bio | website | math.jhu.edu/jmartinezgarcia |
---|---|---|
location | Baltimore, MD | |
age | 29 | |
visits | member for | 5 years, 9 months |
seen | Apr 8 at 23:33 | |
stats | profile views | 1,128 |
I am a JJ Sylvester Assistant Professor at Johns Hopkins University. Previously I did my PhD at the University of Edinburgh.
My interests are in Birational Geometry, constant scalar curvature Kahler metrics, Fano varieties, K-stability and Tian alpha-invariants.
Apr
24 |
awarded | Popular Question |
Apr
2 |
awarded | Notable Question |
Mar
29 |
awarded | Popular Question |
Oct
31 |
awarded | Popular Question |
Jul
2 |
awarded | Curious |
Jun
13 |
accepted | Castelnuovo's rationality criterion on singular surfaces? |
Jun
13 |
comment |
Castelnuovo's rationality criterion on singular surfaces?
Ooops, the du Val case is so obvious, you are right. Glad to see that it is sharp for singularities. Thanks. |
Jun
9 |
awarded | Yearling |
Jun
9 |
asked | Castelnuovo's rationality criterion on singular surfaces? |
Feb
13 |
accepted | Intuitive pictures in characteristic p |
Jan
15 |
awarded | Popular Question |
Nov
15 |
comment |
Existence of constant scalar curvature Kahler metrics on projective manifolds
Awesome answer, thanks. |
Nov
15 |
accepted | Existence of constant scalar curvature Kahler metrics on projective manifolds |
Oct
22 |
asked | Existence of constant scalar curvature Kahler metrics on projective manifolds |
Jun
26 |
awarded | Necromancer |
Jun
25 |
awarded | Excavator |
Jun
25 |
awarded | Suffrage |
Feb
26 |
comment |
Injective morphism from curves to $\mathbb CP^2$
Oh, OK. I thought cusp meant $y^2-x^3$ (never read it anywhere, just from conversations), but I realise that analytically that is also $y^m-x^n$ for $m,n>1$ , which I think is equivalent to what you say. Thanks a lot :) |
Feb
22 |
comment |
Injective morphism from curves to $\mathbb CP^2$
@Jérémy: why must the singularities be cuspidal? |
Dec
22 |
comment |
Intuitive pictures in characteristic p
I particularly like this interpretation. Did you come with it by yourself or is it well known? |