3
$\begingroup$

I am looking for a reference for the assertion in the title.

In more detail, let $\Gamma=\{1,\gamma\}$ be a group of order 2. Let $A$ be a $\Gamma$-module (an abelian group on which $\Gamma$ acts). Then I can compute the cohomology groups $H^k(\Gamma,A)$ using a certain nice periodic resolution.

Let $A_1\overset{\partial}{\longrightarrow} A_0$ be a morphism of $\Gamma$-modules, which I regard as a complex of $\Gamma$-modules of length 2 with $A_0$ in degree 0 and $A_1$ in degree $-1$. I can define the hypercohomology groups ${\Bbb H}^k(\Gamma, A_1\to A_0)$ of this complex in some reasonable way (I would be glad to have a reference for a standard definition!).

Now let $$(\varphi_1,\varphi_0)\colon\ (A_1\overset{\partial}{\longrightarrow} A_0)\,\longrightarrow\, (A_1'\overset{\partial'}{\longrightarrow} A_0')$$ be a quasi-isomorphism, that is, a morphism of complexes such that the induced morphisms $$\ker\partial\to\ker\partial'\qquad\text{and}\qquad {\rm coker\,}\partial\to{\rm coker\,}\partial'$$ are isomorphisms.

Proposition. If $(\varphi_1,\varphi_0)$ above is a quasi-isomorphism, then the induced homomorphism on hypercohomology $$\varphi_*\colon\ {\Bbb H}^k(\Gamma, A_1\to A_0)\,\to\, {\Bbb H}^k(\Gamma, A_1'\to A_0')$$ is an isomorphism.

I am looking for a reference (rather than a proof) for this proposition.

$\endgroup$
5
  • 3
    $\begingroup$ I don’t think this is stated there, but Brown’s book on group cohomology defines cohomology with coefficients in a chain complex and constructs a natural spectral sequence converging to it. Your quasi-isomorphism induces an isomorphism between these spectral sequences. There might be a more explicit reference, but that might do if you can’t find one. $\endgroup$ Sep 1, 2021 at 14:31
  • $\begingroup$ @AndyPutman: Many thanks! The necessary assertion is stated in Brown's book (for group homology rather than cohomology). This is Proposition VII.5.2: Weak equivalence of chain $G$-complexes induces isomorphism on homology. $\endgroup$ Sep 1, 2021 at 15:06
  • $\begingroup$ @AndyPutman: This book also contains sign rules for the differentials in double complexes obtained as ${\rm Hom}(C,C')$ and $C\otimes C'$, where $C$ and $C'$ are given complexes. Many thanks again! $\endgroup$ Sep 1, 2021 at 15:10
  • 1
    $\begingroup$ What's your definition of group hypercohomology? The definition I usually use makes this true by definition... $\endgroup$ Sep 1, 2021 at 16:38
  • $\begingroup$ It sounds like you're interested in crossed modules more generally. In that case, why not look at the classifying space (i.e. the equivalent data of the 2-type)? Then your condition of 'quasi-isomorphism' gives a weak equivalence of 2-types (indeed: the corresponding 2-type is the fiber of a map of 1-types, and your condition is exactly that you have a commuting square and that the induced map on the fiber is an equivalence on pi_1 and pi_2). Now any invariant of the 2-type is a 'quasi-isomorphism invariant'. For example, the cohomology of that classifying space. $\endgroup$ Sep 3, 2021 at 15:08

2 Answers 2

5
$\begingroup$

This is surely not the type of answer you want (I think Andy's comment is). But I think it's worth explaining that there's a general formalism that makes this type of result transparent. I'm sorry if this is clear to you already.

Let $$\Gamma_G : \mathrm{Mod}_{k[G]} \to \mathrm{Mod}_k$$ be the functor of taking $G$-invariants ($k$ is some ground ring). It is left exact, and $\mathrm{Mod}_{k[G]}$ has enough injectives, so it admits a derived functor $$ R\Gamma_G : D^+(\mathrm{Mod}_{k[G]}) \to D^+(\mathrm{Mod}_k).$$ The derived functor goes between derived categories.

Hypercohomology can be expressed in terms of $R\Gamma_G$: if $M^\bullet$ is a cochain complex of $G$-modules, then $H^q(R\Gamma_G(M^\bullet)) = \mathbb H^q(G,M^\bullet)$.

Now by definition quasi-isomorphisms of cochain complexes are isomorphisms in the derived category. So the fact that group hypercohomology takes quasi-isomorphisms to isomorphisms is equivalent, in the language of derived categories, to the statement that $R\Gamma_G$ takes isomorphisms to isomorphisms. That is, the claim that you want is implicit in the statement that hypercohomology is a functor between derived categories.

