Representation theory of finite groups (4/4)

After spending some time in the realm of characters, we make a little detour to the beautiful world of algebraic number theory. Algebraic number theory is, roughly speaking, the study of finite extensions $$K$$ of $$\mathbb{Q}$$ and moreover the ring of integers of K, $$\mathcal{O}_K$$, which is all the elements in $$K$$ that satisifes a monic polynomial with integer coefficients. In such a ring, the amazing property that a non-zero ideal factorize uniquely into prime ideals of $$\mathcal{O}_K$$ holds. I will most definitely try to write on algebraic number theory later, so look out for that. In this article we will prove one of the first things one prove in algebraic number theory, namely that the algebraic integers form a ring.

$$\alpha \in \mathbb{C}$$ is said to be an algebraic integer if $$\alpha$$ satisfies a monic polynomial with integer coefficients. We start by proving a lemma we need:

Lemma 4.1: $$\alpha \in \mathbb{C}$$ is an algebraic integer if and only if $$(1,\alpha,\alpha^2,\alpha^3, \dots)$$ is finitely generated. Here $$(1,\alpha,\alpha^2,\alpha^3, \dots)$$ is the ideal generated by all powers of $$\alpha$$.

Proof. If $$(1,\alpha,\alpha^2,\alpha^3, \dots)$$ is finitely generated, then for some $$n \in \mathbb{N}_{\geq 1}$$ we have for $$a_i \in \mathbb{Z}$$ that $$\alpha^n = a_0+a_1\alpha+a_2\alpha^2+\dots+a_{n-1}\alpha^{n-1}$$ which is the same as saying that $$\alpha$$ satisfy a monic polynomial with integer coefficients. Conversely, if $$\alpha$$ is an algebraic number we can find $$a_i \in \mathbb{Z}$$, not all zero, so that for some $$n \in \mathbb{N}_{\geq 1}$$:  $$\alpha^n = a_0+a_1\alpha+\dots+a_{n-1}\alpha^{n-1}$$ Now we proceed by (strong) induction to prove that $$\alpha^{n+i} \in (1,\alpha,\dots,\alpha^{n-1} )$$. We just proved the base case, so assume $$\alpha^{n+i}\in (1,\alpha,\dots,\alpha^{n-1} )$$ for $$i=0,1,\dots,k$$. Then we have by the induction hypothesis that \begin{alignat*}{2} \alpha^{n+k+1} &=\alpha^{n+k}\alpha = (b_0+b_1\alpha+\dots+b_{n-1}\alpha^{n-1} )\alpha \\ &= b_0\alpha+b_1\alpha^2+\dots+b_{n-1}\alpha^n \\ &= b_0\alpha+b_1\alpha^2+\dots+b_{n-1}(a_0+a_1\alpha+\dots+a_{n-1}\alpha^{n-1}) \\ &= b_{n-1}a_0+(b_0+b_{n-1}a_1)\alpha+ \dots + b_{n-1}a_{n-1}\alpha^{n-1} \end{alignat*} which finishes the proof by induction. $$\blacksquare$$

Using this lemma we can prove that the algebraic integers forms a ring. We will denote this ring by $$\mathbb{A}_{\text{int}}$$.

Theorem 4.2: $$\mathbb{A}_{\text{int}}$$ is a subring of $$\mathbb{C}$$.

Proof. Clearly $$1 \in \mathbb{A}_{\text{int}}$$. What is left to prove is that if $$\alpha,\beta \in \mathbb{A}_{\text{int}}$$ then $$\alpha-\beta, \alpha\beta \in \mathbb{A}_{\text{int}}$$. Let $$\Gamma_{\alpha-\beta} = (1,\alpha-\beta,(\alpha-\beta)^2,(\alpha-\beta)^3)$$ and $$\Gamma_{\alpha\beta} = (1,\alpha\beta,(\alpha\beta)^2,\dots )$$. Defining $$\Gamma_{\alpha},\Gamma_{\beta}$$ similarly, we have by lemma 4.1 that both those are finitely generated, and hence we see that $$\Gamma_{\alpha-\beta}, \Gamma_{\alpha\beta}$$ also has to be finitely generated because these are the subgroups of a finitely generated abelian group. Again by lemma 4.1 that implies $$\alpha-\beta,\alpha\beta \in \mathbb{A}_{\text{int}}$$. $$\blacksquare$$

As one would expect by the name algebraic integers, $$\mathbb{A}_{\text{int}} \cap \mathbb{Q} = \mathbb{Z}$$, an exercise we leave to the reader. Our next goal is to prove that $$\chi(g)$$ is an algebraic integer. For this we use the fact that an invertible matrix of finite order is diagonalizable ( over $$\mathbb{C}$$ ). In the spirit of this article, the proof we present is representation theoretic, moreover it is a corollary of Schur’s lemma.

Lemma 4.3: Let $$A \in GL_n(\mathbb{C})$$ so that $$A^m = I$$ for some $$m \in \mathbb{N}_{\geq 1}$$. Then $$A$$ is diagonalizable with eigenvalues being $$m$$-th roots of unity.

Proof. We study the map $$\rho: \mathbb{Z}/m\mathbb{Z} \to GL_n(\mathbb{C})$$ defined by $$(k \pmod{m}) \mapsto A^{k \pmod{m}}$$. Since $$A^m = I$$, this is a well-defined map, and it is also seen to be a homomorphism, hence a representation. By Maschke’s theorem, we have a complete decomposition: $$\rho \sim \varphi^{(1)} \oplus \cdots \oplus \varphi^{(r)}$$ Since $$\mathbb{Z}/m\mathbb{Z}$$ is abelian, it follows by corollary 1.6 from the first article in this series that $$\varphi^{(i)}$$ is of degree 1, and hence $$\varphi^{(i)}: \mathbb{Z}/m\mathbb{Z} \to \mathbb{C}^{*}$$ and $$r=n$$. Let $$T: \mathbb{C}^n \to \mathbb{C}^n$$ be the equivalence of $$\rho$$ with $$\varphi^{(1)} \oplus \cdots \oplus \varphi^{(r)}$$. Then $$T^{-1}\rho_gT = \begin{pmatrix} \varphi^{(1)}_g & 0 & \cdots & 0 \\ 0 & \varphi^{(2)}_g & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & \varphi^{(n)}_g \end{pmatrix}$$ for all $$g \in \mathbb{Z}/m\mathbb{Z}$$. Letting $$g=1$$ we get that $$T^{-1}AT=\begin{pmatrix} \varphi^{(1)}_1 & 0 & \cdots & 0 \\ 0 & \varphi^{(2)}_1 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & \varphi^{(n)}_1 \end{pmatrix}=\begin{pmatrix}\lambda_1 & 0 & \cdots & 0 \\ 0 & \lambda_2 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & \lambda_n \end{pmatrix}$$ for $$\lambda_i \in \mathbb{C}^*$$ which are the eigenvalues of $$A$$, so $$A$$ is diagonalizable. Furthermore $$\begin{pmatrix}\lambda_1^m & 0 & \cdots & 0 \\ 0 & \lambda_2^m & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & \lambda_n^m \end{pmatrix} =(T^{-1}AT)^m = T^{-1}A^mT = I$$ so all eigenvalues are $$m$$-th roots of unity. $$\blacksquare$$

Clearly an $$m$$-th root of unity is an algebraic integer. In the following we prove that an arbitrary character evaluated in $$g$$ is indeed a sum of $$m$$-th roots of unity and hence an algebraic integer since $$\mathbb{A}_{\text{int}}$$ is a ring.

