A technical question about derivations of sheaves on group schemes - MathOverflow

A technical question about derivations of sheaves on group schemes

2009-12-09T17:51:40Z

Let $G$ be a group scheme (for instance, over $k$ a field of characteristic 0). Let $e$ be its unit. I denote by $O_G$ the structural sheaf of $G$.

Let $D_e : O_{G,e} \to k$ a derivation.

I would like to get directly (ie, without any consideration about the cotangent bundle, or some canonical isomorphisms...) a derivation $D : O_G\to O_G$ that extends $D_e$, and which is compatible with the action of $G$. That is, I would like to get this derivation by the mean of the multiplication map : $m : G \times G \to G$, etc., etc.

I have guessed this question would not be difficult, and would only be a matter of technics, but I can't manage to do it.

Answer by Leonid Positselski for A technical question about derivations of sheaves on group schemes

2009-12-09T18:56:18Z

You interpret your derivation $D_e$ as a distribution on $G$ supported in $e$, and then your derivation $D$ is the convolution with $D_e$ with respect to $m$. I.e., take your local function on $G$, compose it with $m$ to obtain a local function on $G\times G$, and apply $D_e$ along one of the arguments.

Answer by Jonathan Wise for A technical question about derivations of sheaves on group schemes

2009-12-09T19:12:14Z

Your derivation at the origin is a map $\mathrm{Spec} k[\epsilon] / \epsilon^2 \rightarrow G$ whose restriction to $\epsilon = 0$ is the inclusion of the origin. This induces a map

$G \times_{\mathrm{Spec} k} \mathrm{Spec} k[\epsilon]/\epsilon^2 \rightarrow G \times G \rightarrow G$

where the first map is the product with the map described above and the second is the multiplication map. Dually, the composed map gives a map of sheaves of algebras

$\mathcal{O}_G \rightarrow \mathcal{O}_G[\epsilon] / \epsilon^2$

which is the same thing as a derivation from $\mathcal{O}_G$ to itself.