From: baez@galaxy.ucr.edu (John Baez) Subject: Re: Geometric quantization Date: 9 Jul 2000 02:32:16 GMT Newsgroups: sci.physics.research Summary: Cech cohomology Okay, now I'm gonna wrap up one of the loose ends of this thread. For a while now I've been talking about the first Chern class and how it classifies complex line bundles, because this is an important ingredient in geometric quantization. It took a lot longer than I expected, and even now there's a whole lot more to say, probably because I've been so inefficient about explaining it so far. I want to say a bit more. But I won't say all there is to say... I'll just say enough so we can declare victory and get the heck on out of here. So: Toby and I invented a homology theory for triangulated surfaces, whose job was to classify complex line bundles. Toby worked out all the details, and then I asked him to generalize this to triangulated manifolds of arbitrary dimension. Here's what he came up with: >A jchain f is, for each cooriented jcell c of the triangulation, >a map f_c: c -> U(1). Great, that's right.... >But I see a bit of a problem with this. >On a surface, a closed 1chain specifies a line bundle, >but in a manifold of general dimension n, >it's a closed (n-1)chain which specifies a line bundle. >To keep things consistent, we might want to do things this way: >A jcochain f is, for each cooriented cell c of codimension j, >a map f_c: c -> U(1). This forms a cohomology theory. >Then isomorphism classes of line bundles form the 1st cohomology group. Right! Flipping the homology theory upside-down this way was a very good idea! After we've flipped it upside down, we get a cohomology theory. And this cohomology theory is just a slight variant of Cech cohomology with U(1) coefficients! I guess I had better explain that. Let's be general: What's the Cech cohomology of a topological space X with coefficients in a topological abelian group G? Well, first we take our space X and pick an open cover of it, say {U_a}. Associated to this we get a bunch of "cochain groups". The "ith cochain group" is defined to consist of all continuous maps from all (i+1)-fold intersections of these U_a's to our coefficient group G. There is an obvious homomorphism d from the ith cochain group to the (i+1)st cochain group, which looks very much like the kind of "coboundary operator" we've been talking about. We have d^2 = 0 so we get a cochain complex. Then we take the cohomology groups of this cochain complex. Let's call them H({U_a},G). These cohomology groups depend on the cover {U_a}. But if we've got another cover {V_b} which is finer - meaning that each V_b is a subset of some particular U_a - we get a map H({U_a},G) -> H({V_b},G) So we can take a direct limit over all covers. That's the Cech cohomology H(X,G). In fact we often don't need to take a direct limit over all covers - we can use any cover as long it's fine enough! For the details see Bott and Tu's book "Differential Forms in Algebraic Topology". Okay, what does this have to do with what we were talking about? Well, suppose X is a triangulated n-manifold . Then we can take the sets U_a to be the n-simplices in X. Their i-fold intersections will be the simplices of codimension (i-1), if I've got the numbers straight. So in this case, the cohomology groups H({U_a},G) will be the cohomology groups you just invented! - at least when we specialize to the case G = U(1). In fact, the annoying little -1 in the "i-1" of the previous paragraph cancels the annoying little +1 in the "i+1" of an earlier paragraph, so everything works perfectly: your ith cohomology group is exactly what I'm calling H^i({U_a},U(1)). Even better, this is a case where the cover is "fine enough". So the cohomology groups you just invented are just the Cech cohomology groups H^i(X,U(1)). You may have noticed that I pulled a fast one: the n-simplices aren't open sets, so we are bending the usual definition of Cech cohomology to use them as our cover {U_a}. But I promise you, it's okay. >Now, if we can just show that this cohomology theory >is equivalent to singular cohomology over Z removed 1 level! Right: the task is now to show that the Cech cohomology group H^1(X,U(1)) is the same as the singular cohomology group H^2(X,Z)! If we can do this, we'll have seen that H^2(X,Z) classifies complex line bundles over X whenever X is a manifold. (Of course it's true even for any topological space.) Do we actually want to do this??? It's not hard *if* we're feeling comfortable with Cech cohomology, or at least willing to fake it. At this point, that stuff Dave Rusin whispered to you via email would come in handy. He basically gave away the key trick - the relation between U(1) and Z. However, you don't need to go down this road now if you don't want to. You have already seen, in a much more direct way, that H^1(X,U(1)) (as you