9

According to the nLab, Tannaka Duality states that every rigid monoidal category with a fiber functor (whatever that means) is equivalent to the category of modules over some Hopf Algebra. I think I understand why this is the case.

The problem:

Now, fix a Lie Group $G$ and then consider the category of all smooth, real-valued, finite-dimensional linear representations of $G$. This is surely a rigid monoidal category with a fibre functor. If so, what is the corresponding Hopf Algebra?

Further comment:

In some circumstances the aforementioned problem seems to be much simpler.

For instance, if $G$ is just a finite (or even infinite, if considered as a discrete group and we do not care about continuity) group then the sought Hopf Algebra is just $\mathbb R G$, the group algebra of $G$.

Similarly, if $\frak{g}$ is a (real) Lie Algebra, then the sought Hopf Algebra is just $U(\frak g)$, the universal enveloping algebra.

A little more complicated, but still simpler than the smooth case, is the algebraic group case, I think. If $G$ is a (real, complex... does it matter, by any chance?) affine algebraic group, then I guess the Hopf Algebra should just be $\mathbb R [G]$, the algebra of regular functions on $G$.

What I have tried so far:

I have spent countless hours on the nLab, trying to understand this correspondence. I have also tried the literature, in particular Cartier's Primer on Hopf Algebras, which seems to be close to what I need. Is there any specific reference I should look into? It seems to me that everyone prefers spending more time with the other cases I've mentioned, than with the Lie group case.

My first thought was that the Hopf algebra corresponding to the Lie group case would simply be $\mathcal{C}^\infty(G)$, but then I have realised that this isn't a true Hopf algebra (long story short: the comultiplication would make you land in $\mathcal{C}^\infty(G\times G)$, but this is not the same as $\mathcal{C}^\infty(G) \otimes \mathcal{C}^\infty(G)$.

So then I imagined it would be $R(G)$, the algebra of (smooth?) representative functions, which is definitely a Hopf Algebra, but I am unsure if it is the right one. Here and there, I have heard that $R(G)$ makes you recover the category of representations of the envelope of $G$, so this doesn't really answer my question.

Cartier spends a lot of time developing $\mathcal{C}_c ^{-\infty}(G)$, the algebra of distributions with compact support, and even $\mathcal{C}_\mathrm{fin} ^{-\infty}(G)$, the algebra of distributions with finite support, though in the end he doesn't prove the result I expected.

Am I simply hoping for an elegant answer that it is nowhere to be found?

Thanks in advance.

Melanzio
  • 789
  • 2
    @Malanzio - In the "algebraic case" (with a "Tannakian neutral category" $T$ over a field $k$) it follows $T \cong Rep(G)$ with $G$ an affine group scheme over $k$. There is an exposition on this by Deligne/Milne - you may find references here that are interesting. – hm2020 Sep 22 '25 at 11:44
  • The group you are looking for is the pro-algebraic completion (an affine group scheme, generally not of finite type) of $G$. It might not be obvious how to compute it though. – Julian Quast Sep 28 '25 at 14:16
  • @JulianQuast So it's the Hopf Algebra that represents the affine group scheme? – Melanzio Sep 28 '25 at 14:22
  • @JulianQuast Your comment is interesting but it leaves me with even more questions than answers. First of all, I wasn't looking for a group: I was looking for an Algebra which is supposed to exist; and secondly, the pro-algebraic completion of G is universal with respect to all finite dimensional representations of G, true; but it doesn't specifically encode the smoothness of the representations of G, which is what I am interested in; right? – Melanzio Sep 28 '25 at 14:32
  • @Melanzio By "pro-algebraic completion" I mean the limit of all linear algebraic groups $H$ over $\mathbb R$, together with a smooth map $G \to H$. The resulting affine group scheme will be the spectrum of the Hopf algebra you are looking for. This essentially follows, as the category of smooth representations of $G$ is a union of categories generated by a single representation, under tensor products, duals and subquotients. – Julian Quast Sep 28 '25 at 14:45
  • Why would the category be generated by a single representation together with those operations? I think it should be generated by the irreducible representations, and those operations. Right? – Melanzio Sep 29 '25 at 07:56
  • @Melanzio I'm not saying it is generated by a single representation, I'm saying it is a union of categories generated by a single representation. Just take every representation that it contains. This gives a projective system of algebraic groups (via reconstruction), because this union is filtered. It is filtered because the category generated by two representations is also generated by their direct sum. – Julian Quast Sep 30 '25 at 11:19

0 Answers0