# Christoffel Symbols Tensorial Nature and their Transformation

28 Jul 2019 · Comments

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

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:

Both equations are equal, so

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

Using a local basis where

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:

Also

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

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

### Covariant Derivative Transformation

Looking at the definition of the covariant derivative:

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

Expanding equation $\ref{5}$,

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

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

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

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

Using equation $\ref{5}$

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

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