defined it) classifies complex line bundles in the case when X is a triangulated surface. And that may be good enough for now. Of course, we also wanted to understand the differential-geometric side of the story: >>There's one extra little thing we should check to round off the >>story: this first Chern number is the same as the first Chern number >>we computed using a connection on our line bundle - by integrating >>the curvature over M and dividing by 2 pi i. >Well, then let me do that! >First, I need an inner product on the line bundle L. >May I suggest the standard inner product on the local trivialisations? Sure! >Since the transition functions map to U(1) (C* won't do!), >this defines an inner product on all of L. >In equations, <(p,z),(p,z')> = z* z'; >on an edge, <(p,f_e(p)z),(p,f_e(p)z'> = f_e(p)* z* f_e(p) z' >= |f_e(p)|^2 z* z' = 1 z* z' = z* z' = <(p,z),(p,z')>. >Now I need a connection compatible with this inner product. >May I suggest differentiation WRT the local trivialisation coordinates? Sure! >No, I may not! since it wouldn't be well defined on the edges. Oh - whoops! >But I can do this arbitrarily close to the edges, >and I can do it on any discrete point on an edge, such as all vertices. >Thus, all the curvature is concentrated near the edges. >Now, at this point, the region of non0 curvature might look like . /|\ | | | | | | | | | \|/ . >where the central line is the edge e. >But let's in fact move the left boundary of this region >entirely to the edge e, making it look like . |\ | \ | | | / |/ . >The left boundary will still be thought of >as lying in the left triangle, >but the right triangle's trivialisation can be used throughout. >So, using this coordinate, A = 0 on the right boundary of the region. >What is the value of A on the left region? >Well, if v is a vector at a point p on the edge e, >then A_v f = v[f] + A(v) f(p) is the basic equation. >Using a prime to indicate the left triangle's coordinate, >(A_v f)' = v[f'] + A'(v) f'(p). >Now, you get from an unprimed function to a primed function >by multiplying by the function f_e. >Thus, (v[f] + A(v) f(p)) f_e(p) = v[f f_e] + A'(v) (f f_e)(p) >= f(p) v[f_e] + v[f] f_e(p) + A'(v) f(p) f_e(p), >so A(v) = A'(v) + v[f_e]/f_e(p), or A = A' + (d f_e)/f_e. >Since A' = 0 on the left boundary, A = d ln f_e there. >Now, I want to integrate the curvature F = dA over the region, >which by Stokes' theorem is to integrate A over the region's boundary. Excellent!!! Yes, that's the key - Stokes' theorem. As usual! The whole relation between differential forms and topology is practically just a big fat spinoff of Stokes' theorem. >The integral of A over the right boundary is 0; >over the left boundary, it's \int_e d ln f_e. >Therefore, the integral of F over the entire manifold >is the sum over each edge e of \int_e d ln f_e. >Presumably, I could have calculated this a long time ago, >but it would then have been quite a bit less enlightening. Yes - never do a calculation before its time. >Was it Wheeler who said something like >"Never calculate anything until you know the answer."? Yes, it was. He taught Feynman this principle. You can read about this in one of those Feynman biographies - either Gleick's _Genius_, or Jagdish Mehra's _The Beat of a Different Drum_, or Feynman's own collections of anecdotes, _Surely You Must be Joking_ and _What Do You Care What Other People Think?_. I forget which one, but any mathematical physicist worth beans should read all of them. As Wheeler's grad student, Feynman would do calculations, and screw up somewhere in the middle, and get the wrong answer. When he showed these calculations to Wheeler, Wheeler could always spot exactly where Feynman went wrong, because Wheeler knew the right answer ahead of time and could sense when things were screwing up. Eventually Feynman learned how to figure out the answer before doing the calculation. >>The Cech cohomology group is >>the direct limit of the cohomology groups of >>the nerve complexes of the open covers, right? >>So, ... what does that have to do with what we've been doing??? Spanier's definition of Cech cohomology is a bit different from the one I used above (which you can find in Bott and Tu). They both involve 1) starting with some sort of cohomology that depends on an open cover and 2) then taking a direct limit over covers. But the kind of cohomology we're starting with is rather different. It's probably not worth bothering to relate Spanier's stuff to what I just said - it's too much of a digression! But I will say this: they should match when we take the coefficient group G to be discrete. You may enjoy thinking about the Cech cohomology of that wiggly "topologist's circle" that I mentioned in another post. You can see how the trick of taking the direct limit over covers manages to avoid the nastiness that occurs at the wiggly spot.