Ricci curvature of the symplectic group - MathOverflow

Ricci curvature of the symplectic group

2011-06-10T15:01:44Z

Is the Ricci curvature of the compact symplectic group $Sp(n)$ bounded below by $cn$ for some constant $c > 0$ independent of $n$?

For $O(n)$ and $U(n)$ I know many references which state such a bound on Ricci curvature, although none of them actually include complete proofs. (Pointers to such proofs for $O(n)$ and $U(n)$ would also be appreciated.) The usual practice for these groups is merely to refer to Cheeger and Ebin's book, which develops enough general theory of curvature of Lie groups that carrying out the calculations is presumably a straightforward exercise, for those who are on top of such things. (As far as I can see, Cheeger and Ebin don't even state the results in these particular cases.) It's been about ten years since I've been up on such things, which is why I'm hoping someone here knows the answer instead of just trying to work it out myself.

Answer by Claudio Gorodski for Ricci curvature of the symplectic group

2011-06-10T15:11:43Z

Compact Lie groups with bi-invariant metric have nonnegative sectional curvature. In fact, there is an explicit formula $K(X,Y)=c\cdot ||[X,Y]||^2$ for some positive constant $c$ and orthonormal $X$, $Y\in\mathfrak g$. It follows that the Ricci curvature $Ric(X,X)=c \cdot \sum_i ||[X,E_i]||$ where $(E_i)$ is an orthonormal basis of $\mathfrak g$ containing $X$. From this formula you can see that $Ric(X,X)\geq0$ and $Ric(X,X)=0$ if and only if $X$ lies in the center of $\mathfrak g$.

In particular, if $G$ is semisimple, the center of $\mathfrak g$ is zero and $Ric(X,X)>0$ for $X\neq0$. By compactness, you can find a constant $\kappa$ such that $Ric(X,X)\geq\kappa||X||^2$.

Answer by Mark Meckes for Ricci curvature of the symplectic group

2011-06-16T14:02:53Z

The groups $SO(n)$, $SU(n)$, and $Sp(n)$ all have Ricci tensor equal to a constant times the metric tensor. (Note: contrary to what I wrote in the question and what one may find stated in several places in the literature, this is false for $U(n)$. This is easy to see from Claudio's answer: if a Lie group has nontrivial center then its Ricci tensor cannot be nondegenerate.)

With the normalization induced by the standard embedding in $\mathbb{R}^{\beta n^2}$ (where $\beta = 1,2,4$ in the three cases above, respectively), the constant is $$ \frac{\beta(n+2)}{4} - 1. $$ Reference: Appendix F of An Introduction to Random Matrices by Anderson, Guionnet, and Zeitouni.