bio | website | mysbfiles.stonybrook.edu/… |
---|---|---|
location | Simons Center | |
age | 31 | |
visits | member for | 5 years, 1 month |
seen | 15 hours ago | |
stats | profile views | 2,965 |
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 |