I am reading a paper by Berezin, entitled General Concept of Quantization. He writes:
and then:
This ought to be some clever manipulation of matrices and yet I am not able to show how (1.4) follows from (1.1) if $\omega$ is given to be invertible.
Any leads?
If we multiply, $$\omega^{\gamma k}\frac{\partial \omega^{\alpha\beta}}{\partial x^k} +\omega^{\beta k}\frac{\partial \omega^{\gamma\alpha}}{\partial x^k} +\omega^{\alpha k}\frac{\partial \omega^{\beta\gamma}}{\partial x^k}=0$$ by $\omega_{i\alpha}\omega_{j\beta}\omega_{l\gamma}$ we get, $$ \omega_{i\alpha}\omega_{j\beta}\omega_{l\gamma}\omega^{\gamma k}\frac{\partial \omega^{\alpha\beta}}{\partial x^k} +\omega_{i\alpha}\omega_{j\beta}\omega_{l\gamma}\omega^{\beta k}\frac{\partial \omega^{\gamma\alpha}}{\partial x^k} +\omega_{i\alpha}\omega_{j\beta}\omega_{l\gamma}\omega^{\alpha k}\frac{\partial \omega^{\beta\gamma}}{\partial x^k}=0 $$ Since $\omega_{l\gamma}\omega^{\gamma k}=\delta^k_l$, $\omega_{j\beta}\omega^{\beta k}=\delta^k_j$, $\omega_{i\alpha}\omega^{\alpha k}=\delta^k_i$ we have, $$ \omega_{i\alpha}\omega_{j\beta}\delta^k_l\frac{\partial \omega^{\alpha\beta}}{\partial x^k} +\omega_{i\alpha}\omega_{l\gamma}\delta^k_j\frac{\partial \omega^{\gamma\alpha}}{\partial x^k} +\omega_{j\beta}\omega_{l\gamma}\delta^k_i\frac{\partial \omega^{\beta\gamma}}{\partial x^k}=0 $$ or, \begin{equation} \omega_{i\alpha}\omega_{j\beta}\frac{\partial \omega^{\alpha\beta}}{\partial x^l} +\omega_{i\alpha}\omega_{l\gamma}\frac{\partial \omega^{\gamma\alpha}}{\partial x^j} +\omega_{j\beta}\omega_{l\gamma}\frac{\partial \omega^{\beta\gamma}}{\partial x^i}=0 \tag{I} \label{I} \end{equation} Since, $$ \omega_{i\alpha}\omega^{\alpha \beta} = \delta^\beta_i $$ we have, $$ \omega_{i\alpha} \frac{\partial \omega^{\alpha \beta}}{\partial x^l} + \omega^{\alpha \beta} \frac{\partial \omega_{i\alpha}}{\partial x^l}=0 \Rightarrow \omega_{i\alpha} \frac{\partial \omega^{\alpha \beta}}{\partial x^l}=- \omega^{\alpha \beta} \frac{\partial \omega_{i\alpha}}{\partial x^l} $$ In the same way, $$ \omega_{l\gamma} \frac{\partial \omega^{\gamma \alpha}}{\partial x^j}=- \omega^{\gamma\alpha} \frac{\partial \omega_{l\gamma}}{\partial x^j} $$ and, $$ \omega_{j\beta} \frac{\partial \omega^{\beta \gamma}}{\partial x^i}=- \omega^{\beta\gamma} \frac{\partial \omega_{j\beta}}{\partial x^i} $$ So we can rewrite (\ref{I}) as, $$ -\omega_{j\beta}\omega^{\alpha\beta} \frac{\partial \omega_{i\alpha}}{\partial x^l} -\omega_{i\alpha}\omega^{\gamma\alpha} \frac{\partial \omega_{l\gamma}}{\partial x^j} -\omega_{l\gamma}\omega^{\beta\gamma} \frac{\partial \omega_{j\beta}}{\partial x^i} =0 $$ Using the skew-symmetric property of $\omega$ we have, $$ \omega_{j\beta}\omega^{\beta\alpha} \frac{\partial \omega_{i\alpha}}{\partial x^l} +\omega_{i\alpha}\omega^{\alpha\gamma} \frac{\partial \omega_{l\gamma}}{\partial x^j} +\omega_{l\gamma}\omega^{\gamma\beta} \frac{\partial \omega_{j\beta}}{\partial x^i} =0 $$ and since $\omega_{j\beta}\omega^{\beta\alpha}=\delta^\alpha_j$, $\omega_{i\alpha}\omega^{\alpha\gamma}=\delta^\gamma_i$, $\omega_{l\gamma}\omega^{\gamma\beta}=\delta^\beta_l$ we have, $$ \delta^\alpha_j \frac{\partial \omega_{i\alpha}}{\partial x^l} +\delta^\gamma_i \frac{\partial \omega_{l\gamma}}{\partial x^j} +\delta^\beta_l \frac{\partial \omega_{j\beta}}{\partial x^i} =0 $$ or, $$ \frac{\partial \omega_{ij}}{\partial x^l} + \frac{\partial \omega_{li}}{\partial x^j} + \frac{\partial \omega_{jl}}{\partial x^i} =0 $$ Replace $l$ with $k$: $$ \frac{\partial \omega_{ij}}{\partial x^k} + \frac{\partial \omega_{ki}}{\partial x^j} + \frac{\partial \omega_{jk}}{\partial x^i} =0 $$ to get (1.4).