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Right now, I am trying to understand better the cup product structure. I am interested in deriving the ring structure of the group cohomology $H^*(C_2,\mathbb{Z})$ with a nontrivial group action. The group action of a nontrivial element of $C_2$ sends an element $e\in \mathbb{Z}$ to $-e$. The cohomology groups are easily to obtain. The result is that $H^n(C_2,\mathbb{Z})=\mathbb{Z}/2\mathbb{Z}$ for $n$ an odd number and $H^n(C_2,\mathbb{Z})=0$, otherwise. I think, basically, my question can be boiled down to some smaller question.

  1. Is there a cup product structure for nontrivial group actions? Since people normally talk about group cohomology with trivial group actions, I am wondering whether it works when a group action is nontrivial. Do you need to modify the cup product formula?

  2. Is there any good procedure to follow to obtain a cohomology ring from cohomology groups?

  3. It seems that in this case it could be a ring with infinite number of generators with odd degrees and the cup product of any two generators vanishes. Am I correct?

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    $\begingroup$ The cup product of an element of $H^m(G,A)$ and one of $H^n(G,B)$ is an element of $H^{m+n}(G,A\otimes B)$. If $G=C_2$ and $A$ and $B$ are both $\Bbb Z$ with nontrivial action, then $A\otimes B$ is $\Bbb Z$ with trivial action. $\endgroup$ – Lord Shark the Unknown Oct 11 '18 at 19:00
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    $\begingroup$ @HermanChu The point is that $H^*(C_2, \Bbb Z_{sign})$ is not a ring. It carries the structure of a module over $H^*(C_2, \Bbb Z)$, where here $\Bbb Z$ carries the trivial action. $\endgroup$ – user98602 Oct 11 '18 at 22:33
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    $\begingroup$ @HermanChu I'll write something as soon as I get a chance, which may be a while. I suggest editing your question body so that it is about the question (2) you posed in the comments: what is the module structure of $H^*(C_2,\Bbb Z_{sign})$? (1) is too broad / too hard. $\endgroup$ – user98602 Oct 12 '18 at 11:59
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    $\begingroup$ Sorry, the last part of my comment was nonsense. (1) Yes, he meant that $\Bbb Z$ has trivial action. By not telling you the action, the author implicitly means it is trivial. What I was thinking of when I wrote that was "2." in your main question body. $\endgroup$ – user98602 Oct 12 '18 at 17:32
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    $\begingroup$ There is. I assumed you already knew the ring structure of $H^*(C_2; \Bbb Z)$ (as before, $\Bbb Z$ has trivial group action), or maybe $H^*(C_2; \Bbb Z/2)$. Either is more or less good enough. The easiest way for me to see how to do the calculations you want involves knowing products of some low-degree cohomology classes, and using the existing proofs. This could be done with the usual model of cochains. But it would also be relatively easy if we use the definition of the product structure using projective resolutions (eg, as in Ken Brown's book). $\endgroup$ – user98602 Oct 13 '18 at 1:38
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Let me start with a brief correction. For the graded group $H^*(G; M)$ to carry a ring structure, you should demand that $M$ is a ring and, most frequently, that the action of $G$ on $M$ is trivial. However, you do have the following structure more generally. Given $G$-modules $M$ and $N$, there is always a map $$H^*(G;M) \otimes H^*(G; N) \to H^*(G; M \otimes N).$$ So you can ask about what this map is for various values of $M$ and $N$. (If the action of $G$ on $M$ is non-trivial, but there is still a $G$-map $M \otimes M \to M$ satisfying the usual associativity rule, then $H^*(G;M)$ is still a ring.

Of interest to you are $C_2$ and what I will write as $\Bbb Z_+$ and $\Bbb Z_-$, where the nontrivial element acts either as $\times (+1)$ or $\times (-1)$; that is, the action on $\Bbb Z_+$ is trivial and the action on $\Bbb Z_-$ is negation. For computational reasons, we will also be interested in the case of $M = \Bbb Z/2$, the field with two elements and (necessarily) trivial action. Note that the three things we get are (ignoring ordering, which is essentially inconsequential): \begin{align*}H^*(C_2; \Bbb Z_+) \otimes H^*(C_2; \Bbb Z_+) &\to H^*(C_2; \Bbb Z_+)\\ H^*(C_2; \Bbb Z_+) \otimes H^*(C_2; \Bbb Z_-) &\to H^*(C_2; \Bbb Z_-)\\ H^*(C_2; \Bbb Z_-) \otimes H^*(C_2; \Bbb Z_-) &\to H^*(C_2; \Bbb Z_+)\\ \end{align*}

So $H^*(C_2; \Bbb Z_+)$ is a ring, $H^*(C_2; \Bbb Z_-)$ is a module over that ring, and there is a pairing $$H^*(C_2; \Bbb Z_-) \otimes H^*(C_2; \Bbb Z_-) \to H^*(C_2; \Bbb Z_+),$$ which is compatible with all of the module structures in an appropriate sense, but don't worry about that.


I take as a given the following fact: as a ring, we have $H^*(C_2; \Bbb Z/2) = \Bbb Z/2[x]$, where $x$ lives in degree $1$. In particular, $H^n(C_2; \Bbb Z/2) = \langle x^n\rangle$.

