highest weight representations inside tensor product Let $G$ be a semisimple simply connected group over an algebraically closed field $k$ of characteristic zero, $B$ a Borel and $T$ a maximal torus.
Let $\lambda,\mu,\nu$ be dominant characters of $T$.
Let $V(\lambda)$ be the irreducible representation of highest weight $\lambda$.
If $\lambda=\mu+\nu$, then we know that $V(\lambda)$ is the direct factor of multiplicity one inside $V(\mu)\otimes V(\nu)$.
Is it possible to characterize $V(\lambda)$ inside this tensor product ? 
 A: To expand my comment (and Robert Bryant's) a little further, the main point here is that tensor products of finite dimensional irreducible representations behave nicely.  In the equivalent Lie algebra setting, the finite dimensional irreducibles are those whose highest weights are dominant integral in the dual of a Cartan subalgebra $\mathfrak{t}:= \mathrm{Lie} \;T$ of $\mathfrak{g} := \mathrm{Lie} \;G$, relative to some fixed choice of positive (or simple) roots.  
In particular, $V(\mu) \otimes V(\nu)$ has highest weight $\lambda:= \mu + \nu$ with multiplicity 1.   The complete reducibility of this tensor product depends mainly on the fact that the center $Z(\mathfrak{g})$ of the universal enveloping algebra of $\mathfrak{g}$ (generated by "Casimir operators") acts by distinct central characters on the distinct finite dimensional representations.  In other words, the sum of all copies of a typical $V(\pi)$ in the tensor product decomposition is a single "eigenspace" for the action of the center, where the action is given by a "central character" $\chi_\pi$.    
Since the summand $V(\lambda)$ occurs only once in the tensor product (the weight space for $\lambda$ being one dimensional), $V(\lambda)$ may be characterized as the set of all vectors on which $Z(\mathfrak{g})$ acts by the central character $\chi_\lambda$.    (There may of course be many isomorphic summands of smaller highest weights $\pi$, where the center acts by the single character $\chi_\pi$.)    In different language this is essentially Robert's comment.     
