h-cobordism

From Wikipedia, the free encyclopedia
(Redirected from H-cobordism theorem)

In geometric topology and differential topology, an (n + 1)-dimensional cobordism W between n-dimensional manifolds M and N is an h-cobordism (the h stands for homotopy equivalence) if the inclusion maps

are homotopy equivalences.

The h-cobordism theorem gives sufficient conditions for an h-cobordism to be trivial, i.e., to be C-isomorphic to the cylinder M × [0, 1]. Here C refers to any of the categories of smooth, piecewise linear, or topological manifolds.

The theorem was first proved by Stephen Smale for which he received the Fields Medal and is a fundamental result in the theory of high-dimensional manifolds. For a start, it almost immediately proves the generalized Poincaré conjecture.

Background[edit]

Before Smale proved this theorem, mathematicians became stuck while trying to understand manifolds of dimension 3 or 4, and assumed that the higher-dimensional cases were even harder. The h-cobordism theorem showed that (simply connected) manifolds of dimension at least 5 are much easier than those of dimension 3 or 4. The proof of the theorem depends on the "Whitney trick" of Hassler Whitney, which geometrically untangles homologically-tangled spheres of complementary dimension in a manifold of dimension >4. An informal reason why manifolds of dimension 3 or 4 are unusually hard is that the trick fails to work in lower dimensions, which have no room for entanglement.

Precise statement of the h-cobordism theorem[edit]

Let n be at least 5 and let W be a compact (n + 1)-dimensional h-cobordism between M and N in the category C=Diff, PL, or Top such that W, M and N are simply connected. Then W is C-isomorphic to M × [0, 1]. The isomorphism can be chosen to be the identity on M × {0}.

This means that the homotopy equivalence between M and N (or, between M × [0, 1], W and N × [0, 1]) is homotopic to a C-isomorphism.

Lower dimensional versions[edit]

For n = 4, the h-cobordism theorem is true topologically (proved by Michael Freedman using a 4-dimensional Whitney trick) but is false PL and smoothly (as shown by Simon Donaldson).

For n = 3, the h-cobordism theorem for smooth manifolds has not been proved and, due to the 3-dimensional Poincaré conjecture, is equivalent to the hard open question of whether the 4-sphere has non-standard smooth structures.

For n = 2, the h-cobordism theorem is equivalent to the Poincaré conjecture stated by Poincaré in 1904 (one of the Millennium Problems[1]) and was proved by Grigori Perelman in a series of three papers in 2002 and 2003,[2][3][4] where he follows Richard S. Hamilton's program using Ricci flow.

For n = 1, the h-cobordism theorem is vacuously true, since there is no closed simply-connected 1-dimensional manifold.

For n = 0, the h-cobordism theorem is trivially true: the interval is the only connected cobordism between connected 0-manifolds.

A proof sketch[edit]

A Morse function induces a handle decomposition of W, i.e., if there is a single critical point of index k in , then the ascending cobordism is obtained from by attaching a k-handle. The goal of the proof is to find a handle decomposition with no handles at all so that integrating the non-zero gradient vector field of f gives the desired diffeomorphism to the trivial cobordism.

This is achieved through a series of techniques.

1) Handle rearrangement

First, we want to rearrange all handles by order so that lower order handles are attached first. The question is thus when can we slide an i-handle off of a j-handle? This can be done by a radial isotopy so long as the i attaching sphere and the j belt sphere do not intersect. We thus want which is equivalent to .

We then define the handle chain complex by letting be the free abelian group on the k-handles and defining by sending a k-handle to , where is the intersection number of the k-attaching sphere and the (k − 1)-belt sphere.

2) Handle cancellation

Next, we want to "cancel" handles. The idea is that attaching a k-handle might create a hole that can be filled in by attaching a (k + 1)-handle . This would imply that and so the entry in the matrix of would be . However, when is this condition sufficient? That is, when can we geometrically cancel handles if this condition is true? The answer lies in carefully analyzing when the manifold remains simply-connected after removing the attaching and belt spheres in question, and finding an embedded disk using the Whitney trick. This analysis leads to the requirement that n must be at least 5. Moreover, during the proof one requires that the cobordism has no 0-,1-,n-, or (n + 1)-handles which is obtained by the next technique.

3) Handle trading

The idea of handle trading is to create a cancelling pair of (k + 1)- and (k + 2)-handles so that a given k-handle cancels with the (k + 1)-handle leaving behind the (k + 2)-handle. To do this, consider the core of the k-handle which is an element in . This group is trivial since W is an h-cobordism. Thus, there is a disk which we can fatten to a cancelling pair as desired, so long as we can embed this disk into the boundary of W. This embedding exists if . Since we are assuming n is at least 5 this means that k is either 0 or 1. Finally, by considering the negative of the given Morse function, −f, we can turn the handle decomposition upside down and also remove the n- and (n + 1)-handles as desired.

4) Handle sliding

Finally, we want to make sure that doing row and column operations on corresponds to a geometric operation. Indeed, it isn't hard to show (best done by drawing a picture) that sliding a k-handle over another k-handle replaces by in the basis for .

The proof of the theorem now follows: the handle chain complex is exact since . Thus since the are free. Then , which is an integer matrix, restricts to an invertible morphism which can thus be diagonalized via elementary row operations (handle sliding) and must have only on the diagonal because it is invertible. Thus, all handles are paired with a single other cancelling handle yielding a decomposition with no handles.

The s-cobordism theorem[edit]

If the assumption that M and N are simply connected is dropped, h-cobordisms need not be cylinders; the obstruction is exactly the Whitehead torsion τ (W, M) of the inclusion .

Precisely, the s-cobordism theorem (the s stands for simple-homotopy equivalence), proved independently by Barry Mazur, John Stallings, and Dennis Barden, states (assumptions as above but where M and N need not be simply connected):

An h-cobordism is a cylinder if and only if Whitehead torsion τ (W, M) vanishes.

The torsion vanishes if and only if the inclusion is not just a homotopy equivalence, but a simple homotopy equivalence.

Note that one need not assume that the other inclusion is also a simple homotopy equivalence—that follows from the theorem.

Categorically, h-cobordisms form a groupoid.

Then a finer statement of the s-cobordism theorem is that the isomorphism classes of this groupoid (up to C-isomorphism of h-cobordisms) are torsors for the respective[5] Whitehead groups Wh(π), where

See also[edit]

Notes[edit]

  1. ^ "Millennium Problems | Clay Mathematics Institute". www.claymath.org. Retrieved 2016-03-30.
  2. ^ Perelman, Grisha (2002-11-11). "The entropy formula for the Ricci flow and its geometric applications". arXiv:math/0211159.
  3. ^ Perelman, Grisha (2003-03-10). "Ricci flow with surgery on three-manifolds". arXiv:math/0303109.
  4. ^ Perelman, Grisha (2003-07-17). "Finite extinction time for the solutions to the Ricci flow on certain three-manifolds". arXiv:math/0307245.
  5. ^ Note that identifying the Whitehead groups of the various manifolds requires that one choose base points and a path in W connecting them.

References[edit]