# Ricci curvature of the symplectic group

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.

• I'm assuming that you are referring to $Sp(n)$ with a bi-invariant metric. How are you normalising the metric? Unit volume? – José Figueroa-O'Farrill Jun 10 '11 at 15:35
• José: Yes, I want a bi-invariant metric. Being careful about normalizations has always been a weak suit of mine, but I believe unit volume is the normalization I want here. (Bad answer: I want whichever kind normalization makes it true that the curvature of $O(n)$ is $(n-1)/4$.) – Mark Meckes Jun 10 '11 at 15:52
• Actually I think I want the normalization induced by the standard embedding in a Euclidean space... – Mark Meckes Jun 10 '11 at 23:43

## 2 Answers

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.

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 $Ric(X,X)\geq\kappa||X||^2$.

• I want a lower bound $c/n$ where $c$ is independent of $n$ (I'll edit the question to clarify that). Unless I'm missing something this argument isn't strong enough to yield that. – Mark Meckes Jun 10 '11 at 15:20
• Sorry, that should have been $cn$. – Mark Meckes Jun 10 '11 at 15:45
• The explicit formula cited by Claudio is exactly the formula I remembered all too well from Cheeger-Ebin. I believe the constant $c$ is equal to $\frac{1}{4}$. But at this point I don't see how to proceed except to use explicit formulas for a basis of the Lie algebra. But this seems straightforward to me. – Deane Yang Jun 10 '11 at 18:02
• If you use the negative of the Killing form $\beta$, then $Ric(X,X)=-\frac14 trace ad_X^2 =-\frac14 \beta(X,X) =-\frac12 \sum_{\alpha\in\Delta^+}\alpha(X)^2$ using the real root decomposition and assuming $X$ lies in the Cartan subalgebra. For a fixed type, say $C_n$ in the case you are asking, I think you can figure out the dependence on $n$. The positive root system of $C_n$ is $2\theta_i$, $\theta_i\pm\theta_j$. You also need to keep track of the normalization of the metric. – Claudio Gorodski Jun 10 '11 at 19:37
• It seems to me that if you work things out for $O(n)$, you are almost done, since any other compact group is a subgroup. So if you know what $[X,Y]$ is for any two elements $X$ and $Y$ of an orthonormal basis of $o(n)$, then you can deduce information about the sectional and therefore Ricci curvature of any Lie subgroup of $O(n)$. – Deane Yang Jun 11 '11 at 9:07