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Hi there,

Please tell me if I should divide these into individual questions next time.

Short intro:

Spherical Harmonics are a nice collection of functions. They are orthogonal and allow you to take a function $f$ and break it into a linear combination of spherical harmonic functions $Y_n^m$. This is very similar to how the Fourier Transform allows you to break a function into a linear combination, which is the Fourier Series.

Questions about Spherical Harmonics:

  1. What kind of functions can be broken down using this linear combination?

  2. I read in Wikipedia here about having the Laplacian of $f$ equal to zero. How is that connected to this linear combination (or is that an unrelated and has more to do with physics)?

  3. For the following linear combination of $f$: $f=\displaystyle\sum_n^\infty\displaystyle\sum_{m=-n}^n\alpha_{nm}Y_{nm}(\theta,\phi)$ it seems as though $f$ takes only two parameters $\theta$ and $\phi$ but that doesn't make a lot of sense because (I think) there should be a radial parameter $r$ somewhere. Where am I wrong?

Thank you very much, help saves lives :)

Ofer

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Spherical harmonics are a basis of $L^2(S^2)$, where $S^2$ is the unit sphere. You can multiply them by a basis of $L^2(0,\infty)$ to obtain a basis of $L^2(R^3)$. –  Helge Jul 21 '10 at 14:24
    
What does $L^2(0,\infty)$ mean? –  Ofer Jul 21 '10 at 14:40
4  
Check the nice and elementary book Harmonic Function Theory, by Axler, Bourdon, and Ramey; it has some good material on the spherical Laplacian on $\mathbb{S}^n$ and spherical harmonics. –  Pietro Majer Jul 21 '10 at 14:47
    
$L^2(0,\infty)$ = square integrable functions $(0,\infty) \to \mathbb{C}$. The essential point is that using Hilbert space theory most things become semi-trivial. –  Helge Jul 21 '10 at 15:16
5  
You should. That will at least tell you why your first question isn't completely sensible: the spaces of functions are infinite dimensional, but definitions (that I like) of linear combination usually require finitely many objects. In some sense it is false to say that functions can be broken up into linear combination. It is better to say that functions can be approximated arbitrarily well with linear combinations. –  Willie Wong Jul 21 '10 at 15:36

3 Answers 3

If you are only interested in expanding polynomials then you can forgo Hilbert spaces and get an exact expansion into a finite linear combination.

Spherically-invariant case Let $f$ be a polynomial in $x_1,\ldots, x_n$ (the coefficients may be real, complex, or even a field of characteristic zero). If $f$ is invariant under isometries of the $n$-dimensional space, i.e. $f(Ax)=f(x)$ for any $n\times n$ orthogonal matrix $A,$ then $f=g(r^2),$ where $g$ is a polynomial in one variable of degree $\deg(f)/2$ and

$$r^2=x_1^2+x_2^2+\ldots+x_n^2.$$

This is intuitively clear: a rotationally invariant function depends only on the distance to the center and $g(t)=f(t,0,\ldots,0)$ is a polynomial. (For $n\geq 2$ it's sufficient that $f$ be invariant under rotations, i.e. the special orthogonal group $SO(n)$.)

The theory of spherical harmonics, which is closely related to the representation theory of the orthogonal group $O(n),$ generalizes this observations to polynomials (and even more general functions) that are not necessarily spherically invariant.

General case The first key theorem states that a polynomial $f$ can be expanded as

$$ f(x)=\sum_{k=0}^d r^{2k}h_k(x) \qquad (*)$$

where $d\leq\deg(f)/2$ and $h_k$ is a harmonic polynomial of degree at most $\deg(f)-2k.$ Recall that a polynomial is $h$ is harmonic if $\Delta h=0,$ where $\Delta$ is the Laplace operator,

$$\Delta h=\frac{\partial^2 h}{\partial x_1^2}+\frac{\partial^2 h}{\partial x_2^2}+\ldots+\frac{\partial^2 h}{\partial x_n^2}.$$

Moreover, the harmonic polynomials $h_k$ in $(*)$ are uniquely determined. If $f$ is spherically invariant then $h_k$ is simply the coefficient of $t^k$ in the polynomial $g(t)$ from above, viewed as a constant (i.e. degree $0$) polynomial.

So now every polynomial can be expanded in terms of the powers of $r^2$ and harmonic polynomials. What can we say about the latter? Let $\mathcal{H_\ell}$ (respectively, $\mathcal{H}$) denote the space of homogeneous harmonic polynomials of degree $\ell$ (respectively, all harmonic polynomials). Every harmonic polynomial is uniquely decomposed into the sum of homogeneous polynomials of various degrees, and these components are themselves harmonic:

$$\mathcal{H}=\bigoplus_{\ell\geq 0}\mathcal{H}_\ell\qquad (**)$$

The second key theorem states that $\mathcal{H}_\ell$ is an irreducible representation of the orhtogonal group $O(n)$ of dimension $\binom{n+\ell-1}{n-1}-\binom{n+\ell-3}{n-1}$ (for $\ell=0$ or $1$, the second term is zero). For $n\geq 3,$ this representations remains irreducible upon restriction to the special orthogonal group $SO(n)$ (for $n=3$, this is the rotation group).

