Whitehead manifold

First three tori of Whitehead manifold construction

In mathematics, the Whitehead manifold is an open 3-manifold that is contractible, but not homeomorphic to ${ℝ}^3.$ Template:Harvs discovered this puzzling object while he was trying to prove the Poincaré conjecture, correcting an error in an earlier paper Template:Harvtxt where he incorrectly claimed that no such manifold exists.

A contractible manifold is one that can continuously be shrunk to a point inside the manifold itself. For example, an open ball is a contractible manifold. All manifolds homeomorphic to the ball are contractible, too. One can ask whether all contractible manifolds are homeomorphic to a ball. For dimensions 1 and 2, the answer is classical and it is "yes". In dimension 2, it follows, for example, from the Riemann mapping theorem. Dimension 3 presents the first counterexample: the Whitehead manifold.

Construction

Take a copy of $S^3,$ the three-dimensional sphere. Now find a compact unknotted solid torus $T_1$ inside the sphere. (A solid torus is an ordinary three-dimensional doughnut, that is, a filled-in torus, which is topologically the product of a circle and a disk.) The closed complement of the solid torus inside $S^3$ is another solid torus.

A thickened Whitehead link. In the Whitehead manifold construction, the blue (untwisted) torus is a tubular neighborhood of the meridian curve of $T_1$, and the orange torus is $T_2.$ Everything must be contained within $T_1.$

Now take a second solid torus $T_2$ inside $T_1$ so that $T_2$ and a tubular neighborhood of the meridian curve of $T_1$ is a thickened Whitehead link.

Note that $T_2$ is null-homotopic in the complement of the meridian of $T_1.$ This can be seen by considering $S^3$ as ${ℝ}^3\cup \{\mathrm{\infty }\}$ and the meridian curve as the z-axis together with $\mathrm{\infty }.$ The torus $T_2$ has zero winding number around the z-axis. Thus the necessary null-homotopy follows. Since the Whitehead link is symmetric, that is, a homeomorphism of the 3-sphere switches components, it is also true that the meridian of $T_1$ is also null-homotopic in the complement of $T_2.$

Now embed $T_3$ inside $T_2$ in the same way as $T_2$ lies inside $T_1,$ and so on; to infinity. Define W, the Whitehead continuum, to be $W=T_{\mathrm{\infty }},$ or more precisely the intersection of all the $T_k$ for $k=1,2,3,\dots .$

The Whitehead manifold is defined as $X=S^3\setminus W,$ which is a non-compact manifold without boundary. It follows from our previous observation, the Hurewicz theorem, and Whitehead's theorem on homotopy equivalence, that X is contractible. In fact, a closer analysis involving a result of Morton Brown shows that $X\times ℝ\cong {ℝ}^4.$ However, X is not homeomorphic to ${ℝ}^3.$ The reason is that it is not simply connected at infinity.

The one point compactification of X is the space $S^3/W$ (with W crunched to a point). It is not a manifold. However, $({ℝ}^3/W)\times ℝ$ is homeomorphic to ${ℝ}^4.$

David Gabai showed that X is the union of two copies of ${ℝ}^3$ whose intersection is also homeomorphic to ${ℝ}^3.$

Related spaces

More examples of open, contractible 3-manifolds may be constructed by proceeding in similar fashion and picking different embeddings of $T_{i+1}$ in $T_i$ in the iterative process. Each embedding should be an unknotted solid torus in the 3-sphere. The essential properties are that the meridian of $T_i$ should be null-homotopic in the complement of $T_{i+1},$ and in addition the longitude of $T_{i+1}$ should not be null-homotopic in $T_i\setminus T_{i+1}.$

Another variation is to pick several subtori at each stage instead of just one. The cones over some of these continua appear as the complements of Casson handles in a 4-ball.

The dogbone space is not a manifold but its product with ${ℝ}^1$ is homeomorphic to ${ℝ}^4.$

See also

• List of topologies
• Tame manifold

Further reading

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