Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
1  
Be careful: $A^{\ast}$ is only a "formal" Hopf algebra (using its pseudo-compact topology). To be precise, when dualizing $A \otimes_k A \rightarrow A$ one gets a $k$-linear map $A^{\ast} \rightarrow A^{\ast} \widehat{\otimes}_k A^{\ast}$, the completed tensor product taken with respect to the pseudo-compact topology. I see no reason why this should land inside $A^{\ast} \otimes_k A^{\ast}$ (whose natural map to the completed tensor product is injective). These matters, including the long and dry story of pseudo-compactness, are rather thoroughly discussed in SGA3, Expose VII$_{\rm{B}}$. –  BCnrd Jul 9 '10 at 19:32
    
@BCnrd: I think that OP is defining $A^*$ not as the full dual, but as only those linear maps $A \to k$ that factor through some finite-dimensional quotient algebra. This "Hopf dual" I think is a Hopf algebra with the algebraic tensor product, isn't it? I've usually seen it written not as $A^*$, which should mean the full dual, but as something like $A^\circ$. –  Theo Johnson-Freyd Jul 9 '10 at 19:54
    
The language and notation need to be pinned down more precisely. It would help a lot to identify the foundational sources, as BCnrd has done. See also my comment to Theo's answer. –  Jim Humphreys Jul 9 '10 at 20:17
1  
@Chuck: Out of curiosity, is there something specific about reductive groups that you expect to study in this way? There are certainly very powerful ways to work with reductive groups over arbitrary fields without any digressions into hyperalgebras, etc. –  BCnrd Jul 9 '10 at 21:38
1  
Re "it is distressing that this question comes up as the third hit on Google". Rather, it is something to be celebrated. One think the StackOverflow folks did when they wrote this software was to think seriously about how Google Pagerank works, and try to trick it. They more or less succeeded: it is a feature, not a bug, that MO pages turn up very high on Google. (On the other hand, this gives us all the more impetus to make the quality of the pages high, as we're now one of the first places people go with math questions.) –  Theo Johnson-Freyd Jul 10 '10 at 20:57

2 Answers 2

up vote 5 down vote accepted

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.)

share|improve this answer
    
Great, and thanks for the references! –  Chuck Hague Jul 10 '10 at 19:49
1  
The link above seems to be dead, but Cartier's notes are now available here: math.osu.edu/~kerler.2/VIGRE/InvResPres-Sp07/Cartier-IHES.pdf –  Matthew Towers Feb 11 '12 at 14:05
    
Thanks for the updated link (when I try the old one it refuses me permission, which I guess is equivalent to dead). This is often a problem with notes or preprints posted on the internet outside arXiv. –  Jim Humphreys Feb 11 '12 at 18:01

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?.

share|improve this answer
    
In prime characteristic the universal enveloping algebra of the Lie algebra does not work well for a semisimple algebraic group. Here the appropriate substitute is the algebra of distributions (or hyperalgebra) as discussed in the book by Demazure-Gabriel and in I.7 of Jantzen's book Representations of Algebraic Groups. The language here is well suited to the study of more general group schemes. –  Jim Humphreys Jul 9 '10 at 20:14
    
@Jim: Good to know. Yes, I think that OP is using better language than I am versed in. As I said, I really don't know anything that's not over $\mathbb C$. –  Theo Johnson-Freyd Jul 9 '10 at 20:19
    
This is very interesting! I didn't know that Tannaka-Krein duality has this connection to the Hopf dual. –  Chuck Hague Jul 10 '10 at 19:51
    
@Chuck Hague: It shouldn't be too surprising. In each case, you're testing an algebra against all its finite-dimensional representations. (I've been on a TK kick of late, as can be seen e.g. from my recent MO activity.) –  Theo Johnson-Freyd Jul 10 '10 at 20:58
    
references for $k[G]\cong U(g)^o$ in the `nice case' are given in Kassel's Quantum Groups book: Abe, Hopf Algebras, CUP 1980; Hochschild, Basic theory of algebraic groups and Lie algebras (Springer GTM); Joyal and Street, An introduction to Tannakian duality in Springer LNM 1488; Sweedler, Hopf algebras. –  Matthew Towers Mar 6 '12 at 13:09

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.