Reconciling 2 expressions for Riemann curvature tensor

[Shirish]

The whole calculation is wrong because it's based on a wrong assumption: that you can obtain Carroll's equation 3.66 by applying the Riemann curvature tensor to general (non-commuting) vector fields. You can't. You obtain Carroll's Equation 3.66 by applying the Riemann curvature tensor to coordinate basis vector fields only.

(Hope I don't sound argumentative, it's just that I always try to get full clarity) So a couple of points:

1. I was asking about calculation on a standalone basis (the calculation itself doesn't refer to Carroll's equation) - look at it as a generic differential geometry calculation. I'm not sure if (and where) I've gone wrong in that
2. The identity is a result on second order covariant derivatives and the definition is that of a total covariant derivative. Both can be found in Lee's Introduction to Riemannian Manifolds (resp. Proposition 4.21 and Proposition 4.17)

 

[PeterDonis]

I was asking about calculation on a standalone basis

Then I don't understand what you are trying to show with the calculation. I also don't understand what in Carroll you are trying to compare it to, if it isn't Equation 3.66.

(I note, btw, that while you use general vectors $X$, $Y$, and $Z$, you use the coordinate basis vector $\partial_\rho$ instead of a general covector. You should use a general covector.)

 

[Shirish]

Then I don't understand what you are trying to show with the calculation.

So the calculation shows that $$X^{\mu}Y^{\nu}Z^{\sigma}R^{\rho}_{\ \ \sigma\mu\nu}\partial_{\rho}=X^{\mu}Y^{\nu}\big(T_{\mu\nu}^{\ \ \ \ \lambda}(\nabla_{\partial_{\lambda}}Z)^{\rho}+\nabla^2Z(\text{d}x^{\rho},\partial_{\nu},\partial_{\mu})-\nabla^2Z(\text{d}x^{\rho},\partial_{\mu},\partial_{\nu})\big)\partial_{\rho}$$
$$\implies X^{\mu}Y^{\nu}\big(Z^{\sigma}R^{\rho}_{\ \ \sigma\mu\nu}\big)\partial_{\rho}=X^{\mu}Y^{\nu}\big(T_{\mu\nu}^{\ \ \ \ \lambda}(\nabla_{\partial_{\lambda}}Z)^{\rho}+\nabla^2Z(\text{d}x^{\rho},\partial_{\nu},\partial_{\mu})-\nabla^2Z(\text{d}x^{\rho},\partial_{\mu},\partial_{\nu})\big)\partial_{\rho}$$
$$\implies Z^{\sigma}R^{\rho}_{\ \ \sigma\mu\nu}=T_{\mu\nu}^{\ \ \ \ \lambda}(\nabla_{\partial_{\lambda}}Z)^{\rho}+\nabla^2Z(\text{d}x^{\rho},\partial_{\nu},\partial_{\mu})-\nabla^2Z(\text{d}x^{\rho},\partial_{\mu},\partial_{\nu})$$ and this is what I was trying to compare to Carroll's eq. 3.66

 

[PeterDonis]

So the calculation shows that $$X^{\mu}Y^{\nu}Z^{\sigma}R^{\rho}_{\ \ \sigma\mu\nu}\partial_{\rho}=X^{\mu}Y^{\nu}\big(T_{\mu\nu}^{\ \ \ \ \lambda}(\nabla_{\partial_{\lambda}}Z)^{\rho}+\nabla^2Z(\text{d}x^{\rho},\partial_{\nu},\partial_{\mu})-\nabla^2Z(\text{d}x^{\rho},\partial_{\mu},\partial_{\nu})\big)\partial_{\rho}$$
$$\implies X^{\mu}Y^{\nu}Z^{\sigma}\big(R^{\rho}_{\ \ \sigma\mu\nu}\big)\partial_{\rho}=X^{\mu}Y^{\nu}\big(T_{\mu\nu}^{\ \ \ \ \lambda}(\nabla_{\partial_{\lambda}}Z)^{\rho}+\nabla^2Z(\text{d}x^{\rho},\partial_{\nu},\partial_{\mu})-\nabla^2Z(\text{d}x^{\rho},\partial_{\mu},\partial_{\nu})\big)\partial_{\rho}$$
$$\implies R^{\rho}_{\ \ \sigma\mu\nu}=T_{\mu\nu}^{\ \ \ \ \lambda}(\nabla_{\partial_{\lambda}}Z)^{\rho}+\nabla^2Z(\text{d}x^{\rho},\partial_{\nu},\partial_{\mu})-\nabla^2Z(\text{d}x^{\rho},\partial_{\mu},\partial_{\nu})$$ and this is what I was trying to compare to Carroll's eq. 3.66

You still do not get it. It makes no sense to compare what you calculated with Carroll's Equation 3.66. I have already explained why several times. There is no point in explaining it again. You are evidently on a fool's errand and you are not listening when I try to explain that.

 

[PeterDonis]

The OP question is based on a misunderstanding, which has been repeatedly explained. Enough is enough. Thread closed.