say
>> that the *real* Ashtekar connection is a connection on the internal
>> spinor bundle over space. This is an SU(2) bundle, so the holonomy
>> is an element of SU(2) after we trivialize the bundle over the
>> endpoints of our path.
>Okay.. well, one more question about one specific connection: we
>have the Ashtekar connection A^i_a = \Gamma^i_a + \gamma k^i_a where
>\Gamma^i_a is the "spin connection associated to the triad e^i_a".
>What kind of connection is this; what is the bundle and structure group?
This particular spin connection is the same kind of connection as the
real Ashtekar connection itself: a connection on the internal spinor
bundle over space. So it's an SU(2) connection.
>Incidentally, the definition of the spin connection,
>\partial_{[a} e^i_{b]} = \Gamma^i_{[a} e_{b]j}, really seems to be begging
>for a formulation in terms of an exterior derivative (since that
>antisymmetrized partial derivative is in there), but I can't quite see
>how to rewrite that identity in terms of differential forms over some
>bundle.
That's the "definition" of the spin connection? Yuck. It looks like
a horrible index-ridden *formula* for the spin connection, not the
conceptual definition.
>I also see the words "spin connection" a lot, used in seemingly different
>ways. What is the definition of a "spin connection"?
A connection on some sort of spin bundle or spinor bundle. (In some
other reply to your infinite list of questions I described what the
spin bundle is; it's a principal bundle. The spinor bundle is an
associated vector bundle. So there's no big difference between a
connection on a spin bundle and on the associated spinor bundle.)
Greg Egan
greg...@netspace.zebra.net.au (Greg Egan) wrote:
> What I'm curious to know is, is there any reason that this property of dp
> should be obvious? I can only see it by writing out the individual
> commutators and anti-commutators and observing the way dp's mixture of
> linear and anti-linear behaviour makes everything work. Is there a simple
> way of seeing that multiplying by i *must* yield the Hodge dual?
Having pondered this a bit longer, one answer is that in so(3,1) the only
elements that commute with a given element R are real-linear combinations
of R itself, and its Hodge dual, *R:
[R, a R + b *R] = 0
If V is a complex Lie algebra, and phi:V->so(3,1) is a Lie algebra
homomorphism, then:
[v, iv] = 0, for every v in V, by definition of the Lie bracket
[phi(v), phi(iv)] = 0
so phi(iv) = a phi(v) + b *phi(v), for some real a,b.
It seems intuitively reasonable (if not quite obvious) that the case where
phi is dp will yield the simplest possibility, with a=0, b=1 for all v.
--
Greg Egan
Email address (remove name of animal and add standard punctuation):
gregegan netspace zebra net au
john baez
Greg Egan <greg...@netspace.zebra.net.au> wrote:
>I think there's a typo here: su(2) consists of the traceless
>*skew*-hermitian 2x2 complex matrices. I know I'm nit-picking -- the two
>sets are isomorphic, and you just multiply by i to convert between them --
>but the set of hermitian matrices isn't closed under the Lie bracket [...]
You're right, and the distinction is important. Thanks!
>This is slightly off-topic, but it strikes me as quite amazing that the
>same standard double cover from SU(2) to SO(3) [formula omitted]
>extends so easily to the case SL(2,C) to SO(3,1) and sl(2,C) to so(3,1)
>just by adding the identity matrix to the set of spin matrices, to
>represent the time axis. The thing I find most "miraculous" is that a
>factor of i in the sl(2,C) element v gives you the Hodge dual of the
>planes of rotation!
Yes, this is one of those things that makes life in 4 dimensions so fun!
If I were god, I might decide to make spacetime 4-dimensional just for
this reason. And it's not really off-topic, because this miracle has a
lot to do with Ashtekar's formulation of general relativity in terms of
a "self-dual" or "spin" connection - a formulation which only works in 4
dimensions.
The miracle seems to be the collision of two nice general facts:
------------------------------------------------------------------------
1) For any n-dimensional oriented vector space V equipped with a flat
metric, the Hodge star operator maps p-forms to (n-p)-forms. More
precisely, using standard jargon:
*: Lambda^p V* -> Lambda^{n-p} V*
So when the dimension n is even and p = n/2, the Hodge star operator maps
p-forms to p-forms. Depending on n and the signature of the metric, we
then have either
** = 1
or
** = -1.
In the latter case, the Hodge star operator is a "complex structure" -
meaning that we can use it to make Lambda^p V* into a complex vector space,
with the Hodge star operator playing the role of i.
In particular, for a Lorentzian metric, we have ** = -1 on p-forms when
the dimension of V is a multiple of 4.
