Let $K/k$ is a field extension and $G$ an affine group scheme over $K$. What are the Tannakian fundamental groups of these two $k$-tensor categories (with trivial fiber functors over $k$):

1. The category of pairs of finite vector spaces over $k$ with an isomorphism of their extensions to $K$,

2. The category of finite vector spaces $V$ over $k$ with a representation $\rho: G\to \mathrm{GL}(V\otimes_k K)$.


  • $\begingroup$ I think you'll have to specify the fibre functor in the 1st case. Projection on the first or second element of the pair? Not that it matters that much… $\endgroup$
    – jmc
    Jan 22, 2015 at 18:26
  • $\begingroup$ @jmc Yes, you can take either of vector spaces as the fiber functor. $\endgroup$
    – Mostafa
    Jan 22, 2015 at 18:30
  • $\begingroup$ I would guess that the first is $\mathbb{Z}$ and the second is the Weil restriction $Res_{K/k}(G)$, by extrapolating from the case $K=k$... $\endgroup$ Jan 23, 2015 at 6:00
  • $\begingroup$ @JesseSilliman — The first case is trivial if $K = k$ (if I'm not mistaken). Moreover, $\mathbb{Z}$ is not an affine group scheme, right? Finally, I think the first case should also depend on $K$. $\endgroup$
    – jmc
    Jan 23, 2015 at 6:20
  • 2
    $\begingroup$ @JesseSilliman — In the second case $k = \mathbb{R}$, $K = \mathbb{C}$, and $G = \mathbb{G}\mathrm{m}$ should give real vector spaces with a grading after complexification. Your version gives real Hodge structures. And those aren't equivalent. I agree that it should be something like the Weil restriction, though… $\endgroup$
    – jmc
    Jan 23, 2015 at 6:30

1 Answer 1


One possibly useful way of describing these groups is by their universal properties. The first group has the property that homomorphisms from it to any other pro-algebraic group $H$ are in bijection with $H(K) / H(k)$, and the second group has the property that homomomorphisms from it to $H$ are in bijection with homorphisms from $G$ to $H(K)$. So the second one is the reverse of the universal property of the Weil restriction.

We can check these properties easily using Tanaka-Krien duality. Homomorphisms from $G$ to $H$ up to conjugacy are the same as homomorphisms from the category of $H$-reps to the category of $G$-reps, and actual homomorphisms are those plus an isomorphism of fiber functors. Any functor from the category of $H$-reps to the category of pairs of $k$-vector spaces inside the same $K$-space must send an $H$-rep to a pair of vector spaces where the second differs from the first by an element of $H(K)$. This is well-defined up to automorphisms of the second space, which are $H(k)$. A similar argument works for the second one.

Assume $K$ is a Galois extension of $k$ of degree $d$. We can use this unversal property to compute the groups. Let's see what happens when we pull the groups back to $K$. Pullback is adjoint to Weil restriction. So the universal functor now sends $H$ to $\operatorname{Res}_{K/k} H (K) / \operatorname{Res}_{K/k} H (k) = H(K)^d / H(K) = H(K)^{d-1}$. So this functor is equivalent to the free group on $d$ generators. So the group is the free pro-algebraic group on $d-1$ generators, or the pro-algebraic envelope of the free group on $d-1$ generators.

How do we descend this from $K$ to $k$? We need a natural $\operatorname{Gal}(K/k)$ action. From our construction, it is more natural to identify our free group with the subgroup of the free group on $d$ generators that alternates element-inverse-element-inverse, which is free on the $d-1$ generators $x_2x_1^{-1}, x_3x_1^{-1}, \dots, x_d x_1^{-1}$. The $d$ generators correspond to the automorphisms of $K$, so $\operatorname{Gal}(K/k)$ acts on them naturally by left multiplication.

The second one is the same thing except with the pro-algebraic envelope of the free product of all the Galois conjugate copies of $G$.

  • $\begingroup$ I can't see why "Any finite-type group has only finitely many of these". Furthermore a group may be huge but has a good description. $\endgroup$
    – Mostafa
    Jan 25, 2015 at 7:48
  • 1
    $\begingroup$ It's because of Goursat's Lemma. because $PGL_n$ is a simple group, if you have $k$ maps to $PGL_n$ then the induced map to $\left(PGL_n\right)^k$ is surjective. My claim is more that the group is so large that it's not very helpful to describe it. $\endgroup$
    – Will Sawin
    Jan 25, 2015 at 14:32

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