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What are some examples of $4$-manifolds $M$ for which the class $w_3(TM)\in H^3(M;\mathbb{Z}/2)$ is nontrivial? Is there a mapping torus with this property?

Motivation: I am wondering whether any such $4$-manifolds can be "built out of" a $3$-manifold by the mapping torus construction, despite the fact that $w_3$ vanishes on $3$-manifolds. In asking myself this, I realized my that go-to examples of $4$-manifolds -- $\mathbb{R} P^4$, $\mathbb{C} P^2$, and $K3$ -- all have trivial $w_3$.

aaa
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    $\mathbb{R}P^2\times \mathbb{R}P^2$ is a non-orientable $4$-manifold with non-vanishing $w_3$. I have to think about the mapping cylinder. – ThorbenK Dec 04 '18 at 12:29
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    Presumably you mean mapping torus instead of cylinder: the latter is not a very interesting manifold. For the former, you should be able to explicitly calculate the SW classes of $f: M \to M$ in terms of $w_i(M)$ and the map $f^*$. –  Dec 04 '18 at 16:39
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    @ThorbenK Right, by Whitney product. I see. Thanks! – aaa Dec 04 '18 at 18:22
  • @MikeMiller Right, I meant mapping torus. – aaa Dec 04 '18 at 18:22
  • I spent a really long time trying to show that $w_3(T_f) = 0$ and I was not able to do so. So I think it might be productive to try to come up with examples instead. One needs either $M$ to be non-orientable or $f$ to be orientation-reversing. One also needs to demand that the endomorphism $1 - f^*$ of $H^2(M;\Bbb Z/2)$ is not an isomorphism. –  Dec 05 '18 at 00:03
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    @ThorbenK Was this question a related idea? I tried to do a calculation of $w_3$ (defined as the Poincare dual to the locus where two generic sections become linearly dependent) using the obvious vector field as well as a homotopy between a nonvanishing tangent field $X$ on $M$ and $f_* X$. I didn't get anywhere, which is one of the reasons I started thinking the answer was "yes, such a mapping torus exists". –  Dec 05 '18 at 21:36
  • @MikeMiller I added my attempt as an answer as it would exceed the limitations of a comment. I think it is quite similar to your approach. Nevertheless I'm still unsure how fruitful my attempt is. It all boils down to getting a little knowledge on the $\text{Diff}(M)$-action of $\Gamma(TM\setminus 0)$ I think. – ThorbenK Dec 05 '18 at 22:46

2 Answers2

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This is a partial answer, showing what one should demand if they were to try to find an example with $w_3(T_f) \neq 0$: If $M$ is an orientable 4-manifold then $w_3(M) = 0$. So $f$ would have to either be an orientation-reversing diffeomorphism or a diffeomorphism of a non-orientable manifold.

To prove this, use Wu classes. These are the classes $\nu_i \in H^i(M;\Bbb Z/2)$ for which the two maps $H^{n-i}(X;\Bbb Z/2) \to \Bbb Z/2$ given by $\nu_i \cdot x$ and $\text{Sq}^i x$ agree. Wu's theorem is that we have the property $$\sum_{i=0}^{\lfloor k/2\rfloor} \text{Sq}^{k-i} \nu_i = w_i.$$

We see from the definition that $\nu_3 = 0$ because $\text{Sq}^3$ vanishes on classes of degree less than $3$, and we see from orientability that $\nu_1 = 0$, and hence $\nu_2 = w_2$. Therefore we have $w_3 = \text{Sq}^1 w_2$. (In fact, this is true for an arbitrary 4-manifold; one needs to argue that $(\text{Sq}^1)^2 w_1$, which in principle contributes, is always zero). The operation $\text{Sq}^1$ is sometimes better known as the Bockstein map. This map factors as the composite of the integral Bockstein $\beta_{\Bbb Z}: H^2(M;\Bbb Z/2) \to H^3(M;\Bbb Z)$ and reducing coefficients modulo 2, so it suffices to show that if $M$ is an oriented closed 4-manifold, we have $\beta_{\Bbb Z} w_2(M) = 0$.

A really elegant proof of this fact is given in the main proposition of this short note. I will not reproduce it. The essential point is that the Bockstein long exact sequence shows that $\beta_{\Bbb Z} w_2(M) = 0$ if and only if $w_2(M)$ lifts to an integral class, and that note explains how to show that $w_2(M)$ lifts to an integral class.

