vector potential
Benjamin P. Carter
there exist another vector field A such that B = del cross A?
If there are localized sources such that Maxwell's equations are
satisfied, then one can define A and (the scalar potential) phi in the
usual way, using the Green's function for free space. But this won't do
in general. For example an electromagnetic plane wave has no sources, but
it can be obtained from a vector potential (which is also in the form of a
plane wave).
So I am wondering: what is the general theorem for the existence of a
vector potential?
--
Ben Carter
Toby Bartels
>Benjamin P. Carter <b...@netcom.com> wrote:
>>Given a vector field B, such that del dot B = 0, under what conditions does
>>there exist another vector field A such that B = del cross A?
>It is sufficient (but not necessary) that the domain of B be simply
>connected, i.e. that any loop in the domain of B be contractible
>to a point. An example of a non-simply connected region is, for
>example, R^3 with the z-axis removed.
Note that R^3 itself is simply connected.
So, if you believe that space is simply R^3 (no black holes and so forth),
then in any real life situation you can be sure
that your B field has a corresponding A field.
Even if you believe in black holes,
the region around Earth is still simply connected.
However, if in studying B you make certain approximations
and these approximations aren't valid for |z| < epsilon,
then anything you calculate about A may also be invalid --
even if you're only concerned with A in the region |z| >> epsilon!
To get around this, if you can, restrict yourself to a smaller region
which is simply connected and where your approximations are still valid,
and then draw conclusions about A only from the data of B in that region.
-- Toby
to...@ugcs.caltech.edu
A. Garrett Lisi
>
> Given a vector field B, such that del dot B = 0, under what conditions does
> there exist another vector field A such that B = del cross A?
Such an A will exist in the region iff every closed oriented curve in
the region is the boundary of some compact oriented surface.
-Garrett Lisi
Kevin A. Scaldeferri
Benjamin P. Carter <b...@netcom.com> wrote:
>Given a vector field B, such that del dot B = 0, under what conditions does
>there exist another vector field A such that B = del cross A?
I'm going to switch to the 4-vector formulation immediately and talk
in terms of the field strength tensor F and the 4-vector A. (At the
end, I'll actually restrict attention to space only, though, and what
I say there is exactly true for your question too.)
You can always find such an A locally. To determine if such an A
exists globally, you must know something about the topology of
spacetime. Specifically, you must know if the second cohomology group
is trivial.
What does that mean? Well, the second (nth) cohomology group
essentially tells you how many 2(n)-forms (anti-symmetric tensors) w
there are which have dw = 0 but do not have an associated 1(n-1)-form u
such that du = w. (d is exterior differentiation.) This group is
trivial if every such 2-form has such a 1-form associated with it.
Now this forms language can easily be translated into the more familiar
tensor language of electromagnetism. F is really a 2-form, A is
really a 1-form, all those types of vector and tensor derivatives are
exterior differentiation in disguise.
So, we have made the seemingly circular statement that given F which
satisfies the Maxwell equation dF = 0, there exists a vector potential
A such that F = dA if the topology requires that if dF = 0 there existS
A .... But, the thing that makes this useful is that there are ways to
compute the cohomology groups without examining forms and exterior
derivatives. You can piece together simple manifolds you do know the
cohomology of or you can exploit the duality with the homology groups
or you can use other techniques (or you can look them up).
The most readable description of this I know is John Baez and
Javier Muniain's book "Gauge Fields, Knots and Gravity". For more of
the math, check out Nakahara or Nash and Sen or the references
contained in them (which are real math books). Personally, I think
both of these books leave out lots of good stuff, so I'd check out
some math books if you really want to learn this stuff, say
"Differential Topology" by Guillemin and Pollack or, if you are
feeling strong, "Differential Forms in Algebraic Topology" by Bott and
Tu. MTW also has a somewhat odd introduction to differential forms in
Chapter 4 with lots of wacky pictures that may or may not help you
understand and visualize differential forms.
Oh, perhaps you'd like to know the answer in a couple simple
topologies. I'll just talk about the topology of space since
non-trivial topologies involving time-like directions are wierd. So,
just consider the topology of spacetime to be M x R, where M is some
3-manifold. If M is R^3 then the cohomology is trivial and you can
always find A globally. If M is a 3-sphere then the second cohomology
group is trivial (although others aren't) so you can find A globally.
If M is a 3-torus then all the cohomology groups are non-trivial and
you can't find A globally.
--
======================================================================
Kevin Scaldeferri Calif. Institute of Technology
The INTJ's Prayer:
Lord keep me open to others' ideas, WRONG though they may be.
Robert Israel
|> "Benjamin P. Carter" wrote:
|> >
|> > Given a vector field B, such that del dot B = 0, under what conditions does
|> > there exist another vector field A such that B = del cross A?
|>
|> the region is the boundary of some compact oriented surface.
No, that would be the condition for a vector field B with curl 0 to have
a scalar potential. What you want is every closed surface (homeomorphic
image of a sphere) to be the boundary of a region with compact closure.
Then every vector field B that is smooth in the region has a vector
potential in the region.
For example, if the region is the complement of a point P, this is false
(take the closed surface to be a sphere enclosing P), although it does
satisfy your condition. And indeed there is a divergence-free vector
field B in this region (the field of a point charge at P) that doesn't
have a vector potential. On the other hand, if you take the region to
be the complement of an infinite ray starting at P, my condition is
satisfied and a vector potential will exist.
Robert Israel isr...@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia
Vancouver, BC, Canada V6T 1Z2
Benjamin P. Carter
mathematics that are not part of the usual curriculum for physics.
Before I immerse myself in cohomology or exterior differential forms,
I would like to see what can be done by more conventional means.
to...@ugcs.caltech.edu (Toby Bartels) writes:
> ... if in studying B you make certain approximations
>and these approximations aren't valid for |z| < epsilon,
>then anything you calculate about A may also be invalid --
>even if you're only concerned with A in the region |z| >> epsilon!
A concrete example or two would help. I would particularly like to see an
example of a B field that needs to be defined in a multiply-connected
region, such that there is no vector potential defined throughout that
region. It would be especially interesting if the example bears some
relationship to somebody's research.
Another more convincing motivation would be a precise but elementary
statement of a theorem, along with a claim that exterior differential
forms or cohomology or whatever is helpful, or required for, or the most
natural approach to, the proof of the theorem.
Not that I don't appreciate the clues that have been offered so far.
--
Ben Carter
bu...@pac2.berkeley.edu
Benjamin P. Carter <b...@netcom.com> wrote:
>>de Rham cohomology, not the first. A typical example where we can't
>>do what Benjamin Carter wants is when our space is R^3 minus the origin
>>and B is the magnetic field produced by a magnetic monopole at the origin.
>>This space has nonvanishing second de Rham cohomology.
>
>I wonder: is there an elementary way to show the non-existence of a vector
>potential for this example?
How about this? We know from Gauss's law (or by directly evaluating
the integral if you prefer) that the integral of B, taken over the
surface of a sphere of radius r centered on the origin, is 4 pi q (q
being the charge of the magnetic monopole). But if B were the curl of
A, then this integral would have to be zero, by Stokes's theorem:
\int_{surface} (curl A).da = \int_{boundary of surface} A.dl,
and the boundary of that 2-sphere is the null set.
-Ted
Aaron Bergman
wrote:
>ba...@galaxy.ucr.edu (john baez) writes:
>>if we're trying
>>to find a potential whose gradient is a given curl-free vector field, we're
>>guaranteed to succeed if our space is simply connected, but if we're
>>trying to find a vector field whose curl is a given divergence-free
>>vector field, this helps not a whit - we want vanishing of the second
>>de Rham cohomology, not the first. A typical example where we can't
>>do what Benjamin Carter wants is when our space is R^3 minus the origin
>>and B is the magnetic field produced by a magnetic monopole at the origin.