Lemma 4.4: Let $$\chi$$ be a character of a finite group $$G$$, with $$|G|=m$$. Then for any $$g \in G$$, there exists $$\lambda_{g1},\dots,\lambda_{gm}$$, which are $$m$$-th roots of unity so that $$\chi(g)=\lambda_{g1}+\dots+\lambda_{gm}$$.

Proof. Let $$\rho: G \to GL_m(\mathbb{C})$$ be a representation with character $$\chi$$. Fix $$g \in G$$. Then $$g^m=1$$. Hence $$I=\rho_{g^m}=\rho_g^m$$, and so $$\rho_g$$ is diagonalizable by lemma 4.3 – say, with eigenvalues $$m$$-th roots of unity $$\lambda_{g1},\dots,\lambda_{gm}$$. Hence $$\chi(g)=\lambda_{g1}+\dots+\lambda_{gm}$$. $$\blacksquare$$

We need one final lemma before we can prove one of the main goals of this section: the dimension theorem. Without further ado let us jump into it.

Lemma 4.5: Let $$G$$ be a finite group and $$\varphi$$ an irreducible representation of $$G$$ with degree $$d$$. Fix some $$g \in G$$, and let $$h$$ be the order of the conjugacy class of $$g$$ in $$G$$. If $$\chi_{\varphi}$$ is the associated character to $$\varphi$$ then $$\frac{h}{d}\chi_{\varphi}(g)$$ is an algebraic integer.

Proof. Let us first list all the conjugacy classes of $$G$$: $$C_1,\dots,C_k$$. Since characters are class functions we let $$\chi_i$$ be the value that $$\chi_{\varphi}$$ takes on $$C_i$$. If we are able to show that $$\frac{|C_i|}{d}\chi_i$$ is an algebraic integer for each $$i$$ we are done. The proof revolves around one operator: $$T_i = \sum_{x \in C_i} \varphi_x$$ We claim three things:
1) $$T_i = \frac{|C_i|}{d}\chi_i I$$
2) $$T_iT_j = \sum_{\ell=1}^k a_{ij\ell} T_\ell$$ for some $$a_{ij\ell} \in \mathbb{Z}$$.
3) $$\alpha \in \mathbb{C}$$ is an algebraic integer if and only if there exists $$\alpha_1,\dots,\alpha_t$$ not all zeros such that $$\alpha\alpha_i=\sum_{j=1}^t a_{ij}\alpha_j$$ where $$a_{ij} \in \mathbb{Z}$$ for all $$1\leq i \leq t$$.

The last claim is a standard fact from algebraic number theory, and we thus leave it to the reader to take on good faith or prove (for the converse direction, use that eigenvalues of integer matrices are algebraic integers).

If we for a minute assume 1) and 2) is true we get that $$\left ( \frac{|C_i|}{d}\chi_i\right)\left(\frac{|C_j|}{d}\chi_j\right) = \sum_{\ell=1}^k a_{ij\ell} \left( \frac{|C_\ell|}{d}\chi_\ell \right)$$ which by 3) implies that $$\frac{|C_i|}{d}\chi_i$$ is an algebraic integer. 2) is first a routine calculation then checking that the coefficients are independent of the conjugacy class. We omit that here. For 1) we start by observing $$\varphi_g T_i \varphi_{g^{-1}} = \sum_{x \in C_i} \varphi_g\varphi_x\varphi_{g^{-1}} = \sum_{x \in C_i} \varphi_{gxg^{-1}} = T_i$$ where we in the very last line has used that $$C_i$$ is closed under conjugation and that conjugation by $$g$$ induces a bijection of $$C_i$$. Thus $$\varphi_g T_i \varphi_{g^{-1}} = T_i$$ for all $$g \in G$$. We can then conclude by Schur’s lemma that $$T_i = \lambda I$$ for some $$\lambda \in \mathbb{C}$$. Finding $$\lambda$$ is easy: $$d\lambda = \text{Tr}(\lambda I ) = \text{Tr}(T_i) = \text{Tr}\left( \sum_{x \in C_i} \varphi_x \right) = \sum_{x \in C_i} \chi_i = |C_i|\chi_i$$ This finishes the proof. $$\blacksquare$$

Theorem 4.6: (Dimension theorem)
Let $$\varphi$$ be an irreducible representation of $$G$$ with degree $$d$$. Then $$d \mid G$$.

Proof. By the first orthogonality relations (Theorem 2.5) we have $$1 = \frac{1}{|G|} \sum_{ g \in G} \overline{\chi_{\varphi}(g)}\chi_{\varphi}(g) = \frac{d}{|G|} \sum_{ g \in G} \overline{\chi_{\varphi}(g)}\frac{\chi_{\varphi}(g)}{d}$$ Let us list all the conjugacy classes of $$G$$ as before: $$C_1,\dots,C_k$$. Again let $$\chi_i$$ denote the value $$\chi$$ takes on $$C_i$$. From the calculation from the orthogonality relations over we then get $$\frac{|G|}{d} = \sum_{ g \in G} \overline{\chi_{\varphi}(g)}\frac{\chi_{\varphi}(g)}{d} = \sum_{i=1}^k \sum_{ g \in C_i} \overline{\chi_{{\varphi}_i}}\left( \frac{1}{d} \chi_{{\varphi}_i} \right) = \sum_{i=1}^k \overline{\chi_{{\varphi}_i}}\left( \frac{|C_i|}{d} \chi_{{\varphi}_i} \right)$$ Now $$\frac{|C_i|}{d} \chi_{{\varphi}_i}$$ is an algebraic integer by Lemma 4.5 and since the algebraic integers are closed under complex conjugation, so is $$\overline{\chi_{{\varphi}_i}}$$. Since the algebraic integers is a ring it follows that the last sum is an algebraic integer, that is $$\frac{|G|}{d}$$ is an algebraic integer, and hence it an actual integer. In other words $$d \mid |G|$$. $$\blacksquare$$

With the dimension theorem proven we can finally prove what we promised in the start of these blog posts:

Theorem 4.7: Let $$G$$ be a group of order $$p^2$$ for a prime $$p$$. Then $$G$$ is abelian.

