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Jun 25, 2011 at 18:36 comment added Daniel Pomerleano that's what I was trying to say perhaps not in the clearest way... that's why $S^4$ is not a symplectic manifold.
Jun 25, 2011 at 17:56 comment added agt Dear Daniel Pomerleano, excuse me, but, if $\omega$ is a symplectic form on a compact connected manifold $M$, should not its cohomology class be necessarily nonzero? infact $[\omega]=0$ imply $[\omega^n]=[\omega]^n=0$ and this last contradicts $\int_M\omega\neq 0$. Bye.
Jun 24, 2011 at 11:46 history answered Daniel Pomerleano CC BY-SA 3.0