Orthogonality and Characters

First Orthogonality Relation of Characters

Suppose that we have two equivalent elements in group $G$, $g_{a} = g^{-1}g_{b}g$. And consider these elements in some representation $D(G)$. We can show that the trace, the character, of elements in the same equivalence class have the same character.

$$ \begin{align} \chi(g_{a}) &= \mathrm{Tr}D(g_{a}) = \mathrm{Tr}D(g^{-1} g_{b} g) = \mathrm{Tr}\left[ D(g^{-1})D(g_{b})D(g) \right] \\ &= \mathrm{Tr}\left[ D(g_{b}) D(g^{-1}) D(g) \right] \quad \text{(the trace is cyclic)} \\ &= \mathrm{Tr}D(g_{b}) = \chi(g_{b}) \end{align} $$$$ \therefore \space \chi(g_{a}) = \chi(g_{b}) \quad \text{where} \quad g_{a} \simeq g_{b} $$

Using this fact, let’s take the trace of the orthogonality theorem.

$$ \begin{align} \mathrm{Tr}\left[ \sum_{a=1}^n D^{(i)}(g_{a})^{*}_{\alpha\beta}D^{(j)}(g_{b})_{\mu \nu} \right] &= \mathrm{Tr}\left[ \frac{n}{n_{i}} \delta_{ij} \delta_{\alpha \mu} \delta_{\beta \nu} \right] \\ \sum_{a=1}^n D^{(i)}(g_{a})^{*}_{\alpha\alpha} D^{(j)}(g_{a})_{\mu \mu} &= \frac{n}{n_{i}} \delta_{ij} \delta_{\alpha \mu} \\ \sum_{a=1}^n \chi^{(i)}(g_{a})^{*} \chi^{(j)}(g_{a}) &= n \delta_{ij} \end{align} $$

Now consider what we just stated, how the elements within the same equivalence class will have the same character.

Consider a group with $\cal{C}_{r}$ $(1,2,\dots, k)$ equivalence classes, each with $p_{r}$ elements such that $\sum_{r}^k p_{r} = n$. From the relation above, we can find the first orthogonality relation of characters.

$$ \sum_{a=1}^n \chi^{(i)}(g_{a})^{*} \chi^{(j)}(g_{a}) = \sum_{r=1}^{k} p_{r} \chi^{(i)}(\cal{C}_{r})^{*} \chi^{(j)}(\cal{C}_{r}) = n \delta_{ij} $$

Normal Characters

We define normal characters as

$$ \hat{\chi}^{(i)}(\cal{C}_{r}) = \sqrt{ \frac{p_{r}}{n} } \chi^{(i)}(\cal{C}_{r}) $$

Using normal characters we can make explicit the orthogonality of the first relation.

$$ \sum_{r=1}^{k} p_{r} \chi^{(i)}(\cal{C}_{r})^{*} \chi^{(j)}(\cal{C}_{r}) = n \delta_{ij} \implies \sum_{r=1}^k \hat{\chi}^{(i)}(\cal{C}_{r}) \hat{\chi}^{(j)}(\cal{C}_{r}) = \delta_{ij} $$

So, the normal character vectors are orthogonal.

$$ \begin{align} \bf{\hat{\chi}^{(1)}} &\equiv \left( \hat{\chi}^{(1)}(\cal{C}_{1}), \hat{\chi}^{(1)}(\cal{C}_{2}), \dots, \hat{\chi}^{(1)}(\cal{C}_{k}), \right) \\ \bf{\hat{\chi}^{(2)}} &\equiv \left( \hat{\chi}^{(2)}(\cal{C}_{1}), \hat{\chi}^{(2)}(\cal{C}_{2}), \dots, \hat{\chi}^{(2)}(\cal{C}_{k}), \right) \\ &\vdots \\ \bf{\hat{\chi}^{(N_{i})}} &\equiv \left( \hat{\chi}^{(N_{i})}(\cal{C}_{1}), \hat{\chi}^{(N_{i})}(\cal{C}_{2}), \dots, \hat{\chi}^{(N_{i})}(\cal{C}_{k}), \right) \\ \end{align} $$

There are $N_{i}$ normal character vectors, one for each irrep. And, since the dimension of these vectors is $k$, the number of equivalence classes in a group, we are led to conclude that $N_{i} \leq k$. In fact, it can be shown (not here) that

$$ N_{i} = k $$

Therefore, the number of irreps for a group is equal to the number of equivalence classes.

Second Orthogonality Relation of Characters

We can form a $k \times k$ matrix using the normal character vectors.

$$ U = \begin{pmatrix} \bf{\hat{\chi}^{(1)}} \\ \hline \bf{\hat{\chi}^{(2)}} \\ \hline \vdots \\ \hline \bf{\hat{\chi}^{(k)}} \\ \end{pmatrix} = \begin{pmatrix} \hat{\chi}^{(1)}(\cal{C}_{1}) & \hat{\chi}^{(1)}(\cal{C}_{2}) & \cdots & \hat{\chi}^{(1)}(\cal{C}_{k}) \\ \hat{\chi}^{(2)}(\cal{C}_{1}) & \hat{\chi}^{(2)}(\cal{C}_{2}) & \cdots & \hat{\chi}^{(1)}(\cal{C}_{k}) \\ \vdots & \vdots & \ddots & \vdots \\ \hat{\chi}^{(k)}(\cal{C}_{1}) & \hat{\chi}^{(k)}(\cal{C}_{2}) & \cdots & \hat{\chi}^{(k)}(\cal{C}_{k}) \end{pmatrix} $$

Considering the first orthogonality relation, $UU^{\dagger}=1$, meaning $U$ is a unitary matrix. This in turn implies that $U^{\dagger}U = 1$, indicating a second orthogonality relation: the summation over irreps rather than equivalence classes.

$$ \sum_{i=1}^k \chi^{(i)}(\cal{C}_{r})^{*} \chi^{(i)}(\cal{C}_{s}) = \frac{n}{p_{r}} \delta_{rs} $$

Table of Characters

We can visualize the characters of all representations and equivalence classes, along with the orthogonality relations, using a table of characters.

$$ \begin{array}{c|cccc} G & p_{1} \cal{C}_{1} & p_{2}\cal{C}_{2} & \cdots & p_{k} \cal{C}_{k} \\ \hline D^{(1)} & \chi^{(1)}(\cal{C}_{1}) & \chi^{(1)}(\cal{C}_{2}) & \cdots & \chi^{(1)}(\cal{C}_{k}) \\ D^{(2)} & \chi^{(2)}(\cal{C}_{1}) & \chi^{(2)}(\cal{C}_{2}) & \cdots & \chi^{(2)}(\cal{C}_{k}) \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ D^{(k)} & \chi^{(k)}(\cal{C}_{1}) & \chi^{(k)}(\cal{C}_{2}) & \cdots & \chi^{(k)}(\cal{C}_{k}) \end{array} $$

The values of each row are orthonormalized to the order of the group $n$.

$$ \sum_{r=1}^k p_{r} \chi^{(i)}(\cal{C}_{r})^{*} \chi^{(i)}(\cal{C}_{r}) = n $$

The values of each column are orthonormalized to the value $n / p_{r}$.

$$ \sum_{i=1}^k \chi^{(i)}(\cal{C}_{r})^{*} \chi^{(i)}(\cal{C}_{r}) = \frac{n}{p_{r}} $$

The characters, and the table of characters, form the most important constructs in representation theory.