# What is Bordism good for?

I am doing a course in algebraic topology, and recently we have defined bordism in the following way:

Let $$M_1,M_2$$ be connected oriented n-dimensional manifolds, $$f_i:M_i\rightarrow X$$ be continuous maps. $$(M_1,f_1)$$ and $$(M_2,f_2)$$ are (oriented) bordant if there is an oriented compact (n+1)-dimensional manifold $$N$$ s.t. $$\partial N = M_1 \sqcup \bar M_2$$ and a continuous map $$F:N\rightarrow X$$ s.t. it restricts to $$f_{1}$$ and $$f_2$$ on the respective parts of the boundary.

We have used this definition to construct the abelian group $$\Omega _n(X)$$ of the bordism classes (the sum being induced by the disjoint union) and to construct a map $$T_X : \Omega_n(X)\rightarrow H_n(X)$$ defined by $$[(M,F)]\mapsto f_*[M]$$ , where $$f_*$$ is the pushforward in homology and $$[M]$$ is the fundamental class of M.

We have not done much else in class (and we are not following any book), so I was wandering what this might be useful for. First of all: how is the map $$T_X$$ useful? It seems to me that the only reason why we required our manifolds to be orientable and compact is to have a fundamental class, so that we can define this map, hence it must be important.

The second question would be: what is bordism good for? What can I do with it?

Thank you.

• I think in 1952 René Thom was able to show that each cobordism group is finitely generated with 2-torsion (... and probably did a lot more on the subject ;-) ). – user408858 Jul 15 at 11:53

Let me answer your question with another question:

What is any invariant (e.g. homology?) good for?

I really like bordism, because to me it is a very geometric invariant and I have intuition for it. Looking at all singular cycles

Btw, have you ever drawn explicit cycles representing homology classes in surfaces? I encourage you to do this for $$H_1(\Sigma_g;\mathbb Z)$$. Also show that the "waist" of a genus two surface is null-homologues.

Then you might find that any class can be represented by some union of triangulated circles: i.e. any class lies in the image of $$T_X$$: The cycles are represented by $$1$$ dimensional manifolds. This is so much nicer than any singular cycle. In fact the earliest attempts to define homology were actually defining bordisms!

A natural question is then if this is always possible: Can any singular cycle be represented by a singular manifold (i.e. a map from a manifold to your space)? This is Steenrod's problem. The answer is no for $$H_*(X;\mathbb Z)$$, but it is true for $$H_*(X;\mathbb{Z}_2)$$, where you need to use unoriented bordism to define an anogues map. This was proven by Thom.

But Thom did so much more. While the homology of a point is simple: $$H_*(\mathrm{pt};\mathbb{Z}_2)=\mathbb Z_2$$, the unoriented bordism group $$\mathcal{N}_*(\mathrm{pt})$$ is much more complicated: There are many manifolds which are not the boundary of a higher dimensional manifold. An example is $$\mathbb{RP}^2$$, but I do not want to prove this here. Anyway: Thom managed to completely compute the unoriented cobordism ring, by showing that it is equal to a certain homotopy group. His techniques can also be extended to compute many other cobordism rings. I find this one of the most beautiful stories in mathematics.

Finally I want to remark that cobordism groups naturally pop up in other problems. Here is a baby example: Take a smooth map $$f:M\rightarrow N$$ between finite closed manifolds. Sard's theorem says that there are always many regular values. The preimage $$X=f^{-1}(x)$$ is then a closed submanifold of $$M$$. While this submanifold $$X$$ depends on the regular value chosen, it's cobordism class does not. And it also does not depend on the homotopy class of $$f$$. So there is a natural map $$[M,N]\rightarrow \mathcal{N}_*(M)$$ by taking the preimage of a regular value. To use this invariant, you better be able to compute (part of) the right hand side. Since the work of Thom, we understand how this relates to the homology of $$M$$. You can take this idea much further.