Timeline for "Why" is every polynomial representation of SL(2) selfdual?
Current License: CC BY-SA 2.5
21 events
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| Feb 15, 2011 at 19:00 | comment | added | darij grinberg | Remark: Over at M.SE, a trivial particular case ( math.stackexchange.com/questions/21320 ) has received a very interesting coordinate-free explanation. Maybe this generalizes? | |
| Sep 9, 2010 at 18:43 | history | edited | darij grinberg |
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| Sep 9, 2010 at 18:15 | comment | added | Jim Humphreys | Small comment: maybe algebraic-groups is a better tag here than linear-algebra? | |
| Sep 9, 2010 at 18:11 | answer | added | Jim Humphreys | timeline score: 12 | |
| Sep 9, 2010 at 16:10 | comment | added | darij grinberg | Oh, you are right. I wish I had enough knowledge of algebraic groups to avoid this hassle with mundane groups. | |
| Sep 9, 2010 at 16:08 | history | edited | darij grinberg | CC BY-SA 2.5 |
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| Sep 9, 2010 at 16:01 | comment | added | Victor Protsak | Darij, you need to be a lot more clear from the get-go about the category of representations you are interested in. For example, any automorphism $\sigma$ of $K$ induces an automorphism of $SL_2(K)$ viewed as an abstract group; composing it with a representation $\rho$ that occurs in the tensor space gives one that does not, unless $\sigma$ or $\rho$ is trivial. (Borel and Tits proved some amazing results about "abstract" homomorphisms of algebraic groups which may imply that this is the only obstruction in char 0.) Of course, this is impossible in the category of rational representations. | |
| Sep 9, 2010 at 15:08 | answer | added | George McNinch | timeline score: 13 | |
| Sep 9, 2010 at 13:59 | answer | added | Joel Kamnitzer | timeline score: 13 | |
| Sep 9, 2010 at 13:37 | answer | added | Richard Borcherds | timeline score: 27 | |
| Sep 9, 2010 at 13:07 | comment | added | darij grinberg | t3suji: OK, the canonicity is toast. | |
| Sep 9, 2010 at 13:06 | history | edited | darij grinberg | CC BY-SA 2.5 |
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| Sep 9, 2010 at 13:06 | comment | added | darij grinberg | Christopher: ok, in fact I have never met any representation that wasn't locally finite in my life, though Zorn's lemma will surely construct a whole panopticon. To be completely honest I am working with representations of the group while I actually mean representations of the algebraic group $\mathrm{SL}_2$... | |
| Sep 9, 2010 at 12:31 | comment | added | t3suji | @darij grinberg: this does not help for reducible representations: again, consider the action of automorphisms on possible choices. | |
| Sep 9, 2010 at 12:31 | comment | added | Christopher Drupieski | Perhaps you need to restrict yourself to the finite-dimensional linear representations of $\mathrm{SL}_2(K)$? Every finite-dimensional representation of $\mathrm{SL}_2(K)$ is completely reducible, and more generally, so is every rational (i.e., locally finite) representation of $\mathrm{SL}_2(K)$. But problems could arise with infinite-dimensional representations. | |
| Sep 9, 2010 at 12:28 | history | edited | darij grinberg | CC BY-SA 2.5 |
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| Sep 9, 2010 at 12:27 | comment | added | darij grinberg | Uhm, yes. Sorry! I meant a canonical (up to scalar multiplication) choice. | |
| Sep 9, 2010 at 12:23 | comment | added | t3suji | I'm confused. How could it be possible to have a `canonical' choice for the isomorphism? Even if V is irreducible, this can't be done because (scalar) automorphisms of V act on such choices non-trivially. | |
| Sep 9, 2010 at 12:19 | comment | added | darij grinberg | By the way, I am aware that for compact Lie groups, we have a Haar measure and we get an invariant bilinear form by integrating an arbitrary bilinear form. But there are things I don't like here: (1) This works only over $\mathbb R$, and we would still have to prove that all irreducible representations of $\mathrm{SL}_2\left(\mathbb C\right)$ are already defined over $\mathbb Q$. (2) $\mathrm{SL}_2$ is not compact and we would need Weyl's unitary trick to get it compact. I'm not sure whether it really preserves all representations. (3) I hate analysis and want to see it exterminated. :P | |
| Sep 9, 2010 at 12:17 | history | edited | darij grinberg | CC BY-SA 2.5 |
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| Sep 9, 2010 at 12:09 | history | asked | darij grinberg | CC BY-SA 2.5 |