14
$\begingroup$

The following theorem is commonly attributed to Jacques Hadamard.

Assume $\Sigma$ is a smooth locally convex immersed surface in the Euclidean space. Then $\Sigma$ is embedded and bounds a convex set.

Many authors refer to Hadamard's Sur certaines propriétés des trajectoires en Dynamique (1897) (for example, James Stoker in his Über die Gestalt der positiv... (1936)).

Likely the statement is there, but the paper is long, it is in French and often the statements are not clearly marked; I was searching for it for several days. I asked a friend and she said that it was there 20 years ago, but she could not find it; she also said that it was not easy to extract it from what is written ( = one has to think). [For sure the word immersion is not there.]

I hope someone here knows this paper and can help me.

P.S. Now I see it this way: Stoker was the first who had formulated and proved the theorem; at the beginning of his paper he attributed the theorem to Hadamard because it almost follow from item 23 in his paper. After Stoker everyone did the same.

$\endgroup$
1
  • 2
    $\begingroup$ To remain in the spirit of this site, a present-day referee would probably tell Hadamard : "unclear what you're claiming" ! $\endgroup$ – Sylvain JULIEN Nov 22 '18 at 21:28
11
$\begingroup$

I think the relevant location is item 23, page 352, but what Hadamard aims to is stated as follows:

A smooth, co-orientable surface of $\mathbb{R}^3$ with Gauss curvature bounded below by some $\kappa >0$ is simply connected. (implicitly, the surface is compact without boundary)

("Or une surface à deux côtés et sans points singuliers, à courbure partout positive (la valeur zéro et les valeurs infiniment petites étant exclues) est toujours simplement connexe.")

The goal is to use the Gauss-Bonnet Formula to deduce that when curvature is positive, any two closed geodesics must meet (otherwise they would together bound a total curvature 0 region of the surface).

What is not clear from the text of item 23 is whether the surface assumed to be immersed or embedded. He basically says that the normal map is a global diffeomorphism, because positive curvature makes it a covering of the sphere.

It seems the argument does provide the statement attributed to this paper, although it seems not explicitly stated. Second edit: Mohammad Ghomi gives an argument to that effect in comment.

$\endgroup$
8
  • $\begingroup$ The arguments in 23 do not seem to show that immersed sphere is embedded, even informally; am I wrong? [I see also pictures on page 379 which are relevant to a proof I know, but the words around these pictures seem to be irrelevant.] $\endgroup$ – Anton Petrunin Nov 23 '18 at 4:55
  • $\begingroup$ @AntonPetrunin you are probably right but I do not have much time checking in details. I would not be surprised, given the informality of the discussion, if the attribution of this statement would be somewhat of a stretch. In any case, I do not think it was the point Hadamard wanted to make (and he might assume the embedding in the first place). $\endgroup$ – Benoît Kloeckner Nov 23 '18 at 13:58
  • 5
    $\begingroup$ Injectivity of the gauss map implies embeddedness of the surface via convexity. Namely if the gauss map is injective, then it is easy to see that the surface must lie on one side of its tangent planes. If not, then the height function with respect to some tangent plane must have at least 3 critical points, and so at two of these points the normals will be parallel (this is now a well-known argument, and probably was not hard for Hadamard to figure out either). Once the surface is convex, then it must be embedded. $\endgroup$ – Mohammad Ghomi Nov 24 '18 at 15:37
  • $\begingroup$ @MohammadGhomi that is right, but I do not see this argument in the paper. By the way it seems that you apply basic Morse theory which was developed much later. $\endgroup$ – Anton Petrunin Nov 25 '18 at 20:04
  • 2
    $\begingroup$ @AntonPetrunin: my conclusion is that the setting of the paper is not precise enough to contain the statement attributed to it. There seems to be no consideration of embedded versus immersed surfaces. $\endgroup$ – Benoît Kloeckner Nov 26 '18 at 19:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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