We will describe a surprising phenomenon discovered by Zagier on values of quadratic polynomials. We will then provide an explanation using modular forms. Quadratic polynomials are quite simple objects everybody knew since childhood. Modular forms are less elementary but ubiquitous in modern number theory. In just one sentence, modular forms are special functions satisfying many symmetries. In particular they are periodic $f(z+1)=f(z)$ like the trigonometric function $\sin 2\pi z$, and so a modular form has a Fourier expansion $$f(z)=\sum_{n\ge0} a_n q^n,$$ where $q=e^{2\pi i z}$. The most famous example of a modular form is Ramanujan's $\Delta$-function $$\Delta(z)=q\prod_{n\ge1}(1-q^n)^{24}=q-24q^2+252q^3-1472q^4+4830q^5-\cdots.=:\sum \tau(n)q^n.$$ The coefficients $\tau(n)$ are of great arithmetic interest. If you never saw a modular form before, just remember the first few values of $\tau(n)$ for the purpose of this talk.

These are notes prepared for a talk for Columbia Undergraduate Math Society, Fall 2017. Most of the material is drawn from Zagier's beautiful paper [1] (tailored to fit a one-hour talk).

TopSums of values of quadratic polynomials

Let $Q(x)=ax^2+bx+c$ be a quadratic polynomial with $a,b,c\in \mathbb{Z}$. Its discriminant $D=b^2-4ac$ is a always congruent to 0 or 1 modulo 4, which is not a perfect square when $Q(x)$ is irreducible. We will focus on the case $a<0$ and $D>0$. So geometrically it corresponds to a cap-shape parabola intersecting with $x$-axis at two points.

Let us consider the simplest case $D=5$. Zagier came up with the following sum of values of quadratic polynomials with discriminant 5, $$A(x):=\sum_{b^2-4ac=5, a<0}\max\{0, ax^2+bx+c\}.$$ Here taking only positive values makes the sum more likely to converge. Assume there is no convergence issue for the moment. Notice that by construction $A(x+1)=A(x)$ (as $Q(x+1)$ is also a polynomial with $D=5$) and $A(x)=A(-x)$ (change $b$ to $-b$). So $A(x)$ is an even and periodic function (of period 1). Can one make a wild guess what function this is?

Theorem 1 $A(x)$ is the constant function 2.

This is rather surprising. To convince you this is reasonable, let us compute $A(x)$ for some specific $x$. Since $A(x)$ is even and periodic, we can focus on $0\le x\le 1/2$.

Example 1 Let us compute $A(0)$. Then $$A(x)=\sum_{b^2-4ac=5, a<0,c>0}c.$$ Notice $b^2+4|a||c|=4$, so $a=-1$, $c=1$ and $b=\pm1$. Thus there are only two terms for the sum $$Q(x)=-x^2+x+1, \quad Q(x)=-x^2-x+1.$$ In particular, $A(x)=1+1=2$.
Example 2 Let us compute $A(\frac{1}{2})$. Then $$A(x)=\sum_{b^2-4ac=5, a<0, Q(1/2)>0} Q(1/2).$$ We use the identity $$b^2-4ac=(2ax-b)^2-4a(ax^2+bx+c).$$ In particular, for $x=1/2$, we know $$5=(a-b)^2+4|a|\cdot \left|\frac{1}{4}a+\frac{1}{2}b+c\right|.$$ Since the last term is at least $1/4$, we again obtain a bound on $|a|$, and hence $|b|$, and $|c|$. It turns out we find four possible solutions $$(a,b,c)=(-1, 1, 1), (-1,-1,1), (-1, 3, -1), (-5, 5,-1).$$ And so $A(x)=5/4+1/4+1/4+1/4=2$.

Now the same argument generalizes to any $x\in \mathbb{Q}$ to show that the sum is in fact a finite sum. However, this sum can be infinite in general.

Example 3 Let us compute $A(1/\pi=0.3183\ldots)$. Again the previous identity shows that $5>(2a \cdot 1/\pi -b)^2$. So for any fixed $a$, we have a bound on $|b|$, and hence on $|c|$. This allows us to search for solution starting with small values of $|a|$. 
    $(a,b,c)$ & $Q(1/\pi)$\\\hline
    $(-1, 1, 1)$ & 1.217\\\hline
    $(-1,1,1)$ & 0.580\\\hline
    $(-5,5,-1)$ & 0.085\\\hline
    $(-11, 7, -1)$ & 0.114\\\hline
    $(-409, 259, -41)$ & 0.002 \\\hline
    $(-541, 345, -55)$ & 0.002 \\\hline
The convergence is indeed very fast: the first 6 terms already sum up to 2 (precision 1/1000). Of course $A(1/\pi)$ must be an infinite sum: if the sum is a finite sum and $A(x)=2$, then $x$ must satisfy a quadratic equation over $\mathbb{Z}$!

