A canonical reduced basis for a complex lattice

Real-linear independence of ${\omega }_1,{\omega }_2$ gives $\tau \mathrm{\notin }\mathbb{R}$, so after exchanging the two vectors if necessary one has $\mathrm{Im}\tau >0$ and $({\omega }_1,{\omega }_2)$ is an oriented basis of $\mathrm{\Lambda }$; every oriented basis of $\mathrm{\Lambda }$ is $(a{\omega }_1+b{\omega }_2,c{\omega }_1+d{\omega }_2)$ with integers $a,b,c,d$ satisfying $ad-bc=1$, and conversely every such tuple yields an oriented basis, the coefficients being unique because ${\omega }_1,{\omega }_2$ are real-linearly independent.

Writing $\tau =x+iy$ with $x,y\in \mathbb{R}$ and $y>0$, the form $q(a,b):=\mathrm{\mid }a+b\tau {\mathrm{\mid }}^2=a^2+2abx+b^2(x^2+y^2)$ equals $(a+xb{)}^2+y^2{b}^2$ and also $x^2+y^2$ times a square plus $y^2{a}^2\mathrm{/}(x^2+y^2)$, so $q(a,b)\ge y^2{b}^2$ and $q(a,b)\ge y^2{a}^2\mathrm{/}(x^2+y^2)$; hence $q(a,b)\ge {\epsilon }^2\max (a^2,b^2)$ with ${\epsilon }^2:=y^2\mathrm{/}\max (1,x^2+y^2)>0$ for all real $a,b$.

A matrix $A\in {\mathrm{S}\mathrm{L}}_2(\mathbb{Z})$ fixes ${\tau }_0$ exactly when $b{\tau }_0^2+(a-d){\tau }_0-c=0$; if $b=0$ this equation and $ad=1$ give $a=d=\pm 1$, $c=0$, so $A=\pm I$. If $b\mathrm{\ne }0$, the discriminant $(a-d{)}^2+4bc=(a+d{)}^2-4$ is negative, so the trace $t=a+d$ lies in $\{-1,0,1\}$; writing ${\tau }_0=u+iv$ with $v>0$, the fixed point equation reads $b(u^2-v^2)+(a-d)u-c+(2bu+a-d)vi=0$, so $u=(d-a)\mathrm{/}(2b)$ and $v^2=-((a-d{)}^2+4bc)\mathrm{/}(4b^2)=(4-t^2)\mathrm{/}(4b^2)$.

For the basis of step 1.1 the ratio is ${\tau }_A=(c+d\tau )\mathrm{/}(a+b\tau )$ with $A=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\in {\mathrm{S}\mathrm{L}}_2(\mathbb{Z})$, and $\mathrm{Im}{\tau }_A=\mathrm{Im}\tau \mathrm{/}\mathrm{\mid }a+b\tau {\mathrm{\mid }}^2$.

For trace $t=0$, $d=-a$, so step 1.3 gives $u=-a\mathrm{/}b$ and $v^2=1\mathrm{/}{b}^2$. At a reduced fixed point, $\mathrm{\mid }u\mathrm{\mid }\le \frac{1}{2}$ and $\mathrm{\mid }{\tau }_0\mathrm{\mid }\ge 1$, whence $4a^2\le b^2\le a^2+1$. Thus $3a^2\le 1$, forcing the integer $a=0$, then $\mathrm{\mid }b\mathrm{\mid }=1$ and ${\tau }_0=i$. Now $d=0$ and $-bc=1$, giving precisely $A=\pm \left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)=\mp S$, where $S=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$; both fix $i$ directly.

