Sign up ×

When $F = \mathbb{R}, \mathbb{C}$ or $\mathbb{H}$, there are fibrations $$O(k,F)\rightarrow V_k(F^n)\rightarrow G_k(F^n)$$ where $V_k(F^n)$ are Steifel manifolds and $G_k(F^n)$ are Grassmannians. When $k=1$ these reduce to the Hopf fibrations $$S^{d-1}\rightarrow S^{dn-1}\rightarrow FP^{n-1}$$ where $d=\text{dim}F$.

If $S^{dn-1}$ are given round metrics, people make a big deal about the metrics on $FP^{n-1}$ that make the above Riemannian submersions.

You can also describe the metrics on $FP^{n-1}$ with distances between subsets of the round sphere $S^{dn-1}$ that correspond to lines through the origin in $F^n$. This, from what I've read (and agree with), is also the natural metric to put on $G_k(F^n)$ where "lines" has been replaced by "$k$-planes".

I'm being tricked into believing that there should be some nice metric on $V_k(F^n)$ that makes the map $V_k(F^n)\rightarrow G_k(F^n)$ Riemannian and that I must not have been listening when people were making an equally big deal about these metrics on $G_k(F^n)$ as they do for $FP^n$.

Question: What goes wrong? If nothing, why is this not as interesting as the geometry of the Hopf fibrations? If equally interesting, where is a good source to learn more?

share|cite|improve this question

2 Answers 2

up vote 2 down vote accepted

You are right, this is true. You start with the invariant metric on $O(k,F)$. Then the relevant subgroups act isometrically, thus you get Riemannian submersions. The Grassmann manifolds are symmetric spaces, but the Stiefel manifolds are only homogeneous. A very good source which also gives explicitly all geodesics and even the geodesic distance on Grassmannians is:

  • Y. A. Neretin, On jordan angles and the triangle inequality in grassmann manifold, Geometriae Dedicata, 86 (2001).
share|cite|improve this answer
Thank you. I didn't know of this paper. It was interesting and helpful. – horse with no name May 8 '13 at 13:47

Apriori you don't get a Riemannian submersion because the corresponding subsets of the unit spheres in the $k$-planes will not be "constant distance apart" as they are in the symmetric rank 1 case.

share|cite|improve this answer

Your Answer


