## Hermitian symmetric spaces vs Hermitian homogeneous spaces

A Hermitian symmetric space is a connected complex manifold with a hermitian metric on which the group of holomorphic isometries acts transitively, and which satisfies the following extra condition: each point (equivalently, some point) is an isolated fixed point of some (holomorphic, isometric) involution of the space. There is a beautiful classification theory of Hermitian symmetric spaces, and this is the starting point for the study of Shimura varieties.

The first four conditions (connected, complex, hermitian, homogeneous), which define a Hermitian homogeneous space, all seem natural enough to me. My question: why the need for the extra condition regarding the involution?

This is deliberately vague because I'm not quite sure what form the most satisfying answer will take. But here are a couple of more specific versions of the question. (1) What sort of spaces are we excluding by imposing this condition, and why? (2) Are there other conditions which might seem more natural to me which are equivalent (or equivalent some of the time)? For instance if you were to tell me that a Hermitian homogeneous space that's negatively curved and simply-connected always has such an involution, I would be completely satisfied.

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Here is a geometric answer to (2), or more precisely a slight and classical reinterpretation. First, note that Hermitian symmetric spaces fit into the more general concept of riemannian symmetric space; this can helps you find references.

Denote your space by $X$ (assumed to be Riemannian, or Hermitian if you prefer) and take a point $p$. Consider the geodesic symmetry around $p$: it maps a point $q$ to the point $\sigma_p(q)$ so that $\sigma_p(q), p, q$ lie on a constant speed geodesic, at times $-1,0,1$ respectively. This map is at least defined locally. Then $X$ is symmetric if and only if for all $p\in X$, $\sigma_p$ is globally defined and an isometry (ask it to be also holomorphic if your are in the Hermitian setting, but I guess it will automatically be so since it is conjugate to $-\mathrm{Id}$ by the exponential map). Of course, $\sigma_p$ is your involution.

This condition automatically implies that the isometry group of $X$ acts transitively (because any $q$ is mapped to any $q'$ by the map $\sigma_p$ where $p$ is a midpoint of $[q,q']$).

It also gives you an involution $\theta$ on the Lie algebra $\mathfrak{g}$ of the isometry group of $X$. Now remark that $\theta$ is a linear endomorphism, and $\theta^2=1$ implies that $\mathfrak{g}$ decomposes into two components, the eigenspace associated to the eigenvalue $1$ of $\theta$ and the one associated to $-1$. They are usually denoted by $\mathfrak{h}$ and $\mathfrak{p}$; the latter identifies with the tangent space to $X$ at $p$. This is called Cartan decomposition and is fundamental to the study of these spaces.

The most common reference is Helgason's Differential geometry, Lie groups, and symmetric spaces but I find it quite difficult to read. In the nonpositively curved case, I found Eberlein's Geometry of nonpositively curved manifolds useful.

Concerning the motivation, it seems that the interest in symmetric spaces is that they provide examples between constant curvature spaces, which are very constrained, and homogeneous spaces that are very numerous.

Last, a lead concerning (1). You shall find in Berger's A panoramic view of Riemannian geometry, section 15.8.1 page 719 a formula enabling one to compute the curvature of a homogeneous space. I guess it can be used to show the existence of many such spaces, even in the Hermitian world, that are negatively curved but not symmetric.

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Boringly-but-honestly, for me "symmetric space" means "G/K", where G is reductive or semi-simple, and (for non-compact symmetric spaces) K is maximal compact. That is, the "definition" is not "geometric", but artifactual.

The explanation of this viewpoint is yet another one of those stories wherein one finds oneself "a long-distance commuter" from A to B, and then wonders why one doesn't simply live closer to B.

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If you throw in "compact", so that the space is $G/P$ for some parabolic $P$, here's an alternate characterization: the unipotent radical of $P$ should be abelian. Equivalently, $P$ should be maximal, hence correspond to a simple root, and the coefficient of that simple root in the highest root should be $1$. The number of such simple roots is $|Z(G)|-1$, where $G$ is simple simply connected.

One way these "cominuscule" flag manifolds show up is as the closed $G(O)$ orbits on the affine Grassmannian $G(K)/G(O)$, where $O = {\mathbb C}[[t]]$ and $K = {\mathbb C}((t))$. Their homology gives the corresponding "minuscule" representation of the Langlands dual to $G$, this being the nicest case of the geometric Satake correspondence.

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I very much fear this answer will not satisfy you completely. Symmetric spaces comprise a very small subclass of homogeneous spaces, even if when you restrict to the hermitian ones.

One way to try to understand what is going on is to consider already the class of reductive homogeneous spaces. In positive-definite signature, all homogeneous spaces are reductive, but in indefinite signature this is not necessarily the case.)

A theorem of Ambrose and Singer, reformulated by Kostant, states that a riemannian manifold is (locally) reductive homogeneous if and only if there is a metric connection such that both the torsion and curvature are parallel. The (locally) symmetric spaces are precisely those for which the torsion vanishes. This is a very strong condition.

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