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HOW TO COUNT WITH TOPOLOGY How to Count with Topology Jordan S. Ellenberg University of Wisconsin-Madison AMS-MAA Joint Mathematics Meetings January 11, 2013

How to Count with Topology - Quomodocumque

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Page 1: How to Count with Topology - Quomodocumque

HOW TO COUNT WITH TOPOLOGY

How to Count with Topology

Jordan S. EllenbergUniversity of Wisconsin-Madison

AMS-MAA Joint Mathematics MeetingsJanuary 11, 2013

Page 2: How to Count with Topology - Quomodocumque

HOW TO COUNT WITH TOPOLOGY

COUNTING SQUAREFREE INTEGERS

How many squarefree integers are there? (That is: how manyintegers with no square divisor other than 1?)

Boring answer: infinitely many.

1, 2, 3, 5, 6, 7, 10, 11, 13, 14, 15, 17, 19, . . .

Page 3: How to Count with Topology - Quomodocumque

HOW TO COUNT WITH TOPOLOGY

COUNTING SQUAREFREE INTEGERS

How many squarefree integers are there? (That is: how manyintegers with no square divisor other than 1?)

Boring answer: infinitely many.

1, 2, 3, 5, 6, 7, 10, 11, 13, 14, 15, 17, 19, . . .

Page 4: How to Count with Topology - Quomodocumque

HOW TO COUNT WITH TOPOLOGY

COUNTING SQUAREFREE INTEGERS

Better question: How many squarefree integers are therebetween N and 2N? Call this function sf (N).

sage: len([x for x in range(1000,2000)if Integer(x).is_squarefree()])607

Page 5: How to Count with Topology - Quomodocumque

HOW TO COUNT WITH TOPOLOGY

COUNTING SQUAREFREE INTEGERS

Better question: How many squarefree integers are therebetween N and 2N? Call this function sf (N).

sage: len([x for x in range(1000,2000)if Integer(x).is_squarefree()])607

Page 6: How to Count with Topology - Quomodocumque

HOW TO COUNT WITH TOPOLOGY

COUNTING SQUAREFREE INTEGERS

Better question: How many squarefree integers are therebetween N and 2N? Call this function sf (N).

sf (10) = 7sf (100) = 61

sf (1000) = 607sf (10000) = 6077

sf (100000) = 60787

Looks like:The probability that a random integer is squarefree is about61%.

Page 7: How to Count with Topology - Quomodocumque

HOW TO COUNT WITH TOPOLOGY

COUNTING SQUAREFREE INTEGERS

Better question: How many squarefree integers are therebetween N and 2N? Call this function sf (N).

sf (10) = 7sf (100) = 61

sf (1000) = 607sf (10000) = 6077

sf (100000) = 60787

Looks like:The probability that a random integer is squarefree is about61%.

Page 8: How to Count with Topology - Quomodocumque

HOW TO COUNT WITH TOPOLOGY

COUNTING SQUAREFREE INTEGERS

Heuristic (can be made rigorous):The probability that n is squarefree is the probability that it isnot divisible by 4, and not divisible by 9, and not divisible by25, and . . .

= (1− 1/4)(1− 1/9)(1− 1/25) . . .

=∏

p

(1− p−2) = ζ(2)−1 = 6/π2 = 0.6079 . . .

This is an actual theorem: limN→∞N−1sf (N) = ζ(2)−1.

Indeed, sf (N) = ζ(2)−1N + O(√

N).

Page 9: How to Count with Topology - Quomodocumque

HOW TO COUNT WITH TOPOLOGY

WHERE’S THE TOPOLOGY?

Time-honored analogy:I The ring Z of integers;I The ring k[t] of polynomials over a field (e.g. the complex

numbers)Both are Dedekind domains (commutative Noetherian integrallyclosed domains in which every nonzero prime ideal ismaximal) – it turns out that many algebraic facts about Z are infact general theorems about Dedekind domains.

Page 10: How to Count with Topology - Quomodocumque

HOW TO COUNT WITH TOPOLOGY

But the theory of complex polynomials clearly encounterstopology in a way that integer arithmetic (on its face) does not.