------------------------------------------------------------------------
2) For any n-dimensional oriented vector space V equipped with a flat
metric of signature (p,q), we can identify the Lorentz Lie algebra so(p,q)
with the 2-forms on V, or more precisely, Lambda^2 V*. The point is that
Lambda^2 V* consists of the antisymmetric elements of
V* tensor V*
The metric lets us "raise indices" and think of these as elements of
V tensor V*
which are just linear transformations of V. The linear transformations
we get this way are precisely those in so(p,q).
(For example, if V is 3-dimensional and the metric is the Euclidean one,
this is how we think of infinitesimal rotations as skew-symmetric matrices.)
------------------------------------------------------------------------
The case n = 4 is doubly blessed because it's a multiple of 4 and half
of it is 2. This means that we can think of a 2-form in 4 dimensions as
an element of the Lorentz Lie algebra so(3,1), *and* that this Lie algebra
has a complex structure on it given by the Hodge star operator. Actually
we need to scratch our heads a bit to realize the this complex structure
is compatible with the Lie algebra structure:
[ia,b] = i[a,b] (where i is secretly the Hodge star operator)
but it's true: the miracle, having gone this far, doesn't hold out on us
now!
This means that we can think of 2-form on 4-dimensional Minkowski space
as an element of a 6-dimensional real Lie algebra so(3,1), and we can
think of *that* as a 3-dimensional complex Lie algebra, with the Hodge
star operator corresponding to multiplication by i.
>Is there a simple way of seeing that multiplying by i *must* yield the
>Hodge dual?
Hmm. Simple? Does the above argument count as simple? It has the
advantage of describing a miracle as the intersection of two generalities.
It doesn't show that multiplying by i must yield the Hodge dual; rather,
it shows that the Hodge dual must be multiplying by i. Of course, I
skipped the crucial step of showing that this multiplication by i gets
along with the Lie algebra structure. Also, I didn't prove that the
Lie algebra we got was sl(2,C). For that, it would really pay to introduce
spinors - which is ultimately the whole point, of course.
But perhaps what we want is something different: geometrical intuition.
For that, we might try something like this. Every element of so(3,1) is
a linear combination of "simple" elements which are either infinitesimal
rotations in a space-space plane or boosts in a space-time plane. Applying
the Hodge star operator takes one of these "simple" elements and gives back
another "simple" element involving the orthogonal plane. Thus the Hodge
star of an infinitesimal rotation in a space-space plane is a boost in a
space-time plane and vice versa. From this, it's obvious that when you
do the Hodge star operator twice, you're almost back to where you started.
But due to some minus signs in the metric, if you do the Hodge star operator
twice you get back *minus* what you started with: the rotation or boost in
the opposite direction. So it's a complex structure.
This is fun to visualize, but it's pretty hard for me to use this to
visualize how the Hodge star operator gets along with the Lie algebra
structure! Can anyone *see* how [a,ib] = i[a,b], starting from this
picture?
This stuff has been on my mind recently since I'm finishing up a paper
with John Barrett called "The quantum tetrahedron in dimensions 3 and
4." In recent work on quantum gravity, it turns out to be interesting
to treat the space of geometries of a tetrahedron as a classical phase
space, and then *quantize* this phase space. The resulting "quantum
tetrahedron" is subject to the uncertainty principle, so you can't know
everything about its geometry simultaneously. Then the idea is to build
spacetime out of these quantum tetrahedra.... but the fun thing is, one
can describe the tetrahedron classically using 4 elements of Lambda^2 R^4,
and then the quantization procedure takes advantage of a lot of these
wonderful special features of 4 dimensions... and it's all somehow
just a simplicial version of Ashtekar's ideas about quantum gravity.
john baez
Greg Egan <greg...@netspace.zebra.net.au> proved that the Hodge star
operator on 2-forms in Minkowski space makes the Lie algebra so(3,1)
into a complex Lie algebra, by showing that [*X,Y] = *[X,Y]:
> [*J(a,b),J(b,d)]
> = g(a,a) g(b,b) sign(a,b,c,d) [J(c,d),J(b,d)]
> = g(a,a) g(b,b) sign(a,d,c,b) [J(c,d),J(d,b)]
> = g(a,a) g(b,b) g(d,d) sign(a,d,c,b) J(c,b)
> = g(b,b) *J(a,d)
> = *[J(a,b),J(b,d)]
Nice. There should be something slicker, but I'm not sure what it would
be!
