exposing the langlands conjecture: interconnector between the antipodal galois and lie theories...

Post on 25-Dec-2015

215 Views

Category:

Documents

0 Downloads

Preview:

Click to see full reader

TRANSCRIPT

Exposing the Langlands Conjecture: Interconnector

Between the Antipodal Galois and Lie Theories

Megan O’Reilly

Advisor Dr. Simon Quint

Distinguished Student Fellowship Project

Outline

Robert Langlands/Langlands Program

Algebraic Structures Galois Theory Lie Theory First Conjecture

L-functions P-adic numbers Ideles Adeles

Hecke characters Artin’s Abelian Reciprocity

Langlands 2nd Conjecture Langlands 3rd Conjecture

Robert P. Langlands

born on October 6, 1936 in New Westminster, British Columbia, Canada

1960 he earned a Ph.D. from Yale University in New Haven, Connecticut

professor at the Institute for Advanced Study in Princeton

In 2007 he won a Shaw prize , an international award worth $1 million.

He was especially interested in the relations between Galois Theory of Number Fields and the theory of automorphic forms.

These relations formed the basis of a network of interrelated and increasingly precise conjectures known as Langlands Program.

The Langlands Program

Emerged in the late 1960’s Langlands’ conjectures have given rise to predictions

regarding properties of automorphic forms on specific groups.

Conjectures analogous to Langlands conjectures have appeared in algebraic geometry, mathematical physics, representation theory, number theory and harmonic analysis.

Andrew Wiles used parts of Langlands conjectures in the proof of Fermat’s Last Theorem.

Fermat’s Last Theorem is given xn+yn=zn for n greater than 2 there do not exist integers x, y, and z nonzero that satisfy the equation.

Relevant Classes for Introductory Background to Topics

Algebraic StructuresComplex AnalysisAdvanced Calculus Linear Algebra

Algebraic Structures

Group- a set G with operation * is said to be a

group denoted (G,*) if (G,*) satisfies associativity, inverses, identity and closure. *

Ring – Z is a commutative group with addition and multiplication satisfies closure, the associative property, and the distributive property.

A field- Q is a ring plus it is commutative with multiplication, has a multiplicative identity and every element not zero has a multiplicative inverse

Évariste Galois

1811-1832 France He developed his

Galois groups and Galois fields to show there are no formulas for fifth degree or higher polynomials analogous to the quadratic formula.

Galois Theory

For a field E that contains subfield F, E is called an extension of F. *

A Galois field is a special field related to polynomials.

Subfields of the Galois field correspond to subgroups of the Galois group.

Galois Theory continued

The group G(Q(i)/Q)={Identity, Conjugate} is the group of automorphisms (special maps) of Q(i) leaving Q fixed.

1. For the subfield Q of Q(i), the corresponding Galois group is λ(Q)=all the automorphs, denoted σ, of G(Q(i)/Q) such that σ(e)=e for all e in Q.

λ(Q)={Id, Conj} (Given subfield, found the subgroup)

2. For the subgroup H={Id} of G(Q(i)/Q)={Id,Conj} the corresponding subfield is all e in Q(i) such that σ(e)=e for all σ in H.

The subfield is Q(i). (Given the subgroup, found the subfield)

Marius Sophus Lie

1842-1899 Norway In his early twenties

he learned Abel and Galois Theory, much of which would influence his later work.

Best known for his development of Lie Algebras and Lie Groups.

Lie Theory

Certain matrix groups are Lie groups.Bracket Multiplication: Let A, B be nxn

matrices then [A,B]=AB-BAA real Lie Algebra of matrices is a real

vector space closed under bracket multiplication. *

Lie Theory Continued

Lie Algebras correspond to Lie GroupsLie subalgebras of the Lie Algebra of a Lie

group correspond to connected Lie subgroups

Given a Lie Group of matrices one can find the corresponding Lie Algebra via the derivative.

Given a Lie Algebra of matrices one can find the corresponding Lie Group via the exponential map **

First Conjecture

Suppose that G is a Galois group of a nonabelian Galois Field and that R is an irreducible representation with the Artin L-series L(s,R), then there exists an irreducible representation π of GL(n,AQ) for which L(s,π)=L(s,R)

Emil Artin

1898-1962 Vienna, Austria

Drafted into World War I

Many achievements in mathematics including solving Hilbert’s ninth problem.