>>This space has nonvanishing second de Rham cohomology.
>
>I wonder: is there an elementary way to show the non-existence of a vector
>potential for this example?
In the case of a magnetic monopole, it's really pretty easy. Let's examine
a sphere around the origin and assume that A is defined everywhere on it.
Now, we have that the integral of the magnetic flux through this sphere is
non-zero by one of Maxwell's equations. However, we cam also evaluate the
integral by Stokes's theorem as B is actually a curl. As the sphere has no
boundary, this integral must be zero. So, we have a contradiction and it's
impossible to define A over a sphere surrounding the origin.
On an aside -- it seems to me that one should be able to define a higher
dimensional analogue of monodromy here. Can one do this, and subsequently
define some sort of analogue of the universal cover that has trivial
second homology group?
Aaron (hoping he's got it correct this time. sigh.)
john baez
Benjamin P. Carter <b...@netcom.com> wrote:
>ba...@galaxy.ucr.edu (john baez) writes:
>>A typical example where we can't
>>do what Benjamin Carter wants is when our space is R^3 minus the origin
>>and B is the magnetic field produced by a magnetic monopole at the origin.
>>This space has nonvanishing second de Rham cohomology.
>I wonder: is there an elementary way to show the non-existence of a vector
>potential for this example?
Sure. Consider a sphere centered at the origin. The flux of the B field
through this sphere is nonzero. But we can show that it would have to be
zero if we had B = curl A. Here's how. Suppose we had B = curl A. Chop
the sphere into a northern hemisphere and a southern hemisphere and compute
the flux of the B field through each hemisphere using Stokes' theorem.
In each case you get the line integral of A around the equator, but with
opposite signs, so the total flux of the B field - the sum of the fluxes
through the two hemispheres - must vanish.
More generally, this proof shows that if B = curl A, the flux of the
B field through any closed oriented surface must vanish. By a "closed
oriented surface" I mean either a sphere, or a torus, or a 2-handled
torus, or a 3-handled torus, or....
Of course, for surfaces other than a sphere you can't find a "northern
hemisphere" and "southern hemisphere", but you can chop them up into
more pieces if you like, and the proof still works - you use Stokes'
theorem and show that the fluxes of B field through all the pieces must
add up to zero.
(Actually if you are really clever you only need to chop your surface into
*one* piece. We didn't really need to chop the sphere into a northern
and southern hemisphere! But for some reason I think the proof is easier
to follow if one uses at least two pieces.)
There's a somewhat deeper result which says that this criterion is not
only necessary, but sufficient. More precisely: if div B = 0 and the
flux of B through every closed oriented surface equals zero, then there
exists A such that B = curl A.
Perhaps this is the sort of answer you wanted all along, instead of
the erudite muttering about "differential forms" and "de Rham
cohomology" that you actually got. Well, guess what! Everything I've
been saying in this post is really a discussion of differential forms
and de Rham cohomology. The only reason why people resort to fancier
language is that they want to deal with these issues in any number of
dimensions and on spaces of arbitrary topology, not just subspaces of
R^3 as we have been doing here.
Basically, the story is that Poincare invented homology and cohomology
when trying to answer questions like yours, and it was so fun that
people spent the rest of the century studying them ever more deeply.
So when someone asks a question like yours, there's a tendency for experts
to think "sheesh, we've been working on this all century and this guy
just waltzes in oblivious to all that and wants to know the ANSWERS?" :-)
Thus they're tempted to say "Go learn de Rham cohomology, buster!" - but
this makes it sound more intimidating and a lot less fun than it actually
is.
Toby Bartels
>John Baez <ba...@galaxy.ucr.edu> wrote:
>>A typical example where we can't
>>do what Benjamin Carter wants is when our space is R^3 minus the origin
>>and B is the magnetic field produced by a magnetic monopole at the origin.
>I wonder: is there an elementary way to show the non-existence of a vector
>potential for this example?
Well, here's a simple proof:
We have the field B = n/(4 pi r^2),
where n is a unit vector pointing away from the origin.
If you take the surface integral of this vector field
around a sphere of radius 1 centred at the origin,
oriented outwards, you get, quite conveniently, 1.
Now, Stokes's theorem says, for any vector field A,
the surface integral of Del x A around a surface S
equals the line integral of A around the boundary of S.
In our case, S is a sphere, whose boundary is empty.
So, the line integral of A around the boundary of S must be 0.
Thus, the surface integral of Del x A around S is 0.
But the surface integral of B around S is 1, so B can't be Del x A.
You might not think of this as elementary,
because Stokes's theorem is a deep fact
fundamentally intertwined with the theory of de Rham cohomology.
But I think you'll be happy with it,
because Stokes's theorem is still part of
the standard undergraduate physics education.
What studying cohomology theory gets you
is that you'd realize right away, as I did,
that Stokes's theorem is exactly the tactic to use
to prove that there can be no such A.
-- Toby
to...@ugcs.caltech.edu
Robert Israel
>> A typical example where we can't
>>do what Benjamin Carter wants is when our space is R^3 minus the origin
>>and B is the magnetic field produced by a magnetic monopole at the origin.
>>This space has nonvanishing second de Rham cohomology.
>I wonder: is there an elementary way to show the non-existence of a vector
>potential for this example?
Yes: by Stokes's theorem, integral_S (curl A).dS = 0 over any closed surface
(in particular over a sphere centred at the origin). But of course
integral_S B.dS > 0.
A. Garrett Lisi
> |> Such an A will exist in the region iff every closed oriented curve in
> |> the region is the boundary of some compact oriented surface.
>
> No, that would be the condition for a vector field B with curl 0 to have
> a scalar potential. What you want is every closed surface (homeomorphic
> image of a sphere) to be the boundary of a region with compact closure.
> Then every vector field B that is smooth in the region has a vector
> potential in the region.
[Moderator's note: Quoted text trimmed. -P.H.]
Oops. You're right of course, I got them mixed up.
I'm rather new to differential geometry. Thanks for the correction.
-Garrett Lisi
Kevin A. Scaldeferri
Benjamin P. Carter <b...@netcom.com> wrote:
>Another more convincing motivation would be a precise but elementary
>statement of a theorem, along with a claim that exterior differential
>forms or cohomology or whatever is helpful, or required for, or the most
>natural approach to, the proof of the theorem.
Since this is the easy part of your questions, I'll provide an answer
:).
The obvious example is all those theorems from vector calculus.
Stokes' theorem, Green's theorem, the divergence theorem. You can
prove all these using real analysis, but it's messy and you have to
prove each of them seperately. But, differential topology shows us
that these are really all special cases of one theorem (Stokes is the
lucky one who got his name attached to this theorem). This theorem
can be proved (for all (orientable?) differentiable manifolds) rather
simply once you have developed the machinery of differential forms.
The theorem is compactly stated as
int_S dw = int_delS w
where S is a region to be integrated over, w is a form on that region,
delS is the boundary of S and dw is the exterior derivative of w.
People who really want to impress their friends write the theorem in
an even more compact manner:
<S,dw> = <delS,w>
defining the inner product to be integration. This form is slick
because it makes the duality between homology and cohomology obvious.
(Homology is to regions and boundaries what cohomology is to forms and
exterior derivatives.)
--
======================================================================
Kevin Scaldeferri Calif. Institute of Technology
The INTJ's Prayer:
Lord keep me open to others' ideas, WRONG though they may be.
[Moderator's note: Stokes' theorem holds for all compact orientiable
smooth manifolds with boundary. - jb]
Toby Bartels
>Toby Bartels <to...@ugcs.caltech.edu> wrote:
>>If in studying B you make certain approximations
>>and these approximations aren't valid for |z| < epsilon,
>>then anything you calculate about A may also be invalid --
>>even if you're only concerned with A in the region |z| >> epsilon!
