Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 non-degenerate if $\phi ^{*}\gamma$ is non-degenerate. If $\phi$ is non-degenerate $\phi^{*}\gamma$ defines a Kähler metric $g$ on $M$. If, additionally, $\phi$ is a Lagrangian immersion then it induces a torsion-free flat connection $\nabla$ on $M$. These are facts from paper of V. Cortes, Realization of special Kähler 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 non-degenerate?

share|cite|improve this question
up vote 2 down vote accepted

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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.