Hi Bruce,

thank you for your explanation. I think I might have an idea, why the Hessian matrix works in this case, maybe you can confirm or correct me if I'm wrong. Usually, if you're having a 2d surface in space, the principle curvatures are the eigenvalues of the shape operator (which needs calculation of first and second fundamental form), not the Hessian. Only in the particular case of a vanishing gradient, the eigenvalues are the same as those of the Hessian. F.e. for a quadratic function, the gradient vanishes at the minimum (in 1d first derivative is zero). Since you're modelling a quadratic function in a local neighborhood around each vertex, its minimum lies at the vertex, is that correct? if this is so, then the gradient should vanish and Hessian matrix should be fine and my confusion is resolved :). In the literature I found just vague explanations, sometimes mentioning the shape operator, sometimes the Hessian. I just want to make sure I understand this process correctly.

Cheers

Philipp

Date: Thu, 3 Oct 2019 10:03:52 -0400 (EDT)

Hi Philipp

the Hessian is estimated at each vertex by doing a quadratic fit to the
local surface as the height function over the tangent plane of all the
vertices in a 2-neighborhood of that vertex. I'm not sure what the gradient
vanishing is about, but the curvatures are just the eigenvalues of the
Hessian, so I don't think the gradient has anything to do with it. Not that
we also have some discrete tools for computing curvature

make sense?
cheers
Bruce

On Thu, 3 Oct 2019, LOSKE, PHILIPP
(PGR) wrote:

>
> ????????External Email - Use Caution????????
>
> Hi,
>
> I am trying to understand how exactly FreeSurfer estimates the curvature
> values from the white surface. From the mailing lists I understood that the
> white surface is modeled by fitting a second-order polynomial function and
> curvatures are estimated from the Hessian matrix at each vertex (thank you
> Bruce). However, I still have trouble to understand how this works in
> detail. First, as I understand it, curvature can only be derived from the
> Hessian if the gradient vanishes (why is this the case?), and from
> differential geometry, shouldn't instead the shape operator be calculated at
> each vertex on the surface? Second, are the Gaussian and mean curvatures
> then directly calculated from?Hessian/Shape operator or first principle
> curvatures (and are they saved somewhere?). I tried to find a detailed
> explanation in some of the FreeSurfer papers, but couldn't find anything
> really.
>
> Thank you very much in advance!
>
> Cheers
> Philipp