bio  website  mysbfiles.stonybrook.edu/… 

location  Simons Center  
age  31  
visits  member for  5 years, 3 months 
seen  Jul 19 at 17:04  
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Research Asst Professor
Interested in Symplectic geometry, complex Algebraic geometry, GromovWitten theory, Mirror symmetry, CalabiYau threefolds, Enumerative geometry, Abelian surfaces, Moduli space of objects with real structure, ...
1d

awarded  Nice Question 
May 18 
awarded  SelfLearner 
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 simplyconnected 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 $[a(4z^4+w^2)4z^4w^2]/a^2$, with $a=(1+z^4+z^2w^2+w^4)$, 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 $f:\mathbb{P}^1\to \mathbb{P}^{m}$, $[z,w]\to[z^m,z^{m1}w,\ldots, w^m]$? 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 ;)) 