Do symmetric spaces admit isometric embeddings as intersections of quadrics? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T17:28:03Z http://mathoverflow.net/feeds/question/22859 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/22859/do-symmetric-spaces-admit-isometric-embeddings-as-intersections-of-quadrics Do symmetric spaces admit isometric embeddings as intersections of quadrics? José Figueroa-O'Farrill 2010-04-28T14:44:41Z 2010-11-22T00:05:02Z <p>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.</p> <p><strong>Background</strong></p> <p>Recall that a complete, connected and simply connected pseudoriemannian manifold $(M,g)$ is a <em>symmetric space</em> if the Riemann curvature tensor is parallel with respect to the Levi-Civita connection: $\nabla R = 0$.</p> <p>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 $$\sum_{i=1}^p (dx_i)^2 - \sum_{i=1}^q (dx_{p+i})^2.$$</p> <p>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.</p> <p>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 $$-x_0^2 + x_1^2 + \cdots + x_{n-1}^2 - x_n^2 = -1.$$</p> <p>This continues to be the case for other spaces of constant curvature in other signatures.</p> <p>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.</p> <p>The remaining nontrivial lorentzian symmetric spaces -- the $n$-dimensional <a href="http://www.ams.org/mathscinet-getitem?mr=267500" rel="nofollow">Cahen-Wallach spaces</a> -- 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.</p> <p><strong>Question</strong></p> <blockquote> <p>Is this also the case for the other simply-connected (pseudo)riemannian symmetric spaces?</p> </blockquote> <p>Perhaps asking about quadrics is too strong, so perhaps a weaker question is</p> <blockquote> <p>Are simply-connected symmetric spaces always (locally) algebraic?</p> </blockquote> <p>Here by locally algebraic I mean that they are the universal covering space of an algebraic space.</p> http://mathoverflow.net/questions/22859/do-symmetric-spaces-admit-isometric-embeddings-as-intersections-of-quadrics/22878#22878 Answer by Benoît Kloeckner for Do symmetric spaces admit isometric embeddings as intersections of quadrics? Benoît Kloeckner 2010-04-28T16:54:51Z 2010-04-28T16:54:51Z <p>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).</p> http://mathoverflow.net/questions/22859/do-symmetric-spaces-admit-isometric-embeddings-as-intersections-of-quadrics/22968#22968 Answer by David Bar Moshe for Do symmetric spaces admit isometric embeddings as intersections of quadrics? David Bar Moshe 2010-04-29T12:08:20Z 2010-04-29T12:08:20Z <p>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 <a href="http://en.wikipedia.org/wiki/Generalized_flag_variety" rel="nofollow">page</a>), 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 <a href="http://www.mth.kcl.ac.uk/~berndt/taegu02.pdf" rel="nofollow">reference</a> by: Jaurgen Berndt, describing these results.</p> http://mathoverflow.net/questions/22859/do-symmetric-spaces-admit-isometric-embeddings-as-intersections-of-quadrics/46881#46881 Answer by Fran Burstall for Do symmetric spaces admit isometric embeddings as intersections of quadrics? Fran Burstall 2010-11-22T00:05:02Z 2010-11-22T00:05:02Z <p>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.</p> <p>(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 $\lambda$.</p> <p>(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.</p> <p>(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.</p>