Timeline for When does the tensor product of two irreps contain the adjoint representation?
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Jun 21, 2018 at 9:07 | vote | accept | Severin | ||
| Apr 5, 2017 at 14:54 | comment | added | Jim Humphreys | @user44191: Yes, it's better to use the highest weights rather than labels like $\rho, \sigma$, etc. I don't have a specific counterexample in mind, since the decomposition of tensor products gets so complicated. But I'm skeptical until I see a rigorous proof of necessity and sufficiency. | |
| Apr 4, 2017 at 22:09 | comment | added | user44191 | By $\geq$ I'm referring to the usual partial ordering of weights, as referred to in your answer. Each of the three comes from the same type of reasoning as in your answer - $V_\rho \otimes V_\sigma$ must have $V_\gamma$ as a subrepresentation, so $\rho + \sigma \geq \gamma$; $V_\rho \otimes V_\gamma$ must have $V_\sigma^*$ as a subrepresentation, so $\rho + \gamma \geq \sigma^*$. | |
| Apr 4, 2017 at 17:24 | comment | added | Jim Humphreys | @user44191: I don't understand what your symbol $\geq$ means in this context, or why your condtion would be necessary. | |
| Apr 3, 2017 at 23:39 | comment | added | user44191 | If I remember correctly, it should be enough if $\rho + \sigma \geq \gamma, \rho + \gamma \geq \sigma^*, \sigma + \gamma \geq \rho^*$ where $\gamma$ is the adjoins representation, and $^*$ denotes the dual representation, and if their sum is in the root lattice. It is clearly necessary, and is sufficient for $SL_2$; do you have a counterexample for sufficiency? | |
| Mar 28, 2017 at 16:50 | history | edited | Jim Humphreys | CC BY-SA 3.0 |
deleted 89 characters in body
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| Mar 28, 2017 at 16:11 | history | answered | Jim Humphreys | CC BY-SA 3.0 |