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Let $V$ be a finite-dimensional irreducible representation of a finite group G over an algebraically closed field. If $V \otimes V$ has a fixed subspace acted by $G$, why should it be 1-dimensional? I really don't know how to think about this. I tried reasoning that there is a fixed space then $\phi(g)$ must have eigenvalue 1, but it's not contributive.

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    @Akhil, yes, saying fixed is more than invariant. like you said,$G$act trivially on the elements of the subspace2011-02-28

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There need not be a fixed G-invariant subspace of dimension 1. Consider the cyclic group G of order 4 and the representation on V=C given by gn↦sqrt(−1)n. The corresponding representation on V⊗V ≅ C is given by gn↦(−1)n. The fixed subspace has dimension 0, not 1.

On the other hand, if V is irreducible and V⊗V has a fixed G-invariant subspace of dimension larger than 0, then the dimension is exactly 1. This follows from character theory:

$[1,\chi \cdot \chi] = [ \bar \chi, \chi ] = \begin{cases} 0 & \chi \neq \bar \chi \\ 1 & \chi = \bar \chi \end{cases}$

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    Thanks Jack, your answer in combination with this post: http://qchu.wordpress.com/2009/08/30/the-orthogonality-relations-for-representations-of-finite-groups/ helped me figuring out what I needed.2011-02-28