Unitary representation of $SO(3)$

Question

Definition: $\mathcal{H}$ be a Hilbert space and $U(\mathcal{H})$ denote the unitary operators on it, If Unitary representation of a matrix lie group $G$ is just a homomorphism $\Pi:G\rightarrow U(\mathcal{H})$ with the following continuity condition: $A_n\rightarrow A\Rightarrow \Pi(A_n)v\rightarrow\Pi(A)v$

Now could any one help me what is going on here in detail so that I can understand,

"let $\mathcal{H}=L^2(\mathbb{R}^3,dx)$ the space of all square integrable functions on $\mathbb{R}^3$, for each $R\in SO(3)$ we define an operator $[\Pi_1(R)f](x)=f(R^{-1}x)$, since Lebesgue measure is rotationally invariant, $\Pi_1(R)$ is a unitary operator for each $R\in SO(3)(why?)$ , and it is easy to show $R\rightarrow \Pi_1(R)$ is unitary representation." Thank you

Answer 1

We have $$ \int_{\mathbb R^3} [\Pi_1(R) f](x)\overline{[\Pi_1(R) g]}(x) dx = \int_{\mathbb R^3} f(R^{-1} x) \bar g(R^{-1}x) dx. $$ Now make the substitution $u = R^{-1} x$. Since $R \in SO(3)$ the Jacobian of this transformation is 1. So the above is $\int_{\mathbb R^3} f(u) \bar g(u) du$. This shows that each $\Pi_1(R)$ is a unitary operator since it preserves the $L^2$ inner product.

Answer 2

To show that $\Pi_1(R)$ is unitary you have to prove:

1. $\langle \Pi_1(R) f, \Pi_1(R) g \rangle = \langle f, g \rangle$ for each $f, g\in L^2(\mathbb R^3)$
2. $\Pi_1(R)$ is surjective.

For each $f, g\in L^2(R^3)$ we have $$ \langle \Pi_1(R)f, \Pi_1(R)g \rangle := \int_{R^3} f(R^{-1}x)\overline{g(R^{-1}x)} dx = \int_{R^3} f(x)\overline{g(x)} dx =: \langle f, g \rangle $$ Writing the previous equality we used the rotationally invariance of Lebesgue measure.

For each $f\in L^2(R^3)$, let $\tilde f$ be the function $x \to f(R x)$, we have $$ (\Pi_1(R) \tilde f)(x) = \tilde f(R^{-1}x)= f(R R^{-1}x) = f(x) $$ So both requirements are satisfied.