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I am reading Chapter 11 of Dale Husemöller's Elliptic Curves Springer book and I got stuck on Theorem (1.4) (c.f., image below).

Notation and definitions: Let $L$ and $L'$ be two complex lattices spanned, respectively, by periods $\{\psi_1,\psi_2\}$ and $\{\psi_1',\psi_2'\}$. Set $\tau:=\psi_2/\psi_1\in\mathfrak{H}$ and $\tau':=\psi_2'/\psi_1'\in\mathfrak{H}$, where $\mathfrak{H}$ is the upper-half complex plane. Finally, we introduce two tori $T_\tau:=\mathbb{C}/L$ and $T_{\tau'}:=\mathbb{C}/L'$.

By definition, $T_{\tau}$ and $T_{\tau'}$ are said to be isogenic iff there's a complex number $\lambda$ such that $\lambda L'\subset L$.

Below, $GL^+_2(\mathbb{K})$ denotes the group of two-by-two invertible matrices with positive determinant and with entries in the field $\mathbb{K}$.

My question regarding the proof of (2) below is the following: Why do we need to restrict to the subgroup $GL^+_2(\mathbb{Q})$ of $GL^+_2(\mathbb{R})$ and not just take $GL^+_2(\mathbb{R})$ itself? I see no obstruction in replacing $\mathbb{Q}$ by $\mathbb{R}$ in the proof.

Am I missing something here?

enter image description here

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    $\begingroup$ $a,b,c,d$ are integers. $\endgroup$
    – Wojowu
    Commented Apr 9, 2022 at 21:49

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The lattice $L$ is defined by taking integer combinations of the basis vectors $1$ and $\tau$. If you have arbitrary $a,b,c,d\in\mathbb{R}$ and set $\lambda=c\tau+d$, then $\lambda\cdot 1\in\lambda L'$ will be a real combination of $1$ and $\tau$, but it will not be an integer combination unless $c,d$ themselves are integers.

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  • $\begingroup$ Sorry I am not following you. Why would $\lambda$ need to be a integer combination of 1 and $\tau$? We just want to show that the prime lattice is contained in the unprimed one under $\lambda$-scaling for some choice of complex $\lambda$, no ? Where would the need from your condition come from then? $\endgroup$
    – DaveWasHere
    Commented Apr 9, 2022 at 22:50
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    $\begingroup$ $\lambda$ is - a priori - allowed to be any complex number. But the claim that $\lambda L'\subseteq L$ is saying something very specific: it's saying that $\lambda\cdot 1$ and $\lambda\cdot \tau'$ are both in $L$. This means that $\lambda\cdot 1$ and $\lambda\cdot \tau'$ must both be elements of $\{m+n\tau:m,n\in\mathbb{Z}\}$. $\endgroup$
    – Jonathan Love
    Commented Apr 9, 2022 at 22:52
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    $\begingroup$ I think an example could potentially help. Let's take two lattices, $L'$ spanned by $\{1,i\sqrt{2}\}$ (rectangular lattice) and $L$ by $\{1,i\}$ (square lattice). We want to find any $\lambda\in\mathbb{C}$ that satisfies $\lambda L'\subseteq L$. This means $\lambda$ has to take both $1$ and $i\sqrt{2}$ to elements of $\mathbb{Z}[i]$. But the fact that $\lambda\cdot 1\in\mathbb{Z}[i]$ means that we actually must have $\lambda\in\mathbb{Z}[i]$, and this implies $\lambda\cdot i\sqrt{2}$ can't be in $\mathbb{Z}[i]$. (continued) $\endgroup$
    – Jonathan Love
    Commented Apr 9, 2022 at 23:05
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    $\begingroup$ Hence the two corresponding tori are not isogenous, despite the fact that $\begin{pmatrix} \sqrt{2}&0\\0&1\end{pmatrix}\in GL_2^+(\mathbb{R})$ takes $i$ to $i\sqrt{2}$. $\endgroup$
    – Jonathan Love
    Commented Apr 9, 2022 at 23:05
  • $\begingroup$ Thanks a lot! Got your point. :) $\endgroup$
    – DaveWasHere
    Commented Apr 9, 2022 at 23:43

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