Good covering codes from algebraic curves Massimo Giulietti - - PowerPoint PPT Presentation

Good covering codes from algebraic curves

Massimo Giulietti University of Perugia (Italy) Special Semester on Applications of Algebra and Number Theory Workshop 2: Algebraic curves over finite fields Linz, 14 November 2013

covering codes

(Fn

q, d)

d Hamming distance C ⊂ Fn

q

covering codes

(Fn

q, d)

d Hamming distance C ⊂ Fn

q

covering radius of C R(C) := max

v∈Fn

q

d(v, C)

covering codes

(Fn

q, d)

d Hamming distance C ⊂ Fn

q

covering radius of C R(C) := max

v∈Fn

q

d(v, C)

b b b b

R Fn

q

covering codes

(Fn

q, d)

d Hamming distance C ⊂ Fn

q

covering radius of C R(C) := max

v∈Fn

q

d(v, C)

b b b b

R Fn

q

b v

covering density of C µ(C) := #C · size of a sphere of radius R(C) qn

covering codes

(Fn

q, d)

d Hamming distance C ⊂ Fn

q

covering radius of C R(C) := max

v∈Fn

q

d(v, C)

b b b b

R Fn

q

b v

covering density of C µ(C) := #C · size of a sphere of radius R(C) qn ≥ 1

linear codes

k = dim C r = n − k µ(C) = 1 + n(q − 1) + n

2

• (q − 1)2 + . . . +
• n

R(C)

• (q − 1)R(C)

qr

linear codes

k = dim C r = n − k µ(C) = 1 + n(q − 1) + n

2

• (q − 1)2 + . . . +
• n

R(C)

• (q − 1)R(C)

qr ℓ(r, q)R := min n for which there exists C ⊂ Fn

q with

R(C) = R, n − dim(C) = r

linear codes

k = dim C r = n − k µ(C) = 1 + n(q − 1) + n

2

• (q − 1)2 + . . . +
• n

R(C)

• (q − 1)R(C)

qr ℓ(r, q)R,d := min n for which there exists C ⊂ Fn

q with

R(C) = R, n − dim(C) = r, d(C) = d

linear codes

k = dim C r = n − k µ(C) = 1 + n(q − 1) + n

2

• (q − 1)2 + . . . +
• n

R(C)

• (q − 1)R(C)

qr ℓ(r, q)R,d := min n for which there exists C ⊂ Fn

q with

R(C) = R, n − dim(C) = r, d(C) = d R = 2, d = 4 (quasi-perfect codes) R = r − 1, d = r + 1 (MDS codes) q odd

ℓ(3, q)2,4

in geometrical terms...

Σ = Σ(2, q) Galois plane over the finite field Fq Σ

in geometrical terms...

Σ = Σ(2, q) Galois plane over the finite field Fq

b b b b b b b b b b b b b

S Σ S ⊂ Σ is a saturating set if every point in Σ \ S is collinear with two points in S

in geometrical terms...

Σ = Σ(2, q) Galois plane over the finite field Fq

b b b b b b b b b b b b b

S

b

Σ S ⊂ Σ is a saturating set if every point in Σ \ S is collinear with two points in S

in geometrical terms...

Σ = Σ(2, q) Galois plane over the finite field Fq

b b b b b b b b b b b b b

S

b

Σ S ⊂ Σ is a saturating set if every point in Σ \ S is collinear with two points in S

in geometrical terms...

Σ = Σ(2, q) Galois plane over the finite field Fq

b b b b b b b b b b b b b

S

b

Σ S ⊂ Σ is a saturating set if every point in Σ \ S is collinear with two points in S

in geometrical terms...

Σ = Σ(2, q) Galois plane over the finite field Fq

b b b b b b b b b b b b b

S

b

Σ S ⊂ Σ is a saturating set if every point in Σ \ S is collinear with two points in S a complete cap is a saturating set which does not contain 3 collinear points

in geometrical terms...

Σ = Σ(2, q) Galois plane over the finite field Fq

b b b b b b b b b b b b b

S

b

Σ S ⊂ Σ is a saturating set if every point in Σ \ S is collinear with two points in S a complete cap is a saturating set which does not contain 3 collinear points ℓ(3, q)2,4 = minimum size of a complete cap in P2(Fq)

plane complete caps

TLB: #S >

• 2q + 1

plane complete caps

TLB: #S >

• 2q + 1

in P2(Fq) there exists a complete cap S of size #S ≤ D√q logC q (Kim-Vu, 2003)

plane complete caps

TLB: #S >

• 2q + 1

in P2(Fq) there exists a complete cap S of size #S ≤ D√q logC q (Kim-Vu, 2003) for every q prime q < 67000 there exists a complete cap S of size #S ≤ √q log q (Bartoli-Davydov-Faina-Marcugini-Pambianco, 2012)

plane complete caps

TLB: #S >

• 2q + 1

in P2(Fq) there exists a complete cap S of size #S ≤ D√q logC q (Kim-Vu, 2003) for every q prime q < 67000 there exists a complete cap S of size #S ≤ √q log q (Bartoli-Davydov-Faina-Marcugini-Pambianco, 2012)

naive construction method:

naive construction method: S = {P1, P2,

b b

P1 P2

naive construction method: S = {P1, P2, P3,

b b

P1 P2

b P3

naive construction method: S = {P1, P2, P3, P4,

b b

P1 P2

b P3 b

P4

naive construction method: S = {P1, P2, P3, P4, . . . ,

b b

P1 P2

b P3 b

P4

b b b b b b b b b b b b

naive construction method: S = {P1, P2, P3, P4, . . . , Pn}

b b

P1 P2

b P3 b

P4

b b b b b b b b b b b b

naive vs. theoretical

500 1000 1500 2000 ← Naive algorithm 20000 40000 60000 80000 100000 q

500 1000 1500 2000 20000 40000 60000 80000 100000 q √q · (log q)0.75 → ← Naive algorithm TLB √3q + 1/2

cubic curves

X plane irreducible cubic curve

cubic curves

X plane irreducible cubic curve

• Q
• P

G = X (Fq) \ Sing(X )

• P ⊕ Q

T O

• if O is an inflection point of X, then P, Q, T ∈ G are collinear if

and only if P ⊕ Q ⊕ T = O

cubic curves

X plane irreducible cubic curve

• Q
• P

G = X (Fq) \ Sing(X )

• P ⊕ Q

T O

• if O is an inflection point of X, then P, Q, T ∈ G are collinear if

and only if P ⊕ Q ⊕ T = O for a subgroup K of index m with (3, m) = 1, no 3 points in a coset S = K ⊕ Q, Q / ∈ K are collinear

classification (p > 3)

classification (p > 3)

classification (p > 3)

classification (p > 3)

classification (p > 3)

Y = X 3 XY = (X − 1)3

• Y (X 2 − β) = 1

Y 2 = X 3 + AX + B

how to prove completeness?

