Nikita Kalinin (CINVESTAV), Mikhail Shkolnikov (University of Geneva)
New formula for π to 249 anniversary of Gauss
2 X p p q π a2 + b2 + c2 + d2 − (a + c)2 + (b + d)2 = 2 − 2 a,b,c,d∈Z≥0, ad−bc=1
(sums of the squares of the defects in the triangle inequality) Mathematics should be social This question is among the following popular questions: 0 1
3 4 3 4 0 4 • • • toppling 1 3 1 4 •
Denote by φ◦ the result of the relaxation of φ. The size of avalanches is governed by a power law, if we drop additional sand randomly.
Sandpile model
Definition A sandpile is a collection of indistinguishable sand grains distributed 2 among a subset Γ of Z , that is a function φ :Γ → N0. A vertex v is unstable if φ(v) ≥ 4. An unstable vertex can topple by sending one grain of sand to each of 4 neighbors. Sandpile model
Definition A sandpile is a collection of indistinguishable sand grains distributed 2 among a subset Γ of Z , that is a function φ :Γ → N0. A vertex v is unstable if φ(v) ≥ 4. An unstable vertex can topple by sending one grain of sand to each of 4 neighbors.
0 1
3 4 3 4 0 4 • • • toppling 1 3 1 4 •
Denote by φ◦ the result of the relaxation of φ. The size of avalanches is governed by a power law, if we drop additional sand randomly. 2 ◦ Fix Ω ⊂ R , with non-empty interior Ω , different points ◦ 1 2 1 p1, p2,..., pn ∈ Ω . Consider the grid N Z with the mesh N . Let 1 2 Γ = Ω ∩ N Z . P 1 We consider the state φ = h3i + δ[pi ] on Γ. N Look at the relaxation φ◦ of φ. ◦ Define deviation set CN = {x ∈ Γ|φ 6= 3}. • • • Experiments (Caracciolo et al.) suggest that φ◦ ≡ 3 almost everywhere (as long as N is big enough). More precisely, CN is a “thin graph”.
Tropical curves in sandpiles
The maximal stable state is h3i = φ ≡ 3. Experiments (Caracciolo et al.) suggest that φ◦ ≡ 3 almost everywhere (as long as N is big enough). More precisely, CN is a “thin graph”.
Tropical curves in sandpiles
The maximal stable state is h3i = φ ≡ 3. 2 ◦ Fix Ω ⊂ R , with non-empty interior Ω , different points ◦ 1 2 1 p1, p2,..., pn ∈ Ω . Consider the grid N Z with the mesh N . Let 1 2 Γ = Ω ∩ N Z . P 1 We consider the state φ = h3i + δ[pi ] on Γ. N Look at the relaxation φ◦ of φ. ◦ Define deviation set CN = {x ∈ Γ|φ 6= 3}. • • • Tropical curves in sandpiles
The maximal stable state is h3i = φ ≡ 3. 2 ◦ Fix Ω ⊂ R , with non-empty interior Ω , different points ◦ 1 2 1 p1, p2,..., pn ∈ Ω . Consider the grid N Z with the mesh N . Let 1 2 Γ = Ω ∩ N Z . P 1 We consider the state φ = h3i + δ[pi ] on Γ. N Look at the relaxation φ◦ of φ. ◦ Define deviation set CN = {x ∈ Γ|φ 6= 3}. • • • Experiments (Caracciolo et al.) suggest that φ◦ ≡ 3 almost everywhere (as long as N is big enough). More precisely, CN is a “thin graph”. Sandpile on a disk: a point in the center
Extra grain at the cen- The grid is 3 times finer. The grid is 9 times finer. ter. White: 3 grains, Green: 2 grains, Yellow: 1 grain, Red: 0 grains Proof. Indeed, count the number of incoming and outgoing grains.
That is, h is harmonic at (x, y). In fact, h(x, y) is a “piece-wise” linear function.
Explanation: look at the number of topplings
Consider the number h(x, y) of topplings at a point (x, y) during the relaxation. Proposition If the number of sand grains at (x, y) after relaxation is the same as before the relaxation, then h(x − 1, y) + h(x + 1, y) + h(x, y − 1) + h(x, y + 1) − 4h(x, y) = 0. Explanation: look at the number of topplings
Consider the number h(x, y) of topplings at a point (x, y) during the relaxation. Proposition If the number of sand grains at (x, y) after relaxation is the same as before the relaxation, then h(x − 1, y) + h(x + 1, y) + h(x, y − 1) + h(x, y + 1) − 4h(x, y) = 0.
Proof. Indeed, count the number of incoming and outgoing grains.
That is, h is harmonic at (x, y). In fact, h(x, y) is a “piece-wise” linear function. Scaling limit of toppling function Tropical distance function is
ij lΩ(x, y) = inf lΩ(x, y). (i,j)∈Z2\{(0,0)}
Note that {lΩ = 0} = ∂Ω.
Tropical (weighted) distance function
2 Ω ⊂ R is a bounded convex domain. 2 For any (i, j) ∈ Z denote by cij the minimum of ix + jy over (x, y) ∈ Ω.
ij lΩ(x, y) = ix + jy − cij . Note that {lΩ = 0} = ∂Ω.