While we're here, it's worth noting that your argument also applies to
4-dimensional space with its *Euclidean* metric. In that case there's
an isomorphism between the 2-forms and so(4). But in that case, ** = 1,
so we don't get a complex structure on so(4). Instead, ** = 1 implies
that every 2-form X can be uniquely split into a selfdual part and an
antiselfdual part:
X = A + B with *A = A, *B = -B
given explicitly by
A = (X + *X)/2
B = (X - *X)/2
Your argument again shows that [*X,Y] = *[X,Y]. Thus the bracket of
selfdual guys is selfdual, the bracket of antiselfdual guys is
antiselfdual, and the bracket of a selfdual guy and an antiselfdual
guy is zero. So we get a way to write so(4) as a direct sum of two
3-dimensional Lie algebras. And it turns out that both these are
isomorphic to so(3).
In short, the same magic features of dimension 4 that give
so(3,1) = sl(2,C)
also give
so(4) = so(3) + so(3).
One can more easily show the latter using the fact that SU(2) is a
3-sphere, but still, it's nice to see a systematic approach to these
coincidences.
That leaves just one more case to consider. What about a 4-dimensional
space with a flat metric of signature (2,2)? A world with two time and
two space directions may seem a bit odd, but it has a perverse charm
of its own, since it has a symmetry that switches "time" and "space".
People whose names I forget have written extensively about it. In
this world, the Hodge star operator on 2-forms has ** = 1, so the
above argument again lets us split so(2,2) as a direct sum of two
3-dimensional Lie algebras. And it turns out that
so(2,2) = sl(2,R) + sl(2,R)!
Greg Egan
> so(3,1) = sl(2,C)
> so(4) = so(3) + so(3).
> so(2,2) = sl(2,R) + sl(2,R)!
Thanks for completing the classification. Though I've seen these facts
mentioned before, they're only now starting to make much sense to me.
>There should be something slicker [to prove compatibility of Hodge star
>and Lie bracket], but I'm not sure what it would be!
Here's one approach that doesn't use an explicit construction for the
dual. Define a "matrix product" between (0,2) tensors as follows:
(a tensor b)(c tensor d) = g(b,c) (a tensor d)
If you expand the wedge products of 1-forms and apply this rule:
(a^b)(c^d) = g(b,c) (a tensor d) + g(a,d) (b tensor c)
- g(b,d) (a tensor c) - g(a,c) (b tensor d)
and (c^d)(a^b) is the transpose of this, giving a Lie bracket on 2-forms:
[a^b,c^d] = g(b,c) a^d + g(a,d) b^c + g(a,c) d^b + g(b,d) c^a
The inner product on 2-forms can be stated in terms of the matrix product:
<a^b,c^d> = g(a,c) g(b,d) - g(a,d) g(b,c)
= (-1/2) tr((a^b)(c^d))
where tr(a tensor b) = g(a,b). This is always cyclic, tr(uv)=tr(vu), but
if u,v,w are 2-forms it has the additional property:
tr(uvw) = tr(transpose(uvw)) = -tr(wvu)
which in turn implies:
<v,[w,u]> = (-1/2)(tr(vwu)-tr(vuw)) = tr(vuw) (Eqn 1)
The Hodge star of w, *w, is defined by:
v ^ *w = <v,w> vol = (-1/2) tr(vw) vol
Since (**) is just a constant factor, +/-1, and 2-forms commute:
**v ^ w = w ^ **v = <*v,w> vol = v ^ **w = <v,*w> vol
So <*v,w> = <v,*w>, the dual is self-adjoint, and tr((*v)w) = tr(v(*w)).
It takes a bit of work to go a step further [1] and show that:
tr((*v)wu) = tr(v(*w)u) for all 2-forms u,v,w (Eqn 2)
Now, from (Eqn 1) and (Eqn 2) it follows that:
<v,*[w,u]> = <*v,[w,u]> = tr((*v)uw) = tr(vu(*w)) = <v,[*w,u]>
Since this holds for all v, this shows that *[w,u] = [*w,u].
--------------------------------------------------------------------------
[1] It turns out that (**)vw + (*w)(*v) is a multiple of the identity when
viewed as a (1,1) tensor. (This is clear for the case w=v and v simple,
when (**)vv + (*v)(*v) is the sum of the squares of infinitesimal
rotations in two orthogonal planes. Each squared rotation projects vectors
into its plane, and reverses/rescales them by the same factor,
-(**)<v,v>.)
Since u is antisymmetric, tr((**)vwu + (*w)(*v)u), the trace of a multiple
of u, is zero. Thus:
(**)tr(vwu) = - tr((*w)(*v)u) = tr(u(*v)(*w))
Changing v to (*v) yields:
(**)tr((*v)wu) = (**)tr(uv(*w))
tr((*v)wu) = tr(uv(*w)) = tr(v(*w)u)
Alexander Y. Vlasov
<6umo5v$6b5$1...@pravda.ucr.edu>[&]:
>Oh-oh. Think about it. In general relativity the structure group is SO(3,1),
>a particular group of 4x4 matrices. Say I do parallel transport along a
>curve from one point in spacetime to another. Which 4x4 matrix describes
>the resulting holonomy?
