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Something I have wondered idly about from time to time is: If $(M,\omega), (M',\omega')$ are symplectic manifolds, and you "know what the Lagrangians $L \subset M$ resp. $L' \subset M'$ are," can …

Counterexample: set $U$ to be anything, $V := \text{pt}$, and $\Lambda \subset \overline{U} \oplus \text{pt}$ to be a non-Lagrangian subspace. (But maybe true with some hypotheses, e.g. $\Lambda$ ind …

One answer to your question "why one would be interested in Lagrangian submanifolds" is to quote Weinstein's Symplectic Creed (from a 1981 article), which says everything is Lagrangian --- i.e. (1) mo …

I think that the Floer complex decomposes as a direct sum of over the conjugacy classes of $\pi_1(M)$, no? Since the manifold on which we are doing infinite-dimensional Morse theory is of unbased, no …

If $\Sigma_g$ is a genus-$g$ surface, $g \geq 2$, then let $\mathcal{M}(\Sigma_g)$ be its twisted SU(2) representation variety, i.e. $$\mathcal{M}(\Sigma_g) := \{ (A_1, B_1, \ldots, A_g, B_g) \in SU(2 …

Here are the two versions of Arnold's conjecture on Hamiltonian orbits: Let $(M,\omega)$ be a closed symplectic manifold, and let $H: \mathbb{R/Z} \times M \to \mathbb{R}$ be a nondegenerate Hamil …

In a short paper posted last week, Peter Albers and Urs Frauenfelder prove that if $(M,\omega)$ is any closed symplectic manifold, then in any $\mathcal{C}^\infty$-neighborhood of the identity in $\te …