Simple question on parallel transport in dually flat manifolds I just started studying Information Geometry and its applications by Amari. Right in the first chapter, the author talks about parallel transport in Dually flat manifolds. 
Just some quick notation: 
In the first chapter, the author introduces the definition of Manifold $M$ and one of the many coordinate systems on the manifold, say, $\theta$.
The length of the curve from $\boldsymbol {\theta}$ to $\boldsymbol{ \theta + d\theta}$ is given by $$
d s^{2}=2 D_{\psi}[\theta : \theta+d \theta]=\sum g_{i j} d \theta^{i} d \theta^{j}
$$
A tangent vector can be expressed as $$
d \boldsymbol{\theta}=\sum d \theta^{i} \boldsymbol{e}_{i}
$$ where $
\left\{\boldsymbol{e}_{i}\right\}
$, $i \in \{1,.. ,n\}$ are the basis of the tangent space of M at point  $\boldsymbol \theta$. Similarly, the author introduces a dual affine coordinate system whose corresponding basis is $\left\{e^{* i}\right\}$. Therefore,we can write 
$$
d \boldsymbol \theta^{*}=\sum d \theta_{i}^{*} e^{* i}
$$
Now, one can also write the length of the small line vector as $$
d s^{2}=\langle d \boldsymbol{\theta}, d \boldsymbol{\theta}\rangle= g_{i j} d \theta^{i} d \theta^{j}
$$, which is rewritten as $$
d s^{2}=\left\langle d \theta^{i} e_{i}, d \theta^{j} e_{j}\right\rangle=\left\langle e_{i}, e_{j}\right\rangle d \theta^{i} d \theta^{j}
$$
Hence, it is clear that $$
g_{i j}(\boldsymbol{\theta})=\left\langle\boldsymbol{e}_{i}, \boldsymbol{e}_{j}\right\rangle
$$
Similarly, for the dual affine coordinate system $\boldsymbol \theta^*$, we have $$
g^{* i j}(\boldsymbol \theta^*)=\left\langle e^{* i}, e^{* j}\right\rangle
$$
If $\bf G$ is the Jacobian of the transformation from $\boldsymbol \theta$ to $\boldsymbol \theta^*$, then we can write 
$$
\begin{array}{l}{d \boldsymbol{\theta}^{*}=\mathbf{G} d \boldsymbol{\theta}, \quad d \boldsymbol{\theta}=\mathbf{G}^{-1} d \boldsymbol{\theta}^{*}} \\ {d \theta_{i}^{*}=g_{i j} d \theta^{j}, \quad d \theta^{j}=g^{* j i} d \theta_{i}^{*}}\end{array}
$$

Actual doubt: 
If a tangent vector $\mathbf{A} = A^i\mathbf{e_i} $ is transported
  from a point $\boldsymbol{\theta}$ to $\boldsymbol{\theta^{'}}$, the
  components $A^i$ remain the same because $\mathbf{e_i}$ is same
  everywhere in a dually flat manifold. However, in the later
  paragraphs, it is stated that the length of the vector is not constant
  even in this dually flat manifold. If the components and basis remain
  the same, shouldn't it be the case where even the length of the vector
  is also the same when moved from one point to another point? 
Also, why is dual parallel transport any different parallel transport?
  After all, they are just two different coordinate systems. Because, it
  is given that the manifold is dually flat, so both the coordinate
  systems remain the same across all the points. So, why does the
  parallel transport makes the vector invariant under the original basis
  but the same parallel transport changes the vector in the dual basis? 
  Any intuitive explanations, illustrations or examples on how this
  happens?

P.S: Relevant page from the book. 

 A: In information geometry there are two flat connections $\nabla^e$ and $\nabla^m$ that are dual with respect to the Fisher-Rao metric $g$, which means that for any three vector fields $X,Y,Z$ we have
$$X g(Y,Z)=g(\nabla^e_XY,Z)+g(Y,\nabla^m_XZ).$$
A consequence of this is that if $Y$ is e-parallel along $X$ and $Z$ is m-parallel along $X$, then the angle between $Y$ and $Z$ is constant. However, this is not true if both $Y$ and $Z$ are parallel along $X$ wrt. to one connection only.
The duality above is different from the property enjoyed by the Levi-Civita connection $\nabla^{\text{LC}}$, which satisfies
$$X g(Y,Z)=g(\nabla^{\text{LC}}_XY,Z)+g(Y,\nabla^{\text{LC}}_XZ)$$
for all vector fields $X,Y,Z$.
This is known as self-dualilty or metricity of $\nabla^{\text{LC}}$ and can also be written as $\nabla^{\text{LC}}g=0$ since for any arbitrary connection $\nabla$ we have
$$X g(Y,Z)=g(\nabla_XY,Z)+g(Y,\nabla_XZ)+(\nabla_X g)(Y,Z).$$
In contrast, $\nabla^eg$ and $\nabla^mg$ are both non-zero. 
Since $\nabla^{\text{LC}}$ is not flat, parallel transport is complicated in general. On the other hand, $\nabla^e$ and $\nabla^m$ are flat, so parallel transport is easy. There are sets of dual charts $\theta^i$ and $\eta_i$ that are e- and m-affine, respectively. Dual means that if we denote $e_i=\partial_i=\partial_{\theta^i}$ and $e^i=\partial^i=\partial_{\eta^i}$
$$g(e_i, e^j)=\delta^j_i \Leftrightarrow e^i=g^{ij}e_j\Leftrightarrow e_i=g_{ij}e^j.$$
Affine means that all the Christoffel symbols vanish in that chart, i.e. all Christoffel symbols of $\nabla^e$ vanish in the $\theta$ chart and all Christoffel symbols of $\nabla^m$ vanish in the $\eta$ chart. This in turn implies that parallel transport corresponds to constant coefficients in the respective charts. I.e. $\partial_i$ are e-parallel and $\partial^i$ are m-parallel. The property $g(\partial_i,\partial^j)=\delta^j_i$ reflects the above self-duality condition.