>
>This question *doesn't make sense* unless we make some extra choices. For
>example, if we pick an orthonormal basis of tangent vectors at the starting
>point of the curve and at the endpoint of the curve, *then* we can describe
>the holonomy as a 4x4 matrix. But what we're doing here is choosing a
>trivialization at both endpoints of the curve!
Does it really not make a sense? I check my favorite "Foundation of
Differential Geometry" by S. Kobayashi & K. Nomizu --- they do not
introduce notion of trivialisation for holonomy and seems did not used
word "trivialisation" in the book at all. Maybe I missed something?
Unlike of many physical books about GR they introduce principal bundle and
connection on it as the basic objects and only after that they define vector
bundle as associated one. I.e. there is a total space P, action of a group
G on P, and base M ~ P/G. Connection is defined as a split of tangent space
T(P) on two subspaces, vertical T_G and horisontal T_M, in each point of P.
A form of connection is defined as a map from T(P) to Lie algebra of G, that
sometime may be more simple than manipulations with tangent space to *base* M
and "connection as parallel transport" so usual for physics.
The advantage of consideration above with connection on principle bundle as a
map T(P) -> T(P) is absence of direct notion of co-ordinates, such definition
gives an invariant [1] and local [2] way to think about connection [3].
Let us consider example with base space is line and group SO(2). A simple
model is screw-bolt as total space with screw-nut as a fiber and connection
as a thread of the bolt.
nut
+-+ thread ||
=\\\\| |\\\\\\\\\\\=||
+-+ ||
x -----------> bolt
Really, algebra Lie of SO(2) is isomorphic with real numbers and so we have
a local picture for 2D tangent space to the bolt:
-
\ | vertical ^
\ | projection | g = so(2) ~ R
\ | |
|<--->|-
horizontal
projection
Here form of connection is like vertical projection, i.e map to R {isomorphic
with Lie algebras so(2), u(1) }. If we try to do "parallel transport" along
the bolt, then the "thread-connection" produces rotation of the nut and
angle of rotation is holonomy element. It is unique defined without any
trivialisation, i.e. choice of coordinate systems on nuts with different x.
For general case with dimensions of M and G more than 1 a "hyper-nut" [4]
performs specific kind of "hyper-rotations" during motions in different
directions along multidimensional "hyper-bolt", but for any particular
small motion in given direction there is unique defined transformation of the
nut, because the map motion -> transformation is just *form of connection*
with motion described as vector in tangent space of bolt and transformation
of nut is defined by element of Lie algebra. Do I describe all properly?
Then we do not need some trivialisation to define holonomy for motion
along given curve, Exp(Integral...)
A difficulty is the big dimensions of used manifolds --- for example for
GR we should consider 4+16=20D total space P to describe principal bundle
with group GL(4,R) and, next, to consider 20+4=24D space related with
associated bundle to have possibility work with the tangent space to
space-time instead of 20D tangent space of total space P.
To reduce the dimensions of constructions, for vector bundles is used known
language of "covariant derivatives in direction X" on base manifold and for
more special case, if a fiber has the same dimension as base and could be
considered as tangent space, it is used language of Riemann geometry.
Maybe together with a simplification here is some lost of clarity. Say,
to quantize, as an invariant way to describe the Ashtekar connection
was chosen map from *infinitedimensional* loop space to group G. The new
method of quantization was successful. But I would glad to know, does
another choice, say connection as projection from *finitedimensional* space
T(P) to subspace discussed above could produce something useful?
>And like I said...
>
>>> Physicists do this without even mentioning it.
>
>More later...
I did not see any more since ...
Alexander
----------------------------------------------------------------------
Footnotes [%]:
[~] The sign ~ is for isomorphism everywhere in the text.
[1] Compare it with usual 3-indexes expression in GR with nontrivial
coordinate dependence --- they even can be made zero at given point by
coordinate transformation, but anyway it is the same mathematical object.
[2] Pointwise description of connection, curvature, etc.. Compare it with
idea, that a connection appears because for two different spacetime points
coordinate in tangent or internal space could not be compared directly.
[3] Seems it was introduced by Ehresmann many many years ago.
[4] sl., derog., small (usu n-sided) piece of metal with hyperhole
through the center, used for screwing onto a hyperbolt to secure it,
cf. hairy ball, fiber bundle.
[*] I tried, the address with "galaxy" did not work.
[&] The word "pravda" means "truth" on Russian.
