# On Kühnel--Dillen's determinant condition for Weingarten surfaces

Consider the following definition:

A surface is Weingarten if there's a nontrivial functional relation $$\Psi(H,K)=0$$ between its mean and gaussian curvatures $$H$$ and $$K$$, respectively.

Many works out there (see e.g. the introduction of this Dillen's student thesis) say this definition is equivalent to the condition $$\frac{\partial (H,K)}{\partial (s,t)} = 0,$$ where $$\frac{\partial (H,K)}{\partial (s,t)}$$ is the Jacobian matrix of the map $$(H,K):U\to \mathbb{R}^2$$ on some (any) coordinate system for the surface, i.e. $$\frac{\partial (H,K)}{\partial (s,t)} = \det \begin{bmatrix} H_s & H_t\\ K_s & K_t \end{bmatrix}.$$ where the subindexes indicate partial derivatives.

However, I don't find a proof anywhere and can't prove it myself either. Here's what I've tried.

Ok, first of all, I believe we need to say precisely what we mean by "nontrivial functional relation". The most reasonable definition that I could think of is:

$$\Psi$$ is a nontrivial relation if $$\nabla \Psi\neq 0$$, up to a meager (empty interior) set.

Here, $$\nabla \Psi$$ indicates the gradient.

I thought of this because we can have two sceneries otherwise and both don't bring us any information on the surface.

Suppose $$\nabla \Psi \equiv 0$$ on an open $$U\subset \mathbb{R}^2$$. Then, the two sceneries are: the pair $$(H,K)$$ takes or doesn't take a value into $$U$$. If it doesn't, then this region $$U$$ is just superfluous for us, since our interest is in applying $$\Psi$$ to the pair $$(H,K)$$. If it does, then since $$\Psi$$ must be constant on $$U$$ the condition $$\Psi(H,K)\equiv 0$$ implies $$\Psi\equiv 0$$ on $$U$$. But then, this doesn't bring us any information on the curvatures.

Therefore, we "can't" allow the gradient to be zero on an open set.

Allowing it to vanish on an empty interior set is nice though, because this includes for example all the class of polynomial relations.

That being established, let's prove one side of the equivalence:

If a surface is Weingarten, then its mean and gaussian curvatures satisfy $$\det \begin{bmatrix} H_s & H_t\\ K_s & K_t \end{bmatrix} =0$$ for some (any) parametrization on the surface.

Proof. Say the surface is Weingarten with nontrivial relation $$\Psi$$. Then taking any coordinate system on the surface and taking the differential of $$\Psi(H,K) \equiv 0$$, by the chain rule we have $$D_{(s,t)}(\Psi\circ (H,K)) = \nabla \Psi (H(s,t),K(s,t)) \cdot \begin{bmatrix} H_s & H_t\\ K_s & K_t \end{bmatrix}_{(s,t)}.$$

Now if $$\det \begin{bmatrix} H_s & H_t\\ K_s & K_t \end{bmatrix}\neq 0$$ for a single point, the inverse function theorem gives us that $$(H,K)$$ is a diffeomorphism on an open neighborhood and this tells us that $$\nabla \Psi$$ is zero on an open set, contradicting our definition. Therefore, it must be zero everywhere. $$\square$$

Now, if we want to prove the converse, we must assume that determinant is identically zero and then find a nontrivial relation $$\Psi$$ such that $$\Psi(H,K)=0$$. That's quiet more complicated, right?

Well, intuitively the determinant condition says that the pair $$(H(s,t),K(s,t))\subset \mathbb{R}^2$$ is contained on a "curve" (one-dimensional set), even though it depends on the two parameters of the surface. It would be enough the to exhibit a nontrivial relation $$\Psi$$ such that the pair $$(H,K)$$ is contained in its zero-leveled curve.

Then this brings the following question:

Is every curve a level curve of some nontrivial relation?

This seems to be quite unlikely, since an arbitrary curve may be very complicated, full of loops and self intersections.

However, our curve is not any curve. It is given by the mean-gaussian pair $$(H,K)$$ of a surface. How complicated can it be?

We know for example $$H=\frac{1}{2}(k_1+k_2)$$ and $$K=k_1k_2$$ in terms of principal curvatures $$k_1,k_2$$, but what else?

I really have no idea of how to prove the existence or to build $$\Psi$$ in general. A counterxample seems equally hard to find though...

• This question has nothing to do with curvatures. It is the following fact: Given functions $x(s,t), y(s,t), f(x,y)$ such that $$f(x(s,t),y(s,t)) = 0.$$ Then the Jacobian $$\begin{bmatrix} \partial_sx & \partial_tx \\ \partial_sy & \partial_ty \end{bmatrix}$$ is singular. You should prove this first. The statement about $H$ and $K$ for a Weingarten surface is an immediate consequence. Commented May 26 at 16:37
• @Deane I see. But what about the converse? If the Jacobian is singular, is it always possible to find an $f$? Because this is what I see everyone saying, but can't find a proof... Commented May 26 at 16:45
• If the matrix has rank $1$, then the implicit function theorem says that such a function exists. If the matrix vanishes at a point, then perhaps not. Commented May 26 at 16:52
• @Deane Thanks for the helpful comments. However, it seems that even in the case that the rank is constant $1$, we obtain such a function but only locally and we need a global Weingarten relation for our surface. Then I thought about summing up things using a partition of unity argument, but it didn't work very well either... As I commented in the question, I really don't see how to find such a global nontrivial relation. At the same time though, a counterexample to this would be a surface with very weird curvatures behaviour. Commented May 27 at 0:35

ADDED: As @Derso points out, it is not necessarily true that $$\Sigma$$, as defined below, is a smooth curve. If, however, $$\Sigma$$ is assumed to be a smooth curve, then the rest of the proof works.
Let $$S \subset \mathbb{R}^3$$ be a nonempty connected smooth locally Weingarten surface. Let $$\Sigma = \{ (H(p), K(p)):\ p \in S\}.$$ Then $$\Sigma$$ is a smooth connected curve in $$\mathbb{R}^2$$. Fix an orientation on $$\Sigma$$. For each $$(H,K) \in \Sigma$$, there exists an neighbhorhood $$N_{(H,K)}$$ of $$(H,K)$$ in $$\mathbb{R}^2$$ and a smooth function $$\Phi_{(H,K)}: N_{(H,K)} \rightarrow \mathbb{R}$$ such that $$\Phi_{(H,K)}^{-1}(0) = \Sigma\cap N,$$ $$\nabla\Phi_{(H,K)} \ne 0$$ on $$N_{H,K}$$, and $$\nabla\Phi_{(H,K)}(p)$$ is positively oriented at $$(H,K)$$. Let $$O = \bigcup_{(H,K) \in \Sigma} N_{(H,K)}.$$
At this point, I think you can use a partition of unity to define an open $$O'\subset O$$ and a function $$\Phi: O' \rightarrow \mathbb{R}$$ such that $$\Phi^{-1}(0) = \Sigma$$ and $$\nabla\Phi\ne 0$$ on $$\Sigma$$.
• First, thank you again for the help! Yes, this is pretty much what I had in mind. However, I believe the problem is in claiming $\Sigma$ is a smooth connected curve (curve, right? You said surface in your answer). I mean, a priori $\Sigma$ may have loops and self intersections. Then when you try to use the partition of unity, crossing self intersections of $\Sigma$ will mess up the points where we want $\Phi$ to be zero... A drawing makes that clearer. Commented May 27 at 17:11