2
$\begingroup$

I know the trace of the Riemann tensor is defined by contracting over the first and the third indices (equivalent to the second and the fourth), and the trace over the first two or the last two indices give zero. But what about contracting over the first and the fourth indices? I can't find a symmetry argument that makes this vanish.

$\endgroup$
0
5
$\begingroup$

From Wikipedia: $$ R_{abcd}=-R_{abdc} $$ $$ \Rightarrow g^{ad}R_{abcd} \\=-g^{ad}R_{abdc} \\=-R_{bc} $$

Here's a table of all the possible contractions, all derived using three facts:

  • The definition of the Ricci tensor: $R_{bd}=g^{ac}R_{abcd}$
  • The antisymmetry of the Riemann tensor in its first two and last two indices: $R_{abcd}=-R_{bacd}=-R_{abdc}$
  • When totally symmetric indices are contracted with totally antisymmetric indices, the result is zero: $S^{(ab)}A_{[ab]}=0$. Here the metric is the symmetric tensor and the first two and last two indices of the Riemann are antisymmetric.
Contraction Result
$g^{ab}R_{abcd}$ $0$
$g^{ac}R_{abcd}$ $R_{bd}$
$g^{ad}R_{abcd}$ $-R_{bc}$
$g^{bc}R_{abcd}$ $-R_{ad}$
$g^{bd}R_{abcd}$ $R_{ac}$
$g^{cd}R_{abcd}$ $0$
$\endgroup$
1
  • 1
    $\begingroup$ Thank you so much! $\endgroup$ – jane886 Mar 29 at 7:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.