Extensions of $G$-modules parametrized by $H^1$

Question

Let $G$ be a finitely generated group and let $V$, $W$ be one-dimensional representations of $G$ over $\mathbb{F}_q$. (I guess one can think of $V$ and $W$ simply as $G$-modules, which are isomorphic to $\mathbb{F}_q$, as $\mathbb{F}_q$ vector spaces)

Up to scalars, equivalences of extensions \begin{equation} 0\to W \to E \to V\to 0 \end{equation} are parametrized by $\operatorname{Ext}^1_G(V,W):=H^1(\operatorname{Hom}(F',W)^{G})$ where $F'\to V$ is a projective resolution of $V$ over $\mathbb{Z}G$ and $(-)^G$ is the invariants functor.

Is it true that $\operatorname{Ext}^1_G(V,W) = H^1(G,V^{\vee}\otimes_{\mathbb{Z}} W)$, the group cohomology of $G$, with coefficients in $V^{\vee}\otimes_{\mathbb{Z}} W$?

My reasons to believe it is true are This spectral sequence, and this attempt to prove it here on math.stackexchange.com, where I could not have an answer. The problem in my attempt here was that taking a projective resolution of $\mathbb{Z}$ over $\mathbb{Z}G$, and tensoring it wit $\mathbb{F}_q$ is not necessarily a projective resolution of $V$.

Thank you for any help or references.

Answer 1

The correct version of your statement is the following. Let $K$ be a commutative ring and suppose that $V,W$ are (left) $KG$-modules with $V$ finitely generated projective over $K$. Then $$\mathrm{Ext}_{KG}^n(V,W)\cong H^n(G,\hom(V,K)\otimes_K W).$$

The proof is like this. If $A,B$ are $KG$-modules, then $\hom_K(A,B)$ is a $KG$-module with $G$ acting by conjugation. Moreover, $$\hom_{KG}(K,\hom_K(A,B))\cong \hom_K(A,B)^G\cong \hom_{KG}(A,B).$$ Take an injective resolution $W\to I_\bullet$ over $KG$. Then $\hom_{KG}(V,I_\bullet)\cong \hom_K(K,\hom_K(V, I_\bullet))$. Notice that since $\hom_K(V,-)$ is exact because $V$ is $K$-projective, we have that $\hom_K(V,W)\to \hom_K(V,I_\bullet)$ is a resolution. It is straightforward to check that if $I$ is an injective $KG$-module, then there is a natural isomorphism $\hom_{KG}(-,\hom_K(V,I))\cong \hom_{KG}(-\otimes_K V,I)$, and hence this functor is exact, as $V$ is $K$-projective and $I$ is injective. Thus $\hom_K(V,I)$ is injective. It follows that $\hom_K(V,W)\to \hom_K(V,I_\bullet)$ is an injective resolution. Taking cohomology we see that $$\mathrm{Ext}^n_{KG}(V,W)\cong \mathrm{Ext}^n_{KG}(K,\hom_K(V,W))\cong \mathrm{Ext}^n_{KG}(K,\hom(V,K)\otimes_K W)$$ since $V$ is a finitely generated projective $KG$-module.

It remains to observe that if $M$ is a $KG$-module, then $\mathrm{Ext}^n_{KG}(K,M)\cong H^n(G,M)$. For this, take a free resolution $F_\bullet\to \mathbb Z$ over $\mathbb ZG$. Notice that $H_n(K\otimes_{\mathbb Z} F_\bullet)\cong \mathrm{Tor}^{\mathbb Z}_n(K,\mathbb Z)$ since each free $\mathbb ZG$-module is a free $\mathbb Z$-module. We conclude that $H_n(K\otimes_{\mathbb Z}F_\bullet)=0$ for $n\geq 1$ and is $K$ when $n=0$. Thus $K\otimes_{\mathbb Z}F_\bullet$ is a free resolution of $K$ over $KG$. Therefore, $$H^n(G,M)\cong H^n(\hom_{\mathbb ZG}(F_\bullet,M))\cong H^n(\hom_{KG}(K\otimes_{\mathbb Z}F_\bullet,M))\cong \mathrm{Ext}^n_{KG}(K,M).$$

One can generalize this argument to show that if $V$ is a finitely generated projective $K$-module and $U,W$ are $KG$-modules, then $\mathrm{Ext}^n_{KG}(U\otimes V,W)\cong \mathrm{Ext}^n(U,\hom_K(V,K)\otimes W)$, where $U=K$ in the previous case.

Of course, if $K$ is a field then any finite dimensional $KG$-module is $K$-projective.

Answer 2

No. If $G$ is the trivial group and $q =p$ is a prime, $\operatorname{Ext}^1_G(V, W) = \operatorname{Ext}^1_{\mathbb{Z}}(\mathbb{Z}/ p, \mathbb{Z} / p) \cong \mathbb{Z}/p$ but $H^1(G, V^\vee\otimes_\mathbb{Z} W) = 0$.

I suspect this is never true even for nontrivial groups. Consider the restriction of extensions to the trivial subgroup $\operatorname{Ext}^1_G(V, W) \to \operatorname{Ext}^1_{\mathbb{Z}}(\mathbb{Z}/ p, \mathbb{Z} / p) \cong \mathbb{Z}/p$. Should there be a canonical isomorphism $\operatorname{Ext}^1_G(V, W) \cong H^1(G, V^\vee\otimes_\mathbb{Z} W) = \operatorname{Ext}^1_G(\mathbb{Z}, V^\vee\otimes_\mathbb{Z} W)$, it feels strange that one group of extensions is always splitting over $\mathbb{Z}$ and the other is not.

Say $p = 2$, $G = \mathbb{Z}/2$, and $k = \mathbb{F}_2$, there is only one 1-dimensional representation we just call $k$, and hence $V^\vee\otimes_{\mathbb{Z}} W \cong k$. Back-of-the-napkin free resolutions got me $H^1(G, k) = \mathbb{Z} / 2$, while $\operatorname{Ext}^1_{G}(k, k) = \mathbb{Z}/2\oplus\mathbb{Z}/2$, so this is nontrivial counterexample.

There is no Grothendieck spectral sequence as tensoring with $V^\vee$ does not take acyclic to acylic, and for this reason it is unclear to me what we should expect the relationship between cohomology groups to be. But they are very much not isomorphic.