TopThree variants

Variant I. Of course there should be nothing special about $D=5$. One can do experiments for other $D$ and one obtain similar pattern. Namely the function $$A_D(x):=\sum_{b^2-4ac=D, a<0}\max\{0, ax^2+bx+c\}$$ is always a constant function. Denoted this constant by $\alpha_D$. Assuming $A_D(x)=\alpha_D$ we can even figure out the value $\alpha_D$ by computing $$A_D(0)=\sum_{b^2-4ac=D, a<0, c>0}c=\sum_{|b|<\sqrt{D}}\sum_{c|(D-b^2)/4}c=\sum_{|b|<\sqrt{D}} \sigma_1\left(\frac{D-b^2}{4}\right).$$ Here $\sigma_m(n)=\sum_{c|n} c^m$ is the sum of divisor powers. The first few values of $\alpha_D$ are:

    $D$ & 5 & 8 & 12 & 13 & 17\\\hline
    $\alpha_D$ & 2 & 5 & 10 & 10 & 20 \\

Remark 1 For those familiar with the algebraic number theory of real quadratic fields $K=\mathbb{Q}(\sqrt{D})$: this sum can be recognized as the special value $60 \zeta_K(-1)=\frac{\zeta_K(-1)}{2\zeta(-3)}$.

Variant II. Fix $D=5$ but let us change the sum using a higher power $$B(x)=\sum_{b^2-4ac=5, a<0}\max\{0, (ax^2+bx+c)^3\}.$$ (Here we don't take an even power as it makes less sense to pick out the positive values, and the sum would diverge.) This is easier to prove to be converge (since $Q(x)=O(1/a)$ by the previous identity).

The surprise happens again.

Theorem 2 $B(x)$ is again the constant function 2.
Example 4
  • $B(0)=1^3+1^3=1$.
  • $B(1/2)=(5/4)^3+(1/4)^3+(1/4)^3+(1/4)^3=128/64=2$.
  • $B(1/\pi)\approx1.217^3+0.580^3+0.085^3+0.114^3\approx1.802+0.195+0.001+0.001=1.999$.

Again there should be nothing special about $D=5$, in fact the function $$B_D(x)=\sum_{b^2-4ac=D, a<0}\max\{0, (ax^2+bx+c)^3\}$$ should be again a constant $\beta_D$ given by $$\beta_D=B_D(0)=\sum_{|b|<\sqrt{D}}\sigma_3\left(\frac{D-b^2}{4}\right).$$ The first few values are given by 
    $D$ & 5 & 8 & 12 & 13 & 17\\\hline
    $\alpha_D$ & 2 & 11 & 46 & 58 & 164 \\

Remark 2 $\beta_D$ can be recognized as the special value $120\zeta_K(-3)=\frac{\zeta_K(-3)}{2\zeta(-7)}$.

Variant III. Fix $D=5$. Again there should be nothing special about 3rd powers. Let us consider the case of 5th powers $$C(x)=\sum_{b^2-4ac=5, a<0}\max\{0, (ax^2+bx+c)^5\}.$$ We would expect $C(x) $ is again a constant function, given by the sum $\sum_{|b|<\sqrt{D}}\sigma_5(\frac{D-b^2}{4})$, which can be recognized as $\frac{16380}{691}\cdot \zeta_K(-5)=\frac{\zeta_K(-5)}{2\zeta(-11)}$.

However, we get another surprise:

Theorem 3 $C(x) $ is not a constant function.
Example 5 $C(0)=1^5+1^5=2$, while $C(1/2)=(5/4)^5+(1/4)^5+(1/4)^5+(1/4)^5=(3125+3)/1024\ne2$!

You probably already smell something like this, as it is impossible for a fixed collection of rational numbers to have all their $k$-th powers sum up to 2.

TopThe new phenomenon

Let us stare at the 5th power sum $$C_D(x)=\sum_{b^2-4ac=D, a<0}\max\{0, (ax^2+bx+c)^5\}$$ more carefully. Let us subtract the expected value $\gamma_D=\frac{\zeta_K(-5)}{2\zeta(-11)}$ from the constant term, 
    $D$ & 5 & 8\\\hline
    $\gamma_D$ & $\frac{1742}{691}$ & $\frac{23465}{691}$\\\hline
    $C_D(1/2)-\gamma_D$ & $\frac{47205}{88448}$ & $-\frac{47205}{44224}$ \\\hline
    $C_D(1/3)-\gamma_D$ & $\frac{1175960}{4533651}$ & $-\frac{2351920}{4533651}$ \\\hline
    $C_D(1/4)-\gamma_D$ & $-\frac{135855}{22642688}$ & $\frac{135855}{11321344}$ \\\hline

