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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 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.

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  • $\begingroup$ $RP^2\times RP^2$? $\endgroup$ – user43326 Dec 31 '16 at 18:53
  • $\begingroup$ @user43326: Unless I'm mistaken, $w_1(\mathbb{RP}^2\times\mathbb{RP}^2)^2 \neq 0$. $\endgroup$ – Michael Albanese Dec 31 '16 at 19:11
  • $\begingroup$ @MichaelAlbanese You are right, and if we try to kill the class $w_1^2$, we end up killing $w_1$ as well? $\endgroup$ – user43326 Jan 1 '17 at 6:26
  • $\begingroup$ @user43326: Not necessarily. Consider $K\times S^2$ where $K$ is the Klein bottle. $\endgroup$ – Michael Albanese Jan 1 '17 at 12:24
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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 $w_3(Z)z = 1$.

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  • $\begingroup$ Do you mean $M=(RP^2\times RP^2)\# RP^4\# RP^4$. $\endgroup$ – Xiao-Gang Wen Jan 13 '17 at 1:50
  • $\begingroup$ Yes, that's right. $\endgroup$ – Johannes Nordström Jan 13 '17 at 9:06
  • $\begingroup$ Indeed $w_1(M)^2 \not= 0$. But $M$ is only an intermediate step in the construction of $Z$, which is what I claim does have $w_1(Z)^2 = 0$ and $w_3(Z) \not= 0$. $\endgroup$ – Johannes Nordström Jan 16 '17 at 15:43

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