Let $M$ a regular surface and let $P\in M$ a nonplanar point in $M$. The Dupin indicatrix (DI) is defined to be the conic in $T_P M$ given by the equation $\Pi_ P (v)= 1$. The following problem is from Ted Shifrin's notes (Ch. 2.2 exercise 22c); it gives a geometric interpretation of DI. Here is the problem:
Suppose $P$ is the origin, and $M$ is, in a neighborhood of $P$, the graph of a smooth function $f: U \rightarrow \mathbb{R}$ such that $z= f(x,y)=\frac{1}{2} (k_1 x^2 +k_2y^2) + R(x,y)$, where $\lim_{x,y\rightarrow 0}\frac{R(x,y)}{x^2+y^2} =0$, i.e. the tangent plane $T_P M$ is the $xy$-plane, and the $x$ and $y$-axes are the principal directions at P. Show that for small positive values of $c$, the intersection of a sufficiently small neighborhood of $M$ with the plane $z = c$ “looks like” the Dupin indicatrix. How can you make this statement more precise?
It is not clear to me what kind of statement is satisfactory here. Here is my result for the case where DI is simply a circle:
Fix $\delta>0$. There exists $\eta>0$ such that the following hold. Inside $B((0,0),\eta)\subseteq U$, we have $f(x,y)=x^2+y^2+R(x,y)$, where $|R(x,y)|\le \delta (x^2+y^2)$. Let $0<\epsilon<\eta \sqrt{1-\delta}\ $, and consider $C_\epsilon:=\{(x,y)\in U: \ f(x,y)=\epsilon^2\}$. We denote the circles of radius $r$ by $D_r: = \{(x,y)\in \mathbb{R}^2: \ x^2+y^2=r^2\}$. Then,
- $C_\epsilon$ is between $D_{\frac{\epsilon}{\sqrt{1+\delta}}}$, $D_{\frac{\epsilon}{\sqrt{1-\delta}}}$
Intuitively, we need more than just this to argue that $C_\epsilon$ "looks like" $D_\epsilon$. For example, we would like to know that $C_\epsilon$ is connected. So, I also proved the following:
- $C_\epsilon$ is a regular curve (in particular, $\epsilon^2$ is a regular value of $f$).
- There exists $r: [0,2\pi) \rightarrow \mathbb{R}_+$, which is continuous, and such that the curve $\gamma:[0,2\pi)\rightarrow \mathbb{R}^2 $ defined as $\gamma(\theta)=(r(\theta)\cos\theta,r(\theta)\sin\theta )$ has range the $C_\epsilon$, i.e. $\gamma ([0,2\pi))=C_\epsilon.$