I'm having trouble understanding an argument in a proof from Hall's book "Lie Groups, Lie Algebras, and Representations" (it's a step in the proof of Lemma 11.5). Perhaps someone from this community can provide some clarity.
The situation is as follows. Let $T = (S^1)^k$ be a Torus, where $k \in \mathbb{N}$. Let $t=(e^{2\pi\theta_1}, \ldots, e^{2\pi\theta_k})$ be an element of $T$ and let $S$ be the closure of the cyclic subgroup of $T$ generated by $t$, \begin{align*}S = \overline{\{t^n\ |\ n \in \mathbb{Z} \}}.\end{align*} Let $\mathfrak{t}$ be the Lie Algebra of $T$. Consider the preimage of $S$ under (a scaled version of) the exponential, \begin{align*}\Lambda := \{ H \in \mathfrak{t}\ |\ \exp(2 \pi H) \in S \}. \end{align*} Next, let $\Lambda_0$ be the connected component of zero in $\Lambda$.
Now, since $T$ (and thus $\mathfrak{t}$) is commutative, the exponential is a Lie group homomorphism from $(\mathfrak{t},+)$ to $T$. So $\Lambda$ is a closed additive subgroup of $\mathfrak{t}$. From this it follows that $\Lambda_0$ is a closed additive subgroup of $\mathfrak{t}$.
The author claims that $\Lambda_0$ is not just a subgroup of $\mathfrak{t}$, but a subspace of $\mathfrak{t}$. The argument for this is, that $\Lambda_0 = \mathrm{Lie}(\Lambda)$.
The author argues that the last equation follows from the following two points:
- $\mathrm{Lie}(\Lambda) = \mathrm{Lie}(\Lambda_0)$.
- An element $A$ of a connecte Lie group $G$ can be written as $A = e^{X_1}\cdots e^{X_n}$, with each $X_i \in \mathrm{Lie}(G)$.
These two facts are known to me, but I cannot see how to apply them to get the assertions about $\Lambda_0$ being a vector space.
I agree with you there is a gap in the proof.
Trying to make the notation as least misleading as possible, let $\Phi:\text{Lie}(\Lambda)\to\Lambda_0$ be the exponential map. What we know is that $\Phi$ is 1) continuous, 2) locally a diffeomorphism, 3) a group homomorphism in the sense that $\Phi(x+y)=\Phi(x)+\Phi(y)$, where the first addition is in $\text{Lie}(\Lambda)$ and second addition in $\Lambda_0$ - an additive subgroup of $\mathfrak{t}$ which in turn is a vector space of dimension $k$ over $\Bbb{R}$.
First, $\Phi$ is injective. If $\Phi$ had a non-zero kernel, there would be some $x\in\text{Lie}(\Lambda)$, $x\ne 0$ such that $\Phi(x)=0$. Then $\Phi\rvert_{\Bbb{R}x}$ factors through $\Bbb{R}x/\Bbb{Z}x$ which is a compact group. By continuity $\Phi(\Bbb{R}x)$ is a compact additive subroup of $\Lambda_0$. But $\Lambda_0$ is embedded in $k$-dimensional space, and the only compact additive subgroup of $\Bbb{R}^k$ is zero. (It has to be bounded.) This implies that $\Phi(\Bbb{R}x)=\{0\}$. But now, this contradicts $\Phi$ being a local diffeomorphism - because it must map some open neighborhood of zero in $\text{Lie}(\Lambda)$ injectively into $\Lambda_0$. Any neighborhood of zero intersects $\Bbb{R}x$ in infinitely many non-zero points.
Second, as mentioned in the original post, every $y\in\Lambda_0$ can be written as $y=\Phi(x_1)+\cdots+\phi(x_n)$ for some $x_1,\ldots,x_n\in\text{Lie}(\Lambda)$. But $\Phi$ is an additive homomorphism so $y=\Phi(x_1+\cdots+x_n)$ and therefore $\Phi$ is surjective.
Now let $y\in\Lambda_0$ and $n>1$ an integer. Then $y=\Phi(x)$ for some $x\in\text{Lie}(\Lambda)$. Since $n\frac{x}{n}=x$ in $\text{Lie}(\Lambda)$ then by the additivity of $\Phi$, $n\Phi(\frac{x}{n})=y$ in $\Lambda_0$. But $\frac{y}{n}$ exists in $\mathfrak{t}$ as an element of a vector space so $\Phi(\frac{x}{n})=\frac{y}{n}$ and this shows $\frac{y}{n}\in\Lambda_0$. This proves that $\Lambda_0$ is a divisible group. $\Lambda_0$ is also Abelian, torsion-free (being a subgroup of a vector space). A torsion-free divisible Abelian group is a vector space over $\Bbb{Q}$. So $\Lambda_0$ is a $\Bbb{Q}$-linear subspace of $\mathfrak{t}$.
Finally, let $a\in\Bbb{R}$. There exists some sequence of rationals $q_m$ such that $q_m\to a$ as $m\to\infty$. By continuity of $\Phi$, $\Phi(ax)=\Phi(\lim_{m\to\infty} q_m x)=\lim_{m\to\infty} q_m\Phi(x)=a\Phi(x)\in\mathfrak{t}$. This proves that $a\Phi(x)\in\Lambda_0$, and completes the proof that $\Lambda_0$ is a vector sub-space of $\mathfrak{t}$ over $\Bbb{R}$.