Cut locus of a point on a flat circle

Write $q_L(x)=[x]$. In every period chart the metric coefficient is $1$, and chart changes are translations by $kL$, so the metric is well-defined. The explicit curve $t\mapsto [x+t(y-x)]$ connects any two classes, and $[0]$ shows the circle is nonempty. These facts and [F1]-[F2] establish the quotient metric model; [F3] gives its tangent norms. Thus $C_L$ is nonempty, connected, one-dimensional, and boundaryless.

Fix $p=[x]$ and $q=[y]$. Let $\alpha :[0,1]\to C_L$ be any piecewise $C^1$ path from $p$ to $q$, and lift it from $x$ using [F4]. Its endpoint is $y+kL$ for some $k\in \mathbb{Z}$, by the quotient equivalence in [F2]. Choose a finite subdivision $0=a_0<\cdots <a_m=1$ on which $\alpha$ is $C^1$ on each piece; the local inverse charts for the covering make $\tilde{\alpha }$ $C^1$ on those same pieces. Put ${\mathrm{\Delta }}_i=\tilde{\alpha }(a_{i+1})-\tilde{\alpha }(a_i)$. On each piece the lift has the same speed as $\alpha$, and [F5] gives ${\mathrm{\Delta }}_i={\int }_{a_i}^{a_{i+1}}{\tilde{\alpha }}^{\mathrm{\prime }}(t)dt$ and $\mathrm{\mid }{\mathrm{\Delta }}_i\mathrm{\mid }\le {\int }_{a_i}^{a_{i+1}}\mathrm{\mid }{\tilde{\alpha }}^{\mathrm{\prime }}(t)\mathrm{\mid }dt$. Hence the triangle inequality and the length definition give $\mathrm{\mid }y+kL-x\mathrm{\mid }=\mathrm{\mid }\tilde{\alpha }(1)-\tilde{\alpha }(0)\mathrm{\mid }=\mid {\sum }_{i=0}^{m-1}{\mathrm{\Delta }}_i\mid \le {\sum }_{i=0}^{m-1}\mathrm{\mid }{\mathrm{\Delta }}_i\mathrm{\mid }\le {\sum }_{i=0}^{m-1}{\int }_{a_i}^{a_{i+1}}\mathrm{\mid }{\tilde{\alpha }}^{\mathrm{\prime }}(t)\mathrm{\mid }dt=L_g(\alpha )\mathrm{.}$ Thus $L_g(\alpha )\ge \mathrm{\mid }y+kL-x\mathrm{\mid }\ge {\inf }_{j\in \mathbb{Z}}\mathrm{\mid }y+jL-x\mathrm{\mid }$. Taking the infimum over $\alpha$ gives $d_g(p,q)\ge {\inf }_j\mathrm{\mid }y+jL-x\mathrm{\mid }$.

For any point $[x]$ and any tangent vector $u{\mathrm{\partial }}_x$, define $\gamma (t)=[x+ut]$ for all $t\in \mathbb{R}$. In each period chart its coordinate is affine, the metric coefficients are constant, and [F7] gives $\mathrm{\Gamma }=0$ and the geodesic equation. Thus this is a global geodesic with the prescribed initial data. Uniqueness of the maximal geodesic in [F7] implies that every maximal geodesic is defined on all of $\mathbb{R}$. Hence $C_L$ is geodesically complete.

Put $z=(y-x)\mathrm{/}L$ and choose $n=\lfloor z+1\mathrm{/}2\rfloor$ by [F6]. Then $-1\mathrm{/}2\le z-n<1\mathrm{/}2$. For every integer $j$, this makes $n$ a nearest integer to $z$: if $j\ge n+1$ then $j-z>1\mathrm{/}2$, and if $j\le n-1$ then $z-j\ge 1\mathrm{/}2$. Therefore the infimum in step 1.2 is attained and ${\inf }_{j\in \mathbb{Z}}\mathrm{\mid }y+jL-x\mathrm{\mid }=\mathrm{\mid }y-nL-x\mathrm{\mid }\le L\mathrm{/}2.$ For each integer $j$, the projected straight segment ${\sigma }_j(t)=[x+t(y+jL-x)]$, $0\le t\le 1$, joins $p$ to $q$ and has constant speed $\mathrm{\mid }y+jL-x\mathrm{\mid }$. Choosing $j=-n$ attains the lower bound in step 1.2, so $d_g([x],[y])={\min }_{j\in \mathbb{Z}}\mathrm{\mid }y+jL-x\mathrm{\mid }\mathrm{.}$ This also proves $d_g([x],[y])\le L\mathrm{/}2$ for all $p,q$.

The nonempty, connected, boundaryless hypotheses were checked in step 1.1, and step 2.1 gives geodesic completeness. Hopf--Rinow [F8] therefore makes $(C_L,d_g)$ complete, as required by [F9].

By [F3], the unit tangent vectors at $p$ are exactly $v_{+}={\mathrm{\partial }}_x$ and $v_{-}=-{\mathrm{\partial }}_x$. Their radial geodesics are ${\gamma }_{\pm }(t)=[x\pm t]$ by step 2.1. For $0<t<L\mathrm{/}2$, step 2.2 gives $d_g(p,{\gamma }_{\pm }(t))=t$. At $t=L\mathrm{/}2$, both adjacent period representatives have absolute displacement $L\mathrm{/}2$, so the same equality holds and there are two distinct minimizing semicircle segments: $s\mapsto [x+sL\mathrm{/}2]$ and $s\mapsto [x-sL\mathrm{/}2]$, $0\le s\le 1$. They have the same length $L\mathrm{/}2$ and different interior points. If $t>L\mathrm{/}2$, step 2.2 gives $d_g(p,{\gamma }_{\pm }(t))\le L\mathrm{/}2<t$. Thus the positive minimizing-time set in each direction is exactly $(0,L\mathrm{/}2]$, and [F9] gives $c_p(v_{+})=c_p(v_{-})=L\mathrm{/}2$.

The two finite cut endpoints coincide: ${\gamma }_{+}(L\mathrm{/}2)=[x+L\mathrm{/}2]=[x-L\mathrm{/}2]={\gamma }_{-}(L\mathrm{/}2)$. The quotient relation makes this point independent of the representative $x$; it is the antipode. Since [F3] gives exactly the two unit directions and [F9] defines the cut locus as their finite cut endpoints, $\mathrm{Cut}(p)=\{[x+L\mathrm{/}2]\}$. The computation holds for every $p$. The circle is nonempty and one-dimensional, so the empty- and zero-dimensional cases are not instances; in dimension one both unit directions were handled. The degenerate value $t=0$ is excluded from the cut-time supremum, the included endpoint $L\mathrm{/}2$ still minimizes, and every later time fails strictly. The only choice assumption is the declared ${\mathrm{A}\mathrm{C}}_{\omega }$ used through [F7]-[F9]; the quotient lifts are unique from the specified start and the nearest period is fixed by [F6], with no appeal to full AC. There is no iff claim.