The story of the infinite dimensional space of $ \Delta$ is following, we eliminate ourself with compact smooth non-boundary manifold $ M$ with metric $ g$ , then we have Betrami-Laplace operator $ \Delta_g$ . Why do not consider hodge laplace? Because by hodge theorem the 0-form have only trivial one and high dimensional it is finite… anyway consider the eigenvalue problem: $ $ \Delta_g u=\lambda u$ $ A classical way to investigate the eigenvalue problem is according to consider variational principle and max-min principle. We equip the path integral on the function space $ C^{\infty}(M)$ : $ $ E(f)=\frac{\int_M |\nabla u|^2}{\int_M |u|^2 }$ $ Then it have a sequences of eigenvalue, negative of course: $ $ 0<-\lambda_1<-\lambda_2<…<\lambda_k<…$ $

Then things become interesting, the morse theory of infinite space involve, called it $ X$ , so at least, shrink the far place of $ X$ as a point, in physics, this mean, cut off at fix scale then take the scale to infinite small. What I can do is the following, I can proof the eigenvalue function is uniformly distributed in $ L^2(M)$ and the classical well law, but thing beome complicated when I try to consider topology of the infinite space $ X_{M_g}$ ‘s topology, at finite scale, i.e. $ X_{M_g}^{h}$ which is the cut off at scale $ h$ . Among the other thing, I believe the following issue is true, but without ability to proof it:

Problem for every manifold $ M$ and metric $ g$ on $ M$ , the topology of infinite space $ X_{M_g}$ is the same, beside this, the inverse could be true, i.e. If $ X_{M_1},X_{M_2}$ is not homomorphism for some scale $ h$ then $ M_1,M_2$ is not homomorphism.

I think it is depend by the underling manifold’s topology. But I do not have a rigorous proof, I definitely have a non-rigorous one, if ignore the coverage…

As I find this problem when I try to give a proof of well law, I do not check the reference, may be this problem is a classical one? As always, I will appreciate to any interesting comments and answers, thanks a lots!