Background for Artin’s L-function

Character א of a group (G, ©) is a mapping from G into the set C-{0} such that א(g©h)= א(g) א(h) Character א (mod k) is a special type of character

defined on the residue classes (mod k) such (a)אthat א(a)=0 if and only if gcd(a,k)>1 *

A prime in Z is a positive integer other than 1 such that whenever you write it as a product of two integers one of the factors is one.

In Q(i) a prime is any m+ni (integer m+ni in Q(i), m+ni≠ ±1 or ±i) such that if m+ni=product of 2 other integers, one of them is ±1 or ±i. 2=(1+i)(1-i) is not prime, 3 is prime

Background for Artin’s L-function continued

AK=ring of integers of K where an integer is a solution of p(x)=xn+an-1xn-1+…+a0 with coefficients in Z.

For a Galois field K, an ordinary prime p is unramified if the ideal pAK factors uniquely into a finite product ΠPi of ideals each one to the exponent 1; otherwise p is ramified.

G(K,Q), a Galois group. Stability subgroup GP={g in G: g(P)=P}

Background for Artin’s L-function continued

The Frobenius element Frp is a special element of the Galois group. If the Galois group is abelian, then the Frp is the generator of the stability group which is cyclic for ramified primes.

Theorem (for Q(i)) Suppose that p is an unramified rational prime 1) If p=1 mod 4 then ideal (p)=P1P2, distinct prime ideals and Frp=Id (2) If p=3 mod 4 the ideal (p)=P and Frp=Conj *

Riemann Zeta Function and L-Function L(א,s)

ζ(s)=Π(1-1/ps)-1 the product over all rational primes where s=x+yi is a complex number and Re(s)>1.

L-function for a character א (mod k) is L(s, א)= Π(1- א(p)/ps)-1 and Re(s)>1

The norm of z in Q(i), written N(z), is (z)*(Conj(z))

The Zeta Dedekind Function of K a Galois Field is ζK(s)= Π(1-1/N(P)s)-1 over the prime ideals P.**

Artin’s L-Functions for Abelian Number Fields

For a character א of the Galois group G(K,Q) Artin’s L-function is

L(s,א ) = A Π(1- א(FrP) /ps)-1 where the product is over p’s in Z that are

unramified, A is a factor corresponding to ramified primes p, s is a complex number, and the product converges for Re(s)>1 *

Artin’s L-Function for a General Galois Field Over QFor an irreducible n-dimensional

representation R of the Galois group G(K,Q)

L(s,R)= ΠDet[(1- Tr(R(FrP)) /ps)-1]

Tr=normal trace, FrP is a conjugacy class of automorphs, P a prime in K lying over p.

For n=1 R=character, an irreducible representation

P-adic Metric on Q

For q=p1k1p2

k2p3k3…pr

kr, by definition the p-adic absolute value of q is |q|pi=p-ki.

Example: q=12/125=223/53=223(5-3) Then the 2-adic absolute value of q is |q|2=2-2, the 5-adic absolute value of q is |q|5=53, and the 11-adic absolute value is |q|=110=1.

The Field Qp of P-adic Numbers

Qp contains the subfield Q, with the p-adic metric and every nonzero element [s] of Qp is an abstract equivalence class of Cauchy sequences (a special type of sequence).

Reduced Form of a P-adic Number

X=pN(d0+d1p+d2p2+…+dipi+…) where N is a ordinary integer, each di is a ordinary integer 0<=di<p, and the infinite series in the parentheses converges in the p-adic metric.

Now the p-adic absolute value of ANY p-adic number x is |x|p=p-N *

Ideles

The full infinite product of all p-adic fields is Q∞xQ2xQ3xQ4x…

Note: R=Q∞

The elements of this product are of the form (r,x2,x3,x5,x7,…,xp,…)=(xp) with xp in Qp

The group of ideles of Q denoted IQ is the set of all elements (xp) such that all but finitely many of the coordinates xp are p-adic integers of absolute value 1.

(π,3,5,7,11,13,…) is an idele

Adeles

The ring of adeles of Q denoted AQ is the set of all (xp) such that all but finitely many of the coordinates have an absolute value less than or equal to one.

(xp)=(4,2,3,5,7,…) is an adele

Hecke Characters

A Hecke character אH is a map from IQ into C-{0} such that | אH ((x))|=1 for all (x) in IQ.

David Hilbert

Hilbert in his influential 1900 address listed twenty three unsolved problems which he believed would substantially change the world of mathematics.

