how to count with topology - quomodocumque
TRANSCRIPT
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HOW TO COUNT WITH TOPOLOGY
How to Count with Topology
Jordan S. EllenbergUniversity of Wisconsin-Madison
AMS-MAA Joint Mathematics MeetingsJanuary 11, 2013
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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, . . .
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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, . . .
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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
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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
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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%.
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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%.
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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).
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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.
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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
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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.
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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.
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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
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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!)
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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!)
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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
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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
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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
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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?”
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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?”
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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?”
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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) . . .!
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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)
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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
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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.
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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?
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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?
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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?
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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}|
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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.
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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.
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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)
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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
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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)
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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)
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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)