Characters are the integral weights

Let $\chi \in X^{\ast }(T)$. The composite $\chi \circ \exp :\mathfrak{t}\to S^1$ is a continuous homomorphism, so [L3] gives a unique $\mu \in {\mathfrak{t}}^{\ast }$ with $\chi (\exp X)=e^{i\mu (X)}$. Put $\lambda =i\mu$ and extend it $\mathbb{C}$-linearly to ${\mathfrak{t}}_{\mathbb{C}}$. If $Y\in \mathrm{\Lambda }$, then $1=\chi (\exp Y)=e^{\lambda (Y)}$, so $\lambda (Y)\in 2\pi i\mathbb{Z}$; the map $\chi \mapsto \lambda$ is injective because $\exp$ is surjective.

Choose a $\mathbb{Z}$-basis $Y_1,\dots ,Y_r$ of the full lattice $\mathrm{\Lambda }$. By [L1], $\mathrm{\Phi }:(S^1{)}^r⟶T,qquad\mathrm{\Phi }(e^{i{\theta }_1},\dots ,e^{i{\theta }_r})=\exp \text{\hspace{0.17em}⁣}\left({\sum }_{j=1}^r\frac{{\theta }_j}{2\pi }{Y}_j\right)$ is a Lie-group isomorphism. If $\eta :S^1\to T$ is a cocharacter, each coordinate of ${\mathrm{\Phi }}^{-1}\eta$ is $z\mapsto z^{n_j}$ for a unique $n_j\in \mathbb{Z}$ by [L3]. Thus, with $X_{\eta }={\sum }_{j=1}^r\frac{n_j}{2\pi }{Y}_j,$ one has $\eta (e^{i\theta })=\exp (\theta X_{\eta })$ and $2\pi X_{\eta }\in \mathrm{\Lambda }$. Conversely every $X\in \mathfrak{t}$ with $2\pi X\in \mathrm{\Lambda }$ gives the well-defined cocharacter $e^{i\theta }\mapsto \exp (\theta X)$. Hence $X_{\ast }(T)$ is identified with $\frac{1}{2\pi }\mathrm{\Lambda }$, and the displayed formula also shows that $X_{\eta }=d{\eta }_1(i)$.

Conversely let $\lambda \in {\mathrm{Hom}}_{\mathbb{C}}({\mathfrak{t}}_{\mathbb{C}},\mathbb{C})$ satisfy $\lambda (\mathrm{\Lambda })\subseteq 2\pi i\mathbb{Z}$. Since the lattice basis in step 1.2 spans $\mathfrak{t}$ over $\mathbb{R}$, the restriction of $\lambda$ to $\mathfrak{t}$ takes values in $i\mathbb{R}$. Therefore $\chi (\exp X):=e^{\lambda (X)}$ takes values in $S^1$ and is well defined: if $\exp X=\exp X^{\mathrm{\prime }}$ then $X-X^{\mathrm{\prime }}\in \mathrm{\Lambda }$, so $e^{\lambda (X-X^{\mathrm{\prime }})}=1$. It is a continuous homomorphism $T\to S^1$, so it is a character, and its differentiated weight is $\lambda$; hence the constructions are mutually inverse bijections.

Pairing: with $\chi$ and $\lambda$ related as in step 1.1 and $\eta$ with $2\pi X_{\eta }\in \mathrm{\Lambda }$ as in step 1.2, one has $\chi (\eta (e^{i\theta }))=e^{\lambda (\theta X_{\eta })}=e^{i\theta \lambda (X_{\eta })\mathrm{/}i}$, so the integer $n$ with $\chi \circ \eta (z)=z^n$ is $n=\lambda (X_{\eta })\mathrm{/}i$; it is an integer because $2\pi X_{\eta }\in \mathrm{\Lambda }$ and $\lambda (2\pi X_{\eta })\in 2\pi i\mathbb{Z}$, so that $\lambda (X_{\eta })\mathrm{/}i=\lambda (2\pi X_{\eta })\mathrm{/}(2\pi i)\in \mathbb{Z}$.

Perfectness and freeness: choosing a $\mathbb{Z}$-basis $Y_1,\dots ,Y_r$ of $\mathrm{\Lambda }$ identifies $\mathrm{\Lambda }\cong {\mathbb{Z}}^r$, hence also $X_{\ast }(T)\cong \frac{1}{2\pi }\mathrm{\Lambda }\cong {\mathbb{Z}}^r$, while the characters correspond to the dual basis: $\lambda$ is determined by the integers $\lambda (Y_j)\mathrm{/}(2\pi i)$, and conversely every integer vector gives such a $\lambda$ by linear extension, because the basis spans $\mathfrak{t}$ over $\mathbb{R}$. Hence $X^{\ast }(T)\cong {\mathbb{Z}}^r$ is free of rank $r=\dim T$, the pairing is the dot product in these coordinates, and it is perfect.