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I have found several sources that mention that there are eight non-compact Euclidean 3-manifolds, with four orientable and four non-orientable. There are two standard references for this, but unfortunately both are in German:

Nowacki, Werner. “Die euklidischen, dreidimensionalen, geschlossenen und offenen Raumformen.” Commentarii Mathematici Helvetici 7, no. 1 (1934): 81–93.

Hantzsche, Walter, and Hilmar Wendt. “Dreidimensionale euklidische Raumformen.” Mathematische Annalen 110, no. 1 (1935): 593–611.

So what are the eight non-compact Euclidean 3-manifolds?

Jim Belk
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  • Answering my own question here. I had the hardest time looking up the answer, but once I thought about it I figured it out. It seems like a common enough question that it's worth posting the answer. – Jim Belk Jun 16 '15 at 18:24
  • It should be mentioned that we are only talking about complete noncompact flat 3-manifolds. – Dan Asimov Nov 12 '23 at 01:30

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The four orientable non-compact Eulcidean 3-manifolds are:

  • $\mathbb{R}^3$,
  • $\mathbb{R}^2\times S^1$,
  • $\mathbb{R} \times S^1 \times S^1$, and
  • Another manifold $X$.

The manifold $X$ can be described as the quotient of $[0,1]\times[0,1]\times\mathbb{R}$ obtained by making the identifications $$ (0,y,z) \sim (1,y,z) \qquad\text{and}\qquad (x,0,z) \sim (1-x,1,-z) $$ for all $x,y\in[0,1]$ and $z\in\mathbb{R}$. A few notes about this manifold:

  • The image of $[0,1]\times[0,1]\times\{0\}$ is a Klein bottle, and $X$ deformation retracts onto this Klein bottle. Thus $X$ isn't homeomorphic to any of the other three listed possibilities.

  • $X$ can be described as the vector bundle of $2$-forms on the Klein bottle. Note that this vector bundle is nontrivial since it has no nonzero global section (since the Klein bottle is not orientable).

  • $X$ can also be described as a cylinder ($S^1\times\mathbb{R}$) bundle over a circle, where going once around the circle “flips” the cylinder, i.e. rotates the cylinder $180^\circ$ around an axis perpendicular to the axis of the cylinder.

The four non-orientable non-compact Euclidean 3-manifolds are:

  1. $M\times \mathbb{R}$,

  2. $M\times S^1$,

  3. $K\times \mathbb{R}$, and

  4. Another manifold $Y$.

Note that (2) and (3) are different since $M\times S^1$ deformation retracts onto a torus and $K\times\mathbb{R}$ deformation retracts onto a Klein bottle.

The manifold $Y$ can be described as the quotient of $[0,1]\times[0,1]\times\mathbb{R}$ obtained by making the identifications $$ (0,y,z) \sim (1,y,-z) \qquad\text{and}\qquad (x,0,z) \sim (1-x,1,z) $$ or alternatively $$ (0,y,z) \sim (1,y,-z) \qquad\text{and}\qquad (x,0,z) \sim (1-x,1,-z) $$ for all $x,y\in[0,1]$ and $z\in\mathbb{R}$. A few notes about this manifold:

  • The image of $[0,1]\times[0,1]\times\{0\}$ is a Klein bottle, and $Y$ deformation retracts onto this Klein bottle.

  • However, $Y$ is not homeomorphic to $K\times\mathbb{R}$. For example, $Y$ has just one end, whereas $K\times\mathbb{R}$ has two ends.

  • $Y$ can also be described as a Mobius band bundle over a circle, where going once around the circle corresponds to a homeomorphism of the Mobius band that is not isotopic to the identity. (There is only one such homeomorphism up to isotopy, which acts as inversion on the fundamental group.)

  • $Y$ is double covered by the orientable manifold $X$ given above. This also distinguishes it from $K\times \mathbb{R}$, whose orientable double cover is $S^1\times S^1 \times \mathbb{R}$. The deck transformation of $X$ that gives $Y$ is the map $(x,y,z)\mapsto (x+\tfrac12,y,-z)$.

  • Any line bundle over the Klein bottle is homeomorphic to either $K\times\mathbb{R}$, $X$, or $Y$.

Jim Belk
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    What is a Euclidean 3-manifold? – Georges Elencwajg Jun 16 '15 at 21:17
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    @GeorgesElencwajg A 3-manifold is Euclidean if it supports a complete Riemannian metric under which it is locally isometric to $\mathbb{R}^3$. Under this metric, the universal cover of the manifold is isometric to $\mathbb{R}^3$, which means that the manifold can be obtained as a quotient of $\mathbb{R}^3$ by a group of Euclidean isometries acting freely and properly discontinuously. – Jim Belk Jun 16 '15 at 21:27
  • I see. Thanks Jim. – Georges Elencwajg Jun 16 '15 at 21:35
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    I think you might be able to find the description of these Euclidean manifolds somewhere in the literature on Thurston's 8 geometries for 3-manifolds, although the noncompact examples may well be more difficult to pin down in that literature. One terminology for $X$ in that language is a "twisted $\mathbb{R}$-bundle over $K$", or perhaps more precisely the "orientation-twisted $\mathbb{R}$ bundle over $K$". – Lee Mosher Jun 17 '15 at 12:51
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    Is anyone else surprised that a the manifold $X$ is orientable but deformation retracts onto a Klein bottle? – N. Owad Jun 23 '15 at 13:03
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    @N.Owad Similarly, if $M \subset \mathbb{R}^3$ is a Mobius strip, then an $\epsilon$-neighborhood of $M$ is an orientable $3$-manifold that deformation retracts onto $M$. – Jim Belk Jun 23 '15 at 14:27
  • Ah, that makes sense. Thanks for the insight. – N. Owad Jun 23 '15 at 18:22
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    Excuse me, but isn't $K \times S^1$ compact? – Dario Jul 14 '15 at 22:45
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    @Dario That's an excellent point. I have fixed it. – Jim Belk Jul 17 '15 at 18:07
  • How did you determine that every line bundle over the Klein bottle is homeomorphic to either $ K \times \mathbb{R}, X $ or $ Y $? I am very interested. – Ian Gershon Teixeira Apr 08 '22 at 19:24
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    @IanGershonTeixeira I didn't have to, since I was starting with the information that there are exactly eight manifolds in the desired class. However, you can get the result you menion by observing that there are exactly four homomorphisms from $\pi_1(K)\cong\langle a,b \mid b^{-1}ab=a^{-1}\rangle$ to $\mathbb{Z}_2$, and the two homomorphisms for which the image of $a$ is nontrivial give homeomorphic line bundles, since they differ by a Dehn twist of $K$ along $a$. – Jim Belk Apr 11 '22 at 02:07
  • "Is anyone else surprised that a the manifold is orientable but deformation retracts onto a Klein bottle?" I was, but not terribly surprised, because I knew that S^1 × S^2 has an embedded Klein bottle. – Dan Asimov Nov 12 '23 at 01:24