Motivation. Few years ago I constructed a family of internal Hopf algebras in the Loday-Pirashvili tensor category of linear maps which is in a sense a generalization of the algebra of regular functions on the group $GL(n)$. A student of mine (Matija Bašić) has worked out certain internal analogue of a Weyl algebra in that context and one could conceivably make much of the story on algebraic groups and invariant differential operators in that setup. However, we did not continue for a while and this project stalled in a way. Its motivation was to find a geometric theory of integration of Leibniz algebras in characteristic zero, and more generally Lie algebras in Loday-Pirashvili tensor category. There is an Ado theorem in this setup, so it is sufficient to do integration for Lie subalgebras in a Loday-Pirashvili analogue of $gl(n)$ (here $n$ generalizes to data of a linear map between two finite dimensional vector spaces). For the latter there is an internal version (as well as another version) of a universal enveloping Hopf algebra. On the other hand, there is my analogue of the function Hopf algebra on the $GL$ in Loday-Pirashvili category. I have a picture of certain modification of scheme theory to make it into an algebraic group in certain abstract context, but so far this is not fully done. Thus I would like to prove, not from abstract principles but by direct check that the two Hopf algebras are dual one to another.

But trouble: in classical case of the usual Lie groups and enveloping algebras I am aware only of the proofs using what a Lie group is and what its Lie algebra is, while my GL is given as a Hopf algebra somewhat alike usual functions on $GL(n)$, with just a little more exotic relations. So what I should generalize is down to earth explicit proof that $U(gl_n)$ and $\mathcal{O}(GL(n))$ are dual as Hopf algebras. Emphasis is on Hopf. Notice that it is not sufficient to write down the pairing for multiplicative generators, but for the whole vector space basis: the algebras are not free and to predict the pairing between higher order monomials does not follow from knowing pairing just between matrix elements $t^i_j$ and the generators of the Lie algebras $gl_n\subset U(gl_n)$. But it should not be that difficult.

Question: Do you know how to do algebraically prove that we have a Hopf pairing (possibly with some sort of formulas for the pairing) between classical $GL(n)$ and $U(gl_n)$ ? We should never use any knowledge on $GL(n)$ and $gl_n$ except their generators and relations, as most of other facts are nontrivial to generalize to my problem which motivates the question.

Related questions: explicit-isomorphism-between-distributions-and-universal-enveloping-algebra and hopf-algebra-structure-on-the-universal-enveloping-algebra-of-a-leibniz-algebra.

  • $\begingroup$ If you have access to the groups as geometric objects, then the usual argument of course realizes $U(gl_n)$ as distributions supported near the identity in $GL(n)$; but you do not have access to these, and anyway I think the proof I know somewhere uses characteristic=0. In the quantum groups case, we were not able to find a proof even when n=2 for a Hopf pairing between $U_q(sl_2)$ and $SL_q(2)$ (or rather, we were not able to show the corresponding pairing was nondegenerate) in our quantum groups class (unedited notes at math.berkeley.edu/~theojf/QuantumGroups10.pdf ) $\endgroup$ – Theo Johnson-Freyd Mar 15 '11 at 19:46
  • $\begingroup$ Rather, I think that to give the pairing $GL(n) \otimes U(gl(n)) \to k$ in generators and relations is not difficult (where by "$GL(n)$" I mean the Hopf algebra of $k$-valued functions on the algebraic group), but it is hard to show that the pairing has no kernels. $\endgroup$ – Theo Johnson-Freyd Mar 15 '11 at 19:49
  • $\begingroup$ So what is the pairing in terms of monomials on both sides (it is not enough the pairing for generators, it is not multiplicative, instead the Hopf structure controls the value on the products, which should b controlled) ? I can think of kernel later. For quantum SL(n) depending on version there may be a small finite-dimensional kernel, by a result of Takeuchi. $\endgroup$ – Zoran Skoda Mar 15 '11 at 21:04
  • $\begingroup$ I mean one can try to guess the formulas by applying derivatives many times and trying to write general formulas and use induction, but I have not succeeded into controlling this. Of course, eventually I could have a geometric proof in my generalization, but it is good to know if one is on the right track at least. $\endgroup$ – Zoran Skoda Mar 15 '11 at 21:09
  • $\begingroup$ I would like to see this as a homomorphism from the enveloping algebra onto the Schur algebras. Can this be done explicitly? $\endgroup$ – Bruce Westbury Mar 15 '11 at 23:53

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