A matrix manipulation from Berezin's paper

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.

• It should follow from the fact that $d(\omega^{ij}\omega_{jk})=0$. The main issue is how was (1.1) derived (perhaps footnote 1 explains it)? Commented Feb 16 at 10:54
• @TedBlack (1.1) is taken as the definition. The footnote is just a comment on how the repeated index notation will stand for sum. Btw, I did try doing that already, and didn't seem to get anywhere... :/
– Atom
Commented Feb 16 at 10:56

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, $$$$\omega_{i\alpha}\omega_{j\beta}\frac{\partial \omega^{\alpha\beta}}{\partial x^l} +\omega_{i\alpha}\omega_{l\gamma}\frac{\partial \omega^{\gamma\alpha}}{\partial x^j} +\omega_{j\beta}\omega_{l\gamma}\frac{\partial \omega^{\beta\gamma}}{\partial x^i}=0 \tag{I} \label{I}$$$$ Since, $$\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).
• There was a mistake in last derivation; somehow I managed to insert the tensor product in the partials. This is not the case; instead we have to use $\omega_{i\alpha}\omega^{\alpha\beta}=\delta^\beta_i$ to derive a relation between the partial of the covariant and contravariant tensors. Commented Feb 16 at 17:41