Timeline for Generating a reductive real Lie group with finitely many maximal real tori
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 25, 2012 at 18:55 | comment | added | algori | .. on the other hand, I'm wondering if this condition already implies that the group is reductive. | |
| Jan 25, 2012 at 18:12 | comment | added | algori | Jim -- yes, of course! On the other hand, it would suffice to assume there exist an element with all eigenvalues distinct and $\neq 1$. | |
| Jan 25, 2012 at 13:45 | vote | accept | Hugo Chapdelaine | ||
| Jan 25, 2012 at 13:45 | comment | added | Hugo Chapdelaine | So I solved my confusion, so if one takes the one-parameter subgroup $\left(\begin{array}{cc} a & 0 \newline 0 & 1\end{array}\right)$ one might simply conjugate it by an element of the form $\left(\begin{array}{cc} 1 & b \newline 0 & 1\end{array}\right)$ | |
| Jan 24, 2012 at 21:41 | comment | added | Hugo Chapdelaine | But then I don't quite see how you can generate this group with algebraic one-parameter subgroups "generated" by diagonalizable elements... | |
| Jan 24, 2012 at 21:40 | comment | added | Jim Humphreys | @algori: It's safest to stay with reductive groups, since you might have a direct product of a reductive and a unipotent group in which semisimple elements fail to be dense. | |
| Jan 24, 2012 at 21:33 | comment | added | Hugo Chapdelaine | So I guess that the "simplest" example of a non-reductive real algebraic that contains a semi-simple element is the set of invertible matrices of form $\left( \begin{array}{cc} a & b \newline 0 & 1 \end{array} \right) $ | |
| Jan 24, 2012 at 21:20 | comment | added | algori | Hugo -- no, I'm just assuming that $G$ is algebraic and contains a non-trivial diagonalizable element. In this case non-diagonalizable elements are contained in an algebraic subset $\neq G$. | |
| Jan 24, 2012 at 21:13 | comment | added | Hugo Chapdelaine | Thanks algori, I did not think of taking the exponential of diagonalizable element! So for this density result on diagonalizable elements, are you assuming that $G$ is reductive? | |
| Jan 24, 2012 at 20:52 | history | answered | algori | CC BY-SA 3.0 |