The momentum live in the dual lattice in the toriodal compactification (book reading) (D-Branes Clifford V. Johnson section 4.5 )


$G_{mn}=\delta_{ab} e^a_m e^b_n$


$X^a=X^m e_m^a$ where the equivalence of the toriodal compactification $X^a\sim X^a 2\pi e^a_mn^m$ or identify the lattice $\Lambda=\{ e^a_m n^m,n^m\in\mathbb{Z}\}$ .


The question was then to calculate the $p^n$ (I think the book meant $n$ to be $\sim \mu$ curved space index)
where the claimed result was

$p^n =G^{mn} n_m$  (claim 1)

I followed the calculation in the textbook but had trouble to see how it was.

$p\cdot \delta X\in 2\pi \mathbb{Z}$ where $\delta X\in 2\pi \Lambda$

and so to expand
\begin{equation}
\begin{split}
p\cdot \delta X\in 2\pi \mathbb{Z}
\\
\delta_{ab} p^a  2\pi e^b_m n^m \in 2\pi \mathbb{Z}
\\
\delta_{ab} p^a  e^b_m n^m \in \mathbb{Z}
\end{split}
\end{equation}
using the fact that
$p^a =p^n e_n^a$
\begin{equation}
\begin{split}
\delta_{ab} p^n e_n^a e_m^b  n^m  \sim \mathbb{Z} \\
p^n G_{mn} n^m \sim \mathbb{Z}
\end{split}
\end{equation}
So if I plugged in the claimed result
\begin{equation}
\begin{split}
G^{zn} n_z G_{mn} n^m \sim \mathbb{Z}
\\
G^{z}_m n_z  n^m \sim \mathbb{Z}
\end{split}
\end{equation}
However, this was not necessarily true since the general target space $G^z_m$ does not have to be $\delta^a_m$.
The following text also stated that

the momenta live in the dual lattice ...
$A^* \equiv \{ e^{*am} n_m, n_m\in\mathbb{z} \}$ (claim 2)
where the inverse veilbines $e^{*am}n_m$ are defined in the usual way using the inver se metric
$e^{*am}\equiv e^a_m G^{mn}$(def 1),  or $e^{*am}e_m^b =\delta^{ab}$

However, the indice $e^{*am}\equiv e^a_m G^{mn}$ did not match where on the left it was $a$ and m and on the right it was $a$ and $n$, is that an typo?
Further, the claim themselves did not match
$e^{*am} n_m=  e^a_m G^{mn} n_m  \neq G^{mn } n_m $
Could you help me to decode the context in the book, and to see how the (claim 1) and the (claim 2) were true, and the (def 1), please?
If there's a typo, what's the correct version?
 A: In general, let's we denote the generic torus in $\newcommand{\R}{\mathbb{R}}\R^d$, by $\newcommand{\Z}{\mathbb{Z}}\newcommand{\T}{\mathbb{T}}\newcommand{\n}{\mathsf{n}}\T^d_\Lambda :=\R^d/\Lambda$, where $\Lambda$ is a lattice in $\R^d$, namely
$$\Lambda := \bigoplus_{n=1}^d \omega_n\,\Z, \tag{1}$$
$\omega_n$ being linearly independent vectors in $\R^d$.
The torus that is used in the reference is $\T^d \equiv \T^d_{\Z^d}$, i.e. $\omega_n = e_n$, where $e_n$ is a unit vector pointing in the $n$-th direction. (Please note that my conventions and the book's differ by factors of $2\pi$'s. It should be a nice exercise to translate consistently between the two conventions)
For a general lattice, $\Lambda$, like in (1), the usual way to define the dual lattice is
$$ \Lambda^* :=\left\{q\in\R^d\ \middle\vert\ q\cdot x\in\Z,\ \forall x\in\Lambda   \right\}, \tag{2}$$
where $q\cdot x := \delta_{ab} q^a x^b$ is the standard vector inner-product in $\R^d$.
The argument is, then, that single-valuedness of the wave-function implies that
$$\exp(2\pi\, \mathrm{i}\; p \cdot X) \overset{!}{=} \exp(2\pi\, \mathrm{i}\; p \cdot (X+x)),\quad \forall x\in\Lambda,$$
and hence you get immediately that $p\cdot x\in\Z,\ \forall x\in\Lambda$, giving that $p\in\Lambda^*$,  by the definition (2). Notice that for all of this we only need to use the metric $\delta_{ab}$ of $\R^d$.
Now, the rest of your question boils down to the following. Since $\Lambda^*$ is itself a lattice, we can find $\omega^n_*$ such that
$$\Lambda^* = \bigoplus_{n=1}^d \omega^n_*\, \Z.$$
The question is, then, how are $\omega^n_*$ related to $\omega_n$? Since both sets are $d$-linearly independent vectors in $\R^d$, there is a full-rank matrix relating them. Let's call this matrix $Q^{mn}$, i.e.
$$\omega_*^m = Q^{mn} \omega_n.\tag{3}$$
(By the way, the mismatch of the indices in the book is obviously a typo, it should the correct free index on the LHS.)
Now, any $q\in\Lambda^*$ you can write as $q = \omega^n_* \n_n$, where $\n_n\in\Z$. Then, by (1) and (2) you have that
$$ \Z \ni \omega^n_* \n_n \cdot \omega_m \n^m = \n_n\n^m \delta_{ab}{\omega^n_*}^a{\omega_m}^b\quad \forall\n^m.$$ To satisfy that for all $\n^m$, you need that $\omega^n_*\cdot\omega_m\n_n=\n_m\in\Z$. However, (3) instructs that
\begin{align}
Q^{kn}\omega_k\cdot\omega_m\,\n_n &= \n_m, \qquad\text{or, in other words} \\
Q^{kn}\omega_k\cdot\omega_m &= \delta^n_{~m}.
\end{align}
Using the fact that in the book $\omega_n$ are taken to be $e_n$, so $\omega_k\cdot\omega_m = \delta_{km}$, and that $n$ and $m$ indices are raised and lowered with the metric $G_{mn}$ and its inverse we get immediately that
$$Q^{mn}=G^{mn},$$
hence indeed (3) reads as
$$\omega^m_* = G^{mn} \omega_n \quad\Longleftrightarrow\quad {\omega^m_*}^a = G^{mn} {\omega_n}^a.$$
From here on, you can immediately get the other relation
$${\omega^m_*}^a {\omega_m}^b = G^{mn} {\omega_n}^a {\omega_m}^b = \delta^{ab},$$
by the fact that $G^{mn}$ is the inverse of $G_{mn}$.
