MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The following simple theorem is known as Cauchy's mean value theorem. Let $\gamma$ be an immersion of the segment $[0,1]$ into the plane such that $\gamma(0) \ne \gamma(1)$. Then there exists a point such that the tangent line at that point is parallel to the line passing through $\gamma(0)$ and $\gamma(1)$. So the boundary values of an immersion determine a direction such that for any immersion of a segment with given boundary values there exists a tangent line parallel to the direction.

I would like to propose the following conjecture, generalizing this statement. Instead of immersions of a segment we consider immersions of a compact manifold $M^n$ with non-empty boundary $\partial M$ into ${\mathbf R}^{n+1}$. For a map $f\colon \partial M \to {\mathbf R}^{n+1}$ we consider the space $L(f)$ of all immersions $g\colon M \to {\mathbf R}^{n+1}$ such that $g|_{\partial M}=f$.

The conjectural claim is the following: If $f$ is sufficiently generic, then for every connected component $L_0$ of the space $L(f)$ there exists a hyperplane direction $l$ such that for any immersion $g$ from $L_0$ $l$ is parallel to the tangent plane to $g(M)$ at some point.

If the conjecture is true then it is very interesting how $l$ depends on $g$.

share|cite|improve this question
I suspect you need $M$ to be compact. Otherwise, $M=[0,1)$ looks like a counterexample to me. What reasons do you have for the conjecture to be true? The next simplest case after the intervals would be the disk, I think. Have you tried proving your conjecture for that case? (I am not a differential geometer, so I can't contribute to a solution. But I am curious all the same.) – Harald Hanche-Olsen Feb 25 '10 at 2:19
You are right, thank you. M is supposed to be compact, I forget to mention it. I've tried to solve it for a disk, yes, but unsucsessfully. – Petya Feb 25 '10 at 2:25
Is it stated as an "open-problem" somewhere? – Anton Petrunin Feb 25 '10 at 4:13
@Leonid. It does not really work this way. Look at the ruled surfaces... – Anton Petrunin Feb 25 '10 at 4:17
After deleting my answer: I still think there should be a counterexample --- it is kind of h-principle. But I will better check it in the morning. – Anton Petrunin Feb 25 '10 at 5:33
up vote 4 down vote accepted

The answer is no for $n=2$. It is sufficient to construct 3 surfaces with common boundary (say $\Sigma_i$, $i\in\{1,2,3\}$) such that there is no choice of points $p_i\in\Sigma_i$ with pairwise parallel tangent planes.

Let us take a smooth function $f:S^1\to \mathbb R$, $f(t)\approx\sin(2\cdot t)$ with one little bump near zero so it has 3 local minima and maxima. We want to construct three functions $h_1,h_2,h_3$ from unit disc $D$ to $\mathbb R$ such that each has $f$ as boundary values and

  1. if $\nabla h_1(x)=\nabla h_2(y)$ then $\nabla h_1(x)=0$

  2. $\nabla h_3\not=0$ anywhere in the disc.

Then graphs of functions give the needed surfaces. The graphs of $h_1$ and $h_2$ are parts of boundary of convex hull of graph of $f:\partial D\to\mathbb R$; it is easy to check (1).

The graph of $h_3$ is a ruled surface which formed by lines passing through points $(u,f(u))$, $(\phi(u),f(\phi(u))\in\mathbb R^3$, $u\in S^1$ for some involution diffeomorphism $\phi: S^1\to S^1$. To have the property one has to choose $\phi$ with two fixed points (say at global minima of $f$) so that if $f(\phi(x))=f(x)$ for some $x$ then $f'(\phi(x)\cdot f'(x)<0$. The later is easy to arrange, that is the place we need the bump of $f$.

P.S. Hopefully it is correct now :)

share|cite|improve this answer
You can delete it. Your Idea of counterexample was very good! Even if the conjecture is false, the real question is to find a positive statement generalizing Cauchy's m.v. theorem. – Petya Feb 25 '10 at 5:28
I do not understand your construction. The set of normal unit vectors is not, in general, a curve for a ruled surface. – Petya Feb 25 '10 at 12:22
I want to remark the following: Consider two surfaces in 3-space and suppose they do not have parallel tangent planes. Let its intersection contains a compact component. If one surface is a graph of a function on the plane then this component is non homologous to zero in the second surface. I conclude from that observation that both surfaces from a counterexample should be very curved. – Petya Feb 25 '10 at 15:23
"The set of normal unit vectors is not, in general, a curve for a ruled surface." you are right, I will think a bit to correct the answer --- it does not change much. – Anton Petrunin Feb 25 '10 at 15:53
Now it should be correct --- sorry for all this mess. It should be really simple for those who do h-principle... – Anton Petrunin Feb 25 '10 at 19:30

I will consider another counterexample, based on Anton's ideas. I think it looks simpler.

We will construct a closed domain $D$ in a plane and three functions on it with same restrictions to the boundary of the domain such that there is no choice of points on graphs of functions with pairwise parallel tangent planes.

First observation is the following: Let function $f$ on the coordinate plane depends on the coordinate $x$ only and function $g$ depends on the coordinate $y$ only. Then $df(a)=dg(b)$ if and only if $df(a)=dg(b)=0$. (It seems to me that the crucial Anton's idea is to consider functions with very degenerate gradient map, having value on a curve).

Now we take the functions $f(x,y)=-x^2$ and $g(x,y)=P_4(y)$ where $P_4$ is a polynomial of degree $4$ with two zeroes, three pairwise different critical values and leading term $y^4$.

We set the domain $D$ to be a set of solutions to an inequality $f \ge h$. It is a closed subset in a plane diffeomorphic to a unit disk. We define functions now: the first function is $f$, the second function is $g$. Third function is a function without critical points on $D$ coinciding with a restriction of $f$ (or $g$) to the boundary. Such a function exists! (it is an exercise from Morse theory. I mention here that we need a polynomial of degree 4 (degree 2 is unsufficient) to satisfy that extension without critical points property).

share|cite|improve this answer
Yes it is better than mine. I would add that the function $f$ can bee constructed exactly the same way as $h_3$ in my answer. – Anton Petrunin Feb 26 '10 at 16:27
It is not $f$, it is unnamed third function! I see you really like ruled surfaces. – Petya Feb 26 '10 at 22:34
Well, I do not know an other way to see it... – Anton Petrunin Feb 26 '10 at 22:55

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.