>A concrete example or two would help. I would particularly like to see an
>example of a B field that needs to be defined in a multiply-connected
>region, such that there is no vector potential defined throughout that
>region. It would be especially interesting if the example bears some
>relationship to somebody's research.
As you know if you've been following the thread (and you have been),
what we really need is a region with an uncollapsible sphere,
not (necessarily) an uncollapsible circle.
That is, the missing part is r < epsilon, not |z| < epsilon.
(And, in fact, my z was a mistake; I meant distance *from* z axis,
not distance *along* z axis. But that's moot now.)
As you also know, the example is a magnetic monopole at the origin --
or what would appear to be a magnetic monopole at the origin
if the origin weren't missing from the space --
that is B = n/(4 pi r^2), for n a unit outward vector.
Now, what I would like to give you is an example
where, even though the space we're on is R^3,
we use an approximation for B which isn't valid near the origin,
such that the approximate B has no corresponding A.
The best thing I can think of for the B above
is that the *actual* B is really exactly 0
and we *approximate* 0 as n/(4 pi r^2) far from the origin.
This is a valid approximation far from the origin,
so it's technically a valid example,
but of course nobody would ever use that approximation.
I'll try to think of a more realistic example,
but I warn you I'm not very good at this practical stuff.
(I'm a mathematician, after all, not a physicist.)
>Another more convincing motivation would be a precise but elementary
>statement of a theorem, along with a claim that exterior differential
>forms or cohomology or whatever is helpful, or required for, or the most
>natural approach to, the proof of the theorem.
Well, I could state Stokes's theorem, but you probably know that already.
I think that's very natural in the context of cohomology theory,
where it is a special of the generic rule that
the boundary operator is the adjoint of the coboundary operator.
Even if you don't go *that* deep, however, differential forms are
definitely the natural way to understand Stokes's theorem.
There is a general theorem about differrntial forms,
that the integral of a n form w on the boundary of an (n+1) chain C
equals the integral of the exterior derivative of w on all of C.
This theorem covers, as special cases,
Gauss's theorem (the divergence theorem),
Stokes's thorem, Green's theorem (a special case of Stokes's theorem),
the 2nd fundamental theorem of calculus for line integrals,
and the independence of path of path integrals of analytic functions.
Also, it covers the case of more (or less) than 3 dimensions
(useful in relativity theory, for example).
I think forms are becoming more and more standard in physics
(ultimately, they're just a special case of tensors);
cohomology is a bit more far out, but still possibly interesting.
-- Toby
to...@ugcs.caltech.edu
john baez
Toby Bartels <to...@ugcs.caltech.edu> wrote:
>I think forms are becoming more and more standard in physics
>(ultimately, they're just a special case of tensors);
>cohomology is a bit more far out, but still possibly interesting.
When I was an undergraduate, I was interested in taking a course
in algebraic topology, but I had my doubts about its usefulness.
So I got up the nerve to knock on the door of a physics professor
and ask: "Does algebraic topology have a lot of applications to
physics?" And he replied, gruffly: "It doesn't matter whether
it has a lot of applications to physics. What matters is whether
it *will* have a lot of applications to physics. But I'll tell
you this: make sure you study differential forms!"
What excellent advice! I had never heard of differential forms,
so I was taken aback by his vehemence, but I made sure to learn
about them. Sure enough, they turned out to be very important
in all my work in mathematical physics. And I decided that I
would be forward-looking and go along with my inclination to learn
algebraic topology in hopes that it would turn out to have
applications to physics. This was back in 1981, just a bit before
string theory really caught on big and the level of mathematical
sophistication required among theoretical particle physicists
shot up to a new high. By the time I went to grad school, the
interaction between string theory and algebraic topology was going
great guns. If you didn't know differential forms and plenty of
cohomology theory, you would be left straggling while everyone else
was charging ahead into Donaldson theory, Chern-Simons theory, ever
more sophisticated versions of the Atiyah-Singer index theorem, and
so forth. Of course, only a minute fraction of physicists are
interested in any of this stuff - I'm only talking about theoretical
particle physicists, mathematical physicists, and some theoretical
condensed matter physicists. But among that crew, differential forms
and cohomology theory are de rigeur.
Benjamin P. Carter
I should have figured out for myself:
>Now, Stokes's theorem says, for any vector field A,
>the surface integral of Del x A around a surface S
>equals the line integral of A around the boundary of S. ...
>But the surface integral of B around S is 1, so B can't be Del x A.
>You might not think of this as elementary, ...
I do think of this as elementary, and I regret having posted a dumb
question (after having posted a not so dumb question).
By the way, I've heard of Poincare', but who is de Rham?
--
Ben Carter
te...@intex.com
ba...@galaxy.ucr.edu (john baez) wrote:
> In article <73vp9q$n...@gap.cco.caltech.edu>,
> Toby Bartels <to...@ugcs.caltech.edu> wrote:
>
> >I think forms are becoming more and more standard in physics
> >(ultimately, they're just a special case of tensors);
> >cohomology is a bit more far out, but still possibly interesting.
>
> particle physicists, mathematical physicists, and some theoretical
> condensed matter physicists. But among that crew, differential forms
> and cohomology theory are de rigeur.
>
Could someone give an explanation of what cohomology theory is and why
it's so important?
-----------== Posted via Deja News, The Discussion Network ==----------
http://www.dejanews.com/ Search, Read, Discuss, or Start Your Own
Hans Aberg
>Could someone give an explanation of what cohomology theory is and why
>it's so important?
Here is an easy intro:
Suppose one studies complex functions which have a complex derivative: I
skip the technical details -- it should be complexly differentiable, but a
continuous complex derivative satisfies that.
Since the set of complex numbers constitutes a real plane (of real
dimension two), one can study path integrals in that plane of those
functions. It then turns out that such an integral does not change value
if the curve moves, keeping the end points fixed, as long as we do not
move it over a singularity of the function (a place where the derivative
does not exist).
So if the complex function has a complex derivative in the whole plane,
the path integral only depends on the end points, and in particular the
integral over a closed curve (end points coinciding) is always zero, as we
can integrate over the curve which is just a point.
But the integral
int_gamma 1/z dz -- gamma an index of the integral symbol "int"
where gamma is a closed curve may be non-zero: Then it turns out that the
value is always an integer k times the number 2 pi i, and the number k is
the number of times the curve winds around zero, also counting the
direction as a sign.
Then this set of numbers k taken together are the set of the integers Z,
and this is the (Abelian) cohomology group for the real plane minus one
point: The addition is just the composite of setting one integral path
after another.
We could also add another singularity (removing a point from the complex
plane), by studying integrals
int_gamma 1/(z(z - 1)) dz
Thus z = 0, 1 are removed. Then it turns out that this integral has value
2 pi i (k + l), where k and l are two integers, the winding numbers around
the respective points z = 0, 1.
Then this set of integer pairs (k, l) taken together is the product Z x Z,
and this is the cohomology group for the real plane minus two points.
Now, it is possible to generalize this situation, and the result will be
the various different kinds of cohomology theories in use. On the same
time it will explain why these are cohomology theories important.
So the ingredients are some kind of analytical or topological space
(above, the real plane minus some points), some kind of analysis (above,
complex derivatives), and method to take differentials and to integrate.
One can then generalize these methods in various abstractions and compute
the cohomology groups, and these also may have duals called homology
groups.
One generalization is over the dimension of the space, so one gets one
group for each dimension of the space. For a differential manifold, one
can use differential forms to define a cohomology, the de Rham cohomology.
Another generalization is to merely use a topological space, or something
even more abstract such as an algebraic scheme, a group, a ring, or a
category. One may use a stratified topological space or differential
manifold, giving rise to theories about perverse sheaves and D-modules,
and higher dimensional analytical singularities.
