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Let $M$ denote an arbitrary smooth, closed, connected, n-dimensional manifold for $n\geq 4$. For every such $M$, does there exist a closed (not necessarily connected!) codimension two submanifold $S \subset M$ such that $\pi_1(M\setminus S)$ is a free group? If not, what if one replaces $M$ by $M\# N$ for some other $n$-manifold $N$?

A couple of thoughts on this:

  • By playing around with a nice triangulation of $M$ one can easily construct a codimension two subcomplex of $M$ such that its complement has free fundamental group. But of course resolving the singularities seems complicated, this is why the connected sum might be helpful.

  • If $n$ is big enough, I think, that the number of connected components of $S$ bounds the number of relations in $\pi_1(M)$. Because then the complement can be arranged to have a handle decompositon without two-handles and then the only two-handles of $M$ come from the $n-2$ handles in $S$.

  • If $S$ has simply-connected components, then the canonical map $M \to B\pi_1(M)$ factorizes through a complex of geometric dimension two, which is impossible for many $M$.

  • I think if $M$ has dimension three, then this is impossible since there are too few manifolds with free fundamental group.

  • Finally if $M$ is simply-connected, then this is easy.

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    $\begingroup$ As you suspect, in dimension 3 the answer to your first question should be "no". Indeed, if one deletes any collection of circles from $M$ then the complement is a 3-manifold with toroidal boundary. Since the fundamental group is free, no boundary component can $\pi_1$-embed. But then, by Dehn's lemma, every boundary component is compressible. There's something to check here, but I think it follows that the complement is a connect sum of solid tori and $S^1\times S^2$'s, whence $M$ has to be a connect sum of lens spaces and $S^1\times S^2$'s. $\endgroup$
    – HJRW
    Commented Dec 9 at 12:45
  • $\begingroup$ Yes I was expecting a similar argument, since 3-manifolds with free fundamental group are so restricted, so that one has to end up with something that can be build from $S^1 \times S^2$. $\endgroup$
    – ThorbenK
    Commented Dec 9 at 12:49
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    $\begingroup$ Now that I understand your question I see it's very close to the question of asking which manifolds have handle decompositions such that the 2-skeleton is homotopy-equivalent to a wedge of 2-spheres. The main difference is you are concerned with fundamental groups rather than homotopy-type. $\endgroup$
    – Ryan Budney
    Commented Dec 9 at 22:19
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    $\begingroup$ Are you assuming existence of a handle decomposition (which is true in dimensions $\ne 4$)? $\endgroup$
    – Moishe Kohan
    Commented Dec 12 at 16:06
  • $\begingroup$ @MoisheKohan Oh, I see. I never specified the manifolds to be smooth since I didn't see it as necessary. So feel free to assume that they are smooth or have a handlebody decomposition! $\endgroup$
    – ThorbenK
    Commented Dec 12 at 16:10

2 Answers 2

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To your first question, the answer is yes.

Take a $k$-component trivial link in $S^n$, i.e. the boring, linear embedding

$$\sqcup_k S^{n-2} \to S^n$$

that is the boundary of a linear embedding

$$\sqcup_k D^{n-1} \to S^n.$$

The fundamental group of the exterior of this link is the free group on $k$ elements provided $n \geq 3$.

As a manifold the exterior (of an open tubular neighbourhood of this embedding) is diffeomorphic to a connect-sum of $k$ copies of $S^1 \times D^{n-1}$.

Perhaps I'm misreading your question but I don't see how you have ruled out this basic example.

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    $\begingroup$ I think you misread the question and a (manifold) means arbitrary rather than some. $\endgroup$
    – Moishe Kohan
    Commented Dec 9 at 21:13
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    $\begingroup$ The manifold $M$ is fixed. $\endgroup$
    – Dan Petersen
    Commented Dec 9 at 21:13
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    $\begingroup$ Yes exactly! I'm interested in whether this is possible for an arbitrary $M$. For the application I have in mind though it would suffice of $M$ has non-zero simplicial volume. $\endgroup$
    – ThorbenK
    Commented Dec 9 at 21:16
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    $\begingroup$ @ThorbenK: the way your question was asked threw me off, as you describe in some detail why the answer is no for arbirary $M$. So it looked like you were searching for a case where the answer is yes, as the other interpretation has been ruled-out. $\endgroup$
    – Ryan Budney
    Commented Dec 9 at 21:33
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    $\begingroup$ @RyanBudney I changed the wording a bit, maybe it's less ambiguous now. $\endgroup$
    – ThorbenK
    Commented Dec 10 at 7:30
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As you suspect, in dimension 3 the answer to your first question should be "no". Indeed, if one deletes any collection of circles from $M$ then the complement is a 3-manifold with toroidal boundary. Since the fundamental group is free, no boundary component can $\pi_1$ embed. But then, by Dehn's lemma, every boundary component is compressible. It follows that the complement is a connect sum of solid tori and $S^1\times S^2$’s, whence $M$ has to be a connect sum of lens spaces and $S^1\times S^2$’s.

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