In the section about the orientation of surfaces it is stated that:
By fixing a parametrization $x(u,v)$ of a neighborhood of a point p of a regular surface S, we determine an orientation of the tangent plane $T_pS$, namely, the orientation of the associated ordered basis ${x_u,x_v}$. If p belongs to the coordinate neighborhood of another parametrization $x'(u',v')$, the new basis ${x'_{u'},x'_{v'}}$ is expressed in terms of the first one by
$x'_{u'}=x_u \frac{\partial u}{\partial u'}+x_v \frac{\partial v}{\partial u'}$
$x'_{v'}=x_u \frac{\partial u}{\partial v'}+x_v \frac{\partial v}{\partial v'}$
where $u=u(u',v')$ and $v=v(u',v')$ are the expressions of the change of coordinates. The basis ${x_u,x_v}$ and ${x'_{u'},x'_{v'}} $determine, therefor, the same orientation of $T_p(S)$ if and only if the Jacobian $\frac{\partial(u,v)}{\partial (u',v')}$ of the coordinate change is positive
To get to my question, given another parametrization like above how do I calculate the unit normal vector $N(p):=\frac {x_u \times x_v}{|x_u \times x_v|}$ under the new parametrization.
(I am trying to understand why the Jacobian of the coordinate change needs to be positive)
Best Answer
I interpret this as a purely linear-algebraic issue. If you have two bases of $T_pM$, namely $\{x_u,x_v\}$ and $\{x'_{u'},x'_{v'}\}$ and a relation between the bases given by \begin{equation} \begin{split} x'_{u'} = \frac{\partial u}{\partial u'} x_u + \frac{\partial v}{\partial u'} x_v\\ x'_{v'} = \frac{\partial u}{\partial v'} x_u + \frac{\partial u}{\partial u'} x_v\\ \end{split} \end{equation} then they determine the same orientation if and only if the matrix $$M=\left( \begin{matrix} \frac{\partial u}{\partial u'} & \frac{\partial v}{\partial u'}\\ \frac{\partial u}{\partial v'} & \frac{\partial v}{\partial v'} \end{matrix}\right) $$ has positive determinant. Note that taking the vector product $x'_{u'}\times x'_{v'}$ in terms of ${x_u,x_v}$ yields $$ x_{u'}\times x_{v'} = x_u \times x_v \frac{\partial u}{\partial u'} \frac{\partial v}{\partial v'} + x_v\times x_u\frac{\partial v}{\partial u'}\frac{\partial u}{\partial v'} = \det(M) x_u\times x_v $$ and therefore you can easily check the following: $$ N' = \frac{x'_{u'} \times x'_{v'}}{|x'_{u'} \times x'_{v'}|} = \frac{\det(M)}{|\det(M)|} \frac{x_{u} \times x_{v}}{|x_{u} \times x_{v}|} = sign(\det(M)) N$$ and therefore the orientation as seen from the unit normal agrees if and only if the determinant of the jacobian is positive.