Reconciling different expressions for Riemann curvature tensor

Question

[Note: I think this question is more suited to Physics SE rather than Math, since it refers to Carroll's notes and some equations might have inherent Physics-related assumptions

EDIT: This question has also been crossposted to Math SE. Link in comment.]

I'm reading Carroll's GR notes and I'm having trouble deciphering a particular expression for the Riemann curvature tensor. The coordinate-free definition is (eq. 3.71 in Carroll's notes): $$R(X,Y)Z=\nabla_X\nabla_YZ-\nabla_Y\nabla_XZ-\nabla_{[X,Y]}Z\tag{3.71}$$

An index-based expression is also given (eq. 3.66 in Carroll's notes): $$R^{\rho}_{\ \ \sigma\mu\nu}V^{\sigma}=[\nabla_{\mu},\nabla_{\nu}]V^{\rho}+T_{\mu\nu}^{\ \ \ \ \lambda}\nabla_{\lambda}V^{\rho}\tag{3.66}$$

I want to reconcile these two equations. A bit of background on the first equation, based on my reading of "Semi-Riemannian Geometry: The Mathematical Language of General Relativity" by Newman:

$R$ is a map from $\mathfrak{X}(M)^3$ to $\mathfrak{X}(M)$. The book doesn't really treat $R(X,Y)$ as a separate object (as far as I've read). (So far till the point I've read the book) $R$ is just treated as a map of 3 vector fields and with the weird notation $R(X,Y)Z$ instead of $R(X,Y,Z)$. Then there is a theorem showing that $R$ is multilinear in all its arguments and anti-symmetric in its first 2 arguments. And then the following assertion:

Let $(M,\nabla)$ be a smooth manifold with a connection, and let $(U,(x^i))$ be a chart on $M$. Then in local coordinates, $R(\partial/\partial x^{\mu},\partial/\partial x^{\nu})\partial/\partial x^{\sigma}$ can be expressed as: $$R\bigg(\frac{\partial}{\partial x^{\mu}},\frac{\partial}{\partial x^{\nu}}\bigg)\frac{\partial}{\partial x^{\sigma}}=R_{\ \ \sigma\mu\nu}^{\rho}\frac{\partial}{\partial x^{\rho}}$$

Based on the above, if $V\in\mathfrak{X}(M)$, then $$R\bigg(\frac{\partial}{\partial x^{\mu}},\frac{\partial}{\partial x^{\nu}}\bigg)V=R\bigg(\frac{\partial}{\partial x^{\mu}},\frac{\partial}{\partial x^{\nu}}\bigg)V^{\sigma}\frac{\partial}{\partial x^{\sigma}}=V^{\sigma}R\bigg(\frac{\partial}{\partial x^{\mu}},\frac{\partial}{\partial x^{\nu}}\bigg)\frac{\partial}{\partial x^{\sigma}}=V^{\sigma}R_{\ \ \sigma\mu\nu}^{\eta}\frac{\partial}{\partial x^{\eta}}$$

Now if I act this on the $x^{\rho}$ coordinate function, I get $$\bigg(R\bigg(\frac{\partial}{\partial x^{\mu}},\frac{\partial}{\partial x^{\nu}}\bigg)V\bigg)(x^{\rho})=V^{\sigma}R_{\ \ \sigma\mu\nu}^{\eta}\frac{\partial}{\partial x^{\eta}}(x^{\rho})=V^{\sigma}R_{\ \ \sigma\mu\nu}^{\eta}\frac{\partial x^{\rho}}{\partial x^{\eta}}=R_{\ \ \sigma\mu\nu}^{\rho}V^{\sigma}\tag{1}$$

So far, so good. Now in the first equation (eq. 3.71), I can replace $X,Y$ by fields $\partial_{\mu},\partial_{\nu}$ respectively and $Z$ by $V$, then I can get the local coordinates for both sides by acting them on $x^{\rho}$, i.e. (AFAIK to make notation shorter, expressions like $\nabla_{\partial_{\mu}}$ are written as $\nabla_{\mu}$): $$(R(\partial_{\mu},\partial_{\nu})V)(x^{\rho})=([\nabla_{\mu},\nabla_{\nu}]V)(x^{\rho})-(\nabla_{[\partial_{\mu},\partial_{\nu}]}V)(x^{\rho})=([\nabla_{\mu},\nabla_{\nu}]V)(x^{\rho})\tag{2}$$

Comparing (1) and (2), I get $$R_{\ \ \sigma\mu\nu}^{\rho}V^{\sigma}=([\nabla_{\mu},\nabla_{\nu}]V)(x^{\rho})\tag{3}$$

The above equation completely follows from (3.71). Now to reconcile (3.71) with (3.66), I have to show (comparing RHS of (3) and (3.66)): $$([\nabla_{\mu},\nabla_{\nu}]V)(x^{\rho})=[\nabla_{\mu},\nabla_{\nu}]V^{\rho}+T_{\mu\nu}^{\ \ \ \ \lambda}\nabla_{\lambda}V^{\rho}\tag{4}$$

This is where I'm stuck. How do I go about proving eq. (4)? On the other hand, are eq.s (3.71) and (3.66) even meant to be reconciled (are they fundamentally the same equation?)

My assumption is that (3.66) is the index-based version of the coordinate-free equation (3.71). I'd appreciate any help or corrections!

