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ALgebraic Topology Query (Hatcher)  Not Homework 
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#1
Oct1311, 12:36 PM

P: 37

Hi all!
I haven't posted here in some time, and I am in need of the expertise of you fine folks. I am busy doing some work on spin geometry. Now, as you guys know, spin structures exist on manifolds if their second StiefelWhitney class vanishes. This class is an element of the second cohomology group. So, one needs to be versed in calculating cohomology groups in order to start working with these spin manifols. Much to my embarrassment I discovered that, although I know my way around homotopy, my homology knowledge is very lacking. The goto book for algebraic topology is Hatcher's great book, and it is from this book that my question comes. In it, after quite a lot of theoretical buildup, we reach the punch line that cohomology groups can be calculated via the split short exact sequence 0 > Ext(H_{n1}(X),G) > H^{n}(X;G) > Hom(H_{n}(X),G) > 0 This formula appears on page 198. The part I'm having some issues with is the Ext. I don't get an intuitive sense for what it's doing from the discussion on the preceding pages. If anyone is able/willing to shed some light on this for me, I would greatly appreciate it. Thanks in advance! 


#2
Oct1411, 04:10 PM

Sci Advisor
HW Helper
P: 9,453

I have not studied this since the 1960's but as i recall, Ext measures how far away a module is from being projective, or locally free. In particular it vanishes for free modules. So this tells you that the nth cohomology group really is Hom of the nth homology group, if the n1st homology group is free. Moreover it tells you that the extent to which this fails is due to the torsion in the n1st homology. So an nth cohomology element does define a linear function on the homology, but in case there is torsion in the n1st homology, a non zero cohomology element can give the zero functional.
They are not so easy to compute, but the name EXT is shorthand for "extensions", which are short exact sequences, of length n, modulo the split exact ones. So since sequences involving projective, e.g. free, modules, split, Ext is zero for them. You might look in some books on homological algebra for better explanations, and for some actual examples and calculations. 


#3
Oct2211, 01:16 AM

P: 1,197

Ah, the universal coefficient theorem for cohomology. What a cheeky bastard of a theorem.
I don't know this stuff as well as I should, but maybe I will by the time I am done answering. I used to have the whole picture in my head. It sometimes seems as if I am too busy these days to have time to remember anything, except the old thesis. This is a good excuse for me to do a lot of blabbering, so you'll have to excuse me for going off on tangents. My adviser once told me that most people start with the most complicated case first, and that is wrong (Hatcher is a good book, but not always pedagogically sound at every point, IMO). So, as mathwonk noted, I would tend to think of the free case first and worry about the Ext later. Here's a bit of a digression that makes it seem a bit more interesting and palatable for me. To me, one of the big motivations for cohomology is that it arises naturally when integrating differential forms (the other two big motivations being Poincare duality and obstruction theory). Integrating a differential form yields a linear functional from chains into the reals, which is a cochain. Now, suppose we want this functional to induce a functional on the homology with real coefficients. First, you can restrict the integration map to the cycles to get a linear functional on the cycles. Now, which cochains vanish on boundaries? Stokes theorem says the closed forms do (i.e. those whose exterior derivative vanishes). So, if we want our functional to be defined on homology classes (not just cycles), that naturally leads us to consider the closed forms. Furthermore, if we are just integrating on cycles, the result isn't going to be affected by adding an exact form (i.e., one that is an exterior derivative of somethingStokes theorem, again). So we may as well mod out by the exact forms. So, what we get here is a cochain complex where the boundary map is the exterior derivative. We were naturally led to consider cocycles mod coboundariesin other words, cohomology. There's a kind of symmetry herea duality, you could say. The fact that homology is concerned with cycles gave rise to modding out by coboundaries in cohomology. And the fact that we mod out by boundaries in homology gave rise to the idea of considering cocycles in cohomology. You can think of Poincare duality as expressing a geometric duality between homology and cohomology via the intersection pairing. The universal coefficient theorem expresses an algebraic duality via the Kronecker pairing. In the case of de Rham cohomology, what I've described is the Kronecker pairing. In general, you get a bilinear pairing by evaluating a cohomology class on a homology class, and an argument like the one I gave shows that it is welldefined (although, it won't be nondegenerate in general, which is what leads to that pesky Ext term). So, basically, what I have done is that I have tried to motivate the definition of cohomology by the desire for the Kronecker pairing to be welldefined. When the Kronecker pairing is nondegenerate (free case), you get that the cohomology is the dual of homology (get a map from cohomology to the dual of homology by sticking a cohomology class in the first slot of the pairingthis is the map in the short exact sequence whose kernel is the Ext term). So, that concludes my digression. I can post later about Ext. In Davis and Kirk's lecture notes in algebraic topology, when they define the Kronecker pairing, they summarize my whole post with a little remark that says the Kronecker pairing should be thought of as a generalization of integrating differential forms. My longwinded approach seems a little more helpful. 


