Riemann tensor – commutator of rank 2 tensor

Required math: algebra, calculus

Required physics: none

Reference: d’Inverno, Ray, Introducing Einstein’s Relativity (1992), Oxford Uni Press. – Section 6.5; Problem 6.10.

The covariant derivative of a contravariant vector is defined as

$\displaystyle \nabla_{b}V^{a}\equiv V_{\;;b}^{a}\equiv\frac{\partial V^{a}}{\partial x^{b}}+V^{c}\Gamma_{cb}^{a} \ \ \ \ \ (1)$

This is generalized to the covariant derivative of a higher-rank tensor by the formula

$\displaystyle T_{cd...;e}^{ab...}=\partial_{e}T_{cd...}^{ab...}+T_{cd...}^{fb...}\Gamma_{fe}^{a}+T_{cd...}^{af...}\Gamma_{fe}^{b}+...-T_{fd...}^{ab...}\Gamma_{ce}^{f}-T_{cf...}^{ab...}\Gamma_{de}^{f}-... \ \ \ \ \ (2)$

Ordinary partial derivatives, for a continuously differentiable function ${f\left(x^{a}\right)}$ , are commutative, that is

$\displaystyle \frac{\partial}{\partial x^{b}}\left(\frac{\partial f}{\partial x^{a}}\right)=\frac{\partial}{\partial x^{a}}\left(\frac{\partial f}{\partial x^{b}}\right)$

The covariant derivative, however, is not in general commutative, as we can verify by direct calculation. We want to find

$\displaystyle X_{\;\; b;c;d}^{a}-X_{\;\; b;d;c}^{a}$
which is known as the commutator of the tensor ${X_{\;\; b}^{a}}$ . For the first term, we get, using 2

$\displaystyle X_{\;\; b;c;d}^{a}$	$\displaystyle =$	$\displaystyle \partial_{d}X_{\;\; b;c}^{a}+X_{\;\; b:c}^{e}\Gamma_{ed}^{a}-X_{\;\; e:c}^{a}\Gamma_{bd}^{e}-X_{\;\; b:e}^{a}\Gamma_{cd}^{e}$
$\displaystyle$	$\displaystyle =$	$\displaystyle \partial_{d}\left(\partial_{c}X_{\;\; b}^{a}+X_{\;\; b}^{e}\Gamma_{ec}^{a}-X_{\;\; e}^{a}\Gamma_{bc}^{e}\right)+$
$\displaystyle$	$\displaystyle$	$\displaystyle \Gamma_{ed}^{a}\left(\partial_{c}X_{\;\; b}^{e}+X_{\;\; b}^{f}\Gamma_{fc}^{e}-X_{\;\; f}^{e}\Gamma_{bc}^{f}\right)-$
$\displaystyle$	$\displaystyle$	$\displaystyle \Gamma_{bd}^{e}\left(\partial_{c}X_{\;\; e}^{a}+X_{\;\; e}^{f}\Gamma_{fc}^{a}-X_{\;\; f}^{a}\Gamma_{ec}^{f}\right)-$
$\displaystyle$	$\displaystyle$	$\displaystyle \Gamma_{cd}^{e}\left(\partial_{e}X_{\;\; b}^{a}+X_{\;\; b}^{f}\Gamma_{fe}^{a}-X_{\;\; f}^{a}\Gamma_{be}^{f}\right)$

The other term can be obtained by simply swapping the indices ${c}$ and ${d}$ :

$\displaystyle X_{\;\; b;d;c}^{a}$	$\displaystyle =$	$\displaystyle \partial_{c}X_{\;\; b;d}^{a}+X_{\;\; b:d}^{e}\Gamma_{ec}^{a}-X_{\;\; e:d}^{a}\Gamma_{bc}^{e}-X_{\;\; b:e}^{a}\Gamma_{dc}^{e}$
$\displaystyle$	$\displaystyle =$	$\displaystyle \partial_{c}\left(\partial_{d}X_{\;\; b}^{a}+X_{\;\; b}^{e}\Gamma_{ed}^{a}-X_{\;\; e}^{a}\Gamma_{bd}^{e}\right)+$
$\displaystyle$	$\displaystyle$	$\displaystyle \Gamma_{ec}^{a}\left(\partial_{d}X_{\;\; b}^{e}+X_{\;\; b}^{f}\Gamma_{fd}^{e}-X_{\;\; f}^{e}\Gamma_{bd}^{f}\right)-$
$\displaystyle$	$\displaystyle$	$\displaystyle \Gamma_{bc}^{e}\left(\partial_{d}X_{\;\; e}^{a}+X_{\;\; e}^{f}\Gamma_{fd}^{a}-X_{\;\; f}^{a}\Gamma_{ed}^{f}\right)-$
$\displaystyle$	$\displaystyle$	$\displaystyle \Gamma_{dc}^{e}\left(\partial_{e}X_{\;\; b}^{a}+X_{\;\; b}^{f}\Gamma_{fe}^{a}-X_{\;\; f}^{a}\Gamma_{be}^{f}\right)$

Now we need to take the difference. Assuming the ordinary partial derivatives commute and using the product rule, we get

