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  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:

  1. 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 $\mathbb{R}^3$.

Arshak Aivazian
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  • Recall that compact subsets of $\mathbb{R}^3$ are precisely those subsets that are closed and bounded (Heine-Borel). The definition of a submanifolds as only the open subsets is very non-standard. Futhermore, a solid torus with boundary is not open in $\mathbb{R}^3$, and a solid open torus is not compact. You should probably use a more standard definition of a submanifold. See e.g. https://en.wikipedia.org/wiki/Submanifold. – Frederik Mar 15 '21 at 10:37
  • @Frederik It seems to me that in standard terminology a "compact manifold" is not a "manifold" (but it is a manifold with a boundary). I can reformulate the question without using the concept of a submanifold and speaking only about finite unions of closed Cartesian cubes. – Arshak Aivazian Mar 15 '21 at 10:43
  • The standard term for the subspaces of interest to me is "compact regular closed subsets" https://proofwiki.org/wiki/Definition:Regular_Closed_Set – Arshak Aivazian Mar 15 '21 at 10:45
  • I have usually seen compact manifold as meaning a manifold which is compact as a topological space. You even define it as also having the property of being a closed subset in the question. My problem with the formulation of the question is that you require compact submanifolds to be both open and closed as subsets of $\mathbb{R}^3$. But the only closed and open subsets of $\mathbb{R}^3$ is the empty set and $\mathbb{R}^3$. – Frederik Mar 15 '21 at 10:51
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    @Frederik I edited the question. I think it reads better now. – Arshak Aivazian Mar 15 '21 at 10:54
  • @Frederik In the original version of the question, "a compact submanifold" was defined as "a compact subset whose the closure of its interior equals it", and "a submanifold" as an "open subset". In particular, "a compact submanifold" is not "a submanifold". – Arshak Aivazian Mar 15 '21 at 11:03
  • $H_1$ is the abelianization of $\pi_1$ if the space is path-connected, and moreoever, the fundamental group is not necessarily abelian. They are not equivalent. For example, the disjoint union of three circles have $H_1\cong \mathbb{Z}^3$, however the fundamental group is $\mathbb{Z}$ independant of choice of basepoint. (Answered to a deleted comment) – Frederik Mar 15 '21 at 12:25
  • Let us continue this discussion in chat. – Arshak Aivazian Mar 15 '21 at 12:27

1 Answers1

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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.

user901123
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  • Could you elaborate on this "Sphere theorem"? I was suspecting there should be an obstruction living in $H^3( \pi_1(X))$ for embedding any $2$-complex $X$ into $\mathbb{R}^3$, and that this is always nontrivial if $\pi_1= \mathbb{Z}^n$ and $n>2$. – Connor Malin Mar 16 '21 at 15:24