bio | website | web.mit.edu/abhinavk/www |
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location | ||
age | ||
visits | member for | 5 years, 1 month |
seen | 15 hours ago | |
stats | profile views | 1,132 |
Dec 19 |
awarded | Yearling |
Nov 27 |
reviewed | Approve Isomorphism problem for two radical extensions |
Nov 26 |
reviewed | Approve Simplifying triangulations of 3-manifolds |
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 On a particular case of the ``Tumura-Hayman" theorem : |
Jan 30 |
reviewed | Reject 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 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 |