Talk:Second fundamental form/Archive 1

Archive 1

Error in formula for second fundamental form ?

I think in the section "Classical notation" there is something wrong. The definition of the second fundamental form there is in my opinion only correct if the tangent vectors $\mathbf {r} _{v},\mathbf {r} _{u}$ have absolute value 1. I just calculated the second fundamental form of a cylinder and you don't get -1/R and 0 as eigenvalues of the second fundamental form otherwise. Apart from that the eigenvalues should be independent of the parametrization which is apparently not the case. — Preceding unsigned comment added by 77.21.82.90 (talk) 11:42, 7 December 2008 (UTC)

It looks right to me. With the standard parameterization of the cylinder r(u,v)=(Rcosu,Rsinu,v) so

$\mathbf {r} _{uu}=(-R\cos u,-R\sin u,0),\quad \mathbf {r} _{uv}=\mathbf {r} _{vv}=0.$

The outward normal is n=(cosu, sinu, 0), so computing the dot products

$L=-R,M=N=0\quad \implies I\!I={\begin{bmatrix}-R&0\\0&0\end{bmatrix}}$

which is what we expected to get. (The negative R is because the outward normal was chosen). siℓℓy rabbit (talk) 14:13, 7 December 2008 (UTC)

Hi again, I made a mistake in first comment. What you expect are the eigenvalues -1/R and 0 since the curvature of a circle scales as 1/R (small radius gives large curvature). Apart from that the eigenvalues of the second fundamental form should be independent of the parametrization which is not the case according to the given formula. 193.175.8.208 (talk) 12:40, 8 December 2008 (UTC)

I believe the problem here is that it doesn't make sense to talk about the eigenvalues of the second fundamental form, since the second fundamental form takes as input a pair of vectors $II(X,Y)=I(X,SY)$ where S is the shape operator. The operator whose eigenvalues give the principal curvatures is S, which should be (in the $\mathbf {r} _{u},\mathbf {r} _{v}$ basis),

$S={\begin{bmatrix}E&F\\F&G\end{bmatrix}}^{-1}{\begin{bmatrix}L&M\\M&N\end{bmatrix}}$

where $E=\mathbf {r} _{u}\cdot \mathbf {r} _{u}$, $F=\mathbf {r} _{u}\cdot \mathbf {r} _{v}$, $G=\mathbf {r} _{v}\cdot \mathbf {r} _{v}$. In the case of the cylinder, $E=R^{2},F=0,G=1$. siℓℓy rabbit (talk) 12:57, 8 December 2008 (UTC)

I am not an expert in classical differential geometry but the article on principle curvatures in wiki tells that they (the principle curvatures) should be eigenvalues of the second fundamental form. Also, what is the virtue of calculating the second fundamental form if it does not contain information independent of the parametrization. 193.175.8.208 (talk) 13:57, 8 December 2008 (UTC)

It is something called a tensor. Although the components do depend on the parameterization, they transform in a particular way going from one coordinate system to the other. The invariant information that the second fundamental form contains is that as a bilinear form, it does not depend on the choice of parameterization. (Try it: take a pair of tangent vectors X and Y and compute $Y^{T}{\begin{bmatrix}L&M\\M&N\end{bmatrix}}X$ using two different parameterizations. Note that you will also need to change X and Y when you change parameterization, since you need these to be the same tangent vectors.) siℓℓy rabbit (talk) 14:03, 8 December 2008 (UTC)

Also, the eigenvalues of the matrix ${\begin{bmatrix}L&M\\M&N\end{bmatrix}}$ are the principal curvatures only when the basis vectors are orthonormal. Really, the principal curvatures are the eigenvalues of the shape operator. When the basis is orthonormal, the matrix of the shape operator and the second fundamental form coincide. siℓℓy rabbit (talk) 14:06, 8 December 2008 (UTC)