Let π be the normalised Haar measure for πΊ.
We define
(π£|π€)=β«πΊβ¨π(π)π£|π(π)π€β©ππ(π)which is also an inner product on π since
- conjugate symmetry
(π£|π€)=β«πΊβ¨π(π)π£|π(π)π€β©ππ(π)=β«πΊβββββββββ¨π(π)π€|π(π)π£β©ππ(π)=ββββ(π€|π£)
- linear in second argument
(π£|πΌπ€+π½π’)=β«πΊβ¨π(π)π£|πΌπ(π)π€+π½π(π)π’β©ππ(π)=πΌβ«πΊβ¨π(π)π£|π(π)π€β©ππ(π)+π½β«πΊβ¨π(π)π£|π(π)π’β©ππ(π)=πΌ(π£|π€)+π½(π£|π’)
- positive definite
(π£|π£)=β«πΊβ¨π(π)π£|π(π)π£β©β___β___β>0ππ(π)>0Let {π£π} be an Orthonormal basis with respect to β¨ β
| β
β©
and {π€π} be an orthonormal basis with respect to ( β
| β
).
Then there exists an invertible change of basis π :π βπ with ππ€π =π£π,
which is also a Change of inner product with (π£|π€) =β¨ππ£|ππ€β©.
We define
Λπ(π)=ππ(π)πβ1which is equivalent to π, and unitary since
β¨Λπ(π)π£|Λπ(π)π€β©=β¨ππ(π)πβ1π£|ππ(π)πβ1π€β©=(π(π)πβ1π£|π(π)πβ1π€)=β«πΊβ¨π(βπ)πβ1π£|π(βπ)πβ1π€β©ππ(β)=β«πΊβ¨π(β)πβ1π£|π(β)πβ1π€β©ππ(β)=(πβ1π£|πβ1π€)=β¨π£|π€β©as required.2