Riemann Curvature Tensor and Covariant Derivative

This post is is aimed at all those who studied general relativity on the book “A First Course in GENERAL RELATIVY” by Bernard Schutz, and found themselves in trouble reading the chapter 6.5 about the curvature tensor where the author explains the relation between the covariant derivative and the Riemann tensor itself.

The crucial question is the first mathematical relationship: \nabla_\alpha \nabla_ \beta V^ \mu = ({V^ \mu}_{;\beta})_{, \alpha} + {\Gamma^ \mu }_ {\sigma \alpha} {V^ \sigma}_{;\beta } - { \Gamma ^ \sigma }_{ \beta \alpha } {V^ \mu}_{; \sigma }

How did he get it? I spent 2 hours getting out of it! All subsequent steps are based on this first relationship, so it is crucial to understand it. Obviously some readers might accept this relationship and continue anyway, but I believe there is nothing more beautiful than grasping every single passage given the beauty of the subject.

And now let’s start proving this relationship.

First of all, the correct interpretation of \nabla_\alpha \nabla_ \beta V^ \mu is (\nabla_\alpha \nabla_ \beta V)^ \mu, otherwise we would have the simple equivalence: \nabla_\alpha \nabla_ \beta V^ \mu = \frac{ \partial^2V^ \mu }{ \partial x^ \alpha \partial x^ \beta }, which is not so interesting. So we have to prove that:

(\nabla_\alpha \nabla_ \beta V)^\mu = ({V^ \mu}_{;\beta})_{, \alpha} + {\Gamma^ \mu }_ {\sigma \alpha} {V^ \sigma}_{;\beta } - { \Gamma ^ \sigma }_{ \beta \alpha } {V^ \mu}_{; \sigma }

Now consider the tensor field of type (1\;1): T(\tilde{ \omega }, \vec{X}) :=  \tilde{ \omega } (\nabla_{ \vec{X} }  \vec{V}) where:

  • \tilde{ \omega } is a covector field
  • \vec{X} is a vector field
  • \vec{V} is a given vector field

For a tensor of this type the following relation holds (which can be found using Leibniz rules):

{T^ \mu}_{ \nu; \beta } = {T^ \mu}_{ \nu, \beta } + {T^ \alpha}_{ \nu}\;{\Gamma^ \mu}_{ \alpha \beta } -  {T^ \mu}_{ \alpha} \;{\Gamma^ \alpha}_{ \nu \beta }

where {T^ \mu}_{ \nu; \beta } = {(\nabla_{\beta} T)^{\mu}}_{\nu} and {T^ \mu}_{ \nu, \beta } =  \nabla_{\beta} ({T^{\mu}}_{\nu}).

In the case of the tensor just defined, the relation seen becomes:

{({\nabla_\beta T})^\mu}_\nu = {V^\mu}_{;\nu,\beta} +  {V^\alpha}_{;\nu}\; {\Gamma^ \mu}_{ \alpha \beta } -   {V^\mu}_{;\alpha}\; {\Gamma^ \alpha}_{ \nu \beta }

But {({\nabla_\beta T})^\mu}_\nu are also the components of the tensor U, of type (1\;2):

U( \tilde{ \omega }, \vec{X}, \vec{Y}) := (\nabla_Y T)( \tilde{ \omega }, \vec{X} )

In fact {U^\mu}_{\nu\beta} = (\nabla_\beta T)( \tilde{ \omega }^\mu, \vec{e}_\nu ) =  {({\nabla_\beta T})^\mu}_\nu.

The Christoffel symbols (\Gamma's) are 0 in the coordinate system used for parallel transport on which the covariant derivative is based, therefore in that coordinate system:

{U^\mu}_{\nu\beta} =  {({\nabla_\beta T})^\mu}_\nu = {V^\mu}_{;\nu,\beta}

Consider now another tensor Z of type ( 1\;2) as U, defined as folllows:

Z(  \tilde{ \omega }, \vec{X}, \vec{Y} ) := \omega(\nabla_{\vec{Y}}\nabla_{\vec{X}} \vec{V}).