S parametrized by polynomials defined over Fq S = {(f (t), g(t)) | t ∈ Fq} ⊂ A2(Fq)

how to prove completeness?

S parametrized by polynomials defined over Fq S = {(f (t), g(t)) | t ∈ Fq} ⊂ A2(Fq) P = (a, b) collinear with two points in S if there exist x, y ∈ Fq with det   a b 1 f (x) g(x) 1 f (y) g(y) 1   = 0

how to prove completeness?

S parametrized by polynomials defined over Fq S = {(f (t), g(t)) | t ∈ Fq} ⊂ A2(Fq) P = (a, b) collinear with two points in S if there exist x, y ∈ Fq with Fa,b(x, y) = 0, where Fa,b(x, y) := det   a b 1 f (x) g(x) 1 f (y) g(y) 1  

how to prove completeness?

S parametrized by polynomials defined over Fq S = {(f (t), g(t)) | t ∈ Fq} ⊂ A2(Fq) P = (a, b) collinear with two points in S if there exist x, y ∈ Fq with Fa,b(x, y) = 0, where Fa,b(x, y) := det   a b 1 f (x) g(x) 1 f (y) g(y) 1   P = (a, b) collinear with two points in S if the algebraic curve CP : Fa,b(X, Y ) = 0 has a suitable Fq-rational point (x, y)

how to prove completeness?

S parametrized by polynomials defined over Fq S = {(f (t), g(t)) | t ∈ Fq} ⊂ A2(Fq) P = (a, b) collinear with two points in S if there exist x, y ∈ Fq with Fa,b(x, y) = 0, where Fa,b(x, y) := det   a b 1 f (x) g(x) 1 f (y) g(y) 1   P = (a, b) collinear with two points in S if the algebraic curve CP : Fa,b(X, Y ) = 0 has a suitable Fq-rational point (x, y)

cuspidal case: Y = X 3

G is an elementary abelian p-group q = ph

cuspidal case: Y = X 3

G is an elementary abelian p-group q = ph K = {(tp − t, (tp − t)3) | t ∈ Fq}

cuspidal case: Y = X 3

G is an elementary abelian p-group q = ph K = {(tp − t, (tp − t)3) | t ∈ Fq} S = {(

f (t)

• tp − t + ¯

t, (

g(t)

• tp − t + ¯

t)3)

• Pt

| t ∈ Fq}

cuspidal case: Y = X 3

G is an elementary abelian p-group q = ph K = {(tp − t, (tp − t)3) | t ∈ Fq} S = {(

f (t)

• tp − t + ¯

t, (

g(t)

• tp − t + ¯

t)3)

• Pt

| t ∈ Fq} P = (a, b) is collinear with Px and Py if and only if Fa,b(x, y) := a + (xp − x + ¯ t)(y p − y + ¯ t)2 + (xp − x + ¯ t)2(y p − y + ¯ t) − b((xp − x + ¯ t)2 +(xp − x + ¯ t)(y p − y + ¯ t) + (y p − y + ¯ t)2) = 0

cuspidal case: Y = X 3

G is an elementary abelian p-group q = ph K = {(tp − t, (tp − t)3) | t ∈ Fq} S = {(

f (t)

• tp − t + ¯

t, (

g(t)

• tp − t + ¯

t)3)

• Pt

| t ∈ Fq} P = (a, b) is collinear with Px and Py if and only if Fa,b(x, y) := a + (xp − x + ¯ t)(y p − y + ¯ t)2 + (xp − x + ¯ t)2(y p − y + ¯ t) − b((xp − x + ¯ t)2 +(xp − x + ¯ t)(y p − y + ¯ t) + (y p − y + ¯ t)2) = 0 the curve CP then is Fa,b(X, Y ) = 0

applying Segre’s criterion

(Segre, 1962)

if there exists a point P ∈ C and a tangent ℓ of C at P such that ℓ counts once among the tangents of C at P, the intersection multiplicity of C and ℓ at P equals deg(C), C has no linear components through P, then C is irreducible.

applying Segre’s criterion

(Segre, 1962)

if there exists a point P ∈ C and a tangent ℓ of C at P such that ℓ counts once among the tangents of C at P, the intersection multiplicity of C and ℓ at P equals deg(C), C has no linear components through P, then C is irreducible. Fa,b(X, Y ) := a + (X p − X + ¯ t)(Y p − Y + ¯ t)2 + (X p − X + ¯ t)2(Y p − Y + ¯ t) − b((X p − X + ¯ t)2 +(X p − X + ¯ t)(Y p − Y + ¯ t) + (Y p − Y + ¯ t)2) = 0

applying Segre’s criterion

(Segre, 1962)

if there exists a point P ∈ C and a tangent ℓ of C at P such that ℓ counts once among the tangents of C at P, the intersection multiplicity of C and ℓ at P equals deg(C), C has no linear components through P, then C is irreducible. Fa,b(X, Y ) := a + (X p − X + ¯ t)(Y p − Y + ¯ t)2 + (X p − X + ¯ t)2(Y p − Y + ¯ t) − b((X p − X + ¯ t)2 +(X p − X + ¯ t)(Y p − Y + ¯ t) + (Y p − Y + ¯ t)2) = 0 at P = X∞ the tangents are ℓ : Y = β with βp − β + ¯ t = b

