General Theorems on Bounding Manifolds?

[WWGD]

Main Question or Discussion Point

Hi, there is a result that every closed, oriented 3-manifold is the boundary of a 4-manifold that has only 0- and 2- handles. Anyone know other of these "boundary results" for some higher-dimensional manifolds, e.g., every closed, oriented k-manifold is the boundary of a (k+1)- dimensional manifold..... ?
Thanks.

 

[lavinia]

Hi, there is a result that every closed, oriented 3-manifold is the boundary of a 4-manifold that has only 0- and 2- handles. Anyone know other of these "boundary results" for some higher-dimensional manifolds, e.g., every closed, oriented k-manifold is the boundary of a (k+1)- dimensional manifold..... ?
Thanks.

Don't know about the handles but every closed oriented smooth 3 manifold without boundary bounds an oriented 4 manfold.

An oriented k manifold is generally not a boundary. For example, complex projective 2 space is not a boundary. The cobordism groups are known but not sure of a definitive reference.

Non-orientable manifolds may also be boundaries for example the Klein bottle is a boundary. The real projective plane is not.

The general theorem is that a manifold is an unoriented boundary if and only if all of its Stiefel Whitney numbers are zero. In particular, the manifold must have even Euler characteristic. The Euler characteristic of the Klein bottle is zero. The Euler characteristic of the projective plane is 1.

Two oriented manifolds are oriented cobordant if they form the oriented boundary of an oriented manifold. I think that Thom's theorem says that mod torsion the oriented cobordism group is a polynomial algebra generated by the even dimensional complex projective spaces and that Wall has shown that all of the torsion is of order 2. Two oriented manifolds are oriented cobordant if an only if all of their Pontryagin numbers and all of their Stiefel Whiteney numbers are equal. Note that this is an added condition since unoriented cobordism only requires that all of the Stiefel Whitney numbers be equal.

For example consider an orientable flat Riemannian manifold. Since it can be given a metric whose curvature tensor is identically zero, and since the Weil homomorphism proves that all Pontryagin classes can be expressed as differential forms in the curvature 2 form, all of their Pontryagin numbers are zero. So two flat Riemannian manifolds are oriented cobordant if and only if they have the same Stiefel Whitney numbers - which in fact, they always do.

It is important to point out that these theorems apply to smooth manifolds. I don't know what the corresponding theories say about piece wise linear manifolds and non-differentiable manifolds.