# Unitary representation of $SO(3)$

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

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Now that we saw that the transformation formula is of relevance here, the next answerer might address the issues of why $\Pi_1$ is well-defined in the first place, why it is a homomorphism and why it is strongly continuous... –  commenter Oct 10 '12 at 14:30
The point of my previous comment is: the two answers address the issue of why $\Pi_1(R)$ is a unitary operator, provided that it is well-defined (a word needs to be said here). Assuming this, it is then easy to check that $\Pi_1 \colon SO(3) \to U(\mathcal{H})$ is a homomorphism. However, the continuity condition of your first paragraph needs a little more thought. I suggest that you work this out and post it as an answer. As a hint: Prove it first for continuous and compactly supported $v$, then use that the continuous functions of compact support are dense in $L^2(\mathbb{R}^3)$. –  commenter Oct 11 '12 at 0:24

Consider SO(3) generators: $$[X_i,X_j]=i \epsilon^{ijk} X_k\\ X_1=i {\begin{pmatrix} 0& 0&0\\ 0& 0&-1\\0&1&0 \end{pmatrix}}\\ X_2=i {\begin{pmatrix} 0& 0&1\\ 0& 0&0\\-1&0&0 \end{pmatrix}}\\ X_3=i {\begin{pmatrix} 0& -1&0\\ 1& 0&0\\0&0&0 \end{pmatrix}}$$

Note $X_j^\dagger=X_j$.

Write the SO(3) Lie group elements as $$g=e^{i \alpha_j X_j}$$ where $g^{-1}=g^\dagger$. And $g^{-1}g=g^\dagger g=1$.

So $g=e^{i \alpha_j X_j}$ is the unitary Rep of SO(3). (Just make sure my statement here is correct?)

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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.

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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.

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what about continuity condition? –  Une Femme Douce Oct 11 '12 at 6:05