Eigenforms of the Laplacian on Lie groups

Question

I am a bit rusty in my differential geometry and I would like to confirm that my reasoning below holds, and I have some related questions (and all references to related concepts are of interest to me).

Let $G$ be a compact, connected Lie group endowed with the Riemannian metric induced by the Killing form, I would like to show that there is a global basis of the cotangent bundle composed by $1$-eigenforms of the Hodge Laplacian. My argument goes as follows.

1. We can pick a basis $(\xi_i)_i$ of the Lie algebra of $G$ and obtain $1$-forms $\theta_i\in\Omega^1(G)$ by moving the dual elements via the group action.
2. Since both the exterior derivative and the Hodge star commute with the group action, we have that $\Delta\theta_i$ is in the $\mathbb{C}$-span of $(\xi_i)$, meaning that the Hodge Laplacian is a linear automorphism of $\mathrm{span}_\mathbb{C}(\xi_i)_i$.
3. We can now diagonalize this linear map and its eigenvectors $\tilde{\theta}_i$ are the desired basis.

In particular, this tells us that a basis of $1$-eigenforms of the Hodge Laplacian on $G$ is given by the $1$-forms $f_j\tilde\theta_i$ for $(f_j)_j$ a basis of eigenfunctions (by a Stone-Weierstrass type of argument). Similar conclusions can be made about higher degree forms.

Assuming the above is correct, I have a few questions:

• I expect the result above to be well known. Is there any good reference?
• What can be said about the eigenvalues? I would expect them to be related to the curvature of the metric.
• Is all of the diagonalization procedure really necessary? It feels a bit clunky and dependent on the initial choice of basis. E.g. would the original $\theta_i$ already be eigenforms is we take the basis $\xi_i$ to be orthonormal with respect to the Killing metric?

Answer 1

Here are a few brief comments, but, as you suspect, an enormous amount is known about the Laplacian on functions and forms on compact Lie groups.

• Presumably, you know that the Killing form is non-degenerate only in the semi-simple case, i.e., when the center of $G$ is discrete, so I assume that you are only interested in the compact, semi-simple case. Moreover, in that case, you might as well consider the simply-connected cover of $G$, in which case $G$ is the product of its simple factors and the Laplacian 'preserves' the factors, so you might as well assume $G$ is compact, simple, and simply-connected. So I'll assume that henceforth.

• Because the Laplacian preserves the left-invariant 1-forms on $G$, which are an irreducible representation of the right action of $G$, it must be a multiple of the identity, so, in fact, all of the left-invariant 1-forms on $G$ are eigenforms with the same eigenvalue. (The eigenvalue depends on the specific simple Lie group, and this is calculable from the root diagram.) You'll want to look at some of Weyl's original papers on this or at some of the many expositions of this material. A good place to start is S. Helgason's "Differential Geometry, Lie groups, and Symmetric Spaces".

• The eigenfunctions are well-known and can be calculated explicitly, though the combinatorics can be quite involved. See any book that discusses the Peter-Weyl Theorem. The important thing is that the precise eigenvalues and their multiplicities are determined by the Casimir element and combinatorial formulae involving roots and weights. In the simplest case, $\mathrm{SU}(2)$, you can make everything very explicit. (It's a standard exercise in the geometry of Lie groups.)

• Looking at the products $f\theta$ where $f$ comes from an eigenspace of the Laplacian and $\theta$ is a left-invariant $1$-form gets you a start, but these are not (usually) eigenforms themselves. Instead the set of such products spans a (finite-dimensional) tensor product of representations of $G$ and you usually have to decompose this tensor product into its irreducible pieces (or at least determine the possible highest weights and their multiplicities) before you find the actual eigenspaces of the Laplacian on this tensor product. (There's nearly always more than one eigenvalue of the Laplacian in this tensor product.)

• Once you know what you are looking for, you can find a lot of this information in Tony Knapp's "Lie groups: Beyond an Introduction", or you can calculate most of what one wants to know from the material in Knapp's book if he doesn't give it explicitly.