Re: what is cohomology and what it's physics applications

On 16 Nov 1999 21:01:18 GMT, in sci.physics.research tedsung6674@my-deja.com wrote:

>Hi,
>
>Thanks for people's response to my question about fibre bundles.
>I have another along the same lines.  What is a cohomology (or
>cohomology theory) and how is it used in physics?
>
>Thanks,
>
>Ted
On 16 Nov 1999 18:00:01 -0600, in sci.physics.research baez@galaxy.ucr.edu (John Baez) wrote:

>In article <80rr9p$1f7$1@nnrp1.deja.com>,  <tedsung6674@my-deja.com> wrote:
>
>>Thanks for people's response to my question about fibre bundles.
>>I have another along the same lines.  What is a cohomology (or
>>cohomology theory) and how is it used in physics?
>
>I forget how much math you know, so I'll start out by saying some
>incredibly elementary stuff and then shoot forwards rapidly to some
>more sophisticated stuff.  If you know a medium amount of math,
>you'll probably be bored at first and then completely confused.
>Okay?  So: I'll concentrate on what cohomology theory *is* and 
>leave the physics applications for someone else.  
>
>If you have a space, it can have holes of various different 
>dimensions.   A cohomology theory is a way of defining what
>you mean by holes, and it lets you add and subtract these holes.   
>
>First you gotta understand how we keep track of the dimension of
>a hole!  It takes a bit of getting used to, so don't argue, just
>pay close attention:
>
>Consider a doughnut.  We say this has a 1-dimensional hole because
>it's possible to put a circle in a doughnut which encircles the 
>hole of the doughnut, and we can't "pull this circle tight" - 
>contract it to a point - while keeping it in the doughnut.  
>Since a circle is 1-dimensional we say the hole is 1-dimensional.
>A doughnut has no holes of dimensions other than 1.  
>
>Now consider a basketball: just the rubber skin, not the air inside.
>We say this has a 2-dimensional hole because it's possible to put a
>sphere inside the skin of the basketball which wraps around the
>air inside, and we can't "pull this sphere tight" - contract it
>to a point - while keeping it in the skin of the basketball.  
>Since a sphere is 2-dimensional we say the whole is 2-dimensional.
>A basketball has no holes of dimensions other than 2.  
>
>Now consider an inner tube: just the rubber skin, not the air inside.
>This is like a hollow doughnut!  It's more complicated than the previous
>examples because it has more than one kind of hole.  It has a 1-dimensional
>hole corresponding to how you can wrap a circle around it the long way, 
>just like you can with the doughnut.  But it also has a 1-dimensional
>hole corresponding to how you can wrap a circle around it the short way!
>It also has a 2-dimensional hole corresponding to how you can put a
>torus inside the skin of the inner tube.    This 2-dimensional hole
>is a bit like the example of the basketball, except now we're using a
>torus to wrap around the air inside the inner tube, instead of a sphere.
>
>If you're smart you can figure out other ways to wrap a circle around
>the inner tube: it can sort of *spiral* around the inner tube.  But this
>kind of 1-dimensional hole is a linear combination of the 1-dimensional
>holes I already discussed.  If the circle spirals n times around the
