Whitehead manifold

In mathematics, the Whitehead manifold is an open 3-manifold that is contractible, but not homeomorphic to ${\displaystyle \mathbb {R} ^{3}.}$ It was discovered by J. H. C. Whitehead (1935) while trying to prove the Poincaré conjecture, correcting an error in an earlier paper Whitehead (1934, theorem 3) 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.

Take a copy of ${\displaystyle S^{3},}$ the three-dimensional sphere. Now find a compact unknotted solid torus ${\displaystyle T_{1}}$ inside the sphere. (A solid torus is the topological space ${\displaystyle S^{1}\times D^{2}}$. Intuitively, it 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 unknotted solid torus inside ${\displaystyle S^{3}}$ is another solid torus.

Now take a second unknotted solid torus ${\displaystyle T_{2}}$ inside ${\displaystyle T_{1}}$ so that ${\displaystyle T_{2}}$ and a tubular neighborhood of the meridian curve of ${\displaystyle T_{1}}$ is a thickened Whitehead link.

Note that ${\displaystyle T_{2}}$ is null-homotopic in the complement of the meridian of ${\displaystyle T_{1}.}$ This can be seen by considering ${\displaystyle S^{3}}$ as ${\displaystyle \mathbb {R} ^{3}\cup \{\infty \}}$ and the meridian curve as the z-axis together with ${\displaystyle \infty .}$ The torus ${\displaystyle T_{2}}$ 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 ${\displaystyle T_{1}}$ is also null-homotopic in the complement of ${\displaystyle T_{2}.}$

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

The Whitehead manifold is defined as ${\displaystyle X=S^{3}\setminus W,}$ 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 ${\displaystyle X\times \mathbb {R} \cong \mathbb {R} ^{4}.}$ However, X is not homeomorphic to ${\displaystyle \mathbb {R} ^{3}.}$ The reason is that it is not simply connected at infinity.

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