6
$\begingroup$

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 $Hom_G(V,V')=1$.

$\endgroup$
  • $\begingroup$ This is a part of Schur lemma, as a remark. $\endgroup$ – awllower Jun 16 '13 at 13:21
1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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