Interpretation of the algebra of natural endomorphisms of the fiber functor of $\operatorname{Rep}(G)$

Question

Let $G$ be a connected algebraic group over an algebraically closed field $k$ of characteristic zero (I'm mostly interested in the case of a reductive group).

By the Tannakian formalism, $G(k)$ can be identified with the group of tensor automorphisms of the fiber functor $F:\operatorname{Rep}(G)\to \mathrm{Vect}$, i.e. natural automorphisms $(\phi_X)_{X\in \operatorname{Rep}(G)}$ such that $\phi_{X\otimes Y} = \phi_X \otimes \phi_Y$. Similarly, we can identify the Lie algebra $\mathfrak{g}$ with the set of natural endomorphisms $(\phi_X)_{X\in \operatorname{Rep}(G)}$ of $F$ satisfying $\phi_{X\otimes Y} = \phi_X \otimes 1 + 1 \otimes \phi_Y$.

Is there a similar interpretation of the full algebra $\operatorname{End}(F)$ of natural endomorphisms of $F$, without restrictions regarding tensor products?

The algebra $\operatorname{End}(F)$ seems closely related to the universal enveloping algebra $U\mathfrak{g}$. Indeed, there is an obvious homomorphism $U\mathfrak{g}\to \operatorname{End}(F)$ which I believe is always injective (though I don't have a proof of that, I'd be interested to see one) and has dense image by Jacobson's density theorem. However, I'm not sure how to define an intrinsic topology on $U\mathfrak{g}$ which would yield $\operatorname{End}(F)$ as a some kind of completion of $U\mathfrak{g}$.

Example: If $G=\mathbb{G}_m$, then $U\mathfrak{g}=k[t]$, $\operatorname{End}(F)=k^\mathbb{Z}$ and the map $U\mathfrak{g}\to \operatorname{End}(F)$ sends a polynomial $f(t)$ to $(f(k))_{k\in \mathbb{Z}}$.

Answer 1

This is an elaboration on what is in the other answers. First, general categorical arguments can be used to prove the following. Let $K$ be a field and $A$ a $K$-algebra. The profinite completion $\widehat{A}$ of $A$ is the inverse limit

$$\widehat{A} \cong \lim_I A/I$$

over all f.d. quotients of $A$. This is a profinite vector space, equipped with a suitable topology, although I prefer to think of these things categorically as pro-objects. The relevance of this construction to our situation is that any f.d. $A$-module factors through some f.d. quotient of $A$ so the f.d modules at best see the profinite completion, rather than $A$ itself.

Theorem: $\widehat{A}$ is the endomorphism algebra of the forgetful functor $\text{Mod}_f(A) \to \text{Vect}_f$ from f.d. $A$-modules to f.d. vector spaces.

The sketch of the proof is that, if we drop finiteness conditions, the endomorphism algebra is $A$ exactly, because the forgetful functor is representable by $A$ itself so we can use the Yoneda lemma to conclude. With the finiteness conditions, the forgetful functor is no longer representable, but it is "pro-representable" in a suitable sense by the inverse system of all finite quotients $A/I$. See e.g. my old blog post Operations, pro-objects, and Grothendieck's Galois theory for more details and other examples of this idea.

Corollary: If $\mathfrak{g}$ is a Lie algebra, then the endomorphism algebra of the forgetful functor $\text{Mod}_f(\mathfrak{g}) \to \text{Vect}_f$ is the profinite completion $\widehat{U(\mathfrak{g})}$ of the universal enveloping algebra of $\mathfrak{g}$.

Almost the only non-semisimple example where the answer can be written down in any reasonable way, as far as I can tell, is $\mathfrak{g} = K$. Then $U(\mathfrak{g}) \cong K[x]$ is the polynomial ring in one variable, and assuming that $K$ is algebraically closed for simplicity, its profinite completion is the product

$$\widehat{K[x]} \cong \prod_{a \in K} K[[x - a]]$$

of formal power series rings at each point $a \in K$, by the Chinese remainder theorem; this is exactly analogous to the perhaps more familiar decomposition of the profinite integers $\widehat{\mathbb{Z}} \cong \prod_p \mathbb{Z}_p$ (which in this language is the endomorphism ring of the "forgetful functor" from finite abelian groups to abelian groups).

This does not yet directly address your interest in reductive groups unless they're simply connected (and maybe we need other hypotheses too, I'm less familiar with these) so here are some more results that do.