>short way while it spirals m times around the long way, we say the hole
>it represents is nx + my, where x and y form the "basis" of 1-dimensional
>holes described in the previous case.
>
>Okay: in general, the cohomology of a space X is a bunch of abelian
>groups H^n(X), one for each integer n.  The idea is that H^n(X) consists
>of all the n-dimensional holes, and we can add and subtract these holes.
>
>Now, so far I've been describing holes by looking at *manifolds* mapped
>into our space X.  But we could use other things.  It's actually very
>customary to use *simplices*.  Different ways give different cohomology
>theories.  If we use simplices we get something called "singular cohomology
>theory" or "ordinary cohomology theory".  If we use manifolds we get 
>something called "cobordism theory".   There are actually lots of different
>kinds of cobordism theory depending on what sort of manifolds we use:
>there's "oriented cobordism theory" and "unoriented cobordism theory" 
>and "smooth cobordism theory" and "piecewise-linear cobordism theory" 
>and so on.  And there are lots of other cohomology theories, too, like
>K-theory and stable homotopy theory and so on.  These cohomology theories 
>are related in lots of useful ways.   
>
>In physics, perhaps the most popular cohomology theory is "deRham 
>cohomology theory".  Here we use differential forms to keep track 
>of holes in a smooth manifold.  Since differential forms are very 
>important in physics, it's not surprising that deRham theory has lots 
>of interesting applications.   Stokes' theorem, Gauss' theorem, Green's
>theorem - they're all just the tip of the iceberg called deRham
>theory.   You can find an elementary introduction to deRham cohomology 
>in my book "Gauge Fields, Knots and Gravity", and in lots of other places, 
>like Flanders' book "Differential Forms with Applications to the Physical 
>Sciences", or von Westenholz' book "Differential Forms in Mathematical 
>Physics".  If you're a physicist, deRham theory is the place to start!
>
>Okay, finally for some more high-powered stuff.  For every cohomology
>theory H^n, there's a bunch of spaces S(n) called the "classifying spaces"
>of that cohomology theory.  The cohomology group H^n(X) of any space 
>X is really just given by [X,S(n)] - the set of homotopy classes of maps 
>from X to S(n), which turns out to be a group, thanks to some special 
>stuff about S(n).   For example, for ordinary cohomology the classifying
>space is called an "Eilenberg-MacLane space" and denoted K(Z,n).  In
>general, for any cohomology theory, the spaces S(n) fit together into a
>gadget called a "spectrum".   So in a sense, cohomology theory is really
>the study of spectra!   This is how the experts think about it.   But to
>really understand this viewpoint, you gotta understand what's so great
>about spectra.  Ultimately it turns out that spectra are just a super-duper-
>generalization of abelian groups: they are "infinitely categorified, 
>infinitely stabilized abelian groups".  This is why they're so great. 
>Anyone who wants to learn more about this should read Adams' book "Infinite
>Loop Spaces".
>
On Thu, 18 Nov 1999 02:08:03 GMT, in sci.physics.research toby@ugcs.caltech.edu (Toby Bartels) wrote:

