# Global homological dimension of group rings

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 $$k$$?

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.
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).