# Cobordisms and Euler characteristics

I am trying to understand exactly which role the Euler characteristic plays in (smooth) cobordism theory, and especially why the answer seems to depend on the dimensions of the manifolds in question. Suppose $M$ and $N$ are two closed smooth $n$-manifolds and that we have a smooth cobordism $(W;M,N)$ (i.e. $W$ is a smooth compact $(n+1)$-manifold with boundary $\partial W=M\sqcup N$). Poincare-Lefschetz duality then predicts that $H_k(W,N;\mathbb{Z}_2)\cong H_{n+1-k}(W,M;\mathbb{Z}_2)$ for every $k\in \mathbb{Z}$. In particular, if we denote by $\chi(W,M)$ the Euler characteristic of the homology $H_{*}(W,M;\mathbb{Z}_2)$ we see that $\chi(W,N)=(-1)^{n+1}\chi(W,M)$. By additivity of the Euler characteristic we also have $$\chi(W,N)+\chi(N)=\chi(W)=\chi(W,M)+\chi(M).$$ If $n$ is odd we therefore conclude that $\chi(M)=\chi(N)$, so in this case $\chi$ is a cobordism invariant. My question is:

What happens when $n$ is even? Is the Euler useful in any way for understanding cobordisms of even dimensional manifolds? Any interesting statement would be much appreciated, even if additional assumptions such as orientability, spin etc. are needed. (Note that the sphere $S^2$ is null-cobordant, the cobordism given by the 3-ball, so clearly $\chi$ is not a cobordism invariant of even dimensional manifolds...)

Thanks in advance :).

• The Euler characteristic of an odd-dimensional closed manifold is zero. – abx May 16 '17 at 13:21
• Yes. Consider the case when $M=\emptyset=N$. And in even dimensions the mod $2$ Euler number is a cobordism invariant. And, as your $S^2$ example shows, mod $2$ is the best you can do: it does not get better with orientability, spin, ... or even stable parallelizability. – Tom Goodwillie May 16 '17 at 13:33
• @abx Uhh, thanks for pointing out that (for me) long forgotten fact! – MBIS May 16 '17 at 20:06

## 1 Answer

René Thom proved that two closed $n$-manifolds $M$ and $N$ are (unoriented) cobordant iff their Stiefel-Whitney numbers agree: for any partition $i_1 + \dotsb + i_k = n$,

$$[M]\frown w_{i-1}(M)w_{i_2}(M)\dotsm w_{i_k}(M) = [N]\frown w_{i-1}(N)w_{i_2}(N)\dotsm w_{i_k}(N).$$

The mod 2 Euler characteristic is a Stiefel-Whitney number: $\chi(M)\bmod 2 = [M]\frown w_n(M)$. Thus, if $M$ and $N$ are unoriented cobordant, then their Euler characteristics differ by a multiple of 2. (There might be a more direct proof of this using handles.)

You can get rid of the mod 2 assumption if you work with (stably almost) complex cobordism, equipping all manifolds and cobordisms with a complex structure on the stable normal bundle. (In particular, this determines an orientation of your manifold.) Milnor and Novikov showed that $M$ and $N$ are complex cobordant iff their Chern numbers agree: these are defined in the same way as Stiefel-Whitney numbers, but for Chern classes.

The Euler characteristic is a Chern number: $\chi(M) = [M]\frown c_n(M)$ if $M$ is $2n$-dimensional, and $\chi(M) = [M]\frown 0$ if $M$ is odd-dimensional. Thus in any dimension, the Euler characteristic is a cobordism invariant for stably almost complex cobordism.

• Here's a different proof of the first statement, in the even-dimensional case. A handle decomposition induces a sequence of surgeries; at each step, you remove $S^k\times D^{n-k}$ and glue $D^{k+1} \times S^{n-k-1}$. When $n$ is even, the Euler characteristics of the two pieces differ by 2, and the boundary is odd-dimensional, so has Euler characteristics 0. – Marco Golla May 16 '17 at 14:00
• @ArunDebray Thanks a lot for this comment. It will certainly be useful for me. Do you by any chance know a good reference (preferably easily accessible) for learning about complex manifolds? Btw. for $n$ even it also follows form my computation that $\chi(N)=\chi(M)+2\chi(W,M)$ which directly implies $\chi(N)=\chi(M)$ (mod 2). – MBIS May 16 '17 at 20:47
• @MadsR.Bisgaard I don't know much, but here's a reasonable starting point. – Arun Debray May 16 '17 at 21:11
• @ArunDebray Thanks a lot (and as you correctly figured: by "complex manifold" I of course meant "complex bordism") :). – MBIS May 16 '17 at 22:01