#4
Oct2411, 12:28 PM

P: 1,197

ALgebraic Topology Query (Hatcher)  Not Homework
After thinking about it for a while and trying to get some intuition, I've concluded the theorem just isn't that intuitive, at least as far as the kernel part. I have been happy just understanding it in the free case and understanding that if you are not in that case, you get some kernel, but it's just something you can worry about when it comes up. That may be all you need in order to have a working knowledge of the theorem. Just worry about the kernel if there is one, but you know, aside from that kernel that cohomology is the dual of homology. Something I have struggled with in grad school is that I like to understand things fully. But, often, I just don't have time for it, so I settle for a partial understanding. So, you have to decide whether it's really worthwhile or not.
I now recall how the proof goes, but I am not sure how you would come up with it. The best one I've seen so far is in Davis and Kirk's Lecture notes on Algebraic topology, and it's not their proof, so there must be other references (actually, they make you fill in some details as an exercise). I've said I don't know how you would come up with a proof, but here are some ideas that could point you in the right direction. The first observation is that if you take the dual of homology, it kills all the torsion because any homomorphism from a cyclic group to a Z has to be trivial. But cohomology can have torsion. Just cook up an example where the homology in some dimension is Z/ 6Z and see what happens when you dualize the chain complex. Hatcher has an example like that. So, if you want to understand the kernel, you are going to have to worry about what killed the torsion. Torsion in homology is created by boundaries, so you might try to look at the cycles and boundaries, rather than just the homology (the point here is that cycles and boundaries are free groups, and duality works better for free things). That brings up the idea of considering this short exact sequence, which comes from the definition of homology and dualizing it, rather than just dualizing homology: 0→B_n → Z_n → H_n→0 Note that this is actually a free resolution of H_n, so taking the cohomology will yield Ext(H_n), by definition (though the dimension isn't what we want here). If you dualize, you get 0→H_n*→Z_n*→B_n*0 Then, you want to figure out where the map that comes from the Kronecker pairing fits into the picture. It's going from H^n to H_n*. It turns out to sort of factor through a map into Z_n*, in some sense. This is the point at which the proof becomes somewhat mysterious to me, so I will be very sketchy in describing how to finish the proof. The first trick is that map from H^n to Z_n* comes from another short exact seqence: 0→Z_n→C_n→B_n+1→0. What you do is use the long exact sequence in cohomology (zigzag lemma). You get another long exact sequence in cohomology from the other short exact sequence I mentioned: 0→B_n → Z_n → H_n→0. Now, this long exact sequence in cohomology contains that Ext term that you want in the universal coefficient theorm. Essentially, the rest of the proof consists of gluing the two long exact sequences together using a little homological algebra lemma. After a little work, you get the short exact sequence in the universal coefficient theorem. It's also helpful to just be good with Ext, so it may help to just compute some examples. 


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