MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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

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?

share|cite|improve this question
up vote 7 down vote accepted

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.

share|cite|improve this answer
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? – DaveK Jul 28 '12 at 18:20
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. – Misha Jul 28 '12 at 18:41
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. – Allen Hatcher Jul 28 '12 at 20:05
A theorem from 68 is an «old theorem»? :-) – Mariano Suárez-Alvarez Jul 29 '12 at 1:02
Operationally, anything that came before me is old, and anything after is young. – Ryan Budney Jul 29 '12 at 1:12

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.

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.