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Suppose I have a representation $V$ of a finite group $G$, and decompose $V$ into $V = V_1 \oplus V_2$ where $V_1$ and $V_2$ are irreducible representations.

If I understand correctly, it is known that for any other decomposition $V = W_1 \oplus W_2$ into irreducibles $W_1, W_2$, we must have $V_1 \cong W_1$ and $V_2 \cong W_2$ (after reordering if necessary). However, I am having trouble seeing why this must be true.

I believe the idea is to apply Schur's Lemma to the identity map $I \colon V \to V$. The image $I(V_1)$ must be isomorphic to $V_1$ by Schur's Lemma, but I don't see how we can be sure that this image can be identified with one of the $W_i$ as opposed to some new subrepresentation $U \subset W_1 \oplus W_2$.

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Let $W$ be an irreducible subrepresentation of $V_1 \oplus V_2$, where $V_1$ and $V_2$ are not isomorphic. Consider the corresponding projection maps $W \to V_1, W \to V_2$. By Schur's lemma, these maps are either zero or isomorphisms. If $W \to V_1$ is an isomorphism, then $W \to V_2$ must be zero, so $W$ must in fact be the subrepresentation $V_1 \oplus 0$. Similarly for if $W \to V_2$ is an isomorphism.

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  • $\begingroup$ Thank you, that was very clear. I still wonder however about the case when $V_1$ and $V_2$ are isomorphic. $\endgroup$ – GoatsRule Jan 17 '16 at 23:22
  • $\begingroup$ @Goats: in that case it can happen that both maps are isomorphisms. $\endgroup$ – Qiaochu Yuan Jan 18 '16 at 0:36

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