2
$\begingroup$

Let $G$ be an arbitrary group and $k$ a field. Denote by $Rep_{k}(G)$ the category of finite dimensional representations of $G$ over $k$. The usual tensor product and dual operations for representations equip $Rep_{k}(G)$ with the structure of a $k$-linear rigid abelian tensor category. The forgetful functor $Rep_{k}(G) \longrightarrow Vec_{k}$ as fiber functor. So we get a neutral Tannakian category.

My question is :

If we consider the $G$ as constant affine group scheme over $k$, does the Tannakian fundamental group isomorphic to $G$?

$\endgroup$
5
  • 2
    $\begingroup$ If $G$ is infinite, then the constant group scheme $G$ is not affine. $\endgroup$
    – S. Carnahan
    Commented Dec 25, 2012 at 6:15
  • 3
    $\begingroup$ If I'm not mistaken, the group of automorphisms of the fiber functor is the pro-algebraic hull of the original group. $\endgroup$
    – S. Carnahan
    Commented Dec 25, 2012 at 6:18
  • 1
    $\begingroup$ @ S. Carnahan : that is right. You may turn your comment into an answer ? $\endgroup$
    – Niels
    Commented Dec 25, 2012 at 8:57
  • $\begingroup$ @S. Carnahan: could you also define "pro-algebraic hull" (a definition doesn't seem to be immediately available by searching)? $\endgroup$ Commented Dec 25, 2012 at 10:12
  • 2
    $\begingroup$ The pro-algebraic hull is the inverse limit over all finite dimensional representations of G of the Zariski closure of the image of G in GL_n. It's maybe worth pointing out that pro-algebraic hulls can be real beasts. For example, if G=Z, the integers, and k=C, the complexes, then the C-points of pro-algebraic hull is isomorphic to the direct sum of the additive group C and the group of all endomorphisms of the multiplicative group C^*. $\endgroup$
    – ChrisLazda
    Commented Dec 25, 2012 at 23:46

1 Answer 1

3
$\begingroup$

Following Niels's suggestion, I'm turning my comments into an answer.

The first point I want to make is the observation that if $G$ is an infinite group, then the constant group scheme $G_k$ is not affine over $k$. As a scheme, it is an infinite disjoint union of copies of the spectrum of $k$, and the ring of global functions is an infinite product of copies of $k$. If you take the spectrum of that ring, the underlying topological space is not your original discrete space, first of all because it is quasi-compact. I think it has strictly larger cardinality, due to the existence of lots of exotic prime ideals. I imagine there is some relation with Stone-Cech compactification, but I am too lazy to look it up or try to figure it out.

In general the group of automorphisms of the fiber functor is an object called the pro-algebraic hull of $G$. It is an initial object among all homomorphisms from $G$ to pro-algebraic groups, and exists by the completeness of the category of affine group schemes. There is a more concrete description given in the comments by ChrisLazda, and as you can see from his/her example, it can depend strongly on the base field.

As far as I can tell, the notion of pro-algebraic hull was folklore (or perhaps an exercise for graduate students) for a while. Googling yields a description in this paper, together with an application.

$\endgroup$
2
  • $\begingroup$ The (maximal) spectrum is of an $I$-indexed product of copies of a field $k$ contains, and ought to just be equal to, the Stone-Cech compactification $\beta I$ of $I$ (sketch: look at what happens to idempotents); identifying $\beta I$ with the set of ultrafilters on $I$, the quotient by the maximal ideal corresponding to an ultrafilter is the corresponding ultrapower of $k$. $\endgroup$ Commented Dec 26, 2012 at 0:51
  • 1
    $\begingroup$ I'm not sure about the origin of the notion either, but certainly "pro-algebraic hulls" can be at least traced back to Hochschild & Mostow, Pro-affine algebraic groups, Amer. J. Math 1969. $\endgroup$ Commented Dec 26, 2012 at 13:52

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .