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 \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} To fix this we will use the intrinsic definition of the Christoffel symbol \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} 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 \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} Applying the definition from \ref{eq:intchristoffel} to the highlighted terms gives us \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} Expressing the difference (while getting rid of the $$\frac{1}{2}$$ factor) and applying the second differential operator \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} 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.

Symmetry of the Riemann-Christoffel Tensor

Symmetry of the Riemann-Christoffel Tensor In this episode, we delve into some of the symmetric and anti-symmetric properties of the Riemann-Christoffel tensor. This work corresponds to problem 245 in Chapter 12 of 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. $$$$(\nabla_\alpha\nabla_\beta - \nabla_\beta\nabla_\alpha)T^\gamma = R^\gamma_{\cdot\delta\alpha\beta}T^\delta \label{eq:commutator}$$$$ Switching the indices $$\alpha$$ and $$\beta$$ \eqalign { (\nabla_\beta\nabla_\alpha - \nabla_\alpha\nabla_\beta)T^\gamma &= R^\gamma_{\cdot\delta\beta\alpha}T^\delta \cr -(\nabla_\alpha\nabla_\beta - \nabla_\beta\nabla_\alpha)T^\gamma &= R^\gamma_{\cdot\delta\beta\alpha}T^\delta \cr -(\nabla_\alpha\nabla_\beta - \nabla_\beta\nabla_\alpha)T^\gamma &= -R^\gamma_{\cdot\delta\alpha\beta}T^\delta } \label{eq:commutator2} Since this is true for an arbitrary tensor, $$T^\gamma$$, we have $$$$R^\gamma_{\cdot\delta\beta\alpha} = -R^\gamma_{\cdot\delta\alpha\beta} \label{eq:sym1}$$$$

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. \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} 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 \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}

[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 \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} 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 $$\{\alpha,\beta\}\leftrightarrow\{\gamma,\delta\}$$.

Property #3

Now that we have the above symmetries, showing the anti-symmetry of $$\gamma \leftrightarrow \delta$$ is just 3 swaps away. \eqalign { R_{\delta\gamma\beta\alpha} &= R_{\beta\alpha\delta\gamma} && \text{Property #2} \cr &= -R_{\beta\alpha\gamma\delta} && \text{Property #1} \cr &= -R_{\gamma\delta\alpha\beta} && \text{Property #2} \cr }

Values in the 2 Dimensional Riemann-Christoffel Tensor

The symmetries greatly restrict the degrees of freedom of the values in the tensor. Wikepedia tells me that the degrees of freedom from a "simple calculation" can be found to be $$N = \frac{n^2(n^2 - 1)}{12}$$ In our case, $$n = 2$$ so we would expect one independent component. Here are all 16 values where $$x$$ is the only value we can choose. \eqalign { R_{1212} &= x \cr R_{1221} &= -x \cr R_{2112} &= -x \cr R_{2121} &= x \cr &\text{All other elements are 0} }

Sunday, November 5, 2017

Lowering the Index on the Riemann-Christoffel Tensor

Lowering the Index on the Riemann-Christoffel Tensor For the past few months I've been learning about Tensor Calculus. My original goal was to learn about Relativity but there was too much I didn't understand because of the use of tensors. I've always been intrigued by the word "tensors" but not having the foggiest idea what they were I had no ability to go forward.

So, I found a lecture series on YouTube called Tensor Calculus and the Calculus of Moving Surfaces taught by Professor Pavel Grinfeld. I was so intrigued that I bought the text book Introduction to Tensor Analysis and the Calculus of Moving Surfaces and proceeded to do the problems. I've just gotten to the chapter on curvature and figured I'd attempt solutions to some of the problems on-line. (I find the answer key somewhat lacking at times. Dr. Grinfeld has links on the YouTube channel for various resources.) My only quibble about the book is that there are a number of typographical errors. I hope it goes through a second printing.

The other thing I should note before I get started is that there is a lot of stuff on the web about differential geometry and tensors and the notation is all over the map. One of the pleasing things about Grinfeld's treatment is that the notation is fairly simple and consistent. So, I'll be using that notation because of this and also because it's the only one I know!

So, our problem today is to lower the index on the Riemann-Christoffel tensor. I don't pretend that this is the only or most affective or efficient way of doing this. It's the machinery that I employed because of the level of my understanding.

