2024-02-12--notes-RT--number-of-irreducibles.md

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title: The number of irreducible representations of a finite group
date: 2024-02-12

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\newcommand{\trivial}{\mathbf{1}}

The number of irreducible representations of $G$

Recall that $G$ denotes a finite group.

Recall that the space $\mathbb{C}[G]$ of all $\mathbb{C}$-valued functions on $G$ is the vector space underlying the regular representation of $G$.

We introduced the convolution multiplication $\star$ on $\mathbb{C}[G]$ by the rule $$(f_1 \star f_2)(x) = \sum_{yz = x} f_1(y)f_2(z)$$ for $f_1,f_2 \in \mathbb{C}[G]$.

This product makes $\mathbb{C}[G]$ into a (in-general non-commutative) ring. We mostly will avoid invoking general results about rings, and so we define the center $\mathbb{C}[G]$ to be the subspace $$Z = \{ f\in \mathbb{C}[G] \mid f \star h = h \star f \quad \forall h \in \mathbb{C}[G]\}.$$

Proposition: : The subspace $Z$ coincides with the subspace of $\mathbb{C}[G]$ consisting of those functions which are constant on the conjugacy classes of $G$.

In particular, $\dim Z = \# \{\text{conjugacy classes of $G$}\}$.

Proof: : Since $\mathbb{C}[G]$ has a vector space basis consisting of the dirac functions $\delta_g$ for $g \in G$, one immediately sees that $f \in Z$ if and only if $$f \star \delta_g = \delta_g \star f$$ for every $g \in G$.

Note that $\delta_g \star \delta_{g^{-1}} = \delta_1$ is a
*multiplicative identity for the operation $\star$, so that for any $g$,
$\delta_{g^{-1}} = (\delta_g)^{-1}$ is a multiplicative inverse.

Thus 
$$f \star \delta_g = \delta_g \star f \iff f = \delta_g \star f \star \delta_{g^{-1}}.$$


Now, fix $f \in \mathbb{C}[G]$ and $g \in G$, and let's compute the value
of $\delta_g \star f \star \delta_{g^{-1}}$ at an element $h \in G$. We  have

$$(\delta_g \star f \star \delta_{g^{-1}})(h) = \sum_{xyz = h}
\delta_g(x) f(y) \delta_{g^{-1}}(z) = f(g^{-1}hg)$$

We now conclude that $f \in Z$ if and only if $$f(h) = f(g^{-1}hg)
\quad \forall g,h \in G$$ i.e. if and only if $f$ is constant on
the conjugacy classes of $G$.

Since the characteristic functions $\psi_C$ of the conjugacy
classes $C$ of $G$ form a basis for the space of *class
functions*, it follows that $\dim Z$ is the number of conjugacy
classes $C$ of $G$; this completes the proof of the Proposition.

We write $L_1,L_2, \cdots,L_r$ for a complete set of irreducible representations of $G$ on $\mathbb{C}$-vector spaces, no two of which are isomorphic.

Lemma: : Let $z$ be an element of the center $Z \subseteq \mathbb{C}[G]$. For each $i$ there is a scalar $\lambda_i \in \mathbb{C}$ such that for every $v \in L_i$ we have $$z \star v = \lambda_i v.$$

Proof of Lemma: : Note that for each $i$ the mapping "convolution with $z$" -- i.e. the mapping $$\phi:L_i \to L_i \quad \text{given by $\phi(v) = z \star vParseError: KaTeX parse error: Expected 'EOF', got '}' at position 1: }̲$-- is a homomorphism of$G$-representations.

Indeed, note for $g \in G$ that -- since $z \in Z$ -- we have
$$\phi(gv) = \phi(\delta_g \star v)  = z \star \delta_g \star v = 
\delta_g \star z \star v = \delta_g \star \phi(v) = g\phi(v).$$

Now, Schur's Lemma tells us -- since $L_i$ is *irreducible* --
that the endomorphisms of $L_i$ as a $G$-representation identify
with the scalar operators $\mathbb{C} = \mathbb{C} \cdot
\operatorname{id}_{W_i}.$

Thus, there is $\lambda_i \in \mathbb{C}$ such that $$\phi =
\lambda_i \operatorname{id}_{W_i};$$ in other words, $z \star w =
\phi(w) = \lambda_i w$ for $w \in W_i$, as required.

Theorem: : The number $r$ of irreducible representations is equal to $\dim Z$. In particular, $r$ is equal to the number of conjugacy classes in $G$.

Proof: : Write $W_i = \mathbb{C}[G]_{(L_i)}$ for the $L_i$-isotypic component of the regular representation $\mathbb{C}[G]$.

Thus for each $i$, $W_i$ is a direct sum of copies of the
irreducible representation $L_i$, and the quotient representation
$\mathbb{C}[G]/W_i$ contains no irreducible invariant subspace
isomorphic to $L_i$.

You proved for homework that

$$\mathbb{C}[G] = W_1 \oplus W_2 \oplus \cdots \oplus  W_r.$$

In view of this decomposition of $\mathbb{C}[G]$, we may write
$$\delta_1 = f_1 + f_2 + \cdots + f_r$$
for uniquely determined elements $f_i \in W_i$.

Let $z \in Z$. According to the Lemma, there are scalars
$\lambda_i \in \mathbb{C}$ for which $z \star v_i = \lambda_i v_i$
for $v_i \in L_i$.

Since $W_i$ is $L_i$-isotypic, it follows at once that
$$z \star w_i = \lambda_i w_i$$
for each $w_i \in W_i$.	
In particular,
$$z \star f_i = \lambda_i f_i \quad \text{for $i=1,2,\cdots,r$}.$$

Now we notice that
\begin{align*}
z & = z \star \delta_1  = z \star (f_1 + f_2 + \cdots + f_r) \\
& = z \star f_1 + z \star f_2 + \cdots + z \star f_r \\
& = \lambda_1 f_1 + \lambda_2 f_2 + \cdots + \lambda_r f_r	
\end{align*}

This proves that $Z$ is contained in the *span* of the vectors $f_1,f_2,\cdots,f_r$; i.e.
$$Z \subseteq \sum_{i=1}^r \mathbb{C}f_i.$$

We conclude that $$\dim Z \le \dim \sum_{i=1}^r \mathbb{C}f_i  \le r.$$

But on the other hand, we have proved that the *characters*
$\chi_i = \chi_{L_i}$ of the irreducible representations form an
orthonormal -- hence linearly independent -- set of *class functions*
on $G$.

According to the preceding Proposition, $\chi_i \in Z$ for each $i$.
This proves that
$$r = \dim \sum_{i=1}^r \mathbb{C} \chi_i \le \dim Z.$$

We may now conclude that $\dim Z = r$ as required.

Remarks: : With notations as in the proof of the Theorem, note that

- we have an equality $Z = \sum_{i=r}^r \mathbb{C}f_i$ of
  subspaces of $\mathbb{C}[G]$.

- since $\dim Z = r$, conclude that $f_1,f_2,\cdots,f_r$ are *linearly independent* 
  
- Moreover, $f_i \in Z$ for each $i$.

Bibliography

::: {.refs} :::