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Topologically speaking, the compact and connected surfaces are classified into three kinds of surfaces:

  • a sphere
  • a connected sum of tori
  • a connected sum of projective planes.

Also, we know that:

  • the sphere is an ovaloid
  • ovaloids are compact and connected and therefore can be put in one of these categories.

Is there an ovaloid not topologically equivalent to a sphere?

How could I prove that there exist or not an ovaloid that is not topologically equivalent to a sphere?

EvaMGG
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  • An ovaloid is given by the equation $x^2/a^2+y^2/b^2+z^2/c^2=1$. There is a simple change of coordinates that describes an explicit diffeomorphism of an ovaloid with the sphere – Exit path Oct 02 '18 at 16:56
  • @leibnewtz this is the equation of the ellipsoid, an ellipsoid is an ovaloid, but there ovaloids which aren't an ellipsoid. – EvaMGG Oct 02 '18 at 17:00
  • Thanks @SaucyO'Path, it is already edited. – EvaMGG Oct 02 '18 at 17:02
  • @EvaMGG I was just going off the first thing I found on google. What's your definition? – Exit path Oct 02 '18 at 17:02
  • @leibnewtz An ovaloid is a connected and compact surface whose Gauss curvature is always positive. – EvaMGG Oct 02 '18 at 17:03
  • How do you define "ovaloid", and what kind of topological equivalence are you talking about that only gives rise to three classes of surfaces? – Henrik supports the community Oct 02 '18 at 17:04
  • Well apparently Liebmann's theorem answers your question if the surface is closed. – Exit path Oct 02 '18 at 17:07
  • @Henrik I changed the question, maybe this is better. And the definition of ovaloid is a surface that is compact, connected and its Gauss curvature is always positive. – EvaMGG Oct 02 '18 at 17:18
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    The Gauss-Bonnet theorem forces your space to be either a sphere or projective plane: everything else has nonpositive Euler characteristic. –  Oct 03 '18 at 15:28
  • @MikeMiller so I guess I need figure out that ovaloids have positive Euler characteristic...is there a relation between the Euler characteristic and Gauss curvature? – EvaMGG Oct 04 '18 at 19:46
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    The Gauss-Bonnet theorem. –  Oct 04 '18 at 19:48
  • @MikeMiller You should give an official answer. See also https://math.stackexchange.com/q/808203. – Paul Frost Oct 05 '18 at 08:02
  • You can check corollary 5.3.5.1 http://www.mat.ucm.es/~jlafuent/Docencia/cys/cyslc.pdf stating that every ovaloid is homeomorphic to a sphere. Even more, there is a theorem given by hadamard stating that any ovalid is diffeomorphic to a sphere – user1868607 Oct 07 '18 at 16:53

1 Answers1

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You can take as a reference these notes (check corollary 5.3.5.1). The corollary states the following:

If $M$ is a differentiable surface of $\mathbb{R}^3$ connected and compact with Gauss curvature $K \ge 0$ and not identically zero then $M$ is homeomorphic to a sphere.

My understanding is that one requires differentiability to be able to use Gauss curvature.

Now, the ingredients to arrive to this corollary are the following:

  • a classification of topological surfaces which you cite and which is stated at theorem 5.3.5.1 of the document.

  • the Gauss-Bonnet theorem which has been mentionned in the comments and appears in the document at theorem 5.3.4.1.

  • Realize that homeomorphic implies homotopy equivalent.

In fact, the document cites a stronger result by Hadamard in theorem 5.3.6.2:

If $M$ is an ovalid then the Gauss map $\stackrel{\to}{N}: M \to \mathbb{S}^2$ associated with any unitary normal $N$ is a dipheomorphism. In particular, $M$ is dipheomorphic to a sphere.

So as you can see you need quite a bit of machinery to prove the result. However you get to a nice result for ovaloids.

user1868607
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