# Decomposition of a representation into a direct sum of irreducible ones

I'm studying representation theory and in the book (Fulton and Harris) the author makes the following proposition with the following proof:

Proposition: 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.

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

I must confess I didn't understand. The fact that we can decompose $$V$$ like this I do understand that follows from the fact that if $$V$$ has a proper nonzero subrepresentation $$W$$ then there is another subrepresentation $$W'$$ such that $$V = W\oplus W'$$. In that case, if either $$W$$ or $$W'$$ are not irreducible we can apply the same idea to them, until we have the desired decomposition.

Now, this proof of uniqueness I really can't understand. I mean, uniqueness means that if we have

$$V = V_1^{\oplus a_1}\oplus\cdots \oplus V_k^{\oplus a_k}\simeq W_1^{\oplus b_1}\oplus\cdots \oplus W_{r}^{\oplus b_r},$$

then we have $$k = r$$, $$a_i = b_i$$ and $$W_i\simeq V_i$$. I can't understand how this argument the author presents shows all of this.

Indeed the whole point is that $$V$$ has these two decompositions then they are isomorphic, so that there exists one isomorphism

$$\varphi : V_1^{\oplus a_1}\oplus\cdots \oplus V_k^{\oplus a_k}\to W_1^{\oplus b_1}\oplus\cdots \oplus W_r^{\oplus b_r}.$$

If we restrict it to $$V_i$$ we get one isomorphism $$\varphi : V_i\to \varphi(V_i)$$. But why $$\varphi(V_i)=W_j$$ for some $$j$$? I mean, couldn't $$\varphi(V_i)$$ be some other subspace of the direct sum of the $$W_i$$ which is not one of the $$W_i$$ themselves?

So how to understand this proof about the decomposition of a representation? What really is the argument used in this proof?

I think this may be easier to understand if you change the notation a bit. Instead of grouping the direct summands by their isomorphism type, just list them all without grouping. So we have two decompositions $$V=\bigoplus S_m$$ and $$W=\bigoplus T_n$$, where each $$S_m$$ and each $$T_n$$ is irreducible. Given an isomorphism $$\varphi:V\to W$$, let $$\varphi_{mn}:S_m\to T_n$$ be the composition of $$\varphi$$ with the inclusion $$S_m\to V$$ and the projection $$W\to T_n$$. By Schur's lemma, each $$\varphi_{mn}$$ is either an isomorphism or $$0$$.

Now since $$\varphi$$ is injective, for each $$m$$ there must exist some $$n$$ such that $$\varphi_{mn}\neq 0$$. Thus for each $$m$$, there exists some $$n$$ such that $$\varphi_{mn}$$ is an isomorphism, and hence $$T_n\cong S_m$$. Moreover, $$\varphi_{mn}=0$$ for all $$n$$ such that $$T_n\not\cong S_m$$. This means that image of the restriction of $$\varphi$$ to $$S_m$$ is contained in the direct sum of all the $$T_n$$'s which are isomorphic to $$S_m$$.

Now fix an irreducible representation $$R$$ and let $$A\subseteq V$$ be the direct sum of all the $$S_m$$'s that are isomorphic to $$R$$, and let $$B$$ be the direct sum of all the other $$S_m$$'s, so $$V=A\oplus B$$. Similarly, let $$C\subseteq W$$ be the direct sum of all the $$T_n$$'s that are isomorphic to $$R$$, and $$D$$ be the direct sum of all the other $$T_n$$'s, so $$W=C\oplus D$$. The discussion above shows that $$\varphi(A)\subseteq C$$ and $$\varphi(B)\subseteq D$$. Since $$\varphi$$ is surjective, we must have $$\varphi(A)=C$$ and $$\varphi(B)=D$$. Thus $$\varphi$$ gives an isomorphism from $$A$$ to $$C$$. It follows that the number of $$S_m$$'s which are isomorphic to $$R$$ is equal to the number of $$T_n$$'s which are isomorphic to $$R$$, which is exactly what we wanted to prove.

