Skip to main content
MathOverflow

Return to Answer

Added "orientable"
Source Link
Tim Perutz
Tim Perutz
  • 13.2k
  • 1
  • 53
  • 79

The cotangent bundle to $T^2$ is $(\mathbb{C}^\ast)^2$. But a theorem of Totaro (Internat. J. Math. 2 (1991), 563-566; MR1124283) implies that, if $S$ is a closed orientable surface of negative Euler characteristic, there is no diffeomorphism, sending the symplectic orientation to the complex orientation, from $T^*S$ to a smooth affine complex surface.

The cotangent bundle to $T^2$ is $(\mathbb{C}^\ast)^2$. But a theorem of Totaro (Internat. J. Math. 2 (1991), 563-566; MR1124283) implies that, if $S$ is a closed surface of negative Euler characteristic, there is no diffeomorphism, sending the symplectic orientation to the complex orientation, from $T^*S$ to a smooth affine complex surface.

The cotangent bundle to $T^2$ is $(\mathbb{C}^\ast)^2$. But a theorem of Totaro (Internat. J. Math. 2 (1991), 563-566; MR1124283) implies that, if $S$ is a closed orientable surface of negative Euler characteristic, there is no diffeomorphism, sending the symplectic orientation to the complex orientation, from $T^*S$ to a smooth affine complex surface.

Source Link
Tim Perutz
Tim Perutz
  • 13.2k
  • 1
  • 53
  • 79

The cotangent bundle to $T^2$ is $(\mathbb{C}^\ast)^2$. But a theorem of Totaro (Internat. J. Math. 2 (1991), 563-566; MR1124283) implies that, if $S$ is a closed surface of negative Euler characteristic, there is no diffeomorphism, sending the symplectic orientation to the complex orientation, from $T^*S$ to a smooth affine complex surface.