I wish to know if this PDE can be solved (for a real smooth function $\rho$) on a compact complex surface X : $\bar{\partial}\partial \rho \wedge \bar{\partial}\partial \rho + \bar{\partial}\partial \rho \wedge \bar{\partial}\partial \sigma + \bar{\partial}\partial \sigma \wedge \bar{\partial}\partial \sigma = 0$ where $\sigma$ is a given smooth real function on X. (Note that, one has solutions to this in $\mathbb{C}^2$ if $\sigma$ is real analytic). The problem is that the linearisation of this equation is not elliptic. The motivation for solving this is to produce a trivial line bundle with a hermitian metric so that its Chern character forms (not cohomology classes) are zero.
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$\begingroup$ Could you explain the notations ? You are from the community of complex analysis, and people from classical PDEs may not understand. Thanks. $\endgroup$– Denis SerreCommented Oct 1, 2010 at 7:15
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$\begingroup$ Dennis, $\partial (\omega _{IJ} dz^{I} \wedge dz^{\bar{J}} = \frac{\partial \omega _{IJ}}{\partial z^k} dz^k \wedge dz^{I} \wedge dz^{\bar{J}}$ and $\bar{\partial} (\omega _{IJ} dz^{I} \wedge dz^{\bar{J}} = \frac{\partial \omega _{IJ}}{\partial \bar{z}^k} d\bar{z}^k \wedge dz^{I} \wedge dz^{\bar{J}}$ $\endgroup$– VamsiCommented Oct 1, 2010 at 15:03
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$\begingroup$ Einstein summation implied above. Also $\frac{\partial}{\partial z} = \frac{\partial}{\partial x} - i\frac{\partial}{\partial y}$ $\endgroup$– VamsiCommented Oct 1, 2010 at 15:05
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