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 $J$*]). 


2. 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 ?).
3. 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 $M$ ?



Spin^c structures on manifolds with almost complex structure
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](https://www.ma.utexas.edu/users/dafr/DiracNotes.pdf) 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.


