# Finding principal curvatures at points on a surface without parametrization

Mathematics Asked by confusedstudent on August 31, 2020

I have some points in 3D space. I can fit the equation of a surface z = f(x,y) to them either globally or locally.
However:

1. The surface does not pass through the origin
2. I can’t parametrize it.

How can I find principal curvatures at any point on this surface?

I am finding this to be very difficult. All the examples I’ve seen so far parametrize the surface.

I’ve tried a "shortcut" method of finding the normal at each point and looking at how much the angle of the normal changes from a point to its neighbor, and estimating arc length ~ distance between the two points. This didn’t work too well because the points aren’t evenly spaced and I don’t have neighbors in all directions around each point.

Online resources point me to the shape operator, which requires the first and second fundamental forms — which I have no clue how to get from an equation like the one I showed above. I don’t have a curvilinear coordinate system – do I have to have one?

Thank you

If surface $$M$$ defined by the graph of a function $$z=f(x,y)$$, then the function $$(x,y)mapsto (x,y,f(x,y))$$ is a global parameterization $$mathbb{R}^2to M$$. In this parameterization, the shape operator $$s$$ (w.r.t. the positive $$z$$ direction) takes a particularly simple form $$s=left(I-frac{operatorname{grad}f(operatorname{grad}f)^T}{1+|operatorname{grad}f|^2}right)frac{operatorname{Hess}(f)}{sqrt{1+|operatorname{grad}(f)|^2}}$$ Where $$operatorname{Hess}(f)$$ is the matrix of second partial derivatives of $$f$$ and $$operatorname{grad}(f)$$ is the vector of first partials of $$f$$. In index notation, this is $$s^i{}_j=C(partial_ipartial_j f)-C^3(partial_i f)(partial_k f)(partial_kpartial_j f), C=[1+partial_i fpartial_i f]^{-1/2}$$ The principal curvatures of $$M$$ are the eigenvalues of $$s$$.