If I understand correctly, every line bundle $L$ over the (2-dim) torus can be obtained from a quotient of $\mathbb{R}^2 \times \mathbb{C}$ by a $\mathbb{Z}^2$ lattice action. Different line bundles are obtained depending on how the lattice 'twists' the fibers. The connection $D = d + (1/2)(pdq-qdp)$ descends to a connection on $L$, as does the canonical Hermitian pairing.  

My question is this: Suppose we wanted to geometrically quantize the torus by a Kähler polarization. Then we would want to find sections $s$ of $L$ such that $D\_v s = 0$ for antiholomorphic $v$. How can we write down the explicit sections satisfying this equation in the case of the torus with a Hermitian bundle $L$ as above? And how can we prove that for a general compact manifold the dimension of the space of solutions is finite? (I know how to solve the differential equation for the case of $\mathbb{R}^2 \times \mathbb{C}$ and get the Bargmann-Fock space, but it is unclear how one can solve the analogous problem for compact manifolds - at least unclear to me)



Connections on line bundles over the torus
The Kaehler polarisation is a choice of *complex structure*, $I$, on your torus $M\simeq S^1\times S^1$, which is *compatible* with the symplectic form. In other words, your torus becomes a *complex manifold* $X=(M,I)$, and $I$ satisfies $I^t\omega I=\omega$, $g=\omega I>0$. In particular, this tells you that $\omega$ is a positive $(1,1)$-form, i.e., locally 
$\omega = \frac{i}{2}h dz \wedge d\overline{z}$, $h>0$. Of course, since $\dim X=1$, $\omega$ is a $(1,1)$-form for any choice of $I$, but this is not so in higher dimensions.


The connection $D$ can now be decomposed as $D= D^{1,0} + D^{0,1}$. Explicitly, 
$ D^{0,1} = \frac{1}{2}(1+iI)D$ and $ D^{1,0} = \frac{1}{2}(1-iI)D$. Or, if you prefer, choose a local frame $s$ of $L$, and write
$$
D = d + A\_1 dq + A\_2 dp = (\partial + B\_1 dz) + (\overline{\partial} + B\_2 d\overline{z})
$$


Since $\omega$ is of type $(1,1)$, the condition
$D^2 = -i\omega$ translates to
$$
(D^{0,1})^2=0,\quad (D^{1,0})^2=0,\quad D^{0,1}D^{1,0} +D^{1,0} D^{0,1}=-i\omega .
$$
Again, the first two equations hold automatically if $\dim X=1$, but not in general.


This means that on $L$ you get the structure of a *holomorphic* line bundle $\mathcal{L}$ by taking the Dolbeault operator to be $\overline{\partial}\_L= D^{0,1}$. Locally, if you choose a (smooth) frame $s$, this is the above $\overline{\partial}\_L =
\overline{\partial} + B\_2 d\overline{z}$.


Then the "Hilbert space" is $H^0(X,\mathcal{L})= \ker\overline{\partial}\_L \subset A^0(L)$,
where $A^0(L)$ is the (infinite-dimensional) vector space of global smooth sections of $L$. With respect to a local trivialisation, a section $\sigma =fs$ is holomorphic if
$$
\frac{\partial f}{\partial\overline{z}} + B\_2 f=0.
$$


The space $H^0(X,\mathcal{L})$ is finite-dimensional for very general reasons. If you actually want to know how big it is, you look at the (Hirzebruch)-Riemann-Roch formula, which tells you that
$$
\dim H^0(X,\mathcal{L})- \dim H^1(X,\mathcal{L}) = \deg \mathcal{L} + 1-g = 
\deg \mathcal{L}.
$$
Then $\dim H^0(X,\mathcal{L})=\deg \mathcal{L}$ if $\deg\mathcal{L}>0$, $\dim H^0(X,\mathcal{L})= 0$ if $\deg\mathcal{L}<0$ (or if $\deg\mathcal{L}=0$ and $\mathcal{L}$ is nontrivial), and $\dim H^0(X,\mathcal{L})=1$ if $\mathcal{L}=\mathcal{O}\_X$. You can describe the elements of $H^0(X,\mathcal{L})$ quite explicitly
using theta-functions.


Notice that everything - in particular $H^0(X,\mathcal{L})$ - depends on the choice of complex structure $I$, and even the dimension may jump as you vary the complex structure (if $\deg L=0$). But if you choose your prequantum line bundle to be sufficiently positive, the
vector spaces $H^0(X,\mathcal{L})$ "glue" into a vector bundle over the upper half-plane.


