Is $SU(3)/SO(3)$ cobordant with a mapping torus?

Question

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?

Answer 1

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 $\Omega_5^{SO}$.