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 :).

  • 6
    $\begingroup$ The Euler characteristic of an odd-dimensional closed manifold is zero. $\endgroup$
    – abx
    May 16 '17 at 13:21
  • 1
    $\begingroup$ 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. $\endgroup$ May 16 '17 at 13:33
  • $\begingroup$ @abx Uhh, thanks for pointing out that (for me) long forgotten fact! $\endgroup$
    – MBIS
    May 16 '17 at 20:06

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.

  • 2
    $\begingroup$ 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. $\endgroup$ May 16 '17 at 14:00
  • 1
    $\begingroup$ Sorry this is three years late, but I cannot reconcile something basic here: how can the integral Euler number be bordism invariant when $\chi(S^2)\neq 0$, while $S^2$ is nullbordant? And more generally, even without an almost-complex structure, any $SO(2n)$-bundle can formally split into $SO(2)\cong U(1)$-manifolds. One can then define the Euler class in terms of these formal "Chern roots," but again this can't possibly be a bordism invariant. $\endgroup$
    – Elliot G
    Sep 3 '20 at 7:03
  • 2
    $\begingroup$ $S^2$ is nullbordant as an unoriented manifold, but $\mathbb{CP}^1$ is not nullbordant as a stably almost complex manifold; its stably a.c. structure is different from the one arising from $S^2 = \partial B^3$. See here, Example 2.1 and section 5. $\endgroup$ Sep 3 '20 at 18:39
  • 1
    $\begingroup$ @ArunDebray Thanks for the reply and references; they pretty much clear it up. In retrospect, my confusion was from the fact that the Euler characteristic is well defined for $SO(2n)$-bundles, but not an $SO$-bordism invariant. As you said, the definition really just borrows from the stable a.c. setting, so there isn't really any reason it should be an oriented bordism invariant. $\endgroup$
    – Elliot G
    Sep 3 '20 at 21:00
  • 1
    $\begingroup$ @ArunDebray This is probably implicit in your answer, but just to point out that the Euler characteristic is the Chern number $c_n$ only if the stable almost complex structure is induced by an almost complex structure. As you point out, the trivial stable ac structure on $S^2$ has $c_1 = 0$ (and so this is complex nullbordant), so Euler characteristic is not invariant under complex cobordisms between general stably almost complex manifolds; it is true that it is invariant under complex cobordisms between genuine almost complex manifolds. $\endgroup$ Apr 5 at 16:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.