# Ricci Curvature on Grassmannian

Suppose $$G_r(n)$$ is the Grassmannian, which is the collection of all $$r$$ dimensional subspace in $$\mathbb{R}^{n}$$ equipped with the usual invariant metric. Let $$Ricc(G_r(n))$$ be the Ricci curvature tensor. What are the best known constants $$0 such that $$c_{n,r}\leq Ricc(G_r(n))\leq C_{n,r}$$? I can only find results about complex Grassmannian, but not for real Grassmannian.

Both the real and complex Grassmannians are compact irreducible symmetric spaces, and therefore are Einstein with positive Einstein constant. (Thus $$c_{n,r}=C_{n,r}$$, and this constant can be varied by rescaling). A reference is Besse, Einstein manifolds, paragraph 7.75.
• Thank you! However, I'm more interested in how this constant depends on $n$ and $r$ modulo scaling. Is there any reference? – neverevernever May 3 at 20:55
• If $g_1$ and $g_2:=\lambda g_1$ are two invariant metrics on the Grassmannian $G_r(n)$, differing by scaling, then the corresponding constants satisfy $c^1_{n,r}g_1=\operatorname{Ric}(g_1)=\operatorname{Ric}(g_2)=c^2_{n,r}g_2$, so $c^1_{n,r}=\lambda c^2_{n,r}$. Is that your question? – macbeth May 3 at 21:45
• Indeed, one can ask what the Einstein constant is if the metric is scaled to have volume 1 (I believe this is what @RyanBudney means). This is the same as asking what $V^{2/d}$ is, where $V$ is the volume and $d=r(n-r)$ the dimension of $G_r(n)$, if the metric is scaled to have Einstein constant 1. – macbeth May 7 at 14:06
• This can be computed as follows. I believe $V=\operatorname{Vol}(SO_n)/(\operatorname{Vol}(SO_r)\operatorname{Vol}(SO_{n-r}))$, where the three Lie group volumes are taken by considering the Killing form as the metric on each one. Now, there are standard (though complicated) formulas for the volumes of Lie groups with respect to the Killing form metric: see Macdonald, 1980, The volume of a compact Lie group. – macbeth May 7 at 14:12