31

Parameterize a hypersurface $M$ by $r: \Omega \rightarrow \mathbb{R}^n$, and let $T_p M$ denote the tangent space at $p = r(u)$. We define the Weingarten map to be the linear map $L_p : T_p M \rightarrow T_p M$ given by $$L_p(v) = -\partial_v N$$ where $N$ is the unit normal vector field of the hypersurface.

What I'm looking for is perhaps some geometric explanation, pictures or software that will let me sort of see what the Weingarten map looks like for 2d surfaces, and why we might expect that critical points for normal curvature occurs at its eigenvectors. (It could also be the case that this is just due to the magic of linear algebra, but some deeper explanation would be nice).

user901823
  • 353

1 Answers1

14

I realise that this question is over a year old, but thought I will answer it nevertheless to solidify my own understanding.

The Weingarten map is intricately linked with quite a few notions. One way is that it can be looked at as the Derivative of the Gauss map. For example if $ \mathbf{n} : M \to S^2$ is your Gauss map, then the derivative $D\mathbf{n}(x)$ at any point $x \in M$ is a linear map $$D\mathbf{n}(x) : T_x(M) \to T_{\mathbf{n}(x)}(S^2)$$

But the codomain in the last statement is again $T_x(M)$. Now this $D\mathbf{n}(x)$ denoted as $S_x$ is termed the Weingarten map at $x$. This gives for any $Y,Z \in T_x(M)$, the bilinear form $S_x(Y,Z) = \langle S_xY, Z \rangle $.

The latter map is what helps define the shape operator for a surface. It is a symmetric linear operator as one can easily observe.This is quite easy to visualise for 2-d surfaces in $\mathbb{E}^3$ as this would give the variation of the normal field w.r.t a tangent vector at $x$. In other words the Weingarten map serves to describe the extrinsic properties of a surface.

There is more that can be said in terms of the relation to the second fundamental form and also about the symmetry. More importantly there is a nice generalisation of this for Riemannian Manifolds. But I hope this suffices.

Edit: As for your last question, I think its just due to the traits of linear operators and Lagrange multipliers. But I am not sure. Maybe someone else could illuminate a bit more on that topic.

Chip Hogg
  • 103
Vishesh
  • 2,928
  • what is $T_x(M)$? – happymath Nov 02 '14 at 16:14
  • 2
    It is the tangent space to the surface $M$ at the point $x$, in other words, space of tangent vectors to $M$ at $x$. – Vishesh Nov 04 '14 at 02:56
  • 1
    From my understanding of the weingarten map, Dn(x) is the differential of the gauss map not its derivative and the weingarten map is just an other name for the shape operator. Theses points are not clear in your answer. – Cryckx Jan 21 '21 at 14:20