Christoffel Symbols Tensorial Nature and their Transformation
28 Jul 2019
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.