Timeline for Representations of reductive Lie group
Current License: CC BY-SA 2.5
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 13, 2010 at 9:23 | comment | added | Michele Torielli | ok, but in your case, is $\mathfrak{g}$ the Lie algebra of e reductive algebraic group? | |
| May 13, 2010 at 8:40 | vote | accept | Michele Torielli | ||
| May 12, 2010 at 17:34 | comment | added | Bugs Bunny | I am sorry but I find your terminology confusing. An element of the centre of $\mathfrak g$ needs not be diagonalisable as an element of $gl(n)$. Pick your favorite nondiagonalisable $x\in gl(n)$ and consider $Kx \oplus {\mathfrak h}$ as your $\mathfrak g$ where $\mathfrak h$ is any reductive subalgebra of the centraliser of $x$. | |
| May 12, 2010 at 14:28 | comment | added | BCnrd | @Michele: In the (conn'd) ss case in char. 0, strictly speaking B.B. may need to pass to an isogenous cover. Consider graph of $\mathfrak{g}$ in $\mathfrak{g} \times \mathfrak{gl}(V)$. This Lie subalgebra is its own commutator, so by 7.9 in Borel's book on algebraic groups, it arises from unique smooth conn'd subgp $H \subset G \times {\rm{GL}}(V)$, and $H \rightarrow G$ is isogeny (since isom. on Lie alg.). So $H$ is ss and $H \rightarrow {\rm{GL}}(V)$ is a group-lifting of the rep'n. If $G$ were simply connected then $H = G$ and you win; same as in usual Lie theory. | |
| May 12, 2010 at 13:21 | comment | added | Michele Torielli | The answer to my last question is then no, but is it true that the element in the centre of $\mathfrak{g} \subset \mathfrak{gl}(n)$ are diagonalizable? | |
| May 12, 2010 at 12:55 | comment | added | Michele Torielli | @milne:thank you. @Bunny:it's not clear to me what happen if I try to lift a representation of $\mathfrak{g}$. | |
| May 12, 2010 at 11:58 | comment | added | Bugs Bunny | What exactly do you need explaining? | |
| May 12, 2010 at 11:18 | comment | added | JS Milne | There's a complete proof that representations of reductive groups in characteristic zero are semisimple in II, section 5, of the notes on my website. | |
| May 12, 2010 at 9:48 | comment | added | Michele Torielli | Can you explain a bit more in the case in which I start with a representation of the Lie algebra?in my case, I'm using the adjoint representation of $\mathfrak{g}$ in $\mathfrak{gl}(n)$. | |
| May 12, 2010 at 8:32 | history | answered | Bugs Bunny | CC BY-SA 2.5 |