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Let $g >1$ a natural number and $\mathbb{C}^g$ complex vector space which is isomorphic to $\mathbb{R}^{2g}$ is real vector space.

An additive subgroup $\Gamma \subset \mathbb{C}^g$ is called a lattice if there exist $2g$ vectors $\gamma_1,... \gamma_{2g}$, which are linearly independent over $\mathbb{R}$ such that $\Gamma= \mathbb{Z} \gamma_1 + ... + \mathbb{Z} \gamma_{2g}$.

Let $\Gamma, \Gamma' \subset \mathbb{C}^g$ be two lattices with $\Gamma= \mathbb{Z} \gamma_1 + ... + \mathbb{Z} \gamma_{2g}$ and $\Gamma'= \mathbb{Z} \gamma' _1 + ... + \mathbb{Z} \gamma' _{2g}$. Is there a characterization when two quotient groups $\mathbb{C}^g / \Gamma$ and $\mathbb{C}^g / \Gamma'$ are isomorphic as abelian groups in dependence of a certain relation between lattices $\Gamma$ and $\Gamma'$?

My first guess was $\mathbb{C}^g / \Gamma \cong \mathbb{C}^g / \Gamma'$ if and only if there exist a $M \in GL_{2g}(\mathbb{Z})$ with $M \cdot \Gamma = \Gamma' $ and $M \cdot \gamma_i = \gamma_i '$. Or should I require that moreover $M$ lives in $O_{2g}(\mathbb{Z})$, $O_{2g}(\mathbb{Z})$ or even a scalar matrix $c \cdot Id$ with $c \in \mathbb{C} \backslash \{0\}$?

My motivation is my question about Jacobians of Riemann surfaces from Forster's Lectures on Riemann Surfaces. We have a compact Riemann surface $X$ of genus $g$ and Forster's construction of the Jacobian $Jac(X)$ bases on an explicit choice of basis $\omega_1,..., \omega_g$ of the $\mathbb{C}$-space of holomorphic $1$-forms $\Omega (X)$. Forster shows that the subspace of $\mathbb{C}^g$ consisting of all vectors

$$(\int_{\alpha} \omega_1, \int_{\alpha} \omega_2, ... \int_{\alpha} \omega_g)$$

where $α$ runs through the fundamental group $\pi(X)$ form a lattice $\Gamma= \mathbb{Z} \gamma_1 + ... + \mathbb{Z} \gamma_{2g} \operatorname{Per}(\omega_1,..., \omega_g) \subset \mathbb{C}^{g}$ and Jacobian is defined by $Jac(X):= \mathbb{C}^g/ \operatorname{Per}(\omega_1,..., \omega_g)$. At first glace this definition seems to be bad because of a choice of the basis $\omega_1,..., \omega_g$. But Forster remarked also without providing a proof that a choice of a different basis leads to an isomorphic $Jac(X)$.

That is I have to know firstly when two quotients $\mathbb{C}^g / \Gamma$ and $\mathbb{C}^g / \Gamma'$ with lattices $\Gamma$ and $\Gamma'$ are considered as isomorphic abelian groups (I assume that Forster not consider they additionally as compact complex manifolds, or what type of isomorphy Forster consider) and why choosing different basis' gives isomorphic Jacobians?

user267839
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  • It seems that as abelian groups they are all isomorphic? – Arctic Char Aug 23 '20 at 16:00
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    Ok, so if $\Gamma := \mathbb{Z} \gamma_1 + ... + \mathbb{Z} \gamma_{2g} \subset \mathbb{C}^g$ is a lattice and $M \in GL_{2g}(\mathbb{C})$ arbitrary, then every subset $\Gamma' := \mathbb{Z} \gamma'1 + ... + \mathbb{Z} \gamma'{2g} \subset \mathbb{C}^g$ with $\gamma'_j := M \cdot \gamma_j$ is a lattice in usual sense as well. So $\mathbb{C}^g / \Gamma $ and $ \mathbb{C}^g / \Gamma' = \mathbb{C}^g / M \cdot \Gamma $ are always isomorphic as groups and trivially homeomorphic as topological spaces, because both are $g$-tori. – user267839 Aug 23 '20 at 22:11
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    Now I think I found the circumstance which confused me. You mean that arbitrary two tori $\mathbb{C}^g / \Gamma$ and $\mathbb{C}^g / \Gamma'$ are always isomorphic as abelian groups and top spaces. In contrast in https://math.stackexchange.com/questions/316181/when-are-two-tori-biholomorphic and and https://math.stackexchange.com/questions/491437/isomorphism-between-torus for $g=1$ is discused when two tori are isomorphic as complex manifolds, that is when they are biholomorphic. That's the point, right? Since of course in general two tori $\mathbb{C}^g / \Gamma$ and $\mathbb{C}^g / \Gamma'$ – user267839 Aug 23 '20 at 22:19
  • are not biholomorphic. For $g=1$ the important result from linked discussion states that $\mathbb{C}^g / \Gamma $ and $ \mathbb{C}^g / \Gamma'$ are isomorphic if and only if there exist a scalar $\alpha \in \mathbb{C}^*$ with $\alpha \Gamma = \Gamma'$. The interesting question which pops up now is if there is a similar criterion for biholomorphy for $g>1$. – user267839 Aug 23 '20 at 22:21
  • It is a good question but you should not ask it simultaneously here and on Mathoverflow. – Moishe Kohan Aug 25 '20 at 15:39

1 Answers1

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One way to describe the 1-dimensional result is to say that if $f: X\to X'$ is a biholomorphic map of two elliptic curves $X={\mathbb C}/\Gamma, X'= {\mathbb C}/\Gamma'$, then:

Each lift $F$ of $f$ to ${\mathbb C}$ is an invertible complex-affine map $z\mapsto az+b$, equivariant with respect to an isomorphism of free abelian groups $\phi: \Gamma\to \Gamma'$, i.e.: $$ F\circ \gamma= \phi(\gamma)\circ F, \forall \gamma\in \Gamma. $$

Conversely, each affine map $F$ as above descends to a biholomorphic map $f: X\to X'$.

Exactly the same works in higher dimensions when $\Gamma, \Gamma'$ are lattices in ${\mathbb C}^n$, except, of course, invertible complex-affine maps are given by $$ z\mapsto Az + b, A\in GL(n, {\mathbb C}), b\in {\mathbb C}^n. $$ A proof is rather straightforward: Lift $f: X\to X'$ to a biholomorphic map $$ F: {\mathbb C}^n\to {\mathbb C}^n $$ which is then equivariant with respect to an isomorphism $\phi: \Gamma\to \Gamma'$, $$ F\circ \gamma \circ F^{-1}= \phi(\gamma), \forall \gamma\in \Gamma. $$

Differentiating the equivariance condition $$ F\circ \gamma= \phi(\gamma)\circ F, \forall \gamma\in \Gamma, $$ using the Chain Rule we obtain that $$ DF \circ \gamma= DF, \forall \gamma\in \Gamma. $$ Hence, the Jacobian derivative $DF$ is a $\Gamma$-invariant holomorphic mapping ${\mathbb C}^n\to {\mathbb C}^n$, hence, descends to a holomorphic map $X\to {\mathbb C}^n$, which then has to be constant by the compactness of $X$. Thus, the map $z\mapsto DF(z)$ is also constant. In other words, $F$ is a complex-affine map. qed

Traditionally, the equivariance condition is written in its matrix form, by choosing a free basis of $\Gamma$ and specifying its image in $\Gamma'$ under $\phi$. Feel free to rewrite the above equations the same way, personally, I do not like this since it depends on auxiliary choices.

Moishe Kohan
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