# Riemann relations

Posted on November 2, 2017 by Dima

tags: Abelian varieties, positivity

I’ll throw in some coordinate-free-ness, for originality’s sake, and because I hate matrices.

Let $$V$$ be a complex vector space and let $$\Lambda$$ be a finitely generated torsion-free Abelian group such that $$\Lambda \otimes \mathbb{R}\cong V$$ as $$\mathbb{R}$$-vector spaces. The data of $$V$$, $$\Lambda$$ and a Hermitian form $$H: V \otimes V \to \mathbb{C}$$ such that $$H|_{\Lambda \times \Lambda} \subset \mathbb{Z}$$ gives rise to an Abelian variety and a line bundle endowed with a Hermitian metric.

The inclusion $$\Lambda \hookrightarrow V$$ extended linearly to a map of $$\mathbb{R}$$-spaces $$\Lambda \otimes \mathbb{R}\to V$$ is denoted $$\Pi$$.

Let $$A \in GL(\Lambda)$$ be an intergral linear invertible self-map such that $$(Ax, y) = -(x,Ay)$$ wrt the standard inner product on $$\Lambda$$. Via $$\mathbb{R}$$-linear extension this induces a symplectic form on $$V$$, call it $$\omega$$. Define the following Hermitian metric on $$V$$ $H(x,y) = \omega(Ix,y) + i \omega(x,y)$ where $$I$$ is the complex structure on $$V$$. One checks that it is $$\mathbb{C}$$-linear in the first argument, $$\mathbb{C}$$-antilinear in the second. Further, $H(x,y) = \overline{H(y,x)} \qquad \textrm{ iff }\qquad \omega(Ix,Iy) = \omega(x,y)$

Indeed, $H(y,x) = \omega(Iy,x) + i \omega(y,x) = -\omega(x,Iy) - i \omega(x,y)$ and $$-\omega(x,Iy) = \omega(Ix,y)$$ iff $$\omega(Ix,Iy) = \omega(x,y)$$.

Since $$\Pi$$ is an isomorphism and in particular surjective, in order to check that $$H$$ is Hermitian we need to check that $$\omega(I \Pi x, I \Pi y) = \omega(\Pi x, \Pi y)$$ for all $$x,y \in \Lambda \otimes \mathbb{R}$$. Remark now that from definition of $$\omega$$, $$\omega(\Pi x, \Pi y)=(Ax,y)$$.

Pass to the complexification: let $$\Pi \otimes \mathbb{C}: (\Lambda \otimes \mathbb{R}) \otimes_{\mathbb{R}} C \to V \otimes_{\mathbb{R}} \mathbb{C}$$ be the $$\mathbb{C}$$-linear extension of $$\Pi$$. The vector space $$V$$ decomposes into direct sum of eigenspaces of $$I$$: $$V = V^{1,0} \oplus V^{0,1}$$ where $$I$$ acts by $$i$$ on the first piece and by $$-i$$ on the second. Denote $$\Pi' := \pi_{V^{1,0}} \circ \Pi \otimes \mathbb{C}$$. By definition, $$\Pi'$$ is the same as the composition of $$\Pi$$ and identification between $$V \subset V \otimes C$$ and $$V^{1,0}$$ via the projection on $$V^{1,0}$$. When written out in some basis $$\Pi'$$ is called a period matrix. If $$\Lambda \cong \mathbb{Z}^{2g}$$, it is a $$g \times 2g$$ matrix which has the coordinates of the generators of $$\mathbb{Z}^{2g}$$ as its columns.

$\begin{array}{ccccc} \Lambda \otimes \mathbb{R}& \to^\Pi & V & \to & V \otimes \mathbb{C}\\ \downarrow & & \downarrow & \swarrow & \\ \Lambda \otimes \mathbb{C}& \to^{\Pi'} & V^{1,0} & \\ \end{array}$

After restriction to $$V^{1,0}$$ it is clear that $$\omega$$ respects $$I$$ if and only if $$\omega_\mathbb{C}(i \Pi' x, i \Pi'y) = \omega_\mathbb{C}(\Pi' x, \Pi'y)$$ for all $$x,y \in \Lambda \otimes C$$.

