Timeline for volume of compact simple Lie groups under the natural Euclidean embedding
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 10, 2017 at 4:18 | comment | added | John Jiang | @J A S O N: I think you can see it from the Lie algebra $so(n)$, which spans an $n(n-1)/2$ dimensional subspace of $R^{n^2}$. The differential of the submersion corresponds to the projection of $so(n)$ to the first column vector space. Vertical vectors of the said bundle are thus those vectors in $so(n)$ that get mapped to 0, and horizontal ones are orthogonal to the vertical ones, so must be of the form $e_i \wedge e_1$. Now these have length $\sqrt{2}$ each under Euclidean metric of the ambient space. But they get mapped to $e_i$ by $d\pi$, which have length $1$ each! | |
| Jan 9, 2017 at 9:09 | comment | added | JSCB | @RobertBryant: I have been failing to compute the volume of $SO(n)$, and I just came across your comment. I do not understand how the scaling factor $\sqrt 2$ come out from the differential of $\pi$, can you please explain more explicitly? Thank you very much. | |
| Aug 12, 2013 at 10:54 | answer | added | Jim Humphreys | timeline score: 6 | |
| Aug 12, 2013 at 4:31 | answer | added | Peter Forrester | timeline score: 4 | |
| Jan 4, 2012 at 8:48 | answer | added | David Bar Moshe | timeline score: 5 | |
| Jan 4, 2012 at 2:55 | comment | added | John Jiang | By generalized Jacobian map I mean the determinant of the Gram matrix formed by the rows of $d\pi$. | |
| Jan 4, 2012 at 2:54 | comment | added | John Jiang | I guess in general even when $\pi$ is not a Riemannian submersion one could integrate the inverse of the generalized jacobian map along the preimage of $\pi$, and then integrate that over the base manifold? | |
| Jan 4, 2012 at 2:21 | comment | added | Robert Bryant | @Tom: You are quite right. I forgot about the factor of $(\sqrt{2})^{k-1}$ that you have to put in at each level because the natural map $\pi:SO(k)\to S^{k-1}$ is not a Riemannian submersion, but is a Riemannian submersion scaled by a factor of $1/\sqrt{2}$, i.e., each horizontal vector for this bundle is shrunk by a factor of $\sqrt{2}$ by the differential of $\pi$. Thus, the overall factor you need to multiply the answer I gave by is $(\sqrt{2})^{n(n-1)/2}$. The recipe I gave for $U(n)$ is not right either, as the map $\pi:U(k)\to S^{2k-1}$ shrinks horizontal volumes by $2^{k-1}$. Sorry. | |
| Jan 4, 2012 at 1:38 | comment | added | Tom Goodwillie | This is not quite true: $SO(2)$ as a circle in the vector space of $2\times 2$ matrices has radius $\sqrt 2$, in the metric that I am thinking of. | |
| Jan 4, 2012 at 1:29 | comment | added | John Jiang | Oh you are right. Thanks for the observation! | |
| Jan 4, 2012 at 1:26 | comment | added | Robert Bryant | For the orthogonal group $SO(n)$, isn't it just the product of the volumes of the $k$-spheres in their natural embeddings, as $k$ ranges from 1 to $n{-}1$? Similarly, for $U(n)$, it should be the product of the $2k{-}1$-spheres as $k$ ranges from $1$ to $n$, and so on. | |
| Jan 4, 2012 at 1:01 | history | asked | John Jiang | CC BY-SA 3.0 |