applying Segre’s criterion

(Segre, 1962)

if there exists a point P ∈ C and a tangent ℓ of C at P such that ℓ counts once among the tangents of C at P, the intersection multiplicity of C and ℓ at P equals deg(C), C has no linear components through P, then C is irreducible. Fa,b(X, Y ) := a + (X p − X + ¯ t)(Y p − Y + ¯ t)2 + (X p − X + ¯ t)2(Y p − Y + ¯ t) − b((X p − X + ¯ t)2 +(X p − X + ¯ t)(Y p − Y + ¯ t) + (Y p − Y + ¯ t)2) = 0 at P = X∞ the tangents are ℓ : Y = β with βp − β + ¯ t = b Fa,b(X, β) = a − b3

applying Segre’s criterion

(Segre, 1962)

if there exists a point P ∈ C and a tangent ℓ of C at P such that ℓ counts once among the tangents of C at P, the intersection multiplicity of C and ℓ at P equals deg(C), C has no linear components through P, then C is irreducible. Fa,b(X, Y ) := a + (X p − X + ¯ t)(Y p − Y + ¯ t)2 + (X p − X + ¯ t)2(Y p − Y + ¯ t) − b((X p − X + ¯ t)2 +(X p − X + ¯ t)(Y p − Y + ¯ t) + (Y p − Y + ¯ t)2) = 0 at P = X∞ the tangents are ℓ : Y = β with βp − β + ¯ t = b Fa,b(X, β) = a − b3 if P / ∈ X CP is irreducible of genus g ≤ 3p2 − 3p + 1

applying Segre’s criterion

(Segre, 1962)

if there exists a point P ∈ C and a tangent ℓ of C at P such that ℓ counts once among the tangents of C at P, the intersection multiplicity of C and ℓ at P equals deg(C), C has no linear components through P, then C is irreducible. Fa,b(X, Y ) := a + (X p − X + ¯ t)(Y p − Y + ¯ t)2 + (X p − X + ¯ t)2(Y p − Y + ¯ t) − b((X p − X + ¯ t)2 +(X p − X + ¯ t)(Y p − Y + ¯ t) + (Y p − Y + ¯ t)2) = 0 at P = X∞ the tangents are ℓ : Y = β with βp − β + ¯ t = b Fa,b(X, β) = a − b3 if P / ∈ X CP is irreducible of genus g ≤ 3p2 − 3p + 1 CP has at least q + 1 − (6p2 − 6p + 2)√q points

cuspidal case: Y = X 3

G is elementary abelian, isomorphic to (Fq, +)

cuspidal case: Y = X 3

G is elementary abelian, isomorphic to (Fq, +) S = {(L(t) + ¯ t, (L(t) + ¯ t)3)

• Pt

| t ∈ Fq} L(T) =

• α∈M

(T − α), M < (Fq, +), #M = m

cuspidal case: Y = X 3

G is elementary abelian, isomorphic to (Fq, +) S = {(L(t) + ¯ t, (L(t) + ¯ t)3)

• Pt

| t ∈ Fq} L(T) =

• α∈M

(T − α), M < (Fq, +), #M = m P = (a, b) is collinear with Px and Py if and only if Fa,b(x, y) := a + (L(x) + ¯ t)(L(y) + ¯ t)2 + (L(x) + ¯ t)2(L(y) + ¯ t) − b((L(x) + ¯ t)2 +(L(x) + ¯ t)(L(y) + ¯ t) + (L(y) + ¯ t)2) = 0

cuspidal case: Y = X 3

G is elementary abelian, isomorphic to (Fq, +) S = {(L(t) + ¯ t, (L(t) + ¯ t)3)

• Pt

| t ∈ Fq} L(T) =

• α∈M

(T − α), M < (Fq, +), #M = m P = (a, b) is collinear with Px and Py if and only if Fa,b(x, y) := a + (L(x) + ¯ t)(L(y) + ¯ t)2 + (L(x) + ¯ t)2(L(y) + ¯ t) − b((L(x) + ¯ t)2 +(L(x) + ¯ t)(L(y) + ¯ t) + (L(y) + ¯ t)2) = 0 if P / ∈ X CP is irreducible of genus g ≤ 3m2 − 3m + 1

cuspidal case: Y = X 3

G is elementary abelian, isomorphic to (Fq, +) S = {(L(t) + ¯ t, (L(t) + ¯ t)3)

• Pt

| t ∈ Fq} L(T) =

• α∈M

(T − α), M < (Fq, +), #M = m P = (a, b) is collinear with Px and Py if and only if Fa,b(x, y) := a + (L(x) + ¯ t)(L(y) + ¯ t)2 + (L(x) + ¯ t)2(L(y) + ¯ t) − b((L(x) + ¯ t)2 +(L(x) + ¯ t)(L(y) + ¯ t) + (L(y) + ¯ t)2) = 0 if P / ∈ X CP is irreducible of genus g ≤ 3m2 − 3m + 1 CP has at least q + 1 − (6m2 − 6m + 2)√q points

(Sz˝

• nyi, 1985 - Anbar, Bartoli, G., Platoni, 2013)

let P = (a, b) be a point in A2(Fq) \ X; if m <

4

• q/36

then there is a secant of S passing through P.

