Timeline for Ricci curvature of the symplectic group
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 11, 2011 at 18:41 | comment | added | Mark Meckes | @Deane: Yes, I was coming to that conclusion. | |
| Jun 11, 2011 at 9:07 | comment | added | Deane Yang | It seems to me that if you work things out for $O(n)$, you are almost done, since any other compact group is a subgroup. So if you know what $[X,Y]$ is for any two elements $X$ and $Y$ of an orthonormal basis of $o(n)$, then you can deduce information about the sectional and therefore Ricci curvature of any Lie subgroup of $O(n)$. | |
| Jun 10, 2011 at 23:46 | comment | added | Mark Meckes | Like I said, it's been about ten years since I've thought about any of the words you just used. But it's clear that you've given more than enough information that I could grind it out once I remember what they all mean. | |
| Jun 10, 2011 at 19:37 | comment | added | Claudio Gorodski | If you use the negative of the Killing form $\beta$, then $Ric(X,X)=-\frac14 trace ad_X^2 =-\frac14 \beta(X,X) =-\frac12 \sum_{\alpha\in\Delta^+}\alpha(X)^2$ using the real root decomposition and assuming $X$ lies in the Cartan subalgebra. For a fixed type, say $C_n$ in the case you are asking, I think you can figure out the dependence on $n$. The positive root system of $C_n$ is $2\theta_i$, $\theta_i\pm\theta_j$. You also need to keep track of the normalization of the metric. | |
| Jun 10, 2011 at 18:02 | comment | added | Deane Yang | The explicit formula cited by Claudio is exactly the formula I remembered all too well from Cheeger-Ebin. I believe the constant $c$ is equal to $\frac{1}{4}$. But at this point I don't see how to proceed except to use explicit formulas for a basis of the Lie algebra. But this seems straightforward to me. | |
| Jun 10, 2011 at 15:45 | comment | added | Mark Meckes | Sorry, that should have been $cn$. | |
| Jun 10, 2011 at 15:20 | comment | added | Mark Meckes | I want a lower bound $c/n$ where $c$ is independent of $n$ (I'll edit the question to clarify that). Unless I'm missing something this argument isn't strong enough to yield that. | |
| Jun 10, 2011 at 15:11 | history | answered | Claudio Gorodski | CC BY-SA 3.0 |