This question is related to Reconciling different expressions for Riemann curvature tensor, but it's different since it asks for some notational clarification arising out of calculations I did. To not clutter the above link with multiple questions in one post, I'm asking it here.
I'm reading Sean Carroll's notes on GR (here) and eq. $(3.66)$ says $$R^{\rho}_{\ \ \sigma\mu\nu}V^{\sigma}=[\nabla_{\mu},\nabla_{\nu}]V^{\rho}+T_{\mu\nu}^{\ \ \ \ \lambda}\nabla_{\lambda}V^{\rho}\tag{3.66}.$$
To try and figure out how this came about, I started with the general coordinate-free definition of Riemann curvature tensor $$R(X,Y)V=\nabla_X\nabla_YV-\nabla_Y\nabla_XV-\nabla_{[X,Y]}.V$$
Now I express LHS in component form and for the RHS, I use the identity $$\nabla_X(\nabla_YV)=\nabla_{\nabla_XY}V+\nabla^2_{X,Y}V$$ (see Lee, Introduction to Riemannian Manifolds proposition $4.21$).
$$\begin{align*}
X^{\mu}Y^{\nu}V^{\sigma}R^{\rho}_{\ \ \sigma\mu\nu}\partial_{\rho} &= \nabla_{\nabla_XY}V+\nabla^2_{X,Y}V-\nabla_{\nabla_YX}V-\nabla^2_{Y,X}V-\nabla_{[X,Y]}V
\\ &=\nabla_{\nabla_XY}V-\nabla_{\nabla_YX}V-\nabla_{[X,Y]}V+\nabla^2_{X,Y}V-\nabla^2_{Y,X}V
\\ &=\nabla_{T(X,Y)}V+\nabla^2_{X,Y}V-\nabla^2_{Y,X}V
\\ &=X^{\mu}Y^{\nu}T_{\mu\nu}^{\ \ \ \ \lambda}\nabla_{\partial_{\lambda}}V+\nabla^2_{X,Y}V-\nabla^2_{Y,X}V
\\ &=X^{\mu}Y^{\nu}T_{\mu\nu}^{\ \ \ \ \lambda}\nabla_{\partial_{\lambda}}V+\nabla^2V(\text{d}x^{\rho},Y,X)\partial_{\rho}-\nabla^2V(\text{d}x^{\rho},X,Y)\partial_{\rho}
\\ &=X^{\mu}Y^{\nu}T_{\mu\nu}^{\ \ \ \ \lambda}(\nabla_{\partial_{\lambda}}V)^{\rho}\partial_{\rho}+X^{\mu}Y^{\nu}\nabla^2V(\text{d}x^{\rho},\partial_{\nu},\partial_{\mu})\partial_{\rho}-X^{\mu}Y^{\nu}\nabla^2V(\text{d}x^{\rho},\partial_{\mu},\partial_{\nu})\partial_{\rho}
\\ &=X^{\mu}Y^{\nu}\big(T_{\mu\nu}^{\ \ \ \ \lambda}(\nabla_{\partial_{\lambda}}V)^{\rho}+\nabla^2V(\text{d}x^{\rho},\partial_{\nu},\partial_{\mu})-\nabla^2V(\text{d}x^{\rho},\partial_{\mu},\partial_{\nu})\big)\partial_{\rho}.
\end{align*}$$
In the fifth equality I've used the definition of the second order covariant derivative: $$\nabla^2_{X,Y}V(\ldots)=\nabla^2V(\ldots,Y,X).$$ From the above, I get
$$X^{\mu}Y^{\nu}\big(V^{\sigma}R^{\rho}_{\ \ \sigma\mu\nu}\big)\partial_{\rho}=X^{\mu}Y^{\nu}\big(T_{\mu\nu}^{\ \ \ \ \lambda}(\nabla_{\partial_{\lambda}}V)^{\rho}+\nabla^2V(\text{d}x^{\rho},\partial_{\nu},\partial_{\mu})-\nabla^2V(\text{d}x^{\rho},\partial_{\mu},\partial_{\nu})\big)\partial_{\rho}$$
$$\implies V^{\sigma}R^{\rho}_{\ \ \sigma\mu\nu}=T_{\mu\nu}^{\ \ \ \ \lambda}(\nabla_{\partial_{\lambda}}V)^{\rho}+\nabla^2V(\text{d}x^{\rho},\partial_{\nu},\partial_{\mu})-\nabla^2V(\text{d}x^{\rho},\partial_{\mu},\partial_{\nu})$$
Now comparing this to the first equation (eq. $3.66$) in this post, I can make two conclusions:
- The notation $\nabla_{\lambda}V^{\rho}$ actually denotes the $\rho$ component of $\nabla_{\lambda}V$, i.e. $\nabla_{\lambda}V^{\rho}\equiv(\nabla_{\partial_{\lambda}}V)^{\rho}$
- The notation $\nabla_{\mu}\nabla_{\nu}V^{\rho}$ denotes the component of the tensor $\nabla\nabla V$, i.e., $\nabla_{\mu}\nabla_{\nu}V^{\rho}\equiv\nabla\nabla V(\text{d}x^{\rho},\partial_{\nu},\partial_{\mu})$
Even if $X,Y$ are taken to be coordinate basis vector fields (say $X=\partial_{\mu}$ and $Y=\partial_{\nu}$), sure the term involving the commutator involved in Riemann tensor's definition will vanish, but anyways that commutator term gets integrated into the expression for torsion tensor (see third equality in the calculation). So I think this general approach covers the case where $X,Y$ are basis vector fields (please correct me if I'm wrong).
My question is, are the above calculation and conclusions correct?
Best Answer
Indeed, by starting with the definition of the Riemann curvature tensor, then using the quoted identity for second covariant derivatives, and then using the definition of torsion we arrive at \begin{align} R(X,Y)V&=(\nabla\nabla V)(Y,X)-(\nabla\nabla V)(X,Y)+\nabla_{T(X,Y)}V. \end{align} From here, you just plug in $X=\frac{\partial}{\partial x^{\mu}}, Y=\frac{\partial}{\partial x^{\nu}}$, write $V=V^{\lambda}\frac{\partial}{\partial x^{\lambda}}$ (on the LHS), and apply $dx^{\rho}$ to both sides (to extract the $\rho^{th}$ component). This gives the identity that Carrol quotes, after you understand the notation which you correctly did:
Annoying notation with the bracketing ambiguous mess for beginners, but it is what it is.