Morphisms of two fully reducible representations of a group [closed]

Question

I thought this is a simple question and placed it at math.stackexchange.com: https://math.stackexchange.com/questions/3616661/morphisms-of-two-fully-reducible-representations-of-a-group. Since no one there manage to answer this question, I now have to put it here.

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Consider fully reducible representations $\,A(g)\,$ and $\,A^{\,\prime}(g)\,$ of a group $\,G\,$ in vector spaces $\,\mathbb{V}\,$ and $\,{\mathbb{V}}^{\,\prime}\,$, respectively. Let them be intertwined: $$ M~A(g)~=~A^{\,\prime}(g)~M~~. $$ For brevity, I shall denote the kernel $\,\operatorname{Ker} M\,$ simply as $\,\operatorname{Ker}$, the image $\,\operatorname{Im}\,M\,$ as $\,\operatorname{Im}\,$. Being invariant subspaces they support subrepresentations: $$ B(g)\,v~\equiv~A(g)\,v\Bigr{|}_{v\,\in\,\rm{Ker}}\quad,\qquad B^{\,\prime}(g)\,v^{\,\prime}~\equiv~A^{\,\prime}(g)\,v^{\,\prime}\Bigr{|}_{v^{\,\prime}\,\in\,\rm{Im}}\;\;. $$ As $\,A\,$ is fully reducible, any of its subrepresentations has a complementary subrepresentation. E.g., for $\,B\,$ acting in $\,\operatorname{Ker}\,$, its complementary $\,B^{\,\perp}\,$ in $\,{\operatorname{Ker}}^{\perp}\,$ is $$ B^{\perp}(g)\,v\,\equiv\,A(g)\,v\Bigr{|}_{v\,\in\,{\operatorname{Ker}}^{\perp}}\;\;. $$

It is then easy to demonstrate that the same $\,M\,$ intertwines $\,B^{\perp}\,$ and $\,B^{\,\prime}\,$, i.e. $\,M\,B^{\perp}\,=\,B^{\,\prime}\,M\,$. Moreover, if we postulate $\,B^{\,\prime}\,$ to be irreducible, the representations $\,B^{\perp}\,$ and $\,B^{\,\prime}\,$ become equivalent, by Schur's Lemma: $$ B^{\perp}\,\simeq\,B^{\,\prime}\;\;. $$ The inverse is true too: if $\,B^{\perp}\,\simeq\,B^{\,\prime}\,$, there exists a morphism $\,M\,$ intertwining $\,A\,$ and $\,A^{\,\prime}\,$.

To conclude, fully reducible representations $\,A\,$ and $\,A^{\,\prime}\,$ intertwine if and only if they have equivalent subrepresentations. $$ ~~ $$ QUESTION 1: $~~~$In the case of an irreducible $\,A\,$, prove that the multiplicity of $\,A\,$ in $\,A^{\,\prime}\,$ is equal to the dimensionality of the space $\,[A\,,\,A^{\,\prime}]\,$ of all such intertwiners $\,M\,$. $$ ~~ $$ QUESTION 2: $~~~$If $\,\operatorname{dim}\,[A\,,\,A^{\,\prime}]\,=\,\,\infty\,$, would it be right to say that the representations $\,A\,$ and $\,A^{\,\prime}\,$ are equivalent, and their spaces are isomorphic?

Answer 1

Doc, your language is old-fashioned. Such questions become clearer with higher levels of abstraction. You want to go from representations to modules and then further to categories. It leaves unnecessary details out.

Q1 Let $R$ be a simple module (or an irreducible representation in your language). Then the endomorphism ring $[R,R]$ is a division algebra: the multiplication is the composition. For a finite group it needs to be finite-dimensional over the ground field, over which you take your representations. Now $[R,A]$ is a right vector space over $[R,R]$: $$ [R,A]\times [R,R] \rightarrow [R,A], \ (f,g) \mapsto f\circ g , $$ while $R$ is a left vector space over $[R,R]$. The multiplicity of $R$ in $A$ is the dimension of $[R,A]$ over $[R,R]$. The categorical thinking clarifies this statement because the multiplicity is longer a number but the space $[R,A]$ itself. This means the evaluation map gives you a canonical (without choosing any bases anywhere) decomposition $$ \bigoplus_{n} [R_n,A]\otimes_{[R_n,R_n]} R_n \xrightarrow{\cong} A, \ (\alpha_n \otimes r_n)_n \mapsto \sum_n \alpha_n (r_n) $$ where $R_n$ runs over all irreducible representations.

Q2 This is not true. Just pick two distinct irreducibles $S$ and $R$ and take $$A=\infty S, \ A'=A\oplus R$$ where $A$ is a direct sum of infinitely many copies of $S$. Again the categorical thinking allows to write "hom" (the space of intertwiners) canonically via the spaces of linear maps: $$ \oplus_{n} Lin_{[R_n,R_n]} ([R_n,A],[R_n,B]) \xrightarrow{\cong} [A,B], \ (\alpha_n)_n \mapsto \oplus_{n} (\alpha_n \otimes_{[R_n,R_n]} Id_{R_n}) $$ where $Id_{R_n}$ is the identity linear map on $R_n$.