# How to understand the potential energy cost function based on cotan Laplacian operator in 3D mesh deformation

Suppose I have a source triangle mesh surface $$S^0 = (V,E)$$. $$V,E$$ are sets of $$n$$ vertices and $$m$$ edges of the mesh, respectively. $$S^1,S^2,S^3, \ldots, S^l$$ is denoted as a sequence of meshes after each deformation (nonrigid deformation, which means the transformation matrix applied to each vertex is not identical), that tend to finally get close to a target mesh $$T$$. To regularize the deformation, some paper used the Laplacian-based potential energy term, which is given by:

$$G=\sum_{v_i\in V}\left\|\Delta_i {\textbf{x}}^f-\Delta_i {\textbf{x}}^0\right\|^2, \\ \textbf{x}=[v_0,v_1,\ldots, v_n]^T.$$

The Laplace-Beltrami operators $$\Delta$$ acting on the surface are given by:

$$\Delta v_i = \frac{3}{A_i}\sum_{j\in N_i}\frac{\cot{\alpha_{ij}}+\cot{\beta_{ij}}}{2}(v_j-v_i).$$

The definition of angles $$\alpha$$ and $$\beta$$ can be found on this site. My question is if the energy term tries to minimize the Laplacian over the deformed mesh and the Laplacian over the source. Then, it seems that $$\Delta v_i$$ can only be computed from the $$S^0$$, since the $${\textbf{x}}^f$$ is unknown. Or the cotangent weights can be computed from the source mesh, and keep them as constant through the deformation, which probably means keep the angle as curvature features, because the Laplacian has very similar form to the mean curvature?

My question is if the energy term tries to minimize the laplacian over the deformed mesh and the laplacian over the source.

The distance term $$D$$ is deforming the source mesh into the target mesh. The potential term $$G$$ is constraining that deformation to "respect" the original geometry of the source mesh. That's it.

On every iteration the mesh $$R^f$$ is being deformed more and more towards the target by the term $$D$$ of the energy and less and less constrained by the term $$G$$ of the energy as the contribution of $$G$$ is weighted by $$\alpha$$ which decreases asymptotically to zero on each iteration.

There is no "minimization" of laplacian. The laplacian of $$R^0$$ is constant and is the "target" of the $$G$$ term so the laplacian of $$R^f$$ have to be close to the "target" in order to minimize the term.

Note that the Laplacian is translation invariant, so no conflict with the $$D$$ term. Also Laplacian is NOT rotation invariant so "alignment" is needed in order to minimize $$G$$ which is exactly why they choose the Laplacian.

• I sort of understand the function of energy term through the minimization iteration. It meant keeping the geometry properties of the source, such as the order of the neighboring vertex. That is why choosing the source as "target" in G term, instead of choosing target "T". However, I do not understand in this situation, why the laplacian is not rotation invariant? Jun 13, 2023 at 23:36
• This Laplacian operator is vector valued. Vectors are oriented quantities i.e., two vectors of same module are equal only if they have same orientation. That's why it is not rotation invariant. The output of two Laplacians are equal up to rotation. Jun 14, 2023 at 20:52
• In this case, the Laplacian of the mesh can be expressed as a Laplacian matrix dot product x Jun 14, 2023 at 21:05