Proof. [05CC]

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Proof.

First we note that the local potential description of $V$ and $W$ by (1) insures that

(5)

dF_{j}=\frac{\sqrt{-1}}{2}\cdot d\left(\frac{\partial W^{pq}}{\partial u_{j}}\right)\wedge d\eta_{p}\wedge d\bar{\eta}_{q}\\ +\sqrt{-1}\cdot d\left(\frac{\partial V^{ij}}{\partial\eta_{p}}\right)\wedge du_{i}\wedge d\eta_{p}-\sqrt{-1}\cdot d\left(\frac{\partial V^{ij}}{\partial\bar{\eta}_{q}}\right)\wedge du_{i}\wedge d\bar{\eta}_{q}\\ =\frac{\sqrt{-1}}{2}\left(\frac{\partial^{2}W^{pq}}{\partial u_{i}\partial u_{j}}+4\frac{\partial^{2}V^{ij}}{\partial\eta_{p}\partial\bar{\eta}_{q}}\right)du_{i}\wedge d\eta_{p}\wedge d\bar{\eta}_{q}=0.

Moreover, if $\theta_{j}$ denote the coordinates on the torus fiber such that $\partial/\partial\theta_{j}$ are the Hamiltonian vector fields, then the connection 1-forms can be written up to exact forms on $B^{\circ}$ in terms of the local potential $\Phi$ :

(6)

A_{j}=d\theta_{j}+\sqrt{-1}\left(\frac{\partial^{2}\Phi}{\partial u_{j}\partial\eta_{p}}d\eta_{p}-\frac{\partial^{2}\Phi}{\partial u_{j}\partial\bar{\eta}_{q}}d\bar{\eta}_{q}\right).

And one can see explicitly that $F_{j}=dA_{j}$ .

The integrability of the complex structure follows from the fact that the differential ideal generated by $(1,0)$ -forms is closed:

(7)

d(dz_{j})=dV^{ij}\wedge du_{i}+\sqrt{-1}\cdot dA_{j}\\ =-\frac{\partial V^{ij}}{\partial u_{k}}du_{i}\wedge du_{k}-\frac{\partial V^{ij}}{\partial\eta_{p}}du_{i}\wedge d\eta_{p}-\frac{\partial V^{ij}}{\partial\bar{\eta}_{q}}du_{i}\wedge d\bar{\eta}_{q}\\ -\left(\frac{1}{2}\frac{\partial W^{pq}}{\partial u_{j}}d\eta_{p}\wedge d\bar{\eta}_{q}+\frac{\partial V^{ij}}{\partial\eta_{p}}du_{i}\wedge d\eta_{p}-\frac{\partial V^{ij}}{\partial\bar{\eta}_{q}}du_{i}\wedge d\bar{\eta}_{q}\right)\\ =\left(\frac{1}{2}\frac{\partial W^{pq}}{\partial u_{j}}d\bar{\eta}_{q}-2\frac{\partial V^{ij}}{\partial\eta_{p}}du_{i}\right)\wedge d\eta_{p},

where we have only used $\frac{\partial V^{ij}}{\partial u_{k}}=\frac{\partial V^{kj}}{\partial u_{i}}$ .

It is equally easy to verify the Kähler condition:

(8)

d\omega=-du_{j}\wedge dA_{j}+\frac{\sqrt{-1}}{2}\cdot dW^{pq}\wedge d\eta_{p}\wedge d\bar{\eta}_{q}\\ =-\sqrt{-1}\left(\frac{1}{2}\frac{\partial W^{pq}}{\partial u_{j}}du_{j}\wedge d\eta_{p}\wedge d\bar{\eta}_{q}\right)+\frac{\sqrt{-1}}{2}\frac{\partial W^{pq}}{\partial u_{j}}du_{j}\wedge d\eta_{p}\wedge d\bar{\eta}_{q}=0.

Finally, the Ricci-flatness is manifest since $\det(h)=\det V^{-1}\cdot\det W=1$ in the complex coordinates $dz_{j},d\eta_{p}$ . ∎