A kind of surfaces

Is there a way* to prove that a Bonnet Surface $S$ in isothermal coordinates in $R^3$ with mean curvature ($H$) and Gaussian curvature ($K$) both non-constant and where $(H^2-K)=c$, (with c positive constant), must have negative Gaussian curvature?

Thanks in advance for any help! Alex

• Are you omitting some hypotheses? Do you want the surface to be complete or compact? Also, do you really want it to be a Bonnet surface (which is more conditions than just $H^2-K$ being a positive constant)? Mar 6 '17 at 17:14
• Dead prof. Bryant, Thanks you for your answer. I know, but of course I may be wrong, that if a surface is without umbilical points and have $H^2-K$ that is positive, than is a Bonnet's surface. I'm doing it wrong? I have read, more time ago, projecteuclid.org/download/pdf_1/euclid.pjm/1102650843 and I found that if I want that S doesn't have umbilical points and $H^2-K=c$ with c positive constant, and H and K both not constant, than using (10), I have $-2*K*c=|gradH|^2$, than $K$ is negative. I want find another way to get to this result Mar 6 '17 at 18:20
• ...for "for this result", I mean what I found, that $k$ must be negative Mar 6 '17 at 18:40

It is not possible to prove that any surface $S\subset\mathbb{R}^3$ that has both $H$ and $K$ nonconstant but $H^2-K=c^2$ for some constant $c>0$ must have $K$ be negative. In fact, one cannot conclude anything about the sign of $K$ from only these hypotheses, beyond the fact that $K\ge -c^2$.

In fact, it's not difficult to show, using the Cartan-Kähler Theorem, that there exist plenty of such surfaces with $K$ positive. I can supply the details if you want.

Whether adding the condition that the surface be a Bonnet surface could force $K$ to be negative is another question. I think that is probably not the case, but I'd have to work it out to be sure.

Added after the question was changed: There are no Bonnet surfaces $S\subset\mathbb{R}^3$ with the property that $H$ is nonconstant while $H^2-K>0$ is constant (where 'Bonnet surface' means that there is a nontrivial isometric deformation of $S$ in $\mathbb{R}^3$ that preserves the mean curvature $H$). Thus, the OP's question is not sensible.

A proof that such surfaces do not exist can be got by simply differentiating the equation $H^2-K = r^2$ (where $r>0$ is a constant) several times and combining it with the three equations that follow from requiring that the surface admit a nontrvial $1$-parameter family of isometric embeddings into $\mathbb{R}^3$, all with the same nonconstant mean curvature function $H$.

A careful analysis shows that this overdetermined system is incompatible, even locally. The full analysis is too long to input here. However, you can find a discussion of this in Section 7.2 of my introductory lectures on exterior differential systems. If you follow those notes to the discussion on page 50, what this amounts to is seeking a solution to the ODE system on $\mathbb{R}^5$ near the bottom of that page that also satisfies $r_1=0$ while $H_1\not=0$ and $H$ and $r$ are nonzero. Since there are no such solutions, there are no such surfaces.

• Thank you very much for the unswer..can you supply the details?(It's very interesting). If I add that S doesn't have umbilical points? Mar 7 '17 at 1:58
• Because if I add that S doesn't have umbilical points and use the formula for "P" in (10) of the article which I referred above, I think it proves something like $-2*c*k=|gradH|^2$, where $c$ is $(H^2-k)$ Mar 7 '17 at 2:09
• @AlexanderPigazzini: The assumption $H^2-K>0$ already implies that $S$ has no umbilical points. This is because $H^2-K = (\kappa_1-\kappa_2)^2/4$, where $\kappa_i$ are the principal curvatures. The formula for $P$ in (10) of the article that you are citing doesn't help because $A$ and $B$ in the equation above it generally are not zero, so there's no reason to believe that $P$ vanishes. Mar 7 '17 at 9:33
• Thank you prof. Bryant, yes you are right if $H^2-K>0$ already implies that S has no umbilical points, I was confused..sorry! Mar 7 '17 at 10:13
• For the issue of the cited paper, I thought that by the Theorem 1 we have that: “Let $S$ be a piece of an oriented surface in $R^3$ such that it has no umbilic points. Then, $S$ admits a non-trivial isometric deformation preserving the mean curvature function (i.e. is a Bonnet surface) if and only if one of the following conditions holds….ect” The condition are $A=B=0$, or $P=\Delta \alpha=0$. Mar 7 '17 at 10:14