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HOW TO COUNT WITH TOPOLOGY

GOD DEFEATS THE DEVIL

“The ‘classical’ theory (that is,Riemannian) of algebraic functionsover the field of constants of thecomplex numbers is infinitely richer;but on the one hand it is too much so,and in the mass of facts some realanalogies become lost; and above all,it is too far from the theory ofnumbers. One would be totallyobstructed if there were not a bridgebetween the two. And just as Goddefeats the devil: this bridge exists; itis the theory of the field of algebraicfunctions over a finite field ofconstants.” – A. WEIL

Page 12: How to Count with Topology - Quomodocumque

HOW TO COUNT WITH TOPOLOGY

GOD DEFEATS THE DEVIL

Theme: problems of arithmetic statistics over Z (you justmissed a special session about this! But see Wei Ho’s talk in theCurrent Events session) are analogous to problems abouttopology and geometry of moduli spaces over C[t].

But when k is a finite field Fq, these problems retain botharithmetic aspects and geometric aspects.

Page 13: How to Count with Topology - Quomodocumque

HOW TO COUNT WITH TOPOLOGY

GOD DEFEATS THE DEVIL

Theme: problems of arithmetic statistics over Z (you justmissed a special session about this! But see Wei Ho’s talk in theCurrent Events session) are analogous to problems abouttopology and geometry of moduli spaces over C[t].

But when k is a finite field Fq, these problems retain botharithmetic aspects and geometric aspects.

Page 14: How to Count with Topology - Quomodocumque

HOW TO COUNT WITH TOPOLOGY

NUMBERS AND POLYNOMIALS

Z

±1 (units in Z)

squarefree integers

positive integers

primes

absolute value n→ |n|

k[t]

k× (units in k[t])

squarefree polynomials

monic polynomials

monic irreducible polynomials

absolute value P→ cdeg P

Page 15: How to Count with Topology - Quomodocumque

HOW TO COUNT WITH TOPOLOGY

COUNTING SQUAREFREE POLYNOMIALS

Let Fq be a finite field of odd characteristic.

Analogue of [N, 2N]: the set of monic polynomials of degree n.

There are qn of these, so think of N as qn.

What is the probability that a degree-n monic polynomialover Fq is squarefree?

Answer: 1− 1/q.(this can be proved by elementary combinatorics – but seeVakil-Wood 2012 to see the motivic apotheosis of thisapproach!)

Page 16: How to Count with Topology - Quomodocumque

HOW TO COUNT WITH TOPOLOGY

COUNTING SQUAREFREE POLYNOMIALS

Let Fq be a finite field of odd characteristic.

Analogue of [N, 2N]: the set of monic polynomials of degree n.

There are qn of these, so think of N as qn.

What is the probability that a degree-n monic polynomialover Fq is squarefree?

Answer: 1− 1/q.(this can be proved by elementary combinatorics – but seeVakil-Wood 2012 to see the motivic apotheosis of thisapproach!)

Page 17: How to Count with Topology - Quomodocumque

HOW TO COUNT WITH TOPOLOGY

COUNTING SQUAREFREE POLYNOMIALS

Why is 1− 1/q “the same answer” as 6/π2?

Because

ζ(2)−1 =∏

p prime

(1− 1

p2

)= 6/π2

and

ζFq[x](2)−1 =∏

P irreducible

(1− 1

q2degP

)= 1− 1/q

Page 18: How to Count with Topology - Quomodocumque

HOW TO COUNT WITH TOPOLOGY

COUNTING SQUAREFREE POLYNOMIALS

Why is 1− 1/q “the same answer” as 6/π2?

Because

ζ(2)−1 =∏

p prime

(1− 1

p2

)= 6/π2

and

ζFq[x](2)−1 =∏

P irreducible

(1− 1

q2degP

)= 1− 1/q

Page 19: How to Count with Topology - Quomodocumque

HOW TO COUNT WITH TOPOLOGY

COUNTING SQUAREFREE POLYNOMIALS

Why is 1− 1/q “the same answer” as 6/π2?

