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 $P\to M$ 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 In Page 43.


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 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, 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 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.


Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.