Timeline for Homogeneous Spaces and Equivariant Hodge Maps
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 24, 2013 at 13:10 | vote | accept | Mihail Matrix | ||
| Apr 23, 2013 at 10:34 | vote | accept | Mihail Matrix | ||
| Apr 23, 2013 at 10:35 | |||||
| Apr 23, 2013 at 10:33 | comment | added | Mihail Matrix | Great again - Danke! | |
| Apr 23, 2013 at 5:55 | history | edited | Peter Michor | CC BY-SA 3.0 |
added 947 characters in body
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| Apr 22, 2013 at 14:49 | comment | added | Mihail Matrix | @Peter: Great, thanks for the answer. But I am correct in recalling the result $\ast(\phi^k) = g^{-1}(\phi^k,$vol)? | |
| Apr 21, 2013 at 19:10 | comment | added | Robert Bryant | Of course, everything would be fine if we assumed that $G$ preserves a metric and an orientation on $G/H$. | |
| Apr 21, 2013 at 19:07 | comment | added | Robert Bryant | Actually, this is not quite true. Just because the metric is $G$-invariant, it does not follow that there is a $G$-invariant volume form. For example, if $G = \mathrm{O}(n)$ and $H$ is the subgroup that fixes a nonzero vector $v\in\mathrm{R}^n$, then $G/H$ is $S^{n-1}$, and there is a (unique up to constant multiples) $G$-invariant metric on $G/H$, but $G$ does not preserve an orientation on $G/H$, so there is no $G$-invariant volume form. | |
| Apr 21, 2013 at 18:39 | history | answered | Peter Michor | CC BY-SA 3.0 |