The Riemann-Christoffel tensor, $$R^\gamma_{\cdot\delta\alpha\beta}$$ is the interesting components of the following expression; the commutator applied to a contravariant tensor. $$$$(\nabla_\alpha\nabla_\beta - \nabla_\beta\nabla_\alpha)T^\gamma = R^\gamma_{\cdot\delta\alpha\beta}T^\delta \label{eq:commutator}$$$$ Working out all the covariant derivatives you get the following famous expression for the Riemann-Christoffel tensor: $$$$R^\gamma_{\cdot\delta\alpha\beta} = \pard{\Gamma^\gamma_{\beta\delta}}{S^\alpha} - \pard{\Gamma^\gamma_{\alpha\delta}}{S^\beta} + \Gamma^\gamma_{\alpha\omega}\Gamma^\omega_{\beta\delta} - \Gamma^\gamma_{\beta\omega}\Gamma^\omega_{\alpha\delta} \label{eq:rup}$$$$ Lowering of the contravariant index, $$\gamma$$, is achieved by multiplying by the metric tensor. \eqalign{ R_{\gamma\delta\alpha\beta} &= S_{\gamma\epsilon}R^\epsilon_{\cdot\delta\alpha\beta} \cr &= S_{\gamma\epsilon}\left(\pard{\Gamma^\epsilon_{\beta\delta}}{S^\alpha} - \pard{\Gamma^\epsilon_{\alpha\delta}}{S^\beta} + \Gamma^\epsilon_{\alpha\omega}\Gamma^\omega_{\beta\delta} - \Gamma^\epsilon_{\beta\omega}\Gamma^\omega_{\alpha\delta}\right) } \label{eq:rsx} So, the interesting part of this is that we get accustomed to just dropping the indices because that is part of the beauty of tensor notation. The problem comes when you can't do that. The issue here is that multiplication of the metric tensor does not penetrate the partial derivative unscathed as it would for a covariant derivative. We have to be somewhat careful.

Here is what I did. Consider this expression which we can evaluate by the produce rule $$$$\pard{\Gamma_{\gamma,\beta\delta}}{S^\alpha} = \pard{\left(S_{\gamma\epsilon}\Gamma^\epsilon_{\beta\delta}\right)}{S^\alpha} = S_{\gamma\epsilon}\pard{\Gamma^\epsilon_{\beta\delta}}{S^\alpha} + \Gamma^\epsilon_{\beta\delta}\pard{S_{\gamma\epsilon}}{S^\alpha} \label{eq:lc2}$$$$ Note also that the partial derivative of the metric tensor can be found to be $$$$\pard{S_{\gamma\epsilon}}{S^\alpha} = \Gamma_{\gamma,\epsilon\alpha} + \Gamma_{\epsilon,\gamma\alpha} \label{eq:lc3}$$$$ Giving \eqalign{ S_{\gamma\epsilon}\pard{\Gamma^\epsilon_{\beta\delta}}{S^\alpha} &= \pard{\Gamma_{\gamma,\alpha\beta}}{S^\alpha} - \Gamma^\epsilon_{\beta\delta}\pard{S_{\gamma\epsilon}}{S^\alpha} \cr &= \pard{\Gamma_{\gamma,\beta\delta}}{S^\alpha} - \left(\Gamma_{\gamma,\epsilon\alpha} + \Gamma_{\epsilon,\gamma\alpha}\right)\Gamma^\epsilon_{\beta\delta} \cr &= \pard{\Gamma_{\gamma,\beta\delta}}{S^\alpha} - \Gamma_{\gamma,\epsilon\alpha}\Gamma^\epsilon_{\beta\delta} - \Gamma_{\epsilon,\gamma\alpha}\Gamma^\epsilon_{\beta\delta} } \label{eq:lc4} If you look at equation \ref{eq:lc4}, it gives us the value of the first term of equation \ref{eq:rsx} after you multiply out the metric tensor. Doing the same thing to the second term gives us \eqalign{ S_{\gamma\epsilon}\pard{\Gamma^\epsilon_{\alpha\delta}}{S^\beta} &= \pard{\Gamma_{\gamma,\alpha\delta}}{S^\beta} - \Gamma_{\gamma,\epsilon\beta}\Gamma^\epsilon_{\alpha\delta} - \Gamma_{\epsilon,\gamma\beta}\Gamma^\epsilon_{\alpha\delta} } \label{eq:lc5} The 3rd and 4th terms of equation \ref{eq:rsx} are just the Cristoffels with the index lowered. Putting everything together gives us \eqalign{ R_{\gamma\delta\alpha\beta} &= \pard{\Gamma_{\gamma,\beta\delta}}{S^\alpha} - \bbox[5px,border:2px solid red]{\Gamma_{\gamma,\epsilon\alpha}\Gamma^\epsilon_{\beta\delta}} - \Gamma_{\epsilon,\gamma\alpha}\Gamma^\epsilon_{\beta\delta} - \pard{\Gamma_{\gamma,\alpha\delta}}{S^\beta} + \bbox[5px,border:2px solid blue]{\Gamma_{\gamma,\epsilon\beta}\Gamma^\epsilon_{\alpha\delta}} + \Gamma_{\epsilon,\gamma\beta}\Gamma^\epsilon_{\alpha\delta} + \bbox[5px,border:2px solid red]{\Gamma_{\gamma,\alpha\omega}\Gamma^\omega_{\beta\delta}} - \bbox[5px,border:2px solid blue]{\Gamma_{\gamma,\beta\omega}\Gamma^\omega_{\alpha\delta}} } \label{eq:rl1} Careful inspection will reveal that the red boxed terms are the same because the Cristoffel symbols are symmetric in the last two indices and the name of the contracted index doesn't matter. And the same goes for the terms in the blue boxes. It's interesting to note that the original Cristoffel products are disposed of and replaced with a different set. I don't know if there is some deeper meaning there.

The final result is (rearranging to get a more aesthetically pleasing result) \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:final} More coming on the this topic.