Timeline for Nilpotent elements of Lie algebra and unipotent groups
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 22, 2019 at 1:01 | answer | added | Jim Humphreys | timeline score: 3 | |
| Jul 21, 2019 at 17:48 | history | became hot network question | |||
| Jul 21, 2019 at 13:15 | vote | accept | Jdoe | ||
| Jul 21, 2019 at 12:11 | answer | added | YCor | timeline score: 8 | |
| Jul 21, 2019 at 12:04 | comment | added | LSpice | To @YCor's point, I was assuming that your faithful representation of $\mathfrak g$ was the derivative of a representation of $G$. I agree that, without that assumption, it is not intrinsic. | |
| Jul 21, 2019 at 12:02 | comment | added | YCor | @Jdoe no it's obviously not intrinsic to the Lie algebra. Just take the 1-dimensional Lie algebra and $X$ nonzero. As Lie algebra of the 1-dimensional additive group $X$ is nilpotent (not semisimple), and as Lie algebra of the 1-dimensional multiplicative group $X$ is semisimple (not nilpotent). For $G$ semisimple however, it's intrinsic, and $X$ is nilpotent/semisimple iff $\mathrm{ad}(X)$ is. | |
| Jul 21, 2019 at 11:53 | comment | added | LSpice | As to the equivalence: suppose that $\rho(X)$ is nilpotent for some faithful representation, and let $X = X_{\text s} + X_{\text n}$ be the Jordan decomposition of $X$. Then $\rho(X) = \rho(X_{\text s}) + \rho(X_{\text n})$ is the Jordan decomposition of $X$ (Borel Corollary 4.4(2)), but $\rho(X)$ is nilpotent, so $\rho(X_{\text s}) = 0$, so $X_{\text s} = 0$ (by faithfulness), so $X = X_{\text n}$ is nilpotent. | |
| Jul 21, 2019 at 10:21 | answer | added | LSpice | timeline score: 4 | |
| Jul 21, 2019 at 9:56 | history | edited | Jdoe | CC BY-SA 4.0 |
edited body
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| Jul 21, 2019 at 9:55 | comment | added | Jdoe | It means that for any representation $V$ of $G$ with induced representation $\rho : \mathfrak{g} \to \mathfrak{gl}(V)$, $\rho(X)$ is nilpotent is the usual sense. But I imagine it is equivalent to the fact there exists a faithful representation of $\rho$ of $\mathfrak{g}$ such that $\rho(X)$ is nilpotent (and in particular it does not depend on $G$) but I'm not 100% sure. | |
| Jul 21, 2019 at 9:52 | comment | added | YCor | Could you say the definition of "nilpotent element" in the Lie algebra of a linear algebraic group? | |
| Jul 21, 2019 at 9:44 | history | edited | YCor | CC BY-SA 4.0 |
fixed typos
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| Jul 21, 2019 at 9:34 | history | asked | Jdoe | CC BY-SA 4.0 |