The scaled three-torus as a smooth compact flat spatial manifold

The quotient map is open: for any open $U\subseteq {\mathbb{R}}^3$, $q^{-1}(q(U))={\bigcup }_{\lambda \in \mathrm{\Lambda }}(U+\lambda )$ is open, so $q(U)$ is open by the quotient topology. For any $x\in {\mathbb{R}}^3$, choose $0<r<\frac{1}{2}{\min }_i{L}_i$ and let $B=B(x,r)$ be the Euclidean ball. If $y,z\in B$ and $q(y)=q(z)$, then $y-z\in \mathrm{\Lambda }$; each coordinate has absolute value less than $2r<L_i$, so each lattice coordinate is zero and $y=z$. Thus $q{\mathrm{\mid }}_B$ is a continuous open bijection from $B$ to the open set $q(B)$, hence a homeomorphism by [F1]. These neighborhoods give local Euclidean charts.

Every class in ${\mathbb{T}}_L^3$ has a representative in the compact box $C:={\prod }_i[0,L_i]$, since in each coordinate an integer multiple of $L_i$ can be subtracted to leave a value in $[0,L_i)$. Hence $q(C)={\mathbb{T}}_L^3$. The quotient map is continuous by [F1], so [F4] makes its image compact.

If $q(x)\mathrm{\ne }q(y)$, then $x-y\mathrm{\notin }\mathrm{\Lambda }$. Among the lattice vectors with $\mathrm{\mid }x-y-\lambda \mathrm{\mid }\le 1$, each integer coordinate is bounded by $(\mathrm{\mid }x-y\mathrm{\mid }+1)\mathrm{/}{L}_i$, so there are finitely many. Their distances from $x-y$ are all positive, and every other distance exceeds $1$; therefore $\delta :={\inf }_{\lambda \in \mathrm{\Lambda }}\mathrm{\mid }x-y-\lambda \mathrm{\mid }>0$. Choose $0<ϵ<\delta \mathrm{/}3$. If $q(B(x,ϵ))$ and $q(B(y,ϵ))$ met, some $x^{\mathrm{\prime }}\in B(x,ϵ)$ and $y^{\mathrm{\prime }}\in B(y,ϵ)$ would satisfy $x^{\mathrm{\prime }}-y^{\mathrm{\prime }}\in \mathrm{\Lambda }$, implying $\mathrm{\mid }x-y-(x^{\mathrm{\prime }}-y^{\mathrm{\prime }})\mathrm{\mid }\le \mathrm{\mid }x-x^{\mathrm{\prime }}\mathrm{\mid }+\mathrm{\mid }y-y^{\mathrm{\prime }}\mathrm{\mid }<2ϵ<{\inf }_{\lambda \in \mathrm{\Lambda }}\mathrm{\mid }x-y-\lambda \mathrm{\mid }$, a contradiction. These quotient neighborhoods are disjoint, so the torus is Hausdorff. Let $\mathcal{B}$ be the countable rational-box base of [F3]. Since $q$ is open by step 1.1, $\{q(B):B\in \mathcal{B}\}$ is a countable base on ${\mathbb{T}}_L^3$: if $p\in V$ with $V$ open, choose $x\in q^{-1}(p)$ and a rational box $B$ with $x\in B\subseteq q^{-1}(V)$; then $p\in q(B)\subseteq V$. Together with step 1.1 this proves the topological three-manifold conditions in [F2].

The charts $q{\mathrm{\mid }}_B^{-1}$ from step 1.1 form a smooth atlas. In these charts the local representative of $q$ is the identity, so $q$ is a smooth local diffeomorphism. On an overlap, fix lifts $x,y$ of one point with $y-x=\lambda \in \mathrm{\Lambda }$. The distance from $\lambda$ to any other lattice vector is positive because each lattice spacing $L_i$ is positive. Restrict both chart neighborhoods so their lifts are within less than one quarter of that distance from $x$ and $y$. Any pair of lifts of one point in this restricted overlap must then differ by the same $\lambda$; otherwise the triangle inequality would put two distinct lattice vectors less than half their positive separation apart. Thus the transition is locally a fixed translation, smooth by [F5]. The atlas generates the smooth structure of [F2]. Since every chart has open image in ${\mathbb{R}}^3$, there is no boundary. The translation transitions have derivative the identity, so the local one-forms $d{x}^i$ and vector fields ${\mathrm{\partial }}_i$ agree on overlaps and define global smooth fields. A smooth $\mathrm{\Lambda }$-periodic function is constant on quotient fibres; in each chart its representative is a translate of the original smooth function, so it descends smoothly. The product is Hausdorff: $\mathbb{R}$ is Hausdorff because distinct real coordinates have disjoint intervals of radius less than half their distance, and ${\mathbb{T}}_L^3$ is Hausdorff by step 2.1. For distinct product points, separate a differing coordinate by disjoint open neighborhoods and take their products with the full other factor. It is second-countable because the products of a countable rational-interval base of $\mathbb{R}$ and the countable base of step 2.1 form a countable base. Product chart images are open in ${\mathbb{R}}^4$, so they give a boundaryless topological four-manifold. Their transitions are products of identity maps and the smooth translations just proved, hence form a smooth structure. In these charts the displayed $g,N,o$ have the same constant components, so they patch globally; $g$ has signature $(-,+,+,+)$, $N$ is timelike, and $o$ never vanishes. The metric formula [F6] then gives ${\mathrm{\Gamma }}^a{}_{bc}=0$ in every product chart. This proves all claims.