[%] It is not a very good idea to use footnotes in USENET, I hope I prove it.
Greg Egan
greg...@netspace.zebra.net.au (Greg Egan) wrote:
> Here's one approach that doesn't use an explicit construction for the
> dual ...
It probably makes more sense to boil down the compatibility of the Hodge
dual and the Lie bracket to three simple criteria:
(1) The dual is self-adjoint: <v,*w> = <*v,w>
(2) Anything commutes with its dual: [*q,q] = 0
(3) The Lie bracket and the inner product "cycle":
<u,[v,w]> = <v,[w,u]>
>From (2) it follows that you can swap the dual within the Lie bracket:
[*v,w] = [*v,w] + [*v,v] + [*w,w] - [(*v)+(*w), v+w]
= -[*w,v]
= [v,*w]
Compatibility is very easy to show now:
<v,*[w,u]> = <*v,[w,u]> dual is self-adjoint
= <u,[*v,w]> cyclic property
= <u,[v,*w]> swap dual within Lie bracket
= <v,[*w,u]> cyclic property
Hence *[w,u] = [*w,u].
------------------------------------------------------------------------
To establish these three prerequisites in the particular case of 2-forms
in 4 dimensions:
(1) is true from the definition of the Hodge dual and the graded
commutative property of the wedge product:
*v ^ *w = <*v,w> vol = *w ^ *v = <*w,v> vol = <v,*w> vol
(2) If q is simple, its (1,1) tensor version has a 2-dimensional range,
its plane of rotation, which is precisely the null space of *q, since that
has a plane of rotation orthogonal to q's. And vice versa. So:
[*q,q] =(*q)q - q(*q) = 0 - 0 = 0
If q is not simple, it can be split into a linear combination of rotations
in two orthogonal planes:
q = a r + b (*r)
*q = a (*r) + (**)b r
with [*r,r]=0 for the reasons given above. But [r,r]=0 and [*r,*r]=0, so
everything in these sums commutes with everything else, so [*q,q]=0.
(3) is probably best shown via the trace:
ij ji
<u,v> = (1/2) u v = (-1/2) u v = (-1/2) tr(uv)
ij ij
tr(uvw) = tr(transpose(uvw)) = -tr(wvu)
<u,[v,w]> = (-1/2)(tr(uvw)-tr(uwv))
= (-1/2)(tr(vwu)-tr(uwv))
= tr(uwv)
= tr(vuw)
= <v,[w,u]>
john baez
>I've also become very interested in the subject of this discussion, as
>it seems key to understanding both the Ashtekar approach to General Relativity
>and also the recent simplicial versions, relativistic spin networks, and
>spin foam models (i.e., Barrett/Crane, Baez, Barbieri, etc).
I think you're right, this sort of thing seems like the key.
>John Baez:
>> This stuff has been on my mind recently since I'm finishing up a paper
>> with John Barrett called "The quantum tetrahedron in dimensions 3 and 4."
>
>I very much look forward to reading this. Is it Euclidean (SO(4)) or
>does it deal with the Lorentzian case (or does it shine light on the
>relationship between the two, (e.g., as Barrett did for the 3 dimensional
>case).
It's Euclidean, alas. It aims to shed light on the curious fact that
a quantum tetrahedron in 4 dimensions has fewer degrees of freedom than
a quantum tetrahedron in 3 dimensions. A classical tetrahedron in either
3 or 4 dimensions has 6 degrees of freedom: e.g., the lengths of its 6
edges, or the areas of its 4 faces together with the areas of 2 of the
3 parallelograms formed by the midpoints of its edges. (Here of course I'm
talking about the geometry of a tetrahedron in R^3 modulo rotations and
translations.) A quantum tetrahedron in 3 dimensions has only 5 degrees of
freedom because the 2 parallelogram areas do not commute and thus cannot be
simultaneously diagonalized. But a quantum tetrahedron in 4 dimensions has
only 4 degrees of freedom: the areas of the 4 faces. This is why, in the
Barrett-Crane state sum model for 4-dimensional Euclidean quantum gravity,
one only labels the triangular faces of ones triangulated manifold with
spins. At first it really struck me as paradoxical that a tetrahedron in
4 dimensions, even a bizarre "quantum tetrahedron", could have fewer degrees
of freedom than one in 3 dimensions. The problem is thus to get a better
understanding of why this is true. We do so using "geometric quantization" -
a well-developed formalism for studying this sort of issue. Turns out that
it all makes sense!
Of course the Lorentzian aspects are crucial. When I finally get rid of
the backlog of papers I'm supposed to be finishing, I'd like to work on that.
