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I'm trying to find generating manifolds for the cobordism group $\mathit{MO}_5(K(\mathbb Z/2, 2))\cong (\mathbb Z/2)^4$, which can be represented as the cobordism group of closed 5-manifolds $M$ together with a class $B\in H^2(M;\mathbb Z/2)$. I've found three of the four generators; the fourth should be a 5-manifold $M$ such that $w_1(M)w_2(M)\in H^3(M;\mathbb Z/2)$ is nonzero, and here I've gotten stuck.

Such an $M$ cannot be a product of lower-dimensional manifolds, nor can it be the total space of a fiber bundle over the circle.

I'm happy to hear general approaches or ideas as well, such as ways of modifying a manifold to change its cohomology or Stiefel-Whitney classes in useful ways.

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    $\begingroup$ I thought you could get any element in $MO_*K(\mathbb{F}_2, n)$ by starting with (i) elements in $MO_*$ and (ii) the standard embeddings $\mathbb{R}P^k \hookrightarrow \mathbb{R}P^{\infty}$ (as elements of $MO_*K(\mathbb{F}_2, 1)$) and then propagating via the 'Hopf ring' operations: i.e. (under the interpretation you gave) disjoint unions of manifolds with their coh classes, products of manifolds and the cup product in cohomology, and addition of cohomology classes. There's your general statement. $\endgroup$ – Dylan Wilson Sep 18 '17 at 17:50
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    $\begingroup$ So are you sure about the 'no products and not a bundle over S^1'? Because I would've guessed the Dold-Manifold for your missing generator $\endgroup$ – Dylan Wilson Sep 18 '17 at 17:55
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    $\begingroup$ It seems to me that $P(1,n)$ is the mapping cylinder of complex conjugation on $\Bbb{CP}^n$, which is orientable when $n$ is even, so $P(1,2)$ seems suspect. $\endgroup$ – Mike Miller Sep 18 '17 at 19:36
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    $\begingroup$ @DylanWilson I'm confident it's not a product, and pretty confident it's not a circle bundle. I checked the Dold manifold again, using Dold's original calculation of its Stiefel-Whitney classes and got $w_1(P(1, 2)) = 0$. $\endgroup$ – Arun Debray Sep 18 '17 at 19:37
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    $\begingroup$ Indeed, and as we see below, it turns out the condition "w_1w_2 = 0" isn't the one you're looking for. So if you are interested in the generators of $MO_*K(\mathbb{Z}/2, 2)$ in this degree, I go back to my earlier point: you can get them all in the way I indicated. There is also a relation amongst these classes which you can write in terms of the formal group law on $MO_*$ (determined by the coefficients of its canonical logarithm, the coeffs of which are polynomial generators of $MO_*$. I dunno whether those agree with the Dold generators or not offhand, so I don't know explicit mfld reps.) $\endgroup$ – Dylan Wilson Sep 18 '17 at 21:00
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Note that the third Wu class is $\nu_3 = w_1w_2$, so on a closed connected smooth $n$-manifold $M$, $\operatorname{Sq}^3 : H^{n-3}(M; \mathbb{Z}_2) \to H^n(M; \mathbb{Z}_2)$ is given by $\operatorname{Sq}^3(x) = w_1(M)w_2(M)x$.

As $\operatorname{Sq}^i(x) = 0$ if $i > \deg x$, when $n = 5$ we see that $\operatorname{Sq}^3 : H^2(M; \mathbb{Z}_2) \to H^5(M; \mathbb{Z}_2)$ is the zero map. By Poincaré duality, it follows that $w_1(M)w_2(M) = 0$.

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  • $\begingroup$ I can't believe I missed that. Thanks! $\endgroup$ – Arun Debray Sep 18 '17 at 19:40
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    $\begingroup$ (And in case you want the general story: all relations between characteristic classes of n-manifolds were computed by Brown and Peterson in their paper "On relations between characteristic classes". They sort of all come from these Wu-type relations.) $\endgroup$ – Dylan Wilson Sep 18 '17 at 21:12

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