Timeline for When does the tensor product of two irreps contain the adjoint representation?
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 21, 2018 at 9:07 | vote | accept | Severin | ||
| Mar 31, 2017 at 17:32 | comment | added | Venkataramana | The question amounts to asking if there is a $g$ invariant nonzero vector in in the triple tensor product $V^* \otimes W\otimes g$. This is equivalent to whether $V$ occurs in $W\otimes g$. IF $W$ has highest weight $\lambda$ and $g$ has highest weight $\alpha$ (the highest root), then $V=V(\lambda + \alpha)$ the rep with highest weight $\lambda +\alpha$ occurs in $W\otimes g$. Thus $V$ and $W$ are not self dual. | |
| Mar 31, 2017 at 15:54 | comment | added | Venkataramana | If $g$ (the simple Lie algebra ) acts irreducibly on $V$ and $V$ has dimension more than one, then there is an inclusion $g\rightarrow End(V)=V\otimes V^*$. Hence the adjoint representation does embed in $V\otimes V^*$. So at least half of the question has yes as an answer. | |
| Mar 29, 2017 at 9:55 | comment | added | Vincent | Aargh, yes, my bad. 0 or 2 I meant. However the SL2 case may not be very representative for the general case anyway. | |
| Mar 28, 2017 at 16:17 | comment | added | Jim Humphreys | @Vincent: The criterion "at most 2" doesn't seem to work, e.g., when the difference is 1. | |
| Mar 28, 2017 at 16:11 | answer | added | Jim Humphreys | timeline score: 5 | |
| Mar 28, 2017 at 9:41 | comment | added | Vincent | The $SL_2$ case mentioned by user44191 above can be generalized by stating that the difference in dimension between $\sigma$ and $\rho$ must be at most 2 and that this is both necessary and sufficient, see this MSE answer: math.stackexchange.com/a/95882/101420. The general case is harder, I believe. | |
| Mar 28, 2017 at 8:40 | comment | added | user44191 | For a less trivial example, take the 2 and 4 dimensional representations of $SL_2$; their tensor product decomposes into the 3 dimensional representation and the 5 dimensional representation, and the 3 dimensional representation is the adjoint representation. | |
| Mar 28, 2017 at 8:39 | comment | added | user44191 | It is not true that they have to be dual; there is the trivial example where $\rho$ is the trivial representation and $\sigma$ is itself the adjoint representation. In fact, the condition that the two be dual is equivalent to the inclusion of the trivial representation, not the adjoint representation. | |
| Mar 28, 2017 at 8:06 | review | First posts | |||
| Mar 28, 2017 at 8:45 | |||||
| Mar 28, 2017 at 8:05 | history | asked | Severin | CC BY-SA 3.0 |