Frobenius Distributions - PDF Free Download

Frobenius Distributions Edgar Costa (MIT) September 11th, 2018 Massachusetts Institute of Technology Slides available at edgarcosta.org under Research

Polynomials Write f p (x) := f(x) mod p f(x) = a n x n + + a 0 Z[x] Given f p (x) what can we say about f(x)?

Polynomials Write f p (x) := f(x) mod p f(x) = a n x n + + a 0 Z[x] Given f p (x) what can we say about f(x)? factorization of f p (x) factorization of f(x) e.g.: f p(x) irreducible f(x) irreducible factorization of p in Q[x]/f(x) What can we say about f p (x) for arbitrary p? For deg f = 2, quadratic reciprocity gives us that N f (p) := #{α F p : f p (α) = 0} depending only on p mod (f).

Example: Cubic polynomials Theorem (Frobenius) Prob(N f (p) = i) = Prob(g Gal(f) : g fixes i roots),

Example: Cubic polynomials Theorem (Frobenius) Prob(N f (p) = i) = Prob(g Gal(f) : g fixes i roots), ) ( ) ( ) f(x) = x 3 2 = (x 3 2 x 3 2e 2πi/3 x 3 2e 4πi/3 Prob ( N f (p) = k ) 1/3 if k = 0 = 1/2 if k = 1 1/6 if k = 3. g(x) = x 3 x 2 2x + 1 = (x α 1 ) (x α 2 ) (x α 3 ) 2/3 if k = 0 Prob (N g (p) = k) = 1/3 if k = 3.

Example: Cubic polynomials Theorem (Frobenius) Prob(N f (p) = i) = Prob(g Gal(f) : g fixes i roots), ) ( ) ( ) f(x) = x 3 2 = (x 3 2 x 3 2e 2πi/3 x 3 2e 4πi/3 Prob ( N f (p) = k ) 1/3 if k = 0 = 1/2 if k = 1 Gal(f) = S 3 1/6 if k = 3. g(x) = x 3 x 2 2x + 1 = (x α 1 ) (x α 2 ) (x α 3 ) 2/3 if k = 0 Prob (N g (p) = k) = Gal(g) = Z/3Z 1/3 if k = 3.

Elliptic curves E : y 2 = x 3 + ax + b, a, b Z

Elliptic curves E : y 2 = x 3 + ax + b, a, b Z Write E p := E mod p, for p a prime of good reduction What can we say about #E p for an arbitrary p?

Elliptic curves E : y 2 = x 3 + ax + b, a, b Z Write E p := E mod p, for p a prime of good reduction What can we say about #E p for an arbitrary p? Given #E p for many p, what can we say about E?

Elliptic curves E : y 2 = x 3 + ax + b, a, b Z Write E p := E mod p, for p a prime of good reduction What can we say about #E p for an arbitrary p? Given #E p for many p, what can we say about E? studying the statistical properties #E p.

Hasse s bound Theorem (Hasse) #E p = p + 1 a p, a p [ 2 p, 2 p]

Hasse s bound Theorem (Hasse) #E p = p + 1 a p, a p [ 2 p, 2 p] Alternatively, we could also have written the formula above as #E p = L(1), where L(T) = 1 a p T + pt 2 = det(1 T Frob p H 1 (E)) Question What can we say about the error term a p / p as p?

