Christoffel Symbols Tensorial Nature and their Transformation

28 Jul 2019

This post explains the tensorial nature of the components of the connection, aka the Christoffel symbols, by deriving their general transformation law with examples.

The Christoffel symbols, $\Gamma ^{\mu}_{\nu \sigma}$, are in fact the components of the connection (ie $\nabla$)

$$\require{cancel}$$ $$\begin{equation} \nabla _{\sigma}\mathrm {e} _{\nu} = \Gamma^{\mu}_{\nu \sigma} \mathrm{e}_ {\mu} \label{1} \end{equation}$$

where $\mathbf{e}_{\nu}$ is a basis for vector fields.

It important to emphasize that the connection itself (ie $\nabla$) is not a tensor for it is not linear in its argument $Y$ as in $\nabla_{X}{Y}$. In order to find out whether the Christoffel symbols are a component of a tensor or not, their transformation law must be derived first:

$$\begin{equation} \nabla _{\sigma'}\mathrm {e'} _{\nu} =\Gamma^{\mu'}_{\nu' \sigma'} \mathrm {e} _{\mu'} = \Gamma^{\mu'}_{\nu' \sigma'} \Lambda^{\mu}{}_{\mu'}\mathrm {e} _{\mu} \label{2} \end{equation}$$ $$\begin{align*} \nabla _{\sigma'}\mathrm {e'} _{\nu} =& \nabla_{( \Lambda^{\sigma}{}_{\sigma'} \mathrm{e} _{\sigma} )} {( \Lambda^{\nu}{}_{\nu'} \mathrm {e} _{\nu})} \\ =& \Lambda^{\sigma}{}_{\sigma'} \Lambda^{\nu}{}_{\nu'} \nabla_{\sigma}{\mathrm{e} _{\nu}} + \Lambda^{\sigma}{}_{\sigma'} \nabla_{\sigma}(\Lambda^{\nu}_{\nu'}) \mathrm{e} _{\nu} \\ =& \Lambda^{\sigma}{}_{\sigma'} \Lambda^{\nu}{}_{\nu'} \nabla_{\sigma}{\mathrm{e} _{\nu}} + \Lambda^{\sigma}{}_{\sigma'}\mathrm {e} _{\sigma}(\Lambda^{\nu}_{\nu'}) \mathrm{e} _{\nu} \label{eq3} \tag{3} \end{align*}$$

Both equations are equal, so

$$\Gamma^{\mu'}_{\nu' \sigma'} \Lambda^{\mu}{}_{\mu'} \mathrm{e} _{\mu} = \Lambda^{\nu}{}_{\nu'} \Lambda^{\sigma}{}_{\sigma'} \Gamma^{\mu}_{\nu \sigma} \mathrm{e} _{\mu} + \Lambda^{\sigma}{}_{\sigma'} \mathrm{e}_{\sigma} (\Lambda^{\mu}{}_{\nu'}) \mathrm{e}_{\mu}$$

Multiplying both side with $\Lambda^{\mu’}{}_{\mu}$

$$\begin{equation} \Gamma^{\mu'}_{\nu' \sigma'} = \Lambda^{\mu'}{}_{\mu} \Lambda^{\nu}{}_{\nu'} \Lambda^{\sigma}{}_{\sigma'} \Gamma^{\mu}_{\nu \sigma} +\Lambda^{\mu'}{}_{\mu} \Lambda^{\sigma}{}_{\sigma'} \mathrm{e}_{\sigma}(\Lambda^{\mu}{}_{\nu'}) \tag{4} \label{4} \end{equation}$$

