The Poincaré Conjecture, Precisely

Everything in this book so far has been preparation for one sentence. Here it is, with every word defined.

Henri Poincaré asked the question in 1904, at the end of the paper in which he found the homology sphere (7A.8 Homology in Brief). It resisted a century of attempts, became one of the seven Millennium Prize Problems of the Clay Mathematics Institute in 2000, and was proved by Grigori Perelman in three preprints posted in 2002 and 2003, using Richard Hamilton's Ricci flow. This chapter explains what the statement says, why each hypothesis is there, how it relates to the analogous statements in other dimensions, and why a theorem about topology could be proved by a method from differential geometry.

By the end of this chapter you will be able to:

• state the Poincaré conjecture and define every word in it, with a link to where it was built;
• give equivalent formulations, in terms of homotopy spheres;
• explain why each hypothesis is needed, with a counterexample for each;
• describe the analogous statements in other dimensions and who proved them;
• explain why topological and smooth three-manifolds are the same, so that Ricci flow can prove a topological theorem.

9.1The shape of space

9.2Every word

word	meaning	where
three-dimensional manifold	a Hausdorff, second countable space in which every point has a neighbourhood homeomorphic to $\mathbb{R}^3$	7A.3 Manifolds and Surfaces
closed	compact, and without boundary	7A.2 Compactness and Compactification, 7A.3 Manifolds and Surfaces
simply connected	path-connected, and every loop can be continuously shrunk to a point: $\pi_1 = 1$	7A.4 The Fundamental Group
homeomorphic	there is a continuous bijection with continuous inverse	7A.1 Topological Spaces and Quotients
the 3-sphere $S^3$	$\{x \in \mathbb{R}^4 : \lvert x\rvert = 1\}$, equivalently $\mathbb{R}^3\cup\{\infty\}$, or the unit quaternions	7A.2 Compactness and Compactification, 7A.6 Covering Spaces

Equivalent forms. A closed three-manifold $M$ that is homotopy equivalent to $S^3$ is called a homotopy 3-sphere. A homotopy 3-sphere is simply connected, since $\pi_1$ is a homotopy invariant. Conversely, a closed simply connected three-manifold has $\pi_2 = 0$ and $\pi_3 = \mathbb{Z}$, generated by a map $S^3 \to M$ of degree $1$ (7A.8 Homology in Brief, last exercise), and Whitehead's theorem (a map between simply connected cell complexes inducing isomorphisms on all homology groups is a homotopy equivalence) shows that this map is a homotopy equivalence. So the conjecture can be stated as: every homotopy 3-sphere is homeomorphic to $S^3$. It is also equivalent to: every closed three-manifold whose fundamental group is trivial is a sphere; or, using the results of Book 10A, every closed simply connected three-manifold admits a metric of constant positive curvature.

9.3Why each hypothesis is needed

drop	counterexample	why it fails
closed (allow non-compact)	$\mathbb{R}^3$	simply connected and a three-manifold, but not compact, so not $S^3$
closed (allow boundary)	the closed ball $B^3$	compact, simply connected, but it has a boundary
simply connected	$S^2\times S^1$, $\pi_1 = \mathbb{Z}$	closed, but has a non-contractible loop (7A.4 The Fundamental Group)
simply connected	lens spaces $L(p, q)$, $\pi_1 = \mathbb{Z}/p$	closed, with finite non-trivial $\pi_1$ (7A.6 Covering Spaces)
simply connected, replaced by homology	the Poincaré sphere $S^3/I^*$	same homology as $S^3$, but $\pi_1 = I^*$ of order $120$ (7A.8 Homology in Brief)
dimension 3 (dimension 2)	none: true, by classification	the only closed simply connected surface is $S^2$ (7A.5 Computing π₁)

The table shows which conditions do real work. Non-compact simply connected three-manifolds can be very strange: John Henry Constant Whitehead found in 1935 an open subset of $\mathbb{R}^3$, contractible but not homeomorphic to $\mathbb{R}^3$ (the Whitehead manifold), while trying to prove the conjecture. Simple connectivity cannot be weakened to "has the homology of a sphere", by Poincaré's own example. Finite fundamental group is not enough either: the spherical space forms $S^3/\Gamma$ are all closed with finite $\pi_1$, and Perelman's theorem includes the statement that they are the only ones (the elliptization conjecture, 10A.5 Thurston’s Eight Geometries).

9.4Other dimensions

The generalized Poincaré conjecture asks, in each dimension $n$, whether every closed manifold homotopy equivalent to $S^n$ is homeomorphic to $S^n$. (For $n \geq 4$, simply connected is not enough, as $S^2\times S^2$ shows, so the statement uses homotopy equivalence.)

