Diameter of the Grassmannian Just an interesting question that came to my mind while studying(!): Since the Grassmannian $G(k,\mathbb{C}^n)$ is a compact manifold, what do we know about its diameter? Do we know any estimate?
Thank you.
 A: Not a complete answer, but here's the situation for projective space.
On complex projective space, the "obvious" metric is Fubini-Study. To determine the diameter, however, you have to specify a scaling, and three "interesting" scalings come to mind:


*

*An algebraic geometer might scale so that an embedded line has unit area, i.e., the Kähler form generates integral cohomology; in this event, the diameter is $\sqrt{\pi}/2$, same as for a sphere of unit area. (Nifty factlet: For this metric on the projective plane, a smooth curve of degree $n$ has area $n$ by Bézout's theorem and Poincaré duality.)

*A complex  differential geometer might pick the metric of unit holomorphic sectional curvature; in this event, the diameter is $\pi$, same as for a sphere of unit curvature, consequently of area $4\pi$ and "Euclidean radius" equal to unity.

*A Riemannian geometer might choose the scaling for which the Hopf map from the unit $(2n+1)$-sphere is a Riemannian submersion. Since the unit $(2n+1)$-sphere has volume $2\pi(\pi^{n}/n!)$ and a great circle has length $2\pi$, the resulting scaled Fubini-Study metric has volume $\pi^{n}/n!$; each projective line therefore has area $\pi$, curvature $4$, and the diameter is $\pi/2$. (Nifty factlet: For this metric on $\mathbf{CP}^{n}$, the volume is the same as the volume of the unit Euclidean $2n$-ball, which constitutes the top-dimensional cell if the boundary is attached to $\mathbf{CP}^{n-1}$ by the Hopf map. As far as I know, this is a numerological coincidence.)
