Cut locus on a flat rectangular torus from the Dirichlet cell

For $z\in \mathbb{R}$, set $n=\lfloor z+1\mathrm{/}2\rfloor$ using [F2]. Then $-1\mathrm{/}2\le z-n<1\mathrm{/}2$. If $m\ge n+1$, then $\mathrm{\mid }z-m\mathrm{\mid }\ge \mathrm{\mid }z-n\mathrm{\mid }$; if $m\le n-1$, then again $\mathrm{\mid }z-m\mathrm{\mid }\ge \mathrm{\mid }z-n\mathrm{\mid }$. Equality for a second integer occurs exactly at the half-period boundary. Applying this to $z=(y_i-x_i)\mathrm{/}{L}_i$ for $L_1=a,L_2=b$ shows that each coordinate has an attained nearest lattice translate. The squared Euclidean norm is minimized coordinatewise, so ${\min }_{\lambda \in \mathrm{\Lambda }}\mathrm{\mid }y-x+\lambda \mathrm{\mid }=\sqrt{{\delta }_1^2+{\delta }_2^2},{\delta }_i={\min }_{k\in \mathbb{Z}}\mathrm{\mid }{y}_i-x_i+k{L}_i\mathrm{\mid }\mathrm{.}$ This minimum is zero exactly when $y-x\in \mathrm{\Lambda }$.

The quotient projection is open: for open $U\subseteq {\mathbb{R}}^2$, $q^{-1}(q(U))={\bigcup }_{\lambda \in \mathrm{\Lambda }}(U+\lambda ),$ which is open, so [F1] makes $q(U)$ open. The images of rational balls therefore form a countable base. If $q(x)\mathrm{\ne }q(y)$, [F2] and step 1.1 give $\delta ={\min }_{\lambda \in \mathrm{\Lambda }}\mathrm{\mid }y-x+\lambda \mathrm{\mid }>0$. Choose $r>0$ with $2r<\delta$. If $q(B(x,r))$ met $q(B(y,r))$, some $x^{\mathrm{\prime }}\in B(x,r)$ and $y^{\mathrm{\prime }}\in B(y,r)$ would satisfy $x^{\mathrm{\prime }}-y^{\mathrm{\prime }}\in \mathrm{\Lambda }$, yielding a lattice translate of $y-x$ of norm $\mathrm{\mid }{y}^{\mathrm{\prime }}-x^{\mathrm{\prime }}\mathrm{\mid }<2r$, a contradiction. Thus $Q$ is Hausdorff. For $0<\epsilon <\min (a,b)\mathrm{/}2$, each rectangle $U_x=x+(-\epsilon ,\epsilon {)}^2$ has pairwise disjoint lattice translates; $q{\mathrm{\mid }}_{U_x}$ is an open continuous bijection onto the open set $q(U_x)$, hence a chart. On overlaps the coordinate changes are locally translations by elements of $\mathrm{\Lambda }$, so they are smooth. The coordinate metric matrices are the constant identity matrix; [F4] makes this a smooth positive flat metric. The same disjoint-translate description of $q^{-1}(q(U_x))$ shows these charts evenly cover their images, so $q$ is a covering and a local isometry. Projected straight segments join every pair of classes, so $Q$ is path-connected and connected by [F13]. It is nonempty because it contains $q(0)$, and its charts have no boundary.

Fix $x,y\in {\mathbb{R}}^2$ and any piecewise $C^1$ path $\alpha$ in $Q$ from $q(x)$ to $q(y)$. Lift it from $x$ by [F9]. On each path piece lying in a quotient chart, its lift lies in one translated sheet and is the chartwise inverse of $\alpha$; hence the lift is piecewise $C^1$ and has the same coordinate speed. Its endpoint is $y+\lambda$ for some $\lambda \in \mathrm{\Lambda }$. Refine to a common finite subdivision and write ${\mathrm{\Delta }}_j=\tilde{\alpha }(t_{j+1})-\tilde{\alpha }(t_j)$. By [F11], $\mathrm{\mid }{\mathrm{\Delta }}_j\mathrm{\mid }\le {\int }_{t_j}^{t_{j+1}}\mathrm{\mid }{\tilde{\alpha }}^{\mathrm{\prime }}(t)\mathrm{\mid }dt\mathrm{.}$ The increments telescope, the triangle inequality in [F8] bounds their sum, and the local isometry preserves speed. Therefore $\mathrm{\mid }y-x+\lambda \mathrm{\mid }=\mathrm{\mid }\tilde{\alpha }(1)-\tilde{\alpha }(0)\mathrm{\mid }\le {\sum }_j\mathrm{\mid }{\mathrm{\Delta }}_j\mathrm{\mid }\le L_g(\alpha )\mathrm{.}$ By step 1.1 this is at least ${\min }_{\mu \in \mathrm{\Lambda }}\mathrm{\mid }y-x+\mu \mathrm{\mid }$. Taking the infimum over all paths gives the same lower bound for $d_g(q(x),q(y))$.

For every $x\in {\mathbb{R}}^2$ and $w\in T_{q(x)}Q\cong {\mathbb{R}}^2$, define ${\gamma }_{x,w}(t)=q(x+tw)$ for all $t\in \mathbb{R}$. In every quotient chart its coordinates are affine and the metric coefficients are constant, so [F12] gives zero Christoffel symbols and the geodesic equation. This is a global geodesic with the prescribed initial data. Uniqueness in [F12] shows that every maximal geodesic is this one; hence $Q$ is geodesically complete and ${\exp }_{q(x)}(w)=q(x+w)$. This includes $w=0$, whose geodesic is constant.

