Let $M$ be a symplectic manifold with divisor $D$. Then how can we define symplectic reduction for pair $(M,D)$?
0
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
Let me illustrate this using the simplest example $(\mathbb{C}^2,D)$, where $D$ is the conic defined by $xy+1=0$ with $x,y$ the coordinates on $\mathbb{C}^2$. Consider the Hamiltonian $S^1$-action on $\mathbb{C}^2$ defined by $e^{i\theta}(x,y)=(e^{i\theta}x,e^{-i\theta}y)$, then the moment map is $\mu=\frac{|x|^2-|y|^2}{2}$. Notice that the origin is a fixed point of the $S^1$-action, so there is precisely one degenerate $S^1$-orbit, which is a point. On the other hand, the divisor $D\subset\mathbb{C}^2$ is fibered over $\mathfrak{t}^\ast\cong\mathbb{R}$ by the $S^1$ orbits of the Hamiltonian action, all these orbits are regular, which corresponds to the fact that $D$ is an affine curve isomorphic to $\mathbb{C}^\ast$. At this point it's not hard to see the symplectic reduction $(\mathbb{C}^2,D)//S^1$ is $(\mathbb{R}^2,pt)$ with its standard symplectic form.
The above example illustrates something for the general case. Note that we choose $D$ to be $xy+1=0$ simply because $xy$-plane is the $(\mathbb{R}^2,\omega_{std})$ after symplectic reduction with respect to the $S^1$-action, so the picture is like the Hamiltonian action should be compatible with the divisor $D$, otherwise it's hard to make the concept $(M,D)//G$ well-defined.
It's not hard to generalize the example $(\mathbb{C}^2,D)$ considered above to higher dimensions to obtain infinitely many examples with a Hamiltonian torus action, but these are somehow of the same nature mod out some technical difficulties, say you may encounter the cases when some of the reduced spaces $\mu^{-1}(\lambda)//T^k$ are equipped with singular symplectic forms, then you may need to smooth them out using a continuous "symplectomorphism" to obtain regular symplectic reductions.
In the context of mirror symmetry, a Hamiltonian torus action is mirror to a holomorphic map $f$ taking values in the dual complexified torus $T_\mathbb{C}^\vee$. So the problem whether the symplectic reduction can be defined for the pairing $(M,D)$ is mirror to the analytic continuation problem of $f$. For non-abelian symplectic reductions, you need to consider the conjugacy classes of $G_\mathbb{C}^\vee$, where $G^\vee$ denotes the Langlands dual. I hope this will give you some intuition on how the symplectic reduction of $(M,D)$ should be defined.
