Given a compact Riemannian manifold $(M,g)$, is there a subring of $C^{\infty}(M)$ that determines the isomorphism class of $(M,g)$? (In the same way that $C^{\infty}(M)$ determines the diffeomorphism class via the max spectrum.)

Same question for compact metric spaces, more generally.

I am not sure what condition to ask for on these functions... maybe something like $d(f(x), f(y)) \leq d(x,y)$? (Probably this has to be fiddled with to actually get a ring.)

What about $(M, \nabla)$, for $\nabla$ some connection?

If not ... why not?

This is just following the functions on a space determine its geometry philosophy that people in algebraic geometry like.

  • 3
    $\begingroup$ Someone told me that the right notion in this case is something called a spectral triple: IIRC it's the sections of $L^2$ sections of the bundle $\Omega^n(M)$ of forms, along with the operator $d+d^*$ on this, considered as a $C^\infty(M)$ module. I think it is a hard theorem that this recovers the isometry class of the Riemannian manifold. (I don't want to write an answer because I really know nothing about this, just stuff I've heard in passing.) In any case, the keyword 'spectral triple' should help your searches. $\endgroup$ – user98602 Jan 2 '16 at 22:56

This appears to answer the question: https://mathoverflow.net/questions/301869/sheaf-theoretically-characterize-a-riemannian-structure

In particular, the right idea is to consider the sheaf of harmonic functions of the Laplace-Beltrami operator. (Which is only a sheaf of vector spaces, not of rings.)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.