Let $\mathcal{P}:=\mathcal{P}(\mathcal{X})$ be the manifold of all (strictly positive) probability vectors (distributions) on $\mathcal{X}=\{x\_0,\dots,x\_n\}$,


i.e., each $p=(p(x\_0),\dots,p(x\_n))\in \mathcal{P}$ is such that $p(x\_i)>0$ for all $i$ and $\sum\_{i}p(x\_i)=1$ and can be thought of a point in $\mathbb{R}^{n+1}$, i.e., $\mathcal{P}$ is an $n$-dimensional manifold. 


Let $\mathcal{P}=\{p\_{\xi}\}$, where $\xi=(\xi\_1,\dots,\xi\_n)$, be a (global) coordinate system. For example, $\mathcal{P}$ can be parametrized by $\mathcal{P} = \{p\_{\xi}\}$, $\xi = (\xi\_1,\dots,\xi\_n)\in \Xi,$ where
\begin{eqnarray\*}
 \Xi = \{(\xi\_1,\dots,\xi\_n) : \xi\_i>0, \sum\_{i=1}^n\xi\_i < 1\},
\end{eqnarray\*}
with 
\begin{equation}
 p\_{\xi}(x\_i) =
\begin{cases} \xi\_i & \text{for } i = 1,\dots,n
\\
1-\sum\_{i=1}^n\xi\_i &\text{for } i = 0.
\end{cases}
\end{equation}


A Riemannian metric $G(\xi) = [g\_{i,j}(\xi)]$ is defined on $\mathcal{P}$, where 
\begin{eqnarray}
g\_{i,j}(\xi) & = & \sum\_x \frac{\partial}{\partial\xi\_i} (p\_{\xi}(x))~ \frac{\partial}{\partial\xi\_j}(\log p\_{\xi}(x)),
\end{eqnarray}
where the $\log$ is taken coordinatewise.


An affine connection $\nabla$ is defined on $\mathcal{P}$, given by the Christoffel symbols
\begin{eqnarray}
\Gamma\_{ij}^k({\xi}) & = & \sum\_x \frac{\partial}{\partial\xi\_k}(p\_{\xi}(x))~\frac{\partial}{\partial\xi\_i}\left(\frac{\partial}{\partial\xi\_j}\log p\_{\xi}(x)\right).
\end{eqnarray}
Suppose that $\gamma\_t$ is a $\nabla$-geodesic on $\mathcal{P}$. From [THIS](http://projecteuclid.org/euclid.aos/1176345779) article of Amari in Ann. of Statistics, I came to know that the geodesic equation (for this connection) is given by $\ddot l\_t+i\_t=0$, where $l\_t(x)=\log\gamma\_t(x)$, and $i\_t=\sum\_x \dot\gamma\_t(x)\dot l\_t(x)=0$. However, the paper doesn't seem to give any explanation how it is obtained (See Appendix of the paper. $\alpha=1$ corresponds to my question). If anyone can help me derive this geodesic equation, it would be of great help. Thank you.


PS: I asked this question some days ago in math.stackexchange where I couldn't get any response. Here's a [link](https://math.stackexchange.com/questions/796815/geodesic-equation-from-christoffel-symbols) to the question.



Geodesic equation from Christoffel symbols
Let me first re-write your notations as little bit to make it easier for me. 


Let $\xi\in \Xi \subset\mathbb{R}^n$. Define the functions $\pi\_k:\Xi \to\mathbb{R}$ by 


$$ \pi\_k(\xi) = \begin{cases} \xi\_k & k \in \{1, \dots, n\} \\ 1 - \sum\_{1}^n \xi\_i & k = 0 \end{cases} $$


which to me is a more natural way to write your function $p\_\xi: \{0, \ldots, n\} \to \mathbb{R}$. 


Your metric is defined to be 


$$ \mathbf{g} = \sum\_{k = 0}^n \frac{1}{\pi\_k} ~\mathrm{d}\pi\_k \otimes \mathrm{d}\pi\_k = \sum\_{k = 0}^n \mathrm{d}\pi\_k \otimes \mathrm{d} \log \pi\_k $$


What you write as the Christoffel symbols is not really the Christoffel symbol, it is the Christoffel symbol with one index lowered. Note that the Christoffel symbol for the Levi-Civita connection is given by 


$$ g\_{k\ell} \Gamma^{\ell}\_{ij} = \frac12 \left( \partial\_i g\_{kj} + \partial\_j g\_{ki} - \partial\_k g\_{ij}\right) $$


one check indeed


$$ g\_{k\ell} \Gamma^{\ell}\_{ij} = \sum\_{m = 0}^n \partial\_k \pi\_m \partial^2\_{ij} \log\pi\_m = - \sum\_{m = 0}^n \frac{1}{\pi\_m^2} \partial\_k \pi\_m \partial\_i \pi\_m \partial\_j \pi\_m $$


where in the second equality we used that $\pi\_m$ are linear in the coordinates and hence the Hessian vanishes. 


