# Relationships between homology maps of cobordant manifolds

Let $W$ be a cobordism between $n$-manifolds $M$ and $N$, and $f\colon W \mapsto X$ be a map to some manifold $X$.

Does anybody know of any nice examples of general relationships between the images of the maps $g_*\colon H_*(M)\mapsto H_*(X)$ and $h_*\colon H_*(N)\mapsto H_*(X)$ induced by the restrictions $g=f\mathrel{|M}$ and $h=f\mathrel{|N}$, based on some given data about about these manifolds?

For Example, if $M=N$ and $W=[0,1]\times M$, then $f$ is a homotopy between $g$ and $h$, so $g_*=h_*$. Also, if our manifolds are oriented and compact, $X$ is a connected $n$-manifold, and $f$ is smooth, then the degree of $f\mathrel{\mathrel{|}\partial W}=f\mathrel{\mathrel{|}M\coprod N}$ is zero, so the degrees of $g$ and $h$ are equal up to sign, as are the images of $g_*$ and $h_*$ in degree $n$ homology.

• Aren't we supposed to have $g_*[M]=h_*[N]$, where $[M]$ and $[N]$ are fundamental classes? – user43326 Dec 9 '17 at 19:59

## 3 Answers

If we consider cohomology with $\mathbb{Z}/2$-coefficients, if $M$ is a manifold $H^*(M;\mathbb{Z}/2)$ it satisfies Poincaré duality, and is an unstable algebra over the Steenrod algebra. Brown and Peterson in their paper "Algebraic bordism groups" (Annals of maths, 1964) had the wonderful idea to give an algebraic model of geometric bordisms in terms of algebraic bordisms of unstable algebras over the Steenrod algebra.

They showed that this algebraic bordism group $N^{alg}_*$ is isomorphic to Thom's bordism group of smooth manifolds. For any unstable algebra algebra $A$ they define an algebraic bordism group $N^{alg}_*(A)$. In fact, they proved that for any space $K$, Conner, Floyd's bordism group $N_*(K)$ of non-oriented smooth manifold over $K$ is isomorphic to $N^{alg}_*(H^*(K;\mathbb{Z}/2))$. The isomorphism is given by a natural transformation: $$\phi_K:N_*(K)\rightarrow N^{alg}_*(H^*(K;\mathbb{Z}/2))$$ that sends a bordism class $f:M\rightarrow K$ to the algebraic bordism class of the induced map $f^*:H^*(K;\mathbb{Z}/2)\rightarrow H^*(M;\mathbb{Z}/2)$.

Not only thanks to Wu formulas (that use Poincaré duality and Steenrod squares action on $H^*(M;\mathbb{Z}/2)$), you can recover Stiefel-Whitney numbers, and you can determine if two manifolds are cobordant but using the unstable algebra structure of $H^*(K;\mathbb{Z}/2)$ you can recover the geometric bordism group of $K$.

You can say a few things using Poincaré duality and the exact sequence of the pair $(W,\partial W)$. For instance, $M$ is $2k$-dimensional (and everything is oriented), then the signatures of $M$ and $N$ are the same. The standard proof of this says among other things that the rank of the kernel of the inclusion map on $H_{k}(M) \oplus H_{k}(N) \to H_k(W)$ is half of the sum of the ranks of $H_{k}(M)$ and $H_{k}(N)$, and the intersection form vanishes on that kernel. This is a sort of relation between those two maps, ie you couldn't have the inclusion of $H_{k}(M)$ be the $0$ map and that of $H_{k}(N)$ be an injection. There is a similar statement in other dimensions as well.

This idea is sometimes (particularly in reference to $k=1$) described by the slogan `half lives, half dies' as discussed in the answers to this MO question.

I doubt that anyone has written this down in so many words, but I would bet that basically this is about all you can say, at least if you take rational coefficients. In other words, anything that's compatible with Poincaré duality and exact sequences can be realized by a pair of cobordant manifolds. If you want to work over the integers, then there are some more subtle issues relating to torsion classes.

You might find it easier to rephrase the question a little and ask, for a manifold with boundary $(W,\partial W)$, what you can say about the maps induced in homology by the inclusion of the boundary.

You may be able to glean some information from thinking about the Stiefel-Whitney and Pontrjagin numbers of your maps $g$ and $h$. This is explained in Section 17 of Conner and Floyd's "Differentiable Periodic Maps".

Here is the situation in the non-oriented case for mod 2 homology. Since your maps $g:M\to X$ and $h:N\to X$ are bordant, their Stiefel-Whitney numbers all coincide. This means that for any cohomology class $x\in H^k(X;\mathbb{Z}/2)$, and any multi-index $(i_1,\ldots , i_\ell)$ with each $i_j\ge0$ and $i_1 + \cdots + i_\ell = n-k$, we have $$\langle w_{i_1}(M)\cdots w_{i_\ell}(M)g^*(x),[M]\rangle = \langle w_{i_1}(N)\cdots w_{i_\ell}(N)h^*(x),[N]\rangle.$$ It follows that $$\langle g^*(x),w_{i_1}(M)\cdots w_{i_\ell}(M)\cap [M]\rangle = \langle h^*(x),w_{i_1}(N)\cdots w_{i_\ell}(N)\cap[N]\rangle,$$ and that $$\langle x,g_*(w_{i_1}(M)\cdots w_{i_\ell}(M)\cap [M])\rangle = \langle x,h_*(w_{i_1}(N)\cdots w_{i_\ell}(N)\cap[N])\rangle,$$ Since cohomology is dual to homology with mod 2 coefficients, this means that the Poincaré duals of monomials in the Stiefel-Whitney classes have the same image in homology.