Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

-The spherical harmonics $Y_{lm}$ are complete on $L^2(S^2)$. They are also a representation of the (compact) Lie group $SO_3 (\mathbf{R})$.

-The functions $e^{i n x}$ are complete on $L^2([0,2\pi])$. They are also a representation of the (compact) Lie group $U(1).$

My question is basically, how general is this phenomenon? More specifically,

  1. Bessel functions, Hermite polynomials, Legendre polynomials - do each of these represent some Lie group? If so, what is it in each case?

  2. Is there a nice example of some complete functions that represent a non-compact Lie group?

  3. Is there a nice example of some complete functions that do not represent any Lie group at all?

Thanks!

share|improve this question
    
A. Klimyk and N. Vilenkin, Studied this in their 3 volume monography. They also cover $SL(2, \mathbb{R})$ and $SU(1,1)$ which are not compact. –  Sasha Nov 3 '11 at 23:19
    
Thank you for the suggestion. I looked at Klimyk and Vilenkin, and I'm sure that the answers to my questions are in there somewhere. However, these books are too advanced for me and I had trouble finding the answers to my questions. Can anyone help out with a more specific answer, or a more basic reference? –  marlow Nov 11 '11 at 2:46
    
    
I think you guys would find this question math.stackexchange.com/questions/1163032/… interesting –  bolbteppa Feb 24 at 10:31

1 Answer 1

The representations of $SO(3)$ are indexed by numbers $|\ell, m \rangle$ where $\ell \in \frac{1}{2}\mathbb{Z}$ and $|m| \leq \ell$ with $m - \ell \in \mathbb{Z}$.

The matrix elements of Wigner's small matrix give the Legendre polynomials:

$$ \langle \ell, 0 | e^{-i \theta J_z }|\ell, 0 \rangle = P_\ell(\cos \theta)$$

Here $J_z$ is a generator of the $\mathfrak{so}(3)$ Lie Algebra and the exponent is a rotation , $e^{-i \theta J_z } \in SO(3)$.

share|improve this answer

Your Answer

 
discard

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.