Because

ζ(2)−1 =∏

p prime

(1− 1

p2

)= 6/π2

and

ζFq[x](2)−1 =∏

P irreducible

(1− 1

q2degP

)= 1− 1/q

Page 20: How to Count with Topology - Quomodocumque

HOW TO COUNT WITH TOPOLOGY

COHEN-LENSTRA PROBLEM

Every number field K has an class group Cl(K), a finite abeliangroup measuring its failure to satisfy unique factorization.Question: What does the class group of a random number fieldlook like?

d 10001 10002 10003 10005 10006|Cl(Q(

√−d))| 160 64 12 64 86

10007 10009 10010 10011 1001377 96 96 24 96

10014 10015 10018 10019 1002160 54 36 30 52

We know |Cl(Q(√−d))| is around

√d (Brauer-Siegel) and what

powers of 2 divide it (genus theory.)Other than that, is it a “random number?”

Page 21: How to Count with Topology - Quomodocumque

HOW TO COUNT WITH TOPOLOGY

COHEN-LENSTRA PROBLEM

Every number field K has an class group Cl(K), a finite abeliangroup measuring its failure to satisfy unique factorization.Question: What does the class group of a random number fieldlook like?

d 10001 10002 10003 10005 10006|Cl(Q(

√−d))| 160 64 12 64 86

10007 10009 10010 10011 1001377 96 96 24 96

10014 10015 10018 10019 1002160 54 36 30 52

We know |Cl(Q(√−d))| is around

√d (Brauer-Siegel) and what

powers of 2 divide it (genus theory.)Other than that, is it a “random number?”

Page 22: How to Count with Topology - Quomodocumque

HOW TO COUNT WITH TOPOLOGY

COHEN-LENSTRA PROBLEM

Every number field K has an class group Cl(K), a finite abeliangroup measuring its failure to satisfy unique factorization.Question: What does the class group of a random number fieldlook like?

d 10001 10002 10003 10005 10006|Cl(Q(

√−d))| 160 64 12 64 86

10007 10009 10010 10011 1001377 96 96 24 96

10014 10015 10018 10019 1002160 54 36 30 52

We know |Cl(Q(√−d))| is around

√d (Brauer-Siegel) and what

powers of 2 divide it (genus theory.)Other than that, is it a “random number?”

Page 23: How to Count with Topology - Quomodocumque

HOW TO COUNT WITH TOPOLOGY

COHEN-LENSTRA PROBLEM

Every number field K has an class group Cl(K), a finite abeliangroup measuring its failure to satisfy unique factorization.Question: What does the class group of a random number fieldlook like?

d 10001 10002 10003 10005 10006|Cl(Q(

√−d))| 160 64 12 64 86

10007 10009 10010 10011 1001377 96 96 24 96

10014 10015 10018 10019 1002160 54 36 30 52

9 out of 15 divisible by 3.2304 out of the 6077 class numbers with d between 10000 and20000 are divisible by 3: about 38%.

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HOW TO COUNT WITH TOPOLOGY

COHEN-LENSTRA PROBLEM

38% – that sounds like about1− (1− 1/3)(1− 1/9)(1− 1/27)(1− 1/81) . . .!

Page 25: How to Count with Topology - Quomodocumque

HOW TO COUNT WITH TOPOLOGY

In fact, Cohen and Lenstra (1983) make an audacious conjecture assertingthat the maximal p-torsion subgroup of the class group of a randomquadratic field converges to a limiting distribution. When K is an imaginaryquadratic field, Cohen and Lenstra predict:

I a. For all odd primes p, the probability that |Cl(K)| is not a multiple of pis (1− 1/p)(1− 1/p2)(1− 1/p3) . . .

I b. For all odd primes p, the number of elements of exact order p inCl(K) is, on average, 1.

a is not known in any case.

b is known only for p = 3 (Davenport-Heilbronn.)

Now (2012) with explicit error terms! (Bhargava-Shankar-Tsimerman,Thorne-Taniguchi, Y. Zhao in the function field case)

Page 26: How to Count with Topology - Quomodocumque

HOW TO COUNT WITH TOPOLOGY

Theorem: (Ellenberg,Venkatesh,Westerland 2009+2012) TheCohen-Lenstra conjecture is true (in the sense of b., and moregeneral averages) over Fq[x], for all odd p and for all finite fieldsFq which are sufficiently large relative to p and havecharacteristic prime to 2p.

