I will do a seminar which deals with Serre's 'Linear Representations of Finite Groups' and I am a beginner in this field. I came across Schur's lemma (chapter 2.2). It states:

[Let $G$ be a finite group and $V_1$ and $V_2$ two complex vector spaces.]
Let $\rho_1:G \rightarrow \text{GL}(V_1)$ and $\rho_2:G \rightarrow \text{GL}(V_2)$ be two irreducible representations of $G$, and let $F$ be a linear mapping of $V_1$ into $V_2$ such that $\rho_2(s)\circ f = f \circ \rho_1(s)$ for all $s \in G$. Then:
(1) If $\rho_1$ and $\rho_2$ are not isomorphic, we have $f=0$.
(2) If $V_1=V_2$ and $\rho_1=\rho_2$, $f$ is a homothety (a scalar multiple of the identity).

Now my question deals with statement (1). The premise is that $\rho_1$ and $\rho_2$ are not isomorphic. First off, I think it is quite ambiguous what that means. It could mean that both $\rho_1$ and $\rho_2$ are not isomorphic maps in the sense of an isomorphism between $G$ and GL$(V_1)$, resp. GL$(V_2)$. On the other hand it could mean that $\rho_1$ is not isomorphic to $\rho_2$ in the sense that is stated in the accepted answer of this post. I guess, the former case is not meant here since I could not find anything interesting concerning representations that are isomorphic maps. However, there might also be the possibility that these were typos and the author meant to state:

(1) If $V_1$ and $V_2$ are not isomorphic, we have $f=0$.

That is the statement from the Wiki page of Schur's lemma.
Does anyone know what is meant here?


1 Answer 1


On p. 4, Serre defines what it means for two representations to be isomorphic:

Let $\rho$ and $\rho'$ be two representations of the same group $G$ in vector spaces $V$ and $V'$. These representations are said to be similar (or isomorphic) if there exists a linear isomorphism $\tau\colon V\to V'$ which "transforms" $\rho$ into $\rho'$, that is, which satisfies the identity $$ \tau\circ\rho(s) = \rho'(s)\circ \tau\quad\text{for all $s\in G$.}$$

The map $\tau$ is then said to be an isomorphism of the representations $\rho,\rho'$. So what (1) means more explicitly is "if $f$ is not an isomorphism of the representations $\rho_1,\rho_2$, we have $f = 0$."

  • 2
    $\begingroup$ That's not what (1) means. (1) says, exactly, "if there does not exist an isomorphism between $\rho_1$ and $\rho_2$, then $f = 0$." The statement you write is also true but very slightly stronger. $\endgroup$ Commented Sep 4, 2022 at 19:03
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    $\begingroup$ @QiaochuYuan: Yes, thank you for pointing this out. I wanted to draw attention to this. Originally I thought of simply citing the definition Serre gives, and saying that the best way to understand (1) would be take the contrapositive, but ultimately I think given that we have introduced $f$, it would be clearest and most useful to frame it this way. $\endgroup$
    – Alex Ortiz
    Commented Sep 4, 2022 at 20:09

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