• Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.


PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback


(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,047 questions , 2,200 unanswered
5,345 answers , 22,709 comments
1,470 users with positive rep
816 active unimported users
More ...

  Notational meaning of $\nabla_{\lambda}V^{\rho}$ and $\nabla_{\mu}\nabla_{\nu}V^{\rho}$

+ 2 like - 0 dislike

(I'm new here and I hope it's fine to ask a GR/diff geometry question since that is a grad level discussion)

I'm reading Sean Carroll's notes on GR and eq. $(3.66)$ says $$R^{\rho}_{\ \ \sigma\mu\nu}V^{\sigma}=[\nabla_{\mu},\nabla_{\nu}]V^{\rho}+T_{\mu\nu}^{\ \ \ \ \lambda}\nabla_{\lambda}V^{\rho}$$
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 $4.21$)
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}
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:

1. The notation $\nabla_{\lambda}V^{\rho}$ actually denotes the $\rho$ component of $\nabla_{\lambda}V$, i.e. $\nabla_{\lambda}V^{\rho}\equiv(\nabla_{\partial_{\lambda}}V)^{\rho}$
2. 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?

asked Feb 21, 2023 in Mathematics by Shirish (15 points) [ revision history ]

$V^\rho$ is a vector field, thus $\nabla_\lambda V^\rho$ is not the \$rho$ component of $\nabla_\lambda V$. 

$\nabla_\lambda$ = $dx$/$d \lambda$

(I could be mistaking though)

Surely the first halves of 1. and 2. are correct, but the notation in the second halves is somewhat weird.

I didn't check the calculation itself.

@Shirish: I did not check the calculation either, but have you made sure that the notation in your sources is consistent? Have you been careful not to mix notations/conventions in your calculation? For example, $\nabla_\lambda V^\rho=(\nabla_{\partial_\lambda} V)^\rho$ can have two interpretations:  

1) component $\rho$ of the vector field $\nabla_{\partial_\lambda} V$, a scalar (as the components refer to a fixed basis)

2) "abstract index notation" of the tensor $\nabla V$ 

There is a third interpretation: 

$\nabla_\lambda V^\rho=\nabla_\lambda (V^\rho)=\partial_\lambda V^\rho$

That is, the covariant derivative with respect to $\partial_\lambda$ of the component $\rho$ of the vector field $V$, the components referring to a fixed basis as in 1) above. In this case, $V^\rho$ is a scalar.

A combination $\nabla_\lambda V^\rho$ in this sense may easily appear in calculations with iterated covariant derivatives. 

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
Please complete the anti-spam verification

user contributions licensed under cc by-sa 3.0 with attribution required

Your rights