(Sz˝

• nyi, 1985 - Anbar, Bartoli, G., Platoni, 2013)

let P = (a, b) be a point in A2(Fq) \ X; if m <

4

• q/36

then there is a secant of S passing through P. m is a power of p

(Sz˝

• nyi, 1985 - Anbar, Bartoli, G., Platoni, 2013)

let P = (a, b) be a point in A2(Fq) \ X; if m <

4

• q/36

then there is a secant of S passing through P. m is a power of p the points in X \ S need to be dealt with

(Sz˝

• nyi, 1985 - Anbar, Bartoli, G., Platoni, 2013)

let P = (a, b) be a point in A2(Fq) \ X; if m <

4

• q/36

then there is a secant of S passing through P. m is a power of p the points in X \ S need to be dealt with

theorem

if m <

4

• q/36, then there exists a complete cap in A2(Fq) with size

m + q m − 3

(Sz˝

• nyi, 1985 - Anbar, Bartoli, G., Platoni, 2013)

let P = (a, b) be a point in A2(Fq) \ X; if m <

4

• q/36

then there is a secant of S passing through P. m is a power of p the points in X \ S need to be dealt with

theorem

if m <

4

• q/36, then there exists a complete cap in A2(Fq) with size

m + q m − 3 ∼ p1/4 · q3/4

nodal case: XY = (X − 1)3

G is isomorphic to (F∗

q, ·)

G → F∗

q

• v, (v − 1)3

v

• → v

nodal case: XY = (X − 1)3

G is isomorphic to (F∗

q, ·)

G → F∗

q

• v, (v − 1)3

v

• → v

the subgroup of index m (m a divisor of q − 1): K =

• tm, (tm − 1)3

tm

• | t ∈ F∗

q

nodal case: XY = (X − 1)3

G is isomorphic to (F∗

q, ·)

G → F∗

q

• v, (v − 1)3

v

• → v

the subgroup of index m (m a divisor of q − 1): K =

• tm, (tm − 1)3

tm

• | t ∈ F∗

q

• a coset:

S =

• ¯

ttm, (¯ ttm − 1)3 ¯ ttm

• | t ∈ F∗

q

nodal case: XY = (X − 1)3

G is isomorphic to (F∗

q, ·)

G → F∗

q

• v, (v − 1)3

v

• → v

the subgroup of index m (m a divisor of q − 1): K =

• tm, (tm − 1)3

tm

• | t ∈ F∗

q

• a coset:

S =

f (t)

• ¯

ttm ,

g(t)

• (¯

ttm − 1)3 ¯ ttm

• Pt

| t ∈ F∗

q

nodal case: XY = (X − 1)3

G is isomorphic to (F∗

q, ·)

G → F∗

q

• v, (v − 1)3

v

• → v

the subgroup of index m (m a divisor of q − 1): K =

• tm, (tm − 1)3

tm

• | t ∈ F∗

q

• a coset:

S =

f (t)

• ¯

ttm ,

g(t)

• (¯

ttm − 1)3 ¯ ttm

• Pt

| t ∈ F∗

q

• the curve CP:

Fa,b(X, Y ) = a(¯ t3X 2mY m + ¯ t3X mY 2m − 3¯ t2X mY m + 1) −b¯ t2X mY m − ¯ t4X 2mY 2m + 3¯ t2X mY m −¯ tX m − ¯ tY m = 0

(Anbar-Bartoli-G.-Platoni, 2013)

let P be a point in A2(Fq) \ X; if m <

4

• q/36

then there is a secant of S passing through P

(Anbar-Bartoli-G.-Platoni, 2013)

let P be a point in A2(Fq) \ X; if m <

4

• q/36

then there is a secant of S passing through P m is a divisor of q − 1

(Anbar-Bartoli-G.-Platoni, 2013)

let P be a point in A2(Fq) \ X; if m <

4

• q/36

then there is a secant of S passing through P m is a divisor of q − 1 some points from X \ S need to be added to S

(Anbar-Bartoli-G.-Platoni, 2013)

let P be a point in A2(Fq) \ X; if m <

4

• q/36

then there is a secant of S passing through P m is a divisor of q − 1 some points from X \ S need to be added to S

theorem

if m is a divisor of q − 1 with m <

4

• q/36, and in addition (m, q−1

m ) = 1,

then there exists a complete cap in A2(Fq) with size m + q − 1 m − 3

(Anbar-Bartoli-G.-Platoni, 2013)

let P be a point in A2(Fq) \ X; if m <

4

• q/36

then there is a secant of S passing through P m is a divisor of q − 1 some points from X \ S need to be added to S

theorem

if m is a divisor of q − 1 with m <

4

• q/36, and in addition (m, q−1

m ) = 1,

then there exists a complete cap in A2(Fq) with size m + q − 1 m − 3 ∼ q3/4

isolated double point case: Y (X 2 − β) = 1

G cyclic of order q + 1

isolated double point case: Y (X 2 − β) = 1

G cyclic of order q + 1

(Anbar-Bartoli-G.-Platoni, 2013)

if m is a divisor of q + 1 with m <

4

• q/36, and in addition (m, q+1

m ) = 1,

then there exists a complete cap in A2(Fq) with size at most m + q + 1 m

isolated double point case: Y (X 2 − β) = 1

G cyclic of order q + 1

(Anbar-Bartoli-G.-Platoni, 2013)

if m is a divisor of q + 1 with m <

4

• q/36, and in addition (m, q+1

m ) = 1,

then there exists a complete cap in A2(Fq) with size at most m + q + 1 m ∼ q3/4

elliptic case: Y 2 = X 3 + AX + B

if n ∈ [q + 1 − 2√q, q + 1 + 2√q] n ≡ q + 1 (mod p) there exists an elliptic cubic curve X over Fq with #G = n

elliptic case: Y 2 = X 3 + AX + B

if n ∈ [q + 1 − 2√q, q + 1 + 2√q] n ≡ q + 1 (mod p) there exists an elliptic cubic curve X over Fq with #G = n

(Voloch, 1988)

if p does not divide #G − 1, then G can be assumed to be cyclic

elliptic case: Y 2 = X 3 + AX + B

if n ∈ [q + 1 − 2√q, q + 1 + 2√q] n ≡ q + 1 (mod p) there exists an elliptic cubic curve X over Fq with #G = n

(Voloch, 1988)

if p does not divide #G − 1, then G can be assumed to be cyclic problem: no polynomial or rational parametrization of the points of S is possible

elliptic case: Y 2 = X 3 + AX + B

if n ∈ [q + 1 − 2√q, q + 1 + 2√q] n ≡ q + 1 (mod p) there exists an elliptic cubic curve X over Fq with #G = n

