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Reading Princeton Companion I found out that every finitely presented group can be realized as the fundamental group of a 4-manifold.

When starting to write this answer I found this related MO question. However, my question has to do with one of its answers (which is similar to the hint given in the Princeton Companion).

The proceedure for constructing this manifold from a given presentation is first to construct a CW-complex with that fundamental group (by wedge sum of circles puting 2-cells to cover the relations) embedding it in $R^5$ and considering the frontier of a tubular neighborhood.

Two questions come up to me (which are maybe trivial):

1- Why this cannot be done in $R^4$? Or can it be but the result is not the same?

2- Why the resulting manifold has the desired fundamental group?

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(IIRC Stillwell's beautiful book included a detailed construction in the last chapter) –  Mariano Suárez-Alvarez Jul 1 '10 at 22:15
    
Is there a reference for this book? Thanks. –  rpotrie Jul 1 '10 at 22:18
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John StillWell, Classical Topology and Combinatorial Group Theory, GTM 72, Springer-Verlag. –  Mariano Suárez-Alvarez Jul 1 '10 at 22:19
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You cannot expect this construction to be possible for all cases in $\mathbb{R}^4$, because your 2-cells may end up intersecting each other in an unavoidable way. In $\mathbb{R}^5$, you have enough room to avoid intersections. –  S. Carnahan Jul 2 '10 at 0:05
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Other than the general-position machinery not working out, you could make the argument from the opposite direction. Not every finitely presented group is the fundamental group of a 3-manifold. The finite groups that are fundamental groups of 3-manifolds are quire restricted -- they are the finite subgroups of $SO_4$ that act freely on $S^3$. So for example, the symmetric group on $10$ letters isn't the fundamental group of a 3-manifold. –  Ryan Budney Jul 7 '10 at 21:23
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2 Answers

up vote 5 down vote accepted

Related to question (1), suppose you wanted to get a 4-manifold with boundary as a submanifold of $\mathbb{R}^4$ with fundamental group a finitely presented group, with presentation complex $K$. It is a result of Curtis that any 2-complex $K$ is homotopy equivalent to a 2-complex which embeds in $\mathbb{R}^4$ (see also the Stallings reference in the comment on this MO question and Dranishnikov-Repovs - in fact this works if the 2-complex $K$ has only one vertex). However, to thicken up $K\subset \mathbb{R}^4$ to get a tubular neighborhood which is a 4-manifold (with boundary) in $\mathbb{R}^4$ which retracts to $K$, $K$ must be nicely embedded. For example, the 2-cells should be locally flat. I'm not sure if this holds true for the embeddings of Curtis or for the other constructions.

To answer (2), think about what happens when you thicken up the complex. Thickening some points in $\mathbb{R}^5$ gives some 5-balls, whose boundary is a union of 4-spheres. Thickening an interval attached to some points, one gets a 1-handle $D^1\times D^4$, whose boundary removes two 4-balls from the 4-spheres and attaches in $D^1 \times S^3$, giving a 4-manifold with free fundamental group (a connect sum of $S^3\times S^1$'s). The 2-cells thicken up to 2-handles $D^2\times D^3$, which remove $S^1\times D^3$ from the 4-manifold (which doesn't change the fundamental group, since $S^1$ is codimension 3), and replaces it with $D^2 \times S^2$, which has the effect of killing the element in the free group corresponding to the circle by Van Kampen. So you see that this gives the same thing as the construction in Henry Wilton's answer to the other question.

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Taking the boundary of tubular neighborhood in codimension two doesn't preserve fundamental group -- the codimension must be at least three.

For a simple example, if you take the boundary of a tubular neighborhood of a single point in $R^2$, you get a circle, which has nontrivial fundamental group. Similarly, the boundary of a tubular neighborhood of a 2-plane in $R^4$ is the product of a 2-plane with a circle, which again has nontrivial fundamental group. In general, if you start with a 2-manifold in $R^4$, the boundary of a tubular neighborhood will be a circle bundle over the manifold, and will therefore not have the right fundamental group.

This problem is fixed in $R^5$, because you get a sphere bundle instead of a circle bundle, and the 2-sphere has trivial fundamental group.

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Thanks, this is clarifying too! –  rpotrie Jul 9 '10 at 19:01
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