Timeline for Tensor product of simple representations
Current License: CC BY-SA 2.5
15 events
| when toggle format | what | by | license | comment | |
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| Mar 10, 2011 at 23:40 | vote | accept | shenghao | ||
| Mar 10, 2011 at 13:12 | comment | added | George McNinch | @Shengao, again. Sorry, I didn't quite answer the correct question with the previous comment. If $V$ and $W$ are rational representations of the algebraic group $G$, then the image of $G$ in $GL(V) \times GL(W)$ is already closed, and one may as well replace $G$ by the identity component of that image, which is a quotient of $G^0$. After that replacement, $G$ is connected and reductive and one finds the desired semisimplicity when $k$ is assumed to have char. 0. | |
| Mar 10, 2011 at 12:59 | comment | added | George McNinch | @Shenghao: You wrote "I'm just a little worried that if I should distinguish rep. of an abstract group from rational rep". Following the sketch of Chevalley's argument that I provided (as an answer), one sees this isn't a problem. In deciding the semisimplicity, one may as well replace $G$ by its Zariski closure in $GL(V) \times GL(W)$. Then $V, W$ and $V \otimes W$ are rational reps of $G$, and $G$ is reductive. | |
| Mar 10, 2011 at 1:53 | comment | added | fherzig | One way to see that is that the $k$-points of a reduced affine variety are dense in it (by Hilbert's Nullstellensatz). Now use that $W$ is a $G$-submodule (resp. $G(k)$-stable) iff the image of $G$ lands in $P$ (resp. $P(k)$), where $P$ is the parabolic subgroup that stabilises $W$. I think the same works if $k$ is infinite and either $k$ perfect or $G$ reductive (so e.g., in characteristic zero). In both these cases it is known that $G(k)$ is dense in $G$. | |
| Mar 10, 2011 at 1:47 | comment | added | fherzig | Over an algebraically closed field $k$, it's certainly fine: if $G$ is a reduced linear algebraic group over $k$ and $V$ a $G$-module (i.e., given by a homomorphism $G \to GL(V)$ of algebraic groups over $k$), then if $W \subset V$ is $G(k)$-stable it is in fact a $G$-submodule. So is (semi)simple for the $G(k)$-action iff it is (semi)simple as a $G$-module. | |
| Mar 10, 2011 at 0:27 | comment | added | shenghao | I'm just a little worried that if I should distinguish rep. of an abstract group from rational rep. of an algebraic one, as it appears that Chevalley's theorem is for abstract groups... | |
| Mar 10, 2011 at 0:23 | answer | added | George McNinch | timeline score: 12 | |
| Mar 10, 2011 at 0:20 | comment | added | Jim Humphreys | Under your assumptions, see Chevalley's old theorem quoted in the paper by Serre to which Herzig links. For the additive group, see my P.S. above. | |
| Mar 10, 2011 at 0:14 | history | edited | shenghao | CC BY-SA 2.5 |
clarified
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| Mar 10, 2011 at 0:05 | comment | added | shenghao | Thanks Jim. What I had in mind is characteristic 0, and representations are rational representations of algebraic groups (so finite dim). So in this case $V\otimes W$ is semi-simple? And what's the 1-dim module of $\mathbb G_a?$ I thought there was no algebraic homomorphism $\mathbb G_a\to\mathbb G_m.$ | |
| Mar 10, 2011 at 0:04 | comment | added | Jim Humphreys | P.S. My final sentence is nonsense, but the rest of the comment emphasizes the need to focus the question more. | |
| Mar 9, 2011 at 23:59 | answer | added | fherzig | timeline score: 7 | |
| Mar 9, 2011 at 21:37 | comment | added | Jim Humphreys |
The characteristic of the field makes a big difference here, since in prime characteristic tensor products of simple modules are seldom semisimple even if $G$ is (connected) reductive; moreover, such groups often have faithful irreducible representations but only tori are linearly reductive. On the other hand, in characteristic 0 linearly reductive = reductive. (And I guess the modules here are all finite dimensional?) Note too that the additive group has a faithful irreducible 1-dimensional module.
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| Mar 9, 2011 at 21:31 | answer | added | Bruce Westbury | timeline score: 7 | |
| Mar 9, 2011 at 21:06 | history | asked | shenghao | CC BY-SA 2.5 |