Nonsensical proof: decomposition of $N$-fold tensor product of representation of Lie algebra Let $V$ be the a finite dimensional representation of a Lie algebra $\mathfrak{g}$ and consider the space $W=V^{\otimes N}$.
$W$ is also a representation of the symmetric group $S_N$ under the action that permutes the spaces, hence
$$W=\bigoplus_iX_i $$
where $X_i$ are irreducible representations of $S_N$. Now, the action of $A\in\mathfrak{g}$ on W is given by
$$ \rho_W(A)=\sum_{i}\mathbb{1}^{\otimes i-1} \otimes\rho_V(A)\otimes\mathbb{1}^{\otimes N-i-1} $$
Hence $\sigma \circ \rho_W(A)=\rho_W(A)\circ\sigma$ for all $\sigma \in S_N$, but this means $\rho_W(A)$ is a $S_N-$module homomorphism hence by Schur's lemma
$$ \rho_W(A)=\lambda \mathbb{1}$$
But this means that $\rho_W(A)=\bigoplus_i \lambda_i\mathbb{1}_{X_i}$ is diagonal for every $A$ on $W$, hence all the basis vectors of the $X_i$ span one dimensional subrepresentations of $W$ of $\mathfrak{g}$.
Hence we can decompose any $N$-fold tensor product of representations of Lie algebras in one dimensional irreducible representations. 
...But this is absurd, as there are many counterexamples. I'm sure the blatant error has been staring at me for the past hour, but I just can't see it, could anybody point it out?
 A: 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.
