4
$\begingroup$

Let $\mathfrak{g}$ be a complex simple Lie algebra and $\mathfrak{k}$ its complex subalgebra such that $(\mathfrak{g},\mathfrak{k})$ is a Hermitian symmetric pair; $\mathfrak{g}= \mathfrak{k}\oplus\mathfrak{p}$ is the corresponding Cartan decomposition subject to some Cartan involution $\theta$. Moreover, there is a splitting $\mathfrak{p} = \mathfrak{p}^- \oplus \mathfrak{p}^+$.

Problem: Classify all $\theta$-stable parabolic subgroups $\mathfrak{q}=\mathfrak{l}\oplus\mathfrak{u}$ of $\mathfrak{g}$ such that $\mathfrak{l}\subseteq\mathfrak{k}$ and $\mathfrak{p}^+\subseteq\mathfrak{u}$.

Motivation: In the article Dirac operators and Lie algebra cohomology. Represent. Theory 10 (2006), the authors prove that in such a case there is a Hodge decomposition for $\mathfrak{u}$-homology of a unitarizable $(\mathfrak{g},K)$-module. I am interested for which real parabolic subalgebras of some real form of $\mathfrak{g}$ there is a Hodge decomposition.

$\endgroup$
12
  • 1
    $\begingroup$ Isn't $\mathfrak{k}$ only a real Lie subalgebra, not a complex Lie subalgebra? (It's the Lie algebra of a maximal compact subgroup of the underlying connected adjoint semisimple real Lie group, right?) Maybe I am misunderstanding. $\endgroup$
    – user30379
    Jan 3, 2013 at 15:32
  • 1
    $\begingroup$ I've explicitly stated that $\mathfrak{g}$ and $\mathfrak{k}$ are complex algebras. So if $G/K$ is non-compact Hermitian symmetric space, then $\mathfrak{g}$ is the complexification of the Lie algebra of $G$ and similarly for $\mathfrak{k}$. One can realize $G/K$ as a bounded symmetric domain in $\mathfrak{p}^−$ which is diffeomorphic to an open dense subset of $G_\mathbb{C}/P$. I think this is called Harish-Chandra embedding. I hope this clarifies the situation a little bit. $\endgroup$ Jan 3, 2013 at 16:13
  • 1
    $\begingroup$ Aren't the parabolic subalgebras $\mathfrak{q}$ satisfying those properties exactly those of the form $\mathfrak{q}=\mathfrak{q}'\oplus\mathfrak{p}^+$ with $\mathfrak{q}'$ parabolic in $\mathfrak{k}$? $\endgroup$
    – AndreA
    Jan 8, 2013 at 12:41
  • 1
    $\begingroup$ Any $\mathfrak q$ of your form is $\theta$-staple, hence $\mathfrak q= (\mathfrak q\cap\mathfrak k)\oplus (\mathfrak q\cap\mathfrak p)$, moreover $\mathfrak p^+\subseteq\mathfrak u\subseteq q$. If there exists $X\in\mathfrak p^-\cap\mathfrak q$ then there is $Y\in\mathfrak p^+$ such that $X,Y$ are part of a $\mathfrak{sl}_2$ triple, and hence $\mathfrak p^+\not\subset\mathfrak u$. It follows then that $\mathfrak p^+\subseteq\mathfrak q\subseteq\mathfrak k\oplus\mathfrak p^+$. $\endgroup$
    – AndreA
    Jan 10, 2013 at 12:55
  • 1
    $\begingroup$ On the other hand, a subalgebra $\mathfrak q$ is parabolic iff (with the obvious notation) $G/Q$ is compact. Let $\hat{\mathfrak q}=\mathfrak k\oplus\mathfrak p^+$ and $\mathfrak q=\mathfrak q′⊕\mathfrak p^+$. Then $G/Q$ is a bundle over $G/\hat Q$, which is compact, with fiber $\hat Q/Q=K/Q'$, which is compact too. $\endgroup$
    – AndreA
    Jan 10, 2013 at 12:58

1 Answer 1

3
$\begingroup$

The question needs to be made a bit precise. To talk of a Cartan decomposition in your sense, what we need to start with is a real Lie subalgebra ${\mathfrak g}_0$ with a "maximal compact sub-algebra" ${\mathfrak k}_0$ (i.e. Lie algebra of a maximal compact subgroup, assuming that the underlying Lie group $G$ is linear) whose complexifications are $\mathfrak g$ and $\mathfrak k$ respectively. Now there is the extension to $\mathfrak g$ of the Cartan involution $\theta $.

Given this, $\theta $ stable parabolic subalgebras $\mathfrak q$ of the complex Lie algebra $\mathfrak g$ whose nilradical $\mathfrak u$ contains ${\mathfrak p}^+$, are not many! These are exactly the ones such that ${\mathfrak q} \supset {\mathfrak p}^+$ and whose Levi $\mathfrak l$ (which is defined over $\mathbb R$- recall that $G$ is the group of real points of $G({\mathbb C})$) is such that $L\subset K$ (i.e. ${\mathfrak l}\subset {\mathfrak k}$). In terms of the (Vogan-Zuckerman) cohomological representations $A_{\mathfrak q}(0)$, the representation $A_{\mathfrak q}(0)$ is the unique holomorphic discrete series with trivial infinitesimal character.

If you ask that the whole parabolic sub-algebra ${\mathfrak q}$ contains ${\mathfrak p}^+$, this is equivalent to asking that ${\mathfrak u}\cap {\mathfrak p}= {\mathfrak u}\cap {\mathfrak p}^+$, and hence the corresponding $A_{\mathfrak q}(0)$ is a representation of holomorphic type.

I believe this is implicitly contained in a well known paper of Vogan and Zuckerman in Compositio (1984?) on unitary representations with cohomology.

Not too fussed about the "bounty". But this is the "final" answer: every such $q$ is of the form $$ q=m\oplus {\mathfrak p}^+$$ where $m$ is a parabolic subalgebra of $k$.

$\endgroup$
3
  • $\begingroup$ I did not see the additional comments. Andrea has already said that these parabolics are of the form $m\oplus {\mathfrak p}^+$. This happens to be correct. Andrea has more or less given a proof as well. this is the proof. We have the decompositions $$q=l\olpus u$$ and $u=p^+\oplus (u\cap p^{-})$. Since the killing form on $p^+\times p^{-}$ is a perfect pairing, and is degenerate on $u$ it follows that if $u$ contains $p^+$ then it cannot contain any vector in $p^{-}$. By the OP's assumption, $l\subset k$. Now, if $b$ is a Borel subalgebra of $k$, then $b+p^+$ is a Borel subalgebra of g $\endgroup$ Jan 15, 2013 at 2:07
  • $\begingroup$ This implies that if $m$ is a parabolic subalgebra of $k$ then $m\oplus p^+$ is a parabolic subalgebra of $g$. So, these remarks imply that every $\theta $ stable parabolic is of the desired form (I think that Andrea should get the bounty since Andrea has answered the question completely. $\endgroup$ Jan 15, 2013 at 2:09
  • $\begingroup$ Agreed, but since he did not posted an answer I cannot give him the bounty. And thank you for your remarks. $\endgroup$ Jan 16, 2013 at 22:08

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.