Let $M$ be a smooth manifold of dimension at most $3$ and $S \subset M$ a smoothly embedded compact connected codimension $1$ manifold, separating $M$ into two components, $M_1$ and $M_2$. I wonder now if the following is true:

If $\pi_1(M)$ is finitely generated, so is $\pi_1(M_i)$ for $i=1,2$.

One might be inclined to think that this is a purely algebraic matter, but it is quite possible to find an infinitley generated group $A$ and set up a diagram of groups $A \leftarrow B \rightarrow C$ such that the pushout is finitely generated. Hence, there must be some topological restrictions i do not realize. Any help is appreciated.

  • $\begingroup$ I retract my previous bluster, but what is your example of pushout diagrams? $\endgroup$ – user98602 Dec 17 '15 at 2:00
  • $\begingroup$ Your condition that $S$ separate $M$ into two pieces implies $S$ has trivial normal bundle. In particular $M_1$ and $M_2$ are homotopy equivalent to their closure in $M$ which are smooth manifolds with boundary. If $M$ is compact, you are done. $\endgroup$ – PVAL-inactive Dec 17 '15 at 2:04
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    $\begingroup$ But I think the interesting case is when $M$ is non-compact. $\endgroup$ – Daniel Valenzuela Dec 17 '15 at 2:05
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    $\begingroup$ but don't trust me. I'm quite tipsy right now. and I know nothing about non-compact manifolds. $\endgroup$ – Daniel Valenzuela Dec 17 '15 at 2:10
  • $\begingroup$ Yes indeed. The interesting case is when M is non compact, so at least one of the $M_i$ also has to be. $\endgroup$ – H1ghfiv3 Dec 17 '15 at 7:24

This is false in dimension 3. For instance, let $M$ be the Whitehead manifold. Then there exists a compact PL (or smooth as you prefer) submanifold $N$ in $M$ whose complement has infinitely generated fundamental group. This is a special case of the main theorem in

T.W. Tucker, Non-compact 3-manifolds and the missing-boundary problem, Topology 13 (1974), 267-273.

In dimension 2 your conjecture is, of course, true, since surfaces with finitely generated fundamental groups are tame.

  • $\begingroup$ I am working in the smooth category. In the case of $M$ the classic Whitehead manifold, can $N$ be chosen to be a compact codimension $1$ smooth submanifold ? $\endgroup$ – H1ghfiv3 Dec 25 '15 at 12:59
  • $\begingroup$ Yes, in dimension 3, PL is the same as DIFF. $\endgroup$ – Moishe Kohan Dec 25 '15 at 13:07
  • $\begingroup$ Let us just make sure that we are talking about the same "same" here: In dimensions up $3$, every topological manifold admits a unique smooth and a unique PL stucture. This is precisely why we call the $3$ usually distinct categories equivalent , right ? However, in my eyes at least, it is quite possible for a surface $S$ to be embedded in a 3-manifold $M$, such that $S$ is a PL-submanifold of $M$, but not a smooth submanifold. For example, one could take $S$ as the boundary of the unit cube in $\mathbb R^3$. Maybe I am making a mistake here, so feel free to correct me if I am wrong. $\endgroup$ – H1ghfiv3 Dec 25 '15 at 13:14
  • $\begingroup$ Maybe the trick is to take $S$ a $PL$-submanifold as you described and then somehow smoothing the corners to make it a smooth submanifold of $M$ with the same properties. Still, the following question remains open: Can $S$ be chosen as a surface ? $\endgroup$ – H1ghfiv3 Dec 25 '15 at 13:16
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    $\begingroup$ Equivalence of the categories also means that each PL submanifold of a 3-manifold is smoothable. Hence, you can assume that this surface us smooth. $\endgroup$ – Moishe Kohan Dec 26 '15 at 9:33

I think you should be able to show this using Scott's Core theorem. Pick a compact core $C$ in $M$. Now you might want to distinguish some cases, but you should get somthign like the generation of $\pi_1M_i$ by the image of $\pi_1(C\cap M_i)$ and $\pi_1S$. You should also assume properly embeddedness for this to work.

EDIT: So as pointed out in the comments my answer was just too sloppy to understand so I decided to give a proof.

First: how to you come up with the idea of the proof? So my attempt was to look at graphs of groups or groupoids to avoid issues related to base point, connectedness and so on. And now you just start to look at suitable sets covering the core $C$ and the seperating surface $S$ and partition them more and more (and sometimes less) to get nice generating sets for different parts. You get a graph of groups and which contains the relations among the fundamental groups of the sets. Eventually I got multiple partitions of sets which would do the trick. Here is a particularly easy one, to which I arrived by using the above technique, but which actually only consists of two sets.