Proof. By Corollary 3.3 from the previous article we have $$p^2=|G| = d_1^2+\dots+d_k^2$$ where $$d_i$$ are the degrees of the (equivalence classes of) irreducibles representations of $$G$$. By the dimension theorem $$d_i = 1,p,p^2$$ for any $$i$$. The trivial representation has degree $$1$$ and is irreducible, and hence by cardinality $$d_i = 1$$ for all $$i$$. We are now done by Theorem 3.4 $$\blacksquare$$.

Representation theory of finite groups (3/4)

In the previous post we introduced characters and looked at the orthogonality relations. In this post we continue our study of representation by looking at the regular representation – a gem of representation theory. Someone at Discord told me that

It is the whole reason group theory can be transported to operator algebras

which says a lot about its importance already. Maybe another blog post will look more deeply at this representation in the distant future. In this post we will really just touch the surface of the theory behind it.

The (left) regular representation is simply $$G$$ acting on itself by left translation. We make this more precise. Let $$\mathbb{C}G = \left \{ \sum_{g \in G} \alpha_g g \right \}$$ so in other words $$\mathbb{C}G$$ is all formal linear combinations of elements of $$G$$. We put a vector space structure on $$\mathbb{C}G$$ the obvious way, i.e. $$\sum_{g \in G} \alpha_g g + \sum_{g \in G} \beta_g g := \sum_{g \in G} (\alpha_g + \beta_g)g$$ and $$c\sum_{g \in G} \alpha_g g = \sum_{g \in G} (c\alpha_g) g$$. Two elements $$\sum_{g \in G} \alpha_g g, \sum_{g \in G} \beta_g g$$ are equal exactly when $$\alpha_g = \beta_g \enspace \forall g \in G$$. This shows that $$G$$ is indeed a basis for $$\mathbb{C}G$$. Furthermore we make $$\mathbb{C}G$$ into an inner product space with: $$\left \langle \sum_{g \in G} \alpha_g g, \sum_{g \in G} \beta_g g \right \rangle = \sum_{g \in G} \alpha_g\overline{\beta_g}$$ The (left) regular representation $$L : G \to GL(\mathbb{C}G)$$ of $$G$$ is defined by $$L_g \sum_{h \in G} \alpha_h h = \sum_{h \in G} \alpha_h gh$$ We will start with working towards showing that the regular representation decompose very nicely. From now on, whenever we say regular representation, we mean the left regular representation.

Lemma 3.1: The regular representation of $$G$$ is a unitary representation.

Proof. We first show that the regular representation is, well, a representation. Firstly, $$L_g$$ is a linear transformation: $$L_g(\alpha g_1 + g_2) = (\alpha g_1 + g_2)h = \alpha g_1h + g_2h = \alpha L_g(g_1)+L_g(g_2)$$ and hence \begin{alignat*}{2} L_{g_1g_2}\sum_{h \in G} \alpha_h h &= \sum_{h \in G} L_{g_1g_2} \alpha_h h \\ &= \sum_{h \in G} g_1g_2 \alpha_h h \\ &= g_1 \sum_{h \in G} g_2\alpha_h h \\ &= L_{g_1}L_{g_2} \end{alignat*} For unitaryness we compute: \begin{alignat*}{2} \left \langle L_g\sum_{h \in G} \alpha_h h, L_g\sum_{h \in G} \beta_h h \right \rangle &= \left \langle \sum_{h \in G} \alpha_h gh, \sum_{h \in G} \beta_h gh \right \rangle \\ &= \left \langle \sum_{x \in G} \alpha_{xg^{-1}} x, \sum_{x \in G} \beta_{xg^{-1}} x \right \rangle \\ &= \sum_{x \in G} \alpha_{xg^{-1}}\overline{\beta_{xg^{-1}}}\end{alignat*} where we have used the “substitution”-trick mentioned earlier with $$x= gh$$. Now $$\sum_{x \in G} \alpha_{xg^{-1}}\overline{\beta_{xg^{-1}}} = \sum_{h \in G} \alpha_{h}\overline{\beta_h} = \langle \sum_{h \in G} \alpha_h h , \sum_{h \in G} \beta_h h \rangle$$ which finishes the proof. $$\blacksquare$$

We continue our investigation by the regular representation by calculating its character. To do this we start by fixing an ordered basis of $$\mathbb{C}G$$: $$\mathcal{B} = \{g_1,\dots,g_n\}$$. Then $$L_{g}(g_j) = gg_j = g_i$$ for some $$i$$. Hence if $$[L_g]$$ is the matrix of $$L_g$$ with respect to the basis $$\mathcal{B}$$, we see that $$[L_g]_{ij} = \left\{\begin{matrix} 1 & \text{ if } gg_j = g_i \\ 0 & \text{ otherwise } \end{matrix}\right.$$ Now $$gg_i = g_i$$ if and only if $$g=e$$ and so $$\chi_{L}(g)= \left\{\begin{matrix} |G| & \text{ if } g=e \\ 0 & \text{ otherwise } \end{matrix}\right.$$ As the calculation over shows, the character of $$L$$ is very nice to work with. The fact that $$\chi_L(e) = |G|$$ is no surprise for we choosed $$G$$ as a basis to start with. This nevertheless connects $$L$$ to the group in a very specific way: the order of the regular representation is exactly the order of the group. This somewhat trivial insight will be proven very fruitful towards our goal of classifying all groups of order $$p^2$$. This comes down to what is known as the dimension theorem, which we will prove in the next text, after a little detour into the land of algebraic number theory.

Theorem 3.2: Let $$L$$ be the regular representation of $$G$$. Then the decomposition $$L \sim \deg(\varphi^{(1)})\varphi^{(1)} \oplus \cdots \oplus \deg(\varphi^{(s)})\varphi^{(s)}$$ where $$\{\varphi^{(1)}, \dots, \varphi^{(s)} \}$$ is a complete set of representatives of the pairwise inequivalent irreducible representations of $$G$$.

Proof. $$\langle \chi_{L}, \chi_{\varphi^{(i)}} \rangle = \frac{1}{|G|} \sum_{g \in G} \chi_L(g)\overline{\chi_{\varphi{(i)}}(g)} = \frac{1}{|G|}|G| \deg(\varphi^{(i)})$$ and so the theorem follows from theorem 2.7 from the previous article. $$\blacksquare$$

Corollary 3.3: If $$\varphi^{(1)}, \dots, \varphi^{(s)}$$ is a complete list of all inequivalent irreducible representations of $$G$$, then $$|G| = (\deg(\varphi^{(1)}))^2+\dots+(\deg(\varphi^{(s)}))^2$$.

