Is the Inner Product from Group Averaging Positive Definite?

[kakarukeys]

Homework Statement

Does anyone know how to show the inner product obtained from group averaging is positive definite?

Show that
$$\int_Gdg \langle\phi|\hat{U}^{-1}(g)|\phi\rangle \geq 0$$
$$\int_Gdg \langle\phi|\hat{U}^{-1}(g)|\phi\rangle = 0 \Rightarrow \int_Gdg \langle\phi|\hat{U}^{-1}(g) = 0$$

Homework Equations

$$\langle\langle \eta\phi_1, \eta\phi_2\rangle\rangle_{phys} = \int_Gdg \langle\phi_2|\hat{U}^{-1}(g)|\phi_1\rangle$$
$$\eta$$ is the rigging map (not important here, just ignore the L.H.S.)

$$\eta: \Phi \longrightarrow \Phi'$$
$$\eta\phi = \int_Gdg \langle\phi|\hat{U}^{-1}(g)$$

$$\phi, \phi_1, \phi_2\in\Phi\subset H$$ are vectors in a dense subspace of the Hilbert space H.
$$\hat{U}(g)$$ is a unitary representation of G (densely defined) on the Hilbert space H.
$$dg$$ is a bi-invariant measure on G

The Attempt at a Solution

No idea at all. Not sure what kind of maths is needed.

 

[Jhero]

Thank you for your question. To show that the inner product obtained from group averaging is positive definite, we need to prove that \langle\phi|\hat{U}^{-1}(g)|\phi\rangle \geq 0 for all \phi\in\Phi. This can be done using the properties of a unitary representation and the bi-invariant measure on G.

Firstly, we know that \hat{U}(g) is a unitary operator, which means that it preserves the inner product. This can be written as \langle\phi_1|\hat{U}(g)|\phi_2\rangle = \langle\hat{U}(g)\phi_1|\hat{U}(g)\phi_2\rangle for all \phi_1, \phi_2\in H. Therefore, we can rewrite the inner product obtained from group averaging as \langle\phi|\hat{U}^{-1}(g)|\phi\rangle = \langle\hat{U}(g)\phi|\phi\rangle.

Next, we can use the bi-invariant measure on G to rewrite the integral as \int_Gdg \langle\hat{U}(g)\phi|\phi\rangle = \int_Gdg \langle\phi|\hat{U}(g)|\phi\rangle. This is because the bi-invariant measure is invariant under left and right translations on G, and therefore the order of the elements in the integral does not matter.

Now, since \hat{U}(g) is a unitary representation, we know that it preserves the norm of a vector. This means that ||\hat{U}(g)\phi|| = ||\phi|| for all \phi\in H. Therefore, we can rewrite the integral as \int_Gdg \langle\phi|\hat{U}(g)|\phi\rangle = \int_Gdg ||\phi||^2 \geq 0. This shows that the inner product obtained from group averaging is positive definite, as the integral is always greater than or equal to 0.

To show that \int_Gdg \langle\phi|\hat{U}^{-1}(g)|\phi\rangle = 0 \Rightarrow \int_Gdg \langle\phi|\hat{U}^{-1}(g) = 0, we can use the fact that \hat{U}(g)