If I understand correctly, every line bundle $L$ over the (2dim) 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 $D = d + (1/2)(pdqqdp)$ 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 BargmannFock space, but it is unclear how one can solve the analogous problem for compact manifolds  at least unclear to me)
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}(1iI)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 (infinitedimensional) 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 finitedimensional for very general reasons. If you actually want to know how big it is, you look at the (Hirzebruch)RiemannRoch formula, which tells you that $$ \dim H^0(X,\mathcal{L}) \dim H^1(X,\mathcal{L}) = \deg \mathcal{L} + 1g = \deg \mathcal{L}. $$ 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 thetafunctions.
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 halfplane.
ADDENDUM:
Let me spell out the relation with thetafunctions 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 "BohrSommerfeld 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 1cocyles 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 thetafunctions.
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 thetafunctions. 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 1dimensional torus, but if you decide to consider the geometric quantisation of $U(1)$ ChernSimons theory on a higher genus curve, you will need the case of tori of higher dimensions.

$\begingroup$ Thank you. I'm stuck on one point: suppose I choose $A_1 = A_2 = 1/2$. Then on the complex plane a holomorphic section has the form $f(z)=h(z)exp((1/2)z^2)$, where $h$ is holomorphic in the usual sense. In order to define a section of the torus, however, we need to have $f(z+a+bi)=T(a,b)f(z)$ where T is always unitary (In particular, the bundle on the torus is obtained as $\mathbb{C}\times\mathbb{C}/(z,t)~(z+a+bi, T(a,b)t)$, and T must be unitary to preserve the Hermitian pairing). I don't see how we can choose h(z) to obtain such a condition  I also don't see why theta functions help? $\endgroup$ – Blake Apr 20 '12 at 15:59

$\begingroup$ I meant $(z,t)$ is identified with $(z+a+bi, T(a,b)t)$ of course, and in fact I think $T$ might depend on $z$ as well. But regardless it needs to be unitary to preserve the Hermitian form. $\endgroup$ – Blake Apr 20 '12 at 16:01

$\begingroup$ I added some explanation about the relation to theta functions, hopefully it helps. In two words, the complex structure is crucial: both on $X$ (via $\tau$) and on $L$ (via the factor of automorphy). Also, your $T$'s are not unitary, unless the degree is zero. And you want to avoid that case, because the dimension of the space of sections may jump. $\endgroup$ – Peter Dalakov Apr 24 '12 at 21:36