Q:

How do I prove that the Riemann curvature tensor is indeed a tensor?

I know that to prove that something is a tensor I have to show that this thing transforms like a tensor, i.e., like below:
$$ R^{'\alpha}{}_{\gamma\phi\lambda} = \partial_{\beta}x^{\alpha'}\partial_{\sigma'}x^{\gamma}\partial_{\mu'}x^{\phi}\partial_{\nu'}x^{\lambda}R^{\beta}{}_{\gamma\phi\lambda}.$$
But I don't know actually how to start the proof. Do I need to apply this tensor to a vector or something like that? Thanks in advance.

A:

It could be a question in mathematics, but it is also a valid question in gravitation. Mathematicians would probably answer this question in a more formal way such that it could be overcomplicated for a physicist who just would like to, in a sense, demonstrate rather than prove.
You need to start from the definition of Riemann curvature tensor, namely,
$$
R_{\alpha\beta\;\mu}^{\;\;\;\nu} = \partial_\alpha \Gamma_{\beta\;\nu}^{\;\,\mu} - \partial_\beta \Gamma_{\alpha\;\nu}^{\;\,\mu} + 
\Gamma_{\alpha\;\kappa}^{\;\,\mu} \Gamma_{\beta\;\nu}^{\;\,\kappa} 
- \Gamma_{\beta\;\kappa}^{\;\,\mu} \Gamma_{\alpha\;\nu}^{\;\,\kappa}
$$
and then transform the connection $\Gamma$ under a general coordinate transformation. As you might know, the connection is not a tensor and it transforms non-homogeneously. However, the additive parts of the transformed connections would cancel out and make the Riemann curvature transform as tensor