• $n = 1, 2$: true, by the classification of curves and surfaces (7A.3 Manifolds and Surfaces).
• $n \geq 5$: true. Stephen Smale proved it in 1961 (with John Stallings and Christopher Zeeman independently proving cases by other methods), using handle decompositions (7A.10 Morse Theory); there is so much room in high dimensions that handles can be cancelled. Smale received the Fields Medal in 1966.
• $n = 4$: true. Michael Freedman proved it in 1982, with entirely different, topological methods; Fields Medal 1986.
• $n = 3$: true. Perelman, 2002–03.

Smooth versus topological. In some dimensions the smooth version, "homeomorphic" replaced by "diffeomorphic", fails: John Milnor found in 1956 smooth manifolds homeomorphic but not diffeomorphic to $S^7$, the exotic spheres. In dimension $4$ the smooth Poincaré conjecture is open: no one knows whether there is a smooth manifold homeomorphic but not diffeomorphic to $S^4$. In dimension $3$ the question does not arise, as the next section explains. So dimension three was the last case of the topological conjecture to be settled, and in an important sense the hardest: the high dimensions have room for handle-cancelling tricks, dimension four allows Freedman's infinite constructions, and dimension three has neither (Figure 9.1).

9.5Topological and smooth three-manifolds

The Ricci flow is a smooth tool: it differentiates a Riemannian metric, which requires a smooth structure on the manifold (8A.1 Smooth Structures, 9A.1 Riemannian Metrics and Model Spaces). The Poincaré conjecture is a topological statement. The bridge is a theorem special to low dimensions:

Edwin Moise proved in 1952 that every three-manifold can be triangulated, uniquely up to subdivision, and results of James Munkres and J. H. C. Whitehead (around 1960) convert triangulations into smooth structures and back (8A.1 Smooth Structures). So a closed simply connected topological three-manifold can be given a smooth structure, then a Riemannian metric (by a partition of unity, 8A.2 Partitions of Unity), and then the Ricci flow can be run. If the flow shows the smooth manifold is diffeomorphic to $S^3$, it is in particular homeomorphic to $S^3$. In dimension $4$ this bridge fails, which is one reason the smooth four-dimensional case is so different.

9.6History

Poincaré posed the question in 1904. Over the twentieth century many proofs were announced and withdrawn; Whitehead's 1935 attempt produced the Whitehead manifold, and the search drove much of three-manifold topology, from Dehn's lemma (proved by Christos Papakyriakopoulos in 1957) to Thurston's geometrization conjecture (1982), which contains the Poincaré conjecture (10A.5 Thurston’s Eight Geometries). Hamilton introduced the Ricci flow in 1982 and proposed it as a route to geometrization. Perelman posted his three preprints on the arXiv in November 2002, March 2003 and July 2003. Detailed expositions by Bruce Kleiner and John Lott, by John Morgan and Gang Tian, and by Huai-Dong Cao and Xi-Ping Zhu appeared in 2006–08. Perelman was awarded the Fields Medal in 2006 and the Clay Millennium Prize in 2010, and declined both.

9.7Exercises

Solution

$\mathbb{R}^3$ and $S^3\setminus\{\text{point}\} \cong \mathbb{R}^3$ (7A.2 Compactness and Compactification): not compact. $S^2\times S^1$: $\pi_1 = \mathbb{Z}$ (7A.4 The Fundamental Group). $L(5, 1)$: $\pi_1 = \mathbb{Z}/5$ (7A.6 Covering Spaces). $T^3$: $\pi_1 = \mathbb{Z}^3$. Poincaré sphere: $\pi_1 = I^*$, order $120$ (7A.6 Covering Spaces, 7A.8 Homology in Brief).

Solution

(a) $\pi_1$ is a homotopy invariant and $\pi_1(S^3) = 1$. (b) $\pi_1(S^2\times S^2) = \pi_1(S^2)^2 = 1$, but $H_2(S^2\times S^2) = \mathbb{Z}^2$ while $H_2(S^4) = 0$. In dimension $4$ and above, simple connectivity doesn't control the middle homology, so the right hypothesis is that $M$ looks like a sphere to all homotopy invariants. In dimension $3$, Poincaré duality and Hurewicz make simple connectivity enough (7A.8 Homology in Brief).

Solution

Two spheres of radius $R$ whose centres are $L < 2R$ apart meet in a circle, in the plane halfway between them; the points on it are seen from the observer along two different directions (one directly, one as the image of the translated copy), so the same physical points appear on two circles on the sky. The circle is at distance $\frac L2$ from the observer's centre along the axis, so its angular radius $\alpha$ satisfies $\cos\alpha = \frac{L}{2R}$.

Solution

(a) The deck group of the universal cover $S^3 \to M$ is $\pi_1(M)$ (7A.6 Covering Spaces), it acts freely, and since the covering map is a local isometry the deck transformations are isometries of $S^3$. (b) If $\pi_1(M) = 1$, the covering $S^3 \to M$ has one sheet, so it is an isometry.