I have been studying the covariant derivative and I'm confused by the calculation of the Christoffel symbols $\Gamma$. The equation for computing $\Gamma$ is given as: $${\Gamma^c}_{ab} = \frac12 g^{ck} (\partial_a g_{bk} + \partial_b g_{ak} - \partial_k g_{ab} ) $$

This will produce a $\Gamma$ which is symmetric in a and b. And it transforms as: $$\Gamma' = T \Gamma S S + \partial S T$$

Where $S$ is the transformation matrix and $T$ is the inverse. Using this equation, if we are transforming from a flat space where $\Gamma = 0$, then: $$\Gamma' = \partial S T$$

But this equation will not always produce a $\Gamma$ which is symmetric. and so won't give the same answer as the first equation. I've tried this with the following transformation matrix in an $(x,y)$ coordinate system: $$ S = \begin{bmatrix}1 & 0 \\ 0 & x \end{bmatrix} $$

This produces a metric tensor of: $$ g' = gSS = \begin{bmatrix}1 & 0 \\ 0 & x^2 \end{bmatrix}$$

This is the same metric tensor as for polar coordinates (with $x$ instead of $r$). Using the first equation, the Christoffel symbols end up as: $$ \Gamma' = \begin{bmatrix}0 & 0 \\ 0 & -x\end{bmatrix} _{\begin{bmatrix}0 & \frac1x \\ \frac1x & 0\end{bmatrix}}$$

But using the second equation I get: $$ \Gamma' = \partial S T = \begin{bmatrix}0 & 0 \\ 0 & 0\end{bmatrix} _{\begin{bmatrix}0 & 0 \\ \frac1x & 0\end{bmatrix}}$$

Why are they different? I must be missing something, perhaps obvious, but I've researched it and calculated it many times and can't find the problem.

  • $\begingroup$ My guess would be you messed up in matrix calculations. I would highly recommend writing all your equations and doing all calculations in index form. $\endgroup$ – Prof. Legolasov Jun 1 '15 at 22:35
  • $\begingroup$ The transformation law will certainly produce a symmetric connection. Like Hindsight, I don't recommend you use matrices; you have a much higher probability (as if it weren't high already) of messing up. $\endgroup$ – Javier Jun 1 '15 at 23:06
  • $\begingroup$ I have done the calculations in index form and matrix form. I showed them above as matrices as is easier to see visually. In any case, you can see that the $\partial S$ will only ever have one non-zero term (the $\partial x$ of the x). And since T only has entries on its diagonal, the result $\partial S T$ will also only have one non-zero term. How can it end up with three non-zero terms? $\endgroup$ – Paul Jun 2 '15 at 8:01
  • $\begingroup$ If I understand correctly what confuses you is that your example is $\mathbb R^2$, which is not just a manifold but also a vector space. That allows you to write the change of coordinates in the matrix form. But then the transformation matrix in your formulas is the Jacobian of the transformation, which acts on the tangent space, not the same matrix. If you write the transformation as $X=x$ and $Y=xy$ you'll see the difference. $\endgroup$ – MBN Jun 2 '15 at 13:55

Thanks to MBN for the comment - which is also pretty much the answer. Which is basically that I can not make the two equations give the same result, because the $S$ which I was using is invalid.

I was thinking of $S$ in terms of a matrix of coordinate changes in a vector tangent space. The transformation I was trying to achieve was: $$x' = x, y' = xy $$

So then $S$ could be used to transform from $(x, y)$ coordinates to $(x',y')$: $$ \begin{bmatrix}x' \\ y' \end{bmatrix} = S \begin{bmatrix}x \\ y \end{bmatrix} = \begin{bmatrix}1 & 0 \\ 0 & x \end{bmatrix} \begin{bmatrix}x \\ y \end{bmatrix} = \begin{bmatrix}1 & 0 \\ 0 & xy \end{bmatrix} $$

But the $S$ used above in the formula for $\Gamma$ is the Jacobian matrix of partial derivatives:

$$S = \begin{bmatrix} \frac{\partial x'}{\partial x} & \frac{\partial x'}{\partial y} \\ \frac{\partial y'}{\partial x} & \frac{\partial y'}{\partial y} \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ y & x \end{bmatrix} $$

Using that $S$ ends up with a completely different $g$ and $\Gamma$. So my confusion was in not understanding the difference between coordinate transformation matrices and Jacobians, and therefore using an impossible $S$.

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