For a trivial action on the coefficient, we have the following Kuenneth formula for group cohomology:

$$ H^n(G_1 \times G_2; M) \cong [\oplus_{i= 0}^n H^i(G_1;M) \otimes_M H^{n-i}(G_2;M)] \oplus [\oplus_{p =0}^{n+1} \text{Tor}^M(H^p(G_1;M),H^{n+1-p}(G_2;M))] $$ where $G_1$ and $G_2$ are finite groups and/or compact Lie groups --Edit-- and $M$ is a PID such as $Z$?

If the group action on the coefficient $M$ is non-trivial, does the above Kuenneth formula remain valid?

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    $\begingroup$ (I guess $M$ is a ring.) There is not a formula that's quite this simple except in some cases. If $G_1$ and $G_2$ are finite groups, then there is what is called a Tor spectral sequence, which requires contributions from the higher Tor groups over $M$ and is, unfortunately, not quite as simple as all of the Tor groups appearing in a direct sum. If the coefficient actions are not trivial then it gets more complicated. $\endgroup$ Sep 15, 2011 at 2:14
  • $\begingroup$ I am sorry. I should mention that $M$ is a PID, such as $Z$. Also, the action of $G_1\times G_2$ may not be most general. For example, we may assume that only $G_1$ acts non-trivially on $M$. In fact, $G_1\times G_2$ acts "naturally", say, on $H^i(G_1,M)\otimes_Z H^{d−i}(G_2,M)$, Let $a\in H^i(G_1,M)$ and $b\in H^{d−i}(G_2,M)$. We have a group action $(g_1,g_2) \cdot (a\otimes b)=(g_1\cdot a)\otimes (g_2\cdot b)$. $\endgroup$ Sep 15, 2011 at 2:23
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    $\begingroup$ There is a Künneth formula but only when the coefficient is a tensor product $A\bigotimes B$ (and one of them is flat over the base ring). For trivial action and $A=B$ is equal to the base ring we have $A\bigotimes B$ is again equal to the base ring with trivial action. In the general case the action may not factor in that way. However, you seem to be content with $G_1$ acting trivially and then you do have such a tensor product factorisation and everything is OK. $\endgroup$ Sep 15, 2011 at 4:44

1 Answer 1


I learned this through Ken Brown's textbook "Cohomology of Groups" (and studying under him): I basically use the beginning of his Chapter 5 and solve the two exercises in that section.

Let $M$ (resp. $M'$) be an arbitrary $G$-module (resp. $G'$-module), let $F$ (resp. $F'$) be a projective resolution of $\mathbb{Z}$ over $\mathbb{Z}G$ (resp. $\mathbb{Z}G'$), and consider the map $(F\otimes_GM)\otimes(F'\otimes_{G'}M')\rightarrow (F\otimes F')\otimes_{G\times G'}(M\otimes M')$ given by $(x\otimes m)\otimes(x'\otimes m')\mapsto(x\otimes x')\otimes(m\otimes m')$.

Note that $(F\otimes_GM)$ $\otimes(F'\otimes_{G'}M')$ = $(F\otimes M)_G \otimes(F'\otimes M')_G$ ,

which is the quotient of $F\otimes M\otimes F'\otimes M'$ by the subgroup generated by elements of the form $gx\otimes gm\otimes g'x'\otimes m'$. The isomorphism $F\otimes M\stackrel{\cong}{\rightarrow}M\otimes F$ of chain complexes ($M$ in dimension $0$) is given by $x\otimes m\mapsto (-1)^{deg(m)\cdot deg(x)}m\otimes x=m\otimes x$, and so the aforementioned quotient is isomorphic to $F\otimes F'\otimes M\otimes M'$ modulo the subgroup generated by elements of the form $(gx\otimes g'x'\otimes gm\otimes g'm')=(g,g')\cdot(x\otimes x'\otimes m\otimes m')$ where this latter action is the diagonal $(G\times G')$-action. Now this is precisely

$$(F\otimes F'\otimes M\otimes M')_{G\times G'} = (F\otimes F')\otimes_{G\times G'}(M\otimes M')$$ and hence the considered map is an isomorphism.

