Surface all of whose normals intersect at a point

Question

I am new to differential geometry and encountered difficulty when trying to solve the following problem from Dubrovin's Modern Geometry

It's the first problem in exercise 8.4:

Find the surface all of whose normals intersect at a point.

Intuitively I believe that the surface should be a sphere, but I am not sure how to show this formally. Can some one please help me out?

Thanks a lot.

Answer 1

We begin by considering the same problem in the plane: Let $$\gamma:\quad s\mapsto\bigl(x(s),y(s)\bigr)\qquad(-h<s<h)$$ be an arc whose normals all pass through the same point $(0,0)$. We may assume $$x(0)=\rho>0, \quad y(0)=0,\quad \dot x(0)=0,\quad \dot y(0)=1\ .$$ The normal $\nu_s$ corresponding to the point $\gamma(s)$ has the parametric representation $$\nu_s:\quad t\mapsto\bigl(x(s)-t\dot y(s), \>y(s)+t\dot x(s)\bigr)\qquad(-\infty<t<\infty)\ .$$ Looking at the $x$-coordinate of $\nu_s$ we see that this normal passes through $(0,0)$ at the $t$-value $t_s={x(s)\over\dot y(s)}$. This implies that $$y(s)+{x(s)\over\dot y(s)}\>\dot x(s)=0\ ,$$ or $$y(s)\dot y(s)+x(s)\dot x(s)=0\ .$$ Since this has to hold for all $s$ we conclude that $y^2(s)+x^2(s)$ is a constant ($=\rho^2$). It follows that the arc $\gamma$ is in fact a circular arc of radius $\rho$.

Consider now an arbitrary point $p$ of your surface $S$, and let $n$ be the surface normal through $p$. By the foregoing, any plane $\pi$ through $n$ will intersect $S$ in a circular arc, and all these arcs will have the same center $M\in n$ which is the common point of all normals. It follows that $S$ is spherical in the neighborhood of $p$, and if $S$ is connected it is (part of) a sphere.

Answer 2

I've tried to post this as a comment, but it was too long and I couldn't. So, I apologize if someone consider it not an answer.

1) The solution by Christian Blatter rises some interesting aspects. In fact it includes a version for planar curves of my previous answer for planar curves. I just used $\|x\|$ instead of his $\|x\|^2$. My argument shows that a connected hypersurface in $\mathbb R^n$ all of whose normals meet at a point must be open in an $(n-1)$-sphere. For planar curves there's the difficulty that points most surely disconnect. But then the argument applies to see that both sides around that disconnecting point are circular arcs of equal center: for them to meet continuously at the point their radious must coincide and we're done. Also one could be cleverer from the very beginning to use $\|x\|^2$ and no need of dropping any point.

This is all more topological than geometrical: it relies on the property (implied by the normals hypothesis) that spheres with a fixed center are tangent to the surface under consideration.

2) The idea of normal sections involves more geometry. There's only a difficulty at the end: to be sure the circular arcs around $p$ form a neighborhood one must bound from below their length, because otherwise they could very well collapse to not give any open neighborhood (radial-open topology is strictly finer than the usual topology). But this can be solved looking at the geometry. What Christian Blatter shows is that all normal curvatures at the point coincide, in particular the two principal curvatures. Thus all points are umbilic. A classical result is that this only happens for open sets of planes and spheres. But planes doesn't verify the normals condition. A final remark is that the proof of this classical fact is somehow a generalization of the argument in 1).

Answer 3

If all normals of a connected surface $S\subset\mathbb R^3$ meet at a point, the surface is an open subset of a sphere. Indeed, suppose the normals meet at the origin $0$ (by translation). If $0\in S$, we can replace $S$ by $S\setminus\{0\}$ (note that a point never disconnects a surface) to suppose $0\notin S$. Consider the smooth map $F:\mathbb R^3\setminus\{0\}\to\mathbb R:x\mapsto \|x\|$ (drop the origin precisely for differentiability). Fix any $a\in S$, and denote $S_r$ the sphere of radius $r=\|a\|$ and center the origin, which contains the point $a$. Since $F|S_r$ is constant $\equiv r$, its derivative at $a$ vanishes: $d_a(F|S_r)=d_aF|T_aS_r\equiv0$. But the normal to $S$ at $a$ goes through the origin, hence $T_aS=L[a]^\perp=T_aS_r$, so that $d_a(F|S)=d_aF|T_aS=d_aF|T_aS_r\equiv0$. Consequently, the function $F|S$ has all derivatives equal zero, hence, $S$ being connected, $F|S$ is constant, say $\equiv r$. We have thus seen that $S\subset S_r$. Finally, a surface inside another must be open.

Answer 4

If all surface normals do not intersect at a point, the curve has to be twisted at that point, considering the definition of torsion.The magnitude of this deviation can be measured by the scalar curvature, geodesic torsion:

$$ \tau_g = (\kappa_1 -\kappa_2 ) \sin\psi \cos \psi $$

in any direction i.e., for all $ \psi$. So when this vanishes the necessary and sufficient condition is $ \kappa_1 =\kappa_2, $ true only for a sphere.

Answer 5

Consider the field of planes, at each point $Q$ the plane is the one perpendicular to $PQ$. This field has integral surfaces the spheres. It is known that if a field is integrable the maximal integral surfaces are unique.

If we consider the analogous question in the plane: the uniqueness of the integral curve is equivalent to the uniqueness of solutions of differential equations $\frac{dy}{dx}=F(x,y)=$ slope of the line through $(x,y)$.

For the space, finding an integral surface is equivalent to solving a system $\frac{\partial \phi}{\partial x} = F_1(x,y, \phi(x,y))$, $\frac{\partial \phi}{\partial y} = F_2(x,y, \phi(x,y))$ where $F_1(x,y,z)$ and $F_2(x,y,z)$ are the slopes of the plane through the point $(x,y,z)$. Again, the system will have no more than one solution starting with a given value at $(x_0, y_0)$; sometime none ( need extra conditions of integrability). However, in our case, we know we have solutions, the spheres with centers at $P$.