MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Suppose one has a smooth manifold with boundary M and compact on top of it. Is it true that it can always be embedded in an upper half plane such that the boundary is embedded in the hiperplane $x_n=0$? Or are there obstructions to that? If yes, what are those obstructions? Thank you!

share|cite|improve this question
Uhhh, if you Whitney-embed M into a large enough Euclidean space $R^n$, then embed that into $R^n\times R$, doesnt this do it? assuming the hyperplane is part of the "upper half space". – Chris Gerig Feb 8 '12 at 23:39
As Chris mentions, there's no obstruction. You can embed an $n$-dimensional manifold with boundary in a Euclidean half-space of dimension $2n$ "neatly" as you describe. And embeddings are generic if the half-space has dimension $k > 2n$. – Ryan Budney Feb 8 '12 at 23:49
Daniel probably wants to know that the interior can be mapped in the (strictly) upper half space. This is a consequence of the collaring theorem: the boundary of M has a neighborhood homeo (or diffeo) to $\partial M\times [0,1)$. This can extended over $M$ and used as a last coordinate. – Paul Feb 9 '12 at 0:46
Thank you all for the comments. That is exactly what I wanted, Paul! Shall I understand that the proof of Whitney's theorem goes thru without any headaches, other than what you just said? – Daniel Feb 9 '12 at 2:41
That's correct. It's the same proof, just with one (small) extra layer of complexity that doesn't get in the way of any essential steps. – Ryan Budney Feb 9 '12 at 2:44

See Theorem 1.4.3. of

M. Hirsch: Differential Topology, Springer Verlag, 1976

share|cite|improve this answer

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.