1
$\begingroup$

I'm reading the book Introduction to Representation Theory by Pavel Etingof et all, and I have some questions about the remark $3.1.3$. The authors state that by Schur's lemma, any semisimple representation $V$ of $A$ is caninically identified with $$\bigoplus_{X}Hom_{A}(X, V) \otimes X$$ where $X$ runs over all irreducible representations of A. They claim that the function $f: \bigoplus_{X}Hom_{A}(X, V) \otimes X \to V$ given by $g \otimes x \mapsto g(x)$, $x \in X$, $g \in Hom_{A}(X, V)$ is an isomorphism. My first question is where is the Schur's lemma being used? And how can i prove that this function is injective? Thank you in advance

$\endgroup$
2
$\begingroup$

All vector spaces are over some fixed field $k$, that is often omitted. Let us now write (by definition, non-canonically) $V$ as a direct sum of irreducible pieces, $V=Y_1\oplus Y_2\oplus\dots$ and use this below $$ \begin{aligned} &\bigoplus_{X}Hom_{A}(X, V) \otimes X \\ &= \bigoplus_{X}Hom_{A}(X,Y_1\oplus Y_2\oplus\dots ) \otimes X \\ &= \bigoplus_{X}\Big(\ Hom_{A}(X,Y_1)\oplus Hom_{A}(X,Y_2)\oplus\dots\ \Big) \otimes X \\ &= \bigoplus_{X}\Big(\ Hom_{A}(X,Y_1)\otimes X\oplus Hom_{A}(X,Y_2)\otimes X\oplus\dots\ \Big) \\ &= \bigoplus_{X}Hom_{A}(X,Y_1)\otimes X\ \oplus\ \bigoplus_{X} Hom_{A}(X,Y_2)\otimes X\oplus\dots \\ &= \bigoplus_{X\ :\ X=Y_1}Hom_{A}(X,Y_1)\otimes X \ \oplus\ \bigoplus_{X\ :\ X=Y_2} Hom_{A}(X,Y_2)\otimes X\ \oplus\ \dots \text{ (Schur)} \\ &= Hom_{A}(Y_1,Y_1)\otimes Y_1 \ \oplus\ Hom_{A}(Y_2,Y_2)\otimes Y_2\ \oplus\ \dots \\ &= k\otimes Y_1 \ \oplus\ k\otimes Y_2\ \oplus\ \dots \text{ (Schur)} \\ &=Y_1\oplus Y_2\oplus\dots \\ &=V\ . \end{aligned} $$ In the chain of isomorphisms above (denoted by abuse with $=$) only the first and the last step are non-canonical. So, to see that the declared map (induced on each summand by) $g\otimes x\to g(x)$ is canonical, we have only to show that for any $$ \text{choice }V\overset f\longrightarrow \underbrace{Y_1\oplus Y_2\oplus\dots}_W $$ we have the commutativity $\require{AMScd}$ \begin{CD} \bigoplus_{X}Hom_{A}(X, V) \otimes X @>>> \bigoplus_{X}Hom_{A}(X, W) \otimes X \\ @V V V @VV V\\ V @>>f> W \end{CD} thus equivalently for every piece: \begin{CD} Hom_{A}(X, V) \otimes X @>>> Hom_{A}(X, W) \otimes X \\ @V V V @VV V\\ V @>>f> W \end{CD} and on elements, the diagram chasing is: \begin{CD} g \otimes x @>>> f_*(g) \otimes x \\ @V V V @VV V\\ g(x) @>>f> f(g(x)) \end{CD} which is true.

$\endgroup$
  • $\begingroup$ I can't see this: $$\bigoplus_{X} Hom_{A}(X, Y_{1}) \otimes X \oplus ... \oplus \bigoplus_{X} Hom_{A}(X, Y_{n}) \otimes X \cong$$ $$\bigoplus_{X: X \cong Y_{1}}Hom_{A}(X, Y_{1}) \otimes X \oplus ... \oplus \bigoplus_{X: X \cong Y_{n}} Hom_{A}(X, Y_{n}) \otimes X $$ $\endgroup$ – Jaime Grimal Alves Sep 8 at 23:26
  • $\begingroup$ The theorem of Schur, en.wikipedia.org/wiki/…, tells us that the set of homomorfisms between $X$ and $Y_1$, both irreducible, is $0$, when they are not isomorphic. So we can restrict to the case of the only one $X$ in the list under the big direct sum of all irreducible representations which is $\cong Y_1$. $\endgroup$ – dan_fulea Sep 9 at 1:04
  • $\begingroup$ Can these isomorphisms be seen as A-isomorphisms, that is, module isomorphisms? $\endgroup$ – Jaime Grimal Alves Sep 19 at 15:41
1
$\begingroup$

Schur's lemma says, some map under some conditions is an isomorphism.

Your question is to see the obvious map $\bigoplus_X (\text{Hom}_A(X,V)\otimes X)\rightarrow V$ to be an isomorphism.

Suppose for simplicity $V$ is an irreducible representation of $A$.

Hint : How many $A$-linear maps can you think of from $X$ to $V$ if $X$ and $V$ are not isomorphic?

As $V$ is semisimple, it is direct sum of irreducible representations. Suppose $V=V_1\oplus V_2$.

Check : Does the following equality holds? $$\bigoplus_X (\text{Hom}_A(X,V)\otimes X)=\bigg( \bigoplus_X (\text{Hom}_A(X,V_1)\otimes X)\bigg)\bigoplus \bigg( \bigoplus_X (\text{Hom}_A(X,V_2)\otimes X)\bigg)$$ Now, $X$ and $V_1,V_2$ are irreducible representations. Can you see the result?

$\endgroup$

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.