(Voloch, 1988)

if p does not divide #G − 1, then G can be assumed to be cyclic problem: no polynomial or rational parametrization of the points of S is possible Voloch’s solution (1990): implicit description of CP

elliptic case: Y 2 = X 3 + AX + B

if n ∈ [q + 1 − 2√q, q + 1 + 2√q] n ≡ q + 1 (mod p) there exists an elliptic cubic curve X over Fq with #G = n

(Voloch, 1988)

if p does not divide #G − 1, then G can be assumed to be cyclic problem: no polynomial or rational parametrization of the points of S is possible Voloch’s solution (1990): implicit description of CP Voloch’s result would provide complete caps of size ∼ q3/4 for every q large enough

elliptic case: Y 2 = X 3 + AX + B

if n ∈ [q + 1 − 2√q, q + 1 + 2√q] n ≡ q + 1 (mod p) there exists an elliptic cubic curve X over Fq with #G = n

(Voloch, 1988)

if p does not divide #G − 1, then G can be assumed to be cyclic problem: no polynomial or rational parametrization of the points of S is possible Voloch’s solution (1990): implicit description of CP Voloch’s result would provide complete caps of size ∼ q3/4 for every q large enough

?

elliptic case

G cyclic m | q − 1 m prime

elliptic case

G cyclic m | q − 1 m prime Tate-Lichtenbaum Pairing < ·, · >: G[m] × G/K → F∗

q/(F∗ q)m

elliptic case

G cyclic m | q − 1 m prime Tate-Lichtenbaum Pairing < ·, · >: G[m] × G/K → F∗

q/(F∗ q)m

if m2 does not divide #G, then for some T in G[m] < T, · >: G/K → F∗

q/(F∗ q)m

is an isomorphism such that K ⊕ Q → [αT(Q)] where αT is a rational function on X

elliptic case

G cyclic m | q − 1 m prime Tate-Lichtenbaum Pairing < ·, · >: G[m] × G/K → F∗

q/(F∗ q)m

if m2 does not divide #G, then for some T in G[m] < T, · >: G/K → F∗

q/(F∗ q)m

is an isomorphism such that K ⊕ Q → [αT(Q)] where αT is a rational function on X S = {R ∈ G | αT(R) = dtm for some t ∈ Fq}

elliptic case

S = {R ∈ X | α(R) = dtm for some t ∈ Fq}

elliptic case

S = {R ∈ X | α(R) = dtm for some t ∈ Fq}

elliptic case

S = {R ∈ X | α(R) = dtm for some t ∈ Fq} P = (a, b) collinear with two points (x, y), (u, v) ∈ S if there exist x, y, u, v, t, z ∈ Fq with

elliptic case

S = {R ∈ X | α(R) = dtm for some t ∈ Fq} P = (a, b) collinear with two points (x, y), (u, v) ∈ S if there exist x, y, u, v, t, z ∈ Fq with                    y 2 = x3 + Ax + B v 2 = u3 + Au + B (x, y) (u, v) (a, b)

b b b

elliptic case

S = {R ∈ X | α(R) = dtm for some t ∈ Fq} P = (a, b) collinear with two points (x, y), (u, v) ∈ S if there exist x, y, u, v, t, z ∈ Fq with                    y 2 = x3 + Ax + B v 2 = u3 + Au + B α(x, y) = dtm α(u, v) = dzm (x, y) (u, v) (a, b)

b b b

elliptic case

S = {R ∈ X | α(R) = dtm for some t ∈ Fq} P = (a, b) collinear with two points (x, y), (u, v) ∈ S if there exist x, y, u, v, t, z ∈ Fq with                    y 2 = x3 + Ax + B v 2 = u3 + Au + B α(x, y) = dtm α(u, v) = dzm det   a b 1 x y 1 u v 1   = 0 (x, y) (u, v) (a, b)

b b b

elliptic case

S = {R ∈ X | α(R) = dtm for some t ∈ Fq} P = (a, b) collinear with two points (x, y), (u, v) ∈ S if there exist x, y, u, v, t, z ∈ Fq with CP :                    y 2 = x3 + Ax + B v 2 = u3 + Au + B α(x, y) = dtm α(u, v) = dzm det   a b 1 x y 1 u v 1   = 0 (x, y) (u, v) (a, b)

b b b

(Anbar-G., 2012)

if A = 0, then CP is irreducible or admits an irreducible Fq-rational component

(Anbar-G., 2012)

if A = 0, then CP is irreducible or admits an irreducible Fq-rational component if m is a prime divisor of q − 1 with m <

4

• q/64, then there exists a

complete cap in A2(Fq) with size at most m + ⌊q − 2√q + 1 m ⌋ + 31

(Anbar-G., 2012)

if A = 0, then CP is irreducible or admits an irreducible Fq-rational component if m is a prime divisor of q − 1 with m <

4

• q/64, then there exists a

complete cap in A2(Fq) with size at most m + ⌊q − 2√q + 1 m ⌋ + 31 ∼ q3/4

ℓ(r, q)2,4

in geometrical terms...

proposition

ℓ(r, q)2,4 = minimum size of a complete cap in Pr−1(Fq)

in geometrical terms...

proposition

ℓ(r, q)2,4 = minimum size of a complete cap in Pr−1(Fq)

trivial lower bound

#S ≥ √ 2q(N−1)/2 in PN(Fq)

in geometrical terms...

proposition

ℓ(r, q)2,4 = minimum size of a complete cap in Pr−1(Fq)

trivial lower bound

#S ≥ √ 2q(N−1)/2 in PN(Fq)

N = 3

TLB: √ 2 · q

N = 3

TLB: √ 2 · q

(Pellegrino, 1999)

1 2q√q + 2

N = 3

TLB: √ 2 · q

(Pellegrino, 1999)

1 2q√q + 2

(Faina, Faina-Pambianco, Hadnagy 1988-1999)

q2/3

N = 3

TLB: √ 2 · q

(Pellegrino, 1999)

computational results

5000 10000 15000 20000 500 1000 1500 2000 2500 3000 3500

recursive constructions of complete caps

blow-up

S cap in Ar(Fqs)

recursive constructions of complete caps

blow-up

S cap in Ar(Fqs) for each P in S, substitute each coordinate in Fqs with its expansion