Proof. Evaluating the character of $$L$$ in $$e$$ we get that: \begin{alignat*}{2} |G|&=\deg(\varphi^{(1)})\chi_{\varphi^{(1)}}(e) + \dots + \deg(\varphi^{(s)})\chi_{\varphi^{(s)}}(e) \\ &= (\deg(\varphi^{(1)}))^2 + \dots + (\deg(\varphi^{(s)}))^2 \end{alignat*} which is the desired conclusion. $$\blacksquare$$

We will end this post by proving a very nice requirement for when a group is abelian, extending on a result from the first post. Let us fix a complete list of all inequivalent irreducible representations of $$G$$:  $$\varphi^{(1)}, \dots, \varphi^{(s)}$$. In this article, from now on $$\chi_{i}$$ will denote $$\chi_{\varphi^{(i)}}$$  . Furthermore we let $$d_i = \deg(\varphi^{(i)})$$. A function $$f: G \to \mathbb{C}$$ that is constant on conjugacy classes is called a class function. It is clear that the class functions form a subspace of the group algebra, and it can be proven that this is actually the center of the group algebra, and hence we denote them by $$Z(L(G))$$. We will not use this fact, so we only borrow the notation. As we proved earlier, all characters are class functions.  Let $$\text{Cl}(G)$$ be the set of conjugacy classes of $$G$$. Then for $$C \in \text{Cl}(G)$$ we define the function $$\delta_{C}: G \to \mathbb{C}$$ by $$\delta_{C}(g) = \left\{\begin{matrix} 1 & \text{ if } g \in C \\ 0 & \text{ otherwise } \end{matrix}\right.$$ Then as class functions are constant on conjugacy classes we can represent $$f \in Z(L(G))$$ like $$f = \sum_{C \in \text{Cl}(G)} f(C)\delta_C$$ Furthermore the orthogonality $$\langle \delta_C,\delta_{C’} \rangle = \frac{1}{|G|} \sum_{g \in G} \delta_C(g)\overline{\delta_{C’}(g)} = 0$$ whenever $$C \neq C’$$ shows us that $$\{\delta_C \}_{C \in \text{Cl}(G)}$$ is a linearly independent set and so we conclude that $$\{\delta_C\}_{C \in \text{Cl}(G)}$$ is a  basis for $$Z(L(G))$$. This shows that $$\dim(Z(L(G)) = |\text{Cl}(G)|$$.

We know from the first post that if $$G$$ is abelian then all its irreducible representations have degree $$1$$. Conversely if all irreducible representations have degree $$1$$, then Corollary 3.3 implies there are exactly $$|G|$$ irreducible representations. Since the characters of inequivivalent representations are linearly independent by the orthogonality relations, it follows that the space spanned by all the characters of inequivalent representations have at least dimension $$|G|$$. But! All the characters are living inside $$Z(L(G))$$ and we have just seen that $$\dim(Z(L(G))) = |\text{Cl}(G)|$$, so at one side $$\dim(Z(L(G))) \geq |G|$$ but $$|\text{Cl}(G)|\leq|G|\leq \dim(Z(L(G))) = \text{Cl}(G)$$. So we deduce the equality $$\text{Cl}(G)=|G|$$, or in other words – $$G$$ is abelian. Let us state that properly:

Theorem 3.4: Let $$G$$ be a finite group. $$G$$ is abelian if and only if all irreducible representations have degree $$1$$.

Representation theory of finite groups (2/4)

We continue our adventure in the land of representation theory of finite groups. Our goal in this post is to introduce characters and prove (one of) Schur’s orthogonality relations. In the previous post we never really used that we can interpret $$GL(V)$$ as $$GL_n(\mathbb{C})$$ – this will figure more prominently in this text. We start of by concretizing this.

Let $$\rho: G \to GL(V)$$ be a representation, where $$\dim(V)=n$$. Choose a basis $$\mathcal{B} = \{\beta_1,\dots,\beta_n \}$$ of $$V$$. Let $$e_i$$ denote the standard $$i$$-th unit vector of $$\mathbb{C}^n$$. Then the linear map $$T: V \to \mathbb{C}^n$$ defined by $$\beta_1 \mapsto e_i$$ is an isomorphism, and so the representation $$\varphi: G \to GL_n(\mathbb{C})$$ defined by $$\varphi_g = T\rho_g T^{-1}$$ for all $$g \in G$$ is equivalent to $$\rho$$. A direct sum of two representations $$\rho^{(1)}: G \to GL(V), \rho^{(2)}: G \to GL(W)$$ where $$\dim(V)=n,\dim(W)=m$$ can hence be interpreted as $$\rho^{(1)} \oplus \rho^{(2)} : G \to GL_{m+n}(\mathbb{C})$$ so that for all $$g \in G$$ we have $$(\rho^{(1)} \oplus \rho^{(2)})_g = \begin{pmatrix} \rho^{(1)}_g & 0 \\ 0 & \rho^{(2)}_g \end{pmatrix}$$

To study characters we introduce what is called the group algebra of $$G$$. This is denoted as $$L(G)$$ or $$\mathbb{C}^G$$ and defined as $$\{ f | f: G \to \mathbb{C} \}$$, that is all functions from $$G$$ to $$\mathbb{C}$$. Characters live inside this group algebra as we will see towards the end of this post. Defining addition by $$(f_1 + f_2)(g) = f_1(g) + f_2(g)$$ and scalar multiplication by $$(cf)(g) = cf(g)$$ we get a vector space structure on $$\mathbb{C}^{G}$$. Actually we can turn $$\mathbb{C}^G$$ into an inner product space, where the inner product is $$\langle f_1,f_2 \rangle = \frac{1}{|G|}\sum_{g \in G } f_1(g) \overline{f_2(g)}$$ That this is indeed an inner product is a routine calculation and will be left to the reader.

In light of lemma 1.2 from the previous article, we focus on unitary representations. By $$U_n(\mathbb{C})$$ we will mean the group of unitary $$n \times n$$-matrices. Let $$\rho : G \to GL(V)$$ and $$\varphi : G \to GL(W)$$ be representations of $$G$$ and $$T : V \to W$$ be a linear transformation. Then we define $$T^{\sharp}: V \to W$$ by $$T^{\sharp} = \frac{1}{|G|} \sum_{g \in G } \varphi_{g^{-1}}T\rho_g$$ It isn’t hard to see that $$T^{\sharp} \in \text{Hom}_G(\rho,\varphi)$$ and this is a good exercise in the “change of variables”-trick from the previous article. The following is also true.

Lemma 2.1: Using the notation from the paragraph over:

1. If $$T \in \text{Hom}_G(\rho,\varphi)$$, then $$T^{\sharp} = T$$.
2. As a consequence, $$\mathcal{P}: \text{Hom}(V,W) \to \text{Hom}_G(\rho,\varphi)$$ defined by $$T \mapsto T^{\sharp}$$ is a linear map which is onto.

Proof. If $$T \in \text{Hom}_G(\rho,\varphi)$$, then $$\varphi_g T = T \rho_g$$ for all $$g \in G$$. Using this we get $$T^{\sharp} = \frac{1}{|G|} \sum_{g \in G} \varphi_{g^{-1}}T\rho_g = \frac{1}{|G|} \sum_{g \in G} \varphi_{g^{-1}}\varphi_g T = \frac{1}{|G|} \sum_{g \in G} T = T$$ The linearity of $$\mathcal{P}$$ is clear and so is the surjectiveness by 1). $$\blacksquare$$.

