Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
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
up vote 3 down vote accepted

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?)

share|cite|improve this answer

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.

share|cite|improve this answer

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.

share|cite|improve this answer
what about continuity condition? – Un Chien Andalou Oct 11 '12 at 6:05

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.