Merry Christmas!

In the previous post I introduced the idea of an irreducible representation and showed that except in fields of low characteristic, these representations decompose completely. In this post I’ll present Schur’s Lemma at talk about what Schur and Maschke tell us about homomorphisms of representations.

1. Motivation

Fix a group $G$ now, and consider all isomorphism classes of finite-dimensional representations of $G$. We’ll denote this set by $\operatorname{Irrep}(G)$. Maschke’s Theorem tells us that any finite-dimensional representation $\rho$ can be decomposed as $$\bigoplus_{\rho_\alpha \in \operatorname{Irrep}(G)} \rho_{\alpha}^{\oplus n_\alpha}$$ where $n_\alpha$ is some nonnegative integer. This begs the question: what is $n_\alpha$? Is it even uniquely determined by $\rho$?

To answer this I first need to compute $\operatorname{Hom}_G(\rho, \pi)$ for any two distinct irreducible representations $\rho$ and $\pi$. One case is easy.

Lemma 1. Let $\rho$ and $\pi$ be non-isomorphic irreducible representations (not necessarily finite dimensional). Then there are no nontrivial homomorphisms $\phi : \rho \rightarrow \pi$. In other words, $\operatorname{Hom}_G(\rho, \pi) = \{0\}$.

I haven’t actually told you what it means for representations to be isomorphic, but you can guess – it just means that there’s a homomorphism of $G$-representations between them which is also a bijection of the underlying vector spaces.

Proof: Let $\phi : \rho_1 \rightarrow \rho_2$ be a nonzero homomorphism. We can actually prove the following stronger results.

• If $\rho_2$ is irreducible then $\phi$ is surjective.
• If $\rho_1$ is irreducible then $\phi$ is injective.

Exercise. Prove the above two results. (Hint: show that $\operatorname{Im} \phi$ and $\ker \phi$ give rise to subrepresentations.)

Combining these two results gives the lemma because $\phi$ is now a bijection, and hence an isomorphism. $\Box$

2. Schur’s lemma

Thus we only have to consider the case $\rho \simeq \pi$. The result which relates these is called Schur’s Lemma, but is important enough that we refer to it as a theorem.

Theorem 2 (Schur’s Lemma). Assume $k$ is algebraically closed. Let $\rho$ be a finite dimensional irreducible representation. Then $\operatorname{Hom}_{G} (\rho, \rho)$ consists precisely of maps of the form $v \mapsto \lambda v$, where $\lambda \in k$; the only possible maps are multiplication by a scalar. In other words,

$$\operatorname{Hom}_{G} (\rho, \rho) \simeq k$$

and $\dim \operatorname{Hom}_G(\rho, \rho) = 1$.

This is NOT in general true without the algebraically closed condition, as the following example shows.

Example. Let $k = {\mathbb R}$, let $V = {\mathbb R}^2$, and let $G = {\mathbb Z}_3$ act on $V$ by rotating every $\vec x \in {\mathbb R}^2$ by $120^{\circ}$ around the origin, giving a representation $\rho$. Then $\rho$ is a counterexample to Schur’s Lemma.

Proof: This representation is clearly irreducible because the only point that it fixes is the origin, so there are no nontrivial subrepresentations.

We can regard now $\rho$ as a map in ${\mathbb C}$ which is multiplication by $e^{\frac{2\pi i}{3}}$. Then for any other complex number $\xi$, the map “multiplication by $\xi$” commutes with the map “multiplication by $e^{\frac{2\pi i}{3}}$”. So in fact $$\operatorname{Hom}_G(\rho, \rho) \simeq {\mathbb C}$$ which has dimension $2$. $\Box$

Now we can give the proof of Schur’s Lemma.

Proof: Clearly any map $v \mapsto \lambda v$ respects the $G$-action.

Now consider any $T \in \operatorname{Hom}_G(\rho, \rho)$. Set $\rho = (V, \cdot_\rho)$. Here’s the key: because $k$ is algebraically closed, and we’re over a finite dimensional vector space $V$, the map $T$ has an eigenvalue $\lambda$. Hence by definition $V$ has a subspace $V_\lambda$ over which $T$ is just multiplication by $\lambda$.

