Computing fundamental forms of implicit surface [duplicate]

Question

For an implicit surface $F(x,y,z)=0$ we can take $z=f(x,y)$ and use IFT to calculate its fundamental forms. But they are expressed by $dx^2,dxdy,dy^2$.

I wonder if there are other expressions using $dx^2,dy^2,dz^2,dxdy,dydz,dzdx$, just more symmetric.

Also I want to know how to differentiate entirely without taking coordinate parameters $x,y,z$ to get a direct expression by $dF$ with something.

Any help will be appreciated.

Answer 1

I do not know the most efficient answer to the question that you are asking (in fact, the question is a little unclear), but the following addresses the question in the title.

Away from singular points, the gradient of $F$, $\nabla F = F_{x}\frac{\partial}{\partial x} + F_{y} \frac{\partial}{\partial y} + F_{z}\frac{\partial}{\partial z}$, is perpendicular to the surface. Further, at a point $P(a, b, c)$ on the surface $F(x, y, z) =0$, the equation of the tangent plane to the surface is given by $$ F_{x}(P)(x - a) + F_{y}(P)(y - b) + F_{z}(P)(z - c)= 0. $$

It is thus relatively easy to come up with an orthogonal basis for the tangent space to $\mathbb{R}^{3}$ at such points on the surface, but since you only know that one of the components of $\nabla F$ is non-zero, you need to work in cases. For simplicity, work on an open set $U$ where the tangent plane is not horizontal so that you can assume that one of $F_{x}$ or $F_{y}$ is not-zero.

Then you can take the vector fields \begin{align*} \vec{e}_{1} &= -F_{y}\frac{\partial}{\partial x} + F_{x}\frac{\partial} {\partial y}\\ \vec{e}_{2} &= -F_{x}F_{z}\frac{\partial}{\partial x} - F_{y}F_{z}\frac{\partial}{\partial y} + (F_{x}^2 + F_{y}^2)\frac{\partial}{\partial z}\\ \vec{e}_{3} &= F_{x}\frac{\partial}{\partial x} + F_{y} \frac{\partial}{\partial y} + F_{z}\frac{\partial}{\partial z} \end{align*} to be a frame for the tangent space to $\mathbb{R}^{3}$ along the surface, with $\vec{e}_{1}$ and $\vec{e}_{2}$ tangent to the surface and $\vec{e}_{3}$ perpendicular to the surface.

The one-forms dual to the frame above are thus \begin{align*} \omega^{1} &= -\frac{F_{y}}{A} dx + \frac{F_{x}}{A}dy\\ \omega^{2} &= -\frac{F_{x}F_{z}}{AB} dx - \frac{F_{y}F_{z}}{AB}dy + \frac{1}{B}dz\\ \omega^{3} &= \frac{F_{x}}{B}dx + \frac{F_{y}}{B} dy +\frac{F_{z}}{B} dz,\\ \end{align*} where $A = F_{x}^2 + F_{y}^2$ and $B = F_{x}^2 + F_{y}^2 + F_{z}^2$, and the standard inner product on $\mathbb{R}^3$ is then expressed as $$ \mathbf{g} = A \omega^{1}\otimes \omega^{1} + AB \omega^{2}\otimes \omega^{2} + B \omega^{3}\otimes \omega^{3}. $$

Restricted to the surface $S$ in question, one has $$ \mathbf{g}\vert_{S} = A \omega^{1}\otimes \omega^{1} + AB \omega^{2}\otimes \omega^{2}. $$ Substituting in the expressions for $\omega^{1}$ and $\omega^{2}$ one is then able to obtain an expression involving $dx^2, dxdz,$ etc.

Finally, normalizing the gradient vector field $\nabla F = F_{x}\frac{\partial}{\partial x} + F_{y} \frac{\partial}{\partial y} + F_{z}\frac{\partial}{\partial z}$ one obtains the unit normal vector field along the surface expressed by $$ \vec{n} = \frac{\nabla F}{\sqrt{B}} = \frac{F_{x}}{\sqrt{B}}\frac{\partial}{\partial x} + \frac{F_{y}}{\sqrt{B}} \frac{\partial}{\partial y} + \frac{F_{z}}{\sqrt{B}}\frac{\partial}{\partial z}. $$

One should now be able to also express the second fundamental form $II$ relative to the chosen basis by computing $$ II(\vec{e}_{1}, \vec{e}_{1}),\quad II(\vec{e}_{1}, \vec{e}_{2}),\quad \textrm{ and },\quad II(\vec{e}_{2}, \vec{e}_{2}), $$ and obtaining $$ II = II(\vec{e}_{1}, \vec{e}_{1}) \omega^{1} \otimes \omega^{1} + II(\vec{e}_{1}, \vec{e}_{2})\omega^{1} \otimes \omega^{2} + II(\vec{e}_{2}, \vec{e}_{1}) \omega^{2}\otimes \omega^{1} + II(\vec{e}_{2}, \vec{e}_{2})\omega^{2}\otimes \omega^{2}. $$ Substituting in the expressions for $\omega^{1}$ and $\omega^{2}$ one is then able to obtain an expression involving $dx^2, dxdz,$ etc.

Again, it is more than likely that there are more efficient ways to do this.