Two types of elliptic curves a p = p + 1 #E p = Tr Frob p [ 2 p, 2 p] There are two limiting distributions for a p / p

Two types of elliptic curves a p = p + 1 #E p = Tr Frob p [ 2 p, 2 p] There are two limiting distributions for a p / p -2-1 1 2-2 -1 0 1 2

Two types of elliptic curves a p = p + 1 #E p = Tr Frob p [ 2 p, 2 p] There are two limiting distributions for a p / p non-cm End Q E al = Q CM End Q E al = Q( d) -2-1 1 2-2 -1 0 1 2

Two types of elliptic curves a p = p + 1 #E p = Tr Frob p [ 2 p, 2 p] There are two limiting distributions for a p / p non-cm End Q E al = Q CM End Q E al = Q( d) -2-1 1 2-2 -1 0 1 2 Prob(a p = 0)? 1/ p Prob(a p = 0) = 1/2

How to distinguish between the two types? non-cm End Q E al = Q CM End Q E al = Q( d) -2-1 1 2-2 -1 0 1 2 End Q E al End Q E al p Q(Frob p ) a p 0 End Q E p al is a quadratic field

How to distinguish between the two types? non-cm End Q E al = Q CM End Q E al = Q( d) -2-1 1 2-2 -1 0 1 2 End Q E al End Q E al p Q(Frob p ) a p 0 End Q E p al is a quadratic field If E has CM and a p 0, then End Q E al End Q E al p.

Examples E : y 2 + y = x 3 x 2 10x 20 (11.a2) End Q E al 2 Q( 1) End Q E al 3 Q( 11) End Q E al = Q

Examples E : y 2 + y = x 3 x 2 10x 20 (11.a2) End Q E al 2 Q( 1) End Q E al 3 Q( 11) End Q E al = Q E : y 2 + y = x 3 7 (27.a2) p = 2 mod 3 a p = 0 End Q E al p is a Quaternion algebra p = 1 mod 3 End Q E al p Q( 3) End Q E al = Q( 3)

Genus 2 curves/abelian surfaces A := Jac(C : y 2 = a 6 x 6 + + a 0 )

Genus 2 curves/abelian surfaces A := Jac(C : y 2 = a 6 x 6 + + a 0 ) There are 6 possibilities for the real endomorphism algebra: Abelian surface square of CM elliptic curve QM abelian surface square of non-cm elliptic curve CM abelian surface product of CM elliptic curves product of CM and non-cm elliptic curves RM abelian surface product of non-cm elliptic curves generic abelian surface End R A al M 2 (C) M 2 (R) C C C R R R R

Zeta functions and Frobenius polynomials C/Q a nice curve of genus g A := Jac(C) p a prime of good reduction ( ) Z p (T) := exp #C(F p r)t r /r where deg L p (T) = 2g and r=1 Q(t) L p (T) = det(1 t Frob p H 1 (C)) = det(1 t Frob p H 1 (A))

Zeta functions and Frobenius polynomials C/Q a nice curve of genus g A := Jac(C) p a prime of good reduction ( ) Z p (T) := exp #C(F p r)t r /r r=1 where deg L p (T) = 2g and = L p (T) (1 T)(1 pt) L p (T) = det(1 t Frob p H 1 (C)) = det(1 t Frob p H 1 (A)) g = 1 L p (T) = 1 a p T + pt 2 g = 2 L p (T) = 1 a p,1 T + a p,2 T 2 a p,1 pt 3 + p 2 T 4

Endomorphism algebras over finite fields Theorem (Tate) Let A be an abelian variety over F q, given det(1 t Frob H 1 (A)), we may compute rk End(A Fq r ), r 1

Endomorphism algebras over finite fields Theorem (Tate) Let A be an abelian variety over F q, given det(1 t Frob H 1 (A)), we may compute rk End(A Fq r ), r 1 Honda Tate theory = gives us End Q (A Fq r ) up to isomorphism Example If L 5 (T) = 1 2T 2 + 25T 4, then: all endomorphisms are defined over F 25, and A F25 is isogenous to a square of an elliptic curve End Q A al M 2 (Q( 6))

