Answer to Question #147420 in Differential Geometry | Topology for shweta

Question

Calculate the first fundamental form, the second fundamental form, and the

Weingarten matrix of the following surface patches of the unit sphere:

(a) σ(θ, φ) = (cos θ cos φ, cos θ sin φ,sin θ).

(b) σ(u, v) = (sech u cos v,sech u sin v,tanh u).

Answer 1

$$Solution$$

For the parametric surface $r(u,v)$ the first fundamental form is (see http://web.mit.edu/hyperbook/Patrikalakis-Maekawa-Cho/node28.html):

$I=ds^2=Edu^2+2Fdudv+Gdv^2,$ where $E=(r_u,r_u), F=(r_u,r_v),$ $G=(r_v,r_v)$ .

The second fundamental form is: $II=Ldu^2+2Mdudv+Ndv^2$ , where: $L=-(r_u,\text{N}_u),M=-(r_v,\text{N}_u),N=-(r_v,\text{N}_v)$ .

The normal vector is: $\text{N}=r_u\times r_v=\begin{vmatrix} i & j& k \\ (r_u)_1 & (r_u)_2& (r_u)_3\\ (r_v)_1 & (r_v)_2& (r_v)_3 \end{vmatrix}$ , where by $()_j$ we denoted the respective coordinates of the vectors.

NOTE: The Weingarten matrix is the matrix of the shape operator of surface s with respect to variables u and v.

It is given by$\begin{pmatrix} \frac{MF-LG}{EG-F^2} & \frac{LF-ME}{EG-F^2} \\ \frac{NF-MG}{EG-F^2} & \frac{MF-NE}{EG-F^2} \end{pmatrix}$

a). $r_{\theta}=(-sin\,\theta\,cos\varphi,-sin\,\theta\,sin\varphi,cos\,\theta),$

$r_{\varphi}=(-cos\,\theta\,sin\varphi,cos\,\theta\,cos\varphi,0),$

$E=(r_{\theta},r_{\theta})=sin^2\theta\,cos^2\,\varphi+sin^2\theta\,sin^2\,\varphi+cos^2\theta=1$ ,

$F=(r_{\theta},r_{\varphi})=sin\,\theta\,\,cos\,\theta\,cos\varphi\, sin\varphi-sin\,\theta\,\,cos\,\theta\,cos\varphi\, sin\varphi=0,$

$G=(r_{\varphi},r_{\varphi})=cos^2\,\theta\,sin^2\varphi+cos^2\,\theta\,cos^2\varphi=cos^2\,\theta.$

$I=d\theta+cos^2\theta\,d\varphi$ .

$\text{N}=\begin{vmatrix} i & j &k \\ -sin\,\theta\,cos\varphi&-sin\,\theta\,sin\varphi&cos\,\theta\\ -cos\,\theta\,sin\varphi&cos\,\theta\,cos\varphi&0 \end{vmatrix}=i\begin{vmatrix} -sin\,\theta\,sin\varphi&cos\,\theta\\ cos\,\theta\,cos\varphi&0 \end{vmatrix}-j\begin{vmatrix} -sin\,\theta\,cos\varphi&cos\,\theta\\ -cos\,\theta\,sin\varphi&0 \end{vmatrix}+k\begin{vmatrix} -sin\,\theta\,cos\varphi&-sin\,\theta\,sin\varphi\\ -cos\,\theta\,sin\varphi&cos\,\theta\,cos\varphi \end{vmatrix}=$

$=-i\,cos^2\theta\,cos\,\varphi-j\,cos\,\theta\,sin\,\varphi-ksin\,\theta\,cos\,\theta=(-\,cos^2\theta\,cos\,\varphi,-\,cos\,\theta\,sin\,\varphi,-sin\,\theta\,cos\,\theta),$

The second fundamental form is:

$II=Ldu^2+2Mdudv+Ndv^2,$

where $L=-(r_u,\text{N}_u),M=-(r_v,\text{N}_u),N=-(r_v,\text{N}_v)$

The Weingarten matrix is: $\begin{pmatrix} \frac{MF-LG}{EG-F^2} & \frac{LF-ME}{EG-F^2} \\ \frac{NF-MG}{EG-F^2} & \frac{MF-NE}{EG-F^2} \end{pmatrix}$

b). $r_{u}=(-\frac{sinh\,u}{cosh^2\,u}cos\,v,-\frac{sinh\,u}{cosh^2\,u}sin\,v,\frac{1}{cosh^2u}),$

$r_{v}=(-sech\,u\,sin\,v,sech\,u\,cos\,v,0),$

$E=(r_u,r_u)=\frac{sinh^2\,u}{cosh^4\,u}cos^2\,v+\frac{1}{cosh^4\,u}=\frac{cos^2v}{cosh^2u}$ ,

$F=0, G=(r_v,r_v)=\frac{1}{cosh^2u};$

$I=\frac{cos^2v}{cosh^2u}du+\frac{1}{cosh^2u}dv$ .

$\text{N}=\begin{vmatrix} i & j &k \\ -\frac{sinh\,u}{cosh^2\,u}cos\,v&-\frac{sinh\,u}{cosh^2\,u}sin\,v&\frac{1}{cosh^2u}\\ -sech\,u\,sin\,v&sech\,u\,cos\,v&0 \end{vmatrix}=i\begin{vmatrix} -\frac{sinh\,u}{cosh^2\,u}sin\,v&\frac{1}{cosh^2u}\\ sech\,u\,cos\,v&0 \end{vmatrix}-j\begin{vmatrix} -\frac{sinh\,u}{cosh^2\,u}cos\,v&\frac{1}{cosh^2u}\\ -sech\,u\,sin\,v&0 \end{vmatrix}+k\begin{vmatrix} -\frac{sinh\,u}{cosh^2\,u}cos\,v&-\frac{sinh\,u}{cosh^2\,u}sin\,v\\ -sech\,u\,sin\,v&sech\,u\,cos\,v \end{vmatrix}=$

$\text{N}=i\frac{cos\,v}{cosh^3u}+j\frac{sin\,v}{cosh^3u}-k\frac{sinh\,u}{cosh^3\,u}$ .

The second fundamental form is:

$II=Ldu^2+2Mdudv+Ndv^2,$

where $L=-(r_u,\text{N}_u),M=-(r_v,\text{N}_u),N=-(r_v,\text{N}_v).$

The Weingarten Matrix is: $\begin{pmatrix} \frac{MF-LG}{EG-F^2} & \frac{LF-ME}{EG-F^2} \\ \frac{NF-MG}{EG-F^2} & \frac{MF-NE}{EG-F^2} \end{pmatrix}$ .