So let* $K= C\cup \nu S$ and $K_i = K \cap M_i$, where $\nu S$ is a trivial tubular neighborhood (trivial by seperating property and $M$ is connected) small enough that everything looks like a product there (especially transverse intersections with $C$). Let $k^i:S \hookrightarrow K_i$ and $l^i$ the same map post composed to embed $S$ into $M_i$. Now we see that by Van Kampen (here we are again slightly ignoring connectedness of the $K_1,K_2$**) we have $$ \pi_1K = \pi_1K_1 *_{k_*\pi_1S} \pi_1K_2. $$

So this is the pushout $(\pi_1 k^1,\pi_1 k^2)$ and we have a map from it $$c:\pi_1K \twoheadrightarrow \pi_1M = \pi_1 M_1 *_{l_*\pi_1S} \pi_1M_2.$$ But we have more structure to that pushout map, namely that it arises not only from a map $\pi_1 K_i \to \pi_1M$ but that those maps each factor through $\pi_1M_i$. Hence our maps $\pi_1 K_i \to \pi_1M_i$ generate at least the part $\pi_1M_i - l^i_*\pi_1S$. That is because $c_*\pi_1K = c_*\pi_1K_1 \cup c_*\pi_1K_2$ and these two images (or rather the two factors) are always disjoint except for the image of the intersection. But as we have the surface in there from the beginning (algebraically you could also say that we are amalgamating over a larger subgroup already which is preserved, i.e. onto) that is no problem and we see that we get surjections $$\pi_1K_i \twoheadrightarrow\pi_1M_i.$$

So we showed: picking a modified core for $M_i$ gives us a finitely generated fundamental group.

* You might want to check that this proof won't work with $C$ and $C_i$ and that this is not even true in general.

** So by choosing our $K$ we made sure, that every component of $C-(C\cap S)$ touches $\cap S$, so we have that $S$ cuts this component containing $S$ into two components. So you could only look at the connected part and do the other components seperately eventually. This is why I ignore this issue. And another argument might be to just choose $C$ connected, which makes this obsolete as well.

  • $\begingroup$ or rather $\pi_1((C\cap M_i)\cup S)$ $\endgroup$ – Daniel Valenzuela Dec 17 '15 at 2:00
  • $\begingroup$ I was thinking about exactly the same thing $\endgroup$ – H1ghfiv3 Dec 17 '15 at 7:23
  • $\begingroup$ The case when $C$ is contained in, say, $M_1$, already implies that $M_2$ is f.g i think. Any loop $\gamma$ in $M_2$ can be homotoped to lie in $C$. This homotopy must meet $S$, so $\gamma$ must be homotopic to a loop in $S$. The result now follows since $S $ was assumed to be compact. $\endgroup$ – H1ghfiv3 Dec 17 '15 at 7:36
  • $\begingroup$ You get even the generation by $S$! The only case where you might want to be slightly careful is when e.g. $C\cap S$ is not connected and there is a component of $C-(S\cap S)$ which touches $S$ multiple times. By this I mean the case when $\langle \pi_1(C\cap M_i),\pi_1S\rangle \neq \langle\pi_1((C\cap M_i)\cup S)\rangle$ in $\pi_1M_i$ so the RHS is actually bigger, containing a loop in $\pi_1(C,C\cap S)$ closed up in $S$. $\endgroup$ – Daniel Valenzuela Dec 17 '15 at 9:22
  • $\begingroup$ What do you mean by $(\pi_1(C \cap M_i),\pi_1S)$ ? The smallest normal subgroup generated by those groups, when both are regarded as subgroups of $\pi_1(M_i)$ ? $\endgroup$ – H1ghfiv3 Dec 17 '15 at 9:38

Let me make this a new answer, as the old one (particularly the comment section) got really messy. I have been on vacation for a really long time --- excuse my late response.

I am still convinced that the answer should be yes for proper embeddedness and that all the wild excuses kind of miss the point, as they construct wild embeddings. Assume the surface $S$ has a normal bundle $V$ diffeomorphic to $S\times [-1,1]$ s.t. the boundary is also properly embedded. Also assume the compact core $C$ touches $S\times [-1,0]$ and $S\times [0,1]$ (if not you can add a compact $0$-codimensional manifold to $C$).

Then, on fundamental group level, $K=C\cup V$ generates $M$ and $K_i=K\cap M_i$ generate $M_i$. You get this by writing the fundamental groups of $K=K_1\cup K_2$ and $M=M_1\cup M_2$ as pushouts and closely looking at the images $\pi_iK_i\to \pi_1M_i (\to \pi_1M)$.

You can get more details on this by looking at my other answer.

  • $\begingroup$ What exactly do you mean by proper embeddedness in contrast to wild embeddedness ? $\endgroup$ – H1ghfiv3 Jan 23 '16 at 19:36
  • $\begingroup$ It might now really be the time to explain the counterexample in full Detail $\endgroup$ – H1ghfiv3 Jan 24 '16 at 12:10

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