Does there exist a compact submanifold of $\mathbb{R}^3$ whose fundamental group is $\mathbb{Z}^3$?

1. Does there exist a compact submanifold of $\mathbb{R}^3$ whose fundamental group is $\mathbb{Z}^3$ ?

The question in the title is a generalization of the question that really interests me:

2. Does there exist a connected finite set of unit cubes of a Cartesian lattice whose fundamental group is $\mathbb{Z}^3$ ?

It is clear that, for example, $\mathbb{Z}$ and $\mathbb{Z}^2$ are simply realized as a solid torus and a thickened torus, respectively (it is clear that, up to homeomorphism, they are assembled from a finite number of closed unit cubes).

It is clear that it would be sufficient to embed the CW-complex corresponding to the presentation $$\langle a, b, c \mid [a, b] = 1, [b, c] = 1, [a, c] = 1\rangle$$ but (as far as I can see) it doesn't embed in $\mathbb{R}^3$.

Solution 1:

I am surprised to see your question has no answer. There exists no such submanifold of $\Bbb R^3$. There are simpler proofs than the below but perhaps less illuminating.

You may find the following theorem in Hempel's book on 3-manifolds.

If $M$ is a 3-manifold with fundamental group $\Bbb Z^3$, then $M$ is homotopy equivalent to $T^3$, and homeomorphic to $T^3 \# S$ for some homotopy 3-sphere $S$.

Since Perelman, we know that a homotopy 3-sphere is a 3-sphere, so in fact your purported submanifold is homeomorphic to $T^3$.

But not only is $T^3$ not a submanifold of $\Bbb R^3$, in fact, there is no compact submanifold of $\Bbb R^3$ without boundary: this follows from the invariance of domain theorem.

(A shorter proof indicates that your space is aspherical using the "sphere theorem", and then that it must be homotopy equivalent to $K(\Bbb Z^3, 1) \simeq \Bbb Z^3$, and then that such a space has non-trivial third homology, which is only the case for compact 3-manifolds without boundary. Then run the same invariance of domain argument.)

Note that neither argument assumes that your 3-manifold is compact. You prove this.