Let $\gamma:\mathbb{R}\to \Xi$ be a geodesic. The geodesic equation in coordinates can be written as


$$ g\_{k\ell}\circ\gamma~ \ddot{\gamma}^\ell + g\_{k\ell}\circ\gamma~ \Gamma^\ell\_{ij}\circ\gamma ~\dot{\gamma}^i \dot{\gamma}^j = 0$$


plugging in the definitions we have


$$ \sum\_{m = 0}^n (\partial\_k \pi\_m)\circ\gamma \left[ (\partial\_\ell \log \pi\_m) \circ\gamma ~\ddot{\gamma}^{\ell} + (\partial^2\_{ij} \log\pi\_m)\circ\gamma~ \dot{\gamma}^i \dot{\gamma}^j \right] = 0 $$


which we simplify as


$$ \sum\_{m = 0}^n (\partial\_k \pi\_m) \circ \gamma \frac{d^2}{dt^2}( \log\circ\pi\_m\circ \gamma) = 0 $$


Now, $\partial\_k \pi\_m = \delta\_{km}$ if $m > 0$ and $-1$ otherwise. So we have


$$ \frac{d^2}{dt^2}( \log\circ\pi\_k\circ \gamma) - \frac{d^2}{dt^2}(\log\circ\pi\_0 \circ \gamma) = 0 \tag{1}$$




---


This is almost what you want. To finish, we observe the following:


$$ \sum\_{m = 0}^n \pi\_m\circ \gamma = 1 \implies \sum\_{m = 0}^n \frac{d^k}{dt^k} \pi\_m\circ\gamma = 0, k \geq 1 \tag{2}$$


Observe that 


$$ \frac{d^2}{dt^2} \log \circ\pi\_m \circ \gamma = \frac{1}{\pi\_m\circ\gamma} \frac{d^2}{dt^2} \pi\_m\circ\gamma - \frac{1}{(\pi\_m\circ\gamma)^2} (\frac{d}{dt} \pi\_m\circ\gamma)^2 $$


This says that


$$ \sum\_{m = 0}^n \pi\_m\circ\gamma \frac{d^2}{dt^2} \log\circ\pi\_m\circ\gamma = - \sum\_{m = 0}^n \frac{1}{(\pi\_m\circ\gamma)} (\frac{d}{dt} \pi\_m\circ\gamma)^2 \tag{3}$$


Applying (3) to (1) gives 


$$ \sum\_{m = 0}^n \pi\_m\circ\gamma \frac{d^2}{dt^2} \log\circ\pi\_0 \circ\gamma = - \sum\_{m = 0}^n \frac{1}{(\pi\_m\circ\gamma)} (\frac{d}{dt} \pi\_m\circ\gamma)^2 $$


so from (2), and (1) again, we get, for every $k \in \{0,\dots, n\}$, 


$$ \underbrace{\frac{d^2}{dt^2} \log\circ\pi\_k \circ\gamma}\_{\ddot{l}\_t(x\_k)} + \sum\_{m = 0}^n \underbrace{\frac{d}{dt}(\pi\_m\circ\gamma)}\_{\dot{\gamma}\_t(x\_m)} \underbrace{\frac{d}{dt} \log\circ\pi\_m\circ\gamma}\_{\dot{l}\_t(x\_m)} = 0 $$


which is exactly the equation claimed in the question. 




---


I should remark that part of the above answer is reconstructed in differential geometry language from the linked paper. In some ways the original derivation is a bit slicker, especially when it comes to the derivative of what I wrote as equation (3). 


*Addendum to address this final paragraph*:


What you are treating is really the Fisher information metric. Consider the set $\Xi$ which is the interior of the simplex. If you look at the functions $\eta\_i$ such that $(\eta\_i)^2 = \xi\_i$, the simplex is defined as $\sum (\eta\_i)^2 = 1$, with $\eta\_i \in (0,1)$. Notice that this can be interpreted as a quadrant of the unit sphere with $n$ dimensions (so $i$ runs from $0$ to $n$). 


The metric you wrote down in $\xi$ coordinates turns to to be exactly the pull back metric of the standard unit sphere (up to some constant multiples) to $\Xi$ using the mapping $(\eta\_i)^2 = \xi\_i$. 


The geodesics on the unit sphere are the great circles and are thus the intersections of the sphere with hyperplanes. Let $\zeta \in \mathbb{R}^{n+1}$ be a fixed vector, there is a geodesic in $\eta$ coordinates satisfying $\sum \zeta\_i \eta\_i = 0$. We can choose the affine parametrisation $\eta\_i(t)$ to be such that $\sum (\frac{d}{dt}\eta\_i(t))^2 = E$ is constant. It is well-known (since Newton) that in this parametrisation the acceleration of $\eta$ is in the direction of $\eta$ itself, with the proportionality factor $E$: that is


$$ \ddot{\eta} + E \eta = 0. $$


Doing the reverse change of variables you arrive back to the formula for the expression of the geodesics in the $\xi$ coordinates. 


The fact that $\dot{\ell}$ in the language of the paper is a tangent vector, is exactly the statement that in $\eta$ coordinates, any curve along the sphere must satisfy $\sum\_i \eta\_i \dot{\eta}\_i = 0$. By itself it has nothing to do with the fact that $\ell$ is geodesic. The "slick" derivation that I referred to is essentially what I wrote here, but cast in the language of information geometry where for some reason one prefers the coordinates $\xi$ instead of $\eta$. 


