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There's one paper published at siggraph that captured my attention : Steklov Spectral Geometry.

I'm not an expert in geometry processing, but I'm trying to learn as much as I can. You can easily read in the introduction:

In this paper we provide a practical and mathematically justified spectral approach to extrinsics geometry for geometry processing via an extrinsic alternative to the intrinsic Laplace-Beltrami operator.

I read through the paper and, for who like me enjoy signal processing, is a very nice read. It my understanding the operator they propose an alternative of the classic laplace beltrami operator:

$$ \Delta_{\mathcal{S}} \psi(x) = 0 $$

The operator proposed is

$$ \left\{ \begin{array}{ll} \Delta \psi(x) = 0 & x \in \Omega \\ \nabla_n \psi(x) = \lambda \psi(x) & x \in \Gamma \end{array} \right. $$

Which is the "Steklov eigenproblem". The first line of the system above is the laplace equation again (but volumetric) the second line is a condition on the normal derivative.

What I don't understand is why does the former operator represent an "intrinsics" operator while the latter represent an extrinsic operator.

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I'm not entirely sure, but I guess it might have to do with the notions of intrinsic vs extrinsic curvature. The Laplace–Beltrami operator on a manifold is, I think, intrinsic in the sense that it can be constructed from the "internal" geometry within the manifold itself, and doesn't depend on any embedding in an ambient space. The Steklov operator, on the other hand, looks like it's written explicitly in terms of an embedding, so would be extrinsic in that sense.

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  • $\begingroup$ You mentioned exactly the bit I don't get, where is this "embedding" visible in the Steklov operator? Is it in the domain of the differential operator? $\endgroup$ Sep 9, 2019 at 9:03
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    $\begingroup$ The Steklov operator looks like it's defined on a volume in (around?) the actual mesh surface, and using the derivative along the normal to the mesh, ie the whole thing relies not just on points on the mesh/manifold but also the surrounding space into which it's embedded. $\endgroup$ Sep 9, 2019 at 13:36

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