Given a real manifold $M$ with symplectic $2$-form $\omega$,
one can ask whether the cohomology class $[\omega] \in H^2(M;{\mathbb R})$ lies in the image of 
$H^2(M;{\mathbb Z})$. If so, one can ask for a line bundle ${\mathcal L}$
with $c\_1({\mathcal L}) = [\omega]$ (or even better, 
a connection $\alpha$ on $\mathcal L$ whose curvature $curv(\alpha)$ is $\omega$).


In the weakest definition of "geometric quantization", one puts a compatible
almost complex structure on $M$ and uses it to define the pushforward
of $\mathcal L$ to a point in $K$-theory. Call this $Q(M)$.


Are there examples worked out somewhere in which $H^2(M;{\mathbb Z})$
has torsion, so that $\mathcal L$ is not uniquely determined by $\omega$?
Can $Q(M)$ depend on the choice of $\mathcal L$?


I don't have any very good reason for asking this, other than I've felt
it to be a hole in my understanding of geometric quantization. The spaces I care about quantizing never seem to have torsion in $H^2$.



What are the implications of torsion in H^2 for geometric quantization?
If L\_1 and L\_2 are two line bundles on a manifold $M$ that differ by torsion, then their Chern characters
$$ch(L\_1) = 1 + c\_1(L\_1) + \frac{1}{2}c\_1(L\_1)^2 + \cdots$$
$$ch(L\_2) = 1 + c\_1(L\_2) + \frac{1}{2}c\_1(L\_2)^2 + \cdots$$
agree, if only because the right-hand sides of these formulas are taking place in $H^\*(M;\mathbb{Q})$ and there $c\_1(L\_1) = c\_1(L\_2)$. If $L\_1$ and $L\_2$ are prequantizations of a symplectic structure on M, and if we can give M a compatible complex structure, then we have Q(L\_1) = Q(L\_2) by the Riemann-Roch formula
$$Q(L\_1) = \int ch(L\_1) td(M) = \int ch(L\_2) td(M) = Q(L\_2)$$


We do not get a finer equation than $Q(L\_1) = Q(L\_2)$. Below the line is an example of $M$, $L\_1$, and $L\_2$ for which $H^\*(M;L\_1)$ and $H^\*(M;L\_2)$ are not isomorphic, that I thought of first. I'll integrate these answers later.




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I had written an answer with a bogus example of a Kahler manifold with two prequantum line bundles $L\_1$ and $L\_2$ for which $Q(L\_1)$ was different from $Q(L\_2)$. Here's my attempt to repair it--the best I can do is a symplectic manifold with two prequantum line bundles $L\_1$ and $L\_2$ for which $H^0(L\_1) \neq H^0(L\_2)$. But there is definitely some higher cohomology that might cancel this difference when you compute $Q$. I don't have a guess for what happens.


I am using "prequantum line bundle" to mean a line bundle whose real chern class (times $2\pi i$?) is equal to the class of the symplectic form. The premise of my example is that if we have two such line bundles $L\_1$ and $L\_2$ whose integral Chern classes differ by $n$-torsion, then $L\_1$ and $L\_2$ will become isomorphic over some $\mathbb{Z}/n$-cover $Y$ of $M$. (One can see this by thinking about the bundle-theoretic meaning of the Bockstein map $H^1(M;\mathbb{Z}/n) \to H^2(M;\mathbb{Z})$ and the long exact sequence it fits into.).


We can organize the data this way: $f:Y \to M$ is a $\mathbb{Z}/n$-bundle over $M$ with a $\mathbb{Z}/n$-equivariant prequantum line bundle $L$, and $L\_1$ and $L\_2$ are built from $L$ by the formula
$$H^0(U;L\_i) = \alpha\_i\text{-eigenspace of }H^0(f^{-1}(U);L)$$
where $\alpha\_1$ and $\alpha\_2$ are two different characters of $\mathbb{Z}/n$.


So to find such an $M$, we should find a prequantized symplectic manifold $Y$ equipped with a free $\mathbb{Z}/n$-action, such that $\mathbb{Z}/n$ does not act so uniformly on $H^0(Y;L)$ (i.e. so that the different eigenspaces of a generator for $\mathbb{Z}/n$ have different dimensions). The only source of prequantized symplectic manifolds available to me are projective varieties, and the only source of free $\mathbb{Z}/n$-actions are those on projective hypersurfaces.


So, let $Y$ be the degree 5 hypersurface in $\mathbb{P}^3$ given by the equation $x^5 + y^5 + z^5 + w^5 = 0$. If $\eta$ is a 5th root of unity, then the $\mathbb{Z}/5$-action given by $x \mapsto x$, $y \mapsto \eta y$, $z \mapsto \eta^2 z$ and $w \mapsto \eta^3 w$ is free (to check the absence of fixed points, it's important that the eigenvalues of the different letters are distinct). If $L$ is the restriction of $O(1)$ on $P^3$ to $Y$ then $H^0(Y;L) = H^0(P^3;O(1))$, which is the vector space spanned by $x,y,z,$ and $w$, and the generator of $\mathbb{Z}/5$ acts by $x \mapsto x$, $y \mapsto \eta^{-1} y$, $z \mapsto \eta^{-2} z$, and $w \mapsto \eta^{-3}w$.


The conclusion is that if $L\_1$ is the line bundle on $M = Y/(\mathbb{Z}/5)$ corresponding to the eigenvalue 1 (or $\eta^{-1}$ or $\eta^{-2}$ or $\eta^{-3}$) then $H^0(M;L\_1)$ is one-dimensional, and if $L\_2$ is the line bundle on $M$ corresponding to the eigenvalue $\eta^{-4} = \eta$ then $H^0(M;L\_2)$ is zero-dimensional. 


More generally one can take (degree n hypersurface in $\mathbb{P}^r$)/(action of $\mathbb{Z}/n$), but to make the action free one needs $n > r+1$ (and maybe $n$ prime). Taking large $n$ like this creates cohomology in $H^{r-1}(M;L\_1)$ and $H^{r-1}(M;L\_2)$.


