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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

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

where

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

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.