Assuming now that either $M$ or $M'$ is $\mathbb{Z}$-free, we have a corresponding Künneth formula $\bigoplus_{p=0}^nH_p(G,M)\otimes H_{n-p}(G',M')\rightarrow H_n(G\times G',M\otimes M')$

$\rightarrow\bigoplus_{p=0}^{n-1}Tor_1^\mathbb{Z}(H_p(G,M),H_{n-p-1}(G',M'))$ by Proposition I.0.8[Brown]. Note that in order to apply the proposition we needed one of the chain complexes (say, $F\otimes_G M$) to be dimension-wise $\mathbb{Z}$-free (and so with a free resolution $F$ this means we needed $M$ to be $\mathbb{Z}$-free).

Actually, the general Künneth theorem has a more relaxed condition and it suffices to choose $M$ (or $M'$) as a $\mathbb{Z}$-torsion-free module}.

Cohomology Künneth Formula (no proofs, just statements/notes)

Let $M$ (resp. $M'$) be an arbitrary $G$-module (resp. $G'$-module), let $F$ (resp. $F'$) be a projective resolution of $\mathbb{Z}$ over $\mathbb{Z}G$ (resp. $\mathbb{Z}G'$), and consider the cochain cross-product $Hom_G(F,M)\otimes Hom_{G'}(F',M')\rightarrow Hom_{G\times G'}(F\otimes F',M\otimes M')$ which maps the cochains $u$ and $u'$ to $u\times u'$ given by $\langle u\times u',x\otimes x'\rangle=(-1)^{deg(u')\cdot deg(x)}\langle u,x\rangle\otimes\langle u',x'\rangle$. This map is an isomorphism under the hypothesis that either $H_i(G,M)$ or $H_i(G',M')$ is of finite type, that is, the $i$th-homology group is finitely generated for all $i$ (alternatively, we could simply require the projective resolution $F$ or $F'$ to be finitely generated).

For example, if $M=\mathbb{Z}$ then $Hom_G(-,\mathbb{Z})$ commutes with finite direct sums, so we need only consider the case $F=\mathbb{Z}G$. An inverse to the above map is given by $t\mapsto \varepsilon\otimes \phi$, where $\varepsilon$ is the augmentation map and $\phi:F'\rightarrow M'$ is given by $\phi(f')=t(1\otimes f')$. Note that this does not hold for infinitely generated $P=\bigoplus^\infty \mathbb{Z}G$ because $Hom_G(P,\mathbb{Z})\cong\prod^\infty \mathbb{Z}$ is not $\mathbb{Z}$-projective (i.e. free abelian).

Assuming now that either $M$ or $M'$ is $\mathbb{Z}$-free, we have a corresponding Künneth formula $\bigoplus_{p=0}^nH^p(G,M)\otimes H^{n-p}(G',M')\rightarrow H^n(G\times G',M\otimes M')\rightarrow$

$\bigoplus_{p=0}^{n+1}Tor_1^\mathbb{Z}(H^p(G,M),H^{n-p+1}(G',M'))$ by Proposition I.0.8[Brown].

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    $\begingroup$ @ Chris: Thank you very much. I want to make sure that when describing Ku¨nneth formula, do you really mean the exact sequence $0\rightarrow \bigoplus_{p=0}^nH^p(G,M)\otimes H^{n-p}(G',M')\rightarrow H^n(G\times G',M\otimes M')\rightarrow$ $\bigoplus_{p=0}^{n+1}Tor_1^\mathbb{Z}(H^p(G,M),H^{n-p+1}(G',M'))\rightarrow 0$ $\endgroup$ Sep 15, 2011 at 11:18
  • $\begingroup$ @ Chris - Does the above exact sequence for cohomology always split? $\endgroup$ Sep 17, 2011 at 14:41
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    $\begingroup$ Yup, they're split and obey naturality. $\endgroup$ Sep 18, 2011 at 23:15
  • $\begingroup$ Dear Chris: The following paper intlpress.com/HHA/v8/n2/a5/v8n2a5.pdf discussed Ku¨nneth theorem "with local coefficients" (see Theorem 1.7). I wonder if it is equivalent to your result above. $\endgroup$ Dec 23, 2011 at 4:20
  • $\begingroup$ Dear Prof. Wen, that paper has special requirements on the coefficient modules, so if your groups (classifying spaces) and modules fit them, then it's fine. However, the Universal Coefficients Theorem which follows from the Kunneth formula, only holds for trivial group actions, especially as in the case of this paper.. $\endgroup$ Jan 3, 2012 at 1:14

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