Using a local basis where

$$\mathbf{e}_{i} = \frac{\partial}{\partial x^{i}}, \quad i=1,2,\dots ,n$$ $$\Gamma^{\mu'}_{\nu' \sigma'} = \frac{\partial x^{\mu'}}{\partial x^{\mu}} \frac{\partial x^{\nu}}{\partial x^{\nu'}} \frac{\partial x^{\sigma}}{\partial x^{\sigma'}} \Gamma^{\mu}_{ \nu \sigma } + \frac{\partial x^{\mu'}}{\partial x^{\mu}} \frac{\partial x^{\sigma}}{\partial x^{\sigma'}} \frac{\partial}{\partial x^\sigma} \left( \frac{\partial x^\mu}{\partial x^{\nu'}} \right) \Leftrightarrow$$ $$\begin{equation} \tag{5} \boxed{ \Gamma^{\mu'}_{\nu' \sigma'} = \frac{\partial x^{\mu'}}{\partial x^{\mu}} \frac{\partial x^{\nu}}{\partial x^{\nu'}} \frac{\partial x^{\sigma}}{\partial x^{\sigma'}} \Gamma^{\mu}_{ \nu \sigma } + \frac{\partial x^{\mu'}}{\partial x^{\mu}} \frac{\partial^2 x^\mu}{\partial x^{\mu'} \partial x^{\nu'}} } \label{5} \end{equation}$$

Where the last term on the right-hand side was simplified a bit because partial derivatives commute.

It is clear that the first term on the right-hand side transforms like a tensor, but the second inhomogeneous term does not.

Equation $\ref{5}$ has another form which can be derived as follows:

$$\begin{align*} \nabla_{\sigma'}V^{\mu'} =& \frac{\partial x^\sigma}{\partial x^{\sigma'}} \frac{\partial x^{\mu'}}{\partial x^{\mu}} \nabla_{\sigma}V^{\mu} \\ =& \frac{\partial x^\sigma}{\partial x^{\sigma'}} \frac{\partial x^{\mu'}}{\partial x^{\mu}} \frac{\partial}{\partial x^{\sigma}}V^\mu + \frac{\partial x^\sigma}{\partial x^{\sigma'}} \frac{\partial x^{\mu'}}{\partial x^{\mu}} \Gamma^{\mu}_{\nu \sigma} V^\nu \label{6} \tag{6} \\ \end{align*}$$

Also

$$\begin{equation} \tag{7} \nabla_{\sigma'}V^{\mu'} = \frac{\partial x^\sigma}{\partial x^{\sigma'}} \frac{\partial}{\partial x^{\sigma}} \left( \frac{\partial x^{\mu'}}{\partial x^{\mu}} V^\mu \right) + \Gamma^{\mu'}_{ {\nu'} {\sigma'} } \frac{\partial x^{\nu'}}{\partial x^\nu} V^{\nu} \label{7} \end{equation}$$

Equations \ref{6} and \ref{7} are equal:

$$\begin{align*} \frac{\partial x^\sigma}{\partial x^{\sigma'}} \frac{\partial x^{\mu'}}{\partial x^{\mu}} \frac{\partial}{\partial x^{\sigma}}V^\mu + \frac{\partial x^\sigma}{\partial x^{\sigma'}} \frac{\partial x^{\mu'}}{\partial x^{\mu}} \Gamma^{\mu}_{\nu \sigma} V^\nu =& \frac{\partial x^\sigma}{\partial x^{\sigma'}} \frac{\partial}{\partial x^{\sigma}} \left( \frac{\partial x^{\mu'}}{\partial x^{\mu}} V^\mu \right) + \Gamma^{\mu'}_{\nu'\sigma'} \frac{\partial x^{\nu'}}{\partial x^\nu} V^{\nu} \Leftrightarrow \\ \cancel{\frac{\partial x^\sigma}{\partial x^{\sigma'}} \frac{\partial x^{\mu'}}{\partial x^{\mu}} \frac{\partial}{\partial x^{\sigma}}V^\mu} + \frac{\partial x^\sigma}{\partial x^{\sigma'}} \frac{\partial x^{\mu'}}{\partial x^{\mu}} \Gamma^{\mu}_{\nu \sigma} V^\nu =& \frac{\partial x^\sigma}{\partial x^{\sigma'}} \frac{\partial^{2} x^{\mu'}}{\partial x^\sigma \partial x^\mu} V^\mu + \\ & \cancel{\frac{\partial x^\sigma}{\partial x^{\sigma'}} \frac{\partial x^{\mu'}}{\partial x^\mu} \frac{\partial}{\partial x^\sigma} V^\mu} + \Gamma^{\mu'}_{\nu'\sigma'} \frac{\partial x^{\nu'}}{\partial x^\nu} V^\nu \Leftrightarrow \end{align*}$$ $$\Gamma^{\mu'}_{\nu'\sigma'} \frac{\partial x^{\nu'}}{\partial x^\nu} V^{\nu} = \frac{\partial x^\sigma}{\partial x^{\sigma'}} \frac{\partial x^{\mu'}}{\partial x^{\mu}} \Gamma^{\mu}_{\nu \sigma} V^\nu - \frac{\partial x^\sigma}{\partial x^{\sigma'}} \frac{\partial^{2} x^{\mu'}}{\partial x^\sigma \partial x^\mu} V^\mu$$

