# Intuition behind Riemann's bilinear relations

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.

1. What is the geometric meaning behind the fact that we can choose a normalized basis?
2. What is the geometric meaning behind the bilinear relations?
3. 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.

• Sorry but can we do the converse of what the Riemann relations say? That is, given a Riemann surface $X$ of genus $g>0$ and given a basic $(\omega_i)$ for $\Omega(X)$. Can one choose a syplectic basic $(a_i,b_i)$ for $H_1(X,\mathbb{R})$ such that $\int_{a_i}\omega_1=\delta_{i1}$? Thanks. – user49533 Nov 15 '14 at 13:17

Mathwonk has given a good answer to part 1. Here's how I think about part 2.

Let's start with a good old differential geometry computation, nothing complex in sight. Let $X$ be a genus $g$ surface, with $a_1$, ..., $a_g$, $b_1$, ..., $b_g$ a standard basis. Let $\omega$ and $\eta$ be two one-forms. Let $(u_1, u_2, \ldots, u_{2g})$ be the integrals $(\int_{a_1} \omega, \ldots, \int_{a_g} \omega, \int_{b_1} \omega, \ldots, \int_{b_g} \omega)$. Let $(v_1, \ldots, v_{2g})$ be the same integrals for $\eta$. Now, in terms of the $u$'s and the $v$'s, what is $\int_X \omega \wedge \eta$? The answer, which I leave for you to check, is $u_1 v_{g+1} + u_2 v_{g+2} + \cdots + u_g v_{2g} - u_{g+1} v_1 - u_{g+2} v_2 - \cdots - u_{2g} v_g$. It will be notationally convenient to write $J$ for the $(2g) \times (2g)$ block matrix $\left( \begin{smallmatrix} 0 & -\mathrm{Id} \\ \mathrm{Id} & 0 \end{smallmatrix} \right)$. So the above formula is $\int_X u \wedge v = u J v^T$.

Now, this formula is equally correct when $\omega$ and $\eta$ are $\mathbb{C}$-valued $1$-forms (just look at the real and complex parts separately). Let $\omega_1$, ..., $\omega_g$ be a basis for the holomorphic $(1,0)$ forms. Let $P$ be the $g \times (2g)$ matrix whose $k$th row is $(\int_{a_1} \omega_k, \ldots, \int_{a_g} \omega_k, \int_{b_1} \omega_k, \ldots, \int_{b_g} \omega_k)$

Then two interesting things happen:

(1) The form $\omega_k \wedge \omega_{\ell}$ is identically zero, so $\int_X \ \omega_k \wedge \omega_{\ell}=0$. In terms of matrices, this means $P J P^T =0$. Conceptually, the row span of $P$ is Lagrangian for the symplectic form $J$.

(2) For any $(1,0)$ form $\omega$, the form $\mathrm{Im}(\omega \wedge \overline{\omega})$ is real valued and everywhere nonnegative, so $\mathrm{Im} \int_X \ \omega \wedge \overline{\omega} \geq 0$ (and strictly $>0$ unless $\omega=0$). In terms of matrices, this means $\mathrm{Im}\left( P J \overline{P}^T \right)$ is positive definite.

Now, everything I've written is true for an arbitrary basis $\omega_k$, and that's how I like to think about it conceptually. When working computationally, it is convenient to normalize the $\omega$'s to be dual to the $a$'s, so $P$ looks like $\left( \begin{smallmatrix} \mathrm{Id} & Q \end{smallmatrix} \right)$. Translating (1) and (2) into conditions on $Q$ gives you Riemann's bilinear relations.

• This is an excellent answer. I must confess that I never properly understood the Riemann bilinear relations, in spite of using their consequences (Siegel modular space). Thank you for finally demystifying them for me! – Victor Protsak Sep 29 '10 at 5:23

1) I guess in question 1 you are asking why the analytic genus (dimension of holomorphic one forms) equals the topological genus (1/2 the rank of 1st homology). This seems to follow from the hodge estimate giving the upper bound, and the explicit construction of the requisite number of independent forms.

2) , 3) I guess this is related to the positive curvature of projective space and the fact that the existence of a metric with positive curvature and integral cohomology class is thus not only necessary but also sufficient for a variety to be embeddable as a subvariety of projective space, (kodaira embedding theorem).

sorry if this is vague or wrong, it has been over 30 years since I read this stuff. see kodaira-morrow.

oh yes, they are also related to the intersection pairing of curves on topological surfaces.

Remark: David Speyer's answer proves the bilinear relations are true. This is the usual proof, as in say Gunning's Lectures on Riemann surfaces. My answer gives you their geometric consequences. I.e. they verify the hypotheses of Kodaira's embedding theorem, that the Jacobian has a positive integral line bundle as Arapura explained, hence imply the Jacobian is a projective variety.

edit: By the way, there is another important consequence of Riemann's first bilinear relation: it implies the modern version of Abel's theorem. I.e. by Hodge theory, the holomorphic differential forms H^0(K) lie inside the first complex cohomology H^1, and the quotient can be viewed as H^1(O). Since H1 is dual to H^1, the dual space of H^1(O) is isomorphic to the subspace H(0,1) of H1 which is orthogonal to H^0(K). Riemann's 1st bilinear relation says precisely that the Poincare duality isomorphism from H^1 to H1 carries H^0(K) onto that subspace H(0,1), hence the transpose isomorphism carries H^(0,1) ≈ H^1(O) onto H^0(K). Since Poincare duality is defined over the integers, this isomorphism induces an isomorphism from the Picard variety H^1(O)/H^1(Z) to the Albanese variety H^0(K)/H1(Z). This is called the abstract "Abel's theorem".

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.