4
$\begingroup$

I guess it is not. The example I have in mind is: $X^2$ is the spherical suspension of a circle $S^1(t)$ of length $0<t<2\pi$. Then $X$ has constant curvature =1 except at two suspension points, say $N$ and $S$.

But I cannot convince myself, since it seems this manifold can be approximated by a sequence of smooth Riemannian metrics $g_i$ with positive curvature by smoothing arbitrary small neighbourhood of $N$ and $S$, then by Pogorelov, if we assume sufficient smoothness on $g_i$, we get desired embedding by taking limit.

Improve this question
$\endgroup$
11
  • 5
    $\begingroup$ By Alexandrov realization theorem, proved in 1948, every nonnegatively curved $2$-sphere is isometric to the boundary of a convex body. Here "nonnegatively curved" is understood in the comparison sense. The proof strategy is to approximate a given metric by a nonnegatively curved polyhedron, embed the polyhedron, and then pass to a limit of such polyhedra. I suppose one can also smooth the metric, use the PDE method to find an isometric embedding and then pass to a limit but this is not what Alexandrov did because the relevant PDE methos came in later. $\endgroup$
    – Igor Belegradek
    Jan 2, 2016 at 15:08
  • $\begingroup$ @Igor, Do you know any explicit embedding of the suspension $X$ in my post? $\endgroup$
    – user17150
    Jan 2, 2016 at 15:14
  • $\begingroup$ The suspension has an involution fixing the base and also has rotational symmetry. The embedding is unique up to a rigid motion. So the embedding must be given by a surface of revolution obtained by rotating a curve in the $xz$-plane about the $z$-axis, and then reflecting in the plane $z=0$. Find the curve yourself. $\endgroup$
    – Igor Belegradek
    Jan 2, 2016 at 15:25
  • 1
    $\begingroup$ Thanks Igor, I found the expression. Just for reference, it is parametrized by $a(t)=r*\cos(t), b(t)=\int_0^t \sqrt{1-r^2\sin^2{s}}ds$, where the circle has length $2\pi r$ which is the $t$-parameter in O.P.. $\endgroup$
    – user17150
    Jan 2, 2016 at 15:56
  • $\begingroup$ @IgorBelegradek re "relevant PDE methods", I strongly suggest reading Busemann's math review of Nirenberg's paper. $\endgroup$
    – Igor Rivin
    Jan 2, 2016 at 23:26

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.