TITLE: Intuition behind Riemann's bilinear relations 
QUESTION [43 upvotes]: I'm trying to understand Riemann's bilinear relations on the normalized period matrix of a Riemann surface.  Recall that they say the following.  Let $X$ be a compact Riemann surface of genus $g>0$.  Fix a standard basis $\{a_1,b_1,\ldots,a_g,b_g\}$ for $H_1(X,\mathbb{Z})$.  We can then choose a basis $\omega_1,\ldots,\omega_g$ for the space of holomorphic $1$-forms on $X$ with the following property.  Define $A_{i,j}=\int_{a_j} \omega_i$.  Then $A_{i,j} = \delta_{i,j}$.  We call $\omega_1,\ldots,\omega_g$ a normalized basis for the set of holomorphic 1-forms on $X$.
Riemann's bilinear relations say that if $\omega_1,\ldots,\omega_g$ is a normalized basis for the holomorphic 1-forms on $X$ and if we define $B_{i,j}=\int_{b_j} \omega_i$, then the matrix $B=(B_{i,j})$ has the following two properties.  First, it is symmetric.  Second, its imaginary part is positive definite.
I understand the proof of this result, but I feel like I have very little geometric intuition as to why it is true.  This leads to the following three questions.

What is the geometric meaning behind the fact that we can choose a normalized basis?
What is the geometric meaning behind the bilinear relations?
One important consequence of the bilinear relations is that the Jacobian of a Riemann surface is an abelian variety.  What is the geometric intuition behind the relationship between the bilinear relations and the fact that we can make the Jacobian into a variety?

Thank you very much for any help.

REPLY [8 votes]: I don't know if this will make matters better or worse for you, but this can be expressed in
basis free language as follows: A  Hodge structure of type $\{(1,0), (0,1)\}$ consists
of a lattice $H$ together with decomposition $H\otimes \mathbb{C}= H^{10}\oplus H^{01}$
such that $\overline{H^{10}}= H^{01}$.When $X$ is a compact Riemann surface $H=H^1(X,\mathbb{Z})$ carries such a Hodge structure where $H^{10}$ is the space of holomorphic $1$-forms. 
A polarization is a skew symmertric integer valued pairing  $\langle,\rangle$ on $H$ satisfying the Riemann bilinear relations, that is it vanishes on $H^{10}$ (i.e. the space is isotropic) and $-i\langle \alpha,\overline{\alpha}\rangle>0$ for $\alpha\in H^{10}$ nonzero. For a Riemann surface, the cup product pairing
$\langle \alpha,\beta\rangle= \int_X \alpha\wedge \beta$ does the job as explained in David Speyer's answer. 
Note that for any Hodge structure as above, there is a
torus $J=H^{01}/H$ which coincides with the
 Jacobian $H^{01}/H= H^1(X,\mathcal{O}_X)/H^1(X,\mathbb{Z})$ when this comes from  a Riemann surface. In general, there is not much more one can do. But when we have a polarization, we can construct sufficiently many quasi-periodic holomorphic functions (called theta functions)
to embed $J$ into projective space. In other words, $J$ is an abelian variety,
This can be done by explicitly writing down certain Fourier series which
can be shown to converge using the Riemann relations. Alternatively, under the identification
$\wedge^2H\cong H^2(J,\mathbb{Z})$, a polarization defines an integral cohomology class.
The Riemann bilinear relations are exactly the conditions for this to be represented
by a positive $(1,1)$ form, so $J$ can be embedded into projective space by Kodaira's
embedding theorem as Mathwonk indicated.