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 $SU(2)$.



(Co)bordism invariant of Eilenberg–MacLane space becomes vanished
$\newcommand{\Pin}{\mathrm{Pin}}\newcommand{\Sq}{\mathrm{Sq}}$
To simplify notation, I'll denote $(\Pin\_n^\pm\times\mathrm{SU}\_2)/(\mathbb Z/2)$ by $\Pin\_n^{h\pm}$, since these are
analogues of the groups $\mathrm{Spin}\_n^h = (\mathrm{Spin}\_n\times\mathrm{SU}\_2)/(\mathbb Z/2)$. 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$h\pm$-structure on $M$ determines a principal $\mathrm{SO}\_3$-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 $\Omega\_5^{\Pin^{h\pm}}\to\Omega\_5^O(K(\mathbb Z/2, 2))$. 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](https://arxiv.org/pdf/1604.06527.pdf), 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](https://www.math.stonybrook.edu/alumni/2017-Xuan-Chen.pdf)). 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.


