4

Here $ L(S^1) $ is the unique nontrivial line bundle over the circle (the mobius strip).

The manifold $ \mathbb{R}^3 $ (with infinite volume) admits all six of the aspherical Thurston geometries (so not $ S^3 $ or $ S^2 \times E^1 $). Are there other three manifolds (without boundary) that admit all six of the aspherical Thurston geometries?

My guess is that $ S^1 \times \mathbb{R}^2 $ admits all six, but that's just a hunch.

EDIT:

This question

What are the 8 non-compact Euclidean 3-manifolds?

Lists all the noncompact flat 3 manifolds. So it is enough to figure out which of these eight manifolds admit all six of the eight aspherical geometries. Of these manifolds, 4 are already confirmed see comment from Moishe Kohan $$ \mathbb{R}^3, \mathbb{R}^2 \times S^1,\mathbb{R} \times T^2, \mathbb{R} \times L(S^1) $$ One has been ruled out since it has a single end so it cannot admit hyperbolic $ \mathbb{H}^3 $ geometry $$ S^1 \times L(S^1) $$ The last three are line bundles over the Klein bottle and it is unclear which geometries they admit (see comment from Lee Mosher) $$ K^2 \times \mathbb{R},X,Y $$ for description of the manifolds $ X $ and $ Y $ see the question linked above or Flat 3 manifolds and mapping tori of flat surfaces

~End Edit~

Some other background:

  • every closed three manifold admits at most one geometry.

  • A manifold with model geometry $ (X,G) $ has universal cover $ X $. Thus a manifold can only admit multiple model geometries $ (X_1,G_1) $ and $ (X_2,G_2) $ if $ X_1,X_2 $ are homeomorphic.

  • For the six aspherical geometries the models $ X $ are all homeomorphic, so it is possible for one manifold to admit multiple geometries. But for $ S^3,S^2 \times E^1 $ no other model geometry is topologically equivalent to the model geometries $ S^3 $ and $ S^2 \times \mathbb{R} $ thus a manifold with $ S^3 $ or $ S^2 \times E^1 $ geometry cannot admit any of the other Thurston geometries.

  • a finite volume manifold with any one of $ S^3,S^2\times E^1,E^3,Nil,Sol,\mathbb{H}^3 $ geometry only admits that type of geometry

  • There exist finite volume noncompact manifolds which admit both $H^2\times E^1$ and $ \tilde{SL_2} $ geometry. For example $ SL_2(\mathbb{R})/SL_2(\mathbb{Z}) $, which is diffeomorphic to the complement of the Trefoil knot in $ S^3 $, admits both $ \tilde{SL_2} $ and $ H^2\times E^1 $ geometry. See https://math.stackexchange.com/a/73885/758507 Essentially the idea to show it also has $ H^2\times E^1 $ geometry is that the upper half plane model of hyperbolic plane $ H^2 $ quotienting by the action of the modular group $ SL_2(\mathbb{Z}) $ by Moebius transformations gives a two dimensional orbifold $$ \mathcal{D}= \{ z:Im(z)>0, -\frac{1}{2} \leq Re(z)\leq \frac{1}{2}, |z|\geq 1 \} $$ with orbifold points at $ i, e^{\pi i/3},e^{2\pi i/3} $. Using the natural circle bundle over $ H^2 \cong SL_2(\mathbb{R})/SO_2(\mathbb{R}) $ we can form a natural Seifert fiber bundle over this orbifold, which has geometry $ H^2 \times E^1 $.

Ian Gershon Teixeira
  • 10,176
  • 2
    Yes, ${\mathbb R}^2\times S^1$ and also ${\mathbb R}\times T^2$. The thing is, in the isometry group of every aspherical geometry you can find a discrete subgroup isomorphic to ${\mathbb Z}^2$. Then, by inspection, you verify that the quotient is not only homotopy-equivalent but also diffeomorphic to ${\mathbb R}\times T^2$. – Moishe Kohan Apr 08 '22 at 11:55
  • 1
    Your last bullet item is partially wrong: A finite volume hyperbolic manifold cannot admit any other (non-hyperbolic) geometry. – Moishe Kohan Apr 08 '22 at 14:51
  • @MoisheKohan Oh you're right let me change that. Also do you know if $ \mathbb{R}^3, \mathbb{R}^2×S^1,\mathbb{R}×T^2 $ is the complete list of 3 manifolds that admit all six aspherical geometries? Or are there other possibilities like $ K^2×\mathbb{R} ,L(S^1)×\mathbb{R},L(S^1)×S^1$ where $K^2 $ is the Klein bottle and $ L(S^1) $ is the nontrivial line bundle over the circle (the Moebius strip). – Ian Gershon Teixeira Apr 08 '22 at 15:42
  • 1
    $L(S^1)\times S^1$ does not admit a hyperbolic structure (since it is 1-ended). $L(S^1)\times R$ will admit all the geometries (you just need to check the existence of an orientation-preserving fixed-point free isometry for each geometry). As for $K^2\times R$, I am not sure about $Sol$ structure on this manifold, I think, there is none. – Moishe Kohan Apr 08 '22 at 22:56
  • 1
    Actually I think there is a $\text{Sol}$ structure on $K^2 \times \mathbb R$. In the usual model of $\text{Sol}$, namely $\mathbb R^3$ with metric $e^{2t} dx^2 + e^{-2t} dy^2 + dt^2$, the usual $\mathbb Z^2$ action is given by $(m,n) \cdot (x,y,t) = (x+m,y+n,t)$, and this action is isometric and gives $T^2 \times \mathbb R$ in the quotient. But there is also an isometry $(x,y,t) \mapsto (x+\frac{1}{2},-y,t)$ which, together with the $\mathbb Z^2$ action just described, generates a $\pi_1 K^2$-action whose quotient is $K^2 \times \mathbb R$. – Lee Mosher Apr 10 '22 at 17:22

0 Answers0