Search Results

There are probably too many such $(M,g,I_+,I_-)$ to really expect a 'classification'. For instance, consider the case when a complex manifold $(M,I_+)$ has real dimension $4$, and the $I_+$-holomorphi …

The answer is 'no', even in the simplest nontrivial case $n=2$, if you assume that the symplectomorphism is to take $C$ to $C'$. (If you don't assume that the symplectomorphism is supposed to take $C …

The answer is 'yes'. Specifically, the holomorphic version of the Darboux-Weinstein theorem holds, just as it does in the smooth category. In particular, if $L\subset M$ is a holomorphic Lagrangian …

There are obstructions: For example, if $N^n\subset\mathbb{R}^{2n+1}$ is a Legendrian submanifold, then $\nu$, the normal bundle of $N^n$, is isomorphic to $\tau\oplus TN$, where $\tau$ is the trivia …

The injectivity case is well-known and follows quite easily from the statement (usually attributed to Lefschetz) that $L^k_\omega:\Omega^{n-k}(M)\to \Omega^{n+k}(M)$ is a isomorphism for $0\le k\le n$ …

The answer is 'no', the affine symplectic group cannot appear as a Lie subgroup of any compact Lie group. The reason is that the affine symplectic group contains $\mathrm{SL}(2,\mathbb{R})$ as a Lie …

The general solution of your equations in a simply connected domain on which $r_2\not=0$ and $r_1\not=\pm1$ is $$ \beta = \frac12 + \frac1{{(r_1}^2{-}1)}\, \left(\frac{\partial a}{\partial\theta_1}+b( …

The example in $\mathbb{R}^3$ given by $$ \omega = x\,\mathrm{d}y\wedge\mathrm{d}z + y\,\mathrm{d}z\wedge\mathrm{d}x + z\,\mathrm{d}x\wedge\mathrm{d}y, $$ shows that the kernel of $\omega$ need not be …

The answer is 'no, not necessarily'. Consider the following example: Let $M=N=\mathbb{CP}^2$, let $(\omega_0,J_0)$ be the standard Fubini-Study Kähler structure on $M$. Now let $f$ be an arbitrary, …

The answer to your first question is 'no' and the answer to your second question is 'yes'. A simple example, when $n\ge 2$, is to let $M = \mathbb{R}^{2n}/\Lambda$ where $\Lambda\subset \mathbb{R}^{2n …

I'm a bit confused by your question, because I believe that, if one defines an $\omega$-Hamiltonian vector field to be a vector field of the form $X_f = \omega^\#(\mathrm{d}f)$ where $f$ is a (smooth) …

The 'moment conditions' that Ben McKay mentions are simply this: A closed curve $C$ in $\mathbb{C}^n$ bounds a compact Riemann surface (which might be singular) if and only if the integral around $C$ …

A classic example that shows that WLNT doesn't always hold in the holomorphic category is an elliptically fibered $K3$ surface. A K3 surface $S$ is a compact complex symplectic manifold of complex …

Actually, one can always do a (local) surgery to 'resolve' transverse intersection points of Lagrangian submanifolds, and this works in all dimensions, not just dimension $2$. Here is a sketch of an …

The affine group of $(M,\nabla)$ is a Lie group $G$ by Kobayashi's theorem that shows that the automorphism group of any affine connection is a Lie group (see Kobayashi and Nomizu's Foundations of Dif …