The information we get about it is in part global invariants about the
objects, and we can draw conclusions about that. For example, a sphere
have different cohomology from that of a torus, so these cannot be
topologically the same.
But the cohomology also tells when say a differential equation can have a
solution. For example, if a certain linear partial differential equation
is exact, then it always has a local solution which can be obtained by
integration, but it may still not have a global solution, if the
cohomology group is non-zero: The solutions only exist modulo these
cohomological groups.
In modern mathematics, these cohomology theories have been very
successful: The most well-known are the conjectures by Andre Weil, that a
cohomology theory for expressing a Lefschetz fixpoint theorem in a number
theoretic context could be used to prove certain number theory theorems;
Grothendieck then set forth to develop the theoretical framework for such
a cohomology theory, whereas Deligne proved the Weil conjecture. Then,
this framework was used to cover up parts of the proof needed for the
Fermat's last theorem, the last part were settled by Wiles.
Then, turning to physics, one studies various differential equations,
which may be in a context or hidden by gauge theories (that is, one may
reason that certain gauge symmetries are governed by hidden differential
equations which we do not fully see or understand), so it comes natural to
use cohomology for that.
Hans Aberg * Anti-spam: Remove "REMOVE." from email address.
* Email: Hans Aberg <hab...@REMOVE.member.ams.org>
* Home Page: <http://www.matematik.su.se/~haberg/>
* AMS member listing: <http://www.ams.org/cml/>
john baez
>Could someone give an explanation of what cohomology theory is and why
>it's so important?
Crudely speaking, cohomology theory is the study of holes and how curves,
surfaces and so on can wrap around holes.
For example, Maxwell's equations say the divergence of the magnetic field
is zero. Normally this means that the magnetic flux through any sphere
is zero. But if we were able to cut a small hole out of space, or if
Maxwell's equations failed to hold in a small region of space, there
could be a nonzero flux of the magnetic field through a sphere *if* the
sphere wrapped around this hole. In fact it would suffice to remove a
single point from space to get this effect. This example is related to
Benjamin Carter's question. It comes up in the theory of magnetic
monopoles.
Or suppose you had an infinitely long solenoid and considered only the
region of space outside this solenoid. This region of space would have
a hole in it. If we ran a current through the solenoid it would make
a magnetic field inside but not, ideally, in the region of space outside.
Nonetheless if we carry a charged particle in a loop around the solenoid,
its quantum-mechanical phase will change by an amount proportional to
the magnetic flux through the solenoid. This is the Bohm-Aharanov effect.
A crucial aspect of cohomology theory is that holes can have different
"dimensions". In the first example it's important that a 2-dimensional
surface, a sphere, can wrap around the hole. Thus in a certain technical
sense we may say this hole is 2-dimensional - though that way of putting it
may sound confusing at first). In the second example it's important that
a 1-dimensional curve, a loop, can wrap around the hole. Thus we may say
this hole is 1-dimensional.
(Actually we say the first region of space has a "nontrivial 2nd cohomology
class" and the second one has a "nontrivial 1st cohomology class".)
Terry Pilling
dealing with algebraic topology and stuff, I don't
by any means profess to be an expert on this stuff,
but I find it zehr interessant and I hope to convey
a bit of that to (at least some of) you...
Algebraic Topology may not seem very useful at first glance
to physicists, and I can see by this thread so far that many
are quite skeptical since the results given by homotopy,
homology and de Rham cohomolgy theory can be got in other
ways more familiar to the physicist.
Still it is IMHO extremely important right now to see it
done with the beautiful elegance of algebraic topology.
I am certain that you will be hit by a "revelation" that
things that previously seemed to you to be utterly disparate
concepts (or at least only vaguely related) are actually
merely specific cases of an all-encompassing concept.
Let me quote Lewis Ryder from his second edition of
"Quantum Field Theory" (Cambridge Univ. press 1996)
"...Maxwell's equations take on the elegant form dF=0, d*F=J;
the antisymmetry of the field tensor is automatically included!...
..A common reaction of physicists to this type of mathematical
development is one of impatience. After all, they point out,
the equation d*F=J has to be translated into the form \partial_{\mu}
F^{\mu \nu} = j^{\mu} before it can be dealt with...
...This may be true, but in the opinion of a growing number of
physicists the development of notation actually corresponds
to a deepening of our understanding..."
Now I want to show you (in an incomplete sense) some of how far reaching
and enlightening Algebraic topology can be...
The concept that we have been dicussing is homology and cohomology
The first homology group the the set of closed 1-chains (curves) in
a space, modulo the closed 1-chains which are also boundaries.
A closed 1-chain is a boundary if it is homotopic to a point
(can be continuously deformed to a point).
This group is denoted H_1(U) = Z_1(U)/B_1(U)
U is the space, Z = cycles or closed 1-chains and B = 1-boundaries.
For example, on a torus there are 3 types of homotopy classes of
curves, ones that can be contracted to a point, ones that go around
the outside of the hole and ones that go through the hole.
Therefore the 1st homology group of the torus is isomorphic to Z^2
(the free abelian group with 2 generators). In other words, 2-d
vectors with integer entries, the integer being the number
of times you go around each curve (or just a weight attached to it).
The curve that doesn't go around a hole is a boundary and is therefore the
zero element.
The boundary of a boundary is zero. For example, the boundary
of a disk is a circle and the circle has no boundary, a line
on the other hand has a boundary consisting of 2 points.
**Not all curves without a boundary are themselves boundaries!**
In other words if \partial is the boundary operator, there
are curves for which \partial \gamma = 0 but which are not
boundaries themselves, i.e. there is no \sigma such that
\partial \sigma = \gamma.
To see this lets look at the 3 curves we mentioned above
on the torus the curve that is contractible to a point is a boundary,
the boundary of it is zero as required. BUT the other two curves
also have zero boundary, but are NOT themselves bounding any
region of the torus (i.e. they don't split the torus into
two distinct connected components).
These other two curves are exactly the ones that make the
first homology group of the torus non-trivial.
(hint: look up The Mayer-Vietoris sequence for homology and
cohomolgy)
Now what has this to do with Stokes theorem and all that
field theory stuff? EVERYTHING,
There is a duality between 1-forms and 1-chains, and all of
the results above have analogues for 1-forms.
(Like was said in an earlier post, 1-forms are integrated
over 1-chains to give real numbers i.e. an inner product)
The connectedness of the space we are integrating over
defines whether the 1-chains themselves are boundaries or
not.
A 1-form \omega (i.e. f(x)dx) is called CLOSED if d\omega = 0
(where d is the exterior derivative familiar from differential
geometry) Note: this is analogous to the fact that a 1-CHAIN
is closed (a cycle) if its boundary is zero (\partial \gamma = 0).
A 1-form \omega is called EXACT if \omega = df for some 0-form f.
Note: this is analogous to a 1-chain being a boundary.
(a zero form is a continuous function, a 1-form is fdx, a 2-form
is f dx^dy, etc and d is the `coboundary' operator that moves from
0-forms -> 1-forms -> 2-forms... \partial is the boundary operator
which moves you the other direction in chains n -> n-1 -> ...)
Note2: IT IS ALWAYS the case (exactly as above!) that d^2 F = 0
applying the exterior derivative twice is zero, this is
tautological. It is exactly equivalent to the fact that the
curl of a gradient is zero. This is analogous to saying that
the boundary of a boundary is always zero.
The 1st de Rham cohomology group (analogous to the 1st homology
group mentioned above) is the set of closed 1-forms modulo the set
of exact 1-forms. Denoted:
H^1(U) = Z^1(U)/B^1(U)
This group is therefore trivial (only the zero element) if
all closed 1-forms are exact.
Now like we said before, if the space that we are dealing
with is MULTIPLY connected, then there are closed 1-chains
that are NOT themselves boundaries, and like-wise there are
closed 1-forms that are not themselves exact.
