I am trying to clarify certain subtleties regarding prequantization bundles over symplectic manifolds, for which I haven't found any clear explanation so far.
Let me fix some definitions first.
Definition 1: Let $(M, \omega)$ be a symplectic manifold. The symplectic form $\omega$ is called integral if the cohomology class $[\omega]$ of $\omega$ lies in the image of the natural homomorphism $H^2(M; \mathbb{Z}) \to H^2(M; \mathbb{R})$.
Definition 2: An integral symplectic form $\omega$ is called primitive if $\langle [\omega], H_2(M; \mathbb{Z}) \rangle = \mathbb{Z}$.
Definition 3: A prequantization bundle over a symplectic manifold is the data of a principal $S^1$-bundle $\pi : V \to M$ along with a contact form $\alpha$ on $V$ such that the two following points hold:
- $d\alpha = \pi^* \omega$.
- The Reeb vector field $R_{\alpha}$ of $\alpha$ generates the $S^1$-action on $V$.
Two remarks :
First of all, note that a symplectic manifold is the base of a prequantization bundle if and only if its symplectic form is integral.
For any connection $1$-form $\alpha'$ on the total space $V'$ of a principal $S^1$-bundle $\pi' : V' \to M'$ (not necessarily a prequantization bundle), one can show that $d \alpha' = \pi'^* \omega'$, for a certain $2$-form $\omega'$ on $M'$, which is called the curvature form of $\alpha'$. It appears that the cohomology class $[\omega']$ is independent of the choice of $\alpha'$, and is related to the Euler class $eu(\pi')$ of the bundle as follows : $eu(\pi') = - \frac{1}{\int_{S^1}\alpha'} [\omega']$, where $\int_{S^1} \alpha'$ is the integral of $\alpha'$ along any fiber. In particular, we have an identification $$ S^1 \simeq \mathbb{R} / \int_{S^1}\alpha' \mathbb{Z}, $$ and $$ \frac{1}{\int_{S^1}\alpha'} [\omega'] \in H^2(M; \mathbb{Z}). $$
Let us apply this second remark to a prequantization bundle, that is when we are given an integral symplectic form $\omega$, and a connection $1$-form $\alpha$ which satisfies $d\alpha = \pi^* \omega$. In this case, $\alpha$ is a contact form, and its Reeb vector field $R_{\alpha}$ generates the $S^1$-action. Moreover, as above, we have $$ eu(\pi) = - \frac{1}{\int_{S^1}\alpha} [\omega]. $$ In particular, $\frac{1}{\int_{S^1}\alpha} [\omega] \in H^2(M; \mathbb{Z})$.
THE QUESTIONS
1. The uniqueness of the contact form
First, it was mentioned to me that, given that $\omega$ and the principal bundle $\pi$ are fixed, the contact form $\alpha$ is unique. The argument was the following: assume that $\alpha'$ is another contact form on $V$ such that $d\alpha' = \pi^*\omega$ and whose Reeb vector field generates the $S^1$-action on $V$. Then we have $d\alpha' = d\alpha$, and therefore $\alpha' - \alpha = \pi^* \beta$, for a certain $1$-form on $M$. Define $$ \alpha_t := \alpha + \pi^*t\beta, \quad t \in [0,1]. $$ Then $\alpha_t$ is an isotopy of contact forms between $\alpha$ and $\alpha'$, so that the contact manifolds $(V, \xi = \ker \alpha)$ and $(V, \xi' = \ker \alpha')$ are isomorphic. Since moreover both Reeb flows $R_{\alpha}$ and $R_{\alpha'}$ generate the $S^1$-action, both contact forms must be equal.
Two points aren't clear to me in this argument :
Question 1: I'm having trouble showing that $\alpha_t$ is a contact form, which is equivalent to showing that $$ (d\alpha_t)^n \wedge \alpha_t = t((d\alpha')^n \wedge \alpha') + (1-t)((d\alpha)^n \wedge \alpha) \neq 0, \quad \text{where } \dim(V) = 2n + 1. $$ Indeed, both terms are not zero, but can't the sum be?
Question 2: How does the fact that the contact forms are isotopic and their Reeb vector fields generate the $S^1$-action prove that they are equal?
Question 3: if this is true, why isn't it mentioned anywhere ? Isn't it important to know that there can be only one contact structure on a prequantization bundle?
2. The integrality condition
For convenience, let us denote by $\hbar$ the quantity $\frac{1}{\int_{S^1}\alpha}$, which is independent of $\alpha$ (since $eu(\pi) = - \frac{1}{\int_{S^1}\alpha} [\omega]$, and $[\omega]$ is independent of the choice of $\alpha$). I would like to understand what it means that $\frac{1}{\hbar} [\omega] \in H_2(M; \mathbb{Z})$, and the relation between this fact and the fact that $\omega$ is integral.
Question 4: Does "$\omega$ is integral" mean that $\omega$ takes integral values on integral cycles?
Question 5: Is it equivalent to writing $\langle [\omega], H_2(M; \mathbb{Z}) \rangle \subset \mathbb{Z}$?
Question 6: What about torsion elements in $H_2(M; \mathbb{Z})$? How do we apply $[\omega]$ to them?
Question 7: If $\omega$ is integral, and $\frac{1}{\hbar}[\omega] \in H_2(M; \mathbb{Z})$, what can we say about $\frac{1}{\hbar}$: shouldn't it be an integer (see for instance Chapter 6.2 (d) of the book Geometry of Differential Forms by Morita, where the length of the fiber is $2\pi$) ?
Question 8: Is $\frac{1}{\hbar}[\omega]$ a preimage of $[\omega]$ by the homomorphism $H^2(M; \mathbb{Z}) \to H^2(M;\mathbb{R})$?
3. The primitivity assumption and the classification of prequantization bundles
I would also like to understand what it means for $\hbar$ if we assume that $\omega$ is primitive. One source of confusion is that it seems that the very definition of a prequantization bundle is not quite the same everywhere:
- Sometimes, it is defined as a principal bundle whose Euler class is equal to minus the cohomology class of the symplectic form, $eu(\pi) = -[\omega]$, along with the choice of a connection $1$-form $\alpha$ satisfying $d\alpha = \pi^* \omega$ (see for instance https://arxiv.org/pdf/1612.02205.pdf).
- Sometimes, it is defined as I did (see for instance https://arxiv.org/pdf/1308.3224.pdf).
As I unserstand it, the most general definition is the one I gave above, and the condition $eu(\pi) = - [\omega]$ should refer to the case where $\omega$ is primitive, or equivalently that $\hbar = 1$. However, I have trouble making this statement clear: if $\omega$ is primitive, then $$ \langle -\frac{1}{\hbar}[\omega], H_2(M;\mathbb{Z}) \rangle = -\frac{1}{\hbar} \mathbb{Z} = \langle eu(\pi), H_2(M; \mathbb{Z}) \rangle. $$
Question 9: What can we say about $\hbar$ in this situation? Can we use the Euler class to show that it must be equal to $1$?
Question 10: Since principal $S^1$-bundles are entirey determined by their Euler class, and assuming the uniqueness of the contact form as in the first section above, can we classify all the prequantization spaces over a given symplectic manifold $(M, \omega)$ with integral symplectic form? For instance, can we say something like : up to rescaling $\omega$, we can assume that it is primitive, and then all the prequantization spaces over $M$ are classified by $\mathbb{Z} \setminus \{0\}$: they are those principal $S^1$-bundles with Euler class being a non-zero integer multiple of $[\omega]$?
I apologize for the number of questions, and I thank you in advance for you help.
