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Hamiltonian systems, symplectic flows, classical integrable systems
11
votes
When do you go hunting for Lagrangian submanifolds?
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 …
11
votes
1
answer
832
views
Can a symplectic manifold be recovered from its Lagrangians?
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 …
11
votes
1
answer
2k
views
What is known about the strong Arnold conjecture?
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 …
8
votes
1
answer
410
views
How many "elementary" characterizations of twisted SU(2) representation varieties are known?
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 …
3
votes
Square root for Hamiltonian diffeomorphisms
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 …
2
votes
Accepted
If a (linear) relation maps Lagrangian subspaces to Lagrangian subspaces, is it a Lagrangian...
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 …
2
votes
Accepted
floer homology and viterbo's theorem
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 …