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Hello, I am wondering about a statement on this page. Especially about this part:

If $\pi_1$ and $\pi_2$ are irreducible and finite dimensional or unitary, then the intertwining number $c(\pi_1,\pi_2)$ is equal to $1$ or $0$, depending on whether $\pi_1$ and $\pi_2$ are equivalent or not.

$\pi_1$ and $\pi_2$ are representations of a finite group $G$ over a field $K$. Why is there an "or unitary"? Is this true if the representations are just irreducible and finite-dimensional, but $K$ not a splitting field?


As it seams to be wrong, if $\pi_i$ are not absolutely irreducible, how is it possible to determine $\dim_K(\operatorname{Hom}_{K[G]}(\pi) $ if $\pi$ is irreducible, but $K$ not a splitting field. Are there theorems when this dimension is 1?

The background is, that I would like to use Corollary 4.2.

share|improve this question
    
My reading would be that it allows irreducible representations which are non finite dimensional, but intends them always to be irreducible. The statement is not true in general, even for finite groups, if the representations are not absolutely irreducible. –  Geoff Robinson Apr 28 '13 at 17:06
    
Unitary implies complex representations. Every irreducible representation over the complex numbers of a compact group is unitarizable. More generally, this statement holds for unitary reps of locally cpct groups (Schur's lemma). –  plusepsilon.de Apr 29 '13 at 5:34

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