4 ms·
Thanks, I just realized I was also assuming exp to be surjective as well, I stand corrected. The Wikipedia page[1] mentions the issue briefly but I was curious
by chombier 4y ago
Thanks, I just realized I was also assuming exp to be surjective as well, I stand corrected.
The Wikipedia page[1] mentions the issue briefly but I was curious of the counter-examples. IIUC, matrices in SL(2) with trace < -2 have two distinct eigenvalues, one of which is negative[2], and such matrices cannot be reached by exponentiating elements of the Lie algebra sl(2) (traceless matrices).
[1] https://en.wikipedia.org/wiki/Exponential_map_(Lie_theory)#Surjectivity_of_the_exponential https://en.wikipedia.org/wiki/Exponential_map_(Lie_theory)#S...
[2] https://en.wikipedia.org/wiki/SL2(R)#Classification_of_elements https://en.wikipedia.org/wiki/SL2(R)#Classification_of_eleme...
- chombier 4y agoI found this nice accessible proof for the interested: https://planetmath.org/slnrisconnected https://planetmath.org/slnrisconnected
- andi999 4y agoIs this part of the proposition: x=exp X, and x having a double eigenvalue implying that X has a double eigenvalue somehow clear? How do you prove it?