In the case where $$\rho$$ and $$\varphi$$ are irreducible representations, we can describe $$T^{\sharp}$$ very specifically:

Lemma 2.2: Let $$\rho$$ and $$\varphi$$ be as before and also irreducible this time. Furthemore let $$T: V \to W$$ be a linear transformation. If $$\rho \not \sim \varphi$$ then $$T^{\sharp} = 0$$. If $$\rho = \varphi$$ then $$T^{\sharp} = \frac{\text{Tr}(T)}{\deg(\rho)} I$$

Proof. By Schur’s lemma $$T^{\sharp} = 0$$ if $$\rho \not \sim \varphi$$. If $$\rho = \varphi$$ then Schur’s lemma tell us that $$T^{\sharp} = \lambda I$$ for some $$\lambda \in \mathbb{C}$$. Our goal is hence to find $$\lambda$$. We use the well-known concept of trace from linear algebra to do this: $$\lambda\deg(\rho)=\lambda\dim(V)=\lambda \text{Tr}(I)=\text{Tr}(\lambda I)=\text{Tr}(T^{\sharp})$$ What is left to show is that $$\text{Tr}(T)=\text{Tr}(T^{\sharp})$$. By using that $$\text{Tr}(AB)=\text{Tr}(BA)$$ we get that \begin{alignat*}{2} \text{Tr}(T^{\sharp}) &= \frac{1}{|G|} \sum_{g \in G} \text{Tr}(\rho_{g^{-1}}T\rho_g) \\ &= \frac{1}{|G|} \sum_{g \in G } \text{Tr}(T\rho_g\rho_{g^{-1}}) \\ &= \frac{1}{|G|} \sum_{g \in G } \text{Tr}(T) \\ &= \text{Tr}(T) \end{alignat*} This finishes the proof. $$\blacksquare$$

We are closing in on the orthogonality relations. What is left is just one lemma. Before stating and proving it we write down a formula related to matrix multiplication. We will need this in the following proof of the lemma. The formula follows straight away from the definition of matrix multiplication, and here it is in all its glory: Let $$A$$ be a $$r \times m$$-matrix, $$B$$ a $$n \times s$$-matrix and $$E_{ki}$$ a $$m \times n$$-matrix, where $$E_{ki}$$ is the matrix with zeros everywhere except a $$1$$ at $$(k,i)$$. Then if $$A = (a_{ij}), B=(b_{ij}$$, the following formula holds $$(AE_{ki}B)_{\ell j} = a_{\ell k}b_{ij}$$ With this scary-looking formula stated, we prove the last ingredient we need for Schur’s orthogonality relations.

Lemma 2.3: Let $$\rho:G \to U_n(\mathbb{C})$$ and $$\varphi: G \to U_m(\mathbb{C})$$ be unitary represenations of $$G$$. Then $$(E_{ki}^{\sharp})_{\ell j} = \langle \rho_{ij}, \varphi_{k\ell} \rangle$$

Proof. Recall from linear algebra that for a unitary operator $$T$$, we have $$TT^*=I=T^*T$$. Here $$T^*$$ denotes the adjoint of $$T$$. Since $$\varphi_g$$ is unitary we have $$I=\varphi_g\varphi_{g^{-1}}$$ and so $$\varphi_{g^{-1}}=\varphi_g^{-1} = \varphi_g^{*}$$. This implies that $$\varphi_{g^{-1}})_{\ell k} = \overline{(\varphi_{g})_{k\ell}}$$. Using the formula from the paragraph over we get \begin{alignat*}{2} (E_{ki}^{\sharp})_{\ell j} &= \frac{1}{|G|} \sum_{g \in G} (\varphi_{g^{-1}} T \rho_{g})_{\ell j} \\ &= \frac{1}{|G|} \sum_{g \in G} (\varphi_{g^{-1}})_{\ell k}(\rho_g)_{ij} \\ &= \frac{1}{|G|} \sum_{g \in G} \overline{(\varphi_{g})_{k\ell}}(\rho_g)_{ij} \\ &= \langle \rho_{ij}, \varphi_{k\ell} \rangle \end{alignat*} We are done. $$\blacksquare$$

Theorem 2.4: (Schur’s orthogonality relations)
Let $$\rho: G \to U_n(\mathbb{C}), \varphi:G \to U_m(\mathbb{C})$$ be inequivalent unitary representations of $$G$$. Then $$\langle \rho_{ij},\varphi_{k \ell} \rangle = 0$$. Furthermore $$\langle \rho_{ij},\rho_{k\ell}\rangle = \left\{\begin{matrix} 1/n & \text{ if }i=k,\ell=j \\ 0 & \text{otherwise} \end{matrix}\right.$$

Proof. By lemma 2.3, when $$E_{ki}$$ is a $$m \times n$$-matrix, $$(E_{ki}^{\sharp})_{\ell j} = \langle \rho_{ij}, \varphi_{k\ell} \rangle$$, and by lemma 2.2 $$E_{ki}^{\sharp}=0$$, and so $$\langle \rho_{ij},\varphi_{k \ell} \rangle = 0$$. Now, suppose $$E_{ki}$$ is a $$n \times n$$-matrix, so a matrix representing a linear transformation $$V \to V$$. Then by lemma 2.2 we have $$E_{ki}^{\sharp} = \frac{\text{Tr}(E_{ki})}{\deg(\rho)}I = \frac{\text{Tr}(E_{ki})}{n}I$$ By lemma 2.3 we then get $$\langle \rho_{ij}, \varphi_{k\ell} \rangle = (E_{ki}^{\sharp})_{\ell j} = \left(\frac{\text{Tr}(E_{ki})}{n}I \right)_{\ell j} = \left\{\begin{matrix} 1/n & \text{ if } i=k, \ell = j \\ 0 & \text{otherwise} \end{matrix}\right.$$ which was what we wanted to prove. $$\blacksquare$$

With this out of the way we finally introduce characters. To a representation $$\rho: G \to GL(V)$$ we define the character $$\chi_\rho: G \to \mathbb{C}$$ by $$g \mapsto \text{Tr}(\rho_g)$$. This is well-defined as the trace is independent of basis. From a character we can get a lot of information about the representation. For example evaluating the character in the identity element we see $$\chi_\rho (e) = \text{Tr}(\rho_e) = \text{Tr}(I)=\dim(V)=\deg(\rho)$$ Thanks to $$\text{Tr}(AB)=\text{Tr}(BA)$$, characters are preserved by equivalence of representations as the following calculation shows $$\chi_{\rho}(g) = \text{Tr}(\rho_g) = \text{Tr}(T\varphi_gT^{-1})=\text{Tr}(\varphi_gT^{-1}T) = \chi_{\varphi}(g)$$ Another crucial property of the character is that it is constant on conjugacy classes, and this again follows by our good friend  $$\text{Tr}(AB)=\text{Tr}(BA)$$: $$\chi_{\rho}(hgh^{-1}) = \text{Tr}(\rho_{hgh^{-1}}) = \text{Tr}(\rho_h\rho_g\rho_{h^{-1}})=\text{Tr}(\rho_{h^{-1}}\rho_h\rho_g) = \text{Tr}(\rho_g)$$

From Schur’s orthogonality relations we get the first orthogonality relations. This is a central theorem in the theory of representations of finite groups as it shows that the (pairwise inequivalent) irreducible characters (a character is irreducible when its associated representation is irreducible) form an orthonormal set. It also gives a way to check whether a representation is irreducible or not. We will use the theorem to prove the uniqueness of the decomposition provided in Mashcke’s theorem from the previous article.

