Let us consider the Riemann-Christoffel tensor $R^k_{mij}$ defined as
$$ R^k_{mij}=\partial_i\Gamma^k_{jm}-\partial_j\Gamma^k_{im}+\Gamma^k_{in}\Gamma^n_{jm}-\Gamma^k_{jn}\Gamma^n_{im}, $$
where the operator $\partial_i = \frac{\partial (\cdot)}{\partial Z^i}$; $Z^i$ designates the coordinates.
EDITS:
- My idea is to replace all Christoffel symbols because I want the proof to be independent of them up to some point. Indeed, if we take polar or cylindrical coordinates, which can be considered "flat", some Christoffel symbols subsist and thus the vanishing of the Riemann-Christoffel symbol is not intuitive.
- I am assuming a very informal definition of Euclidean space which can be regarded as a space which can accomodate straight lines.
Let us now replace the Christoffel symbols in the first two terms using this definition:
$$ \Gamma^k_{ij}=\mathbf{Z}^k\cdot \partial_j\mathbf{Z}_i $$
where $\mathbf{Z}_i$ and $\mathbf{Z}^i$ represent the covariant and contravariant bases respectively. Thus,
$$ R^k_{mij}= \partial_i \mathbf{Z}^k \cdot \partial_m \mathbf{Z}_j + \mathbf{Z}^k \cdot \frac{\partial^2 \mathbf{Z}_j}{\partial Z^i\partial Z^m} - \partial_j \mathbf{Z}^k \cdot \partial_m \mathbf{Z}_i - \mathbf{Z}^k \cdot \frac{\partial^2 \mathbf{Z}_i}{\partial Z^j\partial Z^m} + \Gamma^k_{in}\Gamma^n_{jm}-\Gamma^k_{jn}\Gamma^n_{im}. $$
Now we use the definition of the derivative of the contravariant basis with respect to the coordinates
$$ \partial_j\mathbf{Z}^k = -\Gamma^k_{ij}\mathbf{Z}^i, $$
on the first and third terms of $R^k_{mij}$. Then,
$$ R^k_{mij}= \Gamma^k_{ni}\Gamma^n_{jm} + \mathbf{Z}^k \cdot \frac{\partial^2 \mathbf{Z}_j}{\partial Z^i\partial Z^m} - \Gamma^k_{jn}\Gamma^n_{im} - \mathbf{Z}^k \cdot \frac{\partial^2 \mathbf{Z}_i}{\partial Z^j\partial Z^m} + \Gamma^k_{in}\Gamma^n_{jm}-\Gamma^k_{jn}\Gamma^n_{im}. $$
Now, the Riemann-Christoffel tensor reduces to:
$$ R^k_{mij}= \mathbf{Z}^k \cdot \left( \frac{\partial^2 \mathbf{Z}_j}{\partial Z^i\partial Z^m} - \frac{\partial^2 \mathbf{Z}_i}{\partial Z^j\partial Z^m} \right). $$
Now, I suppose that the term between parenthesis vanishes in a Euclidean space, but why exactly? Or is it rather that the contravariant basis is perpendicular to the term between parenthesis above? Is this the right way to go in order to prove that the Riemann-Christoffel tensor vanishes in a Euclidean space?