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Tell me if I'm wrong but the following expression from Do Carmo's Riemannian geometry (p.198) can't be correct right?

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(here $[\partial_s, \partial_t] = 0$ and $f$ is a variation which you can think of as a map from two dimensional manifold with coordinates $s,t$ to the manifold of interest $M$).

It is used in calculating the second variation of the energy for curves on Riemannian manifolds. From what I know it should be $$ R\left( \frac{\partial f}{\partial s}, \frac{\partial f}{\partial t} \right) \frac{\partial f}{\partial t} = \nabla_{\partial_s} \nabla_{\partial_t} \frac{\partial f}{\partial t} - \nabla_{\partial_t} \nabla_{\partial_s} \frac{\partial f}{\partial t} - \nabla_{[\partial_s, \partial_t]}\frac{\partial f}{\partial t}$$ where the last term vanishes, giving us $$ \nabla_{\partial_s} \nabla_{\partial_t} \frac{\partial f}{\partial t} = \nabla_{\partial_t} \nabla_{\partial_s} \frac{\partial f}{\partial t} + R\left( \frac{\partial f}{\partial s}, \frac{\partial f}{\partial t} \right) \frac{\partial f}{\partial t}$$ where clearly the derivatives w.r.t $s$ and $t$ as arguments in the riemann tensor switched places in comparison to Do Carmo's expression.

The reason I put this on here is because I have seen this in other articles as well and I am wondering if I am making some dum mistake I don't see. Can anyone point out this mistake or verify this is indeed wrong in Do Carmo's book.

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    What definition of the Riemann curvature tensor is do Carmo using? What definition is the other articles using? Every author seems to use a different convention in my experience, and will lead to these kinds of troubles. – Matt Jun 13 '19 at 21:30
  • Well that seems to solve the problem! Indeed they use another definition. Didn't think there were different ways of defining the curvature tensor. – Tychonoff3000 Jun 13 '19 at 23:08
  • As an addition comment: Regardless of the convention, these situations rarely happen with the Ricci tensor due to the symmetries of the Riemannian curvature tensor (maybe just a sign change). Also, these symmetries are why there are so many different conventions and why they’re all (essentially) equivalent. – Matt Jun 15 '19 at 00:16

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