What does this mean physically?
It means that just because curl(f) = 0 does not mean that
f = grad(g) for some g. Just because Div(B)=0 does not
mean that there is a potential function A such that
B = curl(A).
When is this a big deal?
Well whenever we have a region in our space that the field cannot
penetrate, (ie an infinite solenoid in the case of the Bohm-Aharonov
effect), this makes the space multiply connected!
Green's Theorem, Stokes Theorem and the fundamental theorem
of Calculus are all special cases of the generalized Stoke's
theorem, which can be stated as
\int_{\gamma} d\omega = \int_{\partial \gamma} \omega
Note that if any form A can be written as d \omega
for some \omega (i.e. it is exact) then
\int_{\partial \gamma} A = 0.
And if f = dA for some A, then the integral only depends
on the values of A on the boundary. NOTE: note that if
f is in particular a 1-form (ie f = F dx) then A is
a zero-form (i.e. a function) and since then \gamma
is a 1-chain then \partial \gamma is a set of 2 points,
so that the integral only depends on the value of F at
those 2 points (the end points), which is the fundamental
theorem of calculus. The other theorems arise simply
by considering higher dimensional n-forms and n-chains.
Lets look at some more examples of applications of this:
Gauge theory: We can add an exact 1-form to the vector potential
without changing E and B, which means that E and B depend only
on the de Rham cohomology equivalence class of A and not the
actual A. The Curvature is unaffected by an exact addition to
the connection.
Also we notice that if F = dA (i.e. the curavture is exact), then
Maxwell's equations dF = 0 become trivial!
The inhomogenous Maxwell's equations come from d*F=0 where
the * denotes hodge dual.
How do we calculate stuff?:
If A = f dx is a 1-form in R^3 then
dA = \partial_x A dx^dx + \partial_y A dy^dx
= -\partial_y A dx^dy
d*A = d { A [1/(3-1)!] { \epsilon_{123} dy^dz + \epsilon_{132} dz^dy }}
= d A dy^dz
= \partial_x A dx^dy^dz
Also the adjoint of d is \delta = -*d* and \delta \delta \omega = 0
just like for d.
But now the laplacian on a manifold is Laplacian = (d + \delta)^2
See the excellent paper by Eguchi, Gilkey and Hanson [phys. reports
66, no. 6, 1980. pgs 213-393] for more details on this stuff...
But I am diverging from the point...
More Apps...
Define: The WINDING NUMBER W of a curve \gamma about a point p
is defined as W(\gamma,p) = 1/(2\pi) \int_{\gamma} \omega_{p,\theta}.
where
\omega_{p,\theta} = [-(y-y_0)dx + (x-x_0)dy]/[(x-x_0)^2 + (y-y_0)^2]
and \gamma:[a,b] \rightarrow \mathbb{R}^2 \ {p} is a map from the
closed interval [a,b] to the complement of a point in the plane.
(basically \gamma(t) is a curve in the plane minus p with end points
at \gamma(a) and \gamma(b); minus a point = doubly connected!).
So essentially the winding number counts the angle that a ray drawn
from the point p to the curve traces out when you traverse the curve.
Now if the curve is closed, then the winding number is an integer.
namely it is the number of times you go around the point p.
Notice that your curve could also be the orbits or trajectories of
a vector field on the plane and in this case the point p would
simply be a singular point (a zero) of that vector field.
Now if I write the "angle function" \omega_{p,\theta} in the complex
plane instead of R^2, which is simply the identification (x,y) -> (x+iy)
we see that the angle function becomes \omega -> (1/z) dz and the winding
number is of course still an integer. So by the definition of the
winding number what do we get when we integrate this angle function
around a curve surrounding p?
\int_{\gamma} \omega = 2 \pi i W
the i comes from dz = dx + i dy. I you only go around once then this is
exactly the residue of the function 1/z (i.e the coefficient to 1/z of the
laurent expansion of 1/z is just 1 here)
actually:
Cauchy Integral Theorem:
If \gamma is a closed chain in U whoses winding number around any point
not in U is zero, and f is an analytic function in U, then for any a
in U that is not in the support of \gamma (closure of the region where
\gamma is not zero):
W(\gamma,a) f(a) = 1/(2 \pi i) \int_{\gamma} [f(z)/(z-a)] dz
Residue Theorem:
If f is analytic in U - {a_1,...,a_n} and \gamma is a closed 1-chain
such that W(\gamma,P) = 0 for all P not in U, then
1/(2\pi i) \int_{\gamma}f(z)dz = \sum_{i=1}^r W(\gamma,a_i)Res_{a_i}(f)
(note f(z)dz is a 1-form)
So winding number, which is deeply related to the fact that closed
curves cannot be contracted to a point, leads to the residue theorem.
Which nobody will deny has a few applications in physics! :-)
This also leads to the Euler characteristic of a surface which tells us
(among other things) the number of sources, sinks and saddle points
of a vector field on a surface (the `index'), it is basically an
alternating sum \sum_{i=o}^{N} (-1)^i C_i of a triangulation (closed
graph) drawn on the (N-dimensional) surface and C_i is the number of
i-chains on the graph, for example, on a disk or plane R^2 we have any
graph containing vertices, edges and faces.. the characteristic is then #V
- #E + #F since vertices are zero chains edges are 1 chains etc. Now any
graph in the space will have EXACTLY the SAME euler characteristic! So we
need only compute it once.
Why is this useful? Well because there is a direct relationship
between graphs and vector fields:
Vertices = sources, zeros
Edges = saddles
Faces = sinks
So that any vector field on a disk will have index 1 (since
I can find a single vector field that does, and the index, or Euler
characteristic, is an invariant)
The field I am thinking of is the one that consists of increasing circles
about the origin i.e one singluar point (zero).
On a sphere with g handles, v - e + f = 2 - 2g, so a torus which
is a sphere with 1 handle has v-e+f = 0; Ao you can `comb the
hair' on a torus, i.e there are vector fields on the torus with
no zeros.
Any planet (sphere) will have #Mountains - #passes + #valleys
=2 since any vector field on a sphere has 2 zeros (can't comb the
hair on a sphere!)
In Dynamical Systems (chaos theory) attractors are sinks, etc...
..and the list of applications goes on....
And this is not even touching on covering spaces, G-coverings,
universal covering spaces, principle bundles, Cech cohomology,
and all the goodies that are directly related to homology and
cohomology theory....
Another time perhaps, I am hungry...
I should probably say I will leave it as an excersize
for the reader >:)
-Terry-
"To err is human, but when the eraser wears out ahead of the
pencil, you're overdoing it."
-Josh Jenkins
Benjamin P. Carter
>Algebraic Topology may not seem very useful at first glance
>to physicists, and I can see by this thread so far that many
>are quite skeptical since the results given by homotopy,
>homology and de Rham cohomolgy theory can be got in other
>ways more familiar to the physicist.
I am still trying to make up my mind about this, but here I'll play the
devil's advocate and argue anainst all that high falutin' math.
>The concept that we have been dicussing is homology and cohomology
>The first homology group the the set of closed 1-chains (curves) in
>a space, modulo the closed 1-chains which are also boundaries.
>A closed 1-chain is a boundary if it is homotopic to a point
>(can be continuously deformed to a point).
That's true, but the converse is false. That is to say, a closed 1-chain
need not be homotopic to zero. Consider for example the space that
remains after a small circular hole has been removed from a torus. To be
precise, I will specify that what was removed is homeomorphic to an open
disk. What remains is technically a manifold-with-boundary, which is the
sort of space you must deal with if you are to generalize Stokes's
theorem. In this example, the boundary is homeomorphic to a circle, but
it is not homotopic to zero. (Homology is not homotopy.)
>This group is denoted H_1(U) = Z_1(U)/B_1(U)
>U is the space, Z = cycles or closed 1-chains and B = 1-boundaries.
