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September 10, 2019

The Riemann Hypothesis (Part 2)

Posted by John Baez

Now let’s dig a tiny bit deeper into the Riemann Hypothesis, and the magnificent developments in algebraic geometry it has inspired. My desire to explain this all rather simply is making the story move more slowly than planned, but I guess that’s okay.

Last time I sketched how the function that counts primes x\le x is the sum of a nice smooth increasing function and a bunch of ‘correction terms’, one for each nontrivial zero of the Riemann zeta function. These correction terms grow but also oscillate as xx gets bigger. The Riemann Hypothesis puts a precise bound on how fast they grow. If the real part of all the nontrivial zeros is 1/21/2, as this hypothesis claims, these correction terms are of order x 1/2lnxx^{1/2} \, \ln x for large xx.

The double appearance of the number 1/21/2 here is no coincidence! If the Riemann Hypothesis were false, there’d be a zero with real part β>1/2\beta \, > \, 1/2, and a correction term of order x βlnxx^\beta \,\ln x.

I find this a nice preliminary explanation of what the Riemann Hypothesis means, but a completely terrible explanation of why it should be true.

In struggling to understand this, some of the best algebraic geometers of the last century studied and finally proved a simplified version of the Riemann Hypothesis called the Weil Conjectures. In this simplified version, which deals with some other zeta functions, they found a geometrical reason why the zeros should lie where they do.

So, let’s talk about Weil Conjectures. I’ll try to explain what they’re about with the bare minimum of machinery.

Suppose we have some polynomial equations with integer coefficients in a bunch of variables. For any power q=p nq = p^n of any prime number pp there’s a unique field 𝔽 q\mathbb{F}_q with qq elements. We can count the solutions to our equations where our variables lie in this field. The answer will depend on nn.

For concreteness let’s try a simple example, suggested by Jyrki Lahtonen. We’ll use one equation in two variables:

y 2+y=x 3+x y^2 + y = x^3 + x

and we’ll take p=2p = 2. Here’s how the number of solutions in 𝔽 q\mathbb{F}_q depends on nn:

n q=p n numberofsolutions 1 2 4 2 4 4 3 8 4 4 16 24 5 32 24 6 64 64 7 128 144 8 256 224 9 512 544 10 1024 1024 11 2048 1984 12 4096 4224 \begin{array}{rrr} n & q = p^n \! & \;\; number \; of \; solutions \\ 1 & 2 & 4 \\ 2 & 4 & 4 \\ 3 & 8 & 4 \\ 4 & 16 & 24 \\ 5 & 32 & 24 \\ 6 & 64 & 64 \\ 7 & 128 & 144 \\ 8 & 256 & 224 \\ 9 & 512 & 544 \\ 10 & 1024 & 1024 \\ 11 & 2048 & 1984 \\ 12 & 4096 & 4224 \\ \end{array}

I did this by hand using some tricks, but I’m quite error-prone so please check this if you can.

We instantly see some interesting patterns. The last digit of the number of solutions is always 4 — but that’s too weird to be helpful right now. Also, the number of solutions seems to be growing in a roughly exponential way. Let’s think about that.

Since we have one equation in two unknowns:

y 2+y=x 3+x y^2 + y = x^3 + x

we might expect a one-parameter family of solutions. Here our ‘parameter’ is an element of 𝔽 q\mathbb{F}_q, so we expect roughly q=p nq = p^n solutions.

This is very naive reasoning — it would make sense for linear equations, but we’re dealing with polynomial equations. Still, we can take the actual number of solutions, and subtract off the naively expected number of solutions, namely p np^n, and see what’s left. That’ll be our ‘correction term’:

n correctionterm 1 2 2 0 3 4 4 8 5 8 6 0 7 16 8 32 9 32 10 0 11 64 12 128 \begin{array}{rr} n & correction \; term \\ 1 & 2 \\ 2 & 0 \\ 3 & -4 \\ 4 & 8 \\ 5 & -8 \\ 6 & 0 \\ 7 & 16 \\ 8 & -32 \\ 9 & 32 \\ 10 & 0 \\ 11 & -64 \\ 12 & 128 \\ \end{array}

Nice! We’re seeing powers of 2 show up. If you look at those 0’s you’ll see there’s some sort of “period 4” thing going on. If you write the correction terms in bunches of 4 you see this:

2 0 4 8 8 0 16 32 32 0 64 128 \begin{array}{rrrr} 2 & 0 & -4 & 8 \\ -8 & 0 & 16 & -32 \\ \; 32 & \; \; 0 & -64 & 128 \\ \end{array}

Each row is -4 times the previous row!

