Consider a smooth compact oriented 4-manifold $X$. Although not all 4-manifolds admit a spin structure, they do admit spin-c structures. And if $X$ does admit a spin structure, then there is a canonical spin-c structure. Now if instead $X$ is equipped with a symplectic form $\omega$ then there is also a canonical spin-c structure (corresponding to the spinor bundle which splits with summand $K^{-1}$, the canonical determinant line bundle). This is nice, turning the bijection $Spin^c(X)\simeq H^2(X;\mathbb{Z})$ into a canonical isomorphism.

I would like to question what goes on when I relax $\omega$ to be a near-symplectic form, i.e. a closed 2-form which is symplectic away from its zero-set $Z$ (a finite disjoint union of embedded circles). I still get a bijection $Spin^c(X)\simeq \lbrace \sigma\in H_2(X,Z)\;|\;\partial\sigma=[Z]\in H_1(Z)\rbrace$, but not necessarily canonical.
Are there any conditions I can put on $(X,\omega)$ in order to have a natural origin?


1 Answer 1


When $(X,\omega)$ is a near-symplectic oriented 4-manifold there is always a canonical identification between $\mathrm{Spin}^c(X)$ and the classes in $H_2(X,Z;\mathbb{Z})$ that bound $[Z]$, where $Z=\omega^{-1}(0)$. I call this identification the "Taubes map" $\tau$ in my "Lagrangian matching invariants" papers, since it comes from Taubes's work in near-symplectic geometry. A near-symplectic form $\omega$ can be defined intrinsically as a closed 2-form with $(\omega\wedge \omega) (x)>0$ except at the set $Z$ of points where $\omega(x)=0$; and at those points $\nabla \omega$ has rank 3, hence its image spans a maximal positive-definite subspace of $\Lambda^2 T^*_x X$.

On $X-Z$ one has the canonical $\mathrm{Spin}^c$-structure $\mathfrak{s}_{\mathrm{can}}$ arising from an almost complex structure compatible with $\omega$. An arbitrary $\mathrm{Spin}^c$-structure $\mathfrak{s}$ on $X$ restricts to $X-Z$ as $L\otimes \mathfrak{s}_{\mathrm{can}}$ for a unique line bundle $L$. Define $\tau(\mathfrak{s})\in H_2(X,Z;\mathbb{Z})$ to be the Lefschetz dual to $c_1(L)$. Since restriction $H^2(X;\mathbb{Z})\to H^2(X-Z;\mathbb{Z})$ is injective, the map $\tau\colon \mathrm{Spin}^c(X) \to H_2(X,Z;\mathbb{Z})$ is also injective; and its image is a coset of $H_2(X;\mathbb{Z})$. The structure of $\omega$ near a point of $Z$ is standard, so the image of $\tau$ will be the classes that bound $n[Z]$ for some standard $n$.

That the answer is given by $n=1$ can be visualized by looking at an example on $S^1\times B^3$, with $\omega = dt\wedge \alpha+ \ast \alpha$, where $\alpha$ is a harmonic 1-form on the ball $B$ with one critical point; then gradient flow-lines for $\alpha$ into the critical point, crossed with $S^1$, represent classes in the image of $\tau$, and these precisely bound $Z$.

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    $\begingroup$ Rereading your question, perhaps what you're asking for a canonical spin-c structure on $X$. If so, it's not going to come in any natural way from the near-symplectic structure (unless you deform it to a symplectic structure). $\endgroup$
    – Tim Perutz
    Nov 3, 2012 at 15:01
  • $\begingroup$ You've actually cleared up all questions I (implicitly) had! At first I was thinking if when all the components of $Z$ were null-homologous then perhaps their is a canonical spin-c structure on $X$, somehow by considering the 'minimal surfaces' in $H_2(X,Z)$ with the components as boundary. $\endgroup$ Nov 4, 2012 at 17:15

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