If you try to do this rigorously, there are a few mathematical
prerequisites that were not familiar to many physicists when I was in school:
general (point set) topology, theory of free abelian groups, theory of
manifolds. (What have I left out?)
Contrast the above with the modest prerequisites for the garden-variety
version of Stokes's theorem. Those prerequisites are integral and
differential calculus and vector algebra.
Physicists are likely to look for non-rigorous shortcuts to either version
of Stokes's theorem. For the garden variety, a physicist's proof is to
draw a little square with sides dx and dy, and compute the line integral
of A(x,y) around the boundary of this square as [A(dx,0)-A(0,0)]dy +
[A(0,0)-A(0,dy)]dx, which is the z component of (curl A) dx dy (not
counting terms of higher order in dx and dy). With a little hand-waving,
one argues that no generality was lost in identifying the z axis with the
surface normal, and that a finite piece of a surface can be built up from
little squares (more or less).
There is no need to distinguish covariant from contravariant vectors or
introduce other abstract concepts that students are guaranteed not to
understand the first time around. Why make things more complicated than
is necessary for physical applications?
These are the thoughts that go through my mind when I get stuck in the
middle of some advanced math book.
--
Ben Carter
Rainer Bassus
>Maxwell's equations say the divergence of the magnetic field
>is zero. Normally this means that the magnetic flux through any sphere
>is zero.
Be careful with div E = 0
A spherical vector field with:
E = er Er + eth Eth + eph Eph
E = er 1/r^2
has sources but div E is zero.
div E in Spherical coordinates is:
div E = 1/r^2 d(r^2 Er)/dr + ... +...
Putting in Er in this formula gives as the result zero.
MfG
Rainer Bassus
Rainer...@t-online.de
Toby Bartels
>Terry Pilling <te...@offshell.phys.ndsu.nodak.edu> wrote:
>>A closed 1-chain is a boundary if it is homotopic to a point
>>(can be continuously deformed to a point).
>That's true, but the converse is false. That is to say, a [1-boundary]
>need not be homotopic to zero. Consider for example the space that
>remains after a small circular hole has been removed from a torus. To be
>precise, I will specify that what was removed is homeomorphic to an open
>disk. What remains is technically a manifold-with-boundary, which is the
>sort of space you must deal with if you are to generalize Stokes's
>theorem. In this example, the boundary ... is not homotopic to zero.
>(Homology is not homotopy.)
Right. You can even get an example with a compact boundaryless manifold.
Consider a double torus, a coffee cup with two handles (fixed point font!):
___________ ___________
/ \ / \
/ _ \_____/ _ \
/ / \ I / \ \
| | | I | | |
\ \_/ __I__ \_/ /
\ / \ /
\___________/ \___________/
Now draw a circle around the middle (the "I"s).
This circle is a boundary (of either half)
but it's not homotopic to a point either.
-- Toby
to...@ugcs.caltech.edu
A.J. Tolland
>
> Be careful with div E = 0
> A spherical vector field with:
>
> E = er Er + eth Eth + eph Eph
> E = er 1/r^2
>
> has sources but div E is zero.
Wouldn't that be Div(E) = 0 at every point but the origin, where
we run into problems because r -> 0?
--A.J. Tolland
john baez
Rainer Bassus <Rainer...@t-online.de> wrote:
>Be careful with div E = 0
>A spherical vector field with:
>
>E = er 1/r^2
>
>has sources but div E is zero.
>
>div E in Spherical coordinates is:
>
>div E = 1/r^2 d(r^2 Er)/dr + ... +...
>
>Putting in Er in this formula gives as the result zero.
Actually, a careful computation of the divergence of this electric
field does not give zero; it gives 4 pi times a delta function at
the origin. Of course the delta "function" is not a function, but
rather a distribution. One to be careful manipulating distributions,
and one has to be careful at the point r = 0 when doing calculations
in spherical coordinates, because many of the usual formulas do not
apply at this point.
James Gibbons
> Algebraic Topology may not seem very useful at first glance
> to physicists, and I can see by this thread so far that many
> are quite skeptical since the results given by homotopy,
> homology and de Rham cohomolgy theory can be got in other
In a categorical approach, what we really have are two functors:
DeRham: Diff------> D^b(Vect) precursor of DeRham cohomo.
S: Top------->D^b(Group) precursor of simplical homo.
(Where D^b(Vect) is category of bounded complexes of Vector Spaces,
and D^b(Group) is cat. of bounded complexes of Groups.)
If we consider a Differentiable
manifold as also a topological
space, the two functors are related by integration, (considered
as a product < , >_M : S(M) X DeRham(M) ---> R
where M is a Diff. manifold, and considering varous shifts
of the complexes (ith members).
and where we write :
< C, phi> = Integral_C (phi)
where C belongs to an ith member of D^b(Group)
and phi belongs to an jth member of D^b(Vect).
and was pointed out before, Stokes' theorem holds for all
compact orientable manifolds with boundary. (j=1+i here)
Jim Gibbons
jmf...@aol.com
ba...@galaxy.ucr.edu (john baez) wrote:
>In article <746641$tcn$1...@nnrp1.dejanews.com>, <te...@intex.com> wrote:
>
>>Could someone give an explanation of what cohomology theory is and why
>>it's so important?
>
>Crudely speaking, cohomology theory is the study of holes and how curves,
>surfaces and so on can wrap around holes.
I've been doing the opposite when I think about your stuff. I've
been thinking the analogy (since I still can't talk your lingo)
of turning a sock inside-out. Then, to get more complicated for
contrast, I think about turning a pair of panty hose inside-out--
first with the feet intact and then, with the feet cut off.
One way to turn a sock inside-out is to reach inside the sock,
grab a tiny little piece of the toe so that it's bunched into
a proverbial point, then pull. Now the space that used to be
on the outside is compacted into the inside and visa versa.
I'm just starting on the panty hose but haven't figured out
a way to invert it with just one bunching.
My hare-brained idea (my caveat) is that one could possibly
tell something about an unobservable dimension by looking
at the mechanisms used to invert the observable parts.
/BAH
Subtract a hundred and four for e-mail.
Hans Aberg
>>Crudely speaking, cohomology theory is the study of holes and how curves,
>>surfaces and so on can wrap around holes.
>I've been doing the opposite when I think about your stuff. I've
>been thinking the analogy (since I still can't talk your lingo)
>of turning a sock inside-out. Then, to get more complicated for
>contrast, I think about turning a pair of panty hose inside-out--
>first with the feet intact and then, with the feet cut off.
So using this picture, cohomolgy theory is about socks with holes in it:
If the sock would be infinitely stretchable, it would be the same as a
disc, with some holes in it then, if it has been worn too much by the
universe. Topology and cohomology are theories treating such stretchings
as being the same. So in this context, there would be no point in turning
the sock inside out.
Then, mathematicians use cohomology to patch those holes, that is, one
cannot patch the universe, but one uses the cohomology to understand what
is going on around those holes.
Another popular exercise is to cut a hole in the inner tube of a tire and
turning that inside out.
Benjamin P. Carter
>Another popular exercise is to cut a hole in the inner tube of a tire and
>turning that inside out.
First enlarge the hole until all that is left of the inner tube (yes,
kiddies, tires were not always tubeless) is a couple of circular strips
glued together. Then it is easy to turn the strips inside out.
This sort of topology is intuitive and accessible to high school students.
But something goes wrong when those students get to college and want to
know more about topology. They find that the curriculum requires a detour
through something called "algebra", which is nothing like the algebra they
already know. It seems that before you can get back to the studies of
inner tubes and the like, you have to know all about categories and
homological algebra, which are taught only to graduate students in
mathematics departments. And furthermore, there is something called
"general topology" which is part of something called "real analysis"
which you also have to know as background for "algebraic topology".
