Timeline for Irreducible quotient of $U\otimes V$
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10 events
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| Aug 9, 2013 at 1:49 | comment | added | Yilan Tan | In the prove of it, Chari showed that there is an automorphism $F: V\rightarrow U^t\otimes W$, where $V=\otimes_{i=2} W_{a_i}$ is a highest weight module, $U^t$ is a fundamental representation, and $W$ is an irreducible quotient of $\otimes_{i=1}W_1(a_i)$. Chari claimed that $F(v^+)$ must be a scalar of $u^+\otimes w^+$ without details. It cost me very very long time to figure this out. But my way is too complicated, and I thought it may not be the way Chari thought. Thank you all again for your help. | |
| Aug 9, 2013 at 1:34 | comment | added | Yilan Tan | Thank you all to answer my quetstion. If you are of interest, I tell you why I ask this question. I am doing a research on Weyl module of Yangian of sl_n. The irreducible module of sl_n with fundamental weight is an fundamental representation of Yangian of sl_n, which is called evaluation representation. V.Chari proved, in paper Yangian-their representation and character, that $\bigotimes W_{1}(a_i)$ is a highest weight representation if $a_i-a_j\neq 1$ for $i<j$. Here, $W_{1}(a_i)$ is the irreducible module $V(1)$ of sl_2 pulled back by a automorphism $\tau_a$ of yangian of sl_2. | |
| Aug 8, 2013 at 15:40 | comment | added | Chuck Hague | This argument confuses me. Take $U$, $V$ to be finite-dimensional and simple. Then $W = U \otimes V$ satisfies the desired conditions but in general it will certainly not be irreducible, and it will not have a unique maximal vector - it will have lots of them. | |
| Aug 7, 2013 at 16:06 | comment | added | Jim Humphreys | @Aakumadula: But why should the maximal submodule be unique? | |
| Aug 7, 2013 at 14:04 | comment | added | Venkataramana | any cyclic module will have a unique irreducible quotient: take the maximal submodule not containing the cyclic vector (Zorn's lemma) and go modulo that maximal one. | |
| Aug 7, 2013 at 13:27 | comment | added | Tobias Kildetoft | @YilanTan Yes, but why will the irreducible quotient be unique? | |
| Aug 7, 2013 at 13:22 | comment | added | Yilan Tan | $W$ is irreducible, the maximal vector is unique up to scalar. | |
| Aug 7, 2013 at 11:59 | comment | added | Tobias Kildetoft | But he wrote that he knew that this was the case. The question was about the highest weight of an irreducible quotient of the tensor product (I am still not sure why there should be a unique such). | |
| Aug 7, 2013 at 10:05 | review | First posts | |||
| Aug 7, 2013 at 10:14 | |||||
| Aug 7, 2013 at 9:50 | history | answered | Alexander Premet | CC BY-SA 3.0 |