I'm hoping some beautiful math will appear and make everything fit neatly
into place. The magic of 4 dimensions! Hodge duality as complex structure,
and all that jazz.
Greg Egan
[Lots of great stuff about geometric algebra]
Thanks very much for explaining this! I hadn't heard any of it before,
and it does look incredibly powerful. I've downloaded "Imaginary Numbers
are not Real" and read it through once, but I think it's going to take me
several months to really absorb this.
> (so his J(a,b) is my [ab], which in my picture is = ab = a^b = -b^a = -ba
> if a,b are t,x,y,z and (a !=b))
I'm a little bit wary about simply identifying J(a,b) with [ab] or ab,
because when you apply an infinitesimal rotation to a particular vector,
you have to worry about the triple wedge product that's formed if the
vector has a component perpendicular to the plane of rotation, e.g.
J(x,y)z = 0
but [xy]z = xyz = x^y^z
I can only get a handle on this if I start out by defining things slightly
differently:
J(a,b)e = (1/2)(abe - eab)
where all of these products are geometrical products. Then:
J(x,y)x = (1/2)(xyx - xxy) = -y
J(x,y)y = (1/2)(xyy - yxy) = x
J(x,y)z = (1/2)(xyz - zxy) = 0
Since this definition doesn't care about x and y separately, you might as
well define J for anything:
J(u)w = (1/2)(uw-wu)
Then if you compose two of these Js and take the Lie bracket, you get:
J(u)J(v)w = (1/4)(uvw - uwv - vwu + wvu)
J(v)J(u)w = (1/4)(vuw - vwu - uwv + wuv)
[J(u),J(v)]w = (1/2)((1/2)(uv-vu)w - w(1/2)(uv-vu))
= J((1/2)(uv-vu))w
At this point, having initially defined the Lie bracket by composing these
linear operators J(u), it seems reasonable to define a homomorphic Lie
algebra on the geometric algebra itself:
[u,v] = (1/2)(uv-vu)
I suppose that was a bit circuitous, but it's the only way I can see to
justify the factor of (1/2)!
> then it turns out that:
>
> [J(a,b),J(c,d)] = 1/2([ab],[cd])_ = 1/2(abcd - cdab) [Geometric products!]
>
> = abcd
>
> [where, (a,b )_ is just the commutator ab-ba]
I don't think you can quite say (1/2)(abcd-cdab) = abcd in general,
because, e.g.:
(1/2)(txyz-yztx) = 0
as you'd expect, since J(tx) and J(yz) must commute. You can only do this
if you assume b=c (and a,b,c,d orthonormal):
(1/2)(abbd-bdab) = (1/2)(abbd+badb)
= (1/2)(abbd+abbd)
= abbd
= (b.b) ad
which matches the usual commutator for an orthonormal basis.
But sticking to the general equation, and using * = I = txyz:
*[ab,cd] = I(1/2)(abcd-cdab)
= (1/2)(Iabcd - Icdab)
= (1/2)(Iabcd - cdIab)
= [Iab,cd] = [*ab,cd]
In the third line, I've used the fact that I=txyz commutes with all
bivectors. It anticommutes with single vectors:
txyza = -atxyz
if a is one of t,x,y,z, because it will anticommute with 3 of them and
then commute with itself. Then with bivectors you get a second minus sign
that cancels the first.
john baez
Greg Egan <greg...@netspace.zebra.net.au> wrote:
>It probably makes more sense to boil down the compatibility of the Hodge
>dual and the Lie bracket to three simple criteria:
>
>(1) The dual is self-adjoint: <v,*w> = <*v,w>
>
>(2) Anything commutes with its dual: [*q,q] = 0
>
>(3) The Lie bracket and the inner product "cycle":
>
> <u,[v,w]> = <v,[w,u]>
That's nice - an axiomatic distillation of the whole business!
As you may know, any Lie algebra has a pairing satisfying (3),
called the Killing form. It's defined by:
<v,w> = tr(ad(v)ad(w))
where "ad" stands for the adjoint representation of the Lie
algebra on itself:
ad(v)w = [v,w]
The cyclic property of the trace is what makes (3) work.
The Killing form is nondegenerate for "semisimple" Lie algebras,
which include lots of familiar examples, like sl(n), su(n), and
so(p,q). So the magical features of 4 dimensions lie primarily
in properties (1) and (2), together of course with the fact that
** is 1 or -1.
A basic principle of mathematical physics is that once you come up
with a clever idea, you should milk it for all that it's worth.
There is always the temptation to milk it for *more* than it's worth,
which is why there are so many papers on hep-th. But I don't think
we've reached that point just yet.
So: are there any other nice Lie algebras floating around with a
pairing and a * operator satisfying (1)-(3)?
