2024-02-05--notes-RT.md

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title: Characters of irreducible representations
date: 2024-02-05

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Convolution

We write $\mathbb{C}[G]$ for the space of functions on $G$, viewed as a permutation representation of $G$ (and we suppress the notation for the homomorphism $G \to \operatorname{GL}(\mathbb{C}[G])$).

For functions $f_1,f_2 \in \mathbb{C}[G]$, we define their convolution by the formula $$(f_1 \star f_2)(x) = \sum_{yz = x} f_1(y)f_2(z).$$

If $V$ is a $G$-representation and $f \in \mathbb{C}[G]$, we define $$f \star v = \sum_{g \in G} f(g) gv$$ for $v \in V$.

Remark: : 1. For the basis elements $\delta_g \in \mathbb{C}[G]$ (i.e. the Dirac functions), we have $$\delta_g \star \delta_h = \delta_{gh}.$$

2. The action of $G$ on $\mathbb{C}[G]$ can be described by $$g f =
   \delta_g \star f$$ for $g \in G$ and $f \in \mathbb{C}[G]$.

3. Viewing $\mathbb{C}[G]$ as a $G$-representation, the two
   notions of $\star$ just introduced actually coincide:

   $$f_1 \star f_2 = \sum_{g \in G} f_1(g) \delta_g \star f_2.$$

4. The product $\star$ makes $\mathbb{C}[G]$ into a *ring* (in
   fact, a $\mathbb{C}$-algebra) and $V$ into a
   $\mathbb{C}[G]$-module.  Mostly we won't use this fact - at
   least explicitly - in these notes.


5. Let $W \subseteq \mathbb{C}[G]$ be an invariant subspace.  For
   any $f \in \mathbb{C}[G]$, we have $$f \star f' \in W \quad
   \forall f' \in W.$$

6. The element $\delta_1$ acts as the identity for the $\star$
   operation. Namely, for $f \in \mathbb{C}$ $$f \star \delta_1 =
   \delta_1 \star f.$$ This follows easily from the fact that
   $\delta_1 \star \delta_g = \delta_g \star \delta_1 = \delta_g$
   for all $g \in G$.

Isotypic decomposition

Let $V$ be a $G$-representation and let $L$ be an irreducible $G$-representation.

Consider the set $\mathcal{S}$ of all invariant subspaces $S \subseteq V$ for which $S \simeq L$ as $G$-representation.

Set $$W = \sum_{S \in \mathcal{S}} S;$$ then $W$ is an invariant subspace of $V$.

Proposition: : $W$ is isotypic in the sense that any irreducible invariant subspace of $W$ is isomorphic (as $G$-representation) to $L$.

 Moreover, $[V/W:L] = 0$.

You will prove this in homework.

We write $V_{(L)}$ for the invariant subspace $W$.

You will also prove:

Proposition: : If $L_1,L_2,\cdots,L_r$ is a complete set of non-isomorphic irreducible representations of $G$, then $$V = V_{(L_1)} \oplus V_{(L_2)} \oplus \cdots \oplus V_{(L_r)}.$$

Results about the characters of the irreducible representations

Investigation of certain idempotent elements in $\mathbb{C}[G]$.

Let $L$ be an irreducible representation of $G$ and let $W_1 = \mathbb{C}[G]_{(L)}$.

Use completely reduciblity to write $$\mathbb{C}[G] = W_1 \oplus W_2$$ for some invariant subspace $W_2 \subset \mathbb{C}[G]$.

Note that $[W_2:L] = 0$ by construction.

We now write $$\delta_1 = e_1 + e_2 \quad \text{for $e_1 \in W_1$ and $e_2 \in W_2$}.$$

Proposition: : For $w_1 \in W_1$ and $w_2 \in W_2$ we have

 $$e_1 \star w_1 = w_1,\quad  e_1 \star w_2 = 0,$$
 $$e_2 \star w_1 = 0, \quad \text{and} \quad e_2 \star w_2 = w_2.$$

Proof: : Fix $w_2 \in W_2$. We define a mapping $\phi:W_1 \to W_2$ by the rule $\phi(w_1) = w_1 \star w_2.$

 We note that $\phi$ is a homomorphism of $G$-representations.
 Indeed, recall the action of $g \in G$ on $\mathbb{C}[G]$ is the
 same as that of $\delta_g \star$. Now
 
 $$\phi(\delta_g \star w_1) = (\delta_g \star w_1) \star w_2 =
 \delta_g \star (w_1 \star w_2) = \delta_g \star \phi(w_2).$$

 Since $W_1$ is $L$-isotypic and since $[W_2:L] = 0$, the mapping
 $\phi$ must be 0.
 
 Now conclude that $$0 = w_1 \star w_2 \quad \forall w_1 \in W_1, w_2 \in W_2.$$

 A similar argument shows that $$0 = w_2 \star w_1 \quad \forall
 w_1 \in W_1, w_2 \in W_2.$$ (For $w_1 \in W_1$, one define a
 homomorphism $\psi:W_2 \to W_1$ by the rule $\psi(w_2) = w_2
 \star w_1$ As before, one argues that $\psi = 0$...)