Note that you're right that, for instance, $$\varphi(S_m)$$ might not actually be equal to any of the $$T_n$$. For instance, if $$G$$ is trivial, this is just saying that if you have two bases for the vector space, you can have a vector in one basis that is not a scalar multiple of any single vector in the other basis. But $$\varphi(S_m)$$ is still isomorphic to one of the $$T_n$$. Moreover, $$\varphi(A)$$ is actually equal to $$C$$, or in the language of the question, $$\varphi(V_i^{\oplus a_i})=W_j^{\oplus b_j}$$ for some $$j$$. So while the individual irreducible summands might not map to individual irreducible summands, when you group together all the irreducible summands of a given isomorphism type, they map to the sum of all the irreducible summands of the same isomorphism type.

• Great explanation! Commented Dec 4, 2016 at 3:47
• Is there an easy argument why the number of $S_m$'s isomorphic to $R$ is equal to the number of $T_n$'s isomorphic to $R$? Or equivalently, why do we get for all sets $I,J$ with $\oplus_{i\in I} R\cong\oplus_{j\in J}R$, that $|I|=|J|$ for irreducible $R$? I was able to find a lengthy proof for $C^*$-algebra representations, but this doesn't quite translate to the group representations. Commented Jul 14, 2017 at 2:32
• @Kalua: Just count dimensions. The cardinality $|I|$ is the dimension of $\bigoplus_{i\in I}R$ divided by the dimension of $R$. Commented Jul 14, 2017 at 2:34
• Oh, we're talking finite groups and therefore finite dimensional $R$ here, right? Then it's really that simple. I was coming here from an infinite dimensional ($C^*$-algebra) problem, in that case cardinality counting is not that easy, eg. $2\mathbb{N}\cong 3\mathbb{N}$. But I should open another question for that case. Commented Jul 14, 2017 at 2:40
• Fantastic answer! This really helped, thanks. Commented Aug 29, 2018 at 15:38

Consider the identity map $id: V \to V$. Since $id$ is bijective and commutes with the $G$-action, it is an isomorphism of representations. This implies that for any decomposition $$V= \bigoplus_j W_j^{\oplus b_j}$$ it must be that for each $W_j^{\oplus b_j}$ there exists $V_i^{\oplus a_i}$ such that $id(W_j^{\oplus b_j})=V_i^{\oplus a_i}$, otherwise incurring a violation of Schur's lemma. To see this, note each $W_j$ and $V_i$ are irreducible and hence any homomorphism of representations from one to the other is either the zero map or an isomorphism. But since $id$ is bijective, this means each $W_i$ is isomorphic to its image. This implies that up to a permutation of the factors, the decomposition is unique.

• Thanks for the answer. I must confess, though, that there are two points I still didn't get. First of all if we pick $I : V\to V$ the identity, then $I(v)=v$ so that $I(V_i)=V_i$. So how can $I(V_i)=W_j$? On the other hand if we pick some $\varphi$ map as in my question, why given $V_i$ there's $W_j$ such that $\varphi(V_i)=W_j$? Why can't $\varphi(V_i)$ be some subspace of the direct sum of the $W_i$ which is not any of the $W_i$ themselves?
– Gold
Commented Jun 5, 2016 at 22:52
• If $\varphi(V_i)$ were a proper subspace of $W_i$, then by Schur's lemma it would have to be $0$, as the $V_i$ and the $W_i$ are irreducible. Hence the kernel of $\varphi$ would be nontrivial, contradicting the fact that $V$ is isomorphic to itself Commented Jun 6, 2016 at 23:45
• This answer is wrong. It was explained in the first answer that $\phi(V_i)$ doesn't need to be a subspace of some $W_j$ Commented May 18, 2022 at 10:27
• @PaulJoh Notice it was never claimed in the answer that $\phi(V_i)$ is equal to, or a subspace of, some $W_j$, so there is no contradiction of the accepted answer. The claim is that the isotypic component of type $V_i$ is sent by $\phi$ to the isotypic component of type $W_j$. You are right though that my (now deleted) comment made this error. Commented May 21, 2022 at 5:48