For small $n$, namely $n=2,3,4$, there are standard bases in the spaces $\mathcal{H}_\ell$ that have been tabulated and extensively studied. In particular, for $n=3,$ $\mathcal{H}_\ell$ is a $(2\ell+1)$-dimensional representation of $SO(3)$, or spin $\ell$ representation in physics language, which has a basis indexed by an integer $m,\ -\ell\leq m\leq \ell$ constructed using associate Legendre polynomials $P_\ell^m.$ Depending on your goals, you may want a suitable explicit description of these functions in rectangular or spherical coordinates (the Wikipedia article Spherical harmonics, including the references, is a good starting point).

Let me also mention a common point of confusion related to your question 3: in view of $(*),$ for any polynomial $f$ as above, there exists a unique harmonic polynomial $h$ such that the restrictions of $f$ and $h$ to the unit sphere

$$S^{n-1}: x_1^2+x_2^2+\ldots+x_n^2=1$$ coincide, $f|_{S^{n-1}}=h|_{S^{n-1}}.$ In other words, the space $\mathcal{H}$ of harmonic polynomials can be naturally identified with the space of polynomial functions on the unit sphere $S^{n-1}$ and the decomposition $(**)$ becomes the decomposition of the polynomial functions on the sphere into irreducible representations of $O(n),$ each of which occurs with multiplicity one. These functions are frequently called spherical harmonics.


Addendum Here is a high level view of polynomial spherical harmonics from a representation theory vantage point. The decomposition $(*)$ is related to the representation theory of the Lie algebra $\mathfrak{sl}_2.$ The operators

$$E=r^2/2,\ F=-\Delta/2,\ H=\deg+n/2$$

on the vector space of polynomials in $n$ variables commute with orthogonal transformations and form a representation of $\mathfrak{sl}_2.$ (I've described the skew-symmetric analogue in this answer.) Homogeneous harmonic polynomials are precisely lowest weight vectors for $\mathfrak{sl}_2,$ and the second key theorem amounts to saying that the lowest weight spaces are irreducible representations of $O(n).$ One consequence of this description is that the coefficients $h_k(x)$ in $(*)$ can be found inductively starting with $h_d(x)$ using repeated applications of the Laplace operator (the precise statement is omitted due to bulky notation). Furthermore, the first and second key theorems can be combined into the statement that the space $\mathcal{P}$ of polynomials in $n$ variables has the following decomposition into irreducible components under the joint $O(n)$ and $\mathfrak{sl}_2$ actions:

$$\mathcal{P}=\bigoplus_{\ell\geq 0} \mathcal{H}_\ell \otimes V_\ell, $$

where $V_\ell$ is the lowest weight $\mathfrak{sl}_2$-module with lowest weight $\ell+n/2.$ This is one of the starting points of Roger Howe's theory of reductive dual pairs. In the present case, the reductive dual pair is $(O(n),SL(2,\mathbb{R}))$ over the real numbers.

Remark The group $SL(2)$ is secretely a symplectic group, in fact, the isometry group of a 2-dimensional symplectic vector space, and the theory extends to reductive dual pairs consisting of an orthogonal group and a symplectic group of any rank. The same theory over $\mathbb{Q}$ and its adeles is behind some classical results in the theory of theta functions due to Siegel and Weil. In particular, it explains the importance of theta functions with harmonic coefficients.

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Representation theory uses something similar to the Hilbert space theory needed to understand this problem. So I don't really see the simplification. –  Helge Jul 22 '10 at 9:48
    
Representation theory (or at least the part of it Victor is using) only requires finite dimensional vector spaces. If you think the hard part of Hilbert space theory is questions about convergence, you will find this a simplification. –  David Speyer Jul 22 '10 at 12:15

Spherical harmonics can represent the square-integrable functions, i.e. the functions which, when squared and integrated over the surface of the sphere, produce a finite result.

There's two kinds of spherical harmonics, which accounts for your confusion in your second and third questions. There are two-dimensional spherical harmonics, confined to the surface of a sphere. This is why the expression you give in question three does not include a radial coordinate. For purposes of celestial mechanics, it is useful to extend these to cover the volume outside the sphere, in such a way that the Laplacian is zero. These three-dimensional spherical harmonics provide a basis for the gravitational potential outside a body in space. So the zero Laplacian is because space is empty, and has nothing to do with the linear combination properties of spherical harmonics.

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Regarding the spherical harmonics confined to the surface of a sphere, they take 2 angle parameters (theta, phi) right? Can you use them to create linear combinations (that is, approximations that are arbitrarily close) of square-integrable functions that are parametrized with Cartesian coordinates (x,y) and not polar coordinates (theta, phi)? Thanks a lot! –  Ofer Jul 21 '10 at 19:15
    
What are the cartesian coordinates on the sphere? If you mean the ones inherited from $R^3$, there are two distinct $z$ values (i.e. 2 points) per $(x,y)$. So are you assuming the function is symmetric relative to the $x$-$y$ plane? Also, for any two given sets of coordinates, as long as you have a change of variables formula, you can express a function in either set of coordinates... –  Willie Wong Jul 21 '10 at 21:59

Regarding the second question, the point is that the spherical harmonics are eigenvectors for the Laplacian, and they form an orthogonal (or orthonormal, depending on the normalization) basis for $L^2(S^2)$. This is one of the main reasons why they are useful; using this basis, the Laplacian is diagonal, so the eigenvalues and their multiplicities are evident.

So if $f$ is in the kernel of the Laplacian, that tells you that most of the coefficients are $0$ in the expansion of $f$ in terms of the spherical harmonics (and you will then see that this implies that $f$ is a constant function).

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