$\displaystyle X_{\;\; b;c;d}^{a}-X_{\;\; b;d;c}^{a}$	$\displaystyle =$	$\displaystyle X_{\;\; b}^{e}\left(\partial_{d}\Gamma_{ec}^{a}-\partial_{c}\Gamma_{ed}^{a}\right)-X_{\;\; e}^{a}\left(\partial_{d}\Gamma_{bc}^{e}-\partial_{c}\Gamma_{bd}^{e}\right)+$
$\displaystyle$	$\displaystyle$	$\displaystyle X_{\;\; b}^{f}\left(\Gamma_{ed}^{a}\Gamma_{fc}^{e}-\Gamma_{ec}^{a}\Gamma_{fd}^{e}\right)-X_{\;\; f}^{e}\left(\Gamma_{ed}^{a}\Gamma_{bc}^{f}-\Gamma_{ec}^{a}\Gamma_{bd}^{f}\right)-$
$\displaystyle$	$\displaystyle$	$\displaystyle X_{\;\; e}^{f}\left(\Gamma_{bd}^{e}\Gamma_{fc}^{a}-\Gamma_{bc}^{e}\Gamma_{fd}^{a}\right)+X_{\;\; f}^{a}\left(\Gamma_{bd}^{e}\Gamma_{ec}^{f}-\Gamma_{bc}^{e}\Gamma_{ed}^{f}\right)-$
$\displaystyle$	$\displaystyle$	$\displaystyle \left(\partial_{e}X_{\;\; b}^{a}+X_{\;\; b}^{f}\Gamma_{fe}^{a}-X_{\;\; f}^{a}\Gamma_{be}^{f}\right)\left(\Gamma_{cd}^{e}-\Gamma_{dc}^{e}\right)$

We can now swap the indices ${e}$ and ${f}$ in the first term in the third line (since they are both dummy indices) to get

$\displaystyle X_{\;\; e}^{f}\left(\Gamma_{bd}^{e}\Gamma_{fc}^{a}-\Gamma_{bc}^{e}\Gamma_{fd}^{a}\right)=X_{\;\; f}^{e}\left(\Gamma_{bd}^{f}\Gamma_{ec}^{a}-\Gamma_{bc}^{f}\Gamma_{ed}^{a}\right)$

We can now see that this term cancels the last term on the second line. If we also assume that the affine connections are symmetric, so that $\displaystyle \Gamma_{cd}^{e}=\Gamma_{dc}^{e}$
then the last line disappears and we are left with

$\displaystyle X_{\;\; b;c;d}^{a}-X_{\;\; b;d;c}^{a}$	$\displaystyle =$	$\displaystyle X_{\;\; b}^{e}\left(\partial_{d}\Gamma_{ec}^{a}-\partial_{c}\Gamma_{ed}^{a}\right)-X_{\;\; e}^{a}\left(\partial_{d}\Gamma_{bc}^{e}-\partial_{c}\Gamma_{bd}^{e}\right)+$
$\displaystyle$	$\displaystyle$	$\displaystyle X_{\;\; b}^{e}\left(\Gamma_{fd}^{a}\Gamma_{ec}^{f}-\Gamma_{fc}^{a}\Gamma_{ed}^{f}\right)-X_{\;\; e}^{a}\left(\Gamma_{bc}^{f}\Gamma_{fd}^{e}-\Gamma_{bd}^{f}\Gamma_{fc}^{e}\right)$
$\displaystyle$	$\displaystyle =$	$\displaystyle X_{\;\; b}^{e}\left(\partial_{d}\Gamma_{ec}^{a}-\partial_{c}\Gamma_{ed}^{a}+\Gamma_{fd}^{a}\Gamma_{ec}^{f}-\Gamma_{fc}^{a}\Gamma_{ed}^{f}\right)-$
$\displaystyle$	$\displaystyle$	$\displaystyle X_{\;\; e}^{a}\left(\partial_{d}\Gamma_{bc}^{e}-\partial_{c}\Gamma_{bd}^{e}+\Gamma_{bc}^{f}\Gamma_{fd}^{e}-\Gamma_{bd}^{f}\Gamma_{fc}^{e}\right)$

where again we have swapped ${e}$ and ${f}$ in the second line.

The two terms in parentheses have the same form, and they are known as the Riemann tensor or curvature tensor, defined by

$\displaystyle R_{\;\; edc}^{a}\equiv\partial_{d}\Gamma_{ec}^{a}-\partial_{c}\Gamma_{ed}^{a}+\Gamma_{fd}^{a}\Gamma_{ec}^{f}-\Gamma_{fc}^{a}\Gamma_{ed}^{f}$

In terms of the Riemann tensor, we get for the commutator:

$\displaystyle X_{\;\; b;c;d}^{a}-X_{\;\; b;d;c}^{a}=X_{\;\; b}^{e}R_{\;\; edc}^{a}-X_{\;\; e}^{a}R_{\;\; bdc}^{e}$

This is actually the same result as given in d’Inverno’s problem 6.10, with ${c}$ and ${d}$ swapped around; I just took the original covariant derivatives in the opposite order to d’Inverno and can’t be bothered going through the whole derivation again to change it…

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By Riemann tensor and covariant contraction | Physics tutorials on Monday, 4 March 2013 at 16:24

[...] following through a similar derivation to that given for a rank-2 tensor we find that the factor in the last term can be written in terms of the Riemann [...]

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