7.2: The Weingarten Map

The L-Transformation

Given a vector A in the tangent space to a surface V at a point p, we define another vector L(A) in that same tangent space as follows.

    1. Find a curve X(t) in the surface so that X(t0) = p and X'(t0) = A
    2. Let L(A) = N'(t0), where N(t) is the unit normal vector to the surface at X(t).

In the following demo, we indicate a surface and its tangent plane at p, and, in a separate window, the tangent plane at p viewed perpendicular to the unit normal vector at p. We may select a vector A in the tangent plane and the demonstration shows the image vector L(A).

Exercises

    1. For the sphere of radius R describe the transformation L at any point p. Do the same for a torus of revolution.

    2. For a saddle surface, for which positions of A will L(A) be a scalar multiple of A? For which positions will L(A) be perpendicular to A? Same question for a quadratic function graph in general.

Now consider the transformation L for a parametrized surface X(u,v). Given A = a1Xu(u0,v0) + a2Xv(u0,v0) in the tangent plane of X(t0) we compute

X'(t) = Xu(t0)u'(t0) + Xv(t0)v'(t0)    and
N'(t) = Nu(t0)u'(t0) + Nv(t0)v'(t0)            

for any curve with X'(t0) = A. Thus,

L(A) = N'(t0) = a1Nu(u0,v0) + a2Nv(u0,v0)

Note that the transformation L is a linear transformation on the tangent space at p. Proof. Also note that the definition of L(A) is independent of the parametrization used to define it. Proof.


Coefficients of the Weingarten Map

The normal vector, N(u,v), is perpendicular to both Nu(u,v) and Nv(u,v). To see that this is true, we use the fact that N(u,v) is a unit vector.

N(u,v)·N(u,v) = 1
(N(u,v)·N(u,v))u = 0     
2N(u,v)·Nu(u,v) = 0    

The same argument follows for Nv(u,v).

This means that Nu(u,v) and Nv(u,v) lie in the tangent plane, and can, therefore, be expressed in terms of Xu(u,v) and Xv(u,v). We define the coefficients of the Weingarten map, Lij, by

Nu = -L11Xu - L12Xv
Nv = -L21Xu - L22Xv,

or, written in matrix form,

[Nu] = -[L11    L12][Xu]
[Nv]      [L21   L22][Xv]

The Lij matrix will prove to be a very useful. For example, we can relate it to the Gaussian curvature as follows. We defined the Gaussian curvature, Κ(u,v), by the following condition :

Nu(u,v) × Nv(u,v) = Κ(u,v)Xu(u,v) × X v(u,v).

We then substitute for Nu(u,v) and Nv(u,v) in terms of the coefficients of the Wiengarten map and take the cross-product.

Nu(u,v) × Nv(u,v) = (L11L22-L12L21)Xu(u,v) × X v(u,v)
Nu(u,v) × Nv(u,v) = det(Lij)Xu(u,v)× X v(u,v).

Thus, the Gaussian curvature is equal to the determinant of the Weingarten map:

Κ(u,v) = det(Lij)

Exercises

    1. By a similar substitution into the defining equation for the mean curvature, show that 2H(u,v) = -trace(Lij).


The L Matrix and the Principal Curvatures

Note that the definition of the Weingarten Map in its matrix form describes a linear transformation L that acts on vectors in the tangent plane and sends them to other vectors in the tangent plane. For example, L(Xu) = Nu and L(Xv) = Nv. Since L is a linear map,

L(aXu + bXv) = aNu + bNv.

If we look at the Weingarten Map from a linear algebra perspective, the partial derivative vectors Xu(u,v) and Xv(u,v) form a basis for the tangent space. Furthermore, the matrix of the L-transformation is written with respect to this basis. The eigen values of the L matrix are given by its characteristic equation:

λ2 + trace(L)λ + det(L) = 0

Substituting the results Κ(u,v) = det(L) and 2H(u,v) = -trace(L), yields a familiar equation:

λ2 - 2Hλ + Κ = 0

We saw at the end of Lab 6 that the roots of this quadratic equation are precisely the definitions of the principal curvatures! So, the eigen values of the L matrix are κ1 and κ2.

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