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Assume that M=G/K is a non-compact Hermitian symmetric space, for G the real points of a semisimple (or even simple) algerbaic group and K a maximal compact subgroup. M admits the structure of a complex manifold. Now, if X is a complex subvariety of M then its pre-image in G is a real analytic subvariety of G. My question is: how does one recognize, among all real analytic subvarieties of G, those which project onto complex subvarieties in M?

More concretely, if H is a real Lie subgroup of G (even real algebraic), when is it the case that its image under the projection map (into M=G/K) gives a complex subvariety of M?

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I will explain how to check it in any particular case, but I am not sure whether there is a good classification available (perhaps, it is even easy and I did not think enough about it). Helgason's book and papers of Joe Wolf and Alan Huckleberry may contain some relevant information.

Since the complex structure on $M=G/K$ is homogeneous, the image of $H$ is a complex subvariety if and only if its tangent subspace at $e$ is a complex subspace of $T_e M,$ i.e. invariant under the complex structure operator $J: T_e M\to T_e M.$ Let $\mathfrak{g}=\mathfrak{k}\oplus\mathfrak{p}$ be the Cartan decomposition, then $T_e (G/K)$ may be identified with $\mathfrak{p}$ via the projection onto the second summand and $J=\text{ad}(X_0)$ for a certain element $X_0$ in the center of $\mathfrak{k}$ (if $G$ is simple then this center is one-dimensional). Thus the answer is "yes" if and only if

$$T_e(H/K\cap H)=\mathfrak{h}/\mathfrak{k}\cap\mathfrak{h} \subset \mathfrak{p} \text{ is } \text{ad}(X_0)\text{-invariant.}$$

Subalgebras $\mathfrak{h}$ with this property would need to be classified modulo conjugation by $K$. One can proceed a bit further by complexifying $\mathfrak{g}$, so that $\mathfrak{p}_\mathbb{C}=\mathfrak{p}_{+}\oplus\mathfrak{p}_{-}$ is the eigenspace decomposition for $J_{\mathbb{C}}$ (the eigenvalues are $\pm i$ and $\mathfrak{p}_{\pm}$ are abelian subalgebras of $\mathfrak{g}_{\mathbb{C}}$), keeping in mind that a complex Lie subalgebra of $\mathfrak{g}_\mathbb{C}=\mathfrak{g}\otimes_{\mathbb{R}}\mathbb{C}$ is a complexification of a Lie subalgebra of $\mathfrak{g}$ if and only if it is invariant under the complex conjugation.

If $H$ is a connected semisimple group of Hermitian type and $f:H\to G$ is an embedding, then after a suitable conjugation, the image of a maximal compact subgroup $L<H$ is contained in $K,$ $f(H)=(f(H)\cap K)(f(H)\cap P)$ and $f(H)$ mod $K$ is isomorphic to a Hermitian symmetric space $H/L,$ which is thus realized in $G/K.$ This allows one to construct many examples by starting with (1) a group $H$ as above and (2) a faithful complex representation of $H$ that preserves an indefinite unitary form, so that $G=SU(p,q)$ if the signature is $(p,q).$

A final elementary observation is that there always exist nontrivial subgroups $H$ that do not arise from the previous construction, for example, $NA$ from the Iwasawa decomposition $G=NAK$ projects onto $G/K.$

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Thanks. For the explanation and the references. –  Kobi Jul 20 '10 at 7:33

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