Theorem 2.5: (First orthogonality relations)
Let $$\rho: G \to GL(V)$$, $$\varphi: G \to GL(W)$$ be irreducible representations of $$G$$. Then $$\langle \chi_\rho, \chi_\varphi \rangle = \left\{\begin{matrix} 1 & \text{ if } \rho \sim \varphi \\ 0 & \text{ if } \rho \not\sim \varphi \end{matrix}\right.$$

Proof. Without loss of generalization we can assume, by lemma 1.2 from the previous article, that $$\rho, \varphi$$ are unitary representation, say $$\rho:G \to U_n(\mathbb{C}), \varphi:G \to U_m(\mathbb{C})$$. If $$\rho \not \sim \varphi$$ then Schur’s orthogonality relations yield that $$\langle \rho_{ii},\varphi_{jj} \rangle = 0$$ for all $$i,j$$. If $$\rho \sim \varphi$$, then since $$\chi_{\rho} = \chi_{\varphi}$$, there is no harm done is assuming that $$\rho = \varphi$$. In this case Schur’s orthogonality relations tells us that $$\langle \rho_{ii} , \rho_{jj}\rangle = \frac{1}{n}$$ if $$i = j$$ and $$0$$ if $$i \neq j$$. Now we compute \begin{alignat*}{2} \langle \chi_{\rho},\chi_{\varphi} \rangle &= \frac{1}{|G|} \sum_{g \in G} \chi_{\rho}(g)\overline{\chi_{\varphi}(g)} \\ &= \frac{1}{|G|}\sum_{g \in G} \sum_{i=1}^n \left(\rho_{g}\right)_{ii} \sum_{j=1}^m \overline{\left( \varphi_g \right)_{jj}} \\ &= \sum_{i=1}^n \sum_{j=1}^m \langle \rho_{ii},\varphi_{jj} \rangle \end{alignat*} If $$\rho \not \sim \varphi$$ then this is $$0$$ by the paragraph over. If $$\rho \sim \varphi$$, then we get that the inner product is $$1$$, as desired. $$\blacksquare$$

We end this post by using our new theory to show the uniqueness of the decomposition in Maschke’s theorem. For this we introduce new notation: $$m V$$ and $$m\rho$$ is defined to be the direct sum of $$m$$ copies of $$V$$ and $$\rho$$ respectively. We only need one lemma before attacking the problem at hand.

Lemma 2.6: Let $$\rho: G \to GL_n(\mathbb{C})$$ and $$\varphi: G \to GL_m(\mathbb{C})$$ be representations of $$G$$. Then $$\chi_{\rho \oplus \varphi} = \chi_{\rho} + \chi_{\varphi}$$

Proof. We have $$(\rho \oplus \varphi)_g = \begin{pmatrix} \rho_g & 0 \\ 0 & \varphi_g \end{pmatrix}$$ and so $$\chi_{\rho \oplus \varphi}(g) = \text{Tr}(\rho_g) + \text{Tr}(\varphi_g) = \chi_{\rho}(g)+\chi_{\varphi}(g)$$. $$\blacksquare$$

Hence if $$\rho \sim m_1\varphi^{(1)} \oplus \cdots \oplus m_r\varphi^{(r)}$$ then by induction on lemma 2.6 and evaluating in the identity element of $$G$$ we see that $$\deg(\rho)=m_1\deg(\varphi^{(1)})+\dots+m_r\deg(\varphi^{(r)})$$.

Theorem 2.7: Let $$\varphi^{(1)},\dots,\varphi^{(r)}$$ be a complete list of representatives of the equivalence classes of irreducible representations of $$G$$. Then if $$\rho$$ is a representation of $$G$$, we have $$\rho \sim m_1\varphi^{(1)} \oplus \cdots \oplus m_r\varphi^{(r)}$$ where $$m_i = \langle \chi_{\rho}, \chi_{\varphi^{(i)}} \rangle$$.

Proof. By Mashcke’s theorem we can decompose $$\rho$$ like $$\rho \sim m_1\varphi^{(1)} \oplus \cdots \oplus m_r\varphi^{(r)}$$ By lemma 2.6 we have $$\chi_{\rho} = m_1\chi_{\varphi^{(1)}} + \dots + m_r\chi_{\varphi^{(r)}}$$. By the first orthogonality relations $$\langle \chi_{\varphi^{(i)}}, \chi_{\varphi^{(j)}} \rangle = 1$$ if $$i = j$$ and $$0$$ otherwise. Now $$\langle \chi_{\rho}, \chi_{\varphi^{(i)}} \rangle = m_1\langle \chi_{\varphi^{(1)}}, \chi_{\varphi^{(i)}} \rangle + \dots + m_r \langle \chi_{\varphi^{(r)}}, \chi_{\varphi^{(i)}} \rangle = m_i$$ which is exactly what we wanted. $$\blacksquare$$

This theorem gives us a very nice way to check whether a representation is irreducible or not.

Corollary 2.8: A representation $$\rho: G \to GL(V)$$ is irreducible if and only if $$\langle \chi_{\rho},\chi_{\rho} \rangle = 1$$.

Proof. By theorem 2.7 we have $$\rho = m_1\varphi^{(1)} \oplus \cdots \oplus m_r\varphi^{(r)}$$ for irreducible representations $$\varphi^{(i)}$$ of $$G$$. By the first orthogonality relations $$\langle \varphi^{(i)},\varphi^{(j)} \rangle = 1$$ if and only if  $$i = j$$ and $$0$$ otherwise. Using this we compute the inner product: $$\langle \chi_{\rho}, \chi_{\rho} \rangle = m_1^2+\dots+m_r^2$$. Clearly this is $$1$$ if and only if $$m_i = 1$$ for exactly one $$i$$, and $$0$$ for the rest, and then $$\rho \sim \varphi^{(i)}$$ – that is $$\rho$$ is irreducible. $$\blacksquare$$

Representation theory of finite groups (1/4)

A classical result in a first course in abstract algebra is the following: Let $$G$$ be a group of order $$p^2$$ for a prime $$p$$. Then $$G$$ is abelian. The proof offered often goes by Sylow theory. In this little series of texts we aim to give a different proof via the beautiful representation theory of finite groups. In a nutshell, representation theory tries to transfer problems about algebraic structures to problems in linear algebra, which is often better understood. In the case of a group for example, we represent group elements as invertible linear operators. This let us interpret the group with the machinery of linear algebra, which has proven fruitful over and over again. This series of articles will most likely consist of four entries:

1. Definitions, Maschke’s Theorem, Schur’s Lemma.
2. Character theory.
3. The regular representation.
4. Algebraic numbers, the dimension theorem and groups of order $$p^2$$.

