# Schur's lemma in Serre's 'Linear Representations of Finite Groups'

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?

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 $$f = 0$$."
• 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. Commented Sep 4, 2022 at 19:03
• @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. Commented Sep 4, 2022 at 20:09