# Help understanding proof involving Schur's Lemma.

In the book Representation Theory, A First Course by William Fulton and Joe Harris, they state the following:

Proposition 1.8 For any representation $$V$$ of a finite group $$G$$, there is a decomposition $$V=V_1^{\oplus a_1}\oplus\cdots\oplus V_k^{\oplus a_k},$$ where the $$V_i$$ are distinct irreducible representations. The decomposition of $$V$$ into a direct sum of the $$k$$ factors is unique, as are the $$V_i$$ that occur and their multiplicities $$a_i$$.

Proof. It follows from Schur's lemma that if W is another representation of G, with a decomposition $$W=\oplus W_j^{\oplus b_j}$$, and $$\phi:V\rightarrow W$$ is a map of representations, then $$\phi$$ must map the factor $$V_i^{\oplus a_i}$$ into that factor $$W_j^{\oplus b_j}$$ for which $$W_j\cong V_i$$; when applied to the identity map of $$V$$ to $$V$$, the stated uniqueness follows.

I don't quite follow the proof. I think for the first part, he uses that if $$\phi$$ is a linear map between representations, then $$f$$ restricted to $$W_i$$ is a map between irreducible representations and Schur's lemma tells us that those representations are isomorphic (correct?). Also, I am not sure about the last part "when applied to the identity map of $$V$$ to $$V$$, the stated uniqueness follows".

Edit: I found another proof in REPRESENTATION THEORY FOR FINITE GROUPS by Shaun Tan, but I also do not find it easy to understand that one. Specifically, I do not understand why he says that "If $$j=i$$, then $$\phi\left(V_{i}^{\oplus a_{i}}\right) \neq 0$$ for any $$i$$."

Theorem 4.3. For any finite-dimensional representation $$(\rho, V)$$ of a finite group $$G$$ there is a unique decomposition $$V=V_{1}^{\oplus a_{1}} \oplus V_{2}^{\oplus a_{2}} \oplus \ldots \oplus V_{i}^{\oplus a_{i}}$$ where the $$V_{i}$$ are inequivalent and irreducible with unique multiplicities $$a_{i}$$.

Proof. We suppose $$V=W_{1}^{\oplus b_{1}} \oplus W_{2}^{\oplus b_{2}} \oplus \ldots \oplus W_{j}^{\oplus b_{j}}$$. Then we let $$\phi: V \rightarrow V$$ be the identity map. We use Schur's Lemma. For each irreducible $$V_{i}^{\oplus a_{i}}$$, we restrict the domain of $$\phi$$ to that component. Then, either $$\phi=0$$ or $$\phi$$ is an isomorphism. If $$j=i$$, then $$\phi\left(V_{i}^{\oplus a_{i}}\right) \neq 0$$ for any $$i$$. For each component, $$\phi$$ is an isomorphism such that $$V_{i}^{\oplus a_{i}}$$ maps to $$W_{j}^{\oplus b_{j}}$$ where $$V_{i}$$ is isomorphic to $$W_{j}$$.

• Whose book? "They" indeed... Commented May 6, 2023 at 16:58
• @ArturoMagidin, William Fulton and Joe Harris. Commented May 6, 2023 at 17:05
• Put it in the post, not the comments. Commented May 6, 2023 at 17:08
• @ArturoMagidin, done Commented May 6, 2023 at 17:11

For any representation $$V$$ of a finite group $$G$$ there is a decomposition $$V \cong V_1^{\oplus a_1} \oplus \cdots \oplus V_k^{\oplus a_k},$$ where the $$V_i$$ are distinct irreducible representations. The decomposition of $$V$$ into a direct sum of the $$k$$ factors is unique, as are the $$V_i$$ that occur and their multiplicities $$a_i$$.
Suppose we have an isomorphism $$\phi: V = V_1^{\oplus a_1} \oplus \cdots \oplus V_k^{\oplus a_k} \to W = W_1^{\oplus b_1} \oplus \cdots \oplus W_\ell^{\oplus b_\ell},$$ where $$V_1, \dots, V_k$$ are distinct irreducible representations and so are $$W_1, \dots, W_\ell$$, and $$a_1, \dots, a_k, b_1, \dots, b_\ell$$ are positive integers. For $$1\le i \le k$$ and $$1 \le j \le \ell$$, let $$\phi_{ij}$$ denote the composition $$V_i^{\oplus a_i} \to V \stackrel{\phi}{\longrightarrow} W \to W_j^{\oplus b_j},$$ where the first map is inclusion and the last map is projection. By further restricting to one of the $$a_i$$ $$V_i$$-factors and further projecting to one of the $$b_j$$ $$W_j$$-factors, we get a $$G$$-module map $$V_i \to W_j$$. Now Schur's lemma applies and tells us that this map is zero unless $$V_i \cong W_j$$. Since this holds for each of the $$a_i$$ $$V_i$$-factors and each of the $$b_j$$ $$W_j$$-factors, it follows that $$\phi_{ij} = 0$$ unless $$V_i \cong W_j$$. Therefore $$\phi$$ must map $$V_i^{\oplus a_i}$$ into $$W_{i'}^{\oplus b_{i'}}$$ for some unique index $$i'$$ such that $$V_i \cong W_{i'}$$. The same argument applied to $$\phi^{-1}$$ shows that $$\phi^{-1}$$ maps $$W_{i'}^{b_{i'}}$$ into $$V_i^{\oplus a_i}$$, so $$\phi$$ restricts to an isomorphism $$V_i^{\oplus a_i} \stackrel{\sim}\to W_{i'}^{\oplus b_{i'}}$$. Now by comparing dimensions we get $$a_i \dim V_i = b_{i'} \dim W_{i'}$$, so $$a_i = b_{i'}$$ since $$\dim V_i = \dim W_{i'}$$.
• To be sure, is it correct that $\phi_{ij}=\pi_j\circ \phi \circ \varphi_i$, where $\varphi_i$ is the inclusion map that takes a vector in $V_i^{\oplus a_i}$ and sends it to $V$ as a direct summand and $\pi_j$ is the projection map that takes a vector in $W$ and projects it onto $W_j^{\oplus b_j}$? And restricting it further would mean that we consider $\psi_j\circ\phi_{ij}\circ \chi_i$ where $\chi_i$ takes a vector in $V_i$ and sends it to $V_i^{\oplus a_i}$ and $\psi_j$ takes a vector in $W_j^{\oplus b_j}$ and projects it onto $W_j$? Or am I misunderstanding something? Commented May 7, 2023 at 15:20
• @Jakamay Yes that's correct. But do note that there $a_i$ such maps $\chi_i$ (and similarly $b_j$ such maps $\psi_j$). They are given by $v \mapsto (v,0\dots,0)$, $v \mapsto (0, v,0, \dots,0)$, etc. Commented May 7, 2023 at 18:51