# (Co)bordism invariant of Eilenberg–MacLane space becomes vanished

Consider a (co)bordism invariant $$u_2 Sq^1 u_2+Sq^2 Sq^1 u_2$$ obtained from $$\Omega^5_{O}(K(\mathbb{Z}/2,2)).$$ Here $$u \in H^2(K(\mathbb{Z}/2,2),\mathbb{Z}_2)$$. The $$K(\mathbb{Z}/2,2)$$ is Eilenberg–MacLane space. The $$\mathbb{Z}/2$$ is the finite group of order 2.

Question: Is it true that such a (co)bordism invariant $$u_2 Sq^1 u_2+Sq^2 Sq^1 u_2$$ when being pulled back from the $$\Omega^5_{O}(K(\mathbb{Z}/2,2))$$ to a pulldback new (co)bordism group $$\Omega^5_{\frac{(Pin^\pm \times SU(2))}{\mathbb{Z}/2}}$$ (thus we are allowed to identify more wider classes of manifolds or more (co)bordism invariants by enlarging the cobordant structures of manifolds from
$$\Omega^5_{O}(K(\mathbb{Z}/2,2))$$ to $$\Omega^5_{\frac{(Pin^\pm \times SU(2))}{\mathbb{Z}/2}}$$),

the (co)bordism invariant $$u_2 Sq^1 u_2+Sq^2 Sq^1 u_2$$ becomes 0 in $$\Omega^5_{\frac{(Pin^\pm \times SU(2))}{\mathbb{Z}/2}}$$?

Namely the effective manifold generators (respect of to $$u_2 Sq^1 u_2+Sq^2 Sq^1 u_2$$) in $$\Omega^5_{O}(K(\mathbb{Z}/2,2))$$ becomes trivial or vanished in $$\Omega^5_{\frac{(Pin^\pm \times SU(2))}{\mathbb{Z}/2}}$$?

How to prove this?

• Here $${\frac{(Pin^\pm \times SU(2))}{\mathbb{Z}/2}}$$ is similar to the generalization of Pin$$^c$$ structure where $$U(1)$$ is now replaced by $$SU(2)$$.


This question is actually several questions in one: the argument differs for $$\Pin^{h+}$$ and $$\Pin^{h-}$$, and it's also necessary to specify the maps $$\Omega_5^{\Pin^{h\pm}}\to\Omega_5^O(K(\mathbb Z/2, 2))$$. Any characteristic class of $$M$$ or $$Q$$ in $$H^2(M;\mathbb Z/2)$$ determines a natural map to $$K(\mathbb Z/2,2)$$, and hence a map between these bordism groups. There are several examples, such as $$w_2(Q)$$ and $$w_2(Q) + w_1^2(M)$$, and the answer could in principle differ depending on which class one chooses.

As such, I have a partial answer: in the case of pin$$h+$$ 5-manifolds where the map to $$K(\mathbb Z/2,2)$$ is $$w_2(Q)$$ or $$w_2(Q) + w_1^2(M)$$, the answer is yes, the invariant vanishes.

Since $$u_2\mathrm{Sq}^1 u_2 + \mathrm{Sq}^2\mathrm{Sq}^1u_2$$ is a cobordism invariant, it suffices to check this on a generator of $$\Omega_5^{\Pin^{h+}}$$. This group is isomorphic to $$\mathbb Z/2$$ (this is due to Freed-Hopkins, Theorem 9.89), so we need only to find a single nonbounding pin$$h+$$ 5-manifold and check there.

This generator is the Wu manifold $$W := \mathrm{SU}_3/\mathrm{SO}_3$$, with the principal $$\mathrm{SO}_3$$-bundle $$Q$$ given by the quotient map $$\mathrm{SU}_3\twoheadrightarrow W$$. It's known that $$H^*(W;\mathbb Z/2) = \mathbb Z/2[z_2,z_3]/(z_2^2, z_3^2)$$ with $$|z_i| = i$$; its nonzero Stiefel-Whitney classes are $$w_2(W) = z_2$$ and $$w_3(W) = z_3$$, and that $$w_2(Q) = z_2$$ (this is discussed, for example, at the end of section 6 of Xuan Chen's thesis). In particular, $$w_2(W) = w_2(Q)$$, so $$W$$ admits a pin$$h+$$-structure whose corresponding $$\mathrm{SO}_3$$-bundle is $$Q$$, and because $$\langle w_2(Q)w_3(W), [W]\rangle = 1$$, it doesn't bound as a pin$$h+$$ manifold. Therefore it generates $$\Omega_5^{\Pin^{h+}}$$.

Using the Wu formula, $$\Sq^1z_2 = z_3$$ and $$\Sq^2z_3 = z_2z_3$$, so

$$w_2(Q)\Sq^1w_2(Q) + \Sq^2\Sq^1w_2(Q) = 2z_2z_3 = 0,$$

so we conclude that this invariant vanishes for all pin$$h+$$ 5-manifolds. (Since $$w_1(W)^2 = 0$$, we can also conclude this for the map to $$K(\mathbb Z/2,2)$$ given by $$w_2(Q) + w_1(M)^2$$.)

An analogous argument should be possible in the pin$$h-$$ case, but since $$\Omega_5^{\Pin^{h-}}\cong\mathbb Z/2\oplus\mathbb Z/2$$ (this is again Freed-Hopkins, Theorem 9.89), one would have to find two linearly independent generators and check on them. $$W$$ also admits a pin$$h-$$-structure with corresponding bundle $$Q$$, but I wasn't able to figure out what the other generator is.