# Every representation of a finite groups decomposes into irreducible representations

I'm having trouble understanding this proposition from the first lecture in Fulton and Harris' representation theory book:

Proposition 1.8. For any representation $V$ of a finite group $G$, there is a decomposition $$V = V_1^{\bigoplus a_1}\oplus\cdots\oplus V_k^{\bigoplus a_k}$$ 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 $V_1$ that occur and their multiplicities $a_i$.

How should I understand the $k$ factors in the direct sum? Should I think of $V_i^{\oplus a_i}$ as a specific subspace of $V$? That is, is $V$ uniquely decomposable into these $k$ $G$-invariant subspaces of $V$, whose direct sum is $V$? Or could the corresponding $V_i^{\oplus a_i}$ from two different decompositions merely be isomorphic, rather than identical?

• What are the representations of $G$ when $G$ is a one-element group? What are the irreducible representations? Already in this case you will notice some things. – Orest Bucicovschi Feb 2 '15 at 5:04

## 1 Answer

What is unique is the decomposition of a representation (over a field of characteristic not dividing $|G|$) into its isotypic components

$$V \cong \bigoplus_{\lambda} V_{\lambda}$$

where $\lambda$ denotes an irreducible representation and $V_{\lambda}$ denotes the sum of all subspaces of $V$ isomorphic to $\lambda$. For example, if $G = \mathbb{Z}_2$, then there are two isotypic components, the "even" one corresponding to the trivial representation and the "odd" one corresponding to the sign representation. More generally, if $G = \mathbb{Z}_n$ then there are $n$ isotypic components, one for each $n^{th}$ root of unity, which can be thought of as the eigenspaces of a generator of $G$ acting on $V$.

Each isotypic component $V_{\lambda}$ contains some number $k_{\lambda}$ of copies of the irreducible representation $\lambda$, which is uniquely determined as it must be $\frac{\dim V_{\lambda}}{\dim \lambda}$. The further decomposition of $V_{\lambda}$ into a direct sum of $k$ copies of $\lambda$, however is not unique; it is more or less equivalent to choosing a basis of the multiplicity space $\text{Hom}(\lambda, V)$. For example, if $V$ is trivial, decomposing $V$ as a direct sum of copies of the trivial representation is more or less equivalent to choosing a basis of $V$.

What Fulton and Harris mean by "unique" is the weaker statement that the isomorphism classes of irreducible representations $\lambda$ that occur are uniquely determined, as are their multiplicities $k_{\lambda}$. They do not mean that the subspaces in that decomposition are unique as subspaces, because they aren't in the presence of multiplicities higher than $1$.

• I think you're saying the $\lambda$'s are not unique as subspaces, but what about the $V_{\lambda}$'s themselves? To consider a special case, if all the multiplicities are $1$, does that make the $V_{\lambda}$'s unique as subspaces rather than just up to isomorphism? – Will Nelson Feb 2 '15 at 4:51
• @Will: the $V_{\lambda}$ are always unique as subspaces. "The sum of all subspaces of $V$ isomorphic to $\lambda$" picks out a unique subspace. What is not unique is the decomposition of any particular $V_{\lambda}$ as a direct sum of copies of $\lambda$; this is, as I mentioned above, more or less equivalent to choosing a basis of a vector space called the multiplicity space. If this is still confusing I strongly encourage you to work out the special case of cyclic groups. – Qiaochu Yuan Feb 2 '15 at 4:51