The strong Whitney Embedding Theorem tells us that every smooth n-manifold (n>0) admits a smooth embedding into $\mathbb{R}^{2n}$. Also, every topological $n$-manifold admits an embedding into $\mathbb{R}^{2n+1}$ (Munkres' Topology. Exercise §50.7).

My question now: can this last bound be lowered to $2n$? And if not, which topological $n$-manifold isn't embeddable into $\mathbb{R}^{2n}$? (Whitney's embedding theorem already tells us that such a manifold cannot admit a smooth structure)

  • 2
    $\begingroup$ @Henno aren't knots just $S^1$ as topological manifolds though? $\endgroup$ – Alessandro Codenotti Mar 12 at 16:31
  • 2
    $\begingroup$ mathoverflow.net/questions/34658/… whoops I copied the wrong link earlier. The answer is positive by comments here $\endgroup$ – Alessandro Codenotti Mar 12 at 19:19
  • 2
    $\begingroup$ @AlessandroCodenotti I looked through the papers suggested by the first link in Sergey Melikhov's comment to Andy Putnam's answer and I was not able to find the desired statement. (The second link in Melikhov's comment seems to be broken.) I guess the situation is that the embedding into $\mathbb{R}^{2n}$ somehow follows from the theorems in the Bryant-Mio and Johnston papers, but this isn't obvious to me as a non-expert. $\endgroup$ – Jim Belk Mar 12 at 23:08
  • 2
    $\begingroup$ @JohnSamples It seems strange that such a basic result (not basic in the sense of difficulty of proving) isn't written down anywhere... $\endgroup$ – Léo Mousseau Mar 13 at 9:51
  • 2
    $\begingroup$ It is known that smooth (or more generally PL) manifolds embed into $\mathbb R^{2n}$. See ams.org/journals/tran/1971-157-00/S0002-9947-1971-0278314-4/… $\endgroup$ – Paul Frost Apr 10 at 0:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.