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Let $X$ be a topological space and $R$ is a commutative ring. For $H^*(X)$ we have

$$0\to H^n(X,\mathbb Z)\otimes R\to H^n(X,R)\to \mathrm{Tor}(H^{n+1}(X,\mathbb Z), R)\to0.$$

Is it true that we have inclusion of rings $H^*(X,\mathbb Z)\otimes G\to H^*(X,R)$, i.e. this map preserves multiplicative structure? And what we can say about multiplication on the $\mathrm{Tor}$-component? This is difficult to understand because cohomology ring has geometrical nature, and universal coefficient formula is just homological algebra.

upd Here $X$ needs to be nice e.g. to have the homotopy type of a $CW$-complex with finite number of cells of each dimension, so the $\mathrm{Tor}$-version of UCT holds.

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  • $\begingroup$ If $G$ is not a ring itself, how do you consider $H^*(X,\mathbb{Z})\otimes G$ as a ring? $\endgroup$ Commented Aug 11, 2015 at 7:34
  • $\begingroup$ @archipelago, thank you. firstly i'm interested in case $\mathbb Z_n$ coefficients (so i didn't pay attention) $\endgroup$ Commented Aug 11, 2015 at 7:54
  • $\begingroup$ I have never seen this version of the uct. Could you explain how you obtain this short exact sequence? $\endgroup$ Commented Aug 11, 2015 at 9:19
  • $\begingroup$ @Dan, i apply functor $-\otimes R$ to the chain complex $C^*(X,\mathbb Z)$. maybe, here i need $X$ to be $CW$-complex, or, i don't shure, be compact.. $\endgroup$ Commented Aug 11, 2015 at 9:28
  • $\begingroup$ please correct me, if i was mistaken in what cases this formula really works $\endgroup$ Commented Aug 11, 2015 at 10:50

1 Answer 1

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Now I have some kind of answer.

At first, note that short exact sequence $$0\to H^n(X,\mathbb Z)\otimes R\to H^n(X,R)\to \mathrm{Tor}(H^{n+1}(X,\mathbb Z), R)\to0$$ occurs for finitely-generated (as $\mathbb Z$-module) ring $R$, because of equality $C^*(X,R)=C^*(X,\mathbb Z)\otimes R$. Suggestions about being $X$ a compact or a $CW$-complex are unnecessary.

So, $H^*(X)\otimes R$ really is a subring in $H^*(X,R)$. It follows directly from definition of $\smile$-product on chain complex of $X$: for $r,r'\in R$ and $\alpha,\beta\in C^*(X,\mathbb Z)$ we let $\alpha \otimes r\smile\beta\otimes r'=\alpha\smile\beta\otimes r\cdot r'$. This correctly define multiplication on cohomology, and inclusion $H^n(X)\otimes R\to H^n(X,R)$ preserves this multiplication.

Besides that, I can say something about $\mathrm{Tor}$-component of $H^*(X,R)$, namely that if we have a map $H^m(X)\otimes H^n(X)\to H^{m+n}(X)$ then it canonically defines a map $$H^m(X)\otimes\mathrm{Tor}(H^n(X),R)\oplus\mathrm{Tor}(H^m(X),R)\otimes H^n(X)\to\mathrm{Tor}(H^{m+n}(X),R).\,\,\,({\bf*})$$

Choose resolutions $0\to A_1\to A_0\to H^m(X)\to0$ and $0\to B_1\to B_0\to H^n(X)\to0$. For $H^m(X)\otimes H^n(X)$ we can take a resolution $$0\to C\to A_0\otimes B_1\oplus A_1\otimes B_0\to A_0\otimes B_0\to H^m(X)\otimes H^n(X)\to0.$$ Here $C$ is just kernal of the next arrow. It contains $A_1\otimes B_1$, therefore we can define a map $$A_0/A_1\otimes\mathrm{Tor}(H^n(X),R)\oplus\mathrm{Tor}(H^m(X),R)\otimes B_0/B_1\to \mathrm{Tor}(H^m(X)\otimes H^n(X),R).$$ So, desired map $({\bf*})$ is obtaned by composing with $\mathrm{Tor}(H^m(X)\otimes H^n(X),R)\to\mathrm{Tor}(H^{m+n}(X),R)$, that is given by multiplication in cohomology.

General hypothesis that we can reconstruct $H^*(X,R)$, when we have known $H^*(X)$, is false: for example, $\mathbb RP^3$ and $\mathbb RP^2\vee S^3$ have the same integer cohomology rings, but different ones over $\mathbb Z_2$.

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