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What is a smooth two dimensional real vector bundle $E$ over $S^{2}$ such that the total space $E$, as a smooth four manifold, does not admit a symplectic structure?

To what extent all even dimensional smooth bundles over $S^{2}$ with a symplectic structure are classified? Assume that $E,F$ are two such vector bundles. Does $E\oplus F$ or $E\otimes F$ admit a symplectic structure, too?

The question is motivated by existence of a symplectic structure on the the cotangent bundle of manifolds.

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    $\begingroup$ Every real rank 2 vector bundle over the 2-sphere admits a complex structure, and is an algebraic line bundle over the complex projective line, so is a quasiprojective variety and admits a symplectic structure. $\endgroup$
    – Ben McKay
    Jul 7, 2016 at 13:03
  • $\begingroup$ @BenMcKay Thanks a lot. Is there a proof without algebraic geometry?Can you give an elementary reference about the algebraic geometric aspect of your comment? Do you think that it would be obvious to ask :"Is there a symplectic structure such that each fibre is a Lagrangian submanifold? $\endgroup$
    – Ali Taghavi
    Jul 8, 2016 at 8:40

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