Timeline for Injectivity of Lie group exponential function
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 8, 2014 at 4:09 | comment | added | Oliver Jones | If you give your Lie group a metric then the exponential will be injective outside the tangential cut locus. These have been calculated for the matrix Lie groups by Sakai. I think it should be rather straightforward to see what's going on in the case of $U(2)$. | |
| Jan 28, 2014 at 7:54 | comment | added | YCor | I don't expect anything special, it's just a test case: most likely $SL_2$ is easier, and if the result fails for $SL_2$ then it will certainly fail for $SL_n$. | |
| Jan 28, 2014 at 7:25 | comment | added | Christoph Wockel | I tried $SL_n(\mathbb{R})$ in general, but didn't come to any conclusion. Do you think there is something special about $SL_2(\mathbb{R})$? | |
| Jan 26, 2014 at 21:55 | comment | added | YCor | OK. Have you tried $SL_2(\mathbf{R})$? | |
| Jan 25, 2014 at 8:09 | history | edited | Christoph Wockel | CC BY-SA 3.0 |
added 22 characters in body
|
| Jan 25, 2014 at 8:00 | comment | added | Christoph Wockel | $G$ is any semi-simple finite-dimensional Lie group. $\exp$ is a diffeomorphism onto a closed submanifold of $G$ (in the case of $SL_n(\mathbb{R})$ we have that $\exp$ maps symmetric matrices to positive definite symmetric matices). | |
| Jan 24, 2014 at 22:25 | comment | added | YCor | In your question, what is $G$? any finite-dimensional Lie group? Any semisimple? $SL_n(\mathbb{R})$? Also I don't understand what you mean by "the restriction to $\mathfrak{p}$ is a diffeomorphism", since it is certainly not surjective. Do you mean injective? immersion? closed immersion?... | |
| Jan 24, 2014 at 21:19 | history | edited | Christoph Wockel | CC BY-SA 3.0 |
added 10 characters in body
|
| Jan 24, 2014 at 16:25 | history | asked | Christoph Wockel | CC BY-SA 3.0 |