Timeline for Canonical Metric on Grassmann Manifold
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 15, 2015 at 10:46 | comment | added | Troy Woo | What if I drop the homogeneity condition? i.e. if I talk about invariant metric, but not invariant Finsler metric? Is it possible to characterize such metrics then? | |
| May 15, 2015 at 10:23 | comment | added | Peter Michor | What I said about Finsler was wrong, I forgot that they have to be positive homogeneous. Now my guess is that it is unique (up to a positive constant) and that it is Riemannian. That should follow from invariant theory applied to positively homogeneous continuous functions, in the setting of Verbitzky's answer. | |
| May 15, 2015 at 10:08 | comment | added | Troy Woo | Oh, by the way, is it possible to characterize all invariant Finsler metrics in some way? | |
| May 15, 2015 at 8:54 | comment | added | Peter Michor | Misha Verbitsky's answer gives you a proof in 3 lines. Helgason: "Differential Geometry, Symmetric Spaces, and Lie groups" is a standard reference, but by far too much for this simple question. As a Finsler metric it is not unique: any power of the Riemannian norm is an invariant Finsler metric. | |
| May 15, 2015 at 8:38 | comment | added | Troy Woo | Great!...do you mind providing a reference? i.e. uniqueness of invariant Finsler metric...I'm kinda slow...I am an engineer... | |
| May 15, 2015 at 4:39 | comment | added | Peter Michor | Since it is an irreducible symmetric space of compact type, there is not a lot of choice: any invariant metric is a constant multiple of this one. | |
| May 14, 2015 at 21:53 | comment | added | Troy Woo | So is there any exception to what Yurii has shown? Should any invariant metric on the grassmannian look like a symmetric norm function of Jordan angles? | |
| Aug 30, 2014 at 7:57 | history | answered | Peter Michor | CC BY-SA 3.0 |