Gauss map¶
Given a parametrization \(\pmb{x}: U\subset\mathbb{E}^2\rightarrow S\) of a regular surface at a point \(p\in S\), we could choose a unit normal vector at each point of \(\pmb{x}(U)\) by
Thus we have a differentiable map \(N: U\subset \mathbb{E}^2\rightarrow \mathbb{R}^3\). More generally, if \(V\in S\) is an open set in \(S\) and \(N: V\rightarrow \mathbb{R}^3\) is a differentiable map which associates to each \(p\in V\) a unit normal vector at \(p\), we say that \(N\) is a differentiable field of unit normal vectors on \(V\).
Not all surfaces admit a differentiable field of unit vectors defined on the whole surface. For instance, the Mobius strip.
definition of Gauss map
Let \(S\subset \mathbb{E}^3\) be a regular surface with an orientation \(N\). The map \(N: S\rightarrow \mathbb{R}^3\) takes its values at the unit sphere
thus \(N:S\rightarrow S^2\) is called the Gauss map of \(S\).
Gauss map is differentiable. The differential \(dN_p\) of \(N\) at \(p\in S\) is a linear map from \(T_p(S)\) to \(T_{N(p)}(S^2)\). Since the two are the same space, \(dN_p\) could be looked upon as a linear map on \(T_p(S)\).
Example. check the differential of \(N\) of each surfaces.
(i) Plane. Norm vector is a constant, so \(dN_p\equiv 0\).
(ii) Unit Sphere. Norm vector \(N=(x,y,z)\) and \(dN_p (v)=v\).
(iii) Cylender, i.e. \(x^2+y^2=1\). Norm vector \(N=(x,y,0)\), and
Considering a curve in the surface.
The following is a fact about the differential of Gauss map.
Self-adjoint map of the differential map of Guass map
The differential \(d N_p: T_p (S)\rightarrow T_{p}(S)\) of the Guass map at point \(p\in S\) is a self-adjoint linear map.
We shall show that \(\langle dN_p (w_1), w_2\rangle = \langle w_1, dN_p (w_2)\rangle\).
Now we assume \(\pmb{x}(u,v)\) be a parametrization of \(S\) at \(p\), \(\{\pmb{x}_u, \pmb{x}_v\}\) is the associated basis of \(T_p(S)\). If \(\alpha (t)=\pmb{x}(u(t), v(t))\) is a curve in \(S\), with \(\alpha (0)=p\), then
where \(N_u=dN_p(\pmb{x}_u)\) for \(u\) line, and \(N_v=dN_p (\pmb{x}_v)\) for \(v\) line. So we only need to show that \(\langle dN_p (\pmb{x}_u), \pmb{x}_v\rangle = \langle \pmb{x}_u, dN_p (\pmb{x}_v)\rangle\).
Notice that \(\langle N, \pmb{x}_u \rangle =0\), so taking its derivatives w.r.t \(v\) gives
taking derivatives of \(\langle N, \pmb{x}_v \rangle =0\) w.r.t \(u\) gives \(\langle N_u, \pmb{x}_v\rangle + \langle N, \pmb{x}_{vu}\rangle = 0\).
And we are done.
\(\square\)
Given the above fact, we could associate to \(dN_p\) a quadratic form \(Q\) in \(T_p(S)\), namely \(Q(v)=\langle dN_p(v), v \rangle\) (according to bilinear form \(B(v,w)=\langle dN_p(v), w\rangle\) and \(Q(v)=B(v,v)\)).
Second fundamental form
The quadratic form \(Ⅱ_p\), defined in \(T_p(S)\) by
is called the second fundamental form of \(S\) at \(p\).
We give a geometric interpretation of the above second fundamental form using the normal curvature.
Definition of Normal curvature
Let \(C\) be a regular curve in \(S\) passing through \(p\in S\), \(k\) the curvature of \(C\) at \(p\), with \(\cos \theta=\langle n, N\rangle\) where \(n\) is the normal vector to \(C\) and \(N\) is the normal vector to \(S\) at \(p\). Then the number \(k_n=k\cos \theta\) is called the normal curvature of \(C\subset S\) at \(p\).
Meusnier: geometric interpretation of the second fundamental form
All curves lying on \(S\) and having at a given point \(p\in S\) the same tangent vector share the same normal curvature.
The above proposition allows us to speak of the normal curvature along a given direction at \(p\).