Multiplying by $\frac{\partial x^{\nu}}{\partial x^{\nu’}}$ and relabeling the last dummy index on the right-hand side,

$$\begin{equation} \tag{8} \boxed{ \Gamma^{\mu'}_{\nu'\sigma'} = \frac{\partial x^{\nu}}{\partial x^{\nu'}} \frac{\partial x^\sigma}{\partial x^{\sigma'}} \frac{\partial x^{\mu'}}{\partial x^{\mu}} \Gamma^{\mu}_{\nu \sigma} - \frac{\partial x^{\nu}}{\partial x^{\nu'}} \frac{\partial x^\sigma}{\partial x^{\sigma'}} \frac{\partial^{2} x^{\mu'}}{\partial x^\sigma \partial x^\nu} } \label{8} \end{equation}$$

Covariant Derivative Transformation

Looking at the definition of the covariant derivative:

$$\nabla_{\sigma}V^\mu = \frac{\partial V^\mu}{\partial x^\sigma} + \Gamma^{\mu}_{\nu \sigma} V^\nu$$

Neither one of the terms on the right-hand side above transforms likes the components of a tensor, but their sum does:

Expanding equation $\ref{5}$,

$$\begin{align*} \nabla_{\sigma'}V^{\mu'} =& \frac{\partial x^\sigma}{\partial x^{\sigma'}} \frac{\partial}{\partial x^{\sigma}} \left( \frac{\partial x^{\mu'}}{\partial x^{\mu}} V^\mu \right) + \Gamma^{\mu'}_{ {\nu'} {\sigma'} } \frac{\partial x^{\nu'}}{\partial x^\nu} V^{\nu} \\ =& \frac{\partial x^\sigma}{\partial x^{\sigma'}} \frac{\partial^{2} x^{\mu'}}{\partial x^\sigma \partial x^\mu} V^\mu + \frac{\partial x^\sigma}{\partial x^{\sigma'}} \frac{\partial x^{\mu'}}{\partial x^\mu} \frac{\partial}{\partial x^\sigma} V^\mu + \Gamma^{\mu'}_{\nu'\sigma'} \frac{\partial x^{\nu'}}{\partial x^\nu} V^{\nu} \end{align*}{}$$

Using equation $\ref{8}$ to expand $\Gamma^{\mu’}_{\nu’ \sigma’}$,

$$\begin{align*} \nabla_{\sigma'}V^{\mu'} =& \cancel{ \frac{\partial x^\sigma}{\partial x^{\sigma'}} \frac{\partial^{2} x^{\mu'}}{\partial x^\sigma \partial x^\mu} V^\mu} + \frac{\partial x^\sigma}{\partial x^{\sigma'}} \frac{\partial x^{\mu'}}{\partial x^\mu} \frac{\partial}{\partial x^\sigma} V^\mu + \frac{\partial x^{\nu'}}{\partial x^\nu} \frac{\partial x^{\nu}}{\partial x^{\nu'}} \frac{\partial x^\sigma}{\partial x^{\sigma'}} \frac{\partial x^{\mu'}}{\partial x^{\mu}} \Gamma^{\mu}_{\nu \sigma}V^{\nu} \\ -& \cancel{ \frac{\partial x^{\nu'}}{\partial x^\nu} \frac{\partial x^{\nu}}{\partial x^{\nu'}} \frac{\partial x^\sigma}{\partial x^{\sigma'}} \frac{\partial^{2} x^{\mu'}}{\partial x^\sigma \partial x^\nu} V^{\nu}} \end{align*}$$

