Hopf algebra duality and algebraic groups

Question

Background:

Let $G$ be a linear algebraic group over an algebraically closed field $k$ and let $I \subseteq k[G]$ be the ideal of the identity element. The hyperalgebra $U(G)$ of $G$ is defined to be the subspace of the linear dual of $k[G]$ consisting of all $f$ such that $f(I^n) = 0$ for some $n > 0$; then there is a natural Hopf algebra structure on $U(G)$.

Given any Hopf algebra $A$ over a field $k$, one can also define the Hopf dual $A^*$ of as follows: Let $A^*$ be the subspace of the full linear dual of $A$ consisting of elements that vanish on some two-sided ideal of $A$ of finite codimension. Then $A^*$ has a natural Hopf algebra structure.

EDIT: As pointed out in comments, I am using what might be nonstandard terminology -- perhaps I should write $A^\circ$ instead of $A^*$.

Questions:

I now have two related questions.

1) I assume that in general the hyperalgebra of $G$ is not the same as the Hopf algebra dual of $k[G]$. What is the Hopf dual of $k[G]$?

2) Although one obtains $U(G)$ from $k[G]$ by a dual construction, you can't in general obtain $k[G]$ from $U(G)$ -- for example, if $G$ is reductive, you can't get $k[G]$ from $U(G)$ by some sort of duality since the hyperalgebra doesn't see isogeny. Is there something one can say in the reductive case about the structure of the Hopf dual of $U(G)$? More generally, for which algebraic groups $G$ can we obtain $k[G]$ from $U(G)$ by some sort of duality? And when we can do so, is it as simple as just taking the Hopf dual?

Answer 1

In prime characteristic (or for algebraic groups rather than Lie algebras in general), the comments already posted indicate a need for caution. Jantzen's Part I covers a lot of the ground, but he refers back at a few delicate points to Demazure-Gabriel.

Duality for general Hopf algebras is discussed in section 3.5 of Cartier's 2006 notes

http://inc.web.ihes.fr/prepub/PREPRINTS/2006/M/M-06-40.pdf

Starting with a Hopf algebra $A$, its reduced dual Hopf algebra (denoted by him $R(A)$) lives in the linear dual (a coalgebra in its own right) but is often smaller. So it's complicated to go back and forth. On the other hand, dealing with algebraic groups rather than just Lie algebras makes life a little more complicated, as suggested in his earlier discussion of the algebra of representative functions on a group; see also:

"Remark 3.7.3. Let $k$ be an algebraically closed field of arbitrary characteristic. As in subsection 3.2, we can define an algebraic group over $k$ as a pair $(G,O(G))$ where $O(G)$ is an algebra of representative functions on $G$ with values in $k$ satisfying the conditions stated in Lemma 3.2.1. Let $H(G)$ be the reduced dual Hopf algebra of $O(G)$. It can be shown that $H(G)$ is a twisted tensor product $G \times U(G)$ where $U(G)$ consists of the linear forms on $O(G)$ vanishing on some power $\mathfrak{m}^N$ of the maximal ideal $\mathfrak{m}$ corresponding to the unit element of $G$; here $\mathfrak{m}$ is the kernel of the counit $\epsilon: O(G) \rightarrow k$. If $k$ is of characteristic 0, $U(G)$ is again the enveloping algebra of the Lie algebra $\mathfrak{g}$ of $G$. For the case of characteristic $p >0$, we refer the reader to Cartier [18] or Demazure-Gabriel [32]."

[Note that the \times symbol should be the LaTeX symbol \ltimes, which apparently won't print here.]

ADDED. Given $G$, the hyperalgebra of $G$ is what Cartier denotes by $U(G)$; so it is not quite the reduced dual Hopf algebra of the algebra of regular functions on $G$ in general. In case $G$ is a connected reductive algebraic group over an algebraically closed field of characteristic $p>0$, the hyperalgebra has an explicit "divided power" construction starting with Kostant's integral form of the universal enveloping algebra of the complex analogue of the Lie algebra of $G$ (in the crucial case $G$ semisimple and simply connected), then reducing mod $p$. This is used heavily by Jantzen to investigate the rational representations of $G$ and relevant closed subgroups such as Borel subgroups, etc. (A similar construction is used by Lusztig for the quantum enveloping algebra at a root of unity.)

Answer 2

This isn't at all a complete answer, because I don't know a lot about algebraic groups. But it seemed big enough that I didn't want it to get lost in the comments. And perhaps everything I'm about to say is something you already know. In any case, someone with more expertise than I have should also answer this question.

I assume, since you're talking about the function algebra $k[G]$, that your group $G$ is affine. In any case, that's all I know how to see. In general, for an (affine) algebraic group $G$, its function algebra $k[G]$ is precisely the Hopf algebra reconstructed via Tannaka-Krein from the category of finite-dimensional algebraic representations of $G$. Remember that to any category of finite-dimensional representaitons, TK reconstructs a coalgebra, and to a (symmetric) monoidal category (with duals) it reconstructs a (commutative) bialgebra (with an antipode), i.e. an affine algebraic group. Conversely, let $A$ be any Hopf algebra. Then the Hopf dual $A^\circ$ (I think that's the more standard notation? I tend to see $A^*$ for the full vector-space dual, which is an algebra but not usually a coalgebra. Maybe $A'$ is the standard notation?) is the TK-reconstruction of the category of finite-dimensional $A$-modules (the algebra structure on $A$ defines the category, and hence the coalgebra structure on $A^\circ$; the Hopf structure on $A$ makes the category monoidal with duals, and hence gives the Hopf structure on $A^\circ$). This is all rather trivial after enough definition-unpacking and some theorems about coalgebras.

As for the second question, when $G$ is semisimple, then I think you basically can reconstruct $k[G]$ from $U(G)$. Or, rather, I think that $U(G)$ is the usual universal enveloping algebra, and its Hopf dual $U(G)^\circ$ is the function algebra for the connected simply-connected version of $G$. Certainly, $U(G)$ cannot see past the infinitesimal neighborhood of the identity element of $G$, and so can't tell a group from its connected simply-connected cousin. But I should emphasize the word "think". I'm comfortable that what I said is correct when $k= \mathbb C$ (and hence any algebraically closed field of characteristic zero, because their algebraic theories are the same), but not in general. And as soon as you move away from the semisimple guys, $U(G)^\circ$ is going to be a helluva lot bigger than $k[G]$. See the answers to What algebraic group does Tannaka-Krein reconstruct when fed the category of modules of a non-algebraic Lie algebra?.