When does symmetry force repeated eigenvalues?
Symmetry leaves fingerprints in the Laplace spectrum.
It is well known that continuous symmetries force eigenvalue multiplicities. Less obvious is that purely discrete symmetries already do the same.
If a closed Riemannian manifold admits a symmetry of order greater than 2, its Laplace spectrum cannot be simple.
A single 3-fold rotation is enough.
A Picture
"Superfície não orientável — Bordo trifólio" by Matemateca (IME/USP) / Rodrigo Tetsuo Argenton
Licensed under CC BY-SA 4.0
The surface above has clear 3-fold rotational symmetry.
Any Riemannian manifold admitting this symmetry must have a repeated eigenvalue.
The Mechanism
Let $M$ be closed and connected. The Laplacian decomposes
\[L^2(M) = \bigoplus_{i=0}^\infty E_i, \quad \Delta \varphi = \lambda_i \varphi.\]Key fact:
The Laplacian commutes with isometries.
Thus each eigenspace $E_i$ is preserved by every isometry.
Now fix $k \ge 0$ and consider the first $k+1$ eigenspaces $E_0,\dots,E_k$. Choose orthonormal bases in each of these spaces and let
\[l = \sum_{i=0}^k \dim E_i.\]Define the eigenmap
\[\Phi^k : M \to \mathbb{R}^l\]to be the map whose coordinates are all eigenfunctions spanning $E_0,\dots,E_k$.
For sufficiently large $k$, this map is an embedding.
Now here is the structural point:
Every isometry $\alpha$ satisfies
\[\Phi^k \circ \alpha = U_\alpha \Phi^k\]for some orthogonal matrix
\[U_\alpha \in O(d_0)\times\cdots\times O(d_k).\]Because $\Phi^k$ is an embedding, this assignment $\alpha \mapsto U_\alpha$ is injective.
So we obtain a faithful representation:
\[\mathrm{Isom}(M) \hookrightarrow O(d_0)\times\cdots\times O(d_k).\]Simple Spectrum Forces Order-2 Symmetry
Suppose the spectrum is simple, so $ \dim E_i = 1 $ for all $i$.
Then each block $O(d_i)$ is just ${\pm 1}$.
Therefore
\[\mathrm{Isom}(M) \hookrightarrow \{\pm 1\}^l.\]In particular:
- The isometry group is finite.
- Every element has order at most 2.
So if $M$ admits a symmetry of order greater than 2, the spectrum cannot be simple.
Some eigenvalue must have multiplicity at least 2.
A Geometric Bound
Only finitely many eigenfunctions are needed to embed $M$, with the number bounded in terms of dimension, volume, injectivity radius, and Ricci curvature.
Therefore, in the simple-spectrum case,
\[|\mathrm{Isom}(M)| \le 2^C\]for some geometric constant $C$.
Symmetry constrains eigenfunctions. Eigenfunctions constrain symmetry. And a single 3-fold rotation forces two of them to travel together.