• ver Fq

(x1, x2, . . . , xr) ∈ Ar(Fqs) (x1

1, x2 1, . . . , xs 1, . . . , x1 r , . . . , xs r ) ∈ Ars(Fq)

recursive constructions of complete caps

blow-up

S cap in Ar(Fqs) for each P in S, substitute each coordinate in Fqs with its expansion

• ver Fq

(x1, x2, . . . , xr) ∈ Ar(Fqs) (x1

1, x2 1, . . . , xs 1, . . . , x1 r , . . . , xs r ) ∈ Ars(Fq)

the resulting subset of Ars(Fq) is a cap

recursive constructions of complete caps

blow-up

S cap in Ar(Fqs) for each P in S, substitute each coordinate in Fqs with its expansion

• ver Fq

(x1, x2, . . . , xr) ∈ Ar(Fqs) (x1

1, x2 1, . . . , xs 1, . . . , x1 r , . . . , xs r ) ∈ Ars(Fq)

the resulting subset of Ars(Fq) is a cap

product

S1 cap in Ar(Fq), S2 cap in As(Fq)

recursive constructions of complete caps

blow-up

S cap in Ar(Fqs) for each P in S, substitute each coordinate in Fqs with its expansion

• ver Fq

(x1, x2, . . . , xr) ∈ Ar(Fqs) (x1

1, x2 1, . . . , xs 1, . . . , x1 r , . . . , xs r ) ∈ Ars(Fq)

the resulting subset of Ars(Fq) is a cap

product

S1 cap in Ar(Fq), S2 cap in As(Fq) S1 × S2 is a cap in Ar+s(Fq)

recursive constructions of complete caps

blow-up

S cap in Ar(Fqs) for each P in S, substitute each coordinate in Fqs with its expansion

• ver Fq

(x1, x2, . . . , xr) ∈ Ar(Fqs) (x1

1, x2 1, . . . , xs 1, . . . , x1 r , . . . , xs r ) ∈ Ars(Fq)

the resulting subset of Ars(Fq) is a cap

product

S1 cap in Ar(Fq), S2 cap in As(Fq) S1 × S2 is a cap in Ar+s(Fq) do such constructions preserve completeness?

recursive constructions of complete caps

TN blow-up of a parabola of A2(FqN/2)

recursive constructions of complete caps

TN blow-up of a parabola of A2(FqN/2)

(Davydov-¨ Osterg` ard, 2001)

TN is complete in AN(Fq) ⇔ N/2 is odd.

recursive constructions of complete caps

TN blow-up of a parabola of A2(FqN/2)

(Davydov-¨ Osterg` ard, 2001)

TN is complete in AN(Fq) ⇔ N/2 is odd. Problem: When TN × S is complete?

external/internal points to a segment

external/internal points to a segment

(Segre, 1973)

P, P1, P2 distinct collinear points in A2(Fq)

b b b

P1 P P2 ℓ

external/internal points to a segment

(Segre, 1973)

P, P1, P2 distinct collinear points in A2(Fq)

b b b

P1 P P2 ℓ the point P is internal or external to the segment P1P2 if (x − x1)(x − x2) is a non-square in Fq or not, x, x1, x2 coordinates of P, P1, P2 w.r.t. any affine frame of ℓ.

bicovering and almost bicovering caps

bicovering and almost bicovering caps

let S be a complete cap in A2(Fq).

b b b b b b b

S

bicovering and almost bicovering caps

let S be a complete cap in A2(Fq). a point P / ∈ S is bicovered by S if it is external to a segment P1P2, with P1, P2 ∈ S and internal to another segment P3P4, with P3, P4 ∈ S

b b b b b b b

S

b

P

bicovering and almost bicovering caps

let S be a complete cap in A2(Fq). a point P / ∈ S is bicovered by S if it is external to a segment P1P2, with P1, P2 ∈ S and internal to another segment P3P4, with P3, P4 ∈ S

b b b b b b b

S

b

P P1 P2

bicovering and almost bicovering caps

let S be a complete cap in A2(Fq). a point P / ∈ S is bicovered by S if it is external to a segment P1P2, with P1, P2 ∈ S and internal to another segment P3P4, with P3, P4 ∈ S

b b b b b b b

S

b

P P1 P2 P3 P4

bicovering and almost bicovering caps

let S be a complete cap in A2(Fq). a point P / ∈ S is bicovered by S if it is external to a segment P1P2, with P1, P2 ∈ S and internal to another segment P3P4, with P3, P4 ∈ S

b b b b b b b

S

b

P P1 P2 P3 P4

definition

S is said to be bicovering if for every P / ∈ S is bicovered by S

bicovering and almost bicovering caps

let S be a complete cap in A2(Fq). a point P / ∈ S is bicovered by S if it is external to a segment P1P2, with P1, P2 ∈ S and internal to another segment P3P4, with P3, P4 ∈ S

b b b b b b b

S

b

P P1 P2 P3 P4

definition

S is said to be bicovering if for every P / ∈ S is bicovered by S almost bicovering if there exists precisely one point not bicovered by S

recursive constructions of complete caps

TN blow-up of a parabola in AN(Fq), N ≡ 2 (mod 4) S complete cap in A2(Fq)

recursive constructions of complete caps

TN blow-up of a parabola in AN(Fq), N ≡ 2 (mod 4) S complete cap in A2(Fq)

(G., 2007)

(i) KS = TN × S is complete if and only if S is bicovering

recursive constructions of complete caps

TN blow-up of a parabola in AN(Fq), N ≡ 2 (mod 4) S complete cap in A2(Fq)

(G., 2007)