So, the number of solutions of our polynomial equations is roughly what you’d naively guess, but not exactly: there’s also a ‘correction term’ that grows exponentially in a slower way and also oscillates. In general there could be many such correction terms, but I chose an example where there’s just one — or as we’ll soon see, two, which happen to oscillate at the same rate.

The Weil Conjectures tell us how these oscillating correction terms work.

Now, the equation

y 2+y=x 3+x y^2 + y = x^3 + x

actually describes an elliptic curve. That is, the complex solutions of this equation form a torus if we add one extra point, a ‘point at infinity’.

The Weil Conjectures are especially simple for elliptic curves: in this case, they were conjectured by Emil Artin and proved by Helmut Hasse in 1933 before Weil came along. Here’s one way to say them, not the most fancy way:

Hasse’s Theorem on Elliptic Curves. Given a cubic equation with integer coefficients in two variables that defines an elliptic curve, the number of solutions in 𝔽 q\mathbb{F}_q where q=p nq = p^n is

p nα nβ n p^n - \alpha^n - \beta^n

where α,β\alpha, \beta \in \mathbb{C} have |α|=|β|=p|\alpha| = |\beta| = \sqrt{p}.

Here p np^n is the naive number of solutions, while α n-\alpha^n and β n-\beta^n are the ‘correction terms’. These correction terms grow exponentially like p n/2p^{n/2} but also oscillate.

The fact that |α|=|β|=p|\alpha| = |\beta| = \sqrt{p} is analogous to the Riemann Hypothesis, since it nails down the rate at which the correction terms grow. Indeed it’s very analogous! The Riemann Hypothesis says the number of primes grows like li(x)x/ln(x)li(x) \approx x/\ln(x) plus corrections that grow like x 1/2ln(x)x^{1/2} \ln(x). Hasse’s theorem on elliptic surves says the number of solutions grows like p np^n plus corrections that grow like p n/2p^{n/2}. So there’s a kind of ‘square root sized correction’ thing going on, and that’s no coincidence.

Let’s see what the Hasse–Weil Theorem says about our example. First of all, since the number of solutions has to be real, it turns out we always have β=α¯\beta = \overline{\alpha}. So I could have said

numberofsolutions=p n2Re(α n) number \; of \; solutions = p^n - 2 Re(\alpha^n)

Second of all, we can always figure out this mysterious number α\alpha just by looking at the case n=1n = 1. In our example we count the solutions of

y 2+y=x 3+x y^2 + y = x^3 + x

in the field 𝔽 2\mathbb{F}_2. In this field, regardless of whether x=0x = 0 or x=1x = 1 we have x 3+x=0x^3 + x = 0, and regardless of whether y=0y = 0 or y=1y = 1 we have y 2+y=0y^2 + y = 0. So we get 4 solutions, so in this case

numberofsolutions=2 n2Re(α n) number \; of \; solutions = 2^n - 2 Re(\alpha^n)

says that

4=22Re(α) 4 = 2 - 2 Re(\alpha)

and thus Re(α)=1Re(\alpha) = -1. On the other hand the Hasse–Weil Theorem assures us that |α|=2|\alpha| = \sqrt{2}. So, we get

α=1±i \alpha = -1 \pm i

and if we call one of these solutions α\alpha the other will be β=α¯\beta = \bar{\alpha}. We conclude that

numberofsolutions=2 n(1+i) n(1i) n number \; of \; solutions = 2^n - (-1 + i)^n - (-1 - i)^n

The correction terms here give the funny pattern we saw earlier. You should be imagining a picture of how (1+i) n(-1 + i)^n and (1i) n(-1 - i)^n spiral around in the complex plane, rotating 3/8ths of a turn each time — I drew this picture while figuring this stuff out, but I’m too lazy to draw it here.

If you’re good at computing, you can have fun exploring other examples. Jyrki Lahtonen suggests

y 2+y=x 3 y^2+y=x^3


y 2+xy=x 3+1 y^2+x y=x^3+1

as other fun examples where the corrections conspire to make the number of solutions over 𝔽 q\mathbb{F}_q with q=2 nq = 2^n remain constant for low nn. But you can also look at other cubic equations in two variables, and other prime powers. If you look at more complicated polynomial equations, or equations in more variables, the patterns will become more complicated — but if you subtract off the ‘naively expected’ number of solutions, you may still be able to understand the correction terms.

Now I’m ready to extract the magnificent moral from the story so far — but I’m afraid the calculations I’ve done, which may help some people understand what’s going on, will have made other’s eyes glaze over. So I’ll postpone most of the moral to next time.

