# Spherical harmonic decomposition

1. Sep 26, 2007

### JoAuSc

If you have a function f(x), you can decompose it as an infinite sum of sine waves of the form sin(nx) and cos(nx) for increasing n. If you have a function g(theta,phi), you can decompose it into spherical harmonics; for each order l you have 2l+1 spherical harmonics. Is there anything special about this fact, that for nth order behavior of f(x) you only need 2 functions for each step (sin and cos), but for lth order behavior of g(theta,phi) you need 2l+1 functions?

2. Sep 26, 2007

### D H

Staff Emeritus
You have 2N+1 functions for the sin(nx)/cos(nx) decomposition as well: two for each n from 1 to N plus one constant term, which corresponds to cos(0*x). The sin(0*x) term is identically zero.

3. Sep 26, 2007

### JoAuSc

What I meant is that for each n for f(x), we have two functions, but for each l for g(theta,phi), we have 2l+1. I'm not counting the terms from previous n or l. I guess what I'm asking is this: why do we have one number to describe the order of a Fourier series of f(x) (namely, n), but two numbers (l and m) for spherical harmonics? Is this difference in numbering simply a mathematical convenience, or is there a broader significance?

4. Sep 26, 2007

### D H

Staff Emeritus
The Fourier series addresses a one dimensional problem, hence one index. The spherical harmonics address a two dimensional problem, hence two indices.

5. Sep 26, 2007

### Chris Hillman

World's Shortest Summary of Harmonic Analysis

Consider the Laplacian on the circle $S^1$ and on the sphere $S^2$. Determine the eigenvalues and the eigenspace of eigenfunctions for each eigenvalue. (Seriously--- this is fairly elementary if you know a bit about Sturm-Liouville theory!) Result:

The space of real-valued square integrable functions on the circle, $L^2(S^1)$, decomposes as the orthogonal direct sum of the eigenspaces of the eigenvalues, $-\ell^2, \, \ell \in \mathbold{N}$. For $\ell > 0$ these eigenspaces have dimension two; the eigenfunctions are $\sin(\ell \, \phi), \; \; \cos(\ell \, \phi)$.

The space of real-valued square integrable functions on the sphere, $L^2(S^2)$, decomposes as the orthogonal direct sum of the eigenspaces of the eigenvalues, $-\ell \, (\ell+1)$. For $\ell > 0$ these eigenspaces have dimension $2 \, \ell+1$; the eigenfunctions can be taken to be the Legendre polynomial
$$P(\ell, \cos(\theta))$$
(that's the $1$ in $2 \, \ell + 1$) plus
$$P(\ell, m, \cos(\theta)) \, \cos(m \, \phi), \; \; 1 \leq m \leq \ell$$
$$P(\ell, m, \cos(\theta)) \, \sin(m \, \phi), \; \; 1 \leq m \leq \ell$$
where the $P(\ell, m, \cdot)$ are the associated Legendre functions (that's the $2 \, \ell$ in $2 \, \ell + 1$). That is, the eigenfunctions are the real and imginary parts of the usual spherical harmonics
$$Y^{\ell, m}(\theta, \phi) = P(\ell, m, \cos(\theta)) \, \exp(m \, i \, \phi)$$

See Kenneth I. Gross, "The Evolution of Non-Commutative Harmonic Analysis", Amer. Math. Monthly Aug.-Sept. 1978: 525--548. (Students and academics whose institution subscribes to JSTOR: past issues of the Monthly are available on-line, and past issues back to 1894 are well worth snarfing--- highest recommendation!)

These results can be complexified in the obvious way. I have discussed only the scalar spherical harmonics; there are also vector spherical harmonics and tensor spherical harmonics. Also, these results can be extended with minimal change (Weyl) to the Laplacian associated with compact Lie groups other than $SO(n+1)$ acting on homogeneous spaces other than $S^n$. With more work (Harish-Chandra) to non-compact semi-simple Lie groups. With still more work (Mackey) to infinite-dimensional Lie groups. With still more work (Kirillov, Kostant, etc.) to nilpotent Lie groups.

See also my post #4 in the recently locked thread (thanks, berkeman!) "Representation theory?" https://www.physicsforums.com/showthread.php?t=185965

My point is: yes, there is a broader significance!

Last edited: Sep 26, 2007
6. Sep 26, 2007

### dextercioby

GOOD: 1.Great explanation, Chris !!

BAD: 2.You forgot to mention that the results you presented for the 2 Hilbert spaces are applications of the Peter-Weyl theorem.

OK, solved now.

Last edited: Sep 26, 2007
7. Sep 26, 2007

### Chris Hillman

Peter-Weyl Theorem

Thanks, but it's Ken Gross's explanation, not mine! The paper I cited is well worth reading.

I got distracted before I could add the parenthetical mention of Weyl, known as "Holy Hermann" to his detractors and "Peter" to his friends. Confusingly, Peter of Peter-Weyl is F. Peter, a distinct person and a rare example of a graduate student who made a huge contribution to mathematics but who left mathematics without taking a degree (IIRC--- correct me if I am wrong). Following a natural train of thought, Gleason would be an example of someone who didn't take a degree but who stayed in math and made more than one important contribution. In the spirit in which history is our servant rather than our master, I like to arrogate to mathematical expositors the right of habitually ascribing everything to the most famous partner on any given team. Or to be safe, to simply ascribe everything to Euler That is to say, since it is impossible to put results in their full historical context or to unravel all contributions in a reasonable space or with reasonable effort, we should simply accept that since the history of mathematics is inherently inaccurate and unfair, we shouldn't worry too much about "historical accuracy". (Hmm... this is a rather facist attitude, isn't it?! Maybe I should rethink this...)

Er... did I address your complaint? The paper by Ken Gross does sketch the Peter-Weyl theorem. (EDIT: the answer was "yes".)

Last edited: Sep 26, 2007
8. Sep 29, 2007

### JoAuSc

Thanks for the information, guys. I have a couple more questions.

• Is the fact that the eigenspaces of the laplacian on a sphere have dimension 2l+1 due to the topology being a sphere, or to the fact that we're using the laplacian? In other words, if I were to come up with a similar linear differential operator to use on the sphere, would the eigenfunctions still come in groups of 2l+1 due to the sphere?
• For Fourier analysis, (and correct me if I'm wrong), if we have a periodic function, we would find the Fourier coefficients, but if we have an aperiodic function, we'd find the Fourier tranform. Is there a similar analog for the sphere? Let's say we have a function defined on the sphere. It seems to me that if we were to enlarge the sphere to the point where the radius is infinity, we'd have the 2D plane, and that the analog to finding the coefficients for a spherical harmonic decomposition would be to find the 2D fourier transform, but I don't know if the topology of this infinite-radius sphere being different from a 2D Euclidean space would change the end result.

9. Sep 29, 2007

### Chris Hillman

Hi, JoAuSc, I think it would be best if you followed up by reading some of the sources I cited.

Not every operator induces an orthonormal basis of eigenfunctions. One point I didn't stress is that the Laplacian happens to be a positive definite (or negative-defininite, depending upon sign conventions) and self-adjoint operator, and we use these special features to obtain the ONB of eigenfunctions and other properties.

I think there's some (understandable) confusion about terminology. I was focusing on compact Lie groups (roughly, "closed and bounded", like spheres but unlike planes).

Not sure what you mean by "enlarge" but it sounds like you are talking about a punctured sphere, which does not have the same topology as the original sphere.
Maybe you have in mind a limiting process from finite dimensional Lie groups which transforms some classical Lie groups into others and which can change topology, but if so, that's something else.