bio | website | web.mit.edu/abhinavk/www |
---|---|---|
location | ||
age | ||
visits | member for | 4 years, 7 months |
seen | 13 hours ago | |
stats | profile views | 1,053 |
May 9 |
comment |
Elliptic surfaces with different Kodaira symbols
At least in characteristic not 2 or 3 this is impossible: the elliptic fibration is unique for Kodaira dimension 1. For char 2 or 3 you may have to consider quasi-elliptic fibrations, and I haven't thought through it. |
May 8 |
answered | Lattice polarized K3 surfaces |
May 8 |
comment |
Lattice polarized K3 surfaces
No, the Picard rank can be anywhere between 1 and 20, for a lattice polarized K3 surface. |
Mar 28 |
reviewed | Approve suggested edit on On a particular case of the ``Tumura-Hayman" theorem : |
Jan 30 |
reviewed | Reject suggested edit on Cohomology after completion |
Jan 16 |
comment |
Commutativity of convex hulls and closed balls
I believe I now have a counterexample - see above. |
Jan 16 |
revised |
Commutativity of convex hulls and closed balls
added 1351 characters in body |
Jan 15 |
answered | Commutativity of convex hulls and closed balls |
Jan 14 |
awarded | Good Question |
Jan 12 |
awarded | Custodian |
Jan 12 |
reviewed | Approve suggested edit on Distribution of moduli of quadratic residues |
Jan 2 |
comment |
Determinant and eigenvalues of a specific matrix
If you let $e^{-c}$ be $x$, then the matrix has polynomial entries in $x$, and experiment seems to indicate that the determinant is a product of cyclotomic polynomials (for instance, if $n = 6$, we get $-(x-1)^{15}(x+1)^{15}(x^2+1)^6 (x^2-x+1)^3(x^2+x+1)^3(x^4+1)^2(x^4-x^3+x^2-x+1)(x^4+x^3+x^2+x+1)$. (In particular, the power of $(x \pm 1)$ seems to be $n$ choose $2$.) |
Dec 19 |
awarded | Yearling |
Nov 20 |
comment |
Lattice points and convex bodies
@AntonPetrunin: Good point! |
Nov 20 |
comment |
Lattice points and convex bodies
I would be surprised if this were true even for integer polytopes - that the Erhart polynomial determines the polytope (though I can't seem to find an immediate counterexample by searching online ...). You can certainly do $GL_n(\mathbb{Z})$ transformations without changing the number of integer points. |
Nov 12 |
comment |
Nefness on a K3 surface
Not if the divisor $D$ is reducible - else take $D = C + f$ on a Hirzebruch surface with $C^2 = -2, f^2 = 0, C \cdot f = 1$, and notice $D \cdot C = -1$. If you don't assume $D$ is effective, then Jason's comment shows you that there's no way to distinguish $D$ from its negative. |
Nov 12 |
answered | Nefness on a K3 surface |
Nov 1 |
awarded | Nice Answer |
Nov 1 |
revised |
Rationality of curve does not depend on base change
deleted 1 characters in body |
Nov 1 |
revised |
Rationality of curve does not depend on base change
added 725 characters in body |