Complex quadric as a symmetric space

Question

It is known that a smooth complex quadric is a symmetric space. For example, it is $$\operatorname{Spin}(n+2)/G$$ where $G$ is the maximal parabolic subgroup.

I want a reference for more details and discussions.

Answer 1

So while the projective quadric is indeed a homogeneous space of the form $G/P$ for $P$ parabolic in $G$ (marking it as a generalised flag manifold) this is not the realisation of it as a symmetric space. Indeed this does not even specify a Riemannian (or even pseudo-Riemannian) structure. It does give the quadric a conformal structure so I suppose we could pick a representative metric in that conformal class but still this is not the way to define the symmetric space structure.

To do that we need to take a maximal compact subgroup $K\subset G$. Then with $K_0 := K\cap P$, $K/K_0$ is a symmetric space and is still diffeomorphic to $G/P$. Or looking at it another way $K$ still acts transitively on $G/P$.

Indeed in this way any flag manifold can be given a Riemannian structure and the ones for which this gives a Riemannian symmetric space are called the symmetric R-spaces. These basically comprise the compact Hermitian symmetric spaces and their real forms.

Edit: The default reference would be Helgason's "Differential Geometry, Lie Groups and Symmetric Spaces", particularly the sections on Hermitian symmetric spaces of compact type. Other good sources more generally about symmetric R-spaces include "Basic Transformations of symmetric R-spaces" by Takeuchi, "Geometry of Symmetric R-spaces" by Tanaka. My own PhD thesis: "Isothermic surfaces and their generalisations" goes into detail on all the classical examples of these including the projective quadric although it focuses only on the $G/P$ realisation and not at all on the $K/K_0$ version.