13
$\begingroup$

Let $\mathfrak{g}$ be a Lie algebra and $\mathfrak{g}'$ its subalgebra. Then the universal enveloping algebra $U(\mathfrak{g}')$ can be canonically embedded into $U(\mathfrak{g})$, that of $\mathfrak{g}$.

Now I'm interested in the reverse direction. Given a subalgebra $Y$ of $U(\mathfrak{g})$, under what conditions on $Y$, $Y$ is the universal enveloping algebra $U(\mathfrak{g}')$ of some Lie subalgebra $\mathfrak{g}'$ of $\mathfrak{g}$?

By considering the center of $U(\mathfrak{g})$, not all subalgebras of $U(\mathfrak{g})$ arise as universal enveloping algebras of Lie subalgebras.

I searched the internet but couldn't find anything related to this question. Much appreciated for any answer or reference.

$\endgroup$
5
  • 1
    $\begingroup$ Yoiks, I'd think very few. Did you have in mind any further condition? Stability under some involution, for example? $\endgroup$ Dec 13, 2016 at 1:19
  • $\begingroup$ @paulgarrett, thanks for your comment. I don't know too much about universal enveloping algebra and have no candidate conditions in my mind. $\endgroup$
    – user1832
    Dec 13, 2016 at 1:25
  • $\begingroup$ Ok, well, if a subgroup or subalgebra is defined as the commutator of another subgroup or subalgebra, then there is the fairly-obvious condition on the corresponding subalgebra of the universal enveloping algebra. Being the fixed subalgebra or subgroup under an involution is much less definitive, so far as I can see. Do you contemplate a commutator condition? $\endgroup$ Dec 13, 2016 at 1:28
  • $\begingroup$ @paulgarrett, so you mean if a subalgebra g_1 is the commutator of another g_2, and Y is the commutator of the universal enveloping subalgebra of g_2, then Y is that of g_1? $\endgroup$
    – user1832
    Dec 13, 2016 at 1:48
  • $\begingroup$ IFF it is generated by a subset of $g$. In reality any useful criterion should take into account what you now about your subalgebra, how it is defined, etc... $\endgroup$
    – Bugs Bunny
    Dec 13, 2016 at 13:48

1 Answer 1

15
$\begingroup$

In characteristic $0$, if $Y$ is a Hopf subalgebra of $U(\mathfrak{g})$, then $Y = U(\mathfrak{g}')$ for some Lie subalgebra $\mathfrak{g}'$ of $\mathfrak{g}$ (and conversely, every such subalgebra is a Hopf subalgebra). This is a consequence of Proposition 6.13 and Theorem 5.18 of John W. Milnor and John C. Moore's 1965 paper "On the structure of Hopf algebras."

If you are working over a field of positive characteristic, then the result is also true if you replace the universal enveloping algebra by the restricted enveloping algebra (see Theorem 6.11 of loc. cit.).

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .