Let $(M,\omega)$ be a compact symplectic manifold and the cohomology class $$[\omega]+\frac{1} {2}c\_1(\wedge\_{\mathbb C}^{0,n}(TM, J))\in H\_{dR}^2(M)$$ 
is integral, for some almost complex structure $J$ on $M$ .Then why there exists a [$Spin^c$-structure](http://ncatlab.org/nlab/show/spin%5Ec+structure) $P\to M$ on $M$? AND why a $\text{spin}^c$ structure with determinant $L^{2ω}$ exists on $(M,ω)$.
I saw it in the thesis of [Peter Hochs](http://maths.adelaide.edu.au/peter.hochs/PhD_thesis_Peter_Hochs.pdf) In Page 43.



A question on existence of $Spin^c$-structure $P\to M$
You ask "why does a spin$^c$ structure exist on $(M,\omega)$?" It is a basic fact that every symplectic manifold admits a spin$^c$ structure. I will assume that you are instead interested in what Hochs is claiming in his comment on page 43 of his thesis: a spin$^c$ structure with determinant $L^{2\omega}$ exists on $(M,\omega)$.


The answer is related to [an answer I gave](https://mathoverflow.net/a/149356/21375) to a previous question of yours. Let me quote the relevant part:



> 
> Complex line bundles over a manifold are classified by their first Chern class; we have a bijection
>  $$\{\text{isomorphism classes of complex line bundles on $X$}\} \leftrightarrow H^2(X; \Bbb Z),$$
>  $$L \mapsto c\_1(L).$$
>  The first Chern class is additive with respect to tensor products, so we see that
>  $$c\_1(L^{\otimes 2}) = 2c\_1(L). \tag{$\ast$}$$
> 
> 
> 


On page 43 [Hochs's thesis](http://maths.adelaide.edu.au/peter.hochs/PhD_thesis_Peter_Hochs.pdf), it is claimed that under the integrality condition posed in your question, the line bundle
$$L^{2\omega} \otimes \wedge\_{\Bbb C}^{0, \dim(M)} (TM, J) \longrightarrow M$$
has a square root. Here $L^{2\omega}$ is a line bundle with first Chern class $2[\omega]$.


By my quoted answer, since $[\omega]+ \tfrac{1}{2}c\_1(\wedge\_{\Bbb C}^{0, \dim(M)} (TM, J))$ is integral, there exists a complex line bundle $\mathcal{L} \longrightarrow M$ such that
$$c\_1(\mathcal{L}) = [\omega]+ \tfrac{1}{2}c\_1(\wedge\_{\Bbb C}^{0, \dim(M)} (TM, J)).$$
Furthermore, we have that
\begin{align\*}
c\_1(\mathcal{L}^{\otimes 2}) & = 2[\omega]+ c\_1(\wedge\_{\Bbb C}^{0, \dim(M)} (TM, J)) \\
& = c\_1(L^{2\omega} \otimes \wedge\_{\Bbb C}^{0, \dim(M)} (TM, J)).
\end{align\*}
Hence $\mathcal{L}$ is a square root of $L^{2\omega} \otimes \wedge\_{\Bbb C}^{0, \dim(M)} (TM, J)$ (up to minor annoyances due to $2$-torsion in $H^2(M;\Bbb Z)$, which can be dealt with).


Now, if you look at Proposition D.50 in the [reference](http://math.ucsc.edu/~ginzburg/book-ams.pdf) Hochs mentions, it tells us that on the complex tangent bundle $(TM, J)$, we get a spin$^c$ structure just by choosing a line bundle on $M$. Take $\mathcal{L}$ as described above to be our complex line bundle, and let $\sigma(\mathcal{L})$ denote the associated spin$^c$ structure. Proposition D.50 further says that
$$\det(\sigma(\mathcal{L})) \cong K^\ast \otimes \mathcal{L}^{\otimes 2},$$
where $K^\ast$ is the anticanonical line bundle of $(M,J)$. Since $\mathcal{L}^{\otimes 2} \cong L^{2\omega} \otimes \wedge\_{\Bbb C}^{0, \dim(M)} (TM, J)$ and $K^\ast \otimes \wedge\_{\Bbb C}^{0, \dim(M)} (TM, J)$ is the trivial line bundle, we find that
$$\det(\sigma(\mathcal{L})) \cong L^{2\omega}.$$
The remark in Hochs's thesis you are asking about is just saying that this spin$^c$ structure on $M$ with determinant $L^{2\omega}$ exists.


