Skip to main content
MathOverflow

Return to Answer

Bounty Ended with 100 reputation awarded by user3566
deleted 9 characters in body
Source Link
Misha Verbitsky
Misha Verbitsky
  • 9.2k
  • 1
  • 28
  • 48

Yes. For the proof, see e.g. http://arxiv.org/abs/math/0609617 Theorem 4.1. Here a stronger result is actually proven:

Theorem: Let $(M, \omega)$ be a compact Kahler manifold, and $Z\subset M$ a closed complex submanifold. Denote by $[\omega]\in H^2(M)$ the Kahler class of $M$. Consider a Kahler form $\omega_0$ on $Z$ such that its Kahler class coinsides with the restriction $[\omega]| Z$. Then there exists a Kahler form $\omega'$ on $M$ in the same Kahler class as $\omega$, such that $\omega'\restrict Z=\omega_0$$\omega| Z=\omega_0$.

An additional cohomological assumption is needed, because we build a global extension, and for an extension to a local neighbourhood you don't need it; the positivity (needed in assumption) is achieved by adding a big multiple of a Kaehler form.

Yes. For the proof, see e.g. http://arxiv.org/abs/math/0609617 Theorem 4.1. Here a stronger result is actually proven:

Theorem: Let $(M, \omega)$ be a compact Kahler manifold, and $Z\subset M$ a closed complex submanifold. Denote by $[\omega]\in H^2(M)$ the Kahler class of $M$. Consider a Kahler form $\omega_0$ on $Z$ such that its Kahler class coinsides with the restriction $[\omega]| Z$. Then there exists a Kahler form $\omega'$ on $M$ in the same Kahler class as $\omega$, such that $\omega'\restrict Z=\omega_0$.

An additional cohomological assumption is needed, because we build a global extension, and for an extension to a local neighbourhood you don't need it; the positivity (needed in assumption) is achieved by adding a big multiple of a Kaehler form.

Yes. For the proof, see e.g. http://arxiv.org/abs/math/0609617 Theorem 4.1. Here a stronger result is actually proven:

Theorem: Let $(M, \omega)$ be a compact Kahler manifold, and $Z\subset M$ a closed complex submanifold. Denote by $[\omega]\in H^2(M)$ the Kahler class of $M$. Consider a Kahler form $\omega_0$ on $Z$ such that its Kahler class coinsides with the restriction $[\omega]| Z$. Then there exists a Kahler form $\omega'$ on $M$ in the same Kahler class as $\omega$, such that $\omega| Z=\omega_0$.

An additional cohomological assumption is needed, because we build a global extension, and for an extension to a local neighbourhood you don't need it; the positivity (needed in assumption) is achieved by adding a big multiple of a Kaehler form.

Source Link
Misha Verbitsky
Misha Verbitsky
  • 9.2k
  • 1
  • 28
  • 48

Yes. For the proof, see e.g. http://arxiv.org/abs/math/0609617 Theorem 4.1. Here a stronger result is actually proven:

Theorem: Let $(M, \omega)$ be a compact Kahler manifold, and $Z\subset M$ a closed complex submanifold. Denote by $[\omega]\in H^2(M)$ the Kahler class of $M$. Consider a Kahler form $\omega_0$ on $Z$ such that its Kahler class coinsides with the restriction $[\omega]| Z$. Then there exists a Kahler form $\omega'$ on $M$ in the same Kahler class as $\omega$, such that $\omega'\restrict Z=\omega_0$.

An additional cohomological assumption is needed, because we build a global extension, and for an extension to a local neighbourhood you don't need it; the positivity (needed in assumption) is achieved by adding a big multiple of a Kaehler form.