Whitehead manifold
In mathematics, the Whitehead manifold is an open 3-manifold that is contractible, but not homeomorphic to <math>\R^3.</math> It was discovered by Template:Harvs while 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.<ref name="Gabai">Template:Cite journal</ref>
Construction
Take a copy of <math>S^3,</math> the three-dimensional sphere. Now find a compact unknotted solid torus <math>T_1</math> inside the sphere. (A solid torus is the topological space <math>S^1\times D^2</math>. 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 <math>S^3</math> is another solid torus.
Now take a second unknotted solid torus <math>T_2</math> inside <math>T_1</math> so that <math>T_2</math> and a tubular neighborhood of the meridian curve of <math>T_1</math> is a thickened Whitehead link.
Note that <math>T_2</math> is null-homotopic in the complement of the meridian of <math>T_1.</math> This can be seen by considering <math>S^3</math> as <math>\R^3 \cup \{\infty\}</math> and the meridian curve as the z-axis together with <math>\infty.</math> The torus <math>T_2</math> 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 <math>T_1</math> is also null-homotopic in the complement of <math>T_2.</math>
Now embed <math>T_3</math> inside <math>T_2</math> in the same way as <math>T_2</math> lies inside <math>T_1,</math> and so on; to infinity. Define W, the Whitehead continuum, to be <math>W = T_\infty,</math> or more precisely the intersection of all the <math>T_k</math> for <math>k = 1,2,3,\dots.</math>
The Whitehead manifold is defined as <math>X = S^3 \setminus W,</math> 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 <math>X \times \R \cong \R^4.</math> However, X is not homeomorphic to <math>\R^3.</math> The reason is that it is not simply connected at infinity.
The one point compactification of X is the space <math>S^3/W</math> (with W crunched to a point). It is not a manifold. However, <math>\left(\R^3/W\right) \times \R</math> is homeomorphic to <math>\R^4.</math>
David Gabai showed that X is the union of two copies of <math>\R^3</math> whose intersection is also homeomorphic to <math>\R^3.</math><ref name="Gabai" />
Related spaces
More examples of open, contractible 3-manifolds may be constructed by proceeding in similar fashion and picking different embeddings of <math>T_{i+1}</math> in <math>T_i</math> in the iterative process. Each embedding should be an unknotted solid torus in the 3-sphere. The essential properties are that the meridian of <math>T_i</math> should be null-homotopic in the complement of <math>T_{i+1},</math> and in addition the longitude of <math>T_{i+1}</math> should not be null-homotopic in <math>T_i \setminus T_{i+1}.</math>
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 <math>\R^1</math> is homeomorphic to <math>\R^4.</math>