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Setup

Let $M$ be a closed, orientable, smooth manifold, and $\alpha\in H^2(M)$. Further, let $N\subset M$ be a closed, orientable submanifold that is Poincare dual to $\alpha$, and let $E\rightarrow M$ be the complex line bundle whose Euler class is $\alpha$. Finally, let $s:M\rightarrow E$ be a section transverse to the zero section, and $Z=\{m\in M\mid s(m)=0\}$.

Question

Are $N$ and $Z$ cobordant?

Context

I know that $Z$ is also Poincare dual to $\alpha$, but I am wondering if $Z$ is "unique up to cobordism".

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In a sense, it all boils down to the fact that the same space ($\mathbb CP^\infty$) is both $K(\mathbb Z,2)$ and $BSO(2)$:

  1. Since $\mathbb CP^\infty$ is $K(\mathbb Z,2)$, we have $H^2(X)=[X,\mathbb CP^\infty]$. To get a realization of a class in $H^2(M)$ by a submanifold take the preimage of $\mathbb CP^{N-1}$ under a map $M\to\mathbb CP^N$ corresponding to the class [and transversal to $\mathbb CP^{N-1}$).

  2. Moreover, any codimension 2 cooriented submanifold $Z\subset M$ is a preimage of $\mathbb CP^{N-1}\subset\mathbb CP^N$. Indeed:

    • since $\xi\to\mathbb CP^\infty$ is the universal $SO(2)$-bundle, the normal bundle of $Z$ in $M$ is $f^*\xi$ for some map $f\colon Z\to\mathbb CP^{N-1}$ (for some $N$);
    • choose a small open neighbourhood $U$ of $Z$ isomorphic to the normal bundle of $Z$ and let $Mf$ be the map from $M$ to the 'Thom space' (i.e. 1-point compactification) $M\xi$ of $\xi$ (sending everything outside of $U$ to the 'infinity' and using the natural map $f^*\xi\to\xi$ on $U$);
    • $M\xi=CP^N$ so we have a map $M\to\mathbb CP^N$, and the preimage of $\mathbb CP^{N-1}$ is exactly $Z$.
  3. If we have two such realizations of the same class, we get a homotopy $M\times[0;1]\to CP^N$ — again make this map transversal and take the preimage of $\mathbb CP^{N-1}$ — that's a cobordism you're looking for.

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    $\begingroup$ Ah, this is very nice. And of course this works for $H^1$ too (using $S^1$). $\endgroup$ – Steve D Nov 10 '17 at 20:26
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    $\begingroup$ actually, this doesn't quite answer my question. You've shown that if $N$ is constructed via $\mathbb{CP}^N$, then it is cobordant to $Z$. But there's still the question if every submanifold, Poincare dual to $\alpha$, is cobordant to $Z$ (or I guess equivalently, whether every such submanifold can be realized by your construction). $\endgroup$ – Steve D Nov 11 '17 at 20:29
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    $\begingroup$ Indeed, I was too hasty — I've updated the answer (so now it uses [a special case of] Pontryagin-Thom construction — which is, perhaps, inevitable…). $\endgroup$ – Grigory M Nov 11 '17 at 21:43
  • $\begingroup$ Thank you again! Do you have a reference for where I can learn more about this Thom space/map? $\endgroup$ – Steve D Nov 11 '17 at 23:37
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    $\begingroup$ Well, off the top of my head — Thom's 'Quelques propriétés globales des variétés differentiables' (English translation; AFAIR it's quite readable) or ch. 2 of Stong's book 'Notes on cobordism theory'. But perhaps there are more modern sources with better exposition. $\endgroup$ – Grigory M Nov 12 '17 at 9:50
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Yes, they are equal. Let $M$ be a manifold, We say that two submanifolds $U$ and $V$ of $M$ of dimension $p$ are equivalent if they are cobordant. Denote by $L_p(M)$ the set of equivalent classes. The canonical map $L_p(M)\rightarrow H_p(M)$ is injective for $p\leq 3$, see the reference p.4 section 3.

http://www.neo-classical-physics.info/uploads/3/0/6/5/3065888/thom_-_cobordant_differentiable_manifolds.pdf

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  • $\begingroup$ The dimension of $N$ and $Z$ are $n-2$, so the $\le3$ case is pretty particular. However, that paper does claim the kernel of the map $L_{n-2}(M)\rightarrow H_{n-2}(M)$ is trivial, which is enough to show the answer to my question is "yes". However, the paper you've linked doesn't seem to contain a proof, only says this is a "special case". $\endgroup$ – Steve D Nov 5 '17 at 3:43
  • $\begingroup$ It seems that it is an easy consequence of the theorem 1 quoted above the paragraph which is proved. $\endgroup$ – Tsemo Aristide Nov 5 '17 at 3:45
  • $\begingroup$ I've tried to reason out how this follows, and have not. I don't believe it's an "easy consequence", at any rate. This seems like something so fundamental it should be in a textbook somewhere. Hopefully someone can provide such a reference. $\endgroup$ – Steve D Nov 6 '17 at 16:42

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