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.



Is there a "free abelian group of rank 1" in the category of affine group schemes?
The affine group scheme $G$ you describe is not finite type. It is possible to describe $G$ explicitly.


For $R$ a $k$-algebra, the $R$-points of $G$ are the tensor functorial automorphisms of $\mathrm{Rep}\_k(\mathbb{Z})\to R\mathrm{-mod}$, $V\mapsto V\otimes R$. Let $\varphi$ be such an automorphism.


For $\lambda\in k^\times$, let $V\_\lambda$ be the $1$-dimensiona $\mathbb{Z}$-representation where the generator acts by multiplication by $\lambda$. Then $\varphi\_{V\_\lambda}$ is multiplication by some constant $c\_\lambda\in R^\times$, and the isomorphism $V\_{\lambda}\otimes V\_{\mu}\cong V\_{\lambda\mu}$ implies $\lambda\mapsto c\_\lambda$ must be a group homomorphism $k^\times\to R^\times$.


For $n$ a positive integer, let $W\_n$ be the $n$-dimensional $\mathbb{Z}$-representation where the generator acts by an $n\times n$ Jordan block with 1's on the diagonal. You can check that $\varphi\_{W\_n}$ acts by a matrix of the form
$$
\left[\begin{array}{ccccc}1&b\_1&b\_2&\ldots&b\_{n-1}\\
0&1&b\_1&\ldots&b\_{n-2}\\
\vdots&&\ddots&&\\
0&0&0&\ldots&1
\end{array}\right].
$$
The elements $b\_i\in R$ do not depend on $n$, and you can check that they satisfy the relations satisfied by ${x\choose i}$ for $x$ an indeterminate: for each $i$, $j$, there are integers $c\_{i,j,k}$ such that ${x\choose i}{x\choose j}=\sum\_k c\_{i,j,k}{x\choose k}$ for all $x$, and we will have $$\tag{$\star$}b\_ib\_j=\sum\_k c\_{i,j,k}b\_k.$$


Conversely, given a group homomorphism $k^\times\to R^\times$, $\lambda\mapsto c\_\lambda$ and elements $b\_i\in R$ satisfying the appropriate relations, we can define an automorphism of $V\otimes R$ for each $\mathbb{Z}$-rep $V$ in the following way. We can think of a $\mathbb{Z}$-rep as a square matrix $M$ (giving the action of the generator with respect to some choice of basis). As a function of $n$, the entries of $M^n$ can all be written as $k$-linear combinations of terms of the form $\lambda^n {n\choose j}$, for $\lambda\in k^\times$ and $j\in\mathbb{Z}\_{\geq 0}$. We replace $\lambda^n$ by $c\_\lambda$ and ${n\choose j}$ by $b\_j$ to get a square matrix over $R$, and this is our $R$-automorphism of the forgetful functor.


We have shown that
$$
G(R)=\{(\varphi,(b\_i)):\varphi:k^\times\to R^\times,\,\,b\_i\in R\text{ satisfying ($\star$)}\}.
$$
There is a group homomorphism $\mathbb{Z}\to G(k)$, sending $n\in \mathbb{Z}$ to $(\varphi,(b\_i))$ with $\varphi(\lambda)=\lambda^n$ and $b\_i={n\choose i}$. The coordinate ring of $G$ is
$$
k[k^\times]\otimes k\big[{x\choose i}\big],
$$
where $k[k^\times]$ is the group algebra of $k^\times$, and $k\big[{x\choose i}\big]$ is the algebra generated by formal symbols ${x\choose i}$, subject to the relations $(\star)$. In characteristic $0$, $k\big[{x\choose i}\big]=k[x]$.


