Timeline for Integrals of representations over geodesics
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 2, 2015 at 1:05 | vote | accept | Joonas Ilmavirta | ||
| Feb 26, 2015 at 20:45 | answer | added | Joonas Ilmavirta | timeline score: 1 | |
| Oct 9, 2014 at 7:02 | comment | added | Joonas Ilmavirta | @AliTaghavi, if you take any bi-invariant Riemannian metric on $G$, then the periodic geodesics passing through the identity are precisely the nontrivial homomorphisms $S^1\to G$. Every Lie subgroup is totally geodesic because of this algebraic structure. | |
| Oct 8, 2014 at 19:38 | comment | added | Ali Taghavi | @JoonasIlmavirta In particular, is every torus subgroup, totally geodesic? | |
| Oct 8, 2014 at 19:22 | comment | added | Ali Taghavi | @JoonasIlmavirta I am sorry if my question is trivial. Why the image of $\gamma$ is geodesic? | |
| Jul 21, 2014 at 3:03 | review | First posts | |||
| Jul 21, 2014 at 5:10 | |||||
| Jul 20, 2014 at 17:32 | comment | added | Joonas Ilmavirta | @Venkataramana True, but the second question can have an affirmative answer even for a faithful representation. If $G=SU(3)$ and $\rho$ is the tautological representation, the geodesic $\gamma(w)=\text{diag}(w,\bar w,1)$ gives $I(\rho,\gamma)=\text{diag}(0,0,1)$. Summing three matrices like this gives an invertible one. I don't know how general this phenomenon is. | |
| Jul 20, 2014 at 17:03 | comment | added | Venkataramana | If the representation is faithful, then there is no such geodesic. This follows from Figuera-O'Farrel's answer below, since the integral is zero unless the element $\rho (\gamma) $ is identity. | |
| Jul 20, 2014 at 16:09 | answer | added | José Figueroa-O'Farrill | timeline score: 3 | |
| Jul 20, 2014 at 14:07 | history | asked | Joonas Ilmavirta | CC BY-SA 3.0 |