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Simon Donaldson apparently made the following conjecture: Two closed symplectic 4-manifolds $(X_1,\omega_1)$ and $(X_2,\omega_2)$ are diffeomorphic if and only if $(X_1\times S^2,\omega_1\oplus\omega) …

As an honest question (probably with some subjectivity), how many smooth oriented 4-manifolds are actually symplectic? Can I say half (perhaps under some mild assumptions)? I ask this question because …

The 4-sphere cannot be a symplectic manifold. In particular, it does not admit an atlas whose transition maps are symplectomorphisms $(\mathbb{R}^4,\omega_\text{std})\to(\mathbb{R}^4,\omega_\text{std} …

Given any manifold $M$, we can get a symplectic manifold by taking the cotangent bundle $T^\ast M$ with symplectic form $\omega=\sum dp_i\wedge dq_i$. Given any manifold $M$, we can get a contact man …

Consider a symplectic manifold $(M,\omega)$ and the space of almost complex structures. These are $J:TM\to TM$ with $J^2=-\text{id}$. A given $J$ is $\omega$-tame when $\omega(v,Jv)>0$, and $J$ is $\o …

Consider a smooth compact oriented 4-manifold $X$. Although not all 4-manifolds admit a spin structure, they do admit spin-c structures. And if $X$ does admit a spin structure, then there is a canonic …

Given a Lagrangian submanifold $L\subset(M,\omega)$ of a symplectic manifold, we have Alan Weinstein's celebrated Lagrangian tubular neighborhood theorem. I now look for the analog on Legendrian subma …

Let $(Y,\lambda)$ be a contact manifold, with a codimension-2 contact submanifold $(S,\lambda|_S)$ (this requires $TS\pitchfork\text{Ker}\lambda$). On $Y$ there is a natural vector field, the Reeb fie …