Without further ado, let us jump straight into it.

Throughout this series of texts, we always assume $$G$$ is a finite group. A representation of $$G$$ is a homomorphism $$\rho: G \to GL(V)$$ where $$V$$ is a vector space, here taken to be finite dimensional and over $$\mathbb{C}$$. Hence we can interpret $$GL(V)$$ as $$GL_n(\mathbb{C})$$, the group of invertible $$n \times n$$-matrices over $$\mathbb{C}$$. We often write $$\rho_g(v)$$ or $$\rho_gv$$ for $$\rho(g)(v)$$. The degree of a representation $$\rho: G \to GL(V)$$ is defined to be $$\dim(V)$$. A subspace of $$W$$ of $$V$$ is said to be $$G$$-invariant if $$\rho_g(w) \in W$$ for all $$w \in W, g \in G$$. If we restrict a representation $$\rho_g$$ to a $$G$$-invariant subspace $$W$$ we get a subrepresenation, often written as $$\rho\vert_W$$. Two representations $$\rho: G \to GL(V), \varphi: G \to GL(W)$$ are said to be equivalent if there exists a linear isomorphism $$T: V \to W$$ so that $$T\rho_g = \varphi_g T$$ for all $$g \in G$$. The first result we will prove is Maschke’s theorem – a basic, but important result about how representation of finite groups decompose. For this we need the notion of a direct sum of two representations. There is no surprise in the definition: given two representations $$\varphi^{(1)}: G \to GL(V_1), \varphi^{(2)}: G \to GL(V_2)$$, we define the representation $$\varphi^{(1)} \oplus \varphi^{(2)} : G \to GL(V_1 \oplus V_2)$$ by $$(\varphi^{(1)}\oplus \varphi^{(2)})_g (v_1,v_2) \mapsto (\varphi^{(1)}_g(v_1),\varphi^{(2)}_g(v_2))$$. We will now define three important notions in representation theory. To this end, let $$G$$ be a finite group and $$\rho: G \to GL(V)$$ a represenation of $$G$$.

• $$\rho$$ is irreducible if the only $$G$$-invariant subspace of $$V$$ is $$0$$ and $$V$$,
• $$\rho$$ is completely reducible if $$V = V_1 \oplus \cdots \oplus V_n$$ for $$G$$-invariant subspaces so that $$\rho\vert_{V_1}$$ is irreducible for all $$i=1,\dots,n$$.
• $$\rho$$ is decomposable if $$V=V_1 \oplus V_2$$ for proper non-zero $$G$$-invariant subspaces $$V_1,V_2$$. If $$\rho$$ is not decomposable, it is said to be indecomposable.

Maschke’s theorem is the statement that every representation of a finite group is completely reducible. From the definition of the equivalence of representations, it follows readily that irreducibility, complete reducibility and decomposability is preserved by equivalences. To prove Maschke’s theorem we will take advantage of inner product spaces. To this end, let $$V, \langle \cdot, \cdot \rangle$$ be a (finite-dimensional) inner product space over $$\mathbb{C}$$. In a typical introduction course in linear algebra one studies unitary linear operators $$T: V \to V$$ – surjective operators that satisfy $$\langle Tv, Tw \rangle = \langle v,w \rangle$$ for all $$v, w \in V$$. In the same way a representation $$\rho: G \to GL(V)$$ is said to be unitary if for all $$g \in G$$ $$\langle \rho_gv, \rho_gw \rangle = \langle v, w \rangle$$ for all $$v,w \in W$$ and $$\rho_g$$ is surjective. With this defined we can prove our first lemma towards Mashcke’s theorem.

Lemma 1.1:  A unitary representation of a group $$G$$ is either irreducible or decomposable.

Proof. Let $$\rho: G \to GL(V)$$ be a unitary representation, and suppose that it is reducible. Hence we have some $$G$$-invariant subspace $$W$$ that is non-zero and proper. We have $$V = W \oplus W^{\perp}$$, so if $$W^{\perp}$$ is a $$G$$-invariant subspace, this will show that $$G$$ is decomposable. Let $$v \in W, w \in W^{\perp}$$. Then by $$G$$-invariance of $$W$$ and unitarity we have $$0=\langle w,\rho_{g^{-1}}v \rangle = \langle \rho_{g} w, \rho_{g}\rho_{g^{-1}}v \rangle = \langle \rho_{g}w,v \rangle$$ showing that $$W^{\perp}$$ is $$G$$-invariant and finishing the proof. $$\blacksquare$$

The strength of this lemma comes from the next lemma.

Lemma 1.2: A representation $$\rho:G \to GL(V)$$ of a finite group $$G$$ is equivalent to a unitary representation of $$G$$.

Proof. Let $$\dim(V)=n$$. We construct an inner product on $$\mathbb{C}^n$$ . To do this we introduce the averaging trick, which is a trick often used in representation theory of finite groups. This proof is really more about constructing a space to map into, then the isomorphism itself as we very soon will see. To this end, fix a basis $$\mathcal{B}$$ of $$V$$. Let $$T: V \to \mathbb{C}^n$$ be the canonical isomorphism. For any fixed $$g \in G$$ we let $$\varphi_g = T\rho_g T^{-1}$$. Then this gives a representation $$\varphi:G \to GL_n(\mathbb{C})$$. Let $$\langle \cdot, \cdot \rangle$$ be the the standard inner product on $$\mathbb{C}^n$$. As mentioned in the start, we construct a new inner product: $$(v,w) = \sum_{g \in G} \langle \varphi_g v, \varphi_gw \rangle$$ This is one of the places in the theory where the assumption that $$G$$ is finite is crucial. That this is an actual inner product is a routine calculation, which I leave to the reader. What is more interesting is showing that $$\varphi$$ is actually a unitary representation of $$G$$. For this we introduce another trick that also depends on $$G$$ being finite. We first see that $$( \varphi_h v,\varphi_h w ) = \sum_{g \in G} \langle \varphi_g\varphi_h v, \varphi_g \varphi_h w \rangle = \sum_{g \in G} \langle \varphi_{gh} v, \varphi_{gh} w \rangle$$ Right multiplication by $$h$$ is injective since $$G$$ is a group, and since $$G$$ is finite it is also a bijection. This observation tells us that the change of variables $$x=gh$$, won’t alter the sum – that is $$\sum_{g \in G} \langle \varphi_{gh} v, \varphi_{gh} w \rangle = \sum_{x \in G } \langle \varphi_{x}v, \varphi_{x} w \rangle$$  This trick will be used over and over again. This finishes the proof. $$\blacksquare$$

With this lemma proved, we are ready to prove one of the main theorems of this first article. The proof is very similar to classical proof of the fundamental theorem of arithmetic.