For trace $t=\pm 1$: replacing $A$ by $-A$ changes the sign of $t$ and leaves the fixed points unchanged, so take $t=1$; then $d=1-a$ and $v^2=3\mathrm{/}(4b^2)$, so with $B=\mathrm{\mid }b\mathrm{\mid }\ge 1$ and $\sigma =b\mathrm{/}B\in \{-1,1\}$ one has ${\tau }_0=\sigma (1-2a)\mathrm{/}(2B)+\frac{\sqrt{3}}{2B}i$, and the constraints $\mathrm{\mid }\mathrm{Re}{\tau }_0\mathrm{\mid }\le \frac{1}{2}$ and $\mathrm{\mid }{\tau }_0\mathrm{\mid }\ge 1$ give $\mathrm{\mid }1-2a\mathrm{\mid }\le B$ and $B^2\le a^2-a+1$, hence $3a^2-3a\le 0$ and $a\in \{0,1\}$. For $a=0$ the determinant condition gives $c=-\sigma$ and ${\tau }_0=\sigma \mathrm{/}2+\frac{\sqrt{3}}{2}i$, which lies in $\mathcal{R}$ only for $\sigma =1$, giving ${\tau }_0=e^{i\pi \mathrm{/}3}$ and $A=\left(\begin{array}{cc}0& 1\\ -1& 1\end{array}\right)$; for $a=1$ it gives $c=-\sigma$, ${\tau }_0=-\sigma \mathrm{/}2+\frac{\sqrt{3}}{2}i$, which lies in $\mathcal{R}$ only for $\sigma =-1$, giving again ${\tau }_0=e^{i\pi \mathrm{/}3}$ and $A=\left(\begin{array}{cc}1& -1\\ 1& 0\end{array}\right)$. Together with their negatives and $\pm I$ these six matrices form the stabiliser of $e^{i\pi \mathrm{/}3}$, and both displayed matrices are checked directly to fix $e^{i\pi \mathrm{/}3}$.

Only finitely many values $\mathrm{Im}{\tau }_A$ satisfy $\mathrm{Im}{\tau }_A\ge \frac{1}{2}\mathrm{Im}\tau$: by steps 2.1 and 1.2 that condition implies $\mathrm{\mid }a+b\tau {\mathrm{\mid }}^2\le 2$, hence $a^2+b^2\le 4\mathrm{/}{\epsilon }^2$, which has only finitely many integer solutions $(a,b)$. The value $\mathrm{Im}{\tau }_A=\mathrm{Im}\tau \mathrm{/}\mathrm{\mid }a+b\tau {\mathrm{\mid }}^2$ depends only on $(a,b)$; the determinant equation may have infinitely many solutions $(c,d)$.

For uniqueness let $\tau ,{\tau }^{\mathrm{\prime }}\in \mathcal{R}$ and suppose ${\tau }^{\mathrm{\prime }}=(c+d\tau )\mathrm{/}(a+b\tau )$ with $A=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\in {\mathrm{S}\mathrm{L}}_2(\mathbb{Z})$; replacing $A$ by $A^{-1}=\left(\begin{array}{cc}d& -b\\ -c& a\end{array}\right)$, which also lies in ${\mathrm{S}\mathrm{L}}_2(\mathbb{Z})$ and expresses $\tau$ through ${\tau }^{\mathrm{\prime }}$ in the same form, we may assume $\mathrm{Im}{\tau }^{\mathrm{\prime }}\ge \mathrm{Im}\tau$, and then step 2.1 gives $\mathrm{\mid }a+b\tau \mathrm{\mid }\le 1$; moreover $\mathrm{Im}\tau \ge \sqrt{3}\mathrm{/}2$, because ${\mathrm{Im}}^2\tau =\mathrm{\mid }\tau {\mathrm{\mid }}^2-{\mathrm{Re}}^2\tau \ge 1-\frac{1}{4}$.

Fix an oriented basis $({\omega }_1,{\omega }_2)$ of $\mathrm{\Lambda }$ with reduced ratio ${\tau }_0\in \mathcal{R}$. By step 1.1 the oriented bases of $\mathrm{\Lambda }$ are exactly the $(a{\omega }_1+b{\omega }_2,c{\omega }_1+d{\omega }_2)$ with $A\in {\mathrm{S}\mathrm{L}}_2(\mathbb{Z})$, and by step 2.1 such a basis again has ratio ${\tau }_0$ exactly when $A$ lies in the stabiliser $\mathrm{Stab}({\tau }_0)=\{A\in {\mathrm{S}\mathrm{L}}_2(\mathbb{Z}):(c+d{\tau }_0)\mathrm{/}(a+b{\tau }_0)={\tau }_0\}$; the assignment $A\mapsto (a{\omega }_1+b{\omega }_2,c{\omega }_1+d{\omega }_2)$ is injective, so the oriented bases of $\mathrm{\Lambda }$ with reduced ratio ${\tau }_0$ are in bijection with $\mathrm{Stab}({\tau }_0)$.

