Sign up ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.


By a celebrated theorem of J.Nash, we know that any riemannian manifold with smooth enough metric tensor can be realised as an embedded submanifold of $\mathbb{E}^N$ for some $N$.

Can one hope to be able to embed (compact) manifolds into some space with constant curvature? If the answer is yes and we have the isometric embedding $(M,g) \hookrightarrow Q(k)$, where $Q(k)$ is a space form of constant curvature $k$, do we have any estimates on the largest possible $|k|$? One possible bound comes from boundedness of diameter of positive space forms in terms of their curvature, although, the embedded submanifold might have longer geodesics than the ambient space.

Also, if we let the embedding be merely conformal, that is the induced metric belong to the same conformal class as that of $(M,g)$, what are the possible obstructions?

share|cite|improve this question

1 Answer 1

up vote 8 down vote accepted

There is no bound on curvature, only on the dimension. (The bound for the diameter of ambient space gives no bound for the intrinsic diameter of embedded space.)

In fact, it is straightforward to modify the proof of Nash's theorem to prove the following:

Any $n$-dimensional Riemannian manifold can be embedded into any $m$-dimensional Riemannian manifold if $m\ge N(n)$.

share|cite|improve this answer
Is there a similar statement of sharpness? –  kangdon Oct 1 '11 at 1:59

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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