Tropical (weighted) distance function
2 Ω ⊂ R is a bounded convex domain. 2 For any (i, j) ∈ Z denote by cij the minimum of ix + jy over (x, y) ∈ Ω.
ij lΩ(x, y) = ix + jy − cij . Tropical distance function is
ij lΩ(x, y) = inf lΩ(x, y). (i,j)∈Z2\{(0,0)} Tropical (weighted) distance function
2 Ω ⊂ R is a bounded convex domain. 2 For any (i, j) ∈ Z denote by cij the minimum of ix + jy over (x, y) ∈ Ω.
ij lΩ(x, y) = ix + jy − cij . Tropical distance function is
ij lΩ(x, y) = inf lΩ(x, y). (i,j)∈Z2\{(0,0)}
Note that {lΩ = 0} = ∂Ω. Current estimates: N1/2 < E ≤ N131/208.
Consider f = l{x2+y 2≤N2}. Then the number of lattice points is f −1(0) + f −1(1) + ... . Level sets are governed by the polyhedral structure of f , so we need to find its vertices.
Gauss circle problem
Problem: how many lattice points in a circle?
2 2 2 2 2 |{(i, j ∈ Z ) ∩ {x + y ≤ N }| ∼ πN + E Consider f = l{x2+y 2≤N2}. Then the number of lattice points is f −1(0) + f −1(1) + ... . Level sets are governed by the polyhedral structure of f , so we need to find its vertices.
Gauss circle problem
Problem: how many lattice points in a circle?
2 2 2 2 2 |{(i, j ∈ Z ) ∩ {x + y ≤ N }| ∼ πN + E Current estimates: N1/2 < E ≤ N131/208. Level sets are governed by the polyhedral structure of f , so we need to find its vertices.
Gauss circle problem
Problem: how many lattice points in a circle?
2 2 2 2 2 |{(i, j ∈ Z ) ∩ {x + y ≤ N }| ∼ πN + E Current estimates: N1/2 < E ≤ N131/208.
Consider f = l{x2+y 2≤N2}. Then the number of lattice points is f −1(0) + f −1(1) + ... . Gauss circle problem
Problem: how many lattice points in a circle?
2 2 2 2 2 |{(i, j ∈ Z ) ∩ {x + y ≤ N }| ∼ πN + E Current estimates: N1/2 < E ≤ N131/208.
Consider f = l{x2+y 2≤N2}. Then the number of lattice points is f −1(0) + f −1(1) + ... . Level sets are governed by the polyhedral structure of f , so we need to find its vertices. P f (p)(1−f (p)) Finally, we expect p 2 = π/4. Then, by computer calculations we see that X X f (p)2 = 2 − π/2, and f (p) = 2. p p P n What is p f (p) for n > 2?
Vertices and SL(2, Z)
Lemma. In the quadrant x, y ≥ 0 there is bijection between a, b, c, d ∈ Z≥0, ad − bc = 1 and the vertices p of z = f (x, y). Then, by computer calculations we see that X X f (p)2 = 2 − π/2, and f (p) = 2. p p P n What is p f (p) for n > 2?
Vertices and SL(2, Z)
Lemma. In the quadrant x, y ≥ 0 there is bijection between a, b, c, d ∈ Z≥0, ad − bc = 1 and the vertices p of z = f (x, y). P f (p)(1−f (p)) Finally, we expect p 2 = π/4. P n What is p f (p) for n > 2?
Vertices and SL(2, Z)
Lemma. In the quadrant x, y ≥ 0 there is bijection between a, b, c, d ∈ Z≥0, ad − bc = 1 and the vertices p of z = f (x, y). P f (p)(1−f (p)) Finally, we expect p 2 = π/4. Then, by computer calculations we see that X X f (p)2 = 2 − π/2, and f (p) = 2. p p Vertices and SL(2, Z)
Lemma. In the quadrant x, y ≥ 0 there is bijection between a, b, c, d ∈ Z≥0, ad − bc = 1 and the vertices p of z = f (x, y). P f (p)(1−f (p)) Finally, we expect p 2 = π/4. Then, by computer calculations we see that X X f (p)2 = 2 − π/2, and f (p) = 2. p p P n What is p f (p) for n > 2? Then both identities can be obtained by telescoping.
Φ1(x, y) − Φ1(x, z) − Φ1(y, z) = |x| + |y| − |z|,
2 Φ2(x, y) − Φ2(x, z) − Φ2(y, z) = (|x| + |y| − |z|) .
Mathoverflow proof
Lemma [users: Fedor Petrov and fedja] Let
Φ1(x, y) = |x| + |y| + xy + yx ,
Φ2(x, y) = 2|x||y| − |x|yx − |y|xy ,
where x, y are vectors, xy is the length of the projection of x to y. Mathoverflow proof
Lemma [users: Fedor Petrov and fedja] Let
Φ1(x, y) = |x| + |y| + xy + yx ,
Φ2(x, y) = 2|x||y| − |x|yx − |y|xy ,
where x, y are vectors, xy is the length of the projection of x to y. Then both identities can be obtained by telescoping.
Φ1(x, y) − Φ1(x, z) − Φ1(y, z) = |x| + |y| − |z|,
2 Φ2(x, y) − Φ2(x, z) − Φ2(y, z) = (|x| + |y| − |z|) . Thank you for your attention!
Sam Gitler’s conference, 2016 More formulae
X 1 = 1/24 (a + b)2(c + d)2(a + b + c + d)2 ad−bc=1, 1≤a≤b,1≤c≤d X 1 = 2/3 b2d2(b + d)2 ad−bc=1, 1≤a≤2b,1≤c≤2d