# Is there a “free abelian group of rank 1” in the category of affine group schemes?

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.

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 $0$, $k\big[{x\choose i}\big]=k[x]$.

• Thanks for your concrete and crystal clear answer! It is such a huge group! – YZhou Jul 16 '15 at 6:24

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.

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.