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I wanted to see if I can use a simple idea and derive the Riemann curvature tensor by only using the covariant derivative, because I somehow got the impression that the derivations I read in several books looked overcomplicated. The idea is to take a vector $V_1$ on a loop built of $4$ displacements calculated via the covariant derivative, in the following way:

$$ V_2^n = V_1^n + (\nabla_a{V^{n}_{1}})\Delta{A^a} \\ V_3^n = V_2^n + (\nabla_b{V_2^n})\Delta{B^b} \\ V_4^n = V_3^n - (\nabla_a{V_3^n})\Delta{A^a} \\ V_5^n = V_4^n - (\nabla_b{V_4^n})\Delta{B^a} $$

Where, perhaps not very rigorously, $\Delta{A}^a$ and $\Delta{B}^b$ are small displacements in the chart coordinates derived by the operations $\nabla_a$ and $\nabla_b$. Because of the plus and minus signs, we supposedly end up at the origin point of $V_1$ and then calculate $V_5 - V_1$ to find the difference in the vector induced by this journey.

So, expressing $V_5^n - V_1^n$ only in terms of $V_1^n$ is clearly a lengthy calculation, and yet manageable:

$$ \begin{align*} V_{5}^n - V_1^n = \nabla_a{V_1^n}\Delta{A^a}+V_{1}^n+\nabla_b{(V_{1}^n+ \nabla_a{V_{1}^n\Delta{A^a})}}\Delta{B^b}-\\\Delta{A^a} \nabla_a{(\nabla_a{V_1^n}\Delta{A^a}+V_1^n+ \nabla_b{(V_1^n+\nabla_a{V_1^n}\Delta{A^a})}\Delta{B^b})}-\\\Delta{B^b} \nabla_b{(\nabla_a{V_1^n}\Delta{A^a}+V_1^n+\nabla_b{(V_1^n +\nabla_a{V_1^n\Delta{A^a})}}}\Delta{B^b} -\\ \Delta{A^a}\nabla_a{(\nabla_a{V_1^n}\Delta{A^a}+V_1^n+\nabla_b{(V_1^n+ \nabla_a{V_1^n}\Delta{A^a})\Delta{B^b})}}) - V_1^n \end{align*} $$

Looks formidable, but several terms cancel, and after also neglecting the terms which are of second order in $\Delta{A^a}$ and $\Delta{B^b}$ I got:

$$ V_5^n - V_1^n = (\nabla_{b}\nabla_{a}V_1^n-\nabla_a\nabla_bV_1^n)\Delta{A^a}\Delta{B^b}$$

Which I know is the right result to expect (at least up to sign?), namely we got the commutator $[\nabla_b,\nabla_a]$ acting on $V_1^n$ which is basically where the definition of the Riemann curvature tensor comes from (at least up to torsion).

So my question is mainly whether this procedure is correct, and can be used for deriving the Riemann curvature tensor, or whether there is something to be corrected here.

冥王 Hades
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Amit
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    I'm not going to look through the details, because I did it once, and I'm way too tired to transcribe them into your notation and also type them up. But yes, once you traverse the loop, you'll see that the result after parallel-transporting it around (covariant derivatives are just first order changes, i.e first derivatives, of the parallel-transport) is the identity plus a second order correction involving the curvature tensor. See How to understand $R^{\rho}_{\sigma\mu\nu}$ for constructing geometries? for the statement and pictures. – peek-a-boo Mar 01 '23 at 21:19
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    Indeed, when you traverse the loop, there will be lots of cancellation (the first order terms will vanish, and some of the second order terms $\epsilon^2,\delta^2$ also vanish, and only the $\epsilon\delta$ terms remain) I know some older physics texts define the curvature by this prescription. btw, you can find a relatively concise statement and proof in Lee's Riemannian geometry book. – peek-a-boo Mar 01 '23 at 21:21
  • Thank you @peek-a-boo , that's very helpful – Amit Mar 01 '23 at 21:24
  • This might be helpful: https://math.stackexchange.com/a/4583484/10584 – Deane Mar 01 '23 at 23:08

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