**ADDENDUM:**


Let me spell out the relation with theta-functions a bit more explicitly. 
This is a huge classical topic (see the references at the end), so I'll be a bit sketchy.


Write $V=\mathbb{C}$ and $\Lambda = \mathbb{Z}\oplus \tau \mathbb{Z}$, $\textrm{Im}\tau>0$. Then
$\pi: V\to X=V/\Lambda$ is the universal covering map, and it is a local biholomorphism. 


If your symplectic structure descends from the form $\omega= c\ dq\wedge dp$ on $V=\mathbb{C}$, the "Bohr-Sommerfeld condition" tells you that $c=\frac{2\pi}{\textrm{Im}\tau}n$, 
$n\in \mathbb{Z}$. Notice that $\frac{1}{2\pi}\omega$ is the imaginary part of the hermitian form $H(z,w)=\frac{n}{\textrm{Im}\tau}z\overline{w}$ on $V$ and takes integer values on the lattice.


Suppose $\mathcal{L}$ is as above, with 
$c\_1(\mathcal{L}) =\frac{1}{2\pi}\left[\omega\right]$, $\deg \mathcal{L}=n$.


As any holomorphic line bundle on $V$ is trivial, we can choose a global holomorphic trivialisation 
$\pi^\ast \mathcal{L}\simeq V\times \mathbb{C}$, 
and identify $\mathcal{L}= \pi^\ast \mathcal{L}/\Lambda$
with $V\times\mathbb{C}/\sim$. The equivalence relation is 
$(z,t)\sim (z+\lambda, a(\lambda,z)t)$, $\lambda\in\Lambda=\pi\_1(X)$.
Here $a: \Lambda\times V\to \mathbb{C}^\times$ are the *multipliers* or *factors of automorphy*. They are 1-cocyles of $\Lambda=\pi\_1(X)$ with values in the entire nonvanishing functions $H^0(V,\mathcal{O}^\times\_V)$, i.e., they 
are holomorphic in $z$ and
satisfy
$$
a(\lambda+\mu,z)= a(\mu,\lambda+z)a(\lambda,z).
$$
The multipliers do not, in general, take values in $U(1)$!
Changing the trivialisation of $\pi^\ast \mathcal{L}$ replaces $a$ by a coboundary, and
we have 
$H^1(\Lambda,H^0(V,\mathcal{O}^\times\_V))\simeq H^1(X,\mathcal{O}^\times\_X)$.


Now, we can identify $H^0(X,\mathcal{L})\simeq H^0(V,\mathcal{O}\_V)^{\pi\_1(X)}$.
I.e., we can identify $H^0(X,\mathcal{L})$ with the entire functions which satisfy the
functional equation
$$
\theta(z+\lambda) = a(\lambda,z)\theta(z),
$$
and these are by definition *theta-functions*. 


The explicit structure of the factors of automorphy is known. For instance, in each degree $n$ you have a canonical choice, 
$$
a\_0(\lambda,z) = \exp\left( n\pi i ab \right)\exp \frac{n\pi}{\textrm{Im}\tau}\left(z\overline{\lambda}+ \frac{1}{2}|\lambda|^2\right),\quad \lambda=a+b\tau.
$$
You can get the other bundles of the same degree by multiplying $a\_0$ with characters
$\chi\in \textrm{Hom}(\Lambda,U(1))\simeq Pic^0(X)$.


Finally, a word about the inner product. We can identify $A^0(\mathcal{L})$, the smooth sections of 
$\mathcal{L}$, with smooth functions on $V$, satisfying the same functional equation
as the theta-functions. In terms of this identification the fibrewise hermitian product on $\mathcal{L}$ is
$$
\langle f,g \rangle(z) = f(z)\overline{g(z)}\exp \left(-\frac{n\pi}{\textrm{Im}\tau}|z|^2 \right).
$$
The $L^2$-inner product on $H^0(X,\mathcal{L})$ is obtained by integrating this quantity over $X$. 


Line bundles on a complex torus are described in great detail in Birkenhake and Lange, *Complex Abelian Varieties* (especially Chapter 2), Griffiths and Harris, *Principles of AG*, Chapter 2, Section 6, and
D.Mumford, *Abelian Varieties*, Chapter 1, Section 2.


Here we have been looking at a 1-dimensional torus, but if you decide to consider the geometric quantisation of $U(1)$ Chern-Simons theory on a higher genus curve, you will need the case of tori of higher dimensions.


