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 $\rho: G\to \mathrm{GL}(V\otimes\_k K)$.


Thanks!



Tannakian fundamental group of two explicit tensor categories
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 $G$.


