Global homological dimension of reductive groups Let $k$ be an algebraically closed field. Consider a smooth group scheme  $G$ over $k$. It is well known that the category $\textbf{Rep}_{G}$ is semisimple if and only if one of the following assertions holds.


*

*$G$ is reductive and $\mathrm{char}(k)=0$.

*$G$ is an algebraic torus and $\mathrm{char}(k) = p >0$.


Suppose now that $\textbf{Rep}_{G}$ has finite global homological dimension. 

What can be said about $G$?
What can be said if in addition we assume that $\mathrm{char}(k)=p>0$ and $G$ is reductive?

 A: In positive characteristic the only connected groups of finite homological dimension are the tori.
We need the following result from Jantzen, Representations of algebraic groups. [J, I 5.13],  [J, I 4.6 b].
Let $H$ be a flat subgroup scheme of $G$, such that $G/H$ is an affine scheme. Then $\mathrm {ind}_H^G$ is exact and 
 $H^*(G,\mathrm {ind}_H^G M)\cong H^*(H,M)$ for every $H$-module $M$.
To see that $G$ has infinite homological dimension 
it thus suffices to find an $H$ module $M$ for which $H^*(H,M)$ is big, in the sense that it lives in infinitely many degrees.
We use this principle repeatedly to reduce the problem to smaller groups or group schemes. 
First consider the case that the unipotent radical $G_u$ is nontrivial. In our first reduction we take $H=G_u$.
By chapter 14 of Springer, Linear Algebraic groups, any connected unipotent group contains a normal subgroup 
isomorphic to the additive group $\mathbb G_a$.
So we may next take $H=\mathbb G_a$.
Now for this $H$ the cohomology algebra $H^*(H,k)$ has been computed in [J, I 4.27]. It is big. This settles the nonreductive case.
Now the case that $G$ is reductive, but not a torus.
By [J, I. 9] we may take for $H$ a Frobenius kernel $G_r=\ker F^r:G\to G$.
This is an infinitesimal group scheme of height $r$ and Corollary 5.5 in
Andrei Suslin, Eric M. Friedlander and Christopher P. Bendel,  Infinitesimal 1-Parameter Subgroups and Cohomology
https://www.jstor.org/stable/2152897
shows that $H^*(H,k)$ is big.  Done.