Proposition: If $C$ is an essentially small $K$-linear abelian category equipped with an exact faithful functor $F : C \to \text{Vect}_f$ and $C$ is semisimple in the sense that every object is a finite direct sum of simple objects $V_i$, then $\text{End}(F) \cong \prod_i \text{End}(F(V_i))$.

So, if $C$ is either the category of f.d. representations of a reductive group $G$ or a semisimple Lie algebra $\mathfrak{g}$ (and let's assume $K$ has characteristic $0$ and is algebraically closed) and the simple representations are $V_i$, then

$$\text{End}(F) \cong \prod_i \text{End}_K(V_i) \cong \prod_i M_{\dim V_i}(K)$$

is a product of matrix algebras, as Will said. This specializes to your calculation for $\mathbb{G}_m$ and technically answers your question for reductive groups, although perhaps in a way that makes the answer look boring. However, $\text{End}(F)$ is not just a profinite algebra; if we remember the tensor product of representations $\text{End}(F)$ acquires the structure of a profinite Hopf algebra. So these things have an interesting comultiplication on them.

Finally here is a more general result, although unfortunately I'm not actually sure how to prove it. I think this is a lemma used in an approach to Tannaka-Krein reconstruction via the (co)monadicity theorem. As background, we need to know that profinite vector spaces have continuous duals which are just ordinary vector spaces, and this is a contravariant equivalence of categories, and even a symmetric monoidal equivalence. So it sends profinite algebras to coalgebras, and sends profinite Hopf algebras to Hopf algebras (which are commutative iff the original algebra is cocommutative and vice versa), and sends (continuous) modules to comodules.

Theorem (?): Let $C$ be an essentially small $K$-linear abelian category equipped with an exact faithful functor $F : C \to \text{Vect}_f$ to f.d. vector spaces. Then the dual of the profinite endomorphism algebra $\text{End}(F)$ is a coalgebra $A$, and $F$ exhibits $C$ as the category of f.d. comodules over $A$.

Corollary: If $A$ is a coalgebra and $C = \text{Comod}(A)$, then $\text{End}(F) \cong A^{\ast}$ is the profinite algebra dual to $A$. In particular, if $A = \mathcal{O}_G$ is the Hopf algebra of functions on an affine group scheme $G$ over $K$, so that $\text{Comod}(A) \cong \text{Rep}_f(G)$, then $\text{End}(F) \cong \mathcal{O}_G^{\ast}$.

(Edit: I was being silly here, there's no need to view this as a corollary of that hard theorem. We just need the easier result that f.d. comodules over $A$ can be identified with f.d. (continuous) modules over the dual $A^{\ast}$ regarded as a profinite algebra, and this follows from the general categorical comments earlier. Then we need to know that endomorphisms of the forgetful functor from f.d. (continuous) $A^{\ast}$-modules to vector spaces is $A^{\ast}$ and the argument should be the same as for the profinite completion. But the hard theorem may be of independent interest.)

This is as claimed by Adrien. The dual $\mathcal{O}_G^{\ast}$ is some kind of space of "distributions" on $G$, which naturally suggests a relationship with $U(\mathfrak{g})$, as well as with the group algebra $K[G(K)]$. The tensor automorphisms are sitting inside this dual as the $K$-algebra homomorphisms $\text{Hom}(\mathcal{O}_G, K) \cong G(K)$, while the Lie algebra is sitting inside this dual as the $K$-derivations $\text{Der}(\mathcal{O}_G, K) \cong \mathfrak{g}$ at the identity. Both of these can be recovered abstractly from the profinite Hopf algebra structure dual to the Hopf algebra structure on $\mathcal{O}_G$; these are the grouplike and primitive elements respectively.

Let's see how our results connect together. If $G$ is reductive then we have the Peter-Weyl decomposition $\mathcal{O}_G \cong \oplus_i V_i \otimes V_i^{\ast}$; we can think of the terms of this decomposition as $\text{End}_K(V_i)^{\ast}$, the space of matrix coefficients of the irreducible $V_i$. Then dualizing this gives

$$\text{End}(F) \cong \mathcal{O}_G^{\ast} \cong \prod_i \text{End}_K(V_i)$$

as expected; running this argument in reverse, putting all the claims we've made together actually implies the Peter-Weyl decomposition, by dualizing the calculation of $\text{End}(F)$ in the semisimple case (and using the general fact that because taking duals is a contravariant equivalence between profinite vector spaces and vector spaces, it sends products to coproducts).

