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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.

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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. –  user30379 Jan 3 '13 at 15:32
    
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. –  Vít Tuček Jan 3 '13 at 16:13
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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 $\mathfrak{k}$? –  user175348 Jan 8 '13 at 12:41
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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 $\mathfrak p^+\subseteq\mathfrak q\subseteq\mathfrak k\oplus\mathfrak p^+$. –  user175348 Jan 10 '13 at 12:55
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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. –  user175348 Jan 10 '13 at 12:58

1 Answer 1

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 (18984?) 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 $k$.

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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 –  Venkataramana Jan 15 '13 at 2:07
    
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. –  Venkataramana Jan 15 '13 at 2:09
    
Agreed, but since he did not posted an answer I cannot give him the bounty. And thank you for your remarks. –  Vít Tuček Jan 16 '13 at 22:08

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