Example continued A = Jac(y 2 = x 5 x 4 + 4x 3 8x 2 + 5x 1) (262144.d.524288.1) For p = 5, L 5 (T) = 1 2T 2 + 25T 4, and: all endomorphisms of A 5 are defined over F 25 det(1 T Frob 2 5 H 1 (A)) = (1 2T + 25T 2 ) 2 A 5 over F 25 is isogenous to a square of an elliptic curve End Q A al 5 M 2(Q( 6)) For p = 7, L 7 (T) = 1 + 6T 2 + 49T 4, and: all endomorphisms of A 7 are defined over F 49 det(1 T Frob 2 7 H 1 (A)) = (1 + 6T + 49T 2 ) 2 A 7 over F 49 is isogenous to a square of an elliptic curve End Q A al 7 M 2(Q( 10))

Same L p, different approach We could have looked at the lattice Néron Severi lattice. rk NS(A al ) {1, 2, 3, 4} rk NS(A al p ) {2, 4, 6} Example NS(A al ) NS(A al p ) rk NS(A al 5 ) = rk NS(Aal 7 ) = 4 disc NS(A al 5 ) = 6 mod Q 2 disc NS(A al 7 ) = 10 mod Q 2 rk NS(A al ) 3 Fact By a theorem of Charles, we know that at some point this method will attain a tight upper bound for rk NS(A al ).

Real endomorphisms algebras and Picard numbers Abelian surface End R A al rk NS(A al ) square of CM elliptic curve M 2 (C) 4 QM abelian surface M 2 (R) 3 square of non-cm elliptic curve CM abelian surface C C 2 product of CM elliptic curves product of CM and non-cm elliptic curves C R 2 RM abelian surface R R 2 product of non-cm elliptic curves generic abelian surface R 1

Higher genus Let K be a numberfield such that End A K = End A al, then A K t i=1 An i i, A i unique and simple up to isogeny (over K), B i := End Q A i central simple algebra over L i := Z(B i ), dim Li B i = e 2 i, End Q A K = t i=1 M n i (B i ) Theorem (C Mascot Sijsling Voight) If Mumford Tate conjecture holds for A, then we can compute t {(e i n i, n i dim A i )} t i=1 L i, under an additional assumption (always seems to hold) This is practical and its done by counting points (=computing L p )

Real endomorphisms algebras, {e i n i, n i dim A i } t i=1, and dim L i Abelian surface End R A al tuples dim L i square of CM elliptic crv M 2 (C) {(2, 2)} 2 QM abelian surface M 2 (R) {(2, 2)} 1 square of non-cm elliptic crv CM abelian surface C C {(1, 2)} 4 product of CM elliptic crv {(1, 1), (1, 1)} 2, 2 CM non-cm elliptic crvs C R {(1, 1), (1, 1)} 2, 1 RM abelian surface R R {(1, 2)} 2 prod. of non-cm elliptic crv {(1, 1), (1, 1)} 1, 1 generic abelian surface R {(1, 1)} 1

Example continued A = Jac(y 2 = x 5 x 4 + 4x 3 8x 2 + 5x 1) End Q A al 3 M 2(Q( 3)) End Q A al 5 M 2(Q( 6)) End R A al M 2 (C) Question (262144.d.524288.1) Write B := End Q A al and assume that B is a quaternion algr. Can we guess disc B? If l ell is ramified in B l cannot split in Q(Frob p ) 5, 13, 17 disc B, as they split in Q( 3) 7, 11 disc B, as they split in Q( 6) We can rule out all the primes except 2 and 3 (up to some bnd).

K3 surfaces K3 surfaces are a possible generalization of elliptic curves They may arise in many ways: smooth quartic surfaces in P 3 X : f(x, y, z, w) = 0, deg f = 4 double cover of P 2 branched over a sextic curve X : w 2 = f(x, y, z), deg f = 6 Can we play similar game as before?

