Isometric imbedding of a sphere with positively curved metric QUESTION. Given a Riemannian metric on the sphere $S^n$ with positive sectional survature. Can it be isometrically imbedded into $\mathbb{R}^{n+1}$ (of any class of regularity) as a boundary of a convex set?
This question was motivated by the following three well known facts on isometric imbeddings (please correct me if references are not precise).
1) Given a smooth Riemannian metric with strictly positive Gauss curvature on the 2-sphere $S^2$, it can be isometrically imbedded into $\mathbb{R}^3$ such that its image bounds a convex set (Nirenberg, Pogorelov).
2) On $S^n$ with $n>2$ there exist smooth Riemannian metrics (even with positive sectional curvature) which cannot be smoothly (!) imbedded into $\mathbb{R}^{n+1}$. (Common knowledge.)
3) Any Riemannian metric on $S^n$ can be $C^1$-regularly isometrically imbedded into $\mathbb{R}^{n+1}$. (Special case of Nash's theorem.)
 A: In the smooth category, the problem has a completely different character for $n>2$ than when $n=2$.  
The first obstruction is whether there exists a positive definite quadratic form~$h$ on $S^n$ that satisfies the Gauss equation $ h\circledast h = \mathrm{Riem}(g)$ (where $\circledast$ denotes the so-called Kulkarni product $\circledast:S^2(S^2T^*)\to S^2(\Lambda^2T^*)$).  When $n=3$, such an $h$ always exists (and is unique) when $\mathrm{Riem}(g)$ has positive sectional curvature.  When $n>3$, this places algebraic conditions on $\mathrm{Riem}(g)$; when $\mathrm{Riem}(g)$ has positive sectional curvature and a positive definite $h$ exists satisfying the Gauss equation, it is unique.
The second condition is that the $h$ that is found satisfying the Gauss equation must also satisfy the Codazzi equation $\delta_g h = 0$.  
These conditions are sufficient.
A: Igor is essentially right. If $n > 2$ and the embedding is convex, then the second fundamental form exists almost everywhere. Using the Gauss equations per Robert's answer, it extends uniquely and smoothly to everywhere. The rest is straightforward. I'm omitting it because I'm typing this on an iPhone.
The $C^1$ but non-convex case is due to Nash (who did it in codimension 2) and Kuiper (who did it in codimension 1). It is not a special case of smooth Nash embedding theorem.
ADDED: The argument can be made rigorous by taking a family of smooth convex hypersurfaces converging uniformly to the original convex hypersurface and using the inverse to the map from positive definite second fundamental forms to curvature tenors defined by the Gauss equations to show that the second fundamental form has to converge uniformly to a continuos limit that also weakly solves the Codazzi equations. This can then be integrated to show that the embedding is in fact smooth. Uniqueness then follows by the Cohn-Vossen theorem.
