[proofplan] The exponential map of a compact torus identifies $T$ with the quotient of its [Lie algebra](/page/Lie%20Algebra) $\mathfrak t$ by the lattice $\Lambda_T=\ker\exp_T$. Thus the only issue is whether the function $X\mapsto e^{2\pi i\ell(X)}$ is constant on cosets of $\Lambda_T$. This constancy is equivalent to requiring $e^{2\pi i\ell(Y)}=1$ for every $Y\in\Lambda_T$, which is exactly the condition $\ell(\Lambda_T)\subset\mathbb Z$. Once the formula descends to $T$, linearity of $\ell$ gives the homomorphism property, and surjectivity of $\exp_T$ gives uniqueness. [/proofplan]

[step:Use the exponential quotient description of the torus]We use the standard quotient description of a compact torus: the map $\exp_T:\mathfrak t\to T$ is a surjective continuous [group homomorphism](/page/Group%20Homomorphism) from the additive Lie group $\mathfrak t$ onto the multiplicative group $T$, and for $X,X'\in\mathfrak t$, \begin{align*} \exp_T X=\exp_T X' \end{align*} if and only if \begin{align*} X-X'\in\Lambda_T. \end{align*} Equivalently, the fibres of $\exp_T$ are precisely the cosets $X+\Lambda_T$.[/step]

[guided]The proof is about whether a formula written upstairs on the Lie algebra $\mathfrak t$ descends downstairs to the torus $T$. For a compact torus, the exponential map has exactly the quotient behaviour needed for this descent: every $t\in T$ can be written as $t=\exp_T X$ for some $X\in\mathfrak t$, and two choices $X$ and $X'$ represent the same element of $T$ precisely when their difference lies in the kernel \begin{align*} \Lambda_T=\ker(\exp_T:\mathfrak t\to T). \end{align*} Thus the ambiguity in choosing a logarithm of $t$ is exactly the freedom to replace $X$ by $X+Y$ with $Y\in\Lambda_T$. This is the point at which the torus hypothesis is used. For a compact torus, $\mathfrak t$ is an abelian Lie algebra, $T$ is an abelian Lie group, and the exponential map is a homomorphism from the additive group of $\mathfrak t$ to $T$: \begin{align*} \exp_T(X+X')=\exp_T X\,\exp_T X' \end{align*} for all $X,X'\in\mathfrak t$. The kernel of this homomorphism is exactly $\Lambda_T$, so the [quotient group](/theorems/790) $\mathfrak t/\Lambda_T$ is identified with $T$ through the exponential map.[/guided]

[step:Show that integrality makes the exponential formula well-defined] Assume \begin{align*} \ell(\Lambda_T)\subset\mathbb Z. \end{align*} Define a map $F_\ell:\mathfrak t\to U(1)$ by \begin{align*} F_\ell(X):=e^{2\pi i\ell(X)}. \end{align*} If $X,X'\in\mathfrak t$ satisfy $\exp_T X=\exp_T X'$, then $Y:=X-X'$ belongs to $\Lambda_T$. By the assumed integrality condition, $\ell(Y)\in\mathbb Z$. Since $\ell$ is linear, \begin{align*} \ell(X)=\ell(X')+\ell(Y). \end{align*} Therefore \begin{align*} e^{2\pi i\ell(X)}=e^{2\pi i\ell(X')}e^{2\pi i\ell(Y)}=e^{2\pi i\ell(X')}. \end{align*} The last equality holds because $e^{2\pi i n}=1$ for every $n\in\mathbb Z$. Hence $F_\ell$ is constant on the fibres of $\exp_T$. Since $\exp_T$ is surjective, there is a unique map $\lambda:T\to U(1)$ satisfying \begin{align*} \lambda(\exp_T X)=F_\ell(X)=e^{2\pi i\ell(X)} \end{align*} for every $X\in\mathfrak t$. [/step]

[step:Verify that the descended map is a character] Let $t,t'\in T$. Since $\exp_T$ is surjective, choose $X,X'\in\mathfrak t$ such that \begin{align*} t=\exp_T X \end{align*} and \begin{align*} t'=\exp_T X'. \end{align*} Because $T$ is abelian and $\exp_T$ is a homomorphism from the additive group of $\mathfrak t$, \begin{align*} tt'=\exp_T(X+X'). \end{align*} Using the definition of $\lambda$ and the linearity of $\ell$, we obtain \begin{align*} \lambda(tt')=e^{2\pi i\ell(X+X')}=e^{2\pi i\ell(X)}e^{2\pi i\ell(X')}=\lambda(t)\lambda(t'). \end{align*} Thus $\lambda$ is a group homomorphism. The map $F_\ell:\mathfrak t\to U(1)$ is continuous because $\ell:\mathfrak t\to\mathbb R$ is linear and $a\mapsto e^{2\pi i a}$ is continuous from $\mathbb R$ to $U(1)$. Since $F_\ell=\lambda\circ\exp_T$ and $\exp_T$ is the quotient map for the identification $T\cong\mathfrak t/\Lambda_T$, the descended map $\lambda:T\to U(1)$ is continuous. Hence $\lambda$ is a character. [/step]

[step:Recover integrality from the existence of the character] Conversely, suppose there exists a character $\lambda:T\to U(1)$ such that \begin{align*} \lambda(\exp_T X)=e^{2\pi i\ell(X)} \end{align*} for every $X\in\mathfrak t$. Let $Y\in\Lambda_T$. Then $\exp_T Y=e_T$, where $e_T$ is the identity element of $T$. Since $\lambda$ is a group homomorphism, $\lambda(e_T)=1$. Applying the defining formula at $Y$ gives \begin{align*} e^{2\pi i\ell(Y)}=\lambda(\exp_T Y)=\lambda(e_T)=1. \end{align*} For a real number $a$, the equality $e^{2\pi i a}=1$ holds if and only if $a\in\mathbb Z$. Therefore $\ell(Y)\in\mathbb Z$. Since $Y\in\Lambda_T$ was arbitrary, \begin{align*} \ell(\Lambda_T)\subset\mathbb Z. \end{align*} [/step]

[step:Prove uniqueness from surjectivity of the exponential map] Suppose $\lambda_1:T\to U(1)$ and $\lambda_2:T\to U(1)$ are characters satisfying \begin{align*} \lambda_1(\exp_T X)=e^{2\pi i\ell(X)}=\lambda_2(\exp_T X) \end{align*} for every $X\in\mathfrak t$. Let $t\in T$. By surjectivity of $\exp_T$, choose $X\in\mathfrak t$ such that $t=\exp_T X$. Then \begin{align*} \lambda_1(t)=\lambda_1(\exp_T X)=\lambda_2(\exp_T X)=\lambda_2(t). \end{align*} Thus $\lambda_1=\lambda_2$. Combining the preceding steps proves the stated equivalence and the uniqueness assertion. [/step]