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When studying the representation of the group $SO(3)$ on the space $L^2(\mathbf{S}^2)$, we have the spherical functions as its basis:

$$L^2(\mathbf{S}^2) = \text{span} \left \{ Y^l_m, l \in \mathbf{N}^+, -l \leqslant m \leqslant l \right \}$$ $$f(\theta,\varphi)=\sum_{\ell=0}^\infty \sum_{m=-\ell}^\ell f_\ell^m \, Y_\ell^m(\theta,\varphi)$$

But then suddenly the laplacian comes from nowhere and we says that its eigenfunctions are the spherical functions. I know it has important applications in physics, but from the group-theoretic point of view, is it just an operator like any other else?

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Good question.

The Laplace-Beltrami on $\mathbb S^n$ is not "any" operator, of course. It is related to the group $SO(n)$, like the Euclidean Laplacian is related to the group of isometries of $\mathbb R^n$. The precise mathematical notion that captures this should be that of Casimir operator, but that is something that I do not understand, I must admit.

So I will give you my down-to-earth point of view. The starting point is Schur's lemma of representation theory, according to which, if $\phi\colon V\to V$ is an intertwining map of an irreducible representation of a group, then $\phi$ is a scalar multiple of the identity operator.

So what does it mean? It means that a linear operator $\phi$, that commutes with the action of a group, will be diagonalized by the decomposition of the Hilbert space in irreducible representations (irreps).

And that's precisely what the various Laplacians do: the Laplace-Beltrami commutes with $SO(n)$, therefore the spherical harmonics must be its eigenfunctions, because they are the basis functions of the irreps of $SO(n)$. Similarly, the Euclidean Laplacian commutes with translations, so the plane waves $e_\mathbf{p}(\mathbf{x}):=\exp(i\, \mathbf{x}\cdot\mathbf{ p})$ must be its eigenfunctions.

EDIT (after comments). I have deliberately been sloppy here, because there's a number of technical points that I do not want to address (mostly because I can't, to be honest).

  1. I speak of "action of the group", while the precise term is "representation", as you point out. Which representation, and on which Hilbert space? For both the Laplace-Beltrami and the Euclidean Laplacian, the Hilbert space is $L^2(X)$, with $X=\mathbb S^n$ and $X=\mathbb R^n$, respectively. The group, which we denote by $G$, is $G=SO(n)$ and $G=(\mathbb R^n ,+)$, respectively. And the representation is always given by $$ \rho(g) f (x)=f(g^{-1}\cdot x),$$ where $g^{-1}\cdot x$ reads $x-g$ in the case of $G=(\mathbb R^n, +)$. To say that the Laplacian is an intertwining map means that $$\Delta\rho(g)=\rho(g)\Delta.$$
  2. Schur's lemma says that if $\mathcal H\subset L^2(X)$ is an irreducible subspace for $\rho$ then there exists a scalar $\lambda$ such that $\left.\Delta\right|_{\mathcal H} =\lambda I$. In other words, $$ \Delta f = \lambda f,\qquad \forall f\in\mathcal H.$$ Notice that the eigenvalue $\lambda$ is dependent on $\mathcal H$. Actually, each irreducible subspace is labeled by these eigenvalues. We have therefore realized the irreducible subspaces, that are a rather slippery concept (see definition below), as eigenspaces of the Laplacian, which is much more concrete. Recall that irreducible subspace means that $\rho(g)\mathcal H \subset \mathcal H$ for all $g\in G$ and that if $\mathcal H'$ is a subspace of $\mathcal H$ with the same property then $\mathcal H'$ is either $\{0\}$ or coincides with $\mathcal H$.

I am sweeping something under the rug. Indeed, I am acting as if $\Delta$ were defined on $L^2(X)$, which is not the case: there exist functions whose Laplacian is not defined, or it is not an element of $L^2(X)$. This is especially bad in the case $X=\mathbb R^n$, in which case even the "eigenfunctions" that we discover are not eigenfunctions at all. I am talking about the plane waves $e_\mathbf{p}$, whose $L^2$ norm is $\infty$. There is a lot of theory to study to fully understand how to go around these technical difficulties.

But, whatever, I think that what we did in this post is enough as an illuminating guiding principle.

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  • $\begingroup$ Thanks. In your last paragraph should it be "the Laplace-Beltrami operator commutes with the representation of $SO(n)$ in $\Bbb S^n$"? Does the scalar multiples of the identity operator later become the eigenvalues of the laplacian? Can you elaborate more on the diagonalization of the operator? $\endgroup$ – Ooker Nov 24 '17 at 14:32
  • $\begingroup$ Follow up question: Can any linear operator be viewed as a projection in projective geometry? $\endgroup$ – Ooker Nov 25 '17 at 2:02
  • $\begingroup$ Thanks for your edit. What is $g^{-1}$ in the case of $SO(n)$? I recall that the laplacian has many eigenvalues. Does that contradict to the Schur lemma, which states that the operator is the multiple of only one scalar? And what is the differences between irreducible subspace and invariant subspace? $\endgroup$ – Ooker Nov 26 '17 at 2:00
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    $\begingroup$ If g is a rotation, $g^{-1}$ is its inverse. Concerning the many eigenvalues, I respectfully invite you to read my answer above with more care. As I wrote, we have an eigenvalue for each irreducible subspace, that is, one eigenvalue for each space of spherical harmonics of the same degree. Concerning the last question, that's also written in the text. Moreover it is a basic definition of representation theory, you can find good explanations in pretty much all books on the subject. $\endgroup$ – Giuseppe Negro Nov 26 '17 at 10:30
  • $\begingroup$ I see, I thought the $\lambda$ is fixed for all $\mathcal H$. I think it would be clearer if they have an index, $\lambda_i$ and $\mathcal H_i$. The text also doesn't highlight the difference between the two though. Invariant subspace can still have another invariant subspaces within it (reducible), while irreducible subspace is not. Also, not to defense myself (I agree that I should check it before asking), sometimes getting direct answers rather than having to research is more productive, even when it's just a basic definition. That's why books cannot replace teachers. $\endgroup$ – Ooker Nov 26 '17 at 12:43

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