----- > [!proposition] Proposition. ([[number of irreps equals number of conjugacy classes]]) > Let $G$ be a finite [[group]]. There is [[bijection]] $\{ \text{irreps of $G$ up to iso.} \} \leftrightarrow \{ \chi_{\rho} : \rho \text{ irrep} \} \leftrightarrow \{ \text{conj. classes of }G \}.$ > [!proof]- Proof. ([[number of irreps equals number of conjugacy classes]]) > The first [[bijection]] follows from [[complex group representations are isomorphic iff their characters match]]. We show the second now. > > Suppose $G$ has $h$ [[conjugate|conjugacy classes]]. We know that the [[irreducible group representation|irreducible]] [[character of a representation|characters]] $\{ \chi_{i} \}$ of the [[group representation|irreps]] $\rho_{i}$ form an [[orthonormal]] list in the [[vector space]] $\text{Cl}(G)$ of [[class function|class functions]] from $G$ to [[complex numbers]]. We will show that this is in fact an [[orthonormal basis]] by demonstrating that the only [[class function]] [[orthogonal]] to all $\{ \chi_{i} \}$ is $0$, and hence they [[submodule generated by a subset|generate]] $\text{Cl}(G)$. > > > [!proposition] Lemma. > > Let $\phi$ be a [[class function]] on $G$ that is [[orthogonal]] to every [[character of a representation|character]]. For any [[group representation|representation]] $\rho$ of $G$, $T_{\phi}=\frac{1}{|G|} \sum_{g} \overline{\phi(g)} \rho_{g}$ is the zero [[linear operator|operator]]. > > > [!proof] Proof of Lemma. > > WLOG $\rho$ is [[irreducible group representation|irreducible]] (okay to do because of [[Maschke's Theorem]]). It is easy to show $T_{\phi}$ is [[group-equivariant map|G-equivariant]] using that $\phi$ is a [[class function]]. > > > Let $\chi$ be the [[character of a representation|character]] of $\rho$. The [[trace of a linear operator|trace]] of $T$ is $\text{tr }T = \frac{1}{|G|} \sum_{g} \text{tr } (\overline{\phi(g)}\rho_{g})=\frac{1}{|G|} \sum_{g} \overline{\phi(g)} \chi(g) = \langle \phi, \chi \rangle $ > which equals $0$ by assumption that $\phi$ is [[orthogonal]] to all irreducible characters. By [[Schur's lemma for groups]], $T$ is multiplication by a [[scalar]], $T=cI$. But since $\text{tr }T = 0$, $c=0$ and thus $T=0$. > > > [!proposition] Lemma. > > Suppose that $f$ is a [[class function]] of $G$ with $\langle f, \chi_{\rho} \rangle=0$ for all [[irreducible group representation|irreducible]] $\rho$. Then $T_{f}=0$ for *any* $\rho$ (not necessarily irreducible). > > > [!proof] Proof of Lemma. > > Compute $\text{tr }T_{f}= \sum_{g \in G}\overline{f(g)}\text{tr }\rho_{g}=\langle f, \chi_{\rho} \rangle = \sum_{i} m_{i} \langle f, \chi_{i} \rangle = 0, $ > > where each $\chi_{i}$ is [[irreducible group representation|irreducible]] and we have used [[Maschke's Theorem]]. Using $G$-equivariance of $T_{f}$ we may apply [[Schur's lemma for groups]] to conclude $T_{f}=cI$ for some $c \in \mathbb{C}$, but since its [[trace of a linear operator|trace]] is $0$, $c=0$ and thus $T_{f}=0$. > > Finally, suppose $f$ is a [[class function]] of $G$ with $\langle f, \chi_{\rho} \rangle=0$ for all [[irreducible group representation|irreducible]] $\rho$. In particular, the lemma tells us that, for $\rho^{\text{reg}}$ the [[regular representation]] of $G$, $T_{\rho^{\text{reg}}}=0$. So, $T_{\rho^{\text{reg}}}(x_{e})=\sum_{g \in G} \overline{f(g)}x_{g}=0.$ > The $x_{g}$ are [[linearly independent]], hence it must be the case that $f \equiv 0$. This is what we wanted to show. ----- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```