>John Baez <baez@galaxy.ucr.edu> wrote a lot, including:
>
>>Okay: in general, the cohomology of a space X is a bunch of abelian
>>groups H^n(X), one for each integer n.
>
>In many cohomology theories,
>you can multiply an element of H^m(X) and an element of H^n(X)
>to get an element of H^{m+n}(X).
>Thus, the cohomology groups together form the cohomology ring.
>I just thought I'd mention that.
>(In de Rham cohomology, this is related to multiplying an mform and an nform.)
>
>
>-- Toby
* toby@ugcs.caltech.edu

baez@galaxy.ucr.edu (John Baez) wrote:

Right.  In general, this happens when the spectrum S(n) defining
your cohomology theory is a "ring spectrum".  Ring spectra are to
rings as spectra are to groups: in both cases, we're taking a basic
concept from algebra and then categorifying and stabilizing it 
infinitely many times. 

For an intro to "categorification" and "stabilization", try:

http://math.ucr.edu/home/baez/week121.html

On 25 Nov 1999 23:05:42 -0800, in sci.physics.research baez@galaxy.ucr.edu (John Baez) wrote:

>In article <813tth$7nd$1@rosencrantz.stcloudstate.edu>,
>dale hurliman  <hurliman@nospam.sunlink.net> wrote:
>
>>John Baez wrote:
>
>>> Consider a doughnut.  We say this has a 1-dimensional hole because
>>> it's possible to put a circle in a doughnut which encircles the
>>> hole of the doughnut, and we can't "pull this circle tight" -
>>> contract it to a point - while keeping it in the doughnut.
>
>>All of what you say assumes that your space can be embedded in a
>>Euclidean space of higher dimension, right?  
>
>Actually none of what I said assumes that.  It's an important step,
>when learning about concepts of space, to stop relying on a picture
>of space as embedded in a Euclidean space of higher dimension.  It
>took people about a century of careful thought to realize this, but 
>by the mid-1900s it was quite obvious.
>
>>Is this always possible? Is
>>there a theorem which says that any arbitrary space of N dimensions can
>>always be embedded in a Euclidean space of M>N dimensions?
>
>There are theorems like this, but they don't apply to an "arbitrary
>space".  For example, lots of spaces are infinite-dimensional, and 
>these can't be embedded in any Euclidean space at all - but you can
>still study their cohomology, and some people spend their whole lives
>doing just that.
>
>Here's a theorem along the lines you want: a smooth N-dimensional 
>manifold can always be smoothly embedded in M-dimensional Euclidean 
>space for some M > N.  (In fact there's an explicit upper bound on 
>the necessary M, but I forget what it is.)  Smooth manifolds are 
>the kind of spaces physicists are most likely to want to know the 
>cohomology of - for example, general relativity assumes that spacetime
>is a smooth manifold.  But it turns out to be completely unnecessary,
>and usually downright unhelpful, to think of smooth manifolds as 
>embedded in Euclidean space.
>
* In article <81qqah$dpo$1@nnrp1.deja.com>,
* Jim Heckman  <jheckman@my-deja.com> wrote:
* >In article <81lbg6$iks@charity.ucr.edu>,
* >  baez@galaxy.ucr.edu (John Baez) wrote:
* 
* >> Here's a theorem along the lines you want: a smooth N-dimensional
* >> manifold can always be smoothly embedded in M-dimensional
* >> Euclidean space for some M > N.
* 
* >Yikes!! Really?! 
* 
* YES!!  REALLY!!
* 
* (It's nice to see people getting so excited about theorems.  For
* some reason one rarely sees this on sci.math.research.)
* 
* >>By specifying M-d *Euclidean* space, aren't you
* >implicitly inducing a metric on the N-d manifold? 
* 
* Yes - but of course this metric is completely irrelevant to the
* point of the theorem, which is a theorem about smooth manifolds, 
* not Riemannian manifolds.   
* 
* In other words, I DID NOT SAY THIS:
* 
* If you have a smooth N-dimensional manifold with a given Riemannian 
* metric, there is an isometric embedding of it into M-dimensional 
* Euclidean space with M > N.   
* 
* (Here by "isometric" we mean that the given metric on our manifold 
* is the same as that induced by its embedding in Euclidean space.)
* 
* Now, I COULD have said that, because it's true... but it's not 
* what I actually DID say.
* 
* >But I could have sworn
* >(I find myself saying that a lot, recently :-/) I read it's impossible to
* >embed, say, a 2-d hyberbolic plane (constant negative Gaussian
* >curvature) in *any* Euclidean space. Is this not correct?
* 
* Yes, I'm pretty sure it's not correct, thanks to the theorem I 
* just now stated, which is a much deeper and more surprising theorem 
* than the one I stated before.
* 
* >OTOH, maybe just the *diffeomorphic* structure of the N-d manifold can
* >be replicated in M-d Euclidean space -- not the entire geometry. Is that
* >closer to what you mean?
* 
* Yes, that's what I meant - because I didn't breathe a word about 
* "metrics" or "geometry" until you started talking about that.   Please 
* don't mix up your categories - it's a recipe for misunderstanding!  
* When I say "smooth manifold", I don't mean "Riemannian manifold", and
* when I say "embedding", I don't mean "isometric embedding".

On Fri, 3 Dec 1999, Jim Heckman wrote:

> implicitly inducing a metric on the N-d manifold? But I could have sworn
> (I find myself saying that a lot, recently :-/) I read it's impossible to
> embed, say, a 2-d hyberbolic plane (constant negative Gaussian
> curvature) in *any* Euclidean space. Is this not correct?

