TITLE: Is there a "free abelian group of rank 1" in the category of affine group schemes?
QUESTION [15 upvotes]: Let's fix an algebraically closed field $k$.
The group $\mathbb Z$, as a discrete group scheme, is not affine since it's not quasi-compact. Is there an affine algebraic scheme over $k$ whose representation category is isomorphic to that of $\mathbb Z$?
Here is the motivation for this question. Fix a point $x$ on the circle. The $k$-linear tensor category of local systems of finite dimensional $k$-vector spaces, equipped with the functor $\mathcal F \mapsto \mathcal F_x$, is a Tannakian category. By the Tannakian duality, the category is isomorphic to the representation category of an affine group scheme, possibly not of finite type. This category is isomorphic to the representation category of an affine group scheme $G$. I want to understand what $G$ looks like.

REPLY [4 votes]: Claim : if $k$ is perfect of characteristic $0$, then :
$$\mathbb Z^{k, alg}\simeq  \mathbb G_a\times D_k(k^*)$$
Here $\mathbb Z^{k, alg}$ stands for the pro-algebraic completion of $\mathbb Z$ (the group you are interested in), $\mathbb G_a$ is the additive group, and $D_k(M)$ denotes as usual the diagonalisable $k$-group associated to the abstract abelian group $M$.
Sketch of a proof.
We denote $G=\mathbb Z^{k, alg}$ to simplify. By definition, this is the Tannaka group of the Tannaka category ${\rm Rep}_k \mathbb Z$, that is canonically isomorphic to the category ${\rm Iso_k}$ whose objects are couples $(V,\phi)$, where $V$ is a finite dimensional $k$-vector space, and $\phi\in {\rm GL}(V)$. 
Since $\mathbb Z$ is abelian, so is  $G=\mathbb Z^{k, alg}$, hence since $k$ is perfect, it splits canonically into a product diagonalizable x unipotent :
$G=G^m\times G^u$.  
Moreover $G^m=D_k(\Gamma)$, where
$\Gamma={\rm Hom}_{gr}(G,\mathbb G_m)={\rm Pic}({\rm Rep}_k G)\simeq {\rm Pic}({\rm Iso}_k)\simeq k^*$.
On the other hand $G^u$ is the tannaka group of the category $({\rm Rep}_k G)^u\simeq({\rm Iso}_k)^u={\rm Uni}_k$, where ${\rm Uni}_k$ is the category whose objects are couples $(V,\phi)$, where $V$ is a finite dimensional $k$-vector space, and $\phi$ is a unipotent endomorphism.
Now since $k$ is of characteristic $0$, there is an equivalence
${\rm Rep}_k(\mathbb G_a)\simeq {\rm Uni}_k $ sending  $\rho :\mathbb G_a\rightarrow {\rm GL}(V)$ to $(V,\rho(1))$, compatible with fiber functors, so one deduces  $G^u\simeq \mathbb G_a$.
Complement, probably related to your original interest.
All this is implicitely contained in 
Saavedra Rivano, Neantro
Catégories Tannakiennes. 
Lecture Notes in Mathematics, Vol. 265. Springer-Verlag, Berlin-New York, 1972. 
http://link.springer.com/book/10.1007%2FBFb0059108
last chapter 
Exemples tirés de la géométrie algébrique
p 319 ex 1.2.6 a).
that states that the category of regular connections on $\mathbb P^1\backslash \{0,\infty\} $ admits $\mathbb G_a\times D_k(k/\mathbb Z)$ as Tannaka group, hence, if $k$ is algebraically closed of characteristic $0$, is non-canonically isomorphic to $\mathbb Z^{k, alg}$. This result also shows that the diagonalisable part is not compatible with base change. This explains why one is generally interested only by the unipotent fundamental group.