Define $J = (\Pi\otimes \mathbb{C})^{-1} (I\otimes \mathbb{C}) (\Pi\otimes \mathbb{C})$ Notice that it has the property $$i \Pi' = \Pi' J$$. Also notice that $$\omega_\mathbb{C}(\Pi' x, \Pi' y) = (Ax, y)$$. Then the condition to check becomes $(A J x, J y) = (A x, y)$ or $$J^* A J=A$$. Rewrite the latter identity using the definition of $$J$$, denoting $$\Phi = \Pi \otimes \mathbb{C}$$ $\Phi^* (I\otimes \mathbb{C})^* (\Phi^{-1})^* (A \otimes \mathbb{C}) \Phi^{-1} (I\otimes \mathbb{C}) \Phi = A$ so, using self-adjointness of $$I \otimes \mathbb{C}$$, $I\otimes \mathbb{C}(\Phi^{-1})^* (A \otimes \mathbb{C}) \Phi^{-1} (I\otimes \mathbb{C}) = (\Phi^{-1})^* A \Phi^{-1}$ and further (omitting $$\otimes \mathbb{C}$$ for simplicity of notation) $I (\Phi A^{-1} \Phi^*)^{-1} I = (\Phi A^{-1} \Phi^*)^{-1}$ Given that $$I \otimes \mathbb{C}$$ is semi-simple, the restriction of $$(\Phi A^{-1} \Phi^*)^{-1}$$ to $$V^{1,0}$$ must take values in $$V^{0,1}$$ (if the projection of $$(\Phi A^{-1} \Phi^*)^{-1}(x)$$ on $$V^{1,0}$$ is non-zero element $$y$$, then $$i y i = y$$ which is absurd), and similar reasoning works for $$V^{0,1}$$. We conclude, computing $$\pi_{1,0} \circ (\Phi A^{-1} \Phi^*)^{-1}$$ and $$\pi_{0,1} \circ (\Phi A^{-1} \Phi^*)^{-1}$$, that $\Pi' A^{-1} (\Pi')^* = 0 \textrm{ and } \overline{\Pi'} A^{-1} (\overline{\Pi'})^*$ [the trick is to write down the operator to be inverted as an inverse of block matrix, and observe that the “determinant” term does not matter for vanishing] The second equality is the conjugation of the first, and we are done clarifing the condition for $$H$$ to be Hermitian.

Now let us compute $$H(x,x)$$. As before it will be convenient to pass to the complexification of $$\omega$$ and put $H_\mathbb{C}(x,y) = \omega_\mathbb{C}(I \otimes \mathbb{C}x, y) +i \omega_\mathbb{C}(x,y)$ The second term is defined by the proprety: $$\omega_\mathbb{C}(\Phi x, \Phi y) = (Ax,y)$$, and hence $\omega_\mathbb{C}(x, y) = (A\Phi^{-1}x,\Phi^{-1}y)$ so $\omega_\mathbb{C}(x, y) = ((\Phi^{-1})^* A \Phi^{-1}x, y) = ((\Phi A \Phi^*)^{-1} x, y)$ And so the first term is: $\omega_\mathbb{C}(I \otimes \mathbb{C}x, y) = ((\Phi^* A \Phi)^{-1} (I \otimes \mathbb{C})x, y)$ Assuming $$H$$ to be Hermitian, the restriction of $$(\Phi^* A^{-1} \Phi)^{-1}$$ to $$V^{1,0}$$ writes $$(\Pi' A^{-1} (\overline{\Pi'})^*)^{-1} ) D$$, and its restriction to $$V^{0,1}$$ is $$(\overline{\Pi'} A^{-1} (\Pi')^*)^{-1}$$. Therefore $$H(x,x)$$ is $(2i (\Pi' A^{-1} (\overline{\Pi'})^*)^{-1} x,x)$ on $$V^{1,0}$$. Since positivity depends only on the sign of eigenvalues, passing to the inverse operator does not affect it, and so the condition for positivity of $$H$$ is $(i (\Pi' A^{-1} (\overline{\Pi'})^*)^{-1} x,x) > 0$ for all $$x \in V \hookrightarrow V \otimes \mathbb{C}$$.