mjhjmtu

Consider a (co)bordism invariant
$$
u_2 Sq^1 u_2+Sq^2 Sq^1 u_2
$$
obtained from
$$
Omega^5_O(K(mathbbZ/2,2)).
$$
Here $u in H^2(K(mathbbZ/2,2),mathbbZ_2)$. The $K(mathbbZ/2,2)$ is Eilenberg–MacLane space. The $mathbbZ/2$ 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 $Omega^5_O(K(mathbbZ/2,2))$ to a pulldback new (co)bordism group
$$
Omega^5_frac(Pin^pm times SU(2))mathbbZ/2
$$
(thus we are allowed to identify more wider classes of manifolds or more (co)bordism invariants by enlarging the cobordant structures of manifolds from
$Omega^5_O(K(mathbbZ/2,2))$ to $Omega^5_frac(Pin^pm times SU(2))mathbbZ/2$),

the (co)bordism invariant
$
u_2 Sq^1 u_2+Sq^2 Sq^1 u_2
$ becomes 0 in $Omega^5_frac(Pin^pm times SU(2))mathbbZ/2$?

Namely the effective manifold generators (respect of to $
u_2 Sq^1 u_2+Sq^2 Sq^1 u_2
$) in $Omega^5_O(K(mathbbZ/2,2))$ becomes trivial or vanished in $Omega^5_frac(Pin^pm times SU(2))mathbbZ/2$?

How to prove this?

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

$newcommandPinmathrmPinnewcommandSqmathrmSq$
To simplify notation, I'll denote $(Pin_n^pmtimesmathrmSU_2)/(mathbb Z/2)$ by $Pin_n^hpm$, since these are
analogues of the groups $mathrmSpin_n^h = (mathrmSpin_ntimesmathrmSU_2)/(mathbb Z/2)$. Analogously to how a
spin^$c$-structure on a manifold $M$ determines a line bundle $L$ with $w_2(L) = w_2(M)$, a
pin^$hpm$-structure on $M$ determines a principal $mathrmSO_3$-bundle $Q$ with $w_2(Q) = w_2(M)$ (for
pin^$h+$) or $w_2(Q) = w_2(M) + w_1^2(M)$ (for pin^$h-$).

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 $Omega_5^Pin^hpmtoOmega_5^O(K(mathbb Z/2, 2))$. 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 $u_2mathrmSq^1 u_2 + mathrmSq^2mathrmSq^1u_2$ 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 $W := mathrmSU_3/mathrmSO_3$, with the principal $mathrmSO_3$-bundle $Q$
given by the quotient map $mathrmSU_3twoheadrightarrow W$. 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 $mathrmSO_3$-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

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

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
$Omega_5^Pin^h-congmathbb Z/2oplusmathbb Z/2$ (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.