(i) KS = TN × S is complete if and only if S is bicovering (ii) if S is almost bicovering, then KS ∪ {(a, a2 − z0, x0, y0) | a ∈ FqN/2} is complete for some x0, y0, z0 ∈ Fq

bicovering caps in A2(Fq)

remarks: no probabilistic result is known

bicovering caps in A2(Fq)

remarks: no probabilistic result is known no computational constructive method is known

bicovering caps in A2(Fq)

remarks: no probabilistic result is known no computational constructive method is known in the Euclidean plane, no conic is bicovering or almost bicovering

b

P

bicovering caps in A2(Fq)

remarks: no probabilistic result is known no computational constructive method is known in the Euclidean plane, no conic is bicovering or almost bicovering

b

P

(Segre, 1973)

if q > 13, ellipses and hyperbolas are almost bicovering caps

bicovering caps in A2(Fq)

remarks: no probabilistic result is known no computational constructive method is known in the Euclidean plane, no conic is bicovering or almost bicovering

b

P

(Segre, 1973)

if q > 13, ellipses and hyperbolas are almost bicovering caps let N ≡ 0 (mod 4); if q > 13, then there exists a complete cap of size #TN−2 · [(q − 1) + 1] = q

N 2

how to prove that an algebraic cap is bicovering

S = {(f (t), g(t))

• Pt

| t ∈ Fq}

how to prove that an algebraic cap is bicovering

S = {(f (t), g(t))

• Pt

| t ∈ Fq} P = (a, b) ∈ A2(Fq)

how to prove that an algebraic cap is bicovering

S = {(f (t), g(t))

• Pt

| t ∈ Fq} P = (a, b) ∈ A2(Fq) (1) consider the space curve YP :

• FP(X, Y ) = 0

(a − f (X))(a − f (Y )) = Z 2

how to prove that an algebraic cap is bicovering

S = {(f (t), g(t))

• Pt

| t ∈ Fq} P = (a, b) ∈ A2(Fq) (1) consider the space curve YP :

• FP(X, Y ) = 0

(a − f (X))(a − f (Y )) = Z 2 (2) apply Hasse-Weil to YP (if possible) and find a suitable point (x, y, z) ∈ YP(Fq)

how to prove that an algebraic cap is bicovering

S = {(f (t), g(t))

• Pt

| t ∈ Fq} P = (a, b) ∈ A2(Fq) (1) consider the space curve YP :

• FP(X, Y ) = 0

(a − f (X))(a − f (Y )) = Z 2 (2) apply Hasse-Weil to YP (if possible) and find a suitable point (x, y, z) ∈ YP(Fq) the point P is external to the segment joining Px and Py

how to prove that an algebraic cap is bicovering

S = {(f (t), g(t))

• Pt

| t ∈ Fq} P = (a, b) ∈ A2(Fq) (1) consider the space curve YP,c :

• FP(X, Y ) = 0

(a − f (X))(a − f (Y )) = cZ 2 (2) apply Hasse-Weil to YP (if possible) and find a suitable point (x, y, z) ∈ YP(Fq) the point P is external to the segment joining Px and Py (3) fix a non-square c in F∗

q and repeat for YP,c

bicovering caps from cubic curves

the method works well for S a coset of a cubic X, and P a point off the cubic.

bicovering caps from cubic curves

the method works well for S a coset of a cubic X, and P a point off the cubic. in order to bicover the points on the cubics, more cosets of the same subgroup are needed: the cosets corresponding to a maximal 3-independent subset in the factor group G/K

bicovering caps from cubic curves

the method works well for S a coset of a cubic X, and P a point off the cubic. in order to bicover the points on the cubics, more cosets of the same subgroup are needed: the cosets corresponding to a maximal 3-independent subset in the factor group G/K in the best case bicovering caps of size approximately q7/8 are

• btained

bicovering caps from cubic curves

the method works well for S a coset of a cubic X, and P a point off the cubic. in order to bicover the points on the cubics, more cosets of the same subgroup are needed: the cosets corresponding to a maximal 3-independent subset in the factor group G/K in the best case bicovering caps of size approximately q7/8 are

• btained

for N ≡ 0 (mod 4) complete caps of size approximately q

N 2 − 1 8 are

• btained, provided that suitable divisors of q, q − 1, q + 1 exist

bicovering caps from cubic curves

the method works well for S a coset of a cubic X, and P a point off the cubic. in order to bicover the points on the cubics, more cosets of the same subgroup are needed: the cosets corresponding to a maximal 3-independent subset in the factor group G/K in the best case bicovering caps of size approximately q7/8 are

• btained

for N ≡ 0 (mod 4) complete caps of size approximately q

N 2 − 1 8 are

• btained, provided that suitable divisors of q, q − 1, q + 1 exist

if Voloch’s gap is filled, we will have bicovering caps with roughly q7/8 points for any odd q

the cuspidal case

X : Y − X 3 = 0

(Anbar-Bartoli-G.-Platoni, 2013)

let q = ph, with p > 3 a prime m = ph′, with h′ < h and m ≤

4

√q 4

then there exists an almost bicovering cap contained in X, of size n =          (2√m − 3) q m, if h′ is even m p + √mp − 3 q m, if h′ is odd ∼ q7/8

the nodal case

X : XY − (X − 1)3 = 0

(Anbar-Bartoli-G.-Platoni, 2013)

assume that q = ph, with p > 3 a prime m is an odd divisor of q − 1, with (3, m) = 1 and m ≤

4

√q 3.5

m = m1m2 s.t. (m1, m2) = 1 and m1, m2 ≥ 4 then there exists a bicovering cap contained in X of size n ≤ m1 + m2 m (q − 1)∼ q7/8

the isolated double point case

X : Y (X 2 − β) = 1

(Anbar-Bartoli-G.-Platoni, 2013)

assume that q = ph, with p > 3 a prime m is a proper divisor of q + 1 such that (m, 6) = 1 and m ≤

4

√q 4

m = m1m2 with (m1, m2) = 1 then there exists an almost bicovering cap contained in X of size less than or equal to (m1 + m2 − 3) · q + 1 m + 3 ∼ q7/8

the elliptic case

X : Y 2 − X 3 − AX − B = 0

(Anbar-G., 2012)

assume that q = ph, with p > 3 a prime m is a prime divisor of q − 1, with 7 < m < 1

8

4

√q then there exists a bicovering cap contained in X of size n ≤ 2√m q − 2√q + 1 m

