Orthogonality Theorem

The orthogonal theorem is fundamental to representation theory. Understand this theorem as much as possible, and you will be an academic weapon in this course.

Implication

Orthogonality

First, we can define the meaning of $D^{(i)}(g_{a})_{\alpha\beta}$.

• The group $G = \{ e, g_{1}, g_{2}, \dots, g_{n}\}$ has $n$ elements.
• The group $G$ has at most $N_{i}$ unitary irreps, however, we don’t know $N_{i}$.
• The representation $D^{(i)}(G)$ is the $i^\text{th}$ representation $(i = 1,2,\dots, N_{i})$, with matrices of dimension $n_{i} \times n_{i}$.

So, $D^{(i)}(g_{a})_{\alpha\beta}$ is the matrix element $\alpha\beta$ of the $i^\text{th}$ irrep of $G$, of the element $g_{a} \in G$.

Second, considering that $D^{(i)}(g_{a})_{\alpha\beta}$ is really just a scalar, we can, for all elements of $G$, form an $n$-dimensional complex vector.

$$ D^{(i)}(g_{a})_{\alpha\beta} = (V^{(i)\alpha\beta})_{a} $$

The vector component being specified by

• the group element $a$,
• the representation index $i$,
• and the matrix element index $(\alpha,\beta)$ of the $i^\text{th}$ representation.

Therefore, theorem 2.1 states that these vectors are orthogonal.

Constraint on $N_{i}$ and $n_{i}$

Having stated that $(V^{(i)\alpha\beta})_{a} = D^{(i)}(g_{a})_{\alpha\beta}$, we know that since $1 \leq \alpha,\beta \leq n_{i}$, for each $i$, there are $n_i^{2}$ orthogonal vectors. Furthermore, $i = 1,2,\dots, N_{i}$.

However, each vector has dimension $n$, and since there are at most $n$ mutually orthogonal vectors in an $n$-dimension vector space,

$$ \sum_{i=1}^{N_{i}} n_{i}^{2} \leq n $$

In fact, it can be shown that

$$ \sum_{i=1}^{N_{i}} n_{i}^{2} = n $$

Proof

Reminder

$$ \sum_{a=1}^{n} D^{(i)}(g_{a})^{*}_{\alpha\beta} D^{(j)}(g_{a})_{\mu \nu} = \frac{n}{n_{i}} \delta_{ij} \delta_{\alpha \mu} \delta_{\beta \nu} $$

Part One

Suppose that $i \neq j$, meaning the representations are non-equivalent. The theorem then states

$$ \sum_{a=1} D^{(i)}(g_{a})^{*}_{\alpha\beta} D^{(j)}(g_{a})_{\mu \nu} = 0 \quad \text{where} \quad i \neq j $$

We consider the matrix

$$ M = \sum_{a} D^{(j)}(g_{a}) X D^{(i)}(g_{a}^{-1}) $$

Where $X$ is an $n_{j} \times n_{i}$ matrix, which will be determined later.

We can show that $MD^{(i)}(g) = D^{(j)}(g)M$.

$$ \begin{align} M D^{(i)}(g) &= \sum_{a} \left[ D^{(j)}(g_{a}) X D^{(i)}(g_{a}^{-1}) \right] D^{(i)}(g) \\ &= D^{(j)}(g) D^{(j)}(g^{-1}) \sum_{a} \left[ D^{(j)}(g_{a})X D^{(i)}(g_{a}^{-1}) \right] D^{(i)}(g) \\ &= D^{(j)}(g) \sum_{a} \left[ D^{(j)}(g^{-1}) D^{(j)}(g_{a}) \right] X \left[ D^{(i)}(g_{a}^{-1}) D^{(i)}(g) \right] \\ &= D^{(j)}(g) \sum_{a} \left[ D^{(j)}(g^{-1} g_{a}) \right] X \left[ D^{(i)}(g_{a}^{-1} g) \right] \\ &= D^{(j)}(g) \sum_{a} D^{(j)}(g_{a}) X D^{(i)}(g_{a}^{-1}) \qquad \text{using the rearrangement theorem} \\ &= D^{(j)}(g) M \end{align} $$

Since we imposed that the representations are non-equivalent, according to lemma 3, it must mean that $M = 0$ for any matrix $X$.