Some oriented basis of $\mathrm{\Lambda }$ has maximal imaginary part of its ratio: the set of values $\mathrm{Im}{\tau }_A$ over $A\in {\mathrm{S}\mathrm{L}}_2(\mathbb{Z})$ contains $\mathrm{Im}\tau$ (take $A=I$), so it meets $[\frac{1}{2}\mathrm{Im}\tau ,\mathrm{\infty })$, and by step 3.1 the values in that interval form a nonempty finite set; its maximum is attained at some matrix $A_0$ and dominates every value, because a value outside the interval is $<\frac{1}{2}\mathrm{Im}\tau \le \mathrm{Im}\tau$.

If $b=0$, then $ad=1$ forces $a=d=\pm 1$ and ${\tau }^{\mathrm{\prime }}=(c+a\tau )\mathrm{/}a=\tau +c\mathrm{/}a$; both $\mathrm{Re}\tau$ and $\mathrm{Re}{\tau }^{\mathrm{\prime }}$ lie in $(-\frac{1}{2},\frac{1}{2}]$, so $c\mathrm{/}a=0$, hence ${\tau }^{\mathrm{\prime }}=\tau$.

If $b\mathrm{\ne }0$, then replacing $A$ by $-A$ leaves ${\tau }^{\mathrm{\prime }}$ unchanged, so we may assume $b\ge 1$; by step 3.2, $b\mathrm{Im}\tau \le \mathrm{\mid }a+b\tau \mathrm{\mid }\le 1$, so $b\le 1\mathrm{/}\mathrm{Im}\tau \le 2\mathrm{/}\sqrt{3}<2$ and therefore $b=1$.

Let $({\omega }_1^{\ast },{\omega }_2^{\ast })$ realize the maximum of step 4.1, with ratio ${\tau }^{\ast }$. Replacing ${\omega }_2^{\ast }$ by $k{\omega }_1^{\ast }+{\omega }_2^{\ast }$ changes the ratio to ${\tau }^{\ast }+k$ without changing its imaginary part or orientation. Choose $k=-\lfloor \mathrm{Re}{\tau }^{\ast }+\frac{1}{2}\rfloor$; if the resulting real part is $-\frac{1}{2}$, add one more copy of ${\omega }_1^{\ast }$. Thus we may suppose $-\frac{1}{2}<\mathrm{Re}{\tau }^{\ast }\le \frac{1}{2}$, still with maximal imaginary part.