EVW 09: ”Homological stability for Hurwitz spaces and the Cohen-Lenstraconjecture over function fields,” arXiv:0912.0325EVW 12: ”Homological stability for Hurwitz spaces and the Cohen-Lenstraconjecture over function fields, II,” arXiv:1212.0923

Page 27: How to Count with Topology - Quomodocumque

HOW TO COUNT WITH TOPOLOGY

Theorem: (Ellenberg,Venkatesh,Westerland 2009+2012) TheCohen-Lenstra conjecture is true (in the sense of b., and moregeneral averages) over Fq[x], for all odd p and for all finite fieldsFq which are sufficiently large relative to p and havecharacteristic prime to 2p.

EVW 09: ”Homological stability for Hurwitz spaces and the Cohen-Lenstraconjecture over function fields,” arXiv:0912.0325EVW 12: ”Homological stability for Hurwitz spaces and the Cohen-Lenstraconjecture over function fields, II,” arXiv:1212.0923

(Following work of Friedman-Washington, JK Yu, Achter,. . . )

Typical consequence:

The number of nontrivial 5-torsion points on the Jacobian of a randomhyperelliptic curve over Fq is, on average, 1.

Page 28: How to Count with Topology - Quomodocumque

HOW TO COUNT WITH TOPOLOGY

THE OTHER SIDE OF THE BRIDGE

Bad question: How many monic squarefree complexpolynomials are there of degree n?

OK but uninteresting question: What is the probability that arandom monic degree n polynomial is squarefree?(The answer is 1; Two roots have probability 0 of coinciding!)

Good question: What is the topology (specifically: thecohomology with rational coefficients) of the space of monicsquarefree polynomials of degree n?

Page 29: How to Count with Topology - Quomodocumque

HOW TO COUNT WITH TOPOLOGY

THE OTHER SIDE OF THE BRIDGE

Bad question: How many monic squarefree complexpolynomials are there of degree n?

OK but uninteresting question: What is the probability that arandom monic degree n polynomial is squarefree?(The answer is 1; Two roots have probability 0 of coinciding!)

Good question: What is the topology (specifically: thecohomology with rational coefficients) of the space of monicsquarefree polynomials of degree n?

Page 30: How to Count with Topology - Quomodocumque

HOW TO COUNT WITH TOPOLOGY

THE OTHER SIDE OF THE BRIDGE

Bad question: How many monic squarefree complexpolynomials are there of degree n?

OK but uninteresting question: What is the probability that arandom monic degree n polynomial is squarefree?(The answer is 1; Two roots have probability 0 of coinciding!)

Good question: What is the topology (specifically: thecohomology with rational coefficients) of the space of monicsquarefree polynomials of degree n?

Page 31: How to Count with Topology - Quomodocumque

HOW TO COUNT WITH TOPOLOGY

Weil’s great insight – when a space X can be defined over bothFq and C (technically – when the space can be thought of as ascheme over Spec Z) – then the behavior of the number |X(Fq)|“remembers” the topology of the complex manifold X(C).

{(x, y) ∈ C2 : y2 = x6 + x + 1}

describes a genus 2 surface; thisplaces constraints on thenumber

|{(x, y) ∈ F2q : y2 = x6 + x + 1}|

Page 32: How to Count with Topology - Quomodocumque

HOW TO COUNT WITH TOPOLOGY

Definition: Confn(C), the degree-n configuration space of C, is thespace of squarefree polynomials of degree n. Alternately: thespace of unordered n-tuples of distinct complex numbers, viathe bijection

f (T) = (T − z1)(T − z2) . . . (T − zn) 7→ {z1, . . . , zn}

The configuration space Conf2 is homeomorphic to C× (C− 0):

{z1, z2} 7→ (z1 + z2, (z1 − z2)2)

and up to homotopy, Conf2 is a circle.

Page 33: How to Count with Topology - Quomodocumque

HOW TO COUNT WITH TOPOLOGY

The space Confn is n-dimensional and more and moretopologically complicated as n grows; but from the point ofview of rational cohomology it is beautifully simple. A theoremof Arnol‘d (1969):

The rational cohomology of Confn is stable for n ≥ 2; italways looks like that of a circle.

It follows (after many hidden technicalities) that

The proportion of monic irreducible degree-n polynomialswhich are squarefree is constant for n ≥ 2; it is always1− 1/q.

