Consider a closed connected Riemannian manifold $M$, together with the associated Hilbert space $L^2(M)$ defined with respect to the Riemannian volume density. Let $-\Delta$ be the positive Laplacian $H^2(M)\to L^2(M)$, and let $\{E_j\}_{j\in \mathbb N}$ be the sequence of its eigenvalues, ordered in a non-decreasing way. As the whole space $L^2(M)$ is complicated to study, we can consider instead a sequence of Hilbert spaces $\{V_j\}_{j\in \mathbb N}\subset L^2(M)$ in the limit $j\to \infty$, and try to find out how much this sequence of vector spaces (or an object related to it) tells us about the geometry of $M$.

$\{V_j\}_{j\in \mathbb N}$ could be a filtration of (a dense subspace of) $L^2(M)$, but it does not need to be one. Clearly, Weyl type spectral asymptotics correspond to the choice $$ V_j:=\bigcup_{j'\leq j}\text{Eig}(\Delta,E_{j'}), $$ where $\text{Eig}(\Delta,E_{j'})$ denotes the eigenspace of $\Delta$ corresponding to the eigenvalue $E_{j'}$. My question is: Which other well-known examples of choices for $\{V_j\}_{j\in \mathbb N}$ are there?

Actually, I am new to spectral geometry, and I do not even know what can be said about $M$ geometrically if we know everything about $L^2(M)$, that is I do not know a source which considers the question whether two Riemannian manifolds with isomorphic $L^2$-Hilbert spaces (such that the constant functions are respected) are isometric, diffeomorphic, homeomorphic, homotopy equivalent, etc. In the setting above, this corresponds to the trivial case of choosing $V_j=L^2(M)$ for every $j$. Surely, there must be tons of literature about these questions.

Thank you for your comments and answers.

  • 1
    $\begingroup$ $L^2(M)$ is a separable Hilbert space for any closed connected Riemannian manifold $M$, and any two separable Hilbert spaces are isomorphic. So by itself $L^2(M)$ knows nothing about the geometry of $M$. The general philosophy of spectral theory is that a bare Hilbert space is boring, but a Hilbert space equipped with an operator (or better yet, an algebra of operators) is very interesting. In particular, I know of no source of examples of subspaces $V_j$ that are not eigenspaces of some operator (generally elliptic pseudodifferential). $\endgroup$ Sep 21, 2015 at 21:46

1 Answer 1


Let me explain how to canonically produce a family of $\Delta$-invariant finite dimensional subspaces of $C^\infty(M)$ that completely determines the geometry of $M$.

First, I need to introduce some notation. $\newcommand{\bR}{\mathbb{R}}$ Set $m:=\dim M$. Denote by $g$ the Riemann metric on $M$.

Choose an orthonormal (Hilbert) basis of $L^2(M)$ consisting of eigenvalues $\Psi_j$

$$\Delta \Psi_j =E_j\Psi_j. $$

Next, fix an even smooth function $w:\bR\to[0,\infty)$ with compact support. (You may want think of $w$ as a smooth approximation of $I_{[-1,1]}$, the characteristic function of the interval $[-1,1]$.)

For $\newcommand{\ve}{{\varepsilon}}$ $\ve>0$ denote by $V_\ve$ the finite dimensional subspace of $L^2(M)$ spanned by the eigenfunctions $\Psi_j$ corresponding to the eigenvalues $E_j$ satisfying $\DeclareMathOperator{\supp}{supp}$

$$\ve E_j^{1/2} \in \supp w. $$

We denote by $S(w,\ve)$ this set of eigenvalues. (If you think of $w$ approximating $I_{[-1,1]}$, then $S(w,\ve)$ would consist of eigenvalues satisfying $E_j\leq \ve^{-2}$.)

Now define a smooth map

$$\Phi_\ve: M\to V_\ve \subset L^2(M), \;\; M\ni x\mapsto \Psi_\ve(x)=\sum_j c_j^\ve(x)\Psi_j\in V_\ve,$$


$$c^\ve_j(x):= \ve^{\frac{m+2}{2}}\sqrt{w\bigl(\,\ve E_j^{1/2}\,\bigr)}\;\Psi_j(x). $$

The finite dimensional space $V_\ve$ has a Euclidean structure induced by the inner product

In this paper I prove several things.

  1. For $\ve$ sufficiently small the map $\Phi_\ve$ is an immersion. Denote by $h_\ve$ the metric on $M$ defined by this immersion, i.e., $h_\ve$ is the metric on $M$ obtained by pulling back the Euclidean metric on $V_\ve$ via the map $\Phi_\ve$. (For $\ve$ small, this map is actually an embedding.)
  2. There exists a positive constant $d_m(w)$, depending explicitly on $m$ and $w$ such that as $\ve \to 0$ the metric $\frac{1}{d_m(w)}h_\ve$ converges in the $C^\infty$-topology to the original metric $\cdot g$.$

Loosely speaking, this says that the geometry of the family of finite dimensional subspaces $(V_\ve)$ completely determines the geometry of $M$.

I refer to the paper for more details and of course, a proof.


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