The braid group, analytic number theory, and Weil’s three columns

This post is about a new paper of mine with Akshay Venkatesh and Craig Westerland; but I’m not going to mention that paper in the post. Instead, I want to explain why topological theorems about the stable homology of moduli spaces are relevant to analytic number theory.  If you’ve seen me give a talk about this stuff, you’ve probably heard this spiel before.

We start with Weil’s famous quote about “the three columns”:

“The mathematician who studies these problems has the impression of deciphering a trilingual inscription. In the first column one finds the classical Riemannian theory of algebraic functions. The third column is the arithmetic theory of algebraic numbers.  The column in the middle is the most recently discovered one; it consists of the theory of algebraic functions over finite fields. These texts are the only source of knowledge about the languages in which they are written; in each column, we understand only fragments.”

Let’s see how a classical question of analytic number theory works in Weil’s three languages.  Start with the integers, and ask:  how many of the integers between X and 2X are squarefree?  This is easy:  we have an asymptotic answer of the form

\frac{6}{\pi^2}X + O(X^{1/2}) = \zeta(2)^{-1} X + O(X^{1/2}).

(In fact, the best known error term is on order X^{17/54}, and the correct error term is conjectured to be X^{1/4}; see Pappalardi’s “Survey on k-freeness” for more on such questions.)

So far so good.  Now let’s apply the popular analogy between number fields and function fields, going over to Weil’s column 3, and ask: what’s the analogous statement when Z is replaced by F_q[T]?

(If you’ve never thought about this analogy, Bjorn Poonen supplies a marvelous cheat sheet in section 2.6 of these notes.)

Degree is to F_q[T] as archimedean absolute value is to Z; more precisely, “q^degree” is like archimedean absolute value.  So [X..2X], the set of positive integers whose absolute value is within a constant of X, should be replaced with the set of monic polynomials of a given degree n, where we think of X as comparable to q^n.

The question is now:  of the q^n monic degree-n polynomials in F_q[T], how many are squarefree?  This too is a classical question:  when n > 1, the answer is

q^n – q^{n-1}.

(The absence of an error term is an anomaly having to do with the fact that we’re using a genus-0 function field.  For more general function fields, there will be an error term there; its size is an interesting question which I don’t know if anyone’s thought about.)

This is the same answer we got in the number field case!  If it doesn’t look like the same answer, try writing it as

(1-1/q) q^n = \zeta_{\mathbf{A}^1/\mathbf{F}_q}(2)^{-1} q^n

We’re now ready for Weil’s first column: here we have to replace F_q[T] by C[T].  And now instantly we have a problem; we can’t really count the number of squarefree monic complex polynomials of degree n!

To keep the analogy going, we have to think about the space of such polynomials.  This is nothing other than the configuration space Conf_n(C) of the complex plane:  that is, the moduli space of unordered n-tuples (configurations) of complex numbers.  To get a configuration from a squarefree polynomial, just extract its roots!

Conf_n(C) is a beautiful space:  it’s a K(pi,1), and its fundamental group is Artin’s braid group on n strands, which we denote B_n.  So understanding the topology of this space can be reduced to problems about the braid group  — in particular, to compute the cohomology of Conf_n(C) one needs only compute the group cohomology of B_n.

But this we know!  It’s an old theorem of Arnol’d that

H^0(B_n,Q) = H^1(B_n,Q) = Q

H^i(B_n,Q) = 0   (i > 1)

And I claim this should, again, be seen as the same answer as we obtained in the other two columns.  To see this requires something extra:  that the configuration space Conf^n is actually a scheme over Spec Z, whose F_q-rational points are precisely the squarefree degree-n polynomials in F_q[T].  What’s more, we can identify the ell-adic cohomology of Conf^n / C and Conf^n / F_q.  Thus we have

H^0(Conf^n / F_q, Q_ell) = H^1(Conf^n, F_q, Q_ell) = Q_ell

H^i(Cong^n / F_q, Q_ell) = 0 (i > 1)

And from this (and from knowledge of what the two nonzero cohomology classes actually are) we can compute |Conf^n(F_q)| by the Lefschetz trace formula; the alternating sum of Frobenius traces only has two nonzero terms, and we get (as we’d better!)

q^n – q^{n-1}.
To sum up:  a topological theorem about the cohomology of the braid group implies a theorem about analytic number theory over F_q[T]. That theorem doesn’t in turn imply anything about number fields; but it makes a rather loud suggestion about what should be true in that setting.

The paper with Venkatesh and Westerland pursues the same strategy in a case that turns out to be much harder; in place of configuration spaces, one has Hurwitz spaces, moduli spaces of finite branched covers of the sphere.  More about that in the next post.

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13 thoughts on “The braid group, analytic number theory, and Weil’s three columns

  1. Paul Johnson says:

    Just wanted to say: thanks for mentioning this! I’d sort of been waiting for it, as somehow Hurwitz theory is central to everything I do, but I missed it go up because I tend to check only math.AG and math.CO, not math.NT and math.AT. I suppose I should just check them all.

