On a 3-manifold $Y$, oriented 2-plane fields $\xi$ are oriented rank-2 subbundles of $TY$. Denote the set of such (up to homotopy) by $\Theta=\pi_0\lbrace\xi\rbrace$. What is an explicit canonical map $\Theta\to\mathbb{Z}_2$ ?

In particular, I want to see the canonical mod-2 grading on Seiberg-Witten-Floer Homology, but the classical text of Kronheimer-Mrowka doesn't mention it in terms of 2-plane fields. They construct an isomorphism of $\Theta$ with some abstract set $\mathbb{J}$ based upon "configuration points" $[a]$, and then a canonical map $\mathbb{J}\to\mathbb{Z}_2$ by assigning to $[a]$ an operator and taking its index mod-2. These two maps are tough enough on their own, so I cannot see what it looks like on 2-plane fields... Plus, there should be some purely topological partition of $\Theta$ into two subsets.

Otherwise, perhaps it can be done knowing that oriented 2-plane fields are equivalent to 1-forms of length 1, and are also equivalent to pairs $(\mathfrak{s},\phi)$ where $\mathbb{s}$ is a spin-c structure and $\phi$ is a unit-length spinor. This should then be related to taking a 4-manifold $X$ such that $\partial X=Y$ and with a spin-c structure $\mathfrak{s}_X$ which extends $\mathfrak{s}$.

  • $\begingroup$ I may be guessing incorrectly, but what is wrong with taking the subbundle orthgonal $L$ to $\xi$ (after putting a metric on $Y$). To every (isomorphism class of) line bundle on $Y$ we have its oriantation; send $L$ to $\mathbb{Z}_2$ via its orientation. $\endgroup$ – Somnath Basu Apr 19 '12 at 22:00
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    $\begingroup$ See Kronheimer-Mrowka-Ozsvath-Szabo, "Monopoles and lens space surgeries", section 2.5. $\endgroup$ – Tim Perutz Apr 20 '12 at 5:10
  • $\begingroup$ Perfect! But doesn't this just provide the existence of such a map? I could see trying to construct a map from this, by taking $X$ such that $\partial X=Y$ and then removing a ball from it to get a cobordism $W:Y\to S^3$, giving a parity-relation between $\Theta(Y)$ and $\Theta(S^3)$ via that ``cobordism index'' $\iota(W)$, but then we had to a priori decide on $\Theta(S^3)\to\mathbb{Z}_2$. $\endgroup$ – Chris Gerig Apr 20 '12 at 6:23
  • $\begingroup$ If you only want ONE such map; some of the plane fields split. Some don't. Map the split ones to zero. Admittedly, this is quite boring when Y isn't parallelizable. $\endgroup$ – some guy on the street Apr 21 '12 at 1:10
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    $\begingroup$ @Somanth Basu: Isn't what you construct always zero? We're looking at oriented 2-plane fields on an oriented manifold, so the orthogonal complement (which is canonical) will also be oriented. $\endgroup$ – Dylan Thurston Apr 30 '12 at 18:28

In light of the paper referenced in Tim Perutz' comment, here is the desired map:

An oriented 2-plane field $\xi$ is equivalent to a pair $(\mathfrak{s},\phi)$ on $Y$, where $\mathfrak{s}$ is a spin-c structure and $\phi$ is unit-length spinor. By Proposition 28.1.2 (of Kronheimer-Mrowka's textbook), there exists an oriented 4-manifold $X$ with $\partial X=Y$ and carrying a spin-c structure $\mathfrak{s}_X=(S^+,\rho_X)$ which extends $\mathfrak{s}$, i.e. $\mathfrak{s}\cong(S^+|_Y,\rho_Y)$. Now the relative Euler class $e(S^+,\phi)$ satisfies $e(S^+,\phi)[X,\partial X]$ = $gr_z(X,\mathfrak{s}_X,[a])\in\mathbb{Z}$ for some configuration point $[a]$ associated to $\phi$, and this index is independent of $z$. And by Proposition 28.2.2, this is independent of the choice of $X$ (up to homotopy of $\phi$). This is where the isomorphism $\mathbb{J}(Y)\cong \Theta(Y)$, $\xi\leftrightarrow [a]$, comes from. We can thus write down a map $f:\Theta(Y)\to\mathbb{Z}_2$ given by $$\xi\mapsto e(S^+,\phi)[X,\partial X]\;\text{mod}2$$

To be consistent with the literature: When we apply this to $Y=S^3$ we get the flipped even/odd decomposition of $\widehat{HM}(S^3)$. Simply change $f$ by $Aut(\mathbb{Z}_2)$, i.e. adding a $1$, to get agreement.

Addendum: In fact, this map can be rephrased as the obstruction to finding an almost complex structure $J$ on the above $X$ in such a way that $\xi$ is $J$-invariant.


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