# Computing fundamental forms of implicit surface [duplicate]

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.

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.

• Thanks. Maybe I should make my question clearer ;) Oct 15, 2016 at 11:41
• I mostly just don't understand the following part of the question: 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." You can certainly deal with the exterior derivative of F by treating it as a geometric object on its own merits, but if you want to obtain expressions for the first and second fundamental form of the indicated type, then it would appear to me that you have to use coordinate expressions.
– THW
Oct 17, 2016 at 14:06