All straight curves are geodesics imply Euclidean metric Let $g_{ij}$ be a Riemannian Metric in $\Bbb R^n$ with the following properties: 


*

*$g_{ij}(0)=\delta_{ij}$ , where $ \delta_{ij} =
\begin{cases}
1,  & \text{ $i=j$} \\
0, & \text{ $i \neq j$}
\end{cases}$ 

*For every $q\in \Bbb R^n$ and $1\leq i \leq n,$ the curve $\gamma_i(t)= q + te_i$ is a geodesic of $g$, where $e_i = (\delta_{i1}, ... , \delta_{in}).$
Prove that $g$ is the Euclidean metric.

What I tried: since the above curves are geodesics, I know that $|\gamma_i'(t)|_g^2=\langle e_i,e_i \rangle _g=g_{ii}\;$ is constant, and thus $g_{ii} = 1$ for every $1\leq i \leq n$. I'm having trouble showing that $g_{ij}=0$ for $j \neq i$. Any ideas?
 A: As stated (where you only require lines parallel to the standard axes to be geodesic) the result is false. 
Consider the metric on $\mathbb{R}^2$ given in standard coordinates in matrix form
$$ \begin{pmatrix} (1 + y^2) & xy \\ xy & (1+x^2) \end{pmatrix} $$
The Christoffel symbols can be computed pretty easily to be seen that 
$$ \Gamma_{11}^1 = \Gamma_{22}^{1} = \Gamma_{11}^2 = \Gamma_{22}^2 = 0 $$
and hence all lines parallel to the axes are geodesics. But the metric is obviously not Euclidean. 

In the case where the assumption is that all straight lines are geodesic, one way to prove the result goes something like this:


*

*Again let $x^1, \ldots, x^n$ denote the standard coordinates. Consider all vectors of the form $v = \alpha e_i + \beta e_j$ and the line $\gamma(t) = q + tv$. The geodesic equation in local coordinates implies that for every $k\in \{1, \ldots, n\}$ and $i,j$ fixed (by $v$)
$$ \alpha^2 \Gamma_{ii}^k + 2\alpha\beta \Gamma_{ij}^k + \beta^2 \Gamma_{jj}^k = 0 $$
This equation holds along the curve $\gamma$.

*However, since we can launch the geodesic $\gamma$ from any point $q$ at any direction $v$, the above equation must hold pointwise everywhere. 

*Focus first on the case $\alpha = 1, \beta = 0$. Then this implies that for every $i,k\in \{1, \ldots, n\}$, we have $\Gamma_{ii}^k = 0$. 

*Focus next on the case $\alpha = \beta = 1$. Using the result from point 3, this means that $\Gamma_{ij}^k = 0$ for any $i,j,k\in \{1, \ldots, n\}$. 

*It remains to show (by simple linear algebra) that 
$$ \Gamma^k_{ij} = \frac12 g^{ks} \left( \partial_i g_{sj} + \partial_j g_{si} - \partial_s g_{ij} \right) = 0 $$
for all $i,j,k$ implies that 
$$ \partial_k g_{ij} = 0 $$
for all $i,j,k$. To do so, first notice that the assumed condition, using the non-degeneracy of the metric, implies that 
$$ \partial_i g_{kj} + \partial_j g_{ki} - \partial_k g_{ij} = 0 $$
for all $i,j,k$. Symmetrizing in $k$ and $j$ the second and third terms cancel, but since the metric coefficients are symmetric, the first term remains. 


A non-computational, more conceptual way of making the same argument is this:


*

*If all straight lines are geodesics, then every 2-dimensional plane is totally-geodesic. And hence the corresponding sectional curvature agrees with the induced Gaussian curvature of the plane. 

*It suffices then to show that every 2-dimensional plane is flat. This then implies by point 1 that the induced metric has constant (0) curvature, and hence is locally Euclidean. 

*To show that every 2-dimensional plane is flat, one notes that if straight lines are geodesics, then the "usual triangles" are the geodesic triangles and by the connection between angle defects with curvature this means that the planes are flat. 


Notice that in the direct computation case we are only interested in vectors of the form $v = \alpha e_i + \beta e_j$ in the span of two of the standard basis vectors. In other words, the argument there is also based on proving that sufficiently many of the 2-dimensional planes are flat. To do so we don't really need all straight lines being geodesic, but we only need all straight lines with velocity of the form $v = \alpha e_i + \beta e_j$ are geodesic. 
