# A Transformation properties of the Christoffel symbols

1. Jul 12, 2016

### Chopin

If you want to define a covariant derivative which transforms correctly, you need to define it as $\nabla_i f_j = \partial_i f_j - f_k \Gamma^k_{ij}$, where $\Gamma^k_{ij}$ has the transformation property

$\bar{\Gamma}^k_{ij} = \frac{\partial \bar{x}_k}{\partial x_c}\frac{\partial x_a}{\partial \bar{x}_i}\frac{\partial x_b}{\partial \bar{x}_j}\Gamma^c_{ab} + \frac{\partial \bar{x}_k}{\partial x_c}\frac{\partial x_a}{\partial \bar{x}_i}\frac{\partial^2 x_c}{\partial \bar{x}_a\partial \bar{x}_j}$

This definition doesn't yet assume that $\Gamma^k_{ij}$ is the Levi-Civita connection. Any affine connection needs to have this transformation law in order to make the covariant derivative transform properly.

Now, if we want a torsion-free connection which preserves the metric (i.e. $\Gamma^k_{ij} = \Gamma^k_{ji}$ and $\nabla_k g_{ij} = 0$), then it's straightforward to show that

$\Gamma^k_{ij} = \frac{1}{2}g^{kl}(\partial_i g_{lj} + \partial_j g_{il} - \partial_l g_{ij})$

However, this only shows that if a metric-compatible connection exists, it would have to take the form of that equation. It doesn't necessarily prove that this definition of $\Gamma^k_{ij}$ actually transforms according to the above equation.

I'd like to show this explicitly. I started by substituting in the transformation laws for each component of this definition, and tried to massage it into the form of the above transformation law, but I can't seem to wade my way through the sea of indicies to get everything to line up. I'm guessing that I'm missing some kind of trick for working with tensor transformation laws, but I'm not quite sure what it would be. Can somebody sketch out the method by which you would show this correspondence?

2. Jul 13, 2016

### stevendaryl

Staff Emeritus
It seems fairly straight-forward to me (if a little tedious):

Define $\Gamma^\mu_{\nu \lambda} = \frac{1}{2} g^{\mu \sigma} (\partial_\nu g_{\lambda \sigma} + \partial_\lambda g_{\nu \sigma} - \partial_\sigma g_{\nu \lambda})$

Now, under a coordinate change (using Greek letters as indices for the original coordinate system, and Roman letters as indices for the new coordinate system):

$g_{\mu \nu} \Rightarrow L^i_\mu L^j_\nu g_{ij}$
$g^{\mu \nu} \Rightarrow L^\mu_i L^\nu_j g^{ij}$
$\partial_\mu \Rightarrow L^i_\mu \partial_i$

where $L^i_\mu = \partial_\mu x^i$ and $L^\mu_j = \partial_j x^\mu$.

So if you replace $g_{\mu \sigma}$ etc by $g_{ab}$ in the expression for $\Gamma$, you get the expected thing plus derivatives of $L$. The expected thing is what you would get if $L^i_\mu$ were constant (so it's derivative is zero):

$\Gamma^\mu_{\nu \lambda} = \frac{1}{2} L^\mu_a L^\sigma_b L^c_\nu L^d_\lambda L^e_\sigma g^{ab} (\partial_c g_{de} + \partial_d g_{ce} - \partial_e g_{cd}) = \frac{1}{2} L^\mu_a \delta^e_b L^c_\nu L^d_\lambda g^{ab} (\partial_c g_{de} + \partial_d g_{ce} - \partial_e g_{cd}) = \frac{1}{2} L^\mu_a L^c_\nu L^d_\lambda g^{ab} (\partial_c g_{db} + \partial_d g_{cb} - \partial_b g_{cd}) = L^\mu_a L^c_\nu L^d_\lambda \Gamma^a_{cd}$

where I used $L^\sigma_b L^e_\sigma = \delta^e_b$

If $L^i_\mu$ is not constant, then you get some correction terms:

Correction terms $= \frac{1}{2} g^{\mu \sigma} g_{ab} (\partial_\nu (L^a_\lambda L^b_\sigma) + \partial_\lambda (L^a_\nu L^b_\sigma) - \partial_\sigma (L^a_\nu L^b_\lambda))$

