5
$\begingroup$

Let $\mathfrak{g}_1\subset \mathfrak{g}_2$ be a Lie algebra embedding. Assume both are semisimple. For instance take the standard diagonal embedding $\mathfrak{sl}(2, \mathbb{C})\subset \mathfrak{sl}(3, \mathbb{C})$. This lifts to an embedding $U(\mathfrak{sl}(2, \mathbb{C}))\subset U(\mathfrak{sl}(3, \mathbb{C}))$ of the corresponding enveloping algebras. Consider their centers $Z(\mathfrak{sl}(2, \mathbb{C})),Z(\mathfrak{sl}(3, \mathbb{C}))$. It is relatively easy to see that under the previous embedding we have that $Z(\mathfrak{sl}(2, \mathbb{C}))\cap Z(\mathfrak{sl}(3, \mathbb{C})) = \mathbb{C}$. Is there any reasonable way to "connect" these centers? Let me be a bit more clear. For instance, we can pass the same question to the symmetric algebras of $\mathfrak{g}_1,\mathfrak{g}_2$. Here we have $S(\mathfrak{sl}(2, \mathbb{C})^*)^{SL(2,\mathbb{C})}$ the algebra of $SL(2,\mathbb{C})$-invariant functions on $\mathfrak{sl}(2, \mathbb{C})$ and $S(\mathfrak{sl}(3, \mathbb{C})^{\*} )^{SL(3,\mathbb{C})}$. There is a natural embedding of $S(\mathfrak{sl}(3, \mathbb{C})^{\*} )^{SL(3,\mathbb{C})}$ into $S(\mathfrak{sl}(3, \mathbb{C})^{\*} )^{SL(2,\mathbb{C})}$ and a projection of $S(\mathfrak{sl}(3, \mathbb{C})^{\*} )^{SL(2,\mathbb{C})}$ onto $S(\mathfrak{sl}(2, \mathbb{C})^{\*} )^{SL(2,\mathbb{C})}$ given by restriction. Are there any results that could describe how these centers behave under restrictions?

$\endgroup$

2 Answers 2

4
$\begingroup$

Take a look at "Shifted Schur Functions"

Andrei Okounkov, Grigori Olshanski http://arxiv.org/abs/q-alg/9605042

Section 10: "Coherence property of quantum immanants and shifted Schur polynomials"

In particular formulas 10.4, 10.5 - they discuss "averaging operators" Z(U(gl(n)) -> ZU(gl(N)) , n < N

and later prove certain "good" (coherence) property of special generators of the centers Z(U(gl(k)) which has been studied by the authors and M. Nazarov.

Hope this helps...

What is very interesting for me personally - is try to generalize such things to the case of loop algebras Z(U(\hat gl)). Here certain "good" elements of the centers has been constructed by Talalaev's formula, it is natural to expect that Okounkov-Olshanski-... story can be generalized to loop algebra case

$\endgroup$
2
$\begingroup$

I'm doubtful about getting nice relationships between centers of universal enveloping algebras, if you look at embeddings for arbitrary pairs of semisimple Lie algebras; maybe there are subtle connections in special cases, however. For one thing, centers and algebras of invariant polynomial functions for a given semisimple Lie algebra are related only indirectly in the work of Chevalley and Harish-Chandra. Moreover, embeddings into special linear algebras of unbunded rank occur when you start with a fixed semisimple Lie algebra and consider all of its faithful irreducible representations. But the respective centers of the smaller and larger enveloping algebras are isomorphic to polynomial algebras in the number of indeterminates given by the respective ranks.

If there is a useful way to relate centers, it should be visible in diagonal embeddings for Lie algebras both of small rank such as you describe. But it's not immediately visible to me.

P.S. A tag lie-algebras would be useful.

$\endgroup$
2
  • $\begingroup$ Thank you for the reply. I was asking this question because I'm interested in direct limit Lie algebras. Their enveloping algebras have trivial centers and this of course is a big set back in trying to understand their representation theory. It seems to me there should be some nice generalization of the center in this case, but of course I might be totally wrong about this. P.S. I tried to add a tag with lie-algebras but it didn't let me for some reason. If you can edit the post then you're more than welcome to do so and add the tag. $\endgroup$
    – Alex
    Feb 18, 2013 at 21:19
  • $\begingroup$ @Alex: I've added a tag, but you should be able to edit your question including the tags. Also, it would have been helpful to focus your question more on direct limits, since arbitrary embeddings look pretty hopeless for comparison of centers. But even in the case of direct limits, you have to choose the embeddings explicitly. $\endgroup$ Feb 19, 2013 at 20:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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