Well, the trick of course is to exploit the special features of 4
dimensions. So far we have looked at 4-dimensional *real* vector
spaces with metrics having various signatures, getting nice facts
about so(4), so(3,1), and so(2,2). But what about 4-dimensional
*complex* vector spaces? There are two things like a "metric"
we can put on a complex vector space: a bilinear pairing <.,.>,
or a sesquilinear pairing <.,.> - that is, one that is linear in
one slot and conjugate-linear in the other. The first will give
us a linear Hodge star operator; the second will give a conjugate-
linear Hodge star operator.
If we start with C^4 and give it the bilinear pairing
<(x1,...,x4),(y1,...,y4)> = x1 y1 + ... + x4 y4
the group of linear transformations preserving this pairing is SO(4,C),
with Lie algebra so(4,C). Using the tricks we've developed, we
get a Hodge star operator on so(4,C) satisying properties (1)-(3)
and with ** = 1. As we've already seen in this real case, this
lets us split so(4,C) as a direct sum of two 3-dimensional Lie
algebras. We get:
so(4,C) = so(3,C) + so(3,C)
Unfortunately, this is not too exciting: it's just a complexification
of a fact we already bumped into:
so(4) = so(3) + so(3)
If we start with C^4 and give it the sesquilinear pairing
<(x1,...,x4),(y1,...,y4)> = (x1)* y1 + ... + (x4)* y4
where here * means complex conjugation, not the Hodge star
operator, then the group of transformations preserving this
pairing is SU(4). The Lie algebra of this group is su(4), which
is a real Lie algebra, not a complex one like so(4,C). Using our
familiar tricks we might get a Hodge star operator on su(4)
satisfying properties (1)-(3) and with ** = 1. I say "might"
instead of "do" because there could be problems, coming from the
fact that this * is conjugate-linear rather than linear. But
anyway, suppose it more or less works. What does it do for us?
Well, if it really worked, we could use it to split su(4) as a
direct sum of two Lie algebras. Unfortunately I happen to know
that su(4) does not split this way! So something must go wrong.
I leave it as a puzzle to figure out exactly what.
However, I also happen to know that
su(4) = so(6).
So there is *something* interesting going on with su(4), and it
would be very odd if it weren't related to this magic stuff about
Hodge duals in 4 dimensions.
You may by now think that there is an infinite supply of these
Lie algebra coincidences, but in fact there aren't many more.
Here's one more, which is also related to the magic of 4 dimensions:
sl(4,R) = so(3,3)
This one doesn't involve the Hodge dual. Take R^4 and give it
a volume form. The group preserving this volume form is SL(4,R)
with Lie algebra sl(4,R). Using the volume form we can define a
pairing on 2-forms:
v ^ w = <v,w> vol
This pairing has signature (3,3). Any transformation in SL(4,R)
gives a transformation on 2-forms preserving this pairing, so we
get a homomorphism from SL(4,R) to SO(3,3). This is 2-1, so we
get a 1-1 homomorphism from sl(4,R) to so(3,3), and both these guys
have the same dimension, so it's an isomorphism.
And here's another, which must also be related somehow:
su(2,2) = so(4,2)
This is the basis of twistor theory, but that's another story.
john baez
su(4) = so(6)
and
su(2,2) = so(2,4)
so I might as well say why before I forget.
For the first one, I had the right basic idea but then took a
wrong turn. The basic idea is to start with C^4 and give it
the inner product
<(x1,...,x4),(y1,...,y4)> = (x1)* y1 + ... + (x4)* y4
where * here means complex conjugation. The group of transformations
preserving this pairing is SU(4), whose Lie algebra is su(4). Using
our familiar tricks we get a Hodge star operator on the (complex-valued)
2-forms on C^4. I'll call this operator *. This Hodge star operator is
conjugate-linear and satisfies ** = 1.
Now, whenever we have a conjugate-linear operator * with ** = 1, we can
use it to split the vector space it acts on into a "real part" and an
"imaginary part". That is, we can write any vector as
v = Re(v) + i Im(v)
where
Re(v) = (v + v*)/2
Im(v) = (v - v*)/2i
and all sorts of familiar properties hold.
So far so good. But, unlike other times when we played this game, it's
not true that su(4) can be identified with the antisymmetric rank 2
tensors. After all, su(4) consists of the 4x4 skew-hermitian traceless
matrices, so it's a 15-dimensional real vector space, while the 2-forms
on C^4 form a 6-dimensional complex vector space.