If you expand, you get 6 terms:
Correction terms $= \frac{1}{2} g^{\mu \sigma} g_{ab} ((\partial_\nu L^a_\lambda) L^b_\sigma + (\partial_\nu L^b_\sigma) L^a_\lambda + (\partial_\lambda L^a_\nu) L^b_\sigma + (\partial_\lambda L^b_\sigma) L^a_\nu - (\partial_\sigma L^a_\nu) L^b_\lambda - (\partial_\sigma L^b_\lambda) L^a_\nu)$

Now, let's group like terms:

Correction terms $= \frac{1}{2} g^{\mu \sigma} g_{ab} ((\partial_\nu L^a_\lambda) L^b_\sigma + (\partial_\lambda L^a_\nu) L^b_\sigma + (\partial_\nu L^b_\sigma) L^a_\lambda - (\partial_\sigma L^a_\nu) L^b_\lambda + (\partial_\lambda L^b_\sigma) L^a_\nu - (\partial_\sigma L^b_\lambda) L^a_\nu)$

Since partial derivatives commute, $\partial_\nu L^a_\lambda = \partial_\lambda L^a_\nu$. Also, since $g_{ab} = g_{ba}$, and since $a$ and $b$ are dummy indices, we are free to swap $a$ and $b$.

So we can rewrite the correction terms as:

Correction terms $= \frac{1}{2} g^{\mu \sigma} g_{ab} ((\partial_\nu L^a_\lambda) L^b_\sigma + (\partial_\nu L^a_\lambda) L^b_\sigma + (\partial_\nu L^b_\sigma) L^a_\lambda - (\partial_\nu L^b_\sigma) L^a_\lambda + (\partial_\lambda L^b_\sigma) L^a_\nu - (\partial_\lambda L^b_\sigma) L^a_\nu)$

$= g^{\mu \sigma} g_{ab} (\partial_\nu L^a_\lambda) L^b_\sigma$

$= L^\mu_c L^\sigma_d g^{cd} g_{ab} (\partial_\nu L^a_\lambda) L^b_\sigma$
$= L^\mu_c \delta^b_d g^{cd} g_{ab} (\partial_\nu L^a_\lambda)$ (because $L^\sigma_d L^b_\sigma = \delta^b_d$
$= L^\mu_c g^{cb} g_{ab} (\partial_\nu L^a_\lambda)$
$= L^\mu_c \delta^c_a (\partial_\nu L^a_\lambda)$ (because $g^{cb} g_{ab} = \delta^c_a$)
$= L^\mu_a (\partial_\nu L^a_\lambda)$ (using $\delta$ to rewrite the indices)

Hmm. This is not exactly what you have, but I actually think your correction term is wrong. Your correction term is

$\frac{\partial \bar{x}_k}{\partial x_c} \frac{\partial x_a}{\partial \bar{x}_i} \frac{\partial^2 x_c}{\partial \bar{x}_a \partial \bar{x}_j}$

I don't think it's correct, because if $a$ is an index used for barred coordinates, then it can't also be an index used for unbarred coordinates. I think maybe what was meant was:

$\frac{\partial \bar{x}_k}{\partial x_c} \frac{\partial x_a}{\partial \bar{x}_i} \frac{\partial^2 x_c}{\partial x_a \partial \bar{x}_j}$

We can rewrite $\frac{\partial}{\partial x_a}$ as $\frac{\partial \bar{x}_l}{\partial x_a} \frac{\partial}{\partial \bar{x}_l}$ to get this in the form:
$\frac{\partial \bar{x}_k}{\partial x_c} \frac{\partial x_a}{\partial \bar{x}_i} \frac{\partial \bar{x}_l}{\partial x_a} \frac{\partial^2 x_c}{\partial \bar{x}_l \partial \bar{x}_j}$

and $\frac{\partial x_a}{\partial \bar{x}_i} \frac{\partial \bar{x}_l}{\partial x_a} = \delta^l_i$. So we have:

$\frac{\partial \bar{x}_k}{\partial x_c} \delta^l_i \frac{\partial^2 x_c}{\partial \bar{x}_l \partial \bar{x}_j}$

which simplifies to:

$\frac{\partial \bar{x}_k}{\partial x_c} \frac{\partial^2 x_c}{\partial \bar{x}_i \partial \bar{x}_j}$

which I think is the same as mine, if you rewrite it to:

$L^k_c \partial_i L^c_j$

3. Jul 13, 2016

### Chopin

Ah, yes that's it. I was just about there, but was having trouble commuting the partial derivatives correctly, essentially because of the issue you mentioned with using the wrong coordinate system. Thanks for your help!

Last edited: Jul 13, 2016