Timeline for Geodesics on a Grassmannian
Current License: CC BY-SA 2.5
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 4, 2009 at 8:10 | comment | added | Mariano Suárez-Álvarez | Urgh. Yes, you are right, of course. | |
| Dec 4, 2009 at 8:07 | history | edited | Greg Kuperberg | CC BY-SA 2.5 |
Extended answer
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| Dec 4, 2009 at 8:01 | comment | added | Greg Kuperberg | You have to work harder than that to find the planes. They come from the singular value decomposition of the projection from $V$ to $W$ (or vice-versa). | |
| Dec 4, 2009 at 7:56 | comment | added | Mariano Suárez-Álvarez | Indeed. And looking at it this way it is easy to construct the geodesics: you need only construct the 2-planes $P_1$, ..., $P_k$. You do this as follows: pick a non-zero vector $v_1$ in $V$, and one $w_1$ in $W$, and let $P_1=\langle v_1,w_1\rangle$; next pick $v_2$ in $V$ and $w_2$ in $W$ orthogonal to $P_1$, and put $P_2=\langle v_2,w_2\rangle$, and and so on. | |
| Dec 4, 2009 at 7:41 | history | answered | Greg Kuperberg | CC BY-SA 2.5 |