K3 Surfaces X/Q a K3 surface p rk NS X al p {2, 4,..., 22} This is analogous to studying: p rk End E al p {2, 4}

K3 Surfaces X/Q a K3 surface p rk NS X al p {2, 4,..., 22} This is analogous to studying: p rk End E al p {2, 4} Recall that: rk End E al p = 4 a p = 0? 1 Prob(a p = 0) = p if E is non-cm (Lang Trotter) 1/2 if E has CM by Q( d)

Néron Severi group NS = Néron Severi group of {curves on }/ ρ( ) = rk NS X p := X mod p

Néron Severi group NS = Néron Severi group of {curves on }/ ρ( ) = rk NS X p := X mod p X NS X al ρ(x al ) {1, 2,..., 20}??? X p NS X p al ρ(x p al ) {2, 4,... 22}

Néron Severi group NS = Néron Severi group of {curves on }/ ρ( ) = rk NS X p := X mod p X NS X al ρ(x al ) {1, 2,..., 20}??? X p NS X p al ρ(x p al ) {2, 4,... 22} Theorem (Charles) For infinitely many p we have ρ(x p al ) = min q ρ(x q al ).

The Problem X NS X al ρ(x al ) {1, 2,..., 20}??? X p NS X p al ρ(x p al ) {2, 4,... 22} Theorem (Charles) For infinitely many p we have ρ(x p al ) = min q ρ(x q al ). What can we say about the following: Π jump (X) := { p : ρ(x p al ) > min q ρ(x q al ) }

Two generic K3 surfaces, ρ(x al ) = 1 γ(x, B)? c X B, B

Two generic K3 surfaces, ρ(x al ) = 1 γ(x, B)? c X B, B = Prob(p Π jump (X))? 1/ p

Two generic K3 surfaces, ρ(x al ) = 1 γ(x, B)? c X B, B = Prob(p Π jump (X))? 1/ p Why?

Three K3 surfaces with ρ(x al ) = 2

Three K3 surfaces with ρ(x al ) = 2 No obvious trend

Three K3 surfaces with ρ(x al ) = 2 No obvious trend Could it be related to some integer being a square modulo p?

We can explain the 1/2 Theorem (C, C Elsenhans Jahnel) If ρ(x al ) = min q ρ(x al p ), then there is a d X Z such that: { p > 2 : p inert in Q( } d X ) Π jump (X). In general, d X is not a square.

We can explain the 1/2 Theorem (C, C Elsenhans Jahnel) If ρ(x al ) = min q ρ(x al p ), then there is a d X Z such that: { p > 2 : p inert in Q( } d X ) Π jump (X). In general, d X is not a square. Corollary If d X is not a square: lim inf B γ(x, B) 1/2 X al has infinitely many rational curves. D 3 = 1 5 151 22490817357414371041 3873084974301493372336663588079962607808750567408509842132769703432789353 D 4 =53 2624174618795407 512854561846964817139494202072778341 1215218370089028769076718102126921744353362873 68 D 5 = 1 47 3109 4969 14857095849982608071 445410277660928347762586764331874432202584688016149 658652708525

Experimental data for ρ(x al ) = 2 (again) What if we ignore { p > 2 : p inert in Q( d X ) } Π jump (X)?

Experimental data for ρ(x al ) = 2 (again) What if we ignore { p > 2 : p inert in Q( d X ) } Π jump (X)? ) γ (X ( )?, B c, B Q dx B 1 0.50 γ( ) - 1 0.50 γ( ) - 1 0.50 γ( ) - 0.10 0.05 0.10 0.05 0.10 0.05 100 1000 10 4 10 5 1000 10 4 10 5 100 1000 10 4 10 5

Experimental data for ρ(x al ) = 2 (again) What if we ignore { p > 2 : p inert in Q( d X ) } Π jump (X)? ) γ (X ( )?, B c, B Q dx B 1 0.50 γ( ) - 1 0.50 γ( ) - 1 0.50 γ( ) - 0.10 0.05 0.10 0.10 0.05 0.05 100 1000 10 4 10 5 1000 10 4 10 5 100 1000 10 4 10 5 1 if d X is not a square modulo p Prob(p Π jump (X)) =? 1 otherwise p Why?!?