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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?

Meclassic
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    In coordinates for which the metric tensor is the identity matrix, the Christoffel symbols vanish since they can be expressed in terms of partial derivatives of the metric tensor. – Mason Jul 09 '22 at 22:52
  • Are you trying to show that zero curvature implies Euclidean space? If so, you have show that there exists a change of coordinates such that, using the new coordinates, the terms inside the parentheses vanish. – Deane Jul 09 '22 at 23:54
  • @Mason To me, it makes sense that the Christoffel symbols vanish for affine coordinates for example because the metric tensor has only constant coefficients. The Riemann-Christoffel tensor then vanishes as well. However, does Euclidean space automatically implies that we are working with affine coordinates? – Meclassic Jul 10 '22 at 01:27
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    @Deane I am just trying to come up with a direct proof that the Riemann-Christoffel tensor vanishes in Euclidean spaces, and make sense of it. – Meclassic Jul 10 '22 at 01:44
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    @Meclassic You are free to choose coordinates. In a Euclidean space, you should use the identity map as coordinates. – Mason Jul 10 '22 at 03:02
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    What is the definition of Euclidean space you want to use? Existence of orthonormal coordinates is the standard one. – Deane Jul 10 '22 at 03:56
  • There are many non-flat spaces that "accommodate straight lines." (Some, like the saddle surface, have two lines passing through every point.) Please make explicit how you're using this in your computations. – Ted Shifrin Jul 10 '22 at 19:26
  • @Deane I am assuming that the Euclidean space can accomodate straight-lines (please see edit). – Meclassic Jul 10 '22 at 19:27
  • @Mason Cylindrical or polar coordinates can be regarded as Euclidean, however, note that the Christoffel symbols do not vanish for these examples (their metric is not the identity). – Meclassic Jul 10 '22 at 19:28
  • Please define "can accommodate straight lines." To elaborate slightly on my previous comment, there are various NON-flat surfaces parametrized by $(Z^1,Z^2)$ so that the $Z^1$-curves and $Z^2$-curves are lines. – Ted Shifrin Jul 10 '22 at 19:29
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    "However, does Euclidean space automatically implies that we are working with affine coordinates?" No but it implies that affine coordinates exist. This can be proved starting with the axioms of Euclidean geometry. That and the fact that the curvature tensor is well defined independent coordinates implies that the right side of your last equation vanishes. – Deane Jul 10 '22 at 20:40
  • The informal definition of a Euclidean space given in the edit is quite meaningless. You cannot prove anything with such definition. – Moishe Kohan Jul 10 '22 at 21:04
  • Related. – K.defaoite Jul 11 '22 at 02:02

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