__Introduction to Tensor Analysis and the Calculus of Moving Surfaces__. The answer key says something like "we did this on the final exam." So my goal is to fill that hole.

## Property #1

The first property is the easiest to show and follows directly from the definition and that is that if we switch \(\alpha\) and \(\beta\) (the 3rd and 4th indices) you get the negative value. This is anti-symmetry and it looks like this $$ R^\gamma_{\cdot\delta\beta\alpha} = -R^\gamma_{\cdot\delta\alpha\beta} $$ If you look at the definition, it is evident. $$ \begin{equation} (\nabla_\alpha\nabla_\beta - \nabla_\beta\nabla_\alpha)T^\gamma = R^\gamma_{\cdot\delta\alpha\beta}T^\delta \label{eq:commutator} \end{equation} $$ Switching the indices \(\alpha\) and \(\beta\) $$ \begin{equation} \eqalign { (\nabla_\beta\nabla_\alpha - \nabla_\alpha\nabla_\beta)T^\gamma &= R^\gamma_{\cdot\delta\beta\alpha}T^\delta \cr -(\nabla_\alpha\nabla_\beta - \nabla_\beta\nabla_\alpha)T^\gamma &= R^\gamma_{\cdot\delta\beta\alpha}T^\delta \cr -(\nabla_\alpha\nabla_\beta - \nabla_\beta\nabla_\alpha)T^\gamma &= -R^\gamma_{\cdot\delta\alpha\beta}T^\delta } \label{eq:commutator2} \end{equation} $$ Since this is true for an arbitrary tensor, \(T^\gamma\), we have $$ \begin{equation} R^\gamma_{\cdot\delta\beta\alpha} = -R^\gamma_{\cdot\delta\alpha\beta} \label{eq:sym1} \end{equation} $$

## Property #2

The next symmetric property will require a little more finesse. Now, I don't claim that this is the easiest or most intuitive way to prove this but it does work and it is the one that I came up with. Besides, if you have a better way then it will still be good to see an alternative as that seems to broaden one's understanding. And perhaps you could be so kind as to give me a few clues in the comments.Here we will show that the Riemann-Christoffel tensor with all indices lowered is symmetric if you swap the first two indices with the second two in order. In other words, \(\gamma\leftrightarrow\alpha\) and \(\delta\leftrightarrow\beta\). We start by going to the definition. $$ \begin{equation} \eqalign { R_{\gamma\delta\alpha\beta} &= \pard{\Gamma_{\gamma,\beta\delta}}{S^\alpha} - \pard{\Gamma_{\gamma,\alpha\delta}}{S^\beta} + \Gamma_{\epsilon,\gamma\beta}\Gamma^\epsilon_{\alpha\delta} - \Gamma_{\epsilon,\gamma\alpha}\Gamma^\epsilon_{\beta\delta} } \label{eq:rlower} \end{equation} $$ It's clear that the last term \(\Gamma_{\epsilon,\gamma\alpha}\Gamma^\epsilon_{\beta\delta}\) is symmetric with \(\{\alpha,\beta\}\leftrightarrow\{\gamma,\delta\}\) because the Christoffel symbols are symmetric in the last two indices. $$ \Gamma_{\epsilon,\gamma\alpha}\Gamma^\epsilon_{\beta\delta} = \Gamma_{\epsilon,\alpha\gamma}\Gamma^\epsilon_{\delta\beta} $$ One down, three to go. The third term requires a few additional steps. $$ \eqalign{ \Gamma_{\epsilon,\gamma\beta}\Gamma^\epsilon_{\alpha\delta} &= \left(S_{\omega\epsilon}\Gamma^\omega_{\gamma\beta}\right)\Gamma^\epsilon_{\alpha\delta} && \text{definition of lowered index} \\ &= \Gamma^\omega_{\gamma\beta} \left( S_{\omega\epsilon} \Gamma^\epsilon_{\alpha\delta} \right) && \text{commutativity/associativity} \\ &= \Gamma^\epsilon_{\gamma\beta} \Gamma_{\epsilon,\alpha\delta} && \text{lower the index of second term and rename the contracted index } \omega \rightarrow \epsilon \\ } $$ We see that the result is the same as the initial expression with the required indices swapped. Two down.

