If you have 2 parametrizations of a surface $M$, $x$ and $y$, such that for some point $P \in M$, $x(u_0,v_0)=y(s_0,t_0)=P$. How do you show that the tangent plane of $M$ at $P$ are the same? That is how do I show that $\text{Span}(x_u,x_v)=\text{Span}(y_s,y_t)$.
Prove that two parametrizations of a surface M give the same tangent plane
differential-geometrysurfaces
Related Solutions
It is possible to show that a surface which admits two orthogonal families of geodesic is isometric with a plane (i.e. the Gaussian curvature is zero) in three steps:
- we show the the coordinate lines are orthogonal
- we show that once parametrized the two families of geodesics by coordinate curves $E_v=0$ and $G_u=0$
- we use the fact that when $F=0$, $E_v=0$ and $G_u=0$ the Gaussian curvature is zero
First step
Consider two plane curves $\alpha(t)$ and $\beta(t)$ in the domain of $X$, and let $\gamma(t)=X \circ \alpha$ and $\delta(t)=X \circ \beta$ be the corresponding curves on the regular surface S. Then at the point $p= X(\alpha(t_0))=X(\beta(t_0))$, the angle $\theta$ between the two curves is \begin{equation} \cos{\theta}= \frac{\gamma' \cdot \delta'}{|\gamma'||\delta'|} \end{equation} In the basis ${X_u,X_v}$ the above equation becomes \begin{equation} \cos{\theta}= \frac{X_u \cdot X_V}{|X_u||X_v|}= \frac{F}{E G} \end{equation} Thus, all the coordinate curves of a parametrization are orthogonal if and only if $F=0$ for all $(u,v)$ in the domain of $X$.
Second step
Given the metric tensor $f$ \begin{equation} f=\sqrt{E(u,v)u'^{2}+2 u' v' F(u,v)+G(u,v) v'^{2}} \end{equation} we compute the following partial derivative: \begin{equation} \frac{\partial f}{\partial u}=\frac{1}{2 f} \big(u'^{2} E_u+2 u' v' F_u+v'^2 G_u \big) \end{equation} \begin{equation} \frac{\partial f}{\partial u'}=\frac{1}{f} \big(u' E+v' F \big) \end{equation} \begin{equation} \frac{\partial f}{\partial v}=\frac{1}{2 f} \big(u'^{2} E_v+2 u' v' F_v+v'^2 G_v \big) \end{equation} \begin{equation} \frac{\partial f}{\partial v'}=\frac{1}{f} \big(u' F+ v' G \big) \end{equation} where $E_x$, $F_x$, and $G_x$ are the derivative of $E$, $F$, and $G$ with respect to the coordinate x ={u,v}, while $u'$ and $v'$ are the derivative withy respect to the parameter $t$. On the curve $v=constant$, $u$ can be taken as a parameter, then the curve is $u=t$, $v=c$. We also have that $u'=1$ and $v'=0$. By substituting these values in the previous equations we have: \begin{equation} \frac{\partial f}{\partial u}=\frac{1}{2 \sqrt{E}} E_u \end{equation} \begin{equation} \frac{\partial f}{\partial u'}=\sqrt{E} \end{equation} \begin{equation} \frac{\partial f}{\partial v}=\frac{1}{2 \sqrt{E}} E_v \end{equation} \begin{equation} \frac{\partial f}{\partial v'}=\frac{F}{\sqrt{E}} \end{equation} According to the Euler-Lagrange theorem in the calculus of variations, the parametric equation $(u(t),v(t))$ that optimize the metric $f$ must satisfy the differential equations: \begin{equation} \frac{\partial f}{\partial u}-\frac{d}{dt} \big (\frac{\partial f}{\partial u'} \big)=0 \end{equation} \begin{equation} \frac{\partial f}{\partial v}-\frac{d}{dt} \big (\frac{\partial f}{\partial v'} \big)=0 \end{equation} By substituting the above values, after some calculations we have that the curve $v= constant$ is a geodesic if \begin{equation} E (E_v- 2 F_u) + E_u F =0 \end{equation} This condition is therefore necessary. Conversely if the above condition is satisfied, it is possible to prove that the curve $v= constant$ is a geodesic. We can now repeat the same computation for a curve $u=constant$. In this case, $u=constant$ is a geodesic if and only if \begin{equation} G (G_u- 2 F_v) + G_v F =0 \end{equation}
Third step
We can then conclude that for orthogonal parametrizations two families of geodesics are coordinate curvesmusrt satisfy the conditions $F=0$, $E_v=0$, and $G_u=0$. Thus the metric tensor becomes: \begin{equation} f=\sqrt{E(u)u'^{2}+G(v) v'^{2}} \end{equation} In the situation where one considers an orthogonal parametrization of a neighborhood of a surface, the Gaussian curvature can be written as \begin{equation} K=-\frac{1}{\sqrt{E G}} \big[\frac{\partial }{\partial u} \big(\frac{1}{\sqrt{E}} \frac{\partial \sqrt{G}}{\partial u} \big)+\frac{\partial }{\partial v}\big(\frac{1}{\sqrt{G}} \frac{\partial \sqrt{E}}{\partial v} \big)\big] \end{equation} Since $E=E(u)$ and $G=G(v)$, we have that the Gaussian curvature is zero. The formula I used for the Gaussian curvature can be derived by the usual definition of the Gaussian curvature computed as the radio of the determinants of the second and first fundamental form, by writing the determinal of the second fundamental form by using the triple product. The computation is quite long.
Write $T=\alpha'(t)$. Assume $Y$ is any vector field along $\alpha$ that is tangent to the surface, then $\nabla_TY$ at $\alpha(t)$ is just the projection of $D_TY$ to the tangent plane at $\alpha(t)$; here $D_TY$ is the Euclidean directional derivative.
Now $Y$ is parallel along $\alpha$, iff $\nabla_TY=0$, iff $D_TY$ is everywhere perpendicular to the tangent plane. Now this last property holds for one surface iff it holds for the other.
Best Answer
HINT: Write $x(u,v)=y(s(u,v),t(u,v))$ and use the chain rule to show that $x_u$ and $x_v$ (evaluated at $(u_0,v_0)$) are linear combinations of $y_s$ and $y_t$ (evaluated at $(s_0,t_0)$). This shows $\text{Span}(x_u,x_v)\subset\text{Span}(y_s,y_t)$. Now what?