Next, allow me to make a point. The tensor product up above is natural in the modules $M$ and $N$, in the sense that if you have a $G$-module homomorphism $M \to M'$ and $N \to N'$, it induces maps $H^*(G; M) \to H^*(G; M')$ and similarly with $N'$, and the diagram you get comparing the two products commutes.

OK! Now let's do some computation. Using the resolution given in this post, you can compute the chain complexes computing $H^*(C_2; \Bbb Z_+)$ and $H^*(C_2; \Bbb Z_-)$ are the following, in that order:

$$\cdots \to \Bbb Z \xrightarrow{0} \Bbb Z \xrightarrow{2}\Bbb Z \xrightarrow{0}\Bbb Z \xrightarrow{2}\Bbb Z \xrightarrow{0}\Bbb Z \to 0 $$

$$\cdots \to \Bbb Z \xrightarrow{2}\Bbb Z \xrightarrow{0}\Bbb Z \xrightarrow{2}\Bbb Z \xrightarrow{0}\Bbb Z \xrightarrow{2} \Bbb Z \to 0 $$

Therefore, at the level of cohomology we have, for each $n \geq 0$, \begin{align*}H^{2n+1}(C_2; \Bbb Z_-) &= \Bbb Z/2\\ H^{2n+2}(C_2; \Bbb Z_+) &= \Bbb Z/2\\ H^0(C_2; \Bbb Z_+) &= \Bbb Z \end{align*}

and otherwise zero.

Now, we also need to know the map $H^*(C_2; \Bbb Z_\pm) \to H^*(C_2; \Bbb Z/2)$ induced by $\Bbb Z_\pm \to \Bbb Z/2$ given by reduction by $2$. Using the same (given) resolution, we find that the induced map between chain complexes defining these is precisely given by reducing each $\Bbb Z$ to $\Bbb Z/2$ (which has the result of giving us a complex with no differentials). In particular, the maps $H^*(C_2; \Bbb Z_\pm) \to H^*(C_2, \Bbb Z/2)$ has precisely the result of reduction mod 2. Therefore, the map $H^{2n+1}(C_2; \Bbb Z_-) \to H^{2n+1}(C_2; \Bbb Z/2)$ is an isomorphism for all $n \geq 0$, and similarly $H^{2n+2}(C_2; \Bbb Z_+) \to H^{2n+2}(C_2; \Bbb Z/2)$.


What is this good for? As a warm up, observe that it implies that if we write $y_k \in H^{2k}(C_2; \Bbb Z_+)$ for the nonzero element, we see that $y_k$ reduces to $x^{2k}$; therefore, because $y^k$ reduces to $(x^2)^k$ mod $2$ (because the product structure defined at the top of this post is natural), we see that $y_k = y^k$, and therefore $H^*(C_2; \Bbb Z_+) = \Bbb Z[y]/(2y)$, where $|y| = 2$.

Our next goal is to understand the action of $y$ on $H^*(C_2; \Bbb Z_-)$. Write $z_k \in H^{2k+1}(C_2; \Bbb Z_-)$ for the nontrivial element, as before. Then our computation from the previous section showed that $z_k$ reduces to $x^{2k+1}$, and compatibility of the product structure means that we know $(y^\ell) \cdot (z_k)$ must reduce to $(x^{2\ell}) \cdot (x^{2k+1}) = x^{2\ell + 2k + 1}$; the only thing that reduces to this is $z_{k+\ell}$, so we see that $y^\ell \cdot z_k = z_{k+\ell}$.

Inspired by this, we rewrite $z := z_0$, the element generating $H^1(C_2; \Bbb Z_-)$, and we see that as an $H^*(C_2; \Bbb Z_+)$-module, we may write $H^*(C_2; \Bbb Z_-) = \Bbb Z/2[y] \cdot z;$ that is, every element is of the form $y^k z$ for some $k$, and the action of $y^\ell$ is simply to multiply that power.

Lastly, we want to compute the pairing $$H^*(C_2; \Bbb Z_-) \otimes H^*(C_2; \Bbb Z_-) \to H^*(C_2; \Bbb Z_+).$$

Now we already know that $y^k z$ reduces to $x^{2k+1}$ and $y^\ell z$ reduces to $x^{2\ell+1}$. So by compatibility of product structure, we see that $(y^k z) \cdot (y^\ell z)$ should reduce to $x^{2k+2\ell + 2}$, and so $$(y^k z) \cdot (y^\ell z) \mapsto y^{k+\ell+1}.$$ If you like, you could write this as $z^2 = y$, but this is a little sketchy, since $z$ and $y$ live in different rings, so you're using the pairing above while not making it explicit in notation.

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  • $\begingroup$ In the second sentence I don't think you need the action of $G$ on $M$ to be trivial; you just need it to be a ring action. $\endgroup$ – Qiaochu Yuan Nov 12 '18 at 9:53
  • $\begingroup$ @Qiaochu Ah, good call. I don't know many good examples, but I guess $R[G]$ with translation action (maybe by a subgroup $H$) works. If you know better examples, I would be interested to hear. $\endgroup$ – user98602 Nov 12 '18 at 14:30
  • $\begingroup$ That's not a ring action, but $G$ acting on $R[G]$ by conjugation would work. $\endgroup$ – Qiaochu Yuan Nov 12 '18 at 20:35
  • $\begingroup$ thanks, as is clear I didn't actually check. $\endgroup$ – user98602 Nov 12 '18 at 21:52

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