8
$\begingroup$

Assume $\Sigma_1$ and $\Sigma_2$ are two embedded compact surfaces (say orientable) in an orientable 3-manifold $M$. Assume $\Sigma_1$ and $\Sigma_2$ are homotopic in $M$. Then are they isotopic?

$\endgroup$

2 Answers 2

8
$\begingroup$

No, generally they're not.

For example, there's only one homotopy class $S^2 \to \mathbb R^3$ but there's two isotopy classes of embeddings (given via how the embedding orients the compact 3-manifold it bounds).

edit: I think if your 3-manifold is irreducible and if your maps $S^2 \to M$ are not null homotopic then the answer is likely yes. But if your 3-manifold is say a connect sum of lens spaces then I suspect it's false but I haven't come up with a nice example yet. As Allen points out in the comments below, a connect-sum of lens spaces won't work, at least not when your surface is a sphere.

edit2: As Misha Kapovich points out, for irreducible 3-manifolds and incompressible surfaces homotopy implies isotopy. This is an old theorem of Waldhausen's. "On Irreducible 3-manifolds which are sufficiently large" Ann. of Math (2) 87 (1968) 56--88.

$\endgroup$
7
  • 1
    $\begingroup$ Hi Ryan, thanks! In your $S^2\rightarrow {\mathbb R}^3$ example, if we ignore the orientation, will they be isotopic? I am saying that for embeddings $\iota_1, \iota_2$, there exists an embedding $\iota_3$ so that $\iota_3(S^2)=\iota_1(S^2)$, and $\iota_3$ is isotopic to $\iota_2$. Is there a chance this kind of thing is true in general? $\endgroup$
    – DaveK
    Jul 28, 2012 at 18:20
  • 14
    $\begingroup$ Instead of 2-spheres, consider 2-tori in the 3-space: They are all homotopic, but you have infinitely many isotopy classes corresponding to knot neighborhoods. The right assumption is incompressibility of surfaces and irreducibility of the 3-manifold. Then Waldhausen proved that homotopy implies isotopy. $\endgroup$
    – Misha
    Jul 28, 2012 at 18:41
  • 2
    $\begingroup$ For spheres embedded in 3-manifolds the fact that homotopy implies isotopy is a theorem of Laudenbach in the 1973 Annals. He had to assume the manifolds in question contained no counterexamples to the Poincaré conjecture (i.e. no fake 3-balls) since "homotopic spheres are isotopic" implies the Poincaré conjecture. $\endgroup$ Jul 28, 2012 at 20:05
  • 1
    $\begingroup$ A theorem from 68 is an «old theorem»? :-) $\endgroup$ Jul 29, 2012 at 1:02
  • 1
    $\begingroup$ Operationally, anything that came before me is old, and anything after is young. $\endgroup$ Jul 29, 2012 at 1:12
10
$\begingroup$

If $\Sigma_1 \hookrightarrow M$ is an embedded $\pi_1$-injective surface, then any homotopic embedded surface will be isotopic to $\Sigma$. As Ryan and Allen point out, this is due to Waldhausen for incompressible surfaces of genus $>0$, and to Laudenbach for 2-spheres, together with the Poincare conjecture.

If $\Sigma_1$ is not incompressible in $M$, then there exists $\Sigma_2$ which is homotopic to $\Sigma_1$ but not isotopic. The point is that one may compress $\Sigma_1$ to get a surface $\Sigma'\hookrightarrow M$ which has smaller genus, and then reembed the 1-handle (in the same homotopy class) in a knotted fashion to get a non-isotopic surface $\Sigma_2$. Misha observed this in the comments on Ryans question for tori in $S^3$, but it holds more generally.

There's an intermediate case of $\Sigma_1\hookrightarrow M$ which is incompressible and not $\pi_1$-injective. By the loop theorem, this can only occur if $\Sigma_1$ is 1-sided in $M$, which implies that the surface is non-orientable, so does not fall under the purview of your question. I'm not sure if homotopy implies isotopy in this case - I suspect there are 1-sided Heegaard surfaces which are homotopic but not isotopic, but I don't know examples off the top.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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