# Spin^c structures on manifolds with almost complex structure

Let $M$ be a smooth even-dimensional manifold.

1. Is it true that for each almost-complex structure $J$ on $M$ there exists a canonical spin$^c$ structure $S_J$ associated to $J$ ? (I've read this somewhere but I didn't see the actual construction).

Is there a way to caracterize that spin$^c$ structure ? (What I mean by that is : is there a theorem of the form : $S_J$ is the unique spin$^c$ structure on $M$ satisfying [a property of compability with $J$]).

1. Suppose $M$ admits an almost-complex structure. Does every spin$^c$ structure comes from an almost-complex structure on $M$ (i.e. is it true that every spin$^c$ structure is equal to a $S_J$ for an almost-complex structure $J$ ?). (If not : what happens if we restrict to parallelizable open 4-manifolds ?).

2. Same question as 1. 2. but for spin (and not just spin$^c$) structures. EDIT (to clarifiy this question) : Does every spin structure on $M$ comes from has an associated spin^c structure that is associated to an almost-complex structure $J$ on $M$ ?

Assume that $M$ is oriented throughout. Recall that $M$ has a $\text{Spin}^c$ structure iff the third integral Stiefel-Whitney class $\beta w_2 = W_3 \in H^3(M, \mathbb{Z})$ is trivial. Actually more is true: $\text{Spin}^c$ structures on $M$ are in bijection with trivializations of $W_3$, which are a torsor over $H^2(M, \mathbb{Z})$. So we get a functorial way to associate a $\text{Spin}^c$ structure to an almost complex structure for every choice of nullhomotopy of the composite map

$$BU(n) \to BSO(2n) \xrightarrow{W_3} B^3 \mathbb{Z}.$$

Isomorphism classes of nullhomotopies are a torsor over $H^2(BU(n), \mathbb{Z}) \cong \mathbb{Z}$. I don't know in what sense there's a canonical one, but once you pick one you don't have to make any further choices, and in particular you don't have to make any further choices involving $M$. (Edit #2: But see below.)

As for almost complex structures, there is a canonical fiber bundle

$$SO(2n)/U(n) \to BU(n) \to BSO(2n)$$

and almost complex structures on $M$ correspond to sections of the pullback of this bundle along the classifying map $M \to BSO(2n)$ of the tangent bundle. This admits a map of bundles to the corresponding bundle

$$B^2 \mathbb{Z} \to B \text{Spin}^c(2n) \to BSO(2n)$$

describing $\text{Spin}^c$ structures. The map $SO(2n)/U(n) \to B^2 \mathbb{Z}$ describes a canonical line bundle on the space of linear complex structures on $\mathbb{R}^{2n}$, namely the complex determinant bundle.

This tells us that we already can't expect all $\text{Spin}^c$ structures to come from almost complex structures when $n = 1$: here $SO(2)/U(1)$ is contractible, so there is in a very strong sense a unique almost complex structure on an oriented surface, but $\text{Spin}^c$ structures are a torsor over $H^2(-, \mathbb{Z})$, which can be nontrivial, e.g. on an oriented closed surface.

Edit #2: I believe that there is in fact a distinguished nullhomotopy of the composite map $BU(n) \xrightarrow{W_3} B^3 \mathbb{Z}$, as follows.

We now need the additional assumption that $M$ is Riemannian (although the choice of Riemannian metric won't matter). Then a conceptual description of $W_3 \in H^3(M, \mathbb{Z})$ is that it classifies the bundle of complex Clifford algebras $\text{Cliff}(T_x(M)) \otimes \mathbb{C}$ up to Morita equivalence. Trivializations of this bundle up to Morita equivalence correspond to Clifford module bundles with fiber the unique irreducible representation of $\text{Cliff}(2n) \otimes \mathbb{C}$ (complex spinor bundles).

If $M$ has an almost complex structure, then a distinguished complex spinor bundle can be constructed out of the complex exterior algebra of the tangent bundle. See Exercise 2.1.37 in Freed's Geometry of Dirac Operators for details. The torsor structure over $H^2(M, \mathbb{Z})$ comes from tensoring this bundle with complex line bundles on $M$, and the action of $H^2(BU(n), \mathbb{Z})$ comes from tensoring this bundle with powers of the canonical bundle. It's possible one might want to do this with a particular power; I'm not sure what's going on here exactly, but see Exercise 2.1.54 in Freed.

