Timeline for Which elements lie in a Cartan subalgebra?
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
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| Jul 10 at 12:50 | comment | added | Ido Grayevsky | Thank you very much! | |
| Jul 10 at 5:27 | history | edited | Ido Grayevsky | CC BY-SA 4.0 |
I removed a remark (about Cartan subalgebras coming from regular) elements that was evidently false.
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| Jul 10 at 0:04 | comment | added | YCor | @LSpice the given class of Lie algebra includes all Lie algebras of solvable real linear algebraic groups whose maximal tori are split, and more. In this case the exponential radical is contained in the Lie algebra of the unipotent radical, but the inclusion is often strict. | |
| Jul 9 at 23:58 | comment | added | LSpice | @YCor, re, sorry, by "nilpotent radical" I meant "Lie algebra of the unipotent radical" (a reasonably common abuse of language in my area). Anyway I realise I misread the question; I thought it asked for $\mathfrak g$ to be the full Lie algebra of upper triangular matrices, and so was trying to guess what other kinds of Lie algebra might be considered. | |
| Jul 9 at 23:43 | comment | added | YCor | @LSpice $\mathfrak{r}$ is usually much smaller than the nilpotent radical. For instance, if $\mathfrak{g}$ is nilpotent, then the exponential radical is zero (while the nilpotent radical is the whole algebra), and $\mathfrak{g}$ itself is its unique Cartan subalgebra. | |
| Jul 9 at 23:42 | history | edited | YCor | CC BY-SA 4.0 |
removed capitals from title
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| Jul 9 at 15:22 | comment | added | LSpice | For example, the "$ax + b$ Lie algebra" $\operatorname{Add} \ltimes \operatorname{Add}$ has this property. If $s_1$ and $s_2$ are linearly independent, commuting, semisimple elements of $\mathfrak g \setminus \mathfrak r$, and $X$ is a simultaneous weight vector for them in $\mathfrak r$, then (because the weights are non-$0$) some linear combination of $s_1$ and $s_2$ annihilates $X$, which is a contradiction. So $\mathfrak g/\mathfrak r$ is $1$-dimensional, or $\mathfrak r$ is $0$. | |
| Jul 9 at 15:05 | comment | added | LSpice | It's not clear what you mean by a necessary and sufficient condition on $\newcommand\f{\mathfrak}\f g$ for your statement to hold, since $\f g$ has been fixed and, as @YCor points out, your statement is not true. However, perhaps you mean for $\f g$ now to be a general algebraic Lie algebra whose CSA are (Lie alg. of) tori, and $\f r$ its nilpotent radical. If $\f g/\f r$ is not a torus, then you can lift a non-semisimple element to $\f g \setminus \f r$; it will not belong in a CSA. If $\f g/\f r$ is a torus, then you need every ss element of $\f g \setminus \f r$ to act freely on $\f r$. | |
| Jul 9 at 14:53 | history | edited | LSpice | CC BY-SA 4.0 |
$\textit{regular}$ -> *regular*
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| Jul 9 at 8:33 | comment | added | YCor | Your last statement is not true. If $J$ is a Jordan block of maximal rank, its centralizer has dimension $n$ (the ambient dimension), which equals the rank, but this is not a Cartan subalgebra. | |
| Jul 9 at 8:32 | comment | added | YCor | No. In the example of upper triangular matrices, elements of Cartan subalgebras are diagonalizable matrices. Elements of the exponential radical are nilpotent matrices. So, pick any (upper triangular) matrix that is neither nilpotent nor diagonalizable. | |
| S Jul 9 at 8:26 | review | First questions | |||
| Jul 9 at 9:00 | |||||
| S Jul 9 at 8:26 | history | asked | Ido Grayevsky | CC BY-SA 4.0 |