 Now notice for $w_1 \in W_1$ that $$w_1 = \delta_1 \star w_1 =
 (e_1 + e_2)\star w_1 = e_1 \star w_1 + e_2 \star w_1 = e_1 \star
 w_1$$
 since $e_2 \in W_2 \implies e_2 \star w_1 = 0$ by the preceding results.
 This proves that $e_1 \star w_1 = w_1$ for all $w_1 \in W_1$
 
 Similarly, for $w_2 \in W_2$ we have $$w_2 = \delta_1 \star w_2 =
 (e_1 + e_2)\star w_2 = e_1 \star w_2 + e_2 \star w_2 = e_2 \star
 w_2$$
 since $w_2 \in W_2 \implies e_1 \star w_2 = 0$ by the preceding results.
 
 This completes the proof.

As an immediate consequence, we get:

Corollary: : - $e_1 \star e_1 = e_1$ - $e_2 \star e_2 = e_2$ - $e_1 \star e_2 = e_2 \star e_1 = 0$.

We can actually find a formula expressing $e_1$ in the basis $\{\delta_g\}$ for $\mathbb{C}[G]$:

Proposition: : Let $W_1 = \mathbb{C}[G]_{(L)}$ and suppose that $\mathbb{C}[G] = W_1 \oplus W_2$ for an invariant subspace $W_2$ as before. Write $\delta_1 = e_1 + e_2$ with $e_i \in W_i$. and let $\chi$ be the character of $W_1$. We have $$e_1 = \dfrac{1}{|G|} \sum_{g \in G} \chi(g^{-1}) \delta_g.$$

Proof: : Fix $x \in G$ and define $$\Phi:\mathbb{C}[G] \to \mathbb{C}[G]$$ by the rule $$\Phi(f) = \delta_{x^{-1}} \star e_1 \star f.$$

 We are going to compute the *trace* of $\Phi$ in two different ways.
 
 First, Note that since $e_1 \star w_1 = w_1$ for each $w_1 \in
 W_1$, we see that $\Phi_{\mid W_1}$ is given $$w \mapsto
 \delta_{x^{-1}} \star w$$. Thus $\operatorname{tr}(\Phi_{\mid
 W_1})$ is given by $\chi(x^{-1})$.
 
 Since $e_1 \star W_2 =0$, we conclude that $\Phi_{\mid W_2}) = 0$
 so that $$\operatorname{tr}(\Phi) = \operatorname{tr}(\Phi_{\mid
 W_1}) \oplus \operatorname{tr}(\Phi_{\mid W_2}) =
 \operatorname{tr}(\Phi_{\mid W_1}) = \chi(x^{-1}),$$

 On the other hand, let us express $e_1$ in the basis $\{\delta_g\}$ of $\mathbb{C}[G]$:
 $$e_1  = \sum_{g \in G} \lambda_g \delta_g \quad (\lambda_g \in \mathbb{C}).$$
 
 Let us examine the mapping $$\theta_{x^-1g}:\mathbb{C}[G] \to \mathbb{C}[G]$$
 given by $$w \mapsto \delta_{x^-1 g} \star w.$$ Recall that
 $\mathbb{C}[G]$ is the permutation representation corresponding
 to the action of $G$ on itself by left multiplication. We have
 seen that the trace of the action of an element of $G$ is the
 number of fixed points for that action. We conclude that the trace of 
 $\theta_{x^{-1}g}$ is $|G|$ if $x = g$ and otherwise is 0.
 
 Now, the mapping $\Phi$ is given by
 $$\Phi(w) = \delta_{x^{-1}} \star \left( \sum_{g \in G} \lambda_g \delta_g \right) \star w
  = \sum_{g \in G} \lambda_g \delta_{x^{-1} g} \star w
  = \sum_{g \in G} \lambda_g \theta_{x^{-1} g}(w).$$

 Since the trace is a linear operator, we conclude that
 $$\operatorname{tr}(\Phi)  = \sum_{g \in G} \lambda_g \operatorname{tr}(\theta_{x^{-1}g})
 = \lambda_x |G|.$$

 Now comparing our two computations of $\operatorname{tr}(\Phi)$ we get the formula
 $$\lambda_x |G| = \chi(x^{-1})$$
 i.e. $\lambda_x = \chi(x^{-1})/|G|.$
 
 But then $$e_1 = \sum_{g \in G} \lambda_g \delta_g =
 \dfrac{1}{|G|} \sum_{ g\in G} \chi(g^{-1}) \delta_g$$ as
 required. This completes the proof.

