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Misha Verbitsky
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Question 1. It's not hard to see that $F$ is $\Sigma \times \C P^1$$\Sigma \times {\Bbb C} P^1$ and $\Sigma$ is a torus. Therefore the Albanese map is a projection to a torus.

Question 2. If $f^* \delta=-\delta$, then the effective cycle $f^* \delta+\delta$ is homologous to 0. On a Kahler manifold this is impossible, because an integral of a Kahler form over an effective cycle is always positive.

Question 1. It's not hard to see that $F$ is $\Sigma \times \C P^1$ and $\Sigma$ is a torus. Therefore the Albanese map is a projection to a torus.

Question 2. If $f^* \delta=-\delta$, then the effective cycle $f^* \delta+\delta$ is homologous to 0. On a Kahler manifold this is impossible, because an integral of a Kahler form over an effective cycle is always positive.

Question 1. It's not hard to see that $F$ is $\Sigma \times {\Bbb C} P^1$ and $\Sigma$ is a torus. Therefore the Albanese map is a projection to a torus.

Question 2. If $f^* \delta=-\delta$, then the effective cycle $f^* \delta+\delta$ is homologous to 0. On a Kahler manifold this is impossible, because an integral of a Kahler form over an effective cycle is always positive.

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

Question 1. It's not hard to see that $F$ is $\Sigma \times \C P^1$ and $\Sigma$ is a torus. Therefore the Albanese map is a projection to a torus.

Question 2. If $f^* \delta=-\delta$, then the effective cycle $f^* \delta+\delta$ is homologous to 0. On a Kahler manifold this is impossible, because an integral of a Kahler form over an effective cycle is always positive.