Given any finitely-presented group $G$, there are a few equivalent techniques for constructing smooth/PL 4-manifolds $M$ such that $\pi_1 M$ is isomorphic to $G$. For most constructions of these 4-manifolds, they embed naturally in $S^5$ (as the boundary of regular neighbourhoods of $2$-complexes in $S^5$.)

Question: Are there are any smooth/PL 4-dimensional submanifolds $M$ of $S^4$ such that $\pi_1 M$ has an unsolvable word problem? $M$ would of course have to be a smooth $4$-manifold with non-empty boundary.

I'm aware there are several constructions and obstructions to $2$-complexes embedding in $S^4$. Moreover, I've heard some of the construction techniques fall into the tame topological world and may not be smoothable. The condition given by Kranjc (that $H^2$ of the 2-complex is cyclic) is generally a non-computable condition for a group with non-solvable word problem. Although, perhaps there are many groups with non-solvable word problem and $H^2$ trivial. The closest to references on the subject that I know:

M. Kranjc, "Embedding a 2-complex K in R^4 when H^2(K) is a cyclic group," Pac. J. Math. 150 (1991), 329-339.

A. Shapriro, "Obstructions to the imbedding of a complex in Euclidean space, I. The first obstruction," Ann. of Math., 66 No. 2 (1957), 256--269.

edit: Thanks for the comments people. Now that I'm back in Canada with proper internet (+MathSciNet) access, I did a little digging and came across this:

A. Dranisnikov, D. Repovs, "Embeddings up to homotopy type in Euclidean Space" Bull. Austral. Math. Soc (1993).

They show that any finitely-presented group is the fundamental group of a 2-dimensional polyhedron in $\mathbb R^4$. This was apparently a question of Matthias Kreck's.

And yes, Sam Nead, this question was in part motivated by the concern that 2-knots could have undecidable word problems for the fundamental groups of their complements. I've been thinking about the fundamental groups of 2-knot complements recently, and this is a concern.

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    $\begingroup$ What if one asks the same question requiring only that $\pi_{1}(M)$ fails to be residually finite? $\endgroup$
    – Jon Bannon
    Jul 21, 2010 at 12:19
  • $\begingroup$ Do you want the group to be finitely presented? $\endgroup$ Jul 21, 2010 at 14:04
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    $\begingroup$ You can embed a finite 1-vertex complex in $\mathbb{R}^4$. See: mathoverflow.net/questions/30238/… However, I'm not sure that the complex has a regular neighborhood - some of the 2-cells might have branched points. So it probably won't be $\pi_1$-injective into an open submanifold. $\endgroup$
    – Ian Agol
    Jul 21, 2010 at 15:50
  • $\begingroup$ @Agol: If it's a PL embedding, then what's the problem? It may not have locally flat 2-cells, but it has a refinement with locally flat 2-cells. Isn't that good enough for the question? $\endgroup$ Jul 21, 2010 at 16:04
  • $\begingroup$ @ Greg: I think you're right - I was worried about a neighborhood of a branch point (like a cone on a knot) not having a tubular neighborhood (tubular neighborhoods have obvious retracts to the complex, guaranteeing $\pi_1$-injectivity). But I think one can still take a neighborhood in which the complex will be $\pi_1$-injective using Van Kampen. $\endgroup$
    – Ian Agol
    Jul 30, 2010 at 1:23

1 Answer 1



My memory was quite blurry about this when I originally answered.

See Gonzáles-Acuña, Gordon, Simon, ``Unsolvable problems about higher-dimensional knots and related groups,'' L’Enseignement Mathématique (2) 56 (2010), 143-171.

They prove that any finitely presented group is a subgroup of the fundamental group of the complement of a closed orientable surface in the $4$-sphere, which is much better than I reported.

Original answer:

You most likely would like a finitely presented group, but this might be of interest anyway:

Let $S$ be a recursively enumerable non-recursive subset of the natural numbers and consider the group

$\langle \ a,b,c,d \ | \ a^iba^{-i} = c^idc^{-i} \ \mathrm{for}\ i \in S \rangle$

This has unsolvable word problem. See page 110 of Chiswell's book "A course in formal languages, automata and groups" available on google books (I think it's also in Baumslag's "Topics in Combinatorial Group Theory" but all my books are in boxes at the moment.)

This should be the fundamental group of the complement of a noncompact surface in $\mathbb{R}^4$. You do this in the usual way by beginning with the trivial link on four components and then drawing the movie of the surface in $\mathbb{R}^4$, band summing at each stage to make the conjugates of $b$ and $d$ equal.

I think you end up with a knotted disjoint union of planes. I remember doing this at some point in graduate school when C. Gordon asked me if there were any compact surfaces in the $4$-sphere whose complements have groups with unsolvable word problem.

  • $\begingroup$ To make the question more extreme: Are there two-spheres in $S^4$ whose complements have groups with unsolvable word problems? $\endgroup$
    – Sam Nead
    Jul 21, 2010 at 15:07
  • $\begingroup$ Gordon said that he had a definite reason for wanting to know this, but he also said he couldn't remember what the reason was. I think it's a great question. $\endgroup$ Jul 21, 2010 at 15:13
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    $\begingroup$ It is open whether or not there is an algorithm to detect unknotted two-spheres in $S^4$. Perhaps this is related? $\endgroup$
    – Sam Nead
    Jul 21, 2010 at 15:45
  • $\begingroup$ Yeah, that seems reasonable. $\endgroup$ Jul 21, 2010 at 16:54

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