Theorem 1.3: (Maschke’s theorem)
Every representation of a finite group is completely reducible.

Proof. Let $$\rho: G \to GL(V)$$ be a representation of a finite group $$G$$. By lemma 1.2, $$\rho$$ is equivalent to a unitary representation, and by lemma 1.1 $$\rho$$ is hence either decomposable or irreducible. With this general fact in mind, we proceed by (strong) induction on the dimension $$n$$ of $$V$$. When $$n = 1$$, $$V$$ doesn’t have any non-zero proper subspaces. We assume now the statement is true whenever $$V$$ is a vector space so that $$\dim(V) \leq n$$, and let $$\rho: G \to GL(V)$$ be a representation so that $$\dim(V)=n+1$$. If $$\rho$$ is irreducible we are done, so suppose it is not. Then it is decomposable, so $$V= W \oplus Z$$ for some non-zero proper $$G$$-invariant subspaces. By the induction hypothesis we have $$W=W_1 \oplus \cdots \oplus W_t \qquad Z = Z_{1} \oplus \cdots \oplus Z_{s}$$ for $$G$$-invariant subspaces $$W_i,Z_i$$ so that $$\rho\vert_{W_i}, \rho\vert_{Z_i}$$ are irreducible representations. Hence we get the complete reduction $$V=W_1\oplus \cdots \oplus W_t \oplus Z_1 \oplus \cdots \oplus Z_{s}$$ which finishes the proof. $$\blacksquare$$

While this proves existence of a decomposition, it doesn’t say anything about uniqueness. We will answer the question regarding uniqueness quite specifically later by using the elegant theory of characters. We will take our first step towards the theory of characters, by ending this article with proving Schur’s lemma and a corollary of it. From now on, we will write $$\rho \sim \varphi$$ if $$\rho$$ and $$\varphi$$ are equivalent representations of $$G$$. A homomorphism of representations of $$G$$, from $$\rho: G \to GL(V)$$ to $$\varphi:G \to GL(W)$$, is a linear map $$T:V \to W$$ so that $$T\rho_g = \varphi_g T$$. This is also sometimes called an intertwiner. We are interested in the set of all homomorphism from $$\rho$$ to $$\varphi$$, denoted $$\text{Hom}_G(\rho,\varphi)$$. Schur’s lemma gives a very specific answer in the case where both $$\rho$$ and $$\varphi$$ are irreducible representations. We observe that $$\text{Hom}_G(\rho,\varphi) \subseteq \text{Hom}_G(V,W)$$ and furthermore $$\text{Hom}_G(\rho,\varphi)$$ is a subspace of $$\text{Hom}_G(V,W)$$.

Lemma 1.4: Let $$T:V \to W \in \text{Hom}_G(\rho,\varphi)$$. Then $$\ker(T)$$ is a $$G$$-invariant subspace of $$V$$ and $$\text{Im}(T)$$ is a $$G$$-invariant subspace of $$W$$.

Proof. Let $$v \in \ker(T), g \in G$$. We have $$T\rho_gv = \varphi_g T v = 0$$ so $$\rho_gv \in \ker(T)$$ and hence $$\ker(T)$$ is a $$G$$-invariant subspace of $$V$$. Let $$w \in \text{Im}(T), g \in G$$, then $$Tv = w$$ for some $$v \in V$$. Then $$\varphi_g w = \varphi_g T v = T\rho_g v$$ so $$\text{Im}(T)$$ is a $$G$$-invariant subspace of $$W$$. $$\blacksquare$$

With this lemma we are ready to prove Schur’s lemma. Recall that we are working with vector spaces over the complex numbers. This is the first place where we really use this. What we use is actually the fundamental theorem of algebra applied to this setting: every linear operator on a finite-dimensional complex vector space has an eigenvalue.

Theorem 1.5: (Schur’s lemma)
Let $$\rho: G \to GL(V), \varphi:G \to GL(W)$$ be irreducible representations of $$G$$ and let $$T \in \text{Hom}_G(\rho,\varphi)$$. Then either $$T$$ is invertible or $$T = 0$$. Consequently $$\text{Hom}_G(\rho,\varphi) = 0$$ if $$\rho \not\sim \varphi$$, and $$T = \lambda I$$ for some $$\lambda \in \mathbb{C}$$ if $$\rho = \varphi$$.

Proof. From lemma 1.4, $$\ker(T)$$ is a $$G$$-invariant subspace of $$V$$, but $$\rho$$ is irreducible so either $$\ker(T) = 0$$ or $$\ker(T)=V$$. If $$\ker(T)=V$$ then $$T = 0$$, so assume $$T \neq 0$$. If $$\ker(T)=0$$, then $$T$$ is injective. From lemma 1.4, $$\text{Im}(T)$$ is a $$G$$-invariant subspace of $$W$$, but $$\varphi$$ is irreducible so $$\text{Im}(T)=0$$ or $$\text{Im}(T)=W$$, but $$\ker(T)=0$$ so $$\text{Im}(T) = W$$. Hence $$T$$ is surjective as well, so invertible. It is clear that $$\text{Hom}_G(\rho,\varphi) = 0$$ if $$\rho \not \sim \varphi$$ from this. Finally, let $$\lambda$$ be an eigenvalue of $$T \in \text{Hom}_G(\rho,\rho)$$. Letting $$I:V \to V$$ be the identity operator, we have $$I \in \text{Hom}_G(\rho,\rho)$$. Since $$\text{Hom}_G(\rho,\rho)$$ is a subspace of $$\text{Hom}(V,V)$$ it follows that $$T-\lambda I \in \text{Hom}_G(\rho,\rho)$$. But by definition $$T-\lambda I$$ is not invertible, and so $$T=\lambda I$$ by the previous parts of the theorem. $$\blacksquare$$

A very interesting corollary of Schur’s lemma is the following.

Corollary 1.6: Let G be an abelian group. Then any irreducible representation $$\rho: G \to GL(V)$$ has degree one.

Proof. Fix some $$h \in G$$. Then for all $$g \in G$$ we have $$\rho_h\rho_g = \rho_{hg} = \rho_{gh} = \rho_g\rho_h$$ that is $$\rho_h \in \text{Hom}_G(\rho,\rho)$$, so by Schur’s lemma we have that $$\rho_h = \lambda_h I$$. Now if $$\alpha \in \mathbb{C}$$ and $$v \in V$$ is a non-zero vector then $$\rho_h(\alpha v) = \lambda_h \alpha v \in \mathbb{C}v$$ so $$\mathbb{C}v$$ is a $$G$$-invariant subspace of $$V$$. Since $$\rho$$ is an irreducible representation, this implies that $$\mathbb{C}v=0$$ or $$\mathbb{C}v = V$$, and since $$v$$ is a non-zero vector, we conclude that $$\mathbb{C}v = V$$. Hence $$\dim(V) = 1$$. $$\blacksquare$$