I have often wondered why there is no middle ground, like a popular book
or an undergraduate class on inner tubes, Moebius strips, Klein bottles,
and some of the charming topological problems considered by Euler.
--
Ben Carter
Hans Aberg
hab...@REMOVE.matematik.su.se (Hans Aberg) wrote:
>Another popular exercise is to cut a hole in the inner tube of a tire and
>turning that inside out.
I should have said that a lot of people consider this to be impossible,
because first draw a circle on the outside which is across the tube and
one on the inside which is along the tube. These two circles are in fact
the generators for the cohomology of the torus (an inner tube without a
hole cut in it).
Then before turning the tube inside out, these circles are linked, but
after it has been turned inside out, they are separated. So if you know
how to turn the inner tube inside out, you also know how to separate links
without breaking them.
Now tire inner tubes are not mathematical objects, so I am not sure how
exactly it translates into mathematics.
jmfb...@aol.com
hab...@REMOVE.matematik.su.se (Hans Aberg) wrote:
>In article <haberg-2012...@sl16.modempool.kth.se>,
>hab...@REMOVE.matematik.su.se (Hans Aberg) wrote:
>>Another popular exercise is to cut a hole in the inner tube of a tire and
>>turning that inside out.
>
>I should have said that a lot of people consider this to be impossible,
>because first draw a circle on the outside which is across
I knew I was going to get into trouble [chagrined emoticon here]. I
don't know what you mean by "circle on the ouside which is across
the tube". If I did what I think you said, I would have a pipe
with two open ends rather than a tube with an open circle cut
out of its surface (actually it ends up being an ellipse if one
flattens the sock.
Hans Aberg
>I knew I was going to get into trouble [chagrined emoticon here]. I
>don't know what you mean by "circle on the ouside which is across
>the tube". If I did what I think you said, I would have a pipe
>with two open ends rather than a tube with an open circle cut
>out of its surface (actually it ends up being an ellipse if one
>flattens the sock.
It is easy to draw by hand, but difficult to make a picture in a computer
text file like this one; sorry.
One does not cut the circles; one draws them (so they do not affect the
topology of the surface itself).
Think of the inner tube lying on the table. Then use a pen to draw a
horizontal line on the outside of the tube; if it goes all around the
tube, coming back to the original point after one turn, it is in effect a
circle. This is the direction of one of the circles drawn.
Start at any point of the outside of the tube and draw a line vertically.
If it goes all around, coming back to the original point after one turn,
it will be a circle going into the hole (center) of the tube, down, and up
again to close another circle. This is the direction of the other circle
drawn.
One then takes the thickness of the tube into account and draws them on
the inside and outside of the tube so that from the beginning they are
linked. When the tire has been turned inside out (if possible), they are
no longer linked.
Penny314
co-homology theory not hard to learn.
I select De-rham cohomology which is based on calculus and forms.
A very good introduction to this is in Differential Forms for scientist
and engineers by Harley Flanders.
Basically the properties of a manifold are clarified in terms of the
parameter spaces of the solutions of systems of linear partial differential
equations on it. H1 corresponds to incompressible fluid flow, H2 to Euclidean
electromagnetism and so on.
Hodges old classic on Harmonic integrals is also very interesting.
Its all not so hard as it may sound.
best
penny smith
Hans Aberg
(Sean Case) wrote:
>No, they are still linked. They have, however, changed places: if your
>tube has a nice herringbone or pinstripe pattern woven into it, you will
>find that turning it inside out "reverses" the pattern.
>
> ..___________.. ..___________..
> .._________.. .. \ | / ..
> .._______.. .. \ | / ..
> .._____.. .._____..
>
> a little slice of a the same slice
> striped torus inside out
It is in fact possible to turn a tube inside out in this way, which can be
demonstrated by making two stripes of paper (use equal length) pasted
together as (untwisted) circles and pasted together in a link. This
corresponds to a tube with a very large hole in it; at the start, one
paper stripe should represent the inside of the along the tube (as in the
left picture above), and the other the outside across the tube (as in the
right picture above). One then labels them differently. It is then
possible to turn these paper stripe circles inside out one at a time to
make the labels change place.
The problem though is to define "turning inside out" in a mathematical
way: The torus with a hole sure has two sides, but which is inside and
which is outside, so that one can define the process "turning inside out"?
john baez
>In article <gsc-301298...@wendy73.zip.com.au>,
>g...@zipworld.com.au (Sean Case) wrote:
>>
>>They have, however, changed places: if your
>>tube has a nice herringbone or pinstripe pattern woven into it, you will
>>find that turning it inside out "reverses" the pattern.
>>
>> ..___________.. ..___________..
>> .._________.. .. \ | / ..
>> .._______.. .. \ | / ..
>> .._____.. .._____..
>>
>> a little slice of a the same slice
>> striped torus inside out
>
>I or you are talking about.
There's a nice discussion of turning inner tubes inside out in one
of Martin Gardner's old Scientific American columns. It has pictures
that are a lot easier to understand than anything anybody can draw in
ASCII. Perhaps someone here knows how to get ahold of it.
The basic point of Gardner's column was that if rubber were infinitely
flexible, you could screw off the cap on your inner tube, reach into
the hole, and pull the inner tube inside out. If your inner tube had
circles drawn on the outside going around one way, and circles drawn
on the inside going around the other way, each outside circle would be
linked with each inside circle. And they'd still be linked after you
turned the inner tube inside out, since you can't unlink linked circles
without cutting them.
(To really *prove* you can't unlink these particular circles, the concept
of "linking number" comes in handy. And to define that, it helps to use
some cohomology theory. And hey, that's what this thread is supposedly
about!)
Benjamin P. Carter
>perhaps the best approach is to give the original poster an example of
>co-homology theory not hard to learn.
> I select De-rham cohomology which is based on calculus and forms.
> A very good introduction to this is in Differential Forms for scientist
>and engineers by Harley Flanders.
Several people have recommended this book to me. I finally broke down and
ordered a used copy vie the world wide web. The book is out of print, but
there are plenty of used copies for sale in various obscure book stores.
Thanks to Penny and others for the recommendations. The book arrived
today (just after I had sent an e-mail to the store, asking whether they
had shipped it). I will study it for a while to see how it compares with
a couple of other books I have which cover some of the same topics:
Bishop & Goldberg: "Tensor Analysis on Manifolds";
Loomis & Sternberg: "Advanced Calculus".
--
Ben Carter
john baez
>John Baez wrote:
>>The basic point of Gardner's column was that if rubber were infinitely
>>flexible, you could screw off the cap on your inner tube, reach into
>>the hole, and pull the inner tube inside out.
>This is precisely what I'm trying to do with a stocking appropriately
>modified (stockings are more malleable than socks and don't unravel).
>I am finding that, without cutting the torus such that I can flatten it
>to a rectangle, I cannot do a _single_ "inside-out" maneuver and get a
>torus. To get a torus shape, I need to do two inside-out maneuvers
>but I always end up with the original torus. If I do just one
>inside-out maneuver, I get a very strange pipe form.
Don't get mixed up by the difference between topology and geometry. Alas,
it's probably hopeless discussing this without me actually looking at that
stocking. Some things just don't work well in writing. If you tried to
explain *in writing* how you tie your shoelaces, for me to understand it
would be a major intellectual feat, worthy of an Einstein. It's probably
best to save both our brain cells for something a bit more useful.
(By the way, I've read that Einstein didn't wear socks, and he once
painted his toes black to camouflage some holes in his shoes. The man
was obviously a genius. But I digress....)
>>If your inner tube had
>>circles drawn on the outside going around one way, and circles drawn
>>on the inside going around the other way, each outside circle would be
>>linked with each inside circle. And they'd still be linked after you
>>turned the inner tube inside out, since you can't unlink linked circles
>>without cutting them.
