## Tuesday, November 7, 2017

### Addendum to Symmetry of the Riemann-Christoffel Tensor

Addendum to Symmetry of the Riemann-Christoffel Tensor This is an addendum to the previous post on the symmetries of the Riemann-Christoffel tensor. I received a comment that I should be able to make the arguments purely intrinsicly and without relying on the surface basis which is defined extrinsicly. This is a great comment and I wish that I had thought of it myself. So, here we go.

I think the only part of my original argument that relied on extrinsic characteristics was on the proof of the symmetry with the first two and last two indices. That argument relied on the definition of the surface Christoffel symbols as the dot-product of the partial derivative of the basis with the basis as follows \begin{equation} \eqalign{ \Gamma^\gamma_{\alpha\beta} &= \pard{\vec{S_\alpha}}{S^\beta} \cdot \vec{S^\gamma} \cr \Gamma_{\gamma,\alpha\beta} &= \pard{\vec{S_\alpha}}{S^\beta} \cdot \vec{S_\gamma} } \label{eq:extchristoffel} \end{equation} To fix this we will use the intrinsic definition of the Christoffel symbol \begin{equation} \eqalign{ \Gamma_{\gamma,\alpha\beta} &= \frac{1}{2} \left( \pard{S_{\gamma\beta}}{S^\alpha} + \pard{S_{\gamma\alpha}}{S^\beta} - \pard{S_{\beta\alpha}}{S^\gamma} \right) } \label{eq:intchristoffel} \end{equation} Here again is the definition of the Riemann-Christoffel tensor. The last two terms (the products of the Christoffels) were shown to have the required symmetry in the previous post so I won't repeat that here. I will only be looking at the red-boxed terms \begin{equation} \eqalign{ R_{\gamma\delta\alpha\beta} &= \bbox[5px,border:2px solid red]{\pard{\Gamma_{\gamma,\beta\delta}}{S^\alpha} - \pard{\Gamma_{\gamma,\alpha\delta}}{S^\beta}} + \Gamma_{\epsilon,\gamma\beta}\Gamma^\epsilon_{\alpha\delta} - \Gamma_{\epsilon,\gamma\alpha}\Gamma^\epsilon_{\beta\delta} } \label{eq:rsx} \end{equation} Applying the definition from \ref{eq:intchristoffel} to the highlighted terms gives us \begin{equation} \eqalign{ \pard{\Gamma_{\gamma,\beta\delta}}{S^\alpha} &= \frac{1}{2} \pard{}{S^\alpha} \left( \pard{S_{\gamma\delta}}{S^\beta} + \pard{S_{\gamma\beta}}{S^\delta} - \pard{S_{\delta\beta}}{S^\gamma} \right) \cr \pard{\Gamma_{\gamma,\alpha\delta}}{S^\beta} &= \frac{1}{2} \pard{}{S^\beta} \left( \pard{S_{\gamma\delta}}{S^\alpha} + \pard{S_{\gamma\alpha}}{S^\delta} - \pard{S_{\delta\alpha}}{S^\gamma} \right) } \label{eq:twoterms} \end{equation} Expressing the difference (while getting rid of the $$\frac{1}{2}$$ factor) and applying the second differential operator \begin{equation} \eqalign{ 2 \left( \pard{\Gamma_{\gamma,\beta\delta}}{S^\alpha} - \pard{\Gamma_{\gamma,\alpha\delta}}{S^\beta} \right) &= + \bbox[5px,border:2px solid red] { \pardd{S_{\gamma\delta}}{S^\alpha}{S^\beta} } + \bbox[5px,border:2px solid green]{ \pardd{S_{\gamma\beta}}{S^\alpha}{S^\delta} } - \bbox[5px,border:2px solid blue] { \pardd{S_{\delta\beta}}{S^\alpha}{S^\gamma} } - \bbox[5px,border:2px solid red] { \pardd{S_{\gamma\delta}}{S^\beta}{S^\alpha} } - \bbox[5px,border:2px solid blue] { \pardd{S_{\gamma\alpha}}{S^\beta}{S^\delta} } + \bbox[5px,border:2px solid green]{ \pardd{S_{\delta\alpha}}{S^\beta}{S^\gamma}} } \label{eq:expanded} \end{equation} The two terms in the red boxes are the same because differentiation commutes (assuming continuity, differentiability, etc.). Since we subtract one from the other, they vanish into the mist.

The two terms in the blue boxes are self symmetric because if you make the required index swap you get the same thing you started with because of the symmetry of the metric tensor and commutative property of differentation.

The two terms in the green boxes are mutally symmetric. If you make the required index swap on one of them, you get the other one.

So along with the other items in the other post, we now have an intrinsic proof of the interchange symmetry of the first two and last two indices of the Riemann-Christoffel tensor. Here again is a link to the prior post

Thanks to 'Lemma' who I assume to be Prof. Grinfeld for the comment. I will indulge and give myself a Grinfeld number of #1.