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?
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2 Answers
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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.
answered Jul 28, 2012 at 17:21
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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? $\endgroup$– DaveKCommented Jul 28, 2012 at 18:20
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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. $\endgroup$– MishaCommented Jul 28, 2012 at 18:41
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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. $\endgroup$ Commented Jul 28, 2012 at 20:05
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1$\begingroup$ A theorem from 68 is an «old theorem»? :-) $\endgroup$ Commented Jul 29, 2012 at 1:02
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1$\begingroup$ Operationally, anything that came before me is old, and anything after is young. $\endgroup$ Commented Jul 29, 2012 at 1:12
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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.
