Let $\rho:G\longrightarrow GL(V)$ a finite dimensional representation of G. I was trying to show if $\rho .\rho$ is irreducible then $V$ is one dimensional.

Since $\rho.\rho$ is a representation for the tensor product $V\otimes V$, $\rho.\rho:G\longrightarrow GL(V\otimes V)$, where $\rho.\rho(a)(v\otimes w)=\rho (a)(v)\otimes\rho(a)(w)$. since $V$ is finite dimensional, then $V\otimes V$ is also finite dimensional. Let $W$ be any sub space of $V$. Then clearly $W\otimes W$ is a subspace of $V\otimes V$. And from the irreducibly of $\rho.\rho $, the only invariant subspace are $\{0\}$ and $V\otimes V$. Thus if $W\neq\{0\}$ and $W\neq V$, then I can assume $0\neq x\in W$ where $\rho(a)(x\otimes x)=\rho(a)(x)\otimes\rho(a)(x)\not\subseteq W\otimes W$. Which means, $\rho(a)(W)\not\subseteq W$. How can I continue to sho contradiction on the existence of $W$. Or do you have any other simple way to show the statement?

  • $\begingroup$ I'm thinking about using characters, is your group finite? $\endgroup$
    – M. Van
    Oct 21, 2017 at 13:21

1 Answer 1


Every pure tensor in $V \otimes V$ can be written as the sum of a symmetric part $\frac{1}{2}(v\otimes w + w \otimes v)$ and an antisymmetric part $\frac{1}{2}(v \otimes w - w \otimes v)$, and this decomposition extends to give a direct sum decomposition $V\otimes V = S^2 V \oplus \wedge^2 V$ into these symmetric and antisymmetric parts. It is easy to check that the action of $G$ on $V \otimes V$ preserves this decomposition, and so $V \otimes V$ cannot be irreducible, unless one of these parts is $0$. Now, check that $\wedge^2 V = 0$ iff $V$ has dimension 0 or 1.


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