Extensions of algebraic groups and extensions of fpqc sheaves

Question

There are at least two ways to define $\operatorname{Ext}^p(F,G)$ when $F$ and $G$ are commutative algebraic groups over a field $k$:

1. Pass to the associated fppf sheaves and use an injective resolution of $G$.
2. Work in the category of commutative pro-algebraic groups over $k$, which is abelian and has enough projectives, and take a projective resolution of $F$.

In other words, one can study Yoneda extensions of fppf sheaves or Yoneda extensions of pro-algebraic groups. When $p=1$, these definitions agree because the underlying fppf sheaf of an extension of $G$ by $F$ is an algebraic space with a group structure, hence an algebraic group.

What is the relationship between the two definitions for $p > 1$?

Answer 1

The Yoneda Ext does not in general agree with the derived functor of Hom on fppf sheaves. In "On a nontrivial higher extension of representable abelian sheaves" by L. Breen, it is shown by hand that for a characteristic 2 field $k$, $Ext^2_{fppf}(\alpha_2,\mathbb{G}_m) \neq 0$, while if $k = \bar{k}$, the corresponding Yoneda Ext group vanishes.

You may be interested in Akhil Mathew's recent talk "Multiplicative Polynomial Laws and Commutative Group Schemes" https://www.youtube.com/watch?v=_YF9tnUya-Q. The talk suggests replacing $Ext^*_{fppf}$ by the extension groups in the category of multiplicative polynomial laws, which is smaller than $Ext^*_{fppf}$ but still admits a map from the Yoneda Ext groups. For example, it is stated in Akhil's talk that $Ext_{mult. pol.}^*(G,\mathbb{G}_m)$ is zero for finite $G$ when $\ast > 1$.