But I can’t resist saying this… I’ll explain it better next time, so don’t be worried if it makes no sense now:

It’s really better to include the ‘point at infinity’ among the solutions to our equations: this makes the space of complex solutions into a torus, which is a kind of compact Riemann surface, and it makes the solutions in 𝔽 q\mathbb{F}_q into an elliptic curve, which is a kind of smooth projective algebraic variety. If we count the point at infinity we get

numberofpoints=p nα nβ n+1 number \; of \; points = p^n - \alpha^n - \beta^n + 1

So we have a term growing like p np^n, two terms growing like p n/2p^{n/2} (but also oscillating), and a term growing like p 0p^0 — that is, 1. This should remind you of how a torus can be built from a copy of 2\mathbb{R}^2, two copies of \mathbb{R}, and a copy of 0\mathbb{R}^0 — that is, a point.

So, in this baby version of the Riemann Hypothesis, we are starting to see the geometrical origin of the oscillating correction terms, and why they behave the way they do. All this is part of what Weil hypothesized in a much more general situation — and Grothendieck and Deligne proved!

Posted at September 10, 2019 9:01 AM UTC

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12 Comments & 2 Trackbacks

Re: The Riemann Hypothesis (Part 2)

I verified the calculation for the number of solutions of x 2+y=x 3+x.x^2+y=x^3+x. Here is some gap code:

for n in [1..12] do
    field := GF(2^n);
    count := 0;
    for x in field do 
      for y in field do 
        if y^2+y=x^3+x then count := count+1; fi;
    Print(n, " ", 2^n, " ", count, "\n");

Posted by: The Math Dandy on September 10, 2019 11:40 AM | Permalink | Reply to this

Re: The Riemann Hypothesis (Part 2)

Great! So I didn’t screw up!

Maybe I should learn GAP.

Mathematica and Sage fans out there: does that software support computations in finite fields?

Posted by: John Baez on September 10, 2019 12:51 PM | Permalink | Reply to this

Re: The Riemann Hypothesis (Part 2)

Agreed that the values are correct. Here’s Magma code for the above (can be used in the Magma calculator):

E := EllipticCurve([GF(2)| 0,0,1,1,0 ]);
[ #E(GF(2,n)) - 1 : n in [1..12] ];

Or, if you want to avoid the use of elliptic curves:

[ #[ <x,y> : x,y in GF(2,n) | y^2 + y eq x^3 + x ] : n in [1..12] ];

(One could write explicit loops like the other examples, but that is very much not the Magma way of doing things.)

Posted by: Geoff Bailey on September 10, 2019 2:36 PM | Permalink | Reply to this

Re: The Riemann Hypothesis (Part 2)

Here is the sage code:

for n in range(1, 13):
    field = GF(2^n)
    count = 0
    for x in field:
      for y in field:
        if y^2+y == x^3+x:
            count += 1
    print n, 2^n, count

Posted by: The Math Dandy on September 10, 2019 1:35 PM | Permalink | Reply to this

Re: The Riemann Hypothesis (Part 2)

Thanks! I’ll try to revive my dormant SageMath account and do some calculations like this!

Posted by: John Baez on September 10, 2019 5:29 PM | Permalink | Reply to this

Re: The Riemann Hypothesis (Part 2)

It might also be faster & more handy to actually download SageMath and run it on your laptop. And, just in case you didn’t know, in python range(1,13) means from 1 to 12.

Posted by: jon on September 10, 2019 6:54 PM | Permalink | Reply to this

Re: The Riemann Hypothesis (Part 2)

Since I have some sort of resistance to using software, I’d be really happy if someone checked out the equation

y 3=x 2 y^3 = x^2

counting the solutions in 𝔽 2 n\mathbb{F}_{2^n} for a bunch of low nn, and subtracting off the ‘naive estimate’, namely 2 n2^n. This one is interesting to me because it’s not an elliptic curve: it’s not smooth at the origin! So, Hasse’s Theorem does not apply.

Posted by: John Baez on September 11, 2019 2:00 AM | Permalink | Reply to this

Re: The Riemann Hypothesis (Part 2)

For x 3=y 2x^3=y^2, solutions are easy to count since they are in bijection with the field for any field.

Given tt in the ground field, the solution is (t 2,t 3)(t^2,t^3) and given (x,y)(x,y), you extract tt as the quotient x/yx/y.

Posted by: Peter McNamara on September 11, 2019 3:27 AM | Permalink | Reply to this

Re: The Riemann Hypothesis (Part 2)

Okay, thanks! As you can see, the only way I’ll learn algebraic geometry is by asking lots of questions that make me feel stupid as soon as they’re answered.