But then $V_\lambda$ is a $G$-invariant subspace of $V$! Since $\rho$ is irreducible, this can only happen if $V = V_\lambda$. That means $T$ is multiplication by $\lambda$ for the entire space $V$, as desired. $\Box$

3. Computing dimensions of homomorphisms

Since we can now compute the dimension of the $\operatorname{Hom}_G$ of any two irreducible representations, we can compute the dimension of the $\operatorname{Hom}_G$ for any composition of irreducibles, as follows.

Corollary 3. We have

$$\dim \operatorname{Hom}_G \left( \bigoplus_\alpha \rho_\alpha^{\oplus n_\alpha}, \bigoplus_\beta \rho_\beta^{\oplus m_\beta} \right) = \sum_{\alpha} n_\alpha m_\alpha$$

where the direct sums run over the isomorphism classes of irreducibles.

Proof: The $\operatorname{Hom}$ just decomposes over each of the components as

$$\begin{aligned} \operatorname{Hom}_G \left( \bigoplus_\alpha \rho_\alpha^{\oplus n_\alpha}, \bigoplus_\beta \rho_\beta^{\oplus m_\beta} \right) &\simeq \bigoplus_{\alpha, \beta} \operatorname{Hom}_G(\rho_\alpha^{\oplus n_\alpha}, \rho_\beta^{\oplus m_\beta}) \\ &\simeq \bigoplus_{\alpha, \beta} \operatorname{Hom}_G(\rho_\alpha, \rho_\beta)^{\oplus n_\alpha m_\alpha}. \end{aligned}$$

Here we’re using the fact that $\operatorname{Hom}_G(\rho_1 \oplus \rho_2, \rho) = \operatorname{Hom}_G(\rho_1, \rho) \oplus \operatorname{Hom}_G(\rho_2, \rho)$ (obvious) and its analog. The claim follows from our lemmas now. $\Box$

As a special case of this, we can quickly derive the following.

Corollary 4. Suppose $\rho = \bigoplus_\alpha \rho_\alpha^{n_\alpha}$ as above. Then for any particular $\beta$,

$$n_\beta = \dim \operatorname{Hom}_G(\rho, \rho_\beta).$$

Proof: We have $$\dim \operatorname{Hom}_G(\rho, \rho_\beta) = n_\beta \operatorname{Hom}_G(\rho_\beta, \rho_\beta) = n_\beta$$ as desired. $\Box$

This settles the “unique decomposition” in the affirmative. Hurrah!

It might be worth noting that we didn’t actually need Schur’s Lemma if we were solely interested in uniqueness, since without it we would have obtained $$n_\beta = \frac{\dim \operatorname{Hom}_G(\rho, \rho_\beta)}{\dim \operatorname{Hom}_G(\rho_\beta, \rho_\beta)}.$$ However, the denominator in that expression is rather unsatisfying, don’t you think?

4. Conclusion

In summary, we have shown the following main results for finite dimensional representations of a group $G$.

• Maschke’s Theorem: If $G$ is finite and $\operatorname{char} k$ does not divide $\left\lvert G \right\rvert$, then any finite dimensional representation is a direct sum of irreducibles. This decomposition is unique up to isomorphism.
• Schur’s Lemma: If $k$ is algebraically closed, then $\operatorname{Hom}_G(\rho, \rho) \simeq k$ for any irreducible $\rho$, while there are no nontrivial homomorphisms between non-isomorphic irreducibles.

In the next post I’ll talk about products of irreducibles, and use them in the fourth post to prove two very elegant results about the irreducibles, as follows.

1. The number of (isomorphsim classes) of irreducibles $\rho_\alpha$ is equal to the number of conjugacy classes of $G$.
2. We have $\left\lvert G \right\rvert = \sum_\alpha \left( \dim \rho_\alpha \right)^2$.

Thanks to Dennis Gaitsgory, who taught me this in his course Math 55a. My notes for Math 55a can be found at my website.