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Thank you for your answer, I pursued your recommendations, and now the warning is that it can open file....../COR-.info. What can I do? JC.

El sáb., 5 oct. 2019 a las 12:00, freesurfer-request@nmr.mgh.harvard.edu escribió:

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1. Re: Curvature estimation from Hessian (LOSKE, PHILIPP (PGR))
2. Segmentation faule recon-all (Backhausen, Lea)
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4. Re: Tracula (Yendiki, Anastasia)
5. Question on Z-scoring (Nillo, Ryan Michael R)
6. ERROR gtmseg (Marina Fern?ndez)
7. Re: Question on Z-scoring (Greve, Douglas N.,Ph.D.)
8. Re: ERROR gtmseg (Greve, Douglas N.,Ph.D.)
9. Re: Segmentation faule recon-all (Greve, Douglas N.,Ph.D.)
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15. Re: Errors running TRACULA (data attached) - valid frame numbers (Yendiki, Anastasia)

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Message: 1 Date: Fri, 4 Oct 2019 16:27:51 +0000 From: "LOSKE, PHILIPP (PGR)" p.loske.18@abdn.ac.uk Subject: Re: [Freesurfer] Curvature estimation from Hessian To: "freesurfer@nmr.mgh.harvard.edu" freesurfer@nmr.mgh.harvard.edu Message-ID: < AM0PR0402MB36506BB2F6A775BE1ACFC1F9E89E0@AM0PR0402MB3650.eurprd04.prod.outlook.com

...

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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

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