Remark: : Since $G$ is a finite group, the eigenvalues of the operation of $g \in G$ on any $G$-representation are roots of unity $\zeta$. Notice that $\overline{\zeta} = \zeta^{-1}$ for any root of unity. In particular, if $\chi$ is the character of a representation of $G$, we have $\chi(g^{-1}) = \overline{\chi(g)}.$

Corollary: : Let $\chi$ be the character of $W_1 = \mathbb{C}[G]_{(L)}.$ Then $$\langle \chi,\chi \rangle = \chi(1).$$

Proof: : Note that the Proposition allows us to calculate: $$e_1 \star e_1 = \dfrac{1}{|G|} \sum_{x,g \in G} \chi(g^{-1})\chi(x^{-1}) \delta_{gx}.$$

 The coefficient of $\delta_1$ in this expression is precisely
 $$\dfrac{1}{|G|^2} \sum_{g \in G} \chi(g^{-1}) \chi(g) = 
 \dfrac{1}{|G|^2} \sum_{g \in G} \chi(g) \overline{\chi(g)} = 
 \dfrac{1}{|G|} \langle \chi, \chi \rangle.$$
 
 On the other hand, $e_1 = e_1 \star e_1$ and the coefficient of
 $\delta_1$ in the expression for $e_1$ is $\chi(1^{-1}) = \chi(1)$.
 
 Thus we see that $\chi(1) = \langle \chi,\chi \rangle$ as
 required.

We have so far elided the multiplicities $[\mathbb{C}[G]:L]$ for irreducible representations $L$. We are going to state the result here (and maybe prove it later).

Theorem: : For an irreducible representation $L$ of $G$, the multiplicity $[\mathbb{C}[G]:L]$ is given by $$[\mathbb{C}[G]:L] = \dim_{\mathbb{C}} L.$$

Remark: : If $\chi,\psi$ are the characters of representations of $G$, then $\langle \chi,\psi \rangle = \langle \psi,\chi \rangle.$

 Indeed,
 \begin{align*}
 \langle \chi,\psi \rangle = \dfrac{1}{|G|} \sum_{g \in G}
 \chi(g) \overline{\psi(g)}
 &= \dfrac{1}{|G|} \sum_{g \in G}
 \chi(g) \psi(g^{-1}) \\
 &= \dfrac{1}{|G|} \sum_{x \in G}
 \chi(x^{-1}) \psi(x) \\
 &= \langle \psi,\chi \rangle.
 \end{align*}

Now we are able to prove that main Theorem which shows that the characters of the irreducible representations form an orthonormal set.

Theorem: : Let $U,V$ be irreducible representations of $G$ with characters $\chi,\psi$ respectively. Then $$\langle \chi,\chi \rangle = 1 \quad \text{and} \quad \langle \chi,\psi \rangle = 0.$$

Proof: : Let $W = \mathbb{C}[G]_{(U)}$ be the $U$-isotypic part of $\mathbb{C}[G]$ and let $\chi_W$ be the character of $W$.

 The preceding Theorem tells us that 
 $\chi_W = m\chi$ where $m = \dim U$.
 
 Now, the preceding Corollary tells us that
 $$\langle \chi_W,\chi_W \rangle = \chi_W(1).$$
 
 Note that $$\langle \chi_W,\chi_W \rangle = \langle m \chi, m
 \chi \rangle = m^2 \langle \chi,\chi \rangle.$$
 
 On the other hand,
 $$\chi_W(1) = m\chi(1) = m^2.$$
 
 Thus we find $m^2 \langle \chi,\chi \rangle = m^2$; since $m \ne 0$,
 conclude $\langle \chi,\chi \rangle = 1$ as required.

 Now let $Y = \mathbb{C}[G]_{(U)} + \mathbb{C}[G]_{(V)}$ be the
 sum of the isotypic components.
 
 Note that $\chi_Y = m \chi + n \psi$ where $m = \dim U$ and $n = \dim V$.
 Now we have
 $$\chi_Y(1) = \langle \chi_Y, \chi_Y \rangle.$$
 
 On the one hand,
 $$\chi_Y(1) =m \chi(1) + n \psi(1) = m^2 + n^2.$$
 
 And on the other hand, the preceding corollary shows that
 \begin{align*}
 \langle \chi_Y,\chi_Y \rangle &= \langle m \chi + n \psi,
 m\chi + n \psi \rangle \\ &= m^2 \langle \chi,\chi \rangle + n^2
 \langle \psi,\psi \rangle + 2mn \langle \chi,\psi \rangle \\
 &= m^2 + n^2 + 2mn \langle \chi,\psi \rangle.
 \end{align*}
 
 Thus we find
 $$m^2 + n^2 = m^2 + n^2 + 2mn \langle \chi,\psi \rangle$$
 so that indeed $\langle \chi,\psi \rangle = 0$. This completes the proof.

Remark: : In order to complete the proof that the irreducible characters form an orthonormal basis for the space of class functions on $G$, we still need to prove that the number of distinct irreducible representations is equal to the number of conjugacy classes in $G$.