Timeline for Which Lie groups have adjoint representations that are bounded away from zero?
Current License: CC BY-SA 3.0
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| Mar 27, 2014 at 19:33 | review | Suggested edits | |||
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| Aug 21, 2013 at 10:37 | comment | added | José Figueroa-O'Farrill | @JochenTrumpf: I was simply trying to point out that unimodularity is necessary. Connectedness shows that it is also sufficient. | |
| Aug 21, 2013 at 1:15 | comment | added | Jochen Trumpf | Does this not require the Lie group to be connected (and have a unimodular Lie algebra)? The proof idea given by @Jose seems to require that a group element can be written as a finite product of exponentials. | |
| Aug 20, 2013 at 2:22 | comment | added | Jochen Trumpf | Edited the question to spell out what I mean by "bounded away from zero". Thanks already for the reductive idea. Still evaluating David's answer above. | |
| Aug 19, 2013 at 20:07 | comment | added | user1688 | @Jose: Oh yes right, I was thinking of reductive groups. Sorry. | |
| Aug 19, 2013 at 19:24 | comment | added | Ben McKay | Which orbit of the adjoint representation are we taking? Are we asking that some orbit doesn't have zero in its closure? | |
| Aug 19, 2013 at 18:41 | comment | added | José Figueroa-O'Farrill | Is this not only the case for the Lie groups with unimodular Lie algebras? $1 = \det \operatorname{Ad}(\exp X) = \det \exp \operatorname{ad}(X) = e^{\operatorname{tr} \operatorname{ad}(X)}$, hence $\operatorname{ad}(X)$ must be traceless for all $X$. Unless I'm missing something, this is not the case for every Lie algebra. | |
| Aug 19, 2013 at 14:07 | history | answered | user1688 | CC BY-SA 3.0 |