# 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. Commented Mar 28, 2017 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. Commented Mar 28, 2017 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. Commented Mar 28, 2017 at 9:41
• @Vincent: The criterion "at most 2" doesn't seem to work, e.g., when the difference is 1. Commented Mar 28, 2017 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. Commented Mar 31, 2017 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? Commented Apr 3, 2017 at 23:39
• @user44191: I don't understand what your symbol $\geq$ means in this context, or why your condtion would be necessary. Commented Apr 4, 2017 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 $\rho + \gamma \geq \sigma^*$. Commented Apr 4, 2017 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. Commented Apr 5, 2017 at 14:54