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If $M \subset \mathbb R^n$ is a compact smooth manifold with boundary, and ${M_\varepsilon }$ is the closed $\varepsilon$-neighborhood of $M$ in $\mathbb R^n$, then whether for sufficiently small $\varepsilon$, ${M_\varepsilon }$ is a smooth manifold?

Summer
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1 Answers1

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Yes. This follows from the tubular neighborhood theorem, which you may find in many differential geometry/topology books. See e.g. http://www.google.com/search?q=tubular%20neighborhood%20theorem&um=1&ie=UTF-8&hl=en&tbo=u&tbm=bks

Eric O. Korman
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  • Wouldn't this be rather a comment than an answer? – t.b. Nov 13 '11 at 00:46
  • @t.b.: what's the difference? – Damian Sobota Nov 13 '11 at 01:04
  • @Damian: look at an answer e.g. by Arturo Magidin and look at this which basically says: it is a consequence of the tubular neighborhood theorem, look, here's what Google gives you. Then you should see a fundamental difference (in quality, content, helpfulness, etc...) – t.b. Nov 13 '11 at 01:11
  • I wouldn't have used such a strong word. The link didn't work for me, (it still doesn't), so I stripped out the 's' from https and the rest of the Google noise, sorry that I overlooked the &tbm=bks modifier. – t.b. Nov 13 '11 at 01:27
  • Now it works. Thank you. – t.b. Nov 13 '11 at 01:45
  • @t.b.: ok, thank you, I understand. – Damian Sobota Nov 13 '11 at 02:04
  • Maybe it is not so easy for this situation. The book I am reading says that when $M$ is boundaryless ${M_\varepsilon }$ is indeed a smooth manifold by tubular neighborhood for some $\varepsilon$. But when $M$ has boundary, we can only prove that ${M_\varepsilon }$ is $C^1$ without higher smoothness. That is where I cannot figure out how. – Summer Nov 13 '11 at 14:46