Let $\mathfrak{g}$ be a simple, finite-dimensional, complex Lie algebra and let $V$ be an irreducible nontrivial representation. Is it true that $V \otimes V^*$ always contains copies of $\mathbb{C}$ and $\mathfrak{g}$?
Submodules of the tensor product of a representation and its dual
abstract-algebralie-algebrasrepresentation-theory
Related Solutions
It doesn't work for $U(1)$: take $V$ to be the standard representation, you only get the "nonnegative" representations. Indeed to use Stone Weierstrass in the complex case, the algebra has to be stable under complex conjugation. Maybe consider the $V^n \otimes (V^*)^m$ instead.
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
Yes.
Hint The canonical pairing $$\operatorname{tr} : V \otimes V^* \to \Bbb C$$ is nonzero, so it is a surjection.
Hint On the other hand, as a (nontrivial) representation of $\mathfrak g$ we may view it as a (nonzero) homomorphism $$\mathfrak{g} \to \mathfrak{gl}(V) \cong V \otimes V^* .$$
Remark In a few cases these are the only irreducible subrepresentations. If $\mathfrak{g}$ is $\mathfrak{sl}(n + 1, \Bbb C)$ and $V$ is the standard representation $V := \Bbb C^n$, the tensor product $V \otimes V^*$ decomposes into trace and tracefree parts, and in this case the latter is just adjoint representation $\mathfrak{sl}(n, \Bbb C) = \mathfrak{sl}(V)$: $$V \otimes V^* \cong \mathfrak{gl}(V) \cong \Bbb C \oplus \mathfrak{sl}(V) .$$