Special Kaehler manifolds

Question

If we have complex vector space $V=T^{*}C^{m}$ with standard complex symplectic form $\Omega =\sum_{i=1}^{m}dz^{i}\wedge dw^{i}$, and if $\tau : V\to V$ is standard real structure of $V$ with set of fixed points $V^{\tau }=T^{*}R^{m}$. Then $\gamma =\sqrt{-1}\Omega (.,\tau .)$ defines a Hermitian form. A holomorphic immersion $\phi : M\to V$ of a complex manifold $M$ into $V$ is called nondegenerate if $\phi ^{*}\gamma$ is nondegenerate. If $\phi$ is nondegenerate $\phi^{*}\gamma$ defines a Kaehler metric $g$ on $M$. If, additionaly, $\phi$ is a Lagrangian immersion then it induces a torsionfree flat connection $\nabla$ on $M$. These are facts from paper of V. Cortes, Realization of special Kaehler manifolds as parabolic spheres. So, I tried to understand them by using the simplest example where $m = 2$ but unsuccessfully.

My question is how we get metric $g$ and connection $\nabla$ on $M$, and what means that $\phi^{*}\gamma$ is nondegenerate?

Answer 1

Your questions are answered in section 1.3 of this paper. The basic ideas are as follows.

The form $\phi^\ast \gamma$ is just the pullback of $\gamma$ by $\phi$: For any $m \in M$ and vectors $u, v \in T_m M$, $$(\phi^\ast \gamma)_m(u, v) = \gamma(d\phi_m(u), d\phi_m(v)).$$ So nondegeneracy of $\phi^\ast \gamma$ means nondegeneracy as a form, i.e. $$(\phi^\ast \gamma)(u, v) = 0 \text{ for all $v$ if and only if $u = 0$}.$$

The induced metric $g$ on $M$ is $g = \mathrm{Re}(\phi^\ast \gamma)$.

Section 1.3 of the linked paper explains how to get a flat, torsion-free connection $\nabla$ on $M$ in the case that $\phi$ is a totally complex holomorphic immersion. Proposition 6 tells us that a holomorphic immersion is Lagrangian and nondegenerate if and only if it is Lagrangian and totally complex. Hence we can apply the totally complex case here to get our flat, torsion-free connection.