Whitehead manifold



In mathematics, the Whitehead manifold is an open 3-manifold that is contractible, but not homeomorphic to $$\R^3.$$ discovered this puzzling object while he was trying to prove the Poincaré conjecture, correcting an error in an earlier paper  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 a circle times a disk.) The closed complement of the solid torus inside $$S^3$$ is another solid torus.

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 $$\R^3 \cup \{\infty\}$$ and the meridian curve as the z-axis together with $$\infty.$$  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 $$W = T_\infty,$$ 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, $$\left(\R^3/W\right) \times \R$$ 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 $$\R^4.$$