So in a sense, once you know that group cohomology is a derived functor, then a reference for your statement is any textbook that explains how to talk about derived functors in the language of derived categories, e.g. Gelfand and Manin - Methods of homological algebra, Yekutieli's book "Derived categories", Verdier's thesis, or the Stacks project.

$\endgroup$
5
  • $\begingroup$ Indeed, this is not the type of answer I want by two reasons: (1) My co-author and my potential readers do not know derived categories, and (2) I have in mind nonabelian generalizations to the case when my short complex becomes a crossed module, when there are no derived categories, but there are exact sequences. For a short complex, the spectral sequences that Andy mentions degenerate to exact sequences. $\endgroup$ Sep 1, 2021 at 17:23
  • $\begingroup$ No, this was not clear to me already. This was very helpful. Thank you! $\endgroup$ Sep 1, 2021 at 17:25
  • 1
    $\begingroup$ Just to make it clear, I also think this is the right way to think about this. When writing about group cohomology, there is often the question of what technical point of view to take, and I sometimes struggle to figure out whether to depend on canonical textbooks (like Brown) or more modern points of view. $\endgroup$ Sep 1, 2021 at 18:56
  • 1
    $\begingroup$ @MikhailBorovoi You will probably like this even less, but there are plenty of ways of dealing with your objection (2). Indeed group hypercohomology of crossed modules is just homotopy fixed points of 2-types, and we homotopy theorists have developed plenty of ways of dealing with that. $\endgroup$ Sep 3, 2021 at 17:16
  • 1
    $\begingroup$ @DenisNardin: I know that. Actually, it was I who asked Behrang Noohi to write this paper , where he considers group hypercohomology of crossed modules on different levels. $\endgroup$ Sep 3, 2021 at 18:21
0
$\begingroup$

I give an elementary proof of the fact that a quasi-isomorphism of short complexes (complexes of length 2) of $\Gamma$-modules induces an isomorphism on hypercohomology. Actually, it is very close to Theorem 3.3 on page 16 of my preprint of 1992, where I prove that a quasi-isomorphism of crossed modules with $\Gamma$-action preserves ${\Bbb H}^0$ and ${\Bbb H}^1$ (for any profinite group $\Gamma$).

First I give my elementary cocyclic definitions of cohomology and hypercohomology.

Definition of cohomology

Let $A$ be a $\Gamma$-module, that is, an abelian group written additively, endowed with an action of $\Gamma=\{1,\gamma\}$. We consider the first cohomology group $H^1(\Gamma,A)$. We write $H^1\! A$ for $H^1(\Gamma,A)$. Recall that $$H^1\! A=Z^1\! A/B^1\! A,$$ where $$Z^1\! A=\{a\in A\mid\,^\gamma\! a=-a\}, \quad B^1\! A=\{\,^\gamma\! a'-a'\mid a'\in A\}.$$

We define the second cohomology group $H^2\!A$ by $$ H^2\!A=Z^2\! A/ B^2\! A,$$ where $$Z^2\! A=A^\Gamma:= \{a\in A\mid\, ^\gamma\! a=a\},\quad\ B^2\! A=\{\,^\gamma\! a'+a'\mid a'\in A\}.$$

For $k\in{\mathbb Z}$ we define the coboundary operator $$d^k\colon A\to A,\quad a\mapsto\,^\gamma\! a-(-1)^k a.$$ In other words, $d^k=\gamma-(-1)^k\in {\mathbb Z}[\Gamma]$, where ${\mathbb Z}[\Gamma]={\mathbb Z}\oplus{\mathbb Z}\gamma$ is the group ring of $\Gamma$. We calculate: \begin{align*} d^k\circ d^{k-1}=\big(\gamma-(-1)^k\big)\big(\gamma-(-1)^{k-1}\big)&=\big(\gamma-(-1)^k\big)\big(\gamma+(-1)^k\big)\\ &=\gamma^2-(-1)^{2k}=1-1=0. \end{align*} We see that $d^{k+1}\circ d^k=0$. We define the Tate cohomology groups $H^k_T A$ for all $k\in {\mathbb Z}$ by $$ H^k_T\, A=Z^k\! A/B^k\! A,$$ where \begin{align*} Z^k\! A=\ker d^k=\{a\in A\mid\,^\gamma\! a=(-1)^k a\},\quad B^k\! A={\rm im\,} d^{k-1}=\{\,^\gamma\! a'+(-1)^k a'\mid a'\in A\}. \end{align*} Then clearly $$H^k_T\, A=H^1\! A\quad\text{if $k$ is odd, $\ $ and}\quad H^k_T\, A=H^2\!A\ \ \text{if $k$ is even.}$$

