One approach to the general problem of decomposing a tensor product of irreducible finite-dimensional representations (hence any finite-dimensional representations) into irreducibles is to use the theory of crystals. The crystal of a representation is a colored directed graph associated to that representation. There is a purely combinatorial algorithm for producing the tensor product of two crystals. Then the connected components of the crystal graph correspond to the irreducible representations you're looking for. For many simple Lie algebras, the crystals of the irreducible finite-dimensional representations are described very explicitly. For instance, for $\mathfrak{sl}_n$, they are in terms of semistandard tableaux. So finding the decomposition you seek becomes combinatorics.
To sum up the discussion in the comment into an answer: It is indeed true that for each $A \in \mathfrak g$, $\rho_W(A)$ is a homomorphism of $S_N$-modules. However, Schur's Lemma talks about homomorphisms of simple (a.k.a irreducible) $S_N$-modules, and as OP said, generally $W$ is not irreducible, but decomposes as a direct sum of irreducible $X_i$.
It is still possible to infer something from Schur's Lemma, namely:
For each such irreducible $X_i$, the restriction of $\rho_W(A)$ to $X_i$ is either
- $\rho_W(A)_{\vert X_i} = 0$
or
- $\rho_W(A)_{\vert X_i}$ induces an isomorphism (of $S_N$-modules) to another $X_j$ (where in general $j\neq i$).
The part of Schur's lemma whose sloppy interpretation is that "our endomorphism is a scalar" actually says that the endomorphism ring of a simple $S_N$-module is a skew field, finite dimensional over the ground field $k$ we've tacitly been working with all the time. If that field was $\mathbb C$, then that endomorphims ring is necessarily $\mathbb C$, but if our field was $\mathbb R$, it could in principle be $\mathbb R, \mathbb C$ or $\mathbb H$. However, I think that for the symmetric group $S_N$ it's actually known (cf. MO/10635) that all real (or even rational) representations have Schur index $1$ ($\Leftrightarrow$ Frobenius-Schur indicator $1$), meaning that indeed $End_{\mathbb R[S_N]}(X_i) \simeq \mathbb R$ and hence
- if in the second case above $i=j$, then $\rho_W(A)_{\vert X_i}$ is given by multiplication with some $\lambda_i \in \mathbb R^*$.
To see this in an example, let $k=\mathbb R, \mathfrak g = \mathfrak{sl}_2(\mathbb R), V=$ the standard representation of $\mathfrak g$ on $\mathbb R^2$, and $N=2$. Then $W= V^{\otimes 2}$ as $\mathbb R[S_2]$-module decomposes into four $1$-dimensional components; namely, set $x_i = e_i \otimes e_i $ for $i=1,2$, $x_3 = e_1 \otimes e_2 +e_2\otimes e_1$, and $x_4 = e_1 \otimes e_2 -e_2\otimes e_1 $, and let $X_i := \mathbb Rx_i$. Then $X_4$ is the alternating (sign) representation, the three other $X_i$ are isomorphic to the trivial representation.
Now e.g. for $A=\pmatrix{0 &1\\0&0}$ we have $\rho_W(A)_{\vert X_i} = \begin{cases} 0 \text{ if } i=1,4 \\ x_2 \mapsto x_3 \text{ if } i=2, \text{ giving an iso } X_2 \simeq X_3 \\ x_3 \mapsto 2 x_1 \text{ if } i=3, \text{ giving an iso } X_3 \simeq X_1 \end {cases}$
whereas for $H=\pmatrix{1 &0\\0&-1}$ we have $\rho_W(H)_{\vert X_i} = \begin{cases} x_1 \mapsto 2x_1 \text{ if } i=1, \text{ i.e. } \lambda_1=2 \\x_2 \mapsto -2x_2 \text{ if } i=2, \text{ i.e. } \lambda_2=-2 \\ 0 \text{ if } i=3,4\end {cases}$
etc.
Best Answer
In all generality there is the following basic method: compute the (formal) characters of each representation (using Weyl's character formula or Freudenthal's recursive weight multiplicity formula), multiply the two to find the character of the tensor product, then repeatedly take a highest (remaining) weight in the character, subtract off (the appropriate multiple of) the character of the corresponding irreducible highest weight representation after contributing it to the result, and so on until nothing is left of the character.
There is a better method that needs to develop only the full character of one of the two factors being multiplied; the result is multiplied by the highest weight (only) of the other factor. The result is a non $W$-symmetric formal sum of weights; any non-dominant contributions in it are made more dominant (or annihilated) by anti-reflection in the simple root hyperplanes shifted downwards. This means the non-dominant weights $\mu$ with $\langle\mu,\alpha_i^\vee\rangle=-1$ lie one the shifted hyperplane and are annihilated; weights further down with respect to $\alpha_i^\vee$ are reflected in that shifted hyperplane and their coefficient replaced by its opposite. After repeating thes operations sufficiently often, every remaining term has become dominant, and represents an irreducible component of the tensor product.
There is also a general formula using Littelmann paths, or Lakshmibai-Seshadri paths (which are particular instances), which are techniques related to crystal graphs. They are somewhat more difficult to apply in general, but for classical types there are special instances that can be implemented efficiently, such as the Littelwood-Richardson rule in type $A_n$.