Suppose that $X_{\tau \tau'} = \delta_{\tau \nu} \delta_{\tau'\beta}$. In this case, the element $\mu\alpha$ of $M =0$ is

$$ \begin{align} M_{\mu\alpha} &= \sum_{a} D^{(j)}(g_{a})_{\mu \tau} \delta_{\tau \nu} \delta_{\tau' \beta} D^{(i)}(g_{a}^{-1})_{\beta \alpha} \\ &= \sum_{a} D^{(j)}(g_{a})_{\mu \nu } D^{(i)}(g_{a}^{-1})_{\beta\alpha} \\ &= 0 \end{align} $$

However, since

$$ D^{(i)}(g_{a}^{-1})_{\beta\alpha} = (D^{(i)}(g_{a})^{-1})_{\beta\alpha} = (D^{(i)}(g_{a})^{\dagger})_{\beta\alpha} = D^{(i)}(g_{a})^{*}_{\alpha\beta} $$

We can re-express $M_{\mu\alpha}$ as

$$ M_{\mu\alpha} = \sum_{a=1}^{n} D^{(i)}(g_{a})^{*}_{\alpha\beta} D^{(j)}(g_{a})_{\mu \nu} = 0 \quad \text{where} \quad i \neq j $$

This corresponds to the theorem.

Part Two

Suppose not that $i = j$, meaning the representations are equivalent. We can still re-use the result we found part one:

$$ MD^{(i)}(g) = D^{(i)}(g)M \quad \forall \space g \in G $$

However, since this time the representations are equivalent, the result implies that $M$ is a constant matrix, according to lemmas 2 and 3.

The constant will depend on our choice for $X$. Using, again, $X_{\tau \tau'} = \delta_{\tau \nu} \delta_{\tau' \beta}$, we find that

$$ \begin{align} M_{\mu\alpha} &= \sum_{a} D^{(i)}(g_{a})_{\mu \tau} \delta_{\tau \nu} \delta_{\tau' \beta} D^{(i)}(g_{a}^{-1})_{\beta \alpha} \\ &= \sum_{a} D^{(i)}(g_{a})_{\mu \nu } D^{(i)}(g_{a}^{-1})_{\beta\alpha} \\ &= \lambda_{\nu \beta} \delta_{\mu\alpha} = \lambda_{\nu\beta} \mathbb{1}_{n_{i} \times n_{i}} \end{align} $$$$ \implies \sum_{a} D^{(i)}(g_{a})_{\mu \nu}D^{(i)}(g_{a}^{-1})_{\beta\alpha} = \lambda_{\nu\beta} \delta_{\mu\alpha} $$

Suppose $\mu = \alpha$, on the right-hand side of the expression, the summation $\delta_{\mu \mu} = n_{i}$, considering the dimension of $D^{(i)}(G)$.

On the left-hand side,

$$ \begin{align} D^{(i)}(g_{a})_{\mu \nu} D^{(i)}(g_{a}^{-1})_{\beta \mu} &= D^{(i)}(g_{a}^{-1})_{\beta \mu} D^{(i)}(g_{a})_{\mu \nu} \\ &= (D^{(i)}(g_{a}^{-1})D^{(i)}(g_{a}))_{\beta \nu}\\ &= (D^{(i)}(g_{a}^{-1} g_{a}))_{\beta \nu} \\ &= D^{(i)}(e)_{\beta \nu} = \delta_{\beta \nu} \end{align} $$$$ \implies \sum_{a}^n D^{(i)}(g_{a})_{\mu \nu} D^{(i)}(g_{a}^{-1})_{\beta \mu} = \sum_{a}^n \delta_{\beta \mu} = n \delta_{\beta \nu} $$

We can finally determine $\lambda_{\nu\beta}$.

$$ \lambda_{\nu\beta} \delta_{\mu \mu} = \lambda_{\nu\beta} n_{i} = n \delta_{\beta \nu} \implies \lambda_{\nu\beta} = \frac{n}{n_{i}} \delta_{\nu \beta} \qquad (\delta_{\nu\beta} = \delta_{\beta \nu}) $$

Therefore,

$$ \sum_{a} D^{(i)}(g_{a})_{\mu \nu}D^{(i)}(g_{a}^{-1})_{\beta\alpha} = \frac{n}{n_{i}} \delta_{\nu\beta} \delta_{\mu\alpha} $$

In combination with the result from part one, and considering that $D^{(i)}(g_{a}^{-1})_{\beta\alpha} = D^{(i)}(g_{a})^{*}_{\alpha\beta}$,

$$ \sum_{a=1}^n = D^{(i)}(g_{a})^{*}_{\alpha\beta}D^{(j)}(g_{a})_{\mu \nu} = \frac{n}{n_{i}} \delta_{ij} \delta_{\alpha \mu} \delta_{\beta \nu} $$

And so, we have proven the Orthogonality Theorem.