# Covariant derivative of the Monge-Ampere equation on Kähler manifolds

I am reading D. Joyce book "Compact manifolds with special holonomy" and I have some problems of understanding some computation on page 111, the first line in the proof of Proposition 5.4.6. More specific the following:

Let $$(M,\omega, J)$$ be a compact Kähler manifold with Kähler form $$\omega$$ and complex structure $$J$$. In holomorphic coordinates $$\omega$$ is of the form $$\omega = ig_{\alpha \overline{\beta}}dz^{\alpha} \wedge d\overline{z}^{\beta}$$. Associated to the above data we have the Riemannian metric $$g$$ which may be written in holomorphic coordinates as $$g=g_{\alpha \overline{\beta}}(dz^{\alpha}\otimes d\overline{z}^{\beta} + d\overline{z}^{\beta} \otimes dz^{\alpha})$$. Associated to $$g$$ let $$\nabla$$ be the Levi-Civita connection which also defines a covariant derivative on tensors. For a function $$\phi$$ on $$M$$ one may compute $$\nabla^{k}\phi$$. For example $$\nabla \phi = (\nabla_{\lambda}\phi)dz^{\lambda} + (\nabla_{\overline{\lambda}}\phi)d\overline{z}^{\lambda}=(\partial_{\lambda}\phi)dz^{\lambda} + (\partial_{\overline{\lambda}}\phi)d\overline{z}^{\lambda}$$ (once applied on functions is as the usual $$d$$) and $$\nabla_{\alpha \beta}\phi = \partial_{\alpha \beta} \phi - \partial_{\gamma}\phi \Gamma^{\gamma}_{\alpha \beta}$$, $$\nabla_{\alpha \overline{\beta}}\phi = \partial_{\alpha \overline{\beta}}\phi$$ etc.

In the first sentence of the proof of proposition 5.4.6 Joyce considers the equation $$\det(g_{\alpha \overline{\beta}} + \partial_{\alpha \overline{\beta}}\phi) = e^{f}\det(g_{\alpha \overline{\beta}})$$, where $$f:M\rightarrow \mathbb{R}$$ is a smooth function on $$M$$. After taking the $$\log$$ of this equation he obtains $$\log[\det(g_{\alpha \overline{\beta}} + \partial_{\alpha \overline{\beta}}\phi)] - \log[\det(g_{\alpha \overline{\beta}} )] = f$$ which is obviously a globaly defined equality of functions on $$M$$. Now he takes the covariant derivative $$\nabla$$ of this equation and obtains $$\nabla_{\overline{\lambda}}f = g'^{\mu \overline{\nu}}\nabla_{\overline{\lambda} \mu \overline{\nu}}\phi$$ where $$g'^{\mu \overline{\nu}}$$ is the inverse of the metric $$g'_{\alpha \overline{\beta}} = g_{\alpha \overline{\beta}} + \partial_{\alpha \overline{\beta}}\phi$$ (which he assumes to exists). This last step (when taking the covariant derivative) I do not understant.

In my computation I have the following: When taking the covariant derivative $$\nabla_{\overline{\lambda}}$$ of the equation $$\log[\det(g_{\alpha \overline{\beta}} + \partial_{\alpha \overline{\beta}}\phi)] - \log[\det(g_{\alpha \overline{\beta}} )] = f$$ and using the formula for the derivative of the determinant I obtain $$g'^{\alpha \overline{\beta}}(\partial_{\overline{\lambda}}g_{\alpha \overline{\beta}} + \partial_{\overline{\lambda} \alpha \overline{\beta}}\phi) - g^{\alpha \overline{\beta}}(\partial_{\overline{\lambda}}g_{\alpha \overline{\beta}}) = \partial_{\overline{\lambda}}f = \nabla_{\overline{\lambda}}f$$. This is obviously different to his formula. Moreover the term $$\nabla_{\overline{\lambda}\mu \overline{\nu}}\phi$$ contains not only derivatives of order $$3$$ of $$\phi$$ but it also contains a term with second derivatives of $$\phi$$.

My question is: Where is my mistake? Have I understood something wrong?

• Both expressions are in fact equal! Instead of "This is obviously different to his formula", I would say that "This is non-obviously equal to his formula". It is a very instructive exercise for you. – YangMills Apr 19 at 21:54