# Ricci curvature of the Grassmannian?

Let $G(k, \mathbb{C}^n)$ be the Grassmannian of $k-$dimensional complex linear subspaces of $\mathbb{C}^n.$ We know that the Grassmannian can be embedded to the projective space $(\mathbb{P}^N,\omega_{FS})$ for some $N,$ via the Plucker embedding.

What can we say about the Ricci curvature of the Grassmannian? or even the scalar curvature? (any bounds?)

• The Grassmanian is diffeomorphic to the homogeneous space $U(n)/U(n-k) \times U(k)$. I believe the isotropy representation is irredicuble, and if true, this implies the Grassmanian has a unique homogeneous metric, up to rescaling. So, the question then becomes: is the induced metric on $G(k,\mathbb{C}^n)$ embedded into $\mathbb{P}^N$ homogeneous? (I don't know enough about the Plucker embedding to answer the question). – Jason DeVito Apr 12 '15 at 1:49
• @Joe: As with your question about the diameter, the answer depends on a choice of metric, including overall scaling. (As Jason DeVito notes, the Grassmannian admits a unique homogeneous Kähler metric, which is necessarily Einstein with positive Ricci curvature. The value of the Einstein constant, or of the scalar curvature, depends on the scaling.) – Andrew D. Hwang May 21 '15 at 22:54

The Plücker embedding is an isometry of $G(k,\Bbb C^n)$ to its image in $\Bbb P(\Lambda^k\Bbb C^n)$ with the standard Fubini-Study metric.
In the moving frames notation, for example, the Kähler form on $\Bbb P^N$ is given by $$\frac i2\sum_{j=1}^N \omega_{0\bar j}\wedge\overline\omega_{0\bar j},$$ where $\{f_0;f_1,\dots,f_N\}$ is a unitary frame at the point $[e_0]\in\Bbb P^N$, and $\omega_{0\bar j} = \langle de_0,e_j\rangle$. (That is, $\omega_{0\bar j}$ give an invariant unitary basis for the $(1,0)$ cotangent bundle of $\Bbb P^N$.)
Now, using adapted unitary frames $\{e_\alpha;e_\mu\}$ ($\alpha=1,\dots,k$, $\mu=k+1,\dots,n$) on $G(k,\Bbb C^n)$, the Plücker embedding is given by the mapping $[e_1\wedge\dots\wedge e_k]$, so, taking $f_0=e_1\wedge\dots\wedge e_k$ and differentiating, we see that $$df_0 = \sum_{\alpha,\mu} \omega_{\alpha\bar\mu}e_1\wedge\dots\wedge e_\mu\wedge\dots\wedge e_k,$$ with $e_\mu$ in the $e_\alpha$ slot. That is, the $\omega_{\alpha\bar\mu}$ give the pullback of the unitary coframe on $\Bbb P^N$ ($N=\binom nk$). But they are also the invariant unitary coframe on the hermitian symmetric space $U(n)/\big(U(k)\times U(n-k)\big)$.