# great orthogonality theorem

Theorem:

 Consider two unitary, irreducible matrix representations $\Gamma^{(i)}(G)\,$ and $\Gamma^{(j)}(G)\,$ of a group $G\,$. Then $\sum_{g} \Gamma^{(i)*}(g)_{\alpha\beta} \Gamma^{(j)}(g)_{\gamma\delta} = \frac{[G]}{l_{i}} \delta^{ij} \delta_{\alpha\gamma} \delta_{\beta\delta} \,$, where $[G]\,$ is the order of the group $G\,$, and $l_i\,$ is the dimension of $\Gamma^{(i)}(G)\,$.

Proof:

Define the matrix $M\,$:

$M \equiv \sum_{g\in G} \Gamma^{(i)}(g) X \Gamma^{(j)\dagger}(g)\,$,

where $X\,$ is some unspecified matrix. Then

 $\Gamma^{(i)}(h) M\,$ $= \Gamma^{(i)}(h) \sum_{g} \Gamma^{(i)}(g) X \Gamma^{(j)\dagger}(g)\,$, $= \sum_{g} \Gamma^{(i)}(hg) X \Gamma^{(j)\dagger}(g)\,$, $= \sum_{hg} \Gamma^{(i)}(hg) X \Gamma^{(j)\dagger}(h^{-1}hg)\,$, $= \sum_{g} \Gamma^{(i)}(g) X \Gamma^{(j)\dagger}(h^{-1}g)\,$, (relabeling $hg \to g\,$) $= \sum_{g} \Gamma^{(i)}(g) X \Gamma^{(j)\dagger}(g) \Gamma^{(j)}(h)\,$, $= M \Gamma^{(j)}(h)\,$.

By the converse of Schur's lemma, either $i=j\,$, or $M = 0\,$. If $i = j\,$, then by Schur's lemma $M = m \mathbf{1}\,$, where we find $m\,$ by taking the trace.

$\sum_{g} \Gamma^{(i)}(g) X \Gamma^{(j)\dagger}(g) = m \delta^{ij} \mathbf{1}\,$,
$[G] \operatorname{tr} X = m l_{(i)}\,$. So
$\sum_{g} \Gamma^{(i)}(g) X \Gamma^{(j)\dagger}(g) = \frac{[G]}{l_{i}} \delta^{ij} \operatorname{tr}X \mathbf{1}\,$,

or, with index notation

$\sum_{g} \Gamma^{(i)}(g)_{\alpha\beta} X_{\beta\gamma} \Gamma^{(j)\dagger}(g)_{\gamma\delta} = \frac{[G]}{l_{i}}\delta_{\beta\gamma} X_{\beta\gamma} \delta^{ij} \delta_{\alpha\delta}\,$,

but, since $X_{\beta\gamma}\,$ is arbitrary,

$\sum_{g} \Gamma^{(i)}(g)_{\alpha\beta} \Gamma^{(j)\dagger}(g)_{\gamma\delta} = \frac{[G]}{l_{i}} \delta^{ij}\delta_{\beta\gamma} \delta_{\alpha\delta}\,$,

or, relabeling indices,

$\sum_{g} \Gamma^{(i)*}(g)_{\alpha\beta} \Gamma^{(j)}(g)_{\gamma\delta} = \frac{[G]}{l_{i}} \delta^{ij} \delta_{\alpha\gamma} \delta_{\beta\delta} \,$.

## Consequences

Choosing $\Gamma^{(j)}(G)\,$ to be the identity representation, we get

$\sum_{g} \Gamma^{(i)}(g)_{\alpha\beta} = 0\,$.

Also, the great orthogonality theorem implies an orthogonality between characters.

## References

1. M. Tinkham (1992). Group Theory and Quantum Mechanics. Dover Publications. ISBN 978-0486432472.
2. M. Hamermesh (1989). Group Theory and its Applications to Physical Problems. Dover publications. ISBN 978-0486661810.
3. W. Miller, Jr. (1972). Symmetry Groups and their Applications. Academic Press. ISBN 978-0124974609.
4. J. F. Cornwell (1997). Group Theory in Physics (Three volumes), Volume 1. Academic Press. ISBN 978-0121898007.