I am interested when manifolds $M$ (with boundary) are displaceable, i.e be pushed off themselves. Concretely, I mean there is a diffeomorphism $\phi: M \rightarrow M$ that is identity on the boundary such that $\phi(M_0) \cap M_0 = \emptyset$; here $M_0 \subset M$ is the result of removing a small collar neighborhood of $\partial M$ from $M$. So $M_0$ and $M$ are diffeomorphic. For manifolds to be displaceable, the intersection form needs to vanish.

For example, suppose that $M^{2n}$ is a 2n-manifold with a handlebody decomposition with handles of index at most n. Also, suppose that the intersection form of $M^{2n}$ vanishes. Is it true that $M^{2n}$ is displaceable?

The following manifolds are displaceable: $M = \mathbb{R}^{2n}, \mathbb{R}^2 \times W^{2n-2}, T^*S^k \times W^{2n-2k}$ (for k odd).

  • $\begingroup$ Just to allay my confusion, let me check: assuming $M$ is connected, you are considering only the case that $M$ is noncompact and $\partial M$ is connected. Right? $\endgroup$ – Lee Mosher Aug 9 '17 at 13:52
  • $\begingroup$ M is compact and connected and has connected boundary (which is non empty). $\endgroup$ – user39598 Aug 9 '17 at 17:35
  • $\begingroup$ Just a suggestion: You may first try to solve the problem by a homotopy-equivalence (and, moreover, a map homotopic to the identity rel boundary) rather than a diffeomorphism. If this works, see if there is an argument akin, maybe to the proof of the h-cobordism theorem, to get a diffeomorphism. $\endgroup$ – Moishe Kohan Aug 9 '17 at 21:03

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.