Page 34: How to Count with Topology - Quomodocumque

HOW TO COUNT WITH TOPOLOGY

GEOMETRIC ANALYTIC NUMBER THEORY

I N: A classical problem of analytic number theory:counting problem pertaining to Z.

I F: The function-field analogue of the classical problem:counting problem pertaining to Fq[T].

I T: The geometric analogue of the function field problem:topology problem, computing the cohomology of a familyof moduli spaces over C.

Theorems in T yield theorems in F yield conjectures (andhopefully insights) in N.

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HOW TO COUNT WITH TOPOLOGY

Wonderfully, the topological counterparts to popularconjectures in arithmetic statistics turn out to be statements ofstable cohomology, a hugely active subfield of topology!(Notably: Madsen-Weiss proof of the Mumford conjecture)

Page 36: How to Count with Topology - Quomodocumque

HOW TO COUNT WITH TOPOLOGY

Arithmeticproblem

moduli space

Countingsquarefrees

configurationspace of

unordered points

Cohen-Lenstrafor p-torsion in

class group

moduli ofhyperellipticcurves with

p-level structure(a kind of

Hurwitz space)Counting

degree-d numberfields

classical Hurwitzspaces of d-gonal

curves

Page 37: How to Count with Topology - Quomodocumque

HOW TO COUNT WITH TOPOLOGY

Arithmetic problem moduli spaceVariation of Selmer

groups(Poonen-Rains,

Klagsbrun-Mazur-Rubin,

Bhargava-Shankar –see W Ho at 1!)

moduli spaces ofelliptic surfaces

Batyrev-Maninconjecture

spaces of rationalcurves on Fano

varieties

prime numbertheorem

representation stabilityfor configurationspace of orderedpoints (in the

FI-module sense ofChurch,E,Farb)

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HOW TO COUNT WITH TOPOLOGY

TOPOLOGY AS CONJECTURE MACHINE

We don’t know everything about the topological side, but weknow a lot – and this gives us a principled way to makegeometrically motivated conjectures in arithmetic statistics.

In this way we reproduce many existing conjectures in numbertheory (Cohen-Lenstra, Malle-Bhargava, Boston-Bush-Hajir...)

But sometimes the geometry doesn’t agree with existingconjectures: e.g. the geometric picture suggests that when F is afield containing cube roots of unity, the average square of thenumber of 3-torsion points in the class group should be largerthan the value predicted by Cohen-Lenstra-Martinet.

Fortunately, this was already observed in numerical data byMalle (2010)!

”Repaired” version of Cohen-Lenstra for fields containing pthroots of unity: Garton (2012)

Page 39: How to Count with Topology - Quomodocumque

HOW TO COUNT WITH TOPOLOGY

TOPOLOGY AS CONJECTURE MACHINE

We don’t know everything about the topological side, but weknow a lot – and this gives us a principled way to makegeometrically motivated conjectures in arithmetic statistics.

In this way we reproduce many existing conjectures in numbertheory (Cohen-Lenstra, Malle-Bhargava, Boston-Bush-Hajir...)

But sometimes the geometry doesn’t agree with existingconjectures: e.g. the geometric picture suggests that when F is afield containing cube roots of unity, the average square of thenumber of 3-torsion points in the class group should be largerthan the value predicted by Cohen-Lenstra-Martinet.

Fortunately, this was already observed in numerical data byMalle (2010)!

”Repaired” version of Cohen-Lenstra for fields containing pthroots of unity: Garton (2012)

Page 40: How to Count with Topology - Quomodocumque

HOW TO COUNT WITH TOPOLOGY

TOPOLOGY AS CONJECTURE MACHINE

We don’t know everything about the topological side, but weknow a lot – and this gives us a principled way to makegeometrically motivated conjectures in arithmetic statistics.

In this way we reproduce many existing conjectures in numbertheory (Cohen-Lenstra, Malle-Bhargava, Boston-Bush-Hajir...)

But sometimes the geometry doesn’t agree with existingconjectures: e.g. the geometric picture suggests that when F is afield containing cube roots of unity, the average square of thenumber of 3-torsion points in the class group should be largerthan the value predicted by Cohen-Lenstra-Martinet.

Fortunately, this was already observed in numerical data byMalle (2010)!

”Repaired” version of Cohen-Lenstra for fields containing pthroots of unity: Garton (2012)