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Oct
9
answered When the contraction is a morphism defined over $\overline{\mathbb Q}$
Jul
31
comment K3 surfaces that correspond to rational points of elliptic curves
@LevBorisov Are you aware of the work of Kudla and recent work of Darmon and others? It might have some connection to what you're looking for.
Jul
30
comment K3 surfaces that correspond to rational points of elliptic curves
The field of definition of the point $P$ on $X_0(n)^+$ should probably correspond to the field of the definition of the corresponding divisor on the K3 surface. So saying that it comes from a $\bf{Q}$ point just means that the Picard group of the K3 can be fully realized over a small degree number field (here, probably the $2$-torsion field of the elliptic curves).
Jul
30
comment K3 surfaces that correspond to rational points of elliptic curves
It seems that the K3 is not the Kummer, but is is a double cover (i.e. related by a Shioda-Inose structure to the product abelian surface). See theorem 7.6 of that paper.
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.