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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?

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1 Answer 1

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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.

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Is there an example of such an immersion, that is Lagrangian and totally complex? Is Kaehler immersion same as totally complex? In the paper above everything is pure theory with no examples. –  Novak Djokovic May 1 '13 at 3:54

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