We also have a functor $\text{Rep}_f(G) \to \text{Rep}_f(\mathfrak{g})$ inducing a homomorphism $\widehat{U(\mathfrak{g})} \to \mathcal{O}_G^{\ast}$; in the semisimple case where these guys are just products of matrix algebras corresponding to irreducibles this has the effect of killing the irreducibles of $\mathfrak{g}$ that don't integrate to $G$. In general it does something stranger, e.g. for $\mathbb{G}_m$ we get the map from $\prod_{a \in K} K[[x - a]]$ to $\prod_{\mathbb{Z}} K$ given by evaluation at $a \in \mathbb{Z}$.

Answer 2

For a reductive group the category of representations is semisimple so the algebra of endomorphisms of the fiber functor is just a product of matrix algebras, one for each irreducible representation.

Note that the category of representations over $\mathbb C$ without the tensor structure is a boring invariant of a reductive group - the structure of the category is entirely determined by the cardinality of the set of irreducible representations.

For $G$ semisimple simply-connected, I think one can check that the algebra of endomorphisms of the fiber functor is the completion of $U\mathfrak g$ under the topology generated by two-sided ideals of finite codimension, since quotients by these ideals give finite-dimensional representations of $\mathfrak g$, which extend to finite-dimensional representations of $G$.

Answer 3

In general I believe this should be the full dual algebra of $\mathcal{O}(G)$, or equivalently the profinite completion of the group algebra $k[G]$. If $G$ is simply connected, not necessarily reductive, this is the same as the profinite completion of $U(\mathfrak g)$ as in Will's answer.

Answer 4

I wanted to add this as a comment, but I don't have enough reputation :(

Here you have an algebra $A$ and a locally finite category $\mathscr{C}$ with a fully faithful embedding $\mathscr{C}\to A\text{-mod}$. Under this embedding, $\mathscr{C}$ is stable under taking subobjects, quotients, sums, and tensor products. (Our algebra $A$ here is $U(\mathfrak{g})$ and our category $\mathscr{C}$ is $\operatorname{Rep}(G)$ for connected $G$. The important point is that $\mathscr{C}$ embeds fully faithfully into $A\text{-mod}$.)

From $\mathscr{C}$ we obtain a filtered (or maybe cofiltered?) collection of ideals $I_V$ in $A$, where $I_V=\operatorname{Ann}_A(V)$ for $V$ in $\mathscr{C}$. Note that for any two $V$ and $W$ in $\mathscr{C}$ we have $V$ and $W$ embedding in $V\oplus W$ so that the quotients $A/I_V$ and $A/I_W$ are simultaneous quotients of $A/I_{V\oplus W}$. We therefore get a nice system and can take the limit \begin{equation} \hat{A}_{\mathscr{C}}=\varprojlim_{V\in \operatorname{obj}\mathscr{C}}A/I_V. \end{equation} I think it's fairly easy to argue that the given embedding $\mathscr{C}\to A\text{-mod}$ factors through an equivalence from $\mathscr{C}$ to discrete (topological) $\hat{A}_{\mathscr{C}}$-modules.

Suppose our base field $k$ is algebraically closed. If we consider your $\mathbb{G}_m$ example, each finite-dimensional representation is a sum of characters $V=\oplus_i k(x_i)^{\oplus m_i}$ where $\{x_i:i=1,\dots,n\}$ is a finite collection of distinct points on $\mathbb{Z}$, and we have \begin{equation} A/I_V =\prod_{i=1}^n k(x_i). \end{equation} Hence $\hat{A}_{\mathbb{G}_m}=\hat{A}_{\operatorname{Rep}(\mathbb{G}_m)}$ is the countable product \begin{equation} \hat{A}_{\mathbb{G}_m}=\prod_{x\in \mathbb{Z}}k(x) \end{equation} topologised so that closed ideals are the kernels of the quotients to products over finite collections of integers.

Edit: Here $\mathbb{Z}$ appears as $\operatorname{Rep}(\mathbb{G}_m)$ is identified with the tensor subcategory of semisimple $U(\mathfrak{g})=k[t]$-modules which are supported on the integers.

I don't think it's hard to see that $\hat{A}_{\mathscr{C}}$ is a topological Hopf algebra and one can also show that $\hat{A}_{G}$ is dual to the coalgebra of global functions $\mathscr{O}(G)$ for connected (maybe also smooth) $G$. One also sees that the competion is identified with $\operatorname{End}(F)$ via the natural map which you've already observed.