• + 31
• ∼ q7/8

ℓ(r, q)r−1,r+1

ℓ(r, q)r−1,r+1

Reed-Solomon codes: ℓ(r, q)r−1,r+1 ≤ q + 1

AG codes from elliptic curves

X : Y 2 = X 3 + AX + B 4A3 + 27B2 = 0 O common pole of x and y P1, . . . , Pn rational points of X (distinct from O)

AG codes from elliptic curves

X : Y 2 = X 3 + AX + B 4A3 + 27B2 = 0 O common pole of x and y P1, . . . , Pn rational points of X (distinct from O) Cr = C(D, G)⊥, where G = rO, D = P1 + . . . + Pn, n > r

AG codes from elliptic curves

X : Y 2 = X 3 + AX + B 4A3 + 27B2 = 0 O common pole of x and y P1, . . . , Pn rational points of X (distinct from O) Cr = C(D, G)⊥, where G = rO, D = P1 + . . . + Pn, n > r Cr is an [n, n − r, r + 1]q-MDS-code if and only if for every Pi1, . . . , Pir Pi1 ⊕ . . . ⊕ Pir = O

AG codes from elliptic curves

X : Y 2 = X 3 + AX + B 4A3 + 27B2 = 0 O common pole of x and y P1, . . . , Pn rational points of X (distinct from O) Cr = C(D, G)⊥, where G = rO, D = P1 + . . . + Pn, n > r Cr is an [n, n − r, r + 1]q-MDS-code if and only if for every Pi1, . . . , Pir Pi1 ⊕ . . . ⊕ Pir = O

(Munuera, 1993)

If Cr is MDS then, for n > r + 2, n ≤ 1 2(#X(Fq) − 3 + 2r)

covering radius of elliptic MDS codes

a subset T of an abelian group H is r-independent if for each a1, . . . , ar ∈ T, a1 + a2 + . . . + ar = 0

covering radius of elliptic MDS codes

a subset T of an abelian group H is r-independent if for each a1, . . . , ar ∈ T, a1 + a2 + . . . + ar = 0 {P1, . . . , Pn} maximal r-independent subset of X(Fq)

covering radius of elliptic MDS codes

a subset T of an abelian group H is r-independent if for each a1, . . . , ar ∈ T, a1 + a2 + . . . + ar = 0 {P1, . . . , Pn} maximal r-independent subset of X(Fq) let φr : X → Pr−1 φr = (1 : f1 : . . . : fr−1) with 1, f1, . . . , fr−1 basis of L(rO)

covering radius of elliptic MDS codes

a subset T of an abelian group H is r-independent if for each a1, . . . , ar ∈ T, a1 + a2 + . . . + ar = 0 {P1, . . . , Pn} maximal r-independent subset of X(Fq) let φr : X → Pr−1 φr = (1 : f1 : . . . : fr−1) with 1, f1, . . . , fr−1 basis of L(rO) R(Cr) = r − 1 if and only if each point in Pr−1(Fq) belongs to the hyperplane generated by some φr(Pi1), φr(Pi2), . . . , φr(Pir−1)

(Bartoli-G.-Platoni, 2013)

if

• (X(Fq), ⊕) ∼

= Zm × K cyclic for m > 3 a prime

• S = K ⊕ P covers all the points in A2(Fq) off X
• T ⊃ S is a maximal r-independent subset of X(Fq)

(Bartoli-G.-Platoni, 2013)

if

• (X(Fq), ⊕) ∼

= Zm × K cyclic for m > 3 a prime

• S = K ⊕ P covers all the points in A2(Fq) off X
• T ⊃ S is a maximal r-independent subset of X(Fq)

then almost every point in Pr−1(Fq) belongs to some hyperplane generated by r − 1 points of φr(T)

(Bartoli-G.-Platoni, 2013)

if

• (X(Fq), ⊕) ∼

= Zm × K cyclic for m > 3 a prime

• S = K ⊕ P covers all the points in A2(Fq) off X
• T ⊃ S is a maximal r-independent subset of X(Fq)

then almost every point in Pr−1(Fq) belongs to some hyperplane generated by r − 1 points of φr(T) if m is a prime divisor of q − 1 with m <

4

• q/64, then

ℓ(r, q)r−1,r+1 ≤ (⌈r/2⌉−1)(|S|−1)+2m + 1 r − 2 +2r

(Bartoli-G.-Platoni, 2013)

if

• (X(Fq), ⊕) ∼

= Zm × K cyclic for m > 3 a prime

• S = K ⊕ P covers all the points in A2(Fq) off X
• T ⊃ S is a maximal r-independent subset of X(Fq)

then almost every point in Pr−1(Fq) belongs to some hyperplane generated by r − 1 points of φr(T) if m is a prime divisor of q − 1 with m <

4

• q/64, then

ℓ(r, q)r−1,r+1 ≤ (⌈r/2⌉−1)(|S|−1)+2m + 1 r − 2 +2r∼ (⌈r/2⌉ − 1)q3/4

(Bartoli-G.-Platoni, 2013)

if

• (X(Fq), ⊕) ∼

= Zm × K cyclic for m > 3 a prime

• S = K ⊕ P covers all the points in A2(Fq) off X
• T ⊃ S is a maximal r-independent subset of X(Fq)

then almost every point in Pr−1(Fq) belongs to some hyperplane generated by r − 1 points of φr(T) if m is a prime divisor of q − 1 with m <

4

• q/64, then

ℓ(r, q)r−1,r+1 ≤ (⌈r/2⌉−1)(|S|−1)+2m + 1 r − 2 +2r∼ (⌈r/2⌉ − 1)q3/4

problems

explanation for experimental results

problems

explanation for experimental results Voloch’s proof for plane caps in elliptic cubics

problems

explanation for experimental results Voloch’s proof for plane caps in elliptic cubics bicovering caps in dimensions different from 2

problems

explanation for experimental results Voloch’s proof for plane caps in elliptic cubics bicovering caps in dimensions different from 2 non-recursive constructions in higher dimensions

problems

explanation for experimental results Voloch’s proof for plane caps in elliptic cubics bicovering caps in dimensions different from 2 non-recursive constructions in higher dimensions probabilistic results intrinsic to higher dimensions