Step 1.1 supplies a translate ${\lambda }_{\ast }$ attaining the minimum. The projection of the straight segment from $x$ to $y+{\lambda }_{\ast }$ has constant speed $\mathrm{\mid }y-x+{\lambda }_{\ast }\mathrm{\mid }$, so [F10] gives its length as that norm. Combining it with step 3.1 proves the attained quotient-distance formula $d_g(q(x),q(y))={\min }_{\lambda \in \mathrm{\Lambda }}\mathrm{\mid }y-x+\lambda \mathrm{\mid }=\sqrt{{\delta }_1^2+{\delta }_2^2}\mathrm{.}$

The model is nonempty, connected, boundaryless, and geodesically complete by steps 2.1 and 3.2. Hopf--Rinow [F14], under exactly [A1], makes $(Q,d_g)$ complete. Thus the completeness hypotheses of [F15] and [F16] hold.

Let $u=(u_1,u_2)$ be a unit vector and put $\tau (u)=\min \text{\hspace{0.17em}⁣}(\{\frac{a}{2\mathrm{\mid }{u}_1\mathrm{\mid }}:u_1\mathrm{\ne }0\}\cup \{\frac{b}{2\mathrm{\mid }{u}_2\mathrm{\mid }}:u_2\mathrm{\ne }0\})\mathrm{.}$ The set is nonempty and $0<\tau (u)<\mathrm{\infty }$. For $0<t\le \tau (u)$, $tu\in D$; step 1.1 says zero is a nearest lattice translate, including a tie when a coordinate reaches a face. Steps 3.2 and 4.1 then give $d_g(p,{\exp }_p(tu))=\mathrm{\mid }tu\mathrm{\mid }=t$. If $t>\tau (u)$, choose a nonzero coordinate $u_i$ attaining the minimum. Then $\mathrm{\mid }t{u}_i\mathrm{\mid }>L_i\mathrm{/}2$, where $L_1=a,L_2=b$. Adding the opposite period to that coordinate strictly reduces its absolute value, leaves the other coordinate unchanged, and gives a projected straight competitor of length strictly less than $\mathrm{\mid }tu\mathrm{\mid }=t$. Hence $d_g(p,{\exp }_p(tu))<t$. The minimizing-time set is exactly $(0,\tau (u)]$, so [F15] gives $c_p(u)=\tau (u)$.

For $v\mathrm{\ne }0$, write $v=\mathrm{\mid }v\mathrm{\mid }u$ with $\mathrm{\mid }u\mathrm{\mid }=1$. The formula for $\tau (u)$ in step 5.1 shows $\mathrm{\mid }v\mathrm{\mid }<\tau (u)$ exactly when $\mathrm{\mid }{v}_1\mathrm{\mid }<a\mathrm{/}2$ and $\mathrm{\mid }{v}_2\mathrm{\mid }<b\mathrm{/}2$. Adjoining $v=0$ proves both inclusions $C_p=D^{\circ }$; omitting zero gives the positive domain $D^{\circ }\setminus \{0\}$. For each unit $u$, step 5.1 puts $\tau (u)u$ on $\mathrm{\partial }D$, so every finite cut endpoint lies in $\{[x_0+w]:w\in \mathrm{\partial }D\}$. Conversely, given $w\in \mathrm{\partial }D$, $w\mathrm{\ne }0$; put $u=w\mathrm{/}\mathrm{\mid }w\mathrm{\mid }$. For each nonzero coordinate, its candidate exit time is $L_i\mathrm{\mid }w\mathrm{\mid }\mathrm{/}(2\mathrm{\mid }{w}_i\mathrm{\mid })\ge \mathrm{\mid }w\mathrm{\mid }$, with equality on every face coordinate, so $\tau (u)=\mathrm{\mid }w\mathrm{\mid }$ and $[x_0+w]$ is a cut endpoint. Thus $\mathrm{Cut}(p)=\{[x_0+w]:w\in \mathrm{\partial }D\}$, proving both set inclusions.

In one coordinate, a point strictly between the two half-periods has one nearest period representative, while either endpoint $\pm L_i\mathrm{/}2$ has exactly two, differing by $L_i$. Independence of the two coordinates makes each relative-interior edge point have exactly two nearest lattice lifts and each corner have four. Opposite edges differ by a lattice vector and therefore have the same image. A vertical-edge image and a horizontal-edge image meet only when both coordinates are half-periods; all four corners then give the same corner class. This is the stated branch intersection.

The torus is nonempty and two-dimensional; $a,b>0$ rule out a collapsed period. The zero tangent vector is included in $C_p$ but is not a unit direction. Directions with one zero coordinate are covered by omitting that coordinate's quotient from the minimum in step 5.1. The cut-time supremum is over $t>0$, its endpoint $\tau (u)$ is included and minimizing, and every later time fails strictly. The two set equalities in step 6.1 were proved in both directions. The only choice assumption is [A1] through geodesic uniqueness, Hopf--Rinow [F14] and the cut interfaces [F15, F16]; coordinate rounding is determined by [F2], path lifts are unique from a specified start, and no full AC or family selection is used. No other dimension or iff claim is made. [A1, F2, F12, F14, F15, F16, step 2.1, step 3.2, step 4.2, step 5.1, step 6.1, step 7.1] QED