Answer 1

Working in the abstract index notation, $R(X,Y)V = [\nabla_X, \nabla_Y] V - \nabla_{[X,Y]}V$ boils down to \begin{align} (R(X,Y)V)^A &= X^C\nabla_{C} (Y^D\nabla_{D}V^A) - Y^C\nabla_{C} (X^D\nabla_{D}V^A) - [X,Y]^C{\kern0.1em} \nabla_C V^A \,,\\ &= \big({ X^C \nabla_{C} Y^D {-}\, Y^C \nabla_{C} X^D }\big)(\nabla_{D}V^A) + X^C Y^D ([\nabla_{C},\nabla_{D}]V^A) - [X,Y]^C{\kern0.1em} \nabla_C V^A \,,\\ &= (T(X,Y))^D {\kern0.1em} (\nabla_{D}V^A) + X^C Y^D ([\nabla_{C},\nabla_{D}]V^A) \,. \end{align} To recapitulate, \begin{align} R^A{}_{BCD} V^B X^C Y^D &= ([\nabla_{C},\nabla_{D}]V^A) {\kern0.1em} X^C {\kern0.05em} Y^D + (\nabla_{D}V^A) {\kern0.1em} (T(X,Y))^D \,. \end{align} If we take $X^A = \partial^A{}_\rho$ and $Y^A = \partial^A{}_\sigma$, we find \begin{align} R^\mu{}_{\nu\rho\sigma} V^\nu &= dx^\mu{}_A{\kern0.1em} R^A{}_{BCD} V^B \partial^C{}_\rho{\kern0.05em} \partial^D{}_\sigma \,,\\ &= dx^\mu{}_A{\kern0.05em} ([\nabla_{C},\nabla_{D}]V^A){\kern0.1em} \partial^C{}_\rho{\kern0.05em} \partial^D{}_\sigma + dx^\mu{}_A{\kern0.05em} (\nabla_{D}V^A){\kern0.1em} (T(\partial_\rho,\partial_\sigma))^D \,,\\ &= ([\nabla_\rho,\nabla_\sigma] V)^\mu + (\nabla_\lambda V)^\mu{\kern0.1em} T^\lambda{}_{\rho\sigma} \,, \end{align} where $\nabla_\mu := \nabla_{\partial_\mu}$. Note that by "$\nabla_X V^\mu$" it really means $dx^\mu{}_A (\nabla_X V^A)$; that's what I mean by $(\nabla_\lambda V)^\mu$ and $([\nabla_\rho,\nabla_\sigma] V)^\mu$.

I would say this identity is quite a tricky one. I also got really confused when I first encountered it. I believe abstract index notation helps. Where did the torsion come from? Roughly speaking, we can say $\nabla_X$ also differentiates $Y$ in $\nabla_X(\nabla_YV)$. The torsion term came from that contribution.

Answer 2

I prefer the notation $\nabla_X$ to the apparently more explicit $X^\mu\nabla_\mu$ because in the presence of torsion there is a dangerous ambiguity lurking in the notation $\nabla_\mu$. To see this recall that the curvature definition $$ [\nabla_X,\nabla_Y]Z- \nabla_{[X,Y]}Z= {\bf R}(X,Y) Z $$ holds even when the connection has torsion. Now take $X\to \partial_\mu$ and $Y\to \partial_\nu$ {\it i.e./} $X^\lambda =\delta^\lambda_\mu$, $Y^\lambda =\delta^\lambda_\nu$ so that we are tempted to write $$ \nabla_{\partial_\mu}\stackrel{}{=}\delta^\lambda_\mu \nabla_\lambda\stackrel{?}{=} \nabla_\mu, \quad \nabla_{\partial_\nu}=\delta^\lambda_\nu \nabla_\lambda \stackrel{?}{=}\nabla_\nu. $$ Now, remembering that $[\partial_\mu,\partial_\nu]=0$, our curvature definiition gives us
$$ [\nabla_\mu,\nabla_\nu] Z^\lambda \stackrel{?}{=} Z^\alpha{R^\lambda}_{\alpha \mu\nu}. $$ If, however, we include a Christoffel symbol ${\Gamma^\alpha}_{\nu\mu}$ when $\nabla_\mu$ acts from the left on the index $\nu$ on $\nabla_\nu$ and similarly a ${\Gamma^\alpha}_{\mu\nu}$ when $\nabla_\nu$ is on the left and acting on the index $\mu$, we end up with $$ [\nabla_\mu,\nabla_\nu] Z^\lambda \stackrel{?}{=} Z^\alpha{R^\lambda}_{\alpha \mu\nu}- {T^\sigma}_{\mu\nu} \nabla_\sigma Z^\lambda. $$ Which is equation is correct? With the last version, remembering that
$$ \nabla_X Y-\nabla_Y X= [X,Y] + T(X,Y), $$ we can use Liebnitz rule to find
$$ [X^\mu\nabla_\mu,Y^\nu \nabla_\nu]Z^\lambda = (\nabla_X Y-\nabla_YX)^\sigma \nabla_\sigma Z^\lambda +X^\mu Y^\nu (Z^\alpha{R^\lambda}_{\alpha \mu\nu}- {T^\sigma}_{\mu\nu} \nabla_\sigma Z^\lambda)\nonumber\\ = Z^\alpha{R^\lambda}_{\alpha \mu\nu}X^\mu Y^\nu + [X,Y]^\sigma \nabla_\sigma Z^\lambda, \nonumber $$ which reproduces the curvature definition for general vector fields $X$, $Y$. Both formul\ae\ are correct in their own ways. The difference lies in whether we regard the $\mu$ on $\nabla_\mu$ as a shorthand label specifying the basis vector $ \partial_\mu$, or as an index on the numerical component of a covector.