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 \[\phi:L_i \to L_i \quad \text{given by $\phi(v) = z \star v$}\] – 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 \(i\).

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