Why symmetric spaces? - MathOverflow

Why symmetric spaces?

2010-10-10T16:18:20Z

In Bar-Natan's "Knots at Lunch" seminar at the University of Toronto, we are currently discussing a talk by Alekseev at Montpellier about Rouvière's expansion of the Duflo isomorphism to the setting of symmetric spaces.
We understand the definition of a symmetric space, and we know that people have written books about them; but we I don't understand in what sense symmetric spaces are useful and interesting mathematical objects. In particular:

Are there significant (analytic? geometric? algebraic?) techniques which work for symmetric spaces, but not for more general classes of homogenous spaces?

What I'm trying to understand (at least vaguely) is the role of symmetric spaces in the Kashiwara-Vergne picture, and the conceptual reason one might expect Duflo's isomorphism to generalize to this specific class of mathematical objects. There must be a conceptual explanation why symmetric spaces are the natural class of objects to consider in such contexts.
A closely related question is THIS.

Answer by Tobias Hartnick for Why symmetric spaces?

2010-10-10T16:48:48Z

If one accepts curvature as a measure of complexity of a Riemannian manifold (which one might or might not agree with), then the "simplest" Riemannian manifolds are those of constant curvature. Unfortunately, there are not so many of these; besides Euclidean space the only simply-connected examples are the spheres (constant positive curvature) and the hyperbolic spaces (constant negative curvature). (Of course there is a rich theory of non-simply-connected spaces of constant negative curvature, but never mind.) So, how can one weaken the notion of constant curvature to obtain a larger class of interesting, but not "too complicated" spaces? Well, it seems natural to ask that the covariant derivative of the curvature should be 0. In this case, one has many more 1-connected examples, and these are precisely symmetric spaces. They are nice in various senses:

They can be classified, so there are not too many of them (but still sufficiently many to be "interesting").
They admit a description in terms of Lie groups. This allows for very explicit computations, e.g. of curvature and characteristic classes.
They admit a natural duality (compact/non-neg. curved vs. non-compact/non-pos. curved) if one ignores flat factors. This allows for the transfer of ideas between two different worlds (see e.g. Hirzebruch proportionality).

What more can one ask for? On the other hand, one should stress that they are really rare and special objects, just slightly less rare than manifolds of constant curvature.

Answer by algori for Why symmetric spaces?

2010-10-10T17:21:45Z

Here is one example. When $G$ is a compact Lie group and $H$ is a Lie subgroup, the real cohomology of the homogeneous space $G/H$ is the same as the relative cohomology $H^*(g,h,\mathbf{R})$ where $g$ and $h$ are the Lie algebras of $G$, respectively $H$. This can be proven by averaging, just as in the case when $H$ is trivial. In general the differential in the relative cochain complex is not zero, but when $G/H$ is symmetric, it is, for a simple reason: the symmetric involution acts as $(-1)^d$ on the degree $d$ part of the complex; since this action should commute with the differential, the differential must be 0; for more details see Felix, Halperin, Thomas, Rational homotopy theory, p. 162.

So symmetric spaces are formal. However, in general compact homogeneous spaces need not be formal; e.g. $SU(n)/Sp(n)$ is not formal for $n\geq 5$, see Greub, Halperin, Vanstone, Curvature, connections and cohomology.

Answer by Dan Fox for Why symmetric spaces?

2010-10-11T10:07:34Z

The algebra of invariant differential operators on a symmetric space is commutative, and this is certainly not true for an arbitrary homogeneous space. While it is not true that the commutativity of the algebra $D^{G}$ of $G$ invariant differential operators on a homogeneous space $X = G/H$ implies that $X$ is symmetric, if $G$ is reductive it is true that $D^{G}$ is commutative if and only if $G/H$ is weakly symmetric in the sense defined by Selberg (see E.B. Vinberg's survey in Russ. Math. Surveys 56(1)).