Edit #3: Now that the third question has been clarified, I don't know the answer off the top of my head. In the case of closed oriented surfaces we know that there's a $\mathbb{Z}$'s worth of $\text{Spin}^c$ structures. The unique almost complex structure gives rise to one, and the $2^{2g}$ spin structures also give rise to one (not depending on the spin structure, basically because the Bockstein homomorphism $H^1(M, \mathbb{Z}_2) \to H^2(M, \mathbb{Z})$ is zero in this case). But I don't know if they're the same one; they might differ by a power of the canonical bundle or something (or not, depending on your convention for trivializing $W_3$ as above). In summary,

1. Yes, more or less,
2. No, and
3. I don't know, but I would guess no.
• Thank you for your answer. For 1., the article ncatlab.org/nlab/show/spin%5Ec+structure refers to a "canonical" spin^c structure, but I can't say I understand really. Do you see in there a way to choose a special class of nullhomotopy of the composite map ? Jun 5, 2016 at 20:13
• For 3. I can't see how this answers the question, if M has no spin structure, then the answer to "Does every spin structure on M, etc." is yes. Jun 5, 2016 at 20:14
• @Michael: I think the nLab is ignoring the subtlety I point out above: that although it's true that the pullback of $W_3$ to $BU(n)$ is nullhomotopic, you need to choose a nullhomotopy, and there's more than one. However, again as pointed out above, once you make this choice there are no further choices. For 3 I interpreted the question to be "does every almost complex structure give rise to a canonical spin structure" and the answer is no. If the question you intended to ask is "does every spin structure give rise to a canonical $\text{Spin}^c$ structure" then you should clarify the wording. Jun 5, 2016 at 21:26
• Thank you, so I guess the answer to 1. is NO. But it is YES, up to a nullhomotopy choice. I re-worded question 3. Jun 5, 2016 at 21:47
• @Chris: it depends on what your definition of an equivalence of almost complex structures is. With the (homotopy-theoretic) definition I have in mind, almost complex structures correspond to lifts-up-to-homotopy of the classifying map of the tangent bundle to $BU(n)$, up to homotopy. When $n = 1$ the space of such lifts from $BSO(2)$ is contractible. Jun 6, 2016 at 1:55

A $$\text{Spin}^c$$ structure is equivalent to (a homotopy class of) an almost complex structure on the 2-skeleton of a manifold which extends to the 3-skeleton (except for a surface or when the fiber dimension is odd, where we first need to stabilize the tangent bundle). So in the case of 4-manifolds without 4-handles (in particular 4-manifolds with non-empty boundary) there is a canonical one to one correspondence between $$\text{Spin}^c$$ structures and almost complex structures. The original reference for this result is Gompf's Spin^c structures and homotopy equivalences (https://doi.org/10.2140/gt.1997.1.41), but there is a more expository version available in Surgery on Contact 3-Manifolds and Stein Surfaces by Ozbagci and Stipsicz.

• Thank you for you answer, this answers positively 1 and 2, in the case of 4-manifolds without 3-handles. I was not able to directly see this result in the references you mentioned (I guess it's a direct consequence, but I'm not familiar enough with this subject to see that as trivial). So just so that I understand a little bit better, why having no 3-handles implies that there is a canonical bijection between Spin^c structures and almost-complex structures? Jun 6, 2016 at 15:17
• @Michael I meant without 4-handles. In this case extending an almost complex structure to the 3-skeleton is the same as extending it to $M$.
– PVAL
Jun 7, 2016 at 1:53
• I have to admit I have no intuitive feeling (beyond the definition) as what means geometrically to not have a 4-handle for a 4-manifold. Why is it that extending to the 3-skeleton is the same as extending to M ? Is it because a 4-manifold without 4 handle is equal to its 3 skeleton ? Jun 7, 2016 at 5:46