# When does the tensor product of two irreps contain the adjoint representation?

Let $\rho$ and $\sigma$ be two irreducible representations of the same simple Lie algebra $\mathfrak{g}$. Under which conditions does the decomposition of the tensor product $\rho \otimes \sigma$ into irreducible representations contain the adjoint representation of $\mathfrak{g}$?

My guess is that $\rho$ and $\sigma$ must be dual. But is this true? And if yes, how could one proof it?

• 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. – user44191 Mar 28 '17 at 8:39
• 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. – user44191 Mar 28 '17 at 8:40
• 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. – Vincent Mar 28 '17 at 9:41
• @Vincent: The criterion "at most 2" doesn't seem to work, e.g., when the difference is 1. – Jim Humphreys Mar 28 '17 at 16:17
• 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. – Venkataramana Mar 31 '17 at 15:54

As the comments suggest, there don't seem to be any easily stated necessary and sufficient conditions for the adjoint representation to occur as a summand of the tensor product of two irreducibles (say with highest weights $\lambda$ and $\mu$). Of course, one has to fix a simple system of roots to speak of "highest weight".
However, there is a simple necessary condition. Start with the classical fact that 0 is a weight of an irreducible representation precisely when the highest weight lies in the root lattice. In particular, this applies to the adjoint representation, whose highest weight is the highest root; here the 0 weight space corresponds to a Cartan subalgebra. In turn, classical results show that the highest weights $\nu$ of all irreducible summands lie below the sum of the two given highest weights in the usual partial ordering of weights. It follows immediately that the adjoint representation occurs as a summand of the tensor product only if $\lambda + \mu$ lies in the root lattice.
In some Lie types such as $G_2$, the root lattice equals the weight lattice. But here one sees readily that not all tensor products of two irreducibles have the 14-dimensional adjoint representation as a summand. I don't know of an easily stated sufficient condition in general.
• 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? – user44191 Apr 3 '17 at 23:39
• @user44191: I don't understand what your symbol $\geq$ means in this context, or why your condtion would be necessary. – Jim Humphreys Apr 4 '17 at 17:24
• 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 $\rho + \gamma \geq \sigma^*$. – user44191 Apr 4 '17 at 22:09
• @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. – Jim Humphreys Apr 5 '17 at 14:54