Perspectives on spherical harmonics

Spherical harmonics appear in lots of different places and have different interpretations that at first sight don’t seem to have anything to do with one another. In this post, I’ll try to connect three very common ones (namely as harmonic polynomials, as eigenfunctions of the Laplacian and as irreps of \(\operatorname{SO}(3)\)).

We’re going to define spherical harmonics as homogeneous harmonic polynomials \(\mathbb{R}^3 \to \mathbb{C}\). Let’s break this down:

  • A polynomial of three variables is a finite sum of the form \[\sum_{\alpha} a_\alpha x^{\alpha_x}y^{\alpha_y}z^{\alpha_z}\] over multi-indices \(\alpha \in \mathbb{N}_0^3\). Some examples are \(x^2y + 2z\) or \(xyz + y^2\).
  • The coefficients \(a_\alpha\) can be complex numbers, but we will only plug in real numbers for \(x\), \(y\) and \(z\). That’s why we interpret polynomials as functions \(\mathbb{R}^3 \to \mathbb{C}\).
  • Homogeneous mean that \(\alpha_x + \alpha_y + \alpha_z\) is the same for all the \(\alpha\) we sum over, so all the terms in the sum have the same degree. For example, \(x^2 + 2yz + xz\) is homogeneous, while \(xy + z\) is not.
  • Harmonic means that the Laplacian of the polynomial vanishes: \(p\) is harmonic if \(\Delta p = 0\). Here, \(\Delta = \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2}\).

We will write \(\mathcal{H}_l\) for the space of all homogeneous harmonic polymonials of degree \(l\) (meaning \(\alpha_x + \alpha_y + \alpha_z = l\) for all summands).

If you’ve seen spherical harmonics before (and you presumably have, if you’re reading this post), it’s probably been in the form of functions \(Y_l^m(\theta, \varphi)\) defined on the sphere. So why are we talking about these polynomials on \(\mathbb{R}^3\) instead?

The answer is that every polynomial \(p \in \mathcal{H}_l\) can be written in spherical coordinates as \[p(x, y, z) = r^l Y(\theta, \varphi)\] for some function \(Y: S^2 \to \mathbb{C}\). To see why, write \(x\), \(y\) and \(z\) in spherical coordinates and plug them into the polynomial. They each have a factor of \(r\) and then some factors depending on \(\theta\) and \(\varphi\). So because \(p\) is homogeneous, each summand consists of a factor \(r^l\) times something that depends only on \(\theta\) and \(\varphi\). So we can think of homogeneous polynomials as polynomials defined on the sphere – their continuation to \(\mathbb{R}^3\) is automatically determined by their degree \(l\). Therefore, we won’t really distinguish between homogeneous harmonic polynomials defined on \(\mathbb{R}^3\) and their restrictions to \(S^2\), we will refer to both as spherical harmonics.

This should also explain the name: spherical harmonics are harmonic polynomials living on the sphere.

The functions \(Y_l^m\) that you may have seen are just a particular choice of basis for the vector space of spherical harmonics. If you multiply them by \(r^l\), you get polynomials in \(\mathcal{H}_l\), and \[\{r^l Y_l^m| -l \leq m \leq l\}\] is a basis for \(\mathcal{H}_l\).

Eigenfunctions of the Laplacian

One of the reasons that spherical harmonics are so ubiquitous is that they are the eigenfunctions of the spherical Laplacian \(\Delta_{S^2}\). They key to that is the following fact (which is just a brief calculation): for a function \(Y: S^2 \to \mathbb{C}\), \[\Delta (r^l Y) = r^{l - 2}\left(l(l + 1)Y + \Delta_{S^2}Y\right)\thinspace.\] So \(r^l Y(\theta, \varphi)\) is harmonic if and only if \[\Delta_{S^2}Y = -l(l + 1)Y\thinspace.\] This already proves that spherical harmonics are eigenfunctions of the spherical Laplacian.

But we can say more than that: if we take any eigenfunction \(f: S^2 \to \mathbb{C}\) of the spherical Laplacian and multiply by \(r^l\) (with \(l\) such that \(-l(l + 1)\) gives the eigenvalue1), then \(r^l f(\theta, \varphi)\) must be harmonic. So the eigenfunctions of the spherical Laplacian are in fact in 1-to-1 correspondence with harmonic homogeneous functions on \(\mathbb{R}^3\). It then turns out – and this part is far from obvious – that all such functions are polynomials2! So the spherical harmonics aren’t just eigenfunctions of the spherical Laplacian, they make up all of its eigenfunctions.

Irreducible representations of \(\operatorname{SO}(3)\)

Another famous role that spherical harmonics play is as the irreducible representations of \(\operatorname{SO}(3)\) (more precisely: the (complex) irreducible representations of \(\operatorname{SO}(3)\) are exactly the spaces \(\mathcal{H}_l\)). This is connected to the fact that they are the eigenfunctions of the spherical Laplacian.

That the eigenspaces of the spherical Laplacian are representations of \(\operatorname{SO}(3)\) follows directly from the fact that the Laplacian commutes with rotations: we have a representation \(G \curvearrowright L^2(S^2, \mathbb{C})\) via \[(r \cdot f)(x) := f(r^{-1}x)\] for any rotation \(r\) and \(f \in L^2(S^2, \mathbb{C})\). For an eigenfunction of the Laplacian, we get \[\Delta_{S^2}(r \cdot f) = r \cdot \Delta_{S^2} f = r \cdot \lambda f = \lambda (r \cdot f)\thinspace,\] so each eigenspace is invariant under the action of \(\operatorname{SO}(3)\). Therefore, the representation on \(L^2(S^2)\) can be restricted to each eigenspace, so each \(\mathcal{H}_l\) gives a representation of \(\operatorname{SO}(3)\).

Showing that these representations are in fact irreducible is much more difficult (there’s a proof here for example, if you really want to dive into that). But if we just take that for granted, it’s again easy to show that every irreducible subrepresentation of \(L^2(S^2)\) is a space of spherical harmonics: because the Laplacian is an equivariant map on each such representation, Schur’s Lemma implies that it must be either the zero map (which it isn’t) or multipication by a constant \(\lambda \in \mathbb{C}\). Therefore, each irreducible representation is contained in an eigenspace of the Laplacian. But these eigenspaces are themselves irreducible, so the representation in question must already be equal to the eigenspace.

Finally, it’s possible to show that all irreducible representations of \(\operatorname{SO}(3)\) are subrepresentations of \(L^2(S^2)\). This is again much more difficult and is also a very special fact about \(\operatorname{SO}(3)\) (for example, the Laplacian’s eigenspaces are still irreducible representations in higher dimensions, but they are not the only ones anymore). But combining this with our results from above, the spherical harmonics make up all the irreducible representations of \(\operatorname{SO}(3)\).

  1. I’m skipping over some details here, see for example Claim 4.0.1 here ↩︎

  2. See Corollary 4.0.6 in the same document for a proof ↩︎

Erik Jenner
Erik Jenner
AI Master’s Student