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visits | member for | 3 years, 9 months |
seen | 12 hours ago | |
stats | profile views | 2,037 |
Nov 17 |
comment |
Isometries of hyper-Kähler manifolds
"It is a well-known fact that the isometry group of a compact Ricci-flat space is finite": this is false, take for example a compact flat torus. |
Nov 11 |
answered | For compact complex surfaces $h^{1,0}$ is either $h^{0,1}$ or $h^{0,1} - 1$. Do we need to use the Enriques-Kodaira classification? |
Oct 30 |
answered | Kodaira dimension of co-adjoint orbit |
Oct 28 |
awarded | Enlightened |
Oct 28 |
awarded | Nice Answer |
Aug 21 |
comment |
well known facts on openness condition
A similar question was asked on Stackexchange here:math.stackexchange.com/questions/903891/… Does anybody know the answer to this? |
Aug 14 |
awarded | Good Question |
Jul 16 |
awarded | Popular Question |
Jul 2 |
awarded | Curious |
Feb 22 |
awarded | Yearling |
Sep 1 |
awarded | Necromancer |
Jun 25 |
awarded | complex-geometry |
Jun 25 |
awarded | dg.differential-geometry |
Jun 18 |
comment |
metric scaling for an inequality
Rescaling the metric tensor by $R^{-2}$ rescales distances like $R^{-1}$, which is what you want. |
Jun 17 |
answered | metric scaling for an inequality |
Jun 6 |
comment |
Kahler cone of a product
Think first about tori $M=\mathbb{C}^n/\Lambda$ and $N=\mathbb{C}^m/\Lambda'$, where every Kahler class has a constant representative, and you can compute everything explicitly. You will see that in this case $K_{M\times N}$ is much larger than the cone generated by $K_M\times K_N$, essentially because of Kunneth's formula. |
May 29 |
awarded | Popular Question |
May 17 |
revised |
Dolbeault cohomology
edited tags |
May 15 |
answered | Dolbeault cohomology |
May 8 |
comment |
Converse to Milnor's theorem on manifolds with nonnegative Ricci curvature.
"cook one UP"... |