    And maybe I’ll actually have a question about the math later.

  2. JSE says:

    We realized after the fact that it should have been crossposted to math.AG, but we couldn’t figure out a way to add a category!

  3. Sonia Balagopalan says:

    Cheers for the MO footnote in the paper!

  4. Paul Johnson says:

    So I do have a completely nonmathematical question: how widespread is the use of “slogan” for “pithy sentence summarizing the general idea without worrying about the technicalities”? I’m a big fan of it, but I’m not sure I’ve seen it in a paper before, or really anywhere outside of John Baez’s “This Week’s Finds”.

  5. JSE says:

    I’m not sure, but I certainly grew up on This Week’s Finds so I might have learned it there…

  6. […] is a beautiful blog post by Jordan Ellenberg which explains how to get the analogous version of this statement when is […]

  7. GilYoung Cheong says:

    Hello Pf. Ellenberg. Do you have any references for general k-freeness over finite fields? That is, tweaking what you wrote do we have any known clue for the question: “of the q^n monic degree-n polynomials in F_q[T], how many are” k-free? My friend and I am trying to find it or at least estimate it.

  8. JSE says:

    I’ll bet you can do it along the lines of the methods discussed in this post, but one good way to get the count is to show that the zeta function sum n^{-s}, as n ranges over k-power-frees, is zeta(s) zeta(ks)^{-1}. Then plug in the zeta function of A^1/F_q and work from there. Not sure which of these steps you may already be familiar with, so feel free to ask for expansion.

  9. GilYoung Cheong says:

    Thank you! We will try that first and may ask you further questions.

  10. I like:
    Let a_i be the number of m-power-free polynomials of degree i. We can
    write all monic polynomials as f(x)*g(x)^m uniquely, where
    f(x) is m-power-free. This gives

    q^d=sum_{k\geq 0} a_{d-km} * q^k

    We multiply the above by t^d, where t is a formal variable, and sum
    over d\geq 0:

    \sum_{d\geq 0} q^d t^d = \sum_{d\geq 0} sum_{k\geq 0} a_{d-km}
    t^{d-km} * q^k * t^{km}

    = \sum_{i} a_{i} t^{i} * sum_{k\geq 0} q^k * t^{km}

    That gives

    \sum_{i} a_{i} t^{i} =(1-qt^m)/(1-qt)=(1-qt^m)(1+qt+q^qt^2+q^3t^3+…)

    from which it is easy to see that a_{i}=q^i – q^{i-m+1} for i>=m.

    I think of Jordan’s proof as the “factored” version of the above.

  11. GilYoung Cheong says:

    Thanks Pf. Wood, and I think I was thinking along the similar line after Pf. Ellenberg’s advice and getting hint from the other document (KR) which discussed about 2-free:

    Using Pf. Wood’s notation let a_d = # of n-free polynomials over F_q. We want to show

    Eqn 1: zeta(s) = zeta(ns)sum_(d >= 0) a_d * q^(-ds).

    (Note zeta(s) = sum_(F monic) |F|^(-s) and |F| = q^(degF) => |FG| = q^(deg(FG)) = q^(deg(F) + deg(G)) = q^deg(F) q^(deg(G) = |F||G|.)

    For each degree d >= 0, we have a_d many monic n-free polynomials, so sum_(d >= 0) a_d * q^(-ds) = sum_(F monic, n-free) |F|^(-s).

    Again, as Pf. Wood mentioned, we can write every monic polynomial uniquely in the form f(x)*g(x)^n where f(x) is n-free, so

    zeta(s) = sum_(F monic) |F|^(-s) = sum_(F monic) |Fn|^(-s) * sum_(F monic, n-free) |F|^(-s) = sum_(F monic) |F|^(-ns) * sum_(F monic, n-free) |F|^(-s) = zeta(ns) * sum_(d >= 0) a_d * q^(-ds).

    since we only have finitely many terms.


    Unfortunately, here is where I was stuck. I don’t know how to derive a_d from here and I am still working on it. Feel free to reveal the answer if you have it. Now I am not trying to learn but want to have the right and complete proof.

  12. GilYoung Cheong says:

    Oh, now I see. I somehow could not see we can vary over real numbers (or complex numbers). Thanks for both of you!

  13. […] Fun aside: this computation turns out to be related, via the Grothendieck trace formula, to the fact that the probability that a random monic polynomial over $mathbb{F}_q$ is squarefree is $1 – frac{1}{q}$, which implies that the computation can actually be redone (using the Grothendieck trace formula, together with a comparison theorem between $ell$-adic and singular cohomology) by counting that there are $q^n – q^{n-1}$ monic squarefree polynomials of degree $n$; I learned this from Jordan Ellenberg’s blog post The braid group, analytic number theory, and Weil’s three columns. […]

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