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If there are 2-forms $\omega_1$ and $\omega_2$ in the manifold $M_1$ and $M_2$ respectively, does saying something like $[\omega_1]=[\omega_2]$ make sense if $M_1$ and $M_2$ don't have the same almost-complex structure, where $[\omega_1]$ denotes the cohomology class of $\omega_1$?

Additional info: We have $(M,\omega,J_1)$ and $(M,\omega,J_2)$, where $J_i$ is the almost-complex structure and $J_1$ and $J_2$ are homotopic by a 1-parameter family of diffeomorphisms $\varphi_t:M\rightarrow M$. The homotopy does not necessarily preserve $J_1$. $\omega$ is symplectic and $J_i$ is $\omega$- tame. Then we blow-up symplectically, and obtain respectively $\tilde{\omega}_1,\tilde{J}_1$ and $\tilde{\omega_2},\tilde{J}_2$, where $\tilde{J}_i$ is $\tilde{\omega}_i$-tame. Why are these two 2-forms not necessarily cohomologous?

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Gunthar- This question doesn't make any sense as written. Are $M_1$ and $M_2$ diffeomorphic? –  Ben Webster Feb 2 '11 at 16:36
Yes they are diffeomorphic. –  Gunthar Feb 2 '11 at 16:37
Saying $[\omega_1]=[\omega_2]$ does not make sense at all if $\omega_1$ and $\omega_2$ live on different manifolds. What is the role of the almost-complex structure? Maybe you state your question more precisely. –  Johannes Ebert Feb 2 '11 at 16:39
The edited question still doesn't make sense (surely one needs to fix a diffeomorphism between the two manifolds?). But something that may be relevant is that the symplectic blowup process has a parameter, the volume of the exceptional divisor. –  Tim Perutz Feb 2 '11 at 17:05
I think you are confusing de-Rham cohomology with complex coefficients and Dolbeaut cohomology. If you have two complex forms on a manifold $M$, they are maps $\omega_{1}, \omega_{2}: TM \wedge TM \rightarrow \mathbb{C}$, which are only $\mathbb{R}$-linear. In this case, there is no reference to the a.c. structure, and they can be cohomologous or not. Another thing is if you compute Dolbeaut cohomology, i.e. the cohomology of the operator $\overline{\partial}$: in this case you must fix the complex structure, since the operator $\overline{\partial}$ depends on it –  Fabio Feb 2 '11 at 18:17

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