Please help me understand how, in this introduction to spacetime and fields, the Einstein Cartan equation: $$C^k_{\hspace{2mm} [ji]}-\delta_{[i}^{k}C^l_{\hspace{2mm} j]l}=\frac{\kappa}{2}s_{ij}^{\hspace{2mm}k}, $$ when the starting variation to the Field equation with respect to the contortion tensor ($C$ being the contortion tensor, $s$ being the spin tensor) is derived to be

$$\frac{-1}{\kappa c}\int (C^{kj}_{\hspace{2mm} i}-C^{lj}_{\hspace{2mm} l}\delta_i^k) \sqrt{-\mathfrak{g}}\delta C^i_{\hspace{2mm}jk}\delta \Omega \hspace{3mm} + \frac{1}{2c}\int s_j^{\hspace{2mm} ik}\sqrt{-\mathfrak{g}}\delta C^j_{\hspace{2mm}ik}\delta\Omega=0,$$

which leads directly to the expression $$C^{kj}_{\hspace{2mm} i}-C^{lj}_{\hspace{2mm} l}\delta_i^k = \frac{\kappa}{2}s_i^{\hspace{2mm}jk}. $$

Is there some form of symmetrisation that can reduce this equation into the familiar form above?


Note that the spin tensor is skew-symmetric in its lower indices, $$ s_{ij}{}^k=-s_{ji}{}^k $$

Therefore, we have $s_{ij}{}^k=s_{[ij]}{}^k$. From this, its easy to see that $$ A_{ij}{}^k=s_{ij}{}^k \quad \Leftrightarrow \quad A_{[ij]}{}^k=s_{ij}{}^k $$ as required.

  • 1
    $\begingroup$ Awesome and simple - why I couldn't see this :) Thanks $\endgroup$ – user110235 Mar 3 '16 at 22:18

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