Definition of hypercohomology

Let $A_0\overset{\partial}{\longrightarrow} A_0$ be a short complex of $\Gamma$-modules. For $k\in{\mathbb Z}$ we define a differential $$D^k\colon\, A_1\oplus A_0\to A_1\oplus A_0,\quad\ D(a_1,a_0)=\big(d^{k+1} a_1,\, d^k a_0-(-1)^k a_1\big).$$ We calculate: \begin{align*} D^k\big( D^{k-1}(a_1,a_0)\big)&=D^k\big(\, d^k a_1,\ d^{k-1} a_0-(-1)^{k-1} a_1\,\big)\\ &=\big(d^{k+1}d^k a_1,\ d^k d^{k-1} a_0-(-1)^{k-1}d^k a_1- (-1)^k d^k a_1\big)=0. \end{align*} Thus $D^k\circ D^{k-1}=0$.

We define the $k$-th Tate hypercohomology group ${\mathbb H}^k(A_{1}{\overset\partial\longrightarrow } A_0)$ by $$ {\mathbb H}^k(A_{1}{\overset\partial\longrightarrow } A_0)=Z^k(A_{1}{\overset\partial\longrightarrow } A_0)\,/\,B^k(A_{1}{\overset\partial\longrightarrow } A_0),$$ where \begin{align*} Z^k(A_{1}&{\overset\partial\longrightarrow } A_0)=\ker D^k=\\ &=\{(a_{1},a_0)\in A_{1}\oplus A_0,\,\, \mid\,d^{k+1} a_1=0,\, d^k a_0-(-1)^k\partial a_{1}=0\}, \\ B^k(A_{1}&{\overset\partial\longrightarrow } A_0)={\rm im\,} D^{k-1}=\\ &=\{(d^k a'_1,\ d^{k-1} a'_0-(-1)^{k-1}\partial a'_1\,) \,\mid\, (a'_1, a_0')\in A_1\oplus A_0\}. \end{align*}

Quasi-isomorphism induces isomorphism on hypercohomology

Proposition. If $\varphi=(\varphi_1,\varphi_0)\colon\, (A_1\overset{\partial}{\longrightarrow} A_0)\,\longrightarrow\, (A_1'\overset{\partial'}{\longrightarrow} A_0')$ is a quasi-isomorphism, then the induced homomorphism on hypercohomology $$\varphi_*\colon\ {\Bbb H}^k(\Gamma, A_1\to A_0)\,\to\, {\Bbb H}^k(\Gamma, A_1'\to A_0')$$ is an isomorphism.

Sketch of an elementary proof. From $A_1\overset{\partial}{\longrightarrow} A_0$, using the short exact sequence of complexes $$0\to(\ker\partial\to 0)\to (A_{1}\overset\partial\longrightarrow A_0)\to ({\rm im}\, \partial\hookrightarrow A_0)\to 0,$$ we obtain a long exact sequence $$\dots H^{k-1}{\rm coker}\,\partial\to H^{k+1}\ker\partial\to {\mathbb H}^k(A_{1}\overset\partial\longrightarrow A_0) \to H^k{\rm coker}\,\partial\to H^{k+2}\ker\partial\dots, $$ and similarly, from $A_1'\overset{\partial'}{\longrightarrow} A_0'$ we obtain a long exact sequence $$\dots H^{k-1}{\rm coker}\,\partial'\!\to\! H^{k+1}\ker\partial'\!\to\! {\mathbb H}^k(A_{1}'\!\overset{\partial'}\longrightarrow\! A_0') \!\to\! H^k{\rm coker}\,\partial'\!\to H^{k+2}\ker\partial'\!\dots $$ Our morphism $\varphi=(\varphi_1,\varphi_0)$ induces a morphism of the exact sequences, that is, a commutative diagram with exact rows. Since our morphism $\varphi$ is a quasi-isomorphism, the induced homomorphisms $$ H^{k}\ker\partial\to H^{k}\ker\partial'\quad\text{and} \quad H^k{\rm coker}\,\partial\to H^k{\rm coker}\,\partial' $$ are isomorphisms for all $k$, and by the five lemma we conclude that the induced homomorphism on hypercohomology $$\varphi_*\colon\ {\Bbb H}^k(\Gamma, A_1\to A_0)\,\to\, {\Bbb H}^k(\Gamma, A_1'\to A_0')$$ is an isomorphism, as required.

$\endgroup$
2
  • $\begingroup$ This proof is a version for complexes of length 2 of the proof in the book of Brown. For a complex of lengths 2, the corresponding spectral sequence degenerates into an exact sequence. $\endgroup$ Sep 3, 2021 at 15:54
  • $\begingroup$ @LSpice: Thank you for noticing this. I would prefer to correct this myself $\endgroup$ Sep 3, 2021 at 15:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.