# orthogonality relations

First orthogonality relations:
Let ${\rho}_{\alpha}:G\to {V}_{\alpha}$ and ${\rho}_{\beta}:G\to {V}_{\beta}$ be irreducible representations of a finite group^{} $G$ over the field $\u2102$. Then

$$\frac{1}{|G|}\sum _{g\in G}\overline{{\rho}_{ij}^{(\alpha )}(g)}{\rho}_{kl}^{(\beta )}(g)=\frac{{\delta}_{\alpha \beta}{\delta}_{ik}{\delta}_{jl}}{dim{V}_{\alpha}}.$$ |

We have the following useful corollary.
Let ${\chi}_{1}$, ${\chi}_{2}$ be characters^{} of representations^{} ${V}_{1}$, ${V}_{2}$ of a finite group $G$ over a field $k$ of characteristic^{} $0$. Then

$$({\chi}_{1},{\chi}_{2})=\frac{1}{|G|}\sum _{g\in G}\overline{{\chi}_{1}(g)}{\chi}_{2}(g)=dim(\mathrm{Hom}({V}_{1},{V}_{2})).$$ |

###### Proof.

First of all, consider the special case where $V=k$ with the trivial action of the group. Then ${\mathrm{Hom}}_{G}(k,{V}_{2})\cong {V}_{2}^{G}$, the fixed points. On the other hand, consider the map

$$\varphi =\frac{1}{|G|}\sum _{g\in G}g:{V}_{2}\to {V}_{2}$$ |

(with the sum in $\mathrm{End}({V}_{2})$). Clearly,
the image of this map is contained in ${V}_{2}^{G}$, and it is the identity^{} restricted
to ${V}_{2}^{G}$. Thus, it is a projection^{} with image ${V}_{2}^{G}$. Now, the rank of
a projection (over a field of characteristic 0) is its trace. Thus,

$${dim}_{k}{\mathrm{Hom}}_{G}(k,{V}_{2})=dim{V}_{2}^{G}=\mathrm{tr}(\varphi )=\frac{1}{|G|}\sum {\chi}_{2}(g)$$ |

which is exactly the orthogonality formula for ${V}_{1}=k$.

Now, in general, $\mathrm{Hom}({V}_{1},{V}_{2})\cong {V}_{1}^{*}\otimes {V}_{2}$ is a representation, and ${\mathrm{Hom}}_{G}({V}_{1},{v}_{2})={(\mathrm{Hom}({V}_{1},{V}_{2}))}^{G}$. Since ${\chi}_{{V}_{1}^{*}\otimes {V}_{2}}=\overline{{\chi}_{1}}{\chi}_{2}$,

$${dim}_{k}{\mathrm{Hom}}_{G}({V}_{1},{V}_{2})={dim}_{k}{(\mathrm{Hom}({V}_{1},{V}_{2}))}^{G}=\sum _{g\in G}\overline{{\chi}_{1}}{\chi}_{2}$$ |

which is exactly the relation we desired. ∎

In particular, if ${V}_{1},{V}_{2}$ irreducible, by Schur’s Lemma

$$\mathrm{Hom}({V}_{1},{V}_{2})=\{\begin{array}{cc}D\hfill & {V}_{1}\cong {V}_{2}\hfill \\ 0\hfill & {V}_{1}\ncong {V}_{2}\hfill \end{array}$$ |

where $D$ is a division algebra. In particular, non-isomorphic irreducible
representations have orthogonal^{} characters. Thus, for any representation $V$,
the multiplicities^{} ${n}_{i}$ in the unique decomposition of $V$ into the direct sum^{} (http://planetmath.org/DirectSum)
of irreducibles

$$V\cong {V}_{1}^{\oplus {n}_{1}}\oplus \mathrm{\cdots}\oplus {V}_{m}^{\oplus {n}_{m}}$$ |

where ${V}_{i}$ ranges over irreducible representations of $G$ over $k$, can be determined in terms of the character inner product:

$${n}_{i}=\frac{(\psi ,{\chi}_{i})}{({\chi}_{i},{\chi}_{i})}$$ |

where $\psi $ is the character of $V$ and ${\chi}_{i}$ the character of ${V}_{i}$. In particular, representations over a field of characteristic zero are determined by their character. Note: This is not true over fields of positive characteristic.

If the field $k$ is algebraically closed^{},
the only finite division algebra over $k$ is $k$ itself, so
the characters of irreducible representations form an orthonormal basis for
the vector space^{} of class functions with respect to this inner product.
Since $({\chi}_{i},{\chi}_{i})=1$ for all irreducibles, the multiplicity formula
above reduces to ${n}_{i}=(\psi ,{\chi}_{i})$.

Second orthogonality relations: We assume now that $k$ is algebraically closed. Let $g,{g}^{\prime}$ be elements of a finite group $G$. Then

$$\sum _{\chi}\chi (g)\overline{\chi ({g}^{\prime})}=\{\begin{array}{cc}|{C}_{G}({g}_{1})|\hfill & g\sim {g}^{\prime}\hfill \\ 0\hfill & g\nsim {g}^{\prime}\hfill \end{array}$$ |

where the sum is over the characters of irreducible representations, and ${C}_{G}(g)$ is the centralizer^{} of $g$.

###### Proof.

Let ${\chi}_{1},\mathrm{\dots},{\chi}_{n}$ be the characters of the irreducible representations,
and let ${g}_{1},\mathrm{\dots},{g}_{n}$ be representatives of the conjugacy classes^{}.

Let $A$ be the matrix whose $ij$th entry is $\sqrt{|G:{C}_{G}({g}_{j})|}(\overline{{\chi}_{i}({g}_{j})})$.
By first orthogonality, $A{A}^{*}=|G|I$ (here $*$ denotes conjugate transpose^{}),
where $I$ is the identity matrix^{}. Since left inverses^{} (http://planetmath.org/MatrixInverse) are right ,
${A}^{*}A=|G|I$. Thus,

$$\sqrt{|G:{C}_{G}({g}_{i})||G:{C}_{G}({g}_{k})|}\sum _{j=1}^{n}{\chi}_{j}({g}_{i})\overline{{\chi}_{j}({g}_{k})}=|G|{\delta}_{ik}.$$ |

Replacing ${g}_{i}$ or ${g}_{k}$ with any conjuagate will not change the expression above.
thus, if our two elements are not conjugate^{}, we obtain that ${\sum}_{\chi}\chi (g)\overline{\chi ({g}^{\prime})}=0$.
On the other hand, if $g\sim {g}^{\prime}$, then $i=k$ in the sum above, which reduced to the expression
we desired.
∎

A special case of this result, applied to $1$ is that $|G|={\sum}_{\chi}\chi {(1)}^{2}$, that is, the sum of the squares of the dimensions^{} (http://planetmath.org/Dimension) of the irreducible representations of any finite group is the order of the group.

Title | orthogonality relations |
---|---|

Canonical name | OrthogonalityRelations |

Date of creation | 2013-03-22 13:21:27 |

Last modified on | 2013-03-22 13:21:27 |

Owner | mhale (572) |

Last modified by | mhale (572) |

Numerical id | 16 |

Author | mhale (572) |

Entry type | Theorem |

Classification | msc 20C15 |