With $b=1$ the bound $\mathrm{\mid }a+\tau \mathrm{\mid }\le 1$ of step 3.2 reads $a^2+2a\mathrm{Re}\tau +\mathrm{\mid }\tau {\mathrm{\mid }}^2\le 1$, hence $a(a+2\mathrm{Re}\tau )\le 1-\mathrm{\mid }\tau {\mathrm{\mid }}^2\le 0$, and we distinguish three cases. If $a\ge 1$, then $a+2\mathrm{Re}\tau \le 0$ gives $\mathrm{Re}\tau \le -\frac{1}{2}$, contradicting $\tau \in \mathcal{R}$. If $a=0$, then $\mathrm{\mid }\tau \mathrm{\mid }=1$; the determinant condition $ad-bc=1$ gives $c=-1$ and ${\tau }^{\mathrm{\prime }}=(c+d\tau )\mathrm{/}\tau =d-\bar{\tau }$, and ${\tau }^{\mathrm{\prime }}\in \mathcal{R}$ forces $d=0$, $\tau =i$ or $d=1$, $\tau =e^{i\pi \mathrm{/}3}$, in both cases ${\tau }^{\mathrm{\prime }}=\tau$. If $a\le -1$, then $a+2\mathrm{Re}\tau \ge 0$ gives $\mathrm{Re}\tau \ge -\frac{a}{2}\ge \frac{1}{2}$, so $\mathrm{Re}\tau =\frac{1}{2}$; then $a^2+a+\mathrm{\mid }\tau {\mathrm{\mid }}^2\le 1$ and $\mathrm{\mid }\tau \mathrm{\mid }\ge 1$ give $a\in \{-1,0\}$ and $\mathrm{\mid }\tau \mathrm{\mid }=1$, so $a=-1$, $\tau =e^{i\pi \mathrm{/}3}$, and with $d=-c-1$ one computes ${\tau }^{\mathrm{\prime }}=-c+e^{2\pi i\mathrm{/}3}\in \mathcal{R}$, which forces $c=-1$ and ${\tau }^{\mathrm{\prime }}=\tau \in \mathcal{R}$. Hence ${\tau }^{\mathrm{\prime }}=\tau$ in every case of $b\mathrm{\ne }0$.

If $\mathrm{\mid }{\tau }^{\ast }\mathrm{\mid }<1$, then $(-{\omega }_2^{\ast },{\omega }_1^{\ast })$ is an oriented basis of $\mathrm{\Lambda }$ whose ratio $-1\mathrm{/}{\tau }^{\ast }$ has $\mathrm{Im}(-1\mathrm{/}{\tau }^{\ast })=\mathrm{Im}{\tau }^{\ast }\mathrm{/}\mathrm{\mid }{\tau }^{\ast }{\mathrm{\mid }}^2>\mathrm{Im}{\tau }^{\ast }$, contradicting maximality; hence $\mathrm{\mid }{\tau }^{\ast }\mathrm{\mid }\ge 1$.

The ratio now has positive imaginary part, $-\frac{1}{2}<\mathrm{Re}{\tau }^{\ast }\le \frac{1}{2}$ and $\mathrm{\mid }{\tau }^{\ast }\mathrm{\mid }\ge 1$. It is reduced unless $\mathrm{\mid }{\tau }^{\ast }\mathrm{\mid }=1$ and $\mathrm{Re}{\tau }^{\ast }<0$. In that case the oriented basis $(-{\omega }_2^{\ast },{\omega }_1^{\ast })$ has ratio $-1\mathrm{/}{\tau }^{\ast }=-\overline{{\tau }^{\ast }}$, with real part in $(0,\frac{1}{2})$, modulus $1$ and the same positive imaginary part, hence lies in $\mathcal{R}$.

Steps 4.2 and 5.2 prove that two reduced ratios related by a basis change are equal; with step 7.1 this gives existence and uniqueness of the reduced ratio of $\mathrm{\Lambda }$, and shows it is realized by at least one oriented basis.

By steps 1.3, 2.2 and 2.3 the stabiliser of ${\tau }_0\in \mathcal{R}$ is $\{\pm I,\pm S\}$, of order four, when ${\tau }_0=i$, the six-element set $\{\pm I,\pm \left(\begin{array}{cc}0& 1\\ -1& 1\end{array}\right),\pm \left(\begin{array}{cc}1& -1\\ 1& 0\end{array}\right)\}$ when ${\tau }_0=e^{i\pi \mathrm{/}3}$, and $\{\pm I\}$, of order two, for every other reduced ${\tau }_0$; by step 3.3 these are exactly the numbers of oriented bases of $\mathrm{\Lambda }$ with reduced ratio ${\tau }_0$. Finally $\mathbb{Z}+i\mathbb{Z}$ has the reduced oriented basis $(1,i)$ with ratio $i$, and $\mathbb{Z}+\mathbb{Z}{e}^{i\pi \mathrm{/}3}$ has the reduced oriented basis $(1,e^{i\pi \mathrm{/}3})$ with ratio $e^{i\pi \mathrm{/}3}$, so the exceptional cases occur.