>
>Here is where I'm getting completely confused because my torus form
>doesn't allow an inside circle linked to an outside circle. They are
>separate surfaces. I am considering the nylon threads that form the
>stocking.
Hmm, this at least I should be able to clarify. Forget that stocking
for a minute and look at this:
----------------------------------
| |
| |
| |
----------------------------------
Imagine this is a flat rectangle of very flexible rubber. Okay, now let's
make a torus out of it. Curl it up and attach the top edge to the
bottom edge in a way that makes these two arrows coincide:
--------------->------------------
| |
| |
| |
--------------->------------------
Now you have a long thin cylinder. Next attach the left edge to the
right edge in a way that makes these two arrows coincide:
----------------------------------
| |
V V
| |
----------------------------------
Now you have a torus. Okay, now suppose that before we had done this
we had drawn two lines on our rectangle. On the front we drew a line
like this:
----------------------------------
| |
|----------------------------------|
| |
----------------------------------
On the back we drew a line like this:
----------------------------------
| | |
| | |
| | |
----------------------------------
Now, after we form a torus out of this piece of rubber, these two
lines will become circles! And, better yet, the circles will
be LINKED. Topologically they will look just like this:
/\
/ \ /\ <------- a lousy ASCII attempt to draw
/ \ \ two circles linked in the simplest
/ / \ \ possible way, which also happens to
\ \ / / be the Feynman diagram for linked
\ \ / anyonic vacuum bubbles in a 3-dimensional
\ / \/ quantum field theory, hinting that maybe
\/ all this actually does have something to
do with physics after all.
although geometrically they will look rather different. In other
words, they'll look different, but if you wiggled around that perfectly
flexible rubber torus enough, you could get them to look just like
this picture.
>I promise. This is my last post :-).
Okay. Further discussion of this might drive us both insane.
Sean Case
> I am finding that, without cutting the torus such that I can flatten it
> to a rectangle, I cannot do a _single_ "inside-out" maneuver and get a
> torus. To get a torus shape, I need to do two inside-out maneuvers
> but I always end up with the original torus. If I do just one
> inside-out maneuver, I get a very strange pipe form.
If by "pipe" you mean a long thin tube with a double wall... then it's
your inside-out torus. Turning a torus inside-out is a topological
thing; you don't get exactly the same shape, but it is still a torus.
Sean Case
--
Sean Case g...@zipworld.com.au
Code is an illusion. Only assertions are real.
Terry Pilling
>
> On the back we drew a line like this:
>
> ----------------------------------
> | | |
> | | |
> | | |
> ----------------------------------
>
> Now, after we form a torus out of this piece of rubber, these two
> lines will become circles! And, better yet, the circles will
> be LINKED. Topologically they will look just like this:
>
> >I promise. This is my last post :-).
>
> Okay. Further discussion of this might drive us both insane.
>
>
Hold it John!
At the risk of advocating insanity, I have to resurect this thread
for a second.
In fact, the links *do* become unlinked when the torus is turned
inside out.
To see this clearly, take a piece of paper and identify the edges
as on your diagram. Then glue a piece of string vertically on one
side of the paper and another horozontally on the other side.
Now label one side "side A" and the other side "side B".
You will see that there are two distinct ways to make the torus,
and one way will be the "inside out" version of the other.
AND one way will have the rings linked, while the in the other
they will be unlinked.
In topology (unlike knot theory?) you are allowed to make cuts
(which must be done to turn the torus inside out in 3 dimensions)
and then do whatever moving around you want as long as you reconnect
the edges at the end in the same orientation that they started in.
Notice that two torii that are linked have exactly the same topology,
Euler characteristic, fundamental group, &c. as the same two torii
unlinked. The topology depends on the fact that they are disconnected,
not on *how* they are disconnected.
(Although there may be some pathological "Alexander horned-sphere"-type
surfaces that throw a wrench in this, but I don't really think so...)
-Terry-
Omnia disce, videbis postea nihil esse superfluum.
Hans Aberg
>Don't get mixed up by the difference between topology and geometry. Alas,
>it's probably hopeless discussing this without me actually looking at that
>stocking. Some things just don't work well in writing. If you tried to
>explain *in writing* how you tie your shoelaces, for me to understand it
>would be a major intellectual feat, worthy of an Einstein. It's probably
>best to save both our brain cells for something a bit more useful.
Speaking about tying your shoelaces, I and some friend was able to develop
an interestiong thoery why some tie good knots and other don't. Perhaps
some physicist can give the full theoretical explanation:
If one carefully pulls in the shoelace loops of the tied shoelace knots
out so that the result is a knot (instead of untying the shoelaces), then
there are two variations that can occur, the reef-knot and the granny's
knot. The former shoelace knot will hold well, but the other will not.
So here is an explanation why every second ties a good shoelace knot and
every second generation ties bad one: When mom teaches the child to tie
the knot, she stands in front of the child; the child first ties the
simple crossing of laces, but when mom shows the child doing the
complicated shoelace knot loops, the child is taught the mirror reversed
variation. So every second generation it switches between a reef-knot and
a granny's knot.
>(By the way, I've read that Einstein didn't wear socks, and he once
>painted his toes black to camouflage some holes in his shoes. The man
>was obviously a genius. But I digress....)
I have not tried this; it probably generalizes to covering up holes in
socks as well. But a mathematican would probably prefer to use cohomology
theory. (As in an earlier article in this thread, where cohomology was
likened with studying the behavior around holes, and socks with holes were
likened topological surfaces with holes.)
Hans Aberg
>> socks as well. But a mathematican would probably prefer to use cohomology
>> theory. (As in an earlier article in this thread, where cohomology was
>> likened with studying the behavior around holes, and socks with holes were
>> likened topological surfaces with holes.)
>>
>
> One way to define "homology" and "homotopy" as well is to ask what question(s)
> each are trying to answer. I believe that I read somewhere several years ago
> the following definitions:
> "Homotopy seeks to answer the question: 'Can be continuously deformed to...'"
> while "Homology seeks to answer the question: '...Together make up the
> boundary of...' "
One can use homotopy as well; instead of cohomology/homology, one then
gets algebraic K-theory, formulated by Quillen. Whereas cohomology is
relatively "simple" to work with, homotopy and algebraic K-theory have
been notoriously difficult. The reason is that the homotopy groups contain
much more information than the cohomology groups, so it is more difficult
to compute.
It has also been more unclear in the past exactly what information one
gets out of homotopy and algebraic K-theory in terms of analysis, and the
latter would then relate to physics. Perhaps this is better understood by
now.
But if one uses topology and cohomology in any circumstances, it is also
natural to try to start using homotopy and algebraic K-theory. So there
are probably a people trying that too.
> "Homotopy seeks to answer the question: 'Can be continuously deformed to...'"
> while "Homology seeks to answer the question: '...Together make up the
> boundary of...' "
While is is natural to think of homology as the theory of holes, it can
also be used to study continuos deformations, by combining with with
toplogical arguments. This is how the Lefshcetz fixed point theorem, which
in an algebraic version led to the proof of the Weil conjectures, is done.
It need not be the case that the holes are explicit: If one takes a torus
viewed as a topolical space, and not embedded in R^3, then there are no
explicit holes in it; the torus is the whole universe, and there is no way
to jump outside it to see the holes in it. There the cohomology becomes
useful, because it can give an algebraic meaning to what one mean with a
"hole"; then, when working with analysis, these algebraic classes show up
as obstructions to solving certain differential equations.
I recall that the cohomology can be computed from the homotopy, but I do
not recall exactly how: Is there a spectral sequence for this? It becomes
more complicated, because one can also use cohomology with coefficients;
for example, when using group bundales as used in gauge thoery. Then, one
can add "infinitesimal holes", as in QM, to a surface without the surace
itslef having a hole. (I do not have a book here to look this up, and I do
not recall the details immediately.)