Structure of compact connected abelian Lie groups

By [A1], the countable-choice hypotheses of [L1] and [L2] hold. The exponential map of an abelian Lie group is therefore a homomorphism by [L1], and by [L2] it maps the open neighbourhood $U$ of $0$ diffeomorphically onto the open neighbourhood $V$ of $e$. Hence $\exp (\mathfrak{t})$ contains $V$, and since $\exp (\mathfrak{t})$ is a subgroup containing a neighbourhood of the identity, it is open; an open subgroup of a topological group is closed and its cosets partition the group, so connectedness of $T$ forces $\exp (\mathfrak{t})=T$, that is, $\exp$ is surjective.

The kernel $\mathrm{\Lambda }={\exp }^{-1}(e)$ is a subgroup of $\mathfrak{t}$ and is discrete: by [L2], if $\mathrm{\Lambda }\cap U$ contained a nonzero point then $U$ would contain two distinct points with the same image under the diffeomorphism $\exp {\mathrm{\mid }}_U$. It is closed in $\mathfrak{t}$ as the preimage of the closed singleton $\{e\}$ under the continuous map $\exp$.

Construct the smooth quotient explicitly. For a closed discrete subgroup $D$ of a finite-dimensional real vector space $E$, the quotient map $q:E\to E\mathrm{/}D$ is open, since $q^{-1}(q(O))={\bigcup }_{d\in D}(O+d)$ for open $O$. Choose a ball $B$ about $0$ with $(B-B)\cap D=\{0\}$. The restrictions of $q$ to translates of $B$ are homeomorphisms onto open sets and supply coordinate charts. On overlaps, the two lifts differ locally by a fixed element of $D$, so chart transitions are translations and are smooth. Distinct cosets have disjoint small chart neighbourhoods: if $x-y\mathrm{\notin }D$, closedness of $D$ gives a ball about $x-y$ disjoint from $D$. Images of a countable Euclidean base give a countable base for the quotient. Thus these charts define a Hausdorff, second-countable smooth manifold, and addition and inversion are smooth, being locally addition and negation followed by translations. This is the unique smooth structure for which $q$ is a local diffeomorphism. Apply the construction to $E=\mathfrak{t}$ and $D=\mathrm{\Lambda }$. The map $\stackrel{}{\exp }:\mathfrak{t}\mathrm{/}\mathrm{\Lambda }\to T$, $X+\mathrm{\Lambda }\mapsto \exp X$, is a well-defined bijective homomorphism by steps 1.1 and 2.1. Shrink $B$ into the neighbourhood $U$ of [L2]. In the resulting quotient chart, $\stackrel{}{\exp }$ is exactly $\exp {\mathrm{\mid }}_B$, hence a local diffeomorphism; translations give this property everywhere. A bijective local diffeomorphism has smooth local inverses that form its global inverse. Therefore $\stackrel{}{\exp }$ is a Lie-group isomorphism.

The kernel $\mathrm{\Lambda }$ is a lattice in its real span. We prove the lattice assertion by induction; when the span is zero the subgroup is $\{0\}$ with the empty basis. If $V:=\mathrm{span}(\mathrm{\Lambda })$ is nonzero, fix a Euclidean norm and choose $X_1\in \mathrm{\Lambda }\setminus \{0\}$ of least norm; such an element exists because a closed discrete subset meets every compact ball in a finite set. Subtracting a nearest integer multiple of $X_1$ shows $\mathrm{\Lambda }\cap \mathbb{R}{X}_1=\mathbb{Z}{X}_1$. Let $p:V\to X_1^{\perp }$ be orthogonal projection. For each $X\in \mathrm{\Lambda }$, subtract an integer multiple of $X_1$ so that the $X_1$-coordinate lies in $[-1\mathrm{/}2,1\mathrm{/}2]$. The resulting representatives with $\mathrm{\parallel }p(X)\mathrm{\parallel }\le 1$ lie in a compact cylinder, whose intersection with $\mathrm{\Lambda }$ is finite. If that finite set has nonzero projected norms, their minimum is positive; otherwise $p(\mathrm{\Lambda })$ has no nonzero point of norm at most $1$. In either case $0$ is isolated in $p(\mathrm{\Lambda })$, and translation makes $p(\mathrm{\Lambda })$ discrete. Such a subgroup is also closed: near any point in its closure a ball of diameter smaller than its positive separation contains at most one subgroup point, forcing the limit to equal that point. Induction on $\dim V$ gives a $\mathbb{Z}$-basis of $p(\mathrm{\Lambda })$ whose lifts $X_2,\dots ,X_m\in \mathrm{\Lambda }$, together with $X_1$, generate $\mathrm{\Lambda }$ and are linearly independent over $\mathbb{R}$. Thus $\mathrm{\Lambda }$ is free abelian of rank $m=\dim V$ with a $\mathbb{Z}$-basis that is an $\mathbb{R}$-basis of $V$.

The rank equals $n:=\dim \mathfrak{t}$: if $V=\mathrm{span}(\mathrm{\Lambda })⊊\mathfrak{t}$, then the map $X+\mathrm{\Lambda }\mapsto X+V$ is well defined and continuous by the quotient topology and gives a surjection $\mathfrak{t}\mathrm{/}\mathrm{\Lambda }\to \mathfrak{t}\mathrm{/}V\cong {\mathbb{R}}^{n-m}$ with $n-m>0$; its image is compact by [L5], while ${\mathbb{R}}^{n-m}$ is not compact, a contradiction. Hence $m=n$, so $\mathrm{\Lambda }$ has a $\mathbb{Z}$-basis $X_1,\dots ,X_n$ that is an $\mathbb{R}$-basis of $\mathfrak{t}$; in particular $\mathrm{\Lambda }$ is a full lattice and $T\cong \mathfrak{t}\mathrm{/}\mathrm{\Lambda }$ by step 3.1.

The linear isomorphism ${\mathbb{R}}^n\to \mathfrak{t}$, $(a_1,\dots ,a_n)\mapsto {\sum }_i{a}_i{X}_i$, carries ${\mathbb{Z}}^n$ onto $\mathrm{\Lambda }$. In the quotient charts of step 3.1 it and its inverse remain smooth, so it induces a Lie-group isomorphism ${\mathbb{R}}^n\mathrm{/}{\mathbb{Z}}^n\to \mathfrak{t}\mathrm{/}\mathrm{\Lambda }$. By [A1] the countable-choice hypothesis of [L3] holds. Equip $S^1=\mathbb{R}\mathrm{/}\mathbb{Z}$ with the quotient charts of step 3.1, using open intervals of length less than $1$. The coordinate map $[a_1,\dots ,a_n]\mapsto ([a_1],\dots ,[a_n])$ is a bijective homomorphism from ${\mathbb{R}}^n\mathrm{/}{\mathbb{Z}}^n$ to $(S^1{)}^n$. On products of these intervals its coordinate expression and inverse are the identity, so it is a Lie-group isomorphism for the product smooth structure. Composing gives $T\cong (S^1{)}^r$ with $r=n$. If $n=0$, surjectivity of $\exp :\{0\}\to T$ gives $T=\{e\}$, $\mathrm{\Lambda }=\{0\}$, and the same conclusion uses the empty product. No choice beyond [A1] is required by the finite lattice construction or quotient charts.