2,317 reputation
732
bio website mysbfiles.stonybrook.edu/…
location Simons Center
age 31
visits member for 5 years, 1 month
seen 15 hours ago

Research Asst Professor

Interested in Symplectic geometry, complex Algebraic geometry, Gromov-Witten theory, Mirror symmetry, Calabi-Yau threefolds, Enumerative geometry, Abelian surfaces, Moduli space of objects with real structure, ...


May
18
awarded  Self-Learner
May
10
comment manifold branched covering space for orbifolds
@ Dylan: en.wikipedia.org/wiki/Branched_covering
May
8
revised manifold branched covering space for orbifolds
edited body
May
8
comment manifold branched covering space for orbifolds
@ Ariyan: Think this way, if X is a non simply-connected manifold, instead of orbifold, then such M is simply a finite covering of X.
Apr
28
awarded  Nice Question
Apr
20
revised manifold branched covering space for orbifolds
edited body
Apr
20
revised manifold branched covering space for orbifolds
added 247 characters in body
Apr
20
asked manifold branched covering space for orbifolds
Apr
10
awarded  Yearling
Apr
6
awarded  Popular Question
Mar
30
comment Symplectic form/Kahler metric on a toric manifold
Even for m=2 case of example above, it seems to me that the equality you want does not hold. The coefficient of $dz\wedge d\bar{z}$ in $f^*w_{FS}$ is equal to $[a(4|z|^4+|w|^2)-4|z|^4|w|^2]/a^2$, with $a=(1+|z|^4+|z|^2|w|^2+|w|^4)$, which is different from the corresponding coefficient of $dz\wedge d\bar{z}$ in $w_{FS}$ of $\mathbb{P}^1$.
Mar
30
comment Symplectic form/Kahler metric on a toric manifold
Have you checked this for $f:\mathbb{P}^1\to \mathbb{P}^{m}$, $[z,w]\to[z^m,z^{m-1}w,\ldots, w^m]$? here, everything is explicitly checkable.
Mar
28
comment Symplectic form/Kahler metric on a toric manifold
are not they equal? up to scaling?
Feb
18
revised Triviality of holomorphic vector bundles over contractible Stein manifolds
deleted 2 characters in body
Jan
31
accepted A question on compact sets
Jan
31
comment A question on compact sets
This proof readily extends to compact sets inside any metrizable topological space, instead of $\mathbb{R}^n$. Do you see a way of changing the proof that does not involve the use of metric.
Jan
28
comment A question on compact sets
N is just a number.
Jan
28
reviewed Approve A question on compact sets
Jan
28
comment A question on compact sets
:) YES (I knew someone would say that ;))
Jan
28
asked A question on compact sets