In mathematics, the Whitehead manifold is an open 3-manifold that is contractible, but not homeomorphic to ${\displaystyle \mathbb {R} ^{3}.}$ J. H. C. Whitehead (1935) discovered this puzzling object while he was 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.^[1]

Construction

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 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 ${\displaystyle S^{3))$ is another solid torus.

Now take a second 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 ${\displaystyle \mathbb {R} ^{4}.}$

David Gabai showed that X is the union of two copies of ${\displaystyle \mathbb {R} ^{3))$ whose intersection is also homeomorphic to ${\displaystyle \mathbb {R} ^{3}.}$^[1]

Related spaces

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

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 ${\displaystyle \mathbb {R} ^{1))$ is homeomorphic to ${\displaystyle \mathbb {R} ^{4}.}$