Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

While preparing a seminar I gave today, the following question arose. I asked the seminar participants, but nobody knew the answer. Hence I'm asking it here in MO.

Background

Recall that a complete, connected and simply connected pseudoriemannian manifold $(M,g)$ is a symmetric space if the Riemann curvature tensor is parallel with respect to the Levi-Civita connection: $\nabla R = 0$.

Typical examples are the (simply-connected) spaces of constant curvature: sphere, hyperbolic space, (anti) de Sitter spacetimes,... all of which admit local isometric embeddings as quadrics in some flat pseudoriemannian manifold $\mathbb{R}^{p,q}$. Recall that the flat metric on $\mathbb{R}^{p,q}$ is given in flat coordinates by $$\sum_{i=1}^p (dx_i)^2 - \sum_{i=1}^q (dx_{p+i})^2.$$

For example, the sphere with unit radius of curvature embeds in $\mathbb{R}^{n+1}$ as the quadric $$x_0^2 + x_1^2 + \cdots + x_n^2 = 1,$$ whereas the hyperbolic space embeds in $\mathbb{R}^{1,n}$ as one sheet of the quadric $$-x_0^2 + x_1^2 + \cdots + x_n^2 = -1,$$ again for unit radius of curvature.

Similarly, and again for unit radii of curvature, $n$-dimensional de Sitter spacetime is the universal covering space of the quadric $$-x_0^2 + x_1^2 + \cdots + x_n^2 = 1$$ in $\mathbb{R}^{1,n}$, whereas $n$-dimensional anti de Sitter spacetime is the universal covering space of the quadric $$-x_0^2 + x_1^2 + \cdots + x_{n-1}^2 - x_n^2 = -1.$$

This continues to be the case for other spaces of constant curvature in other signatures.

Other riemannian symmetric spaces, such as the grassmannians, can also admit isometric embeddings, this time in projective spaces, whose image is the intersection of a number of quadrics. This is the celebrated Plücker embedding. Notice that grassmannians do not (generally) have constant sectional curvature.

The remaining nontrivial lorentzian symmetric spaces -- the $n$-dimensional Cahen-Wallach spaces -- can also be locally embedded isometrically in $\mathbb{R}^{2,n}$ as the intersection of two quadrics. In particular this shows that all the indecomposable lorentzian symmetric spaces (in dimension $>1$, at least), which are the (anti) de Sitter and Cahen--Wallach spacetimes, can be locally embedded isometrically as the intersection of quadrics in some pseudoeuclidean space.

Question

Is this also the case for the other simply-connected (pseudo)riemannian symmetric spaces?

Perhaps asking about quadrics is too strong, so perhaps a weaker question is

Are simply-connected symmetric spaces always (locally) algebraic?

Here by locally algebraic I mean that they are the universal covering space of an algebraic space.

share|improve this question
1  
Can P(n), the space of n X n positive definite matrices, be expresed in this manner ? This is a "simple" example of a simply connected symmetric space. –  Suresh Venkat Apr 29 '10 at 12:23

3 Answers 3

According to a theorem by: Dirk Ferus: Symmetric submanifolds of Euclidean space, Math. Ann. 247, 81-93 (1980); the symmetric spaces that admit isometric embedding into Rn are the symmetric R-spaces: (see a reference in the following Wikipedia page), which are both symmetric spaces and real flag manifolds. These spaces consist of the Hermitian symmetric spaces and their non-compact duals. Here is a reference by: Jaurgen Berndt, describing these results.

share|improve this answer
    
Many thanks for this! This paper of Berndt's is useful. I'll have to ask him in person next time I see him. –  José Figueroa-O'Farrill Apr 29 '10 at 12:40
    
More the symmetric $R$-spaces are indeed intersections of quadrics being, as flag manifolds, projective highest weight orbits and so an intersection of quadrics by a result of Kostant. –  Fran Burstall Jun 21 '10 at 22:57
    
Fran, could you please elaborate? –  José Figueroa-O'Farrill Jun 26 '10 at 22:41

Not so much an answer to the original question as a very late (5 months—sorry: only just noticed!) response to José's request for more on symmetric $R$-spaces as intersections of quadrics.

(1) Flag manifolds (thus $G/P$ for $G$ semisimple and $P$ parabolic) arise in lots of ways as projective highest weight orbits: this is a generalisation of the Plücker embedding of Grassmannians, for example. If $V$ is an irrep of $G$ with highest weight $\lambda$ and ($1$-dimensional) highest weight space $L\in\mathbb{P}V$, the orbit $G\cdot L$ is a copy of $G/P$ for some parabolic $P$, and all $G/P$ arise this way for many $\lambda$.

(2) These orbits are intersections of quadrics: first note that $L^2$ is a highest weight space for the weight $2\lambda$ in $S^2 V$ so that the image under the Veronese map $\mathbb{P}V\to \mathbb{P}(S^2V)$ of $G\cdot L$ lies in $\mathbb{P}W$ where $W$ is the irreducible submodule of $S^2V$ with that highest weight, $W$ is called the Cartan square of $V$. Kostant's result (in J. Alg, I think) is that the converse is true: the image of the Veronese map intersects $\mathbb{P}W$ in exactly this orbit. Otherwise said, the flag manifold is cut out by quadratic equations $p(v^2)=0$ for $v\in V$ and $p:S^2 V\to S^2 V/W$ the natural projection.

(3) Example: The Grassmannian of $2$-planes in $\mathbb{C}^4$ is the highest weight orbit in $\mathbb{P}(\bigwedge^2\mathbb{C^4})$. Meanwhile $S^2\bigwedge^2\mathbb{C}^4=W\oplus\bigwedge^4\mathbb{C}^4=W\oplus\mathbb{C}$ so that the Grassmannian is cut out by one quadric equation and we see the celebrated identification with the Klein quadric.

share|improve this answer

For rank one riemannian symmetric spaces, I thinks that you can embbed them just like the real hyperbolic spaces (simply use hermitian products, and be careful when dealing with quaternions and, worse, octonions).

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.