# Christoffel symbols: symmetry

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 17; Box 17.3.

The Christoffel symbols are defined in terms of the basis vectors in a given coordinate system as:

$\displaystyle \boxed{\frac{\partial\mathbf{e}_{i}}{\partial x^{j}}=\Gamma_{\; ij}^{k}\mathbf{e}_{k}} \ \ \ \ \ (1)$

Remember that the basis vectors ${\mathbf{e}_{i}}$ are defined so that

 $\displaystyle ds^{2}$ $\displaystyle =$ $\displaystyle d\mathbf{s}\cdot d\mathbf{s}\ \ \ \ \ (2)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(dx^{i}\mathbf{e}_{i}\right)\cdot\left(dx^{j}\mathbf{e}_{j}\right)\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \mathbf{e}_{i}\cdot\mathbf{e}_{j}dx^{i}dx^{j}\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle \equiv$ $\displaystyle g_{ij}dx^{i}dx^{j} \ \ \ \ \ (5)$

In a locally flat frame using rectangular spatial coordinates, the basis vectors ${\mathbf{e}_{i}}$ are all constants, so from 1, all the Christoffel symbols must be zero: ${\Gamma_{\; ij}^{k}=0}$.

Now let’s look at the second covariant derivative of a scalar field ${\Phi}$:

 $\displaystyle \nabla_{i}\nabla_{j}\Phi$ $\displaystyle =$ $\displaystyle \nabla_{i}\left(\partial_{j}\Phi\right)\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \partial_{i}\partial_{j}\Phi-\Gamma_{ij}^{k}\partial_{k}\Phi \ \ \ \ \ (7)$

where in 6 we used rule 1 for the covariant derivate: the covariant derivative of a scalar is the same as the ordinary derivative.

In the locally flat frame, this equation reduces to

$\displaystyle \nabla_{i}\nabla_{j}\Phi=\partial_{i}\partial_{j}\Phi \ \ \ \ \ (8)$

Since the covariant derivative is a tensor, this is a tensor equation, and since ordinary partial derivatives commute, this equation is the same if we swap the indices ${i}$ and ${j}$. Tensor equations must have the same form in all coordinate systems, so this implies that 7 must also be invariant if we swap ${i}$ and ${j}$. This means that the Christoffel symbols are symmetric under exchange of their two lower indices:

$\displaystyle \boxed{\Gamma_{ij}^{k}=\Gamma_{ji}^{k}} \ \ \ \ \ (9)$

At first glance, this seems wrong, since from the definition 1 this symmetry implies that

$\displaystyle \frac{\partial\mathbf{e}_{i}}{\partial x^{j}}=\frac{\partial\mathbf{e}_{j}}{\partial x^{i}} \ \ \ \ \ (10)$

In 2-D polar coordinates, if we take the usual unit vectors ${\hat{\mathbf{r}}}$ and ${\hat{\boldsymbol{\theta}}}$ then both these vectors are constants as we change ${r}$ and both of them change when we change ${\theta}$, so it’s certainly not true that ${\partial\hat{\mathbf{r}}/\partial\theta=\partial\hat{\boldsymbol{\theta}}/\partial r}$, for example. However, remember that the basis vectors we’re using are not the usual unit vectors; rather they are defined so that condition 4 is true. In polar coordinates, we have

$\displaystyle ds^{2}=dr^{2}+r^{2}d\theta^{2} \ \ \ \ \ (11)$

so

 $\displaystyle \mathbf{e}_{r}$ $\displaystyle =$ $\displaystyle \hat{\mathbf{r}}=\cos\theta\hat{\mathbf{x}}+\sin\theta\hat{\mathbf{y}}\ \ \ \ \ (12)$ $\displaystyle \mathbf{e}_{\theta}$ $\displaystyle =$ $\displaystyle r\hat{\boldsymbol{\theta}}=-r\sin\theta\hat{\mathbf{x}}+r\cos\theta\hat{\mathbf{y}} \ \ \ \ \ (13)$

For the derivatives, we have

 $\displaystyle \frac{\partial\mathbf{e}_{r}}{\partial\theta}$ $\displaystyle =$ $\displaystyle -\sin\theta\hat{\mathbf{x}}+\cos\theta\hat{\mathbf{y}}=\hat{\boldsymbol{\theta}}\ \ \ \ \ (14)$ $\displaystyle \frac{\partial\mathbf{e}_{\theta}}{\partial r}$ $\displaystyle =$ $\displaystyle -\sin\theta\hat{\mathbf{x}}+\cos\theta\hat{\mathbf{y}}=\hat{\boldsymbol{\theta}} \ \ \ \ \ (15)$

Thus the condition 10 is actually satisfied here.