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Let $G$ be a real semisimple Lie group and $\mathfrak{g}$ be its complexified Lie algebra. We have the flag variety $\mathcal{B}$ of $\mathfrak{g}$ which is the set of all Borel subalgebras of $\mathfrak{g}$. $\mathcal{B}$ is naturally a complex manifold so at each point $\mathfrak{b}\in \mathcal{B} $ the complexified tangent space $ T^{\mathbb{C}}_{\mathfrak{b}}\mathcal{B}$ decomposes into the holomorphic and anti-holomorphic subspaces. $$ T^{\mathbb{C}} _{\mathfrak{b}}\mathcal{B}=T^{\text{hol}} _{\mathfrak{b}}\mathcal{B}\oplus T^{\text{anti-hol}} _{\mathfrak{b}}\mathcal{B}. $$

The real group $G$ acts on $\mathcal{B}$ by conjugation at each Borel subalgebra. We know that in general this action is not transitive. Let $S$ be an orbit of this action. It is a real submanifold of $\mathcal{B}$ so at a point $\mathfrak{b}\in S\subset \mathcal{B} $ we have $$ T^{\mathbb{R}} _{\mathfrak{b}}S\subset T^{\mathbb{R}} _{\mathfrak{b}}\mathcal{B} $$ hence there complexifications also have $$ T^{\mathbb{C}} _{\mathfrak{b}}S\subset T^{\mathbb{C}} _{\mathfrak{b}}\mathcal{B}. $$

My question is: is it true that $T^{\text{hol}} _{\mathfrak{b}}\mathcal{B}\subset T^{\mathbb{C}} _{\mathfrak{b}}S$, both considered as complex subspace of $T^{\mathbb{C}} _{\mathfrak{b}}\mathcal{B}$?

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Am I right in thinking that $SL_2({\mathbb R})$ acting on ${\mathbb RP}^1$ inside ${\mathbb CP}^1$ is already a counterexample? –  Allen Knutson Sep 16 '12 at 3:27

1 Answer 1

No, actually $T^{\text{hol}} _{\mathfrak{b}}\mathcal{B}\subset T^{\mathbb{C}} _{\mathfrak{b}}S$ if and only if $S$ is an open orbit.

In fact, $T^{\mathbb{C}} _{\mathfrak{b}}S$ is invariant under conjugation, because it is the complexification of $T^{\mathbb{R}} _{\mathfrak{b}}\mathcal{B}$. Thus it contains $T^{\text{hol}} _{\mathfrak{b}}\mathcal{B}$ if and only if it contains $T^{\text{anti-hol}} _{\mathfrak{b}}\mathcal{B}$, i.e. if and only if it is equal to $T^{\mathbb{C}} _{\mathfrak{b}}\mathcal{B}$.

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