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?
 A: 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.      
