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Let $(S^n, g)$ be an $n$-dimensional positively curved sphere. Assume the smoothness of the metric, does it admits an isometric embedding into $\mathbb R^{n+1}$?

for $n=2$ it is proved by A.D Alexandrov, also by H. Weyl

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    $\begingroup$ Just a comment, a $C^1$ isometric embedding will always exist by Nash theorem (since the standard embedding of the sphere will be short after a suitable shrinking). $\endgroup$ – Thomas Richard Oct 4 '13 at 12:42
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When $n\geq 3$, the condition for a metric to be even locally isometrically embeddable in $R^{n+1}$ is nontrivial, so you may wish to add such an additional condition. Then this paper "A Priori Bounds for Co-dimension One Isometric Embeddings" by Yanyan Li and Gilbert Weinstein states as a conjecture:

Conjecture. Let $g$ be a smooth metric of non-negative sectional curvature and positive scalar curvature on $S^n$ which is locally isometrically embeddable in $R^{n+1}$. Then $(S^n,g)$ admits a smooth global isometric embedding $X:(S^n,g)\rightarrow R^{n+1}$.

Section 5 of the paper discusses the positive sectional curvature case, and in particular Theorem 7 gives a result in the case $n=3$ for strictly positive curvature.

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The answer of j.c. given prior to mine is of course correct but let me give a trivial reason why (in big dimensions) every Riemannian metric after an arbitrary small perturbation is not isometrically embeddable in $R^{n+1}$, even locally.

It is known that the curvature tensor of a metric induced from an imbedding in $R^{n+1}$ has the following form $$R_{ijkl} =h_{il} h_{jk} - h_{ik} h_{jl}, \ \ \ \ (*) $$ where $h$ is the second fundamental form. The formula $(*)$ is a very strong restriction on the curvature tensor, in big dimensions most $(0,4)$- tensors having the algebraic symmetries of the curvature tensor can not be represented in the form $(*)$. Now, by a arbitrary small perturbation of an arbitrary metric one can achive that its curvature does not satisfies $(*)$; the existence of such a perturbation follows from the fact that at a point for any (0,4)-tensor $T$ satisfying the algebraic symmetries of the curvature tensor there exists a metric whose curvature tensor at this point coincides with $T$. Thus, one can slightly perturb an arbitrary metric such that the result is not imbeddable.

Since a small perturbation of a metric of positive sectional curvature again has positive sectional curvature, the answer on your question is negative.

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