- #1

nonequilibrium

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But that would mean [itex]\bar \partial \omega = \frac{i}{2} \sum \frac{\partial h_{ij}}{\partial \bar z_k} d \bar z^k \wedge dz^i \wedge d z^j = 0[/itex] since [itex]\frac{\partial h_{ij}}{\partial \bar z_k} = \mathcal O(z)[/itex] is zero at our point [itex]z = 0[/itex].

I must be making a stupid mistake. What is it?