Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Question: Let $G$ be a group. For any two representations $V,V'$ of $G$ over $\mathbb C$, let $Hom_G (V,V')$ denote the space of all linear maps $h: V\rightarrow V'$ such that $h\rho'_g = \rho_g h\forall g\in G$. I want to prove that if $V$ and V' are irreducible and $V\cong V'$ then $Hom_G(V,V')$ is 1-dimensional.

Please tell me if I have got it right: By Schur's lemma, Any element $f\in Hom_G(V,V)$ is of the form $\lambda.Id_V$, where $\lambda\in \mathbb C$, so dim $Hom_G(V,V) = 1.$ Now, let $T:V\rightarrow V'$ be an isomorphism, so that $T\rho'_g = \rho_g T$, hence $T\neq0$, and $T\in Hom_G(V,V')$. So $Hom_G(V,V')\neq0.$ Now, for any $h\in Hom_G(V,V')$, we have$$V\xrightarrow{h}V'\xrightarrow{T^{-1}}V$$, so that $T^{-1}h\in Hom_G(V,V)$, and hence $T^{-1}h=\alpha.Id_V$ for some $\alpha\in\mathbb C\Rightarrow h=\alpha.T\circ Id_V=\alpha T$. Hence any element in $Hom_G(V,V')$ is a scalar multiple of $T$, and so dim $Hom_G(V,V')=1$.

share|improve this question
    
This is a part of Schur lemma, as a remark. –  awllower Jun 16 '13 at 13:21

1 Answer 1

up vote 1 down vote accepted

Looks fine. $\phantom{space filler to have more space}$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.