12
$\begingroup$

The cobordism group of 5-dimensional closed oriented manifolds is $\Omega_5^{SO}=Z_2$, which is generated by $SU(3)/SO(3)$. A mapping torus is a fiber bundle over $S^1$. Can $\Omega_5^{SO}$ be generated by a 5-dimensional mapping torus?

$\endgroup$
18
$\begingroup$

The mapping torus $T$ of the complex-conjugation-map $\mathbb{C}P^2 \rightarrow \mathbb{C}P^2$ does the job.

For example by running the Serre spectral sequence with local coefficients, you obtain that the integral cohomology $H^0(T,\mathbb{Z})$, $H^1(T,\mathbb{Z})$, $H^2(T,\mathbb{Z})$, ... of this mapping torus are given by $\mathbb{Z}, \mathbb{Z}, 0, \mathbb{Z}/2, \mathbb{Z}, \mathbb{Z}$. The mod $2$ cohomology $H^0(T,\mathbb{Z}_2)$, $H^1(T,\mathbb{Z}_2)$, $H^2(T,\mathbb{Z}_2)$, ... are therefore given by $\mathbb{Z}/2, \mathbb{Z}/2, \mathbb{Z}/2, \mathbb{Z}/2, \mathbb{Z}/2, \mathbb{Z}/2$, where the two middle classes are connected by a nontrivial action of $Sq^1$, and $Sq^1$ acts trivially everywhere else.

The class in degree $3$ has a nontrivial action of $Sq^2$. To see this, note that this class is in the image of the homomorphism induced by $T\rightarrow (T,\mathbb{C}P^2)$. Now the latter pair is homeomorphic to the pair associated to the trivial bundle, but there you get the statement about $Sq^2$ from the fact that $Sq^2$ commutes with suspension (or something related). Similarly, the class in degree $2$ has a nontrivial action of $Sq^2$.

This determines the action of the Steenrod algebra on $T$ completely. Now by the usual Wu class arguments, you can get from that that $w_2$ and $w_3$ are nontrivial, and by again invoking that $Sq^2$ acts nontrivial on the degree $3$-cohomology, you can use the Wu formula to get that $w_2w_3$ is nonzero.

This is the Stiefel-Whitney number which detects the generator of $\Omega_5^{SO}$.

$\endgroup$
  • $\begingroup$ Thank you very much for the detailed answer! I have a very simple question: is the degree (−1)-map $\mathbb{C}P^2 \rightarrow \mathbb{C}P^2$ reverse the orientation of $\mathbb{C}P^2$ that makes the corresponding mapping torus un-oriented? Here we need an oriented manifold to generate $\Omega_5^{SO}$. $\endgroup$ – Xiao-Gang Wen Apr 29 '14 at 12:01
  • $\begingroup$ I also have a related more general question: what are the characteristic classes that can detect the cobordism group of mapping tori (in some low dimensions). See mathoverflow.net/questions/164513/… . Your $w_2w_3$ is one of the characteristic classes in 5-dimension. Do you know other characteristic classes? $\endgroup$ – Xiao-Gang Wen Apr 29 '14 at 12:13
  • 1
    $\begingroup$ The map $\mathbb{C}P^2\rightarrow\mathbb{C}P^2$ induced by complex conjugation (which is the one we used) still preserves the orientation, as there is no orientation-reversing self-diffeomorphism of $\mathbb{C}P^n$ for $n$ even. $\endgroup$ – Achim Krause Apr 29 '14 at 18:11

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.