# How do I prove that the Riemann curvature tensor is indeed a tensor? [duplicate]

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.

• Would Mathematics be a better home for this question? – Qmechanic Apr 29 '18 at 19:48
• In a word - differential geometry. – Mozibur Ullah Apr 29 '18 at 20:11
• Maybe, the thing is I am studying this subject in a general relativity course, in physics department, so in my head I should ask here. – Luh Apr 29 '18 at 20:30
• You could prove $[\nabla_a,\,\nabla_b]V_c=R_{abcd}V^d$. – J.G. Apr 29 '18 at 21:27
• Physics.SE is the proper home for this question based on the form of the question -- a mathematician would prefer to say that tensors are multilinear maps and when you compute $[\nabla_a,\nabla_b](\alpha V^c)$ the "cross terms" on the scalar field $\alpha$ cancel leaving $\alpha [\nabla_a, \nabla_b] V^c$ and proving that this is a multilinear map of a $[1,0]$-tensor to a $[1,2]$ tensor, therefore it must be a $[1,3]$-tensor $R^a_{bcd}$. That one is asking about how it behaves under coordinate transforms indicates one is instead taking the physicist's approach to such things. – CR Drost Apr 29 '18 at 23:26

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
• I tried to prove using the definition of a tensor and how It transforms but it went bad. I just proved that $[\nabla_{a},\nabla_{ b}] V_{c} = R_{abcd} V^{d}$ but I don't quite understand why does it prove that R is a tensor – Luh May 2 '18 at 14:15
• That proof can provide that if you know the left-hand side is a tensor, then you would be sure the right-hand side is tensor too. In order to do that, you need to make sure that the covariant derivative operator $\nabla_a$ is a tensor (which is the reason that we use it). For the other way that I mentioned in my answer, can you show what went wrong? What is the definition that you used? – Oktay Doğangün May 2 '18 at 18:57