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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.

John Rennie
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Luh
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    Would [math.se] 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
  • If you define the Riemann tensor in a nice way, the tensorial property is evident. All you then have to do is expand out the definition to confirm that it is equal to the standard definition. I would consult Dirac's book "General Relativity" (available as a PDF online) as it takes an economical approach. – user1379857 Apr 29 '18 at 20:46
  • @user1379857: I've looked though Diracs book - he does a bad job of explaining what tensors are. To be honest he just takes them for granted. – Mozibur Ullah Apr 29 '18 at 20:59
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    You could prove $[\nabla_a,,\nabla_b]V_c=R_{abcd}V^d$. – J.G. Apr 29 '18 at 21:27
  • I've added the add homework-and-exercises tag. In the future, please use this tag on questions of this type. –  Apr 29 '18 at 22:34
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    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

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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

Oktay Doğangün
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  • 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 02 '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 02 '18 at 18:57