bio | website | math.princeton.edu/~wsawin |
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location | Princeton, NJ | |
age | 20 | |
visits | member for | 2 years, 11 months |
seen | 3 hours ago | |
stats | profile views | 18,579 |
I am a graduate student at Princeton studying arithmetic algebraic geometry.
Aug 16 |
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Points on the intersection of an affine quadric and cubic over a finite field
What you want may relate to Chevalley's theorem: en.wikipedia.org/wiki/Chevalley%E2%80%93Warning_theorem |
Aug 16 |
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Does this “modeling relationship” occur in mathematics (Galois connections, relation algebra, category theory)?
I don't think modeling relationships form a category because there is no way to compose them and no natural identity element. |
Aug 13 |
answered | Preimage of smooth curves under morphism of smooth varieties |
Aug 12 |
answered | Proofs of the Chevalley-Warning Theorem |
Aug 10 |
answered | Interpretation of $H_1(A_\mathbb{C}^{top},\mathbb{Q})$ |
Aug 10 |
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support of embedded points in a curve
$l_p$ is necessarily a singular point of $C$.... |
Aug 6 |
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Explicit formula for the Poincare dual of a CM endomorphism of an elliptic curve
Because it counts solutions to the equation $x = E(x)$ where $E$ is multiplication by $\sqrt{-D}$. For each such $x$, $(x,x)$ is a transverse intersection point of the two curves. |
Aug 6 |
answered | Explicit formula for the Poincare dual of a CM endomorphism of an elliptic curve |
Aug 6 |
awarded | Enlightened |
Aug 6 |
awarded | Nice Answer |
Aug 5 |
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If a graph invariant is NP-Hard, is its “deck ratio” NP-Hard as well?
I don't understand your example $\phi_1$. If $G$ has an even number of vertices, then $\phi_1(G-v)=-1$, right? So the deck ratio is just minus $\phi_1$. So how can one be NP-hard and the other NP-easy? I think it might be sufficient to mandate nonzero values. |
Aug 4 |
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Connections between Standard, Hodge and Tate conjectures on algebraic cycles?
For the Lefschetz conjecture, we need a certain ring to contain a left inverse or right inverse of an element. In other words we are looking for solutions of the equation $XY=I$, for $Y$ defined over $\mathbb Q$. If we have such a solution after tensoring with $\mathbf Q_\ell$, we must have it before tensoring. |
Aug 4 |
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Connections between Standard, Hodge and Tate conjectures on algebraic cycles?
@ACL: For the Kunneth conjecture, the cycles we are searching for are idempotents in a certain ring. If upon tensoring with $\mathbf Q_\ell$ an algebra has a certain idempotent, then I believe that before tensoring it has corresponding elements whose eigenvalues are in $\mathbf Q_\ell$, that is, they are in an algebraic number field that splits at $\ell$. Combining all $\ell$ I think we get the desired statement. So maybe one needs Tate at each $\ell$. |
Aug 2 |
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Derived categories of curves equivalent then the curves are isomorphic
Can you get a uniform proof using Torelli's theorem? It says that if two curves have the same integral Hodge structure (including Poincare duality) then they are isomorphic. |
Aug 1 |
answered | Cyclotomic character in class field theory |
Aug 1 |
answered | Circles avoiding rational points of height $\le h$ |
Jul 31 |
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Escape the zombie apocalypse
As long as area of each sector grows exponentially, the probability of not having any zombies in the sector falls superexponentially. Since the number of segments grows exponentially, by the union bound you're the probability of each segment having a zombie is close to 1. |
Jul 31 |
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Newton polygons of modular polynomials
I worked out in more detail my thoughts about the bottom corner. I think this is more related to Heegner points of conductor $1$, because these are the CM points associated to split ideals in $\mathbb Z[\omega]$, not in a subring. |
Jul 31 |
revised |
Newton polygons of modular polynomials
added 1285 characters in body |
Jul 30 |
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Connections between Standard, Hodge and Tate conjectures on algebraic cycles?
$H^{2i}(X,\mathbb Q_\ell)$ and $H^{2d-2i}(X,\mathbb Q_\ell)$ are dual Galois representations. If we have a cycle class in the first one that gives a cyclotomic subrepresentation, so we get a cyclotomic quotient representation of the other one. I think the Tate conjecture is usually taken to include semisimplicity (or it implies that?), so this gives you a cyclotomic subrepresentation that is dual to the first one. |