The last part of showing this symmetry is a bit more harrowing as we will need to consider the remaining two terms in tandem. We start by rewriting the first two terms of the definition of \(R\) in terms of the definitions of the Christoffel symbols of the surface. The definitions are $$ \begin{equation} \eqalign{ \Gamma^\gamma_{\alpha\beta} &= \pard{\vec{S_\alpha}}{S^\beta} \cdot \vec{S^\gamma} \cr \Gamma_{\gamma,\alpha\beta} &= \pard{\vec{S_\alpha}}{S^\beta} \cdot \vec{S_\gamma} } \label{eq:christoffel} \end{equation} $$

[EDIT: I've amended this proof with a fully intrinsic version of this portion of the proof. If this extrinsic argument gives you the willies, you can go here

Where \(\vec{S_\gamma}\) is the covariant basis of the surface. Rewriting the first two terms with these definitions and expanding via the product rule $$ \begin{equation} \eqalign { \pard{\Gamma_{\gamma,\beta\delta}}{S^\alpha} - \pard{\Gamma_{\gamma,\alpha\delta}}{S^\beta} &= \pard{}{S^\alpha} \left( \pard{\vec{S_\beta}}{S^\delta} \cdot \vec{S_\gamma} \right) - \pard{}{S^\beta} \left( \pard{\vec{S_\alpha}}{S^\delta} \cdot \vec{S_\gamma} \right) \cr &= \frac{\partial^2 \vec{S_\beta}}{\partial S^\alpha \partial S^\delta} \cdot \vec{S_\gamma} + \bbox[5px,border:2px solid red]{\pard{\vec{S_\beta}}{S^\delta} \cdot \pard{\vec{S_\gamma}}{S^\alpha}} - \frac{\partial^2 \vec{S_\alpha}}{\partial S^\beta \partial S^\delta} \cdot \vec{S_\gamma} - \bbox[5px,border:2px solid blue]{\pard{\vec{S_\alpha}}{S^\delta} \cdot \pard{\vec{S_\gamma}}{S^\beta}} } \label{eq:first2terms} \end{equation} $$ Exchanging indices \(\gamma\leftrightarrow\alpha\) and \(\delta\leftrightarrow\beta\) in the term boxed in blue yields the same two terms in reverse order. So, enough of that one.

The term boxed in red requires the realization that because of their definition the partial derivatives of the basis vectors have the property that \( \pard{\vec{S_\omega}}{S^\epsilon} = \pard{\vec{S_\epsilon}}{S^\omega} \). So, $$ \eqalign { \pard{\vec{S_\beta}}{S^\delta} \cdot \pard{\vec{S_\gamma}}{S^\alpha} &= \pard{\vec{S_\delta}}{S^\beta} \cdot \pard{\vec{S_\alpha}}{S^\gamma} } $$ Which is the same as swapping the required indices. Lastly, we examine the remaining two terms. "Factoring" out the derivative with respect to \( S^\delta \) gives us $$ \frac{\partial^2 \vec{S_\beta}}{\partial S^\alpha \partial S^\delta} \cdot \vec{S_\gamma} - \frac{\partial^2 \vec{S_\alpha}}{\partial S^\beta \partial S^\delta} \cdot \vec{S_\gamma} = \pard{}{S^\delta} \left( \pard{\vec{S^\beta}}{S^\alpha} - \pard{\vec{S^\alpha}}{S^\beta} \right) $$ By the same argument as the previous step, the difference inside the parentheses is identically zero. Now that all the terms are accounted for, we can state that the Riemann-Christoffel tensor is symmetric in \(\{\alpha,\beta\}\leftrightarrow\{\gamma,\delta\}\).

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