It's not correct.  But there is a subtlety: local versus global plays a
role here!  You can embedd in E^3 a whole bunch of surfaces which are
-locally- indistinguishable from H^2 (have constant negative curvature)
but are -globally- quite different.  Take a small neighborhood of one of
these: that would be a "local embedding of (a piece of) H^2 in E^2".  By
going into higher dimensions you can get a -global- embedding on H^2 in
some E^d.  Similar local versus global distinctions arise for embedding
semi-Riemannian maninfolds in higher dimensional E^(p,q).
 
Chris Hillman

Home Page: http://www.math.washington.edu/~hillman/personal.html

On Fri, 3 Dec 1999 04:34:02 GMT, in sci.physics.research toby@ugcs.caltech.edu (Toby Bartels) wrote:

>John Baez <baez@galaxy.ucr.edu> wrote:
>
>>Here's a theorem along the lines you want: a smooth N-dimensional 
>>manifold can always be smoothly embedded in M-dimensional Euclidean 
>>space for some M > N.  (In fact there's an explicit upper bound on 
>>the necessary M, but I forget what it is.)
>
>The upper bound on M is 2N.
>It's fairly easy to prove an upper bound of 2N + 1
>(although I'm not sure I could do it off the top of my head);
>2N, I'm told, is harder.
>
>BTW, you want to say the manifold is *connected*,
>at least in the case N = 0
>(and possibly other cases if you don't require paracompactness).
>
>
>-- Toby
On Fri, 10 Dec 1999 04:15:38 GMT, in sci.physics.research toby@ugcs.caltech.edu (Toby Bartels) wrote:

>John Baez <baez@galaxy.ucr.edu> wrote:
>
>>Toby Bartels <toby@ugcs.caltech.edu> wrote:
>
>>>John Baez <baez@galaxy.ucr.edu> wrote:
>
>>>>Here's a theorem along the lines you want: a smooth N-dimensional 
>>>>manifold can always be smoothly embedded in M-dimensional Euclidean 
>>>>space for some M > N.  (In fact there's an explicit upper bound on 
>>>>the necessary M, but I forget what it is.)
>
>>>The upper bound on M is 2N.
>>>It's fairly easy to prove an upper bound of 2N + 1
>>>(although I'm not sure I could do it off the top of my head);
>>>2N, I'm told, is harder.
>
>>Thanks.  The trick to remembering this stuff is to remember 
>>that generically, an N-dimensional manifold mapped into an 
>>M-dimensional one intersects itself along a submanifold of 
>>dimension N + N - M.   This is just by counting degrees of
>>freedom.  So M = 2N + 1 should certainly work.  On the other
>>hand, M = 2N might give you isolated points of self-intersection.
>>So in this case, you'll need to do some topology involving 
>>intersection numbers to prove you can get rid of the intersection 
>>points by cleverly cancelling them in pairs.  Or something like
>>that.
>
>It would have to be more than just cancelling in pairs.
>Suppose you're trying to embed S^1 in R^2
>and you foolishly end up with a figure "8".
>This is generic in that all intersections are transversal,
>which I believe is the setting for the formula N + N - M,
>so you have the 0D intersection as you should,
>but you can't straighten it out by cancelling in pairs.
>
>>Right, we definitely need paracompactness for these theorems.  
>>Without it, we can always make a "long line", a nonparacompact
>>1-manifold, that can't be embedded in R^n for the simple reason 
>>that it has more points than R^n does!  This is one more reason 
>>people like to tuck paracompactness into the definition of "manifold".   
>
>I forgot about the long line, one of my favorite counterexamples.
>For those who don't know it, let A be the smallest ordinal
>larger than the cardinality of the continuum,
>and put A copies of the interval [0,1[ end to end.
>
>
>-- Toby
>   toby@ugcs.caltech.edu