  • Here's another proof that $(Sq^1)^2w_1 = 0$, or more generally, that $(Sq^1)^2:H^1(X;\mathbb{Z}/2\mathbb{Z})\rightarrow H^3(X;\mathbb{Z}/2\mathbb{Z})$ is always the zero map. Namely, since $\mathbb{R}P^\infty$ represents $H^1(X;\mathbb{Z}/2\mathbb{Z})$, it's enough to verify it in just that case. Writing $H^\ast(\mathbb{R}P^\infty;\mathbb{Z}/2\mathbb{Z})\cong \mathbb{Z}/2\mathbb{Z}[x]$, we have $Sq^1(Sq^1 x) = Sq^1(x^2) = (Sq^1 x)x + x(Sq^1 x) = 0$. – Jason DeVito - on hiatus Dec 06 '18 at 01:34
  • @JasonDeVito That makes one proof, since I didn't give one! –  Dec 06 '18 at 01:57
  • Ha! I need more caffeine. I seem to recall that $(Sq^1)^2 = 0$ (not just on $H^1$ but on any $H^\ast$). I'll try to remember the reference or proof... – Jason DeVito - on hiatus Dec 06 '18 at 02:32
  • @JasonDeVito I'm sure Hatcher proves it somewhere. (As usual, I'm too lazy to find the actual reference: I have said the previous sentence too many times.) –  Dec 06 '18 at 02:42
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    @JasonDeVito Since $Sq^1 = \rho_2 \beta_{\mathbb{Z}}$, where $\rho_2$ is mod 2 reduction, for any class $\alpha$ the class $Sq^1 \alpha$ has an integral lift (namely the integral Bockstein of $\alpha$). Since the integral Bockstein vanishes on classes with integral lifts, we have $Sq^1 Sq^1 \alpha = 0$. Symbolically, $Sq^1 Sq^1 = (\rho_2 \circ \beta_{\mathbb{Z}}) \circ (\rho_2 \circ \beta_{\mathbb{Z}}) = \rho_2 \circ (\beta_{\mathbb{Z}} \circ \rho_2) \circ \beta_{\mathbb{Z}} = 0$. – Aleksandar Milivojević Dec 06 '18 at 23:52
  • @AleksandarMilivojevic Nice, I was trying to do something like that but was stuck for whatever reason. –  Dec 07 '18 at 00:05
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Here is an idea I came up but I was only able essentially reduce it to this question. I had hoped to prove that the mapping torus of a diffeomorphism of an orientable $3$-manifold is zero using the fact that orientable $3$-manifolds have trivial tangent bundles and the definition of Stiefel-Whitney classes as obstruction classes. In the end this gave a good method to produce counterexamples I think.

Let $M$ denote an orientable $3$-manifold and $f$ and orientation preserving diffeomorphism. I will denote the mapping Torus by $T_f$ and the inclusion of some fiber by $\iota\colon M\to T_f$. Fix a trivialization of $T\iota(M)$, which is possible by the aforementioned fact. Now we know that $\iota^*(w_i(S_f))=w_i(\iota^*TS_f)=w_i(TM\oplus \mathbb{R})=0$ we conclude that $w_i(S_f)$ comes from some class in $H^i(S_f,\iota(M);\mathbb{Z}/2\mathbb{Z})$. Using excision the inclusion of the pair $(M\times I,M \times \partial I)\to (S_f,\iota(M))$ induces an isomorphism on cohomology. Therefore we have to understand how the Stiefel-Whitney classes of $(M\times I,\partial M\times I)$ behave.

Since $T(M\times I)\cong \pi^* TM\oplus \mathbb{R}$, where $\pi$ denotes the projection $M\times I\to M$ and this splitting respects the fixed framing at $M\times \partial I$, we have to understand $w_3(\pi^*TM,\pi^*TM|_{M\times \partial I})$. Note that this is the mod $2$ reduction of the relative Euler class. Furthermore note that if we fix some non-vanishing section $\phi$ of $\pi^*TM|_{M\times \{0\}}$ then the section at $\pi^*TM|_{M\times \{1\}}$ is given by $f_* \phi((x,1))=Df_{f^{-1}(x)}(\phi(f^{-1}(x))$. All in all this should imply that $w_3$ is the mod $2$ reduction of the obstruction class for a homotopy between $\phi$ and $f_*(\phi)$.

Therefore we are left with the question how do homotopy classes of non-vanishing vector fields on orientable $3$-manifolds behave under diffeomorphisms of said manifold, which is exactly the aforementioned question. Nevertheless note that $[M,S^2]$, which is the set of homotopy classes of vector fields, surjects quite naturally to $H^2(M;\mathbb{Z})$. So maybe it is possible to deduce the existence of a vector field $\phi$ and a diffeomorphism $f$ such that the obstruction class for a homtopy between $f_* \phi$ and $\phi$ is non-zero mod $2$ using the action of $f$ on $H^2(M)$, but I'm tired right now so I will think about this last part tomorrow.

ThorbenK
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  • The natural surjection I see is to $H_1(M;\Bbb Z)$, taking the preimage of a generic point. I definitely think this should work - I was trying to show $w_3$ is zero this way and when I couldn't, I decided it probably wasn't. I think a nice manifold with a good vector field is what we need. –  Dec 05 '18 at 22:45
  • Yes you are right (As I said, I'm tired). There is a natural surjection to $H^2(M;\mathbb{Z})$ since $S^2$ is the $3$-skeleton of $\mathbb{C}P^\infty$. Like I said maybe it is enough to consider the action of $f$ on $H^2(M)$, but I'm complettly unable to check this right now.I will attempt this tomorrow. – ThorbenK Dec 05 '18 at 22:50
  • It's a very good idea to my eye (which I similarly am too tired and busy to carry out). –  Dec 05 '18 at 22:50