The exponential map of a flat torus is not injective

By [F4], the standard list $(e_i{)}_{i<n}$ is a basis of ${\mathbb{R}}^n$, so this space has dimension $n$. The independent set $\{{\lambda }_i:i<n\}$ extends to a basis, again by [F4], but an independent subset has at most $n$ elements; the extension therefore adds no vector, and the supplied set is already a basis. Hence the linear map $A$ with $A(e_i)={\lambda }_i$ for every $i<n$ is invertible and sends ${\mathbb{Z}}^n$ onto $\mathrm{\Lambda }$. Apply the bound in [F4] to $A^{-1}$, obtaining $K_0\ge 0$, and put $K=K_0+1>0$; then $\mathrm{\mid }{A}^{-1}z\mathrm{\mid }\le K\mathrm{\mid }z\mathrm{\mid }$ for every $z$. A nonzero integer vector has Euclidean norm at least $1$, so every nonzero $\lambda =Am\in \mathrm{\Lambda }$ satisfies $\mathrm{\mid }\lambda \mathrm{\mid }\ge 1\mathrm{/}K$. In particular $\mathrm{\Lambda }$ is uniformly discrete. The maps $A,A^{-1}$ are continuous by the same bound. They induce inverse bijections $\bar{A}:{\mathbb{R}}^n\mathrm{/}{\mathbb{Z}}^n\to {\mathbb{R}}^n\mathrm{/}\mathrm{\Lambda }$, $[x]\mapsto [Ax]$, and $\overline{A^{-1}}$. If $V$ is open in ${\mathbb{R}}^n\mathrm{/}\mathrm{\Lambda }$, then $q_{\mathbb{Z}}^{-1}[{\bar{A}}^{-1}[V]]=A^{-1}[q_{\mathrm{\Lambda }}^{-1}[V]]$ is open; the quotient-topology definition in [F4] therefore makes ${\bar{A}}^{-1}[V]$ open. The identical calculation for $A^{-1}$ proves continuity of the inverse. We verify the quotient manifold properties directly. For distinct orbits $[x]\mathrm{\ne }[y]$, write $d_{\lambda }=\mathrm{\mid }x-y-\lambda \mathrm{\mid }>0$. Only finitely many $\lambda =Am$ can have $d_{\lambda }\le 1$: such an $m$ satisfies $\mathrm{\mid }m\mathrm{\mid }\le K(\mathrm{\mid }x-y\mathrm{\mid }+1)$, leaving finitely many integer vectors. Thus the minimum of $1$ and these finitely many positive distances is a number $d>0$. The quotient images of $B(x,d\mathrm{/}3)$ and $B(y,d\mathrm{/}3)$ are disjoint, proving Hausdorffness. The images of rational Euclidean balls form a countable basis because the quotient map is open. Take $0<2\epsilon <1\mathrm{/}K$. The quotient map $q:{\mathbb{R}}^n\to T$ is injective on each $B(x,\epsilon )$: two points there differ by a lattice vector of norm less than $2\epsilon$, hence by zero. It is open because $q^{-1}q(U)={\bigcup }_{\lambda \in \mathrm{\Lambda }}(U+\lambda )$ is open. Thus these restrictions are smooth quotient charts. Their transitions on overlap components are translations by lattice vectors, so they are smooth with identity derivative; the local Euclidean tensors agree and define the flat metric by [F2], with matrix $I_n$ in every such chart.

Since the local metric matrix is constant, [F3] gives zero Levi–Civita symbols. For any $x,v\in {\mathbb{R}}^n$ the curve $\gamma (t)=q(x+tv)$, $t\in \mathbb{R}$, is smooth and, within every quotient chart, has coordinate velocity $v$ and acceleration zero. Hence [F3] makes it a geodesic for every real $t$. Its initial point is $[x]$ and its initial tangent is the vector identified with $v$.

By the uniqueness in [F1], the geodesic of step 2.1 is the maximal geodesic with that initial data: it already has domain $\mathbb{R}$. Therefore $1$ belongs to the domain for every $v$ and ${\exp }_{[x]}(v)=\gamma (1)=[x+v]$. The result is independent of the representative $x$, since replacing $x$ by $x+\mu$ with $\mu \in \mathrm{\Lambda }$ leaves $[x+v]$ unchanged and translations have identity derivative on tangent coordinates.

Full rank and $n\ge 1$ provide a nonzero lattice vector $\lambda$. The tangent-coordinate vectors $v$ and $v+\lambda$ are distinct, but $[x+v+\lambda ]=[x+v]$; thus the formula of step 3.1 proves periodicity and noninjectivity. For $n=0$, there is only the zero tangent vector and the fibrewise map is injective, so the positive-dimensional hypothesis is necessary. The only choice assumption inherited by this example is the declared ${\mathrm{A}\mathrm{C}}_{\omega }$ used in [F1]; constructing the displayed geodesic itself uses no choice.