So, to get our isomorphism
su(4) = so(6)
here is what we do. The 2-forms on C^4 have an inner product arising
naturally from the inner product on C^4. Because it's defined *naturally*,
without any arbitrary choices, this inner product is preserved by the action
of U(4) on the 2-forms. Also, because the Hodge star operator is defined
naturally, it commutes with this action of U(4) on 2-forms. This means
that U(4) acts on any "real" 2-form to give another "real" one.
The space of "real" 2-forms is a 6-dimensional *real* vector space equipped
with a *real* inner product coming from the inner product on 2-forms. The
above paragraph thus gives us a homomorphism
SU(4) -> SO(6)
This is 2-1 so we get a 1-1 homomorphism of Lie algebras
su(4) -> so(6)
and both spaces are 15-dimensional so this is an isomorphism.
The same sort of trick gives us an isomorphism
su(2,2) = so(4,2)
except that we start with a sesquilinear form of signature (2,2) on
C^4:
<(x1,...,x4),(y1,...,y4)> = (x1)* y1 + (x2)* y2 - (x3)* y3 - (x4)* y4
which gives a sesquilinear form of signature (2,4) on the 2-forms.
This last one is important because so(4,2) is the Lie algebra of the
group of conformal transformations of 4-dimensional Minkowski space.
So: the magic of 4 dimensions has given us
so(4) = so(3) + so(3)
so(3,1) = sl(2,C)
so(2,2) = sl(2,R) + sl(2,R)
sl(4,R) = so(3,3)
su(4) = so(6)
su(2,2) = so(2,4)
There is one gaping hole left: su(3,1). This winds up being related
to the quaternions so I won't discuss it --- there's too much talk
of quaternions here already.
Alan Katz
Greg Egan writes:
> I'm a little bit wary about simply identifying J(a,b) with [ab] or ab,
> because when you apply an infinitesimal rotation to a particular vector,
> you have to worry about the triple wedge product that's formed if the
> vector has a component perpendicular to the plane of rotation, e.g.
>
> J(x,y)z = 0
> but [xy]z = xyz = x^y^z
The bivector algebra reproduces the (Lie) ALGEBRA of INFINITESIMAL rotations.
As usual, for a finite rotation, you have to exponentiate these. So,rotating
a vector a in the x-y plane you get something like:
(1 - J(x,y) + ...) a (1 + J(x,y) +...) = a + (a,J(x,y))_ + ...
and if a=z, the commutator vanishes and you get just z as expected.
(actually there are factors of 1/2 or something in there somewhere).
In the Geometric Algebra framework, finite rotatations are represented
by the EVEN sub-algebra and are called spinors (!) (because they
ARE spinors; it provides a nice way of "seeing" what spinors are).
I forgot to elaborate on this in my message. You may have
noticed that I had a reference note [3], but no reference to it.
I should have mentioned that it is well known that
the bivector algebra reproduces the SO(n,m) Lie algebras, that
Hestenes conjectured that you could represent all Lie algebras
that way, and [3] refered to a paper claiming to do this, but
I haven't looked at it ([3] is reproduced below again).
> > then it turns out that:
> >
> > [J(a,b),J(c,d)] = 1/2([ab],[cd])_ = 1/2(abcd - cdab) [Geometric products!]
> >
> > = abcd
> >
> > [where, (a,b )_ is just the commutator ab-ba]
>
> I don't think you can quite say (1/2)(abcd-cdab) = abcd in general,
Right.
I left out a bit of reasoning; partly because I wanted to
emphasize how *([ab],[cd]) -> (*[ab],[cd]) and partly because I forgot.
I should have said the 1/2 the commutator is = abcd if exactly two
of a,b,c,d are equal to each other, otherwise its zero (in other
words it's abcd or 0). a != b and c != d, else its not a bivector
(its 1) and more than two can't be equal (e.g. a=c AND b=d) or
else its 0 and if all are different, you get I-I=0.
Sorry.
----
[3] In Chapter 8 of [2], Hestenes talks about representing Lie Algebras
by bi-vector algebras. He makes a "Most Interesting Conjecture (MIC),"
that every Lie algebra is isomorphic to a bivector algebra. At the
above mentioned web site, I also found the following, which I have
not had a chance to look at:
C. J. L. Doran, D. Hestenes, F. Sommen and N. van Acker.
Lie Groups as Spin Groups
J. Math. Phys. 34(8), 3642-3669 (1993).
Abstract: It is shown that every Lie algebra can be represented as a
bivector algebra; hence every Lie group can be represented as a spin group.
Thus, the computational power of geometric algebra is available to simplify
the analysis and applications of Lie groups and Lie algebras. The spin
version of the general linear group is thoroughly analyzed, and an
invariant method for constructing real spin representations of other
classical groups is developed. Moreover, it is demonstrated that every
linear transformation can be represented as a monomial of vectors in
geometric algebra.