2
$\begingroup$

I was reading an article about Einstein-Cartan universes and the beginning he defines the contortion tensor as $$ K^a\vphantom{K}_{bc}=S^a\vphantom{S}_{bc}+S_{bc}\vphantom{S}^a+S_{cb}\vphantom{S}^a $$ with $$S^a\vphantom{S}_{bc}=\Gamma^a\vphantom{S}_{[bc]}.$$

Then he writes two equalities that I cannot understand:

$$K_{(a|b|c)}=-2S_{(ac)b}$$ and $${ }K_{[a|b|c]}=-S_{bac}. $$

What are the definition of $K_{(a|b|c)}$ and ${ }K_{[a|b|c]}$?

$\endgroup$
2
$\begingroup$

Obviously, $K_{abc} = g_{ad}{K^d}_{bc}$. And I'm guessing you know that $()$ means symmetrization and $[]$ means antisymmetrization over whatever indices are contained between them.

The vertical bars are there to "suspend" the symmetrization or anti-symmetrization that is happening on the other indices. That is, the $b$ in each case is not involved: \begin{gather} K_{(a|b|c)} = \frac{1}{2} \left( K_{abc} + K_{cba} \right), \\ K_{[a|b|c]} = \frac{1}{2} \left( K_{abc} - K_{cba} \right). \end{gather}

So, for example, it's straightforward to compute \begin{align} K_{(a|b|c)} &= \frac{1}{2} \left( K_{abc} + K_{cba} \right) \\ &= \frac{1}{2} \left( S_{abc} + S_{bca} + S_{cba} + S_{cba} + S_{bac} + S_{abc} \right) \\ &= S_{abc} + S_{b(ca)} + S_{cba} \\ &= S_{abc} + S_{cba} \\ &= -(S_{acb} + S_{cab}) \\ &= -2S_{(ac)b} \end{align} In the third line, I eliminated $S_{b(ca)}$ because it is totally antisymmetric on the last two indices (by definition), so the symmetric part of those two indices is zero. In the fourth line, I swapped the last two indices, and flipped the sign, because of that same antisymmetry.

Similarly, we have \begin{align} K_{[a|b|c]} &= \frac{1}{2} \left( K_{abc} - K_{cba} \right) \\ &= \frac{1}{2} \left( S_{abc} + S_{bca} + S_{cba} - S_{cba} - S_{bac} - S_{abc} \right) \\ &= \frac{1}{2} \left( S_{bca} - S_{bac} \right) \\ &= S_{b[ca]} \\ &= -S_{b[ac]} \\ &= -S_{bac} \end{align} All of these manipulations should become very obvious once you get a little practice with index gymnastics.

$\endgroup$
  • $\begingroup$ Computing the second equation I found out that is equal to $1/2(S_{bca}-S_{bac})$ that differs from $-S_{bac}$. Where am I wrong? $\endgroup$ – raskolnikov Mar 16 '17 at 14:15
  • $\begingroup$ You did the computation right, you just need to go one step further. What you got is the same as $-S_{bac}$. Look at the definition of $S$: it's antisymmetric under exchange of its last two indices. Thus, your first term $S_{bca}$ is equal to $-S_{bac}$. $\endgroup$ – Mike Mar 16 '17 at 14:24
  • 1
    $\begingroup$ Please be careful: index gymnastics with the contorsion tensor may seriously harm your fingers :-) $\endgroup$ – magma Mar 18 '17 at 1:54

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.