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I am learning Schur's Lemma from page 4 here. It says

Schur's Lemma 1. If $(\rho_1, V_1)$ and $(\rho_2, V_2)$ are irreducible representations of a group $G$, then any nonzero homomorphism $\phi : V_1 \mapsto V_2$ is an isomorphism.

Proof. Assuming $\phi$ is nonzero, we can write $v_2 = \phi(v_1) \in V_2$ for some $v_1 \in V_1$. We can then say that $\rho_2(g)(v_2) = \rho_2(g)(\phi(v_1))$, which by the intertwining property of maps between representations, gives that $\rho_2(g)(v_2) = \phi( \rho_1 (g)(v_1)) \in \phi(V_1)$ ...

Why the $\phi$ can commute with $\rho_1$ and $\rho_2$ in the expression above?

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    $\begingroup$ That's the definition of a homomorphism being applied to $\phi$. $\endgroup$ – Jyrki Lahtonen Apr 24 '14 at 13:31
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A homomorphism between representations $(V_{1}, \rho_{1}), (V_{2}, \rho_{2})$ is exactly a linear map $\phi : V_{1} \rightarrow V_{2}$ such that for all $g \in G$,

$$\phi(\rho_{1}(g)(v_{1})) = \rho_{2}(g) \phi(v_{1}).$$

This is exactly what is going on in the proof above. $\phi$ is assumed to be a homomorphism and hence the action of $g$ commutes with the application of $\phi$.

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  • $\begingroup$ Thanks. It seems there are multiple usage of homomorphism, I am confused on what are definitions and what are derived properties. $\endgroup$ – ahala Apr 24 '14 at 20:33
  • $\begingroup$ You're welcome. That's not uncommon the first time you read some topic. Just go back to the definitions when you get stuck and try to parse through what the proof said. $\endgroup$ – Siddharth Venkatesh Apr 24 '14 at 20:36

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