Hilbert’s ninth Problem is: Reciprocity in Algebraic Number Fields. His problem set as a goal to put quadratic reciprocity, cubic reciprocity, and all known reciprocity laws into one format.

Quadratic Residues & Reciprocity

Suppose that p is a rational prime and that a is a rational integer relatively prime to p, then a is a quadratic residue mod p if there exists a rational integer x such that x2≡a mod p.

1. Suppose that at least one of p and q (where both p and q are odd) is congruent to 1 mod 4, then p is a quadratic residue mod q iff q is a quadratic residue mod p

2. Suppose that both p and q are congruent to 3 mod 4, then one of them, say p, is a quadratic residue mod the other one, say q, iff q is NOT a quadradic residue mod of p. *

Artin’s Reciprocity for Galois Fields

Theorem: Suppose that G is the Galois group of an abelian Galois field and that R is a character of that group with the Artin L-series L(s,R). Then there exists a Hecke character אH of the ideles IQ for which L(s, אH)=L(s,R).

Every Hecke character has an L-series and every congruence character gives rise to a Hecke character and those series match.

After Artin proved his reciprocity theorem- relating his L-functions for abelian Galois fields with Hecke’s L-functions- he then realized that his theorem solved Hilbert’s ninth problem.

Example of Abelian Reciprocity

For a quadratic number field K, and a congruence character אD (D the discriminant) ζK(s)= ζ(s)L(s, אD) L(s, אD )=L(s, אH) ζK(s)=LK/Q(Id,s)LK/Q(R,s)

Using Artin’s Reciprocity LK/Q(R,s)=L(אH,s) LK/Q(Id,s)= ζ(s)=L(s,I) Hecke L-function

On the Lie Side

For the abelian group case Emil Artin’s reciprocity theorem applies.

For the nonabelian case:

If R is an n-dimensional irreducible representation of a Galois group of order n, then the Langlands conjecture #1 says that there is an irreducible infinite dimensional representation πA of the group GL(n,AQ)-- which has an L-function L(s, π) – such that L(s, π)=L(s,R).

On the Galois Side

For the Galois group of a Galois field, we need irreducible representations to build the Artin L-functions. When the Galois group is abelian, then the irreducible representation is a character.

Langlands 2nd Conjecture

For Lie groups G each such G has an L-group LG0, a special Lie group. Conjecture 2 involves L-series associated with these groups and finite dimensional representations of them.

G LG0

GL(n) GL(n,C)

SL(n) GL(n,C)/center

Langlands 3rd Conjecture

Now couples Galois theory and Lie theory for a Galois field K. One can build from G(K,Q) and LG0 (the L-group of G), LG (the Galois form of the L-group) from notions in Lie Theory and there is a relation between the L-series of G and LG.

Internet Sites

Autobiography of Robert Langlands. The Shaw Prize. 10 April 2008 <http://www.shawprize.org/en/laureates/2007/mathematical/Langlands_Taylor/autobiography/Langlands.html>

Bernstein, J. and S Gelbart. Ed. “Introduction to the Langlands Program.” Bulletin of the American Mathematical Society. 30 January 2004: Volume 41, Number 2, pages 257-266. 10 April 2008 <http://www.ams.org/bull/2004-41-02/S0273-0979-04-01007-9/S0273-0979-04-01007-9.pdf>.

Biographies of Mathematicians - Évariste Galois. November 1998. Berrien County MSC. 10 April 2008 <http://www.andrews.edu/~calkins/math/biograph/biogaloi.htm>.

Lie, Sophus." Online Photograph. Encyclopædia Britannica Online. 20  Mar.  2008  <http://www.britannica.com/eb/art-12613>.

O’Connor, JJ and EF Robertson. “Marius Sophus Lie.” February 2000. 10 April 2008. <http://www-groups.dcs.st-and.ac.uk/~history/Biographies/Lie.html>.

“The Galois Story.” Online Photograph. Ivers Peterson’s Math Trek. 4 April 2008 <http://images.google.com/imgres?imgurl=http://www.maa.org/mathland/galois.jpg&imgrefurl=http://www.maa.org/mathland/mathtrek_3_1_99.html&h=326&w=268&sz=16&hl=en&start=15&tbnid=LlgAE-2rB9XFNM:&tbnh=118&tbnw=97&prev=/images%3Fq%3DGalois%26gbv%3D2%26hl%3Den>

top related