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Suppose that we are given a CW complex X in terms of the cells and the gluing maps. My understanding is that computing the cup product of the singular cohomology ring from this information is a non-trivial task. I know of two basic strategies that one might take:

1) If the X is homotopy equivalent to a closed oriented manifold, then we can translate from cup product into intersection product and the problem becomes easier to visualize.

2) If X is not too complicated, then we can try to find a simple presentation of X as a finite simplicial complex and compute the cup product explicitly for all the cochains.

My question is: what are other techniques/tricks that can be used to find the cup product?

Surely there must be some general approaches beyond the naive ones I mentioned. Feel free to strengthen the hypotheses or consider specific situations, as I don't expect there to be one trick which works for everything.

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  • $\begingroup$ When I actually have to compute anything like this, I tend to use (1) or and sometimes $(\alpha,\beta)\mapsto \int\alpha\wedge \beta$ in de Rham. (2) is likely to lead to a mess. $\endgroup$ May 11, 2011 at 15:42
  • $\begingroup$ In my experience, any manifold that I understand well enough to be able to compute the de Rham algebra of is probably a manifold that I understand well enough to work out the intersection product on. Would you happen to know of an example which doesn't fit this description? $\endgroup$
    – user332
    May 11, 2011 at 17:04
  • $\begingroup$ Strategy (1) does not need $X$ to be homotopy equivalent to a manifold. See R. Fenn, Techniques of geometric topology'', Chapter 1 and S. Buoncristiano, C. P. Rourke, B. J. Sanderson A geometric approach to homology theory'', Chapter 2. $\endgroup$ May 11, 2011 at 23:41
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    $\begingroup$ I wouldn't say (1) is naive, since much of the work of completing the task is left un-done, in that you still have to perturb the homology classes to be transverse. This isn't always easily done. Morse theory can be useful in these situations. $\endgroup$ May 14, 2011 at 21:54

4 Answers 4

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This is going to be a perhaps tendentious diatribe. But it is what is. Naturality, dimensional arguments, and Poincare duality give a reservoir of elementary examples such as spheres and projective spaces.

In practice, to go from there to more serious examples, one uses spectral sequences to bootstrap up, and then one uses still more spectral sequences to bootstrap up to still more serious examples. The dirty secret is that modern algebraic topologists rarely if ever try to compute cup products by use of cochains, which means that they rarely if ever use cochains for serious calculations. The huge range of known calculations show how well this works.

The actual diagonal map $X\longrightarrow X\times X$ can only be helpful in the very simplest examples, for the obvious reason that explicit calculations must use cellular cochains (not singular, which are far too large for explicit computation), and the diagonal map is never cellular: it takes the n-skeleton to the 2n-skeleton. To compute cup products with cellular cochains, one must find a cellular map homotopic to the diagonal map. While such a cellular approximation always exists, it is rarely an easy task to write one down explicitly.

Peter May

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  • $\begingroup$ This has always been my experience too. Do you know of any situation where cup products can be computed at the chain level but cannot be computed by more conceptual means? $\endgroup$ May 15, 2011 at 4:47
  • $\begingroup$ Would you happen to know a place where I can see this bootstrapping argument worked out carefully for a serious example? $\endgroup$
    – user332
    May 15, 2011 at 12:06
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    $\begingroup$ You can also get the examples that Johannes mentions below by using the Serre SS, this might be what Peter is referring to. $\endgroup$ May 16, 2011 at 9:03
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One of the most useful tools is the Gysin sequence: if $f:E \to B$ is an oriented $S^{n-1}$-bundle, there is a long exact sequence $H^{\ast} (B) \stackrel{f^{\ast}}{\to} H^{\ast} (E) \stackrel{f_{!}}{\to} H^{\ast-n+1}(B) \stackrel{\chi}{\to} H^{\ast+1} \ldots $, where the last map is multiplication by the Euler class. This is derived from the Thom isomorphism, and the main ingredient for the proof is the Kuenneth theorem, but no nontrivial cup structure is needed. The Gysin sequence is strong enough to compute the cohomology rings of $CP^n$, $CP^{\infty}$, $BU(n)$, $U(n)$, $BZ/n$ and some other spaces.

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You can find maps $X\rightarrow Y$ or $Y\rightarrow X$ for some space $Y$ where the cup product is already known, and use the fact that cohomology induces a map of rings.

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The cup product is ultimately the map induced by the diagonal $\Delta: X\to X\times X$. You can get lots of information by studying the actual map.

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    $\begingroup$ Interesting. Are there a few examples you could give to flesh this out a bit? $\endgroup$
    – Dr Shello
    May 11, 2011 at 14:38

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