3
$\begingroup$

If $G$ is an affine group scheme over a field $k$, then we have the forgetful functor $\omega$ from the category ${\rm Rep}_k(G)$ of finite representations of $G$ to the category of finite dimensional $k$-vector spaces. Let $S$ be a $k$-scheme, with the structure map $p: S\to k$. Then ${\rm Aut}_S(\omega)$ (the tensor automorphisms the non-neutral fibre functor $p^*\omega$) is an $S$-group scheme. In fact it is the diagonal of the groupoid scheme ${\rm Aut}_k(\omega)$ (see Deligne's article "catégories tannakiennes" in The Grothendieck Festschrift Volume II, 1.11 and 1.12 for notations and statements).

My question is what is the relation between the $S$-grp sch ${\rm Aut}_S(\omega)$ and the $S$-grp sch $G\times_kS$? are they equal?

$\endgroup$

1 Answer 1

6
$\begingroup$

Yes they are.

In fact the formation of ${\rm Aut}_S(\omega)$ commutes by definition with base change, so you can reduce to $S=k$, and $S\to k={\rm id}$.

Then the fact that the $k$-group morphism $G\to {\rm Aut}_k(\omega)$ is an isomorphism is the reconstruction theorem in Tannaka theory, see Proposition 2.8 in the paragraph

"Recovering an affine group scheme from its representations"

in

Tannakian Categories

P. Deligne and J. S. Milne

in the volume

Hodge cycles, motives, and Shimura varieties.

Lecture Notes in Mathematics, 900.

https://doi.org/10.1007/978-3-540-38955-2_4

or

https://www.jmilne.org/math/xnotes/tc.html

ps : non-neutral fibre functor sounds very strange. A tannakian category is said neutral precisely when there exists a fibre functor ... Also, my advice would be to read Deligne-Milne's article before Deligne's article.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.