In general (for a generic frame):

Z(\omega^{\mu}, \vec{e}_{\nu}, \vec{e}_{\beta}) = {Z^{\mu}}_{\nu \beta} = \omega^{\mu}(\nabla_{\beta}\nabla_{\nu} \vec{V}) = dx^{\mu}(\nabla_{\beta}({V^{\alpha}}_{;\nu}\; \frac{ \partial }{ \partial x^{\alpha}})) =  dx^{\mu}( {V^{\alpha}}_{;\nu,\beta}\; \frac{ \partial }{ \partial x^{\alpha}} +  {V^{\alpha}}_{;\nu}\; \nabla_{\beta}  \frac{ \partial }{ \partial x^{\alpha}} ) = dx^{\mu}( {V^{\alpha}}_{;\nu,\beta}\; \frac{ \partial }{ \partial x^{\alpha}} +  {V^{\alpha}}_{;\nu}\; {\Gamma^{\sigma}}_{\alpha\beta}\; \frac{ \partial }{ \partial x^{\sigma}} ) =  dx^{\mu}( {V^{\alpha}}_{;\nu,\beta}\; \frac{ \partial }{ \partial x^{\alpha}} +  {V^{\sigma}}_{;\nu}\; {\Gamma^{\alpha}}_{\sigma\beta}\; \frac{ \partial }{ \partial x^{\alpha}} ) =  dx^{\mu}( ({V^{\alpha}}_{;\nu,\beta} +  {V^{\sigma}}_{;\nu}\; {\Gamma^{\alpha}}_{\sigma\beta})\; \frac{ \partial }{ \partial x^{\alpha}} )  =  {V^{\mu}}_{;\nu,\beta} +  {V^{\sigma}}_{;\nu}\; {\Gamma^{\mu}}_{\sigma\beta}

Its components in the coordinate system used as the parallel transport for the covariant derivative computation:

{Z^{\mu}}_{\nu \beta}  =  {V^{\mu}}_{;\nu,\beta} +  {V^{\sigma}}_{;\nu}\; {\Gamma^{\mu}}_{\sigma\beta} = {V^{\mu}}_{;\nu,\beta}

because \Gamma's are 0.

Ultimately {U^\mu}_{\nu\beta} =  {Z^{\mu}}_{\nu \beta} in this special coordinate system, and because they are tensors, it follows that their components are the same in any other coordinate system:

(\nabla_\beta \nabla_ \nu V)^ \mu  =  {V^\mu}_{;\nu,\beta} +  {V^\alpha}_{;\nu}\; {\Gamma^ \mu}_{ \alpha \beta } -   {V^\mu}_{;\alpha}\; {\Gamma^ \alpha}_{ \nu \beta }

But here I have a problem:

{Z^{\mu}}_{\nu \beta}  =  (\nabla_\beta \nabla_ \nu V)^ \mu = {V^{\mu}}_{;\nu,\beta} +  {V^{\alpha}}_{;\nu}\; {\Gamma^{\mu}}_{\alpha\beta}

in the general case, as seen a few rows up. And this would imply that:

{V^\mu}_{;\alpha}\; {\Gamma^ \alpha}_{ \nu \beta } = 0

Surely I’ve done some mistake…

The problem is that Z is not a tensor; indeed it’s not linear on the second argument (‘a’ is a scalar field):

Z(  \tilde{ \omega }, a \vec{X}, \vec{Y} ) =  \omega(\nabla_{\vec{Y}}\nabla_{a\vec{X}} \vec{V}) =  \omega(\nabla_{\vec{Y}}(a\nabla_{\vec{X}} \vec{V})) =  \omega((\nabla_{\vec{Y}}a)\;\nabla_{\vec{X}} \vec{V} +a  (\nabla_{\vec{Y}}  \nabla_{\vec{X}} \vec{V} ) ) =   \omega((\nabla_{\vec{Y}}a)\;\nabla_{\vec{X}} \vec{V}) + a \omega( \nabla_{\vec{Y}}  \nabla_{\vec{X}} \vec{V} )

Therefore, U and Z are not comparable!

The only correct expression is:

(\nabla_\beta \nabla_ \nu V)^ \mu  =   {V^{\mu}}_{;\nu,\beta} +  {V^{\sigma}}_{;\nu}\; {\Gamma^{\mu}}_{\sigma\beta}