The non-tensorial term in equation $\ref {8}$ cancels out the one from the partials, and the final result indeed transforms like a tensor.

$$\begin{align*} \therefore{} \nabla_{\sigma'}V^{\mu'} =& \frac{\partial x^\sigma}{\partial x^{\sigma'}} \frac{\partial x^{\mu'}}{\partial x^\mu} \frac{\partial}{\partial x^\sigma} V^\mu + \frac{\partial x^\sigma}{\partial x^{\sigma'}} \frac{\partial x^{\mu'}}{\partial x^{\mu}} \Gamma^{\mu}_{\nu \sigma}V^{\nu} \\ =& \frac{\partial x^\sigma}{\partial x^{\sigma'}} \frac{\partial x^{\mu'}}{\partial x^\mu} \left( \frac{\partial}{\partial x^\sigma} V^\mu + \Gamma^{\mu}_{\nu \sigma}V^{\nu} \right)\\ \frac{\partial}{\partial x^{\sigma'}} V^{\mu'} + \Gamma^{\mu'}_{\nu \sigma'}V^{\nu} =& \frac{\partial x^\sigma}{\partial x^{\sigma'}} \frac{\partial x^{\mu'}}{\partial x^\mu} \left( \frac{\partial}{\partial x^\sigma} V^{\mu} + \Gamma^{\mu}_{\nu \sigma}V^{\nu} \right) \\ T_{\sigma'}^{\mu'} =& \frac{\partial x^\sigma}{\partial x^{\sigma'}} \frac{\partial x^{\mu'}}{\partial x^\mu} T_{\sigma}^{\mu} \\ \end{align*}$$

This resembles the transformation law of a second rank mixed tensor.

In polar coordinates $(r, \theta)$ where $e_r = \frac{\partial}{\partial r}, e_\theta = \frac{\partial}{\partial \theta}$, basic calculation shows that

$$\begin{equation} \tag{9} \Gamma^{\theta}_{r \theta} = \frac{1}{r} \label{9} \end{equation}$$

This is another fact that the symbols $\Gamma ^{\mu}_{\nu \sigma}$ cannot be the components of a tensor for the components vanish in Cartesian coordinates but not in others, such as polar coordinates.

Is it possible to recover $\ref{9}$ from Cartesian coordinates using the general transformation law?

The answer is yes, in a little complicated way:

Knowing that

$$\theta(x,y) = \mathrm{arctan}\left(\frac{y}{x}\right) \\ r(x,y) = \sqrt{x^2 + y^2} \\ x(r,\theta) = r \mathrm{cos}(\theta) \\ y(r, \theta) = r \mathrm{sin}(\theta) \\$$

Using equation $\ref{5}$

$$\begin{align*} \Gamma^{\theta}_{r \theta} =& \cancelto{0}{\frac{\partial x^{\theta}}{\partial x^{\mu}} \frac{\partial x^{\nu}}{\partial x^r} \frac{\partial x^{\sigma}}{\partial x^{\theta}} \Gamma^{\mu}_{ \nu \sigma }} + \frac{\partial x^{\theta}}{\partial x^{\mu}} \frac{\partial^2 x^\mu}{\partial x^{\theta} \partial x^{r}} \\ =& \frac{\partial \theta}{\partial x} \frac{\partial^2 x}{\partial \theta \partial r} + \frac{\partial \theta}{\partial y} \frac{\partial^2 y}{\partial \theta \partial r} \\ =& \frac{x \mathrm{cos}(\theta) + y \mathrm{sin}(\theta)}{x^2 + y^2} \\ =& \frac{1}{r} \end{align*}$$

The first term vanishes since $\Gamma^{\mu}_{ \nu \sigma }$ vanish in Cartesian coordinates. The second term is expanded using Einstein summation convention.