Global homological dimension of group rings

Question

In all that follows, let $k$ be a field and $G$ be a finite group.

It is well-known that the order of $G$ is invertible in $k$ iff the group ring $k[G]$ is semisimple, which is equivalent, inter alia, to the fact that $\operatorname{Ext}^1_{k[G]}(V,W)$ vanish for all $V,W$ left $k[G]$-modules (= $k$-linear representations of $G$), or indeed, $\operatorname{Ext}^i_{k[G]}(V,W)$ for all $i\geq 1$, viꝫ. $k[G]$ has (left) global dimension zero.

Today I learned that this is also equivalent to the (a priori weaker) condition that $k[G]$ be (left-)hereditary, which is equivalent, inter alia, to the fact that $\operatorname{Ext}^2_{k[G]}(V,W)$ vanish for all $V,W$ left $k[G]$-modules. (See, e.g., Dicks, “Hereditary Group Rings”, J. London Math. Soc. 20 (1979) 27–38, theorem 1.)

This suggests the following question: what can be said about $G$ and $k$ if $\operatorname{Ext}^{d+1}_{k[G]}(V,W)$ vanish for all $V,W$ left $k[G]$-modules for a given $d\geq 2$? In other words, if we assume $k[G]$ has (left) global dimension $\leq d$? Does $k[G]$ having finite global dimension imply that the order of $G$ is invertible in $k$?

Answer 1

Here is another proof that the global dimension is infinite that is specific to groups and explicitly identifies a module of infinite projective dimension. Let $G$ be a finite group and suppose that the characteristic $p$ of $k$ divides the order of $G$. Then I claim that the trivial $kG$-module has infinite projective dimension. That is, the group $G$ has infinite mod $p$ cohomological dimension. First of all this follows when $G$ is a cyclic group of order $p$ from the very well known resolution of the trivial module (which can be obtained topologically using infinite lens spaces). If $t$ is the generator, you have a resolution where each module is $kG$ and you alternate between multiplying by $t-1$ and $1+t+\cdots+t^{p-1}$ (except for the augmentation $kG\to k$ at the beginning). When you hom into the trivial module $k$ you end up with a resolution with all the vector spaces $k$ and where all the maps are zero since $p$ is the characteristic of the field $k$ and so while $t-1$ always becomes zero after mapping into a trivial module, $1+t+\cdots+t^{-1}$ becomes multiplication by $p$, which is $0$, after mapping into a trivial module. This shows that $$H^n(C,k)=\mathrm{Ext}^n_{kG}(k,k)\cong k$$ for all $n\geq 0$.

Next assume that $p\mid |G|$. Then $G$ has a cyclic subgroup $C$ of order $p$. Shapiro's lemma now implies that $$\mathrm{Ext}^n_{kG}(k,\mathrm{Coind}_C^G k)=H^n(G,\mathrm{Coind}_C^G k)\cong H^n(C,k)\neq 0$$ so again the trivial module has infinite projective dimension.

Answer 2

If $kG$ is not semisimple, it is a non-semisimple Frobenius algebra and has infinite global dimension always in that case, see for example the books of Lam on rings and modules where the global dimension of a Frobenius algebra is determined. So yes, finite global dimension is equivalent to being semisimple (global dimension zero). For a proof that nonsemisimple Frobenius algebras have infinite global dimension one can use that the syzygy functor $\Omega^1$ is a stable equivalence and thus $\Omega^i(M)$ is always non-zero for all $i>0$ if $\Omega^1(M)$ is not projective (this shows in fact the stronger statement that the finitistic dimension is zero, which implies that the global dimension is infinite when the algebra is not semisimple).