Is there a 4-manifold whose Stiefel-Whitney classes satisfy $w\_1\neq 0$, $w\_3 \neq 0$, and $w\_1^2=0$?


This question follows on from [this one](https://math.stackexchange.com/q/2071685/39599) where the condition $w\_3 \neq 0$ is replaced by $w\_2 \neq 0$. On a closed four-manifold $\operatorname{Sq}^1(w\_2) = w\_3$ so $w\_3 \neq 0$ implies $w\_2 \neq 0$. Note however that not every example with $w\_2 \neq 0$ will have $w\_3 \neq 0$; this is the case for both answers to the previous question.



A 4-manifold with Stiefel-Whitney classes $w_1\neq 0$, $w_3 \neq 0$, and $w_1^2=0$?
Let's begin by reformulating the question a bit. Note that any orientable 4-manifold is spin-c, so in particular has $w\_3 = 0$. The condition that $w\_1 \not= 0 $ is thus redundant.


Wu's theorem gives that on a closed 4-manifold


* $w\_1 = v\_1$, so $Sq^1(c) = w\_1 c$ for any $c \in H^3$ ($\mathbb{Z}/2$ coefficients everywhere)
* $w\_2 = v\_2 + v\_1^2$, so $v\_2 = w\_2 + w\_1^2$. Thus $(w\_2 + w\_1^2)z = z^2$ for any $z \in H^2$
* $w\_3 = v\_3 + Sq^1(v\_2) + Sq^2(v\_1)$. Because $Sq^2(v\_1) = 0$ and $v\_3 = 0$ for degree reasons, this implies $w\_3 = Sq^1(w\_2)$.


We also have universally that $w\_3 = Sq^1(w\_2) + w\_1 w\_2$, so for closed 4-manifolds we get $w\_1 w\_2 = 0$. Putting these together we find


$$w\_3 x = Sq^1(w\_2) x = Sq^1(w\_2 x) + w\_2 Sq^1(x) = w\_1 w\_2 x + w\_2 x^2 = x^4$$


on any closed 4-manifold with $w\_1^2 = 0$. Hence the question is equivalent to looking for a closed 4-manifold with $w\_1^2 = 0$ and some $x \in H^1$ such that $x^4 = 1$.


We can now approach the problem by the following strategy. Look for closed 4-manifolds with the right characteristic numbers. If there are none, then the answer to the question is negative. If there is one, then try to obtain a solution to the problem by surgery.


In this case we need a closed 4-manifold with a distinguished class $x \in H^1$, and the relevant characteristic numbers are $x^4 = 1$, $w\_1^4 = w\_1^3x = w\_1x^2 = 0$; actually one condition is redundant because $w\_1^3 x = Sq^1(w\_1^2 x) = w\_1^2 x^2$.


$RP^2 \times RP^2$ has $w\_1^4 = 0$, and if we take $x$ to be the generator from one of the two factors then $w\_1^2 x^2 = 1$ while $x^4 = 0$. On the other hand $RP^4$ has $w\_4 = 1$, and if we take $x$ to be the generator then $w\_1^2 x^2 = x^4 = 1$, while if $x = 0$ then of course $w\_1^2 x^2 = x^4 = 0$. Hence if we let $M$ be the connected sum of $RP^2 \times RP^2$ and two copies of $RP^4$ then $M$ is a closed 4-manifold with a class $x \in H^4(M)$ such that $w\_1^4 = w\_1^2 x^2 = 0$ and $x^4 = 1$.


$H^1(M) \cong H\_1(M)$ has rank 4. Any $\alpha \in H\_1(M)$ with $w\_1(\alpha) = 0$ can be represented by an embedded circle with trivial normal bundle. Doing surgery on this circle cuts down the rank of $H^1$ by 1. If in addition $x(\alpha) = 0$, then the result of the surgery will still have a class $x$ with the desired properties $x^4 = 1$, $w\_1^4 = w\_1^2 x^2 = 0$. So by doing surgery twice on $M$ we obtain an $X$ with the right characteristic numbers, but $H^1(X)$ of rank 2 generated by the essential classes $w\_1$ and $x$.


In particular $w\_1^3$ and $w\_1^2 x$ both pair to 0 with all elements of $H^1(X)$, so must vanish. Thus, if we let $S \subset N$ be an embedded smooth surface Poincare dual to $w\_1^2$ then $w\_1$ and $x$ both vanish on $S$. It would be nice if we could deduce that $S$ can be taken to be an embedded sphere so that it can be killed by surgery, but that's not obvious to me. 


However, we can cut out a tubular neighbourhood of $S$ to obtain a 4-manifold $Y$ with boundary. Then $x \in H^1(X)$ has a preimage $y \in H^1(Y, \partial Y)$. Now glue $Y$ to itself along the boundary to obtain a closed 4-manifold $Z$, and let $z \in H^1(Z)$ be the image of $y$ from one side. Then $w\_1(Z)^2 = 0$ and $z^4 = 1$, so $w\_3(Z)z = 1$.