So here the naive estimate (of qq points, or q+1q+1 counting the point at infinity) is exactly right! The other kind of irreducible plane cubic is the one with a double point, like

y 2=x 2(x+1) y^2 = x^2(x+1)

and here I guess we get q1q-1 points, or qq if we count the point at infinity, since this is parametrized by

t(t 21,t(t 21)) t \mapsto (t^2-1, t(t^2-1))

Now I’m wondering if as motives, the cubic with a cusp is isomorphic to the projective line, while the one with a double point is isomorphic to the affine line. The second question here seems more risky, since “removing a point” feels different than “a curve crossing over itself in a double point”—but motives may not care about the difference.

Posted by: John Baez on September 11, 2019 4:43 AM | Permalink | Reply to this

Re: The Riemann Hypothesis (Part 2)

This is a bit of a tangent, but since you mentioned motives of curves…

Deligne’s theory of 1-motives is exactly designed to make sense of this type of question, by providing an explicit and unconditional construction of the fragment of the conjectural theory of mixed motives which is (additively) generated by motives of curves. It is explained in Deligne’s paper “Théorie de Hodge III”, Section 10.

Let XX be an algebraic curve over the complex numbers (possibly non-compact and singular). One can associate to XX a 1-motive H M 1(X,)H^1_M(X,\mathbb{Z}), which is a gadget built from complex commutative algebraic groups (lattices, abelian varieties, complex tori) and whose structure reflects the structure of H 1(X(),)H^1(X(\mathbb{C}),\mathbb{Z}) considered together with its mixed Hodge structure. More precisely, a 1-motive is a morphism [LG][L\to G] where L nL\simeq \mathbb{Z}^n is a lattice and GG is a semi-abelian variety, i.e. an extension of an abelian variety by a complex torus. The category of 1-motives (tensored with \mathbb{Q}) is an abelian category.

This 1-motive only depends on the semi-normalisation of XX. In particular for XX a cuspidal cubic it is the same as for 1\mathbb{P}^1, hence trivial. For XX a nodal cubic, one has H M 1(X,)=[0]H^1_M(X,\mathbb{Z})=[\mathbb{Z}\to 0], reflecting the fact that the H 1H^1 of XX is one-dimensional and of weight 00. For XX an elliptic curve, one has H M 1(X,)=[0E]H^1_M(X,\mathbb{Z})=[0\to E]. More generally, for XX a Riemann surface, one has H M 1(X,)=[0Jac(X)]H^1_M(X,\mathbb{Z})=[0\to \mathrm{Jac}(X)] with Jac(X)\mathrm{Jac}(X) the Jacobian of XX.

The 1-motive H M 1(X,)H^1_M(X,\mathbb{Z}) also makes sense for a curve over general field, e.g. a finite field, and it then unifies the \ell-adic cohomology groups of XX for all primes \ell.

While the general story of mixed motives is complicated and still largely conjectural, Deligne’s concrete definition for curves is known to be compatible with the best guesses we have for mixed motives, namely the constructions of Voevodsky on the one hand and of Nori on the other, by work of Orgogozo, Barbieri-Viale, Kahn, Ayoub, …

Posted by: Simon Pepin Lehalleur on September 11, 2019 7:35 PM | Permalink | Reply to this

Re: The Riemann Hypothesis (Part 2)

Thanks, Simon — this is really interesting! You’ll see that in the comments to Part 3 I started asking about the motives we get from taking the motives of elliptic curves over 𝔽 q\mathbb{F}_q and subtracting off 1+𝕃1 + \mathbb{L}, where 𝕃\mathbb{L} is the Lefschetz motive. I don’t know if what you’re saying can help me understand these better… but it probably would if I thought hard enough.

Posted by: John Baez on September 12, 2019 10:21 AM | Permalink | Reply to this

Re: The Riemann Hypothesis (Part 2)

tt should be yx\frac{y}{x}. (For completeness, let’s prove that if x 3=y 2x^3=y^2 and t=yxt=\frac{y}{x} then y=t 3y=t^3 and x=t 2x=t^2. Indeed t 3=y 3x 3=yy 2x 3=yt^3=\frac{y^3}{x^3} = y \frac{y^2}{x^3} =y and t 2=y 2x 2=x 3x 2=xt^2=\frac{y^2}{x^2}=\frac{x^3}{x^2}=x ; very easy but not that immediate to guess for my stupid brain.)

Posted by: Béranger on September 11, 2019 12:40 PM | Permalink | Reply to this
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