Notice the pattern: the second column is actually $-2$ times the first column. More numerical data strongly suggest that $$C_D(x)-\gamma_D= \delta_D\cdot \Phi(x),$$ where $\Phi(x)$ is independent of $D$ (the previous table shows $\delta_8/\delta_5=-2$). Moreover, $\Phi(x)$ (of period 1) has average value 0. So its Fourier expansion has the form $$\Phi(x)=\sum_{n\ge1} c_n \cos (2\pi n x).$$ All these $c_n$'s are transcendental numbers, but the ratio $c_n/c_1$ turns out to be very close to rational numbers numerically: $$c_2/c_1=-\frac{3}{256}=-\frac{3}{2^8},\quad c_3/c_1=\frac{28}{19683}=\frac{28}{3^9},\quad c_4/c_1=-\frac{23}{65536}=-\frac{23}{4^8}.$$ Now the miracle happens if we multiply $c_n/c_1$ by $n^{11}$: $$2^{11} c_2/c_1=-3\cdot 2^3=-24,\quad 3^{11} c_3/c_1=28\cdot9=252, \quad 4^{11} c_4/c_1=-23\cdot 4^3=-1472.$$ These are exactly Ramanujan's $\tau(n)$ for $n=2,3,4$! At the end we have produced the modular form $\Delta(z)$ out of (the Fourier expansion of) the 5th power sum of values of quadratic polynomials!

TopWhat is going on?

As we have seen, the function $\Phi(x)$ (normalized version of $C(x) $) is closely related to Ramanujan's $\Delta(z)$, a modular form for the group $\SL_2(\mathbb{Z})$ of weight 12. More precisely,

Definition 1 A modular form of weight $k$ is a holomorphic function on the complex half-plan $\{z\in \mathbb{C}: \im(z)>0\}$ such that $f(z)=(cz+d)^{-k}f(\frac{az+b}{cz+d})$ for all $\left(\begin{smallmatrix}a & b\\ c & d\end{smallmatrix}\right)\in SL_2(\mathbb{Z})$ and holomorphic when $z\rightarrow i\infty$. Since $SL_2(\mathbb{Z})$ is generated by two elements $\left(\begin{smallmatrix}1 & 1 \\ 0 &1 \end{smallmatrix}\right)$ (translation) and $\left(\begin{smallmatrix} 0 & -1 \\ 1 & 0\end{smallmatrix}\right)$ (inversion), this condition is equivalent to $$f(z)=f(z+1), z^k f(z)=f(-1/z).$$ We say $f(z)$ is a cusp form such that $f(i\infty)=0$ (equivalently the constant term $a_0=0$ in its Fourier expansion). The space of modular forms (resp. cusp forms) of weight $k$ is usually denoted by $\mathcal{M}_k$ and $\mathcal{S}_k$.

A fundamental theorem in the theory of modular forms is the space of modular forms of weight $k$ is finite dimensional. Its dimension can be easily computed. 
    $k$ & 2 & 4 & 6 & 8 & 10 & 12 & 14 & 16 & 18\\\hline
    $\dim \mathcal{M}_k$ &0 & 1 & 1 & 1 & 1 & 2 & 1 & 2 & 2 \\
    $\dim \mathcal{S}_k$ & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 1

One can see weight 12 is the smallest weight for a cusp form to exist, which is exactly spanned by Ramanujan's $\Delta(z)$.

More generally, consider the function $$F_{k,D}(x)=\sum_{b^2-4ac=D, a<0}\max\{0, (ax^2+bx+c)^{k-1}\}.$$ So $A_D(x)=F_{2,D}(x)$, $B_D(x)=F_{4,D}(x)$ and $C_D(x)=F_{6, D}(x)$. Then the normalized version of $F_{k,D}(x)$ is closely related to the modular forms of level $2k$. The reason $C_D(x)$ (or more generally $k\ge6$) behaves differently from $k=2, 4$ is exactly due to the fact that there are cusp forms of weight $2k$ for $k\ge6$ but not for $k=2, 4$.

TopThe modular connection

To briefly explain the connection with modular forms, let us come back to the case of $A(x)$.

Example 6 The key is to show $A(x)=2$ is to make the sum more symmetric: 
  x^2A_D(1/x)-A_D(x)&=\sum_{a<0}\max\{0, a+bx+cx^2\}-\sum_{a<0}\max\{0, ax^2+bx+c\}\\
  &=\sum_{c<0}\max\{0, ax^2+bx+c\}-\sum_{a<0}\max\{0, ax^2+bx+c\}\\
  &=\sum_{c<0<a}\max -\sum_{a<0<c}\max\ (\text{the terms with $a,c<0$ cancel})\\
  &=\sum_{c<0<a}\max+\sum_{c<0<a}\min\ (\text{negation in the second term})\\
  & =\sum_{c<0<a}ax^2+bx+c.
It is clearly that the last term sum to $\alpha_D x^2- \alpha_D=\alpha_D(x^2-1)$. This implies that the function $$A_D^0(x):=A_D(x)-\alpha_D$$ satisfies the functional equation $$x^2 A_D^0(1/x)=A_D^0(x).$$ Using this together with the fact that $A_D^0(x)$ is even, period one, and $A_D^0(0)=0$, and it follows that $A_D^0(x)=0$ for any $x\in \mathbb{Q}$! Then by a continuity argument we know $A_D^0(x)=0$ as desired.

For more general $F_{k,D}(x)$, we similarly consider $$P(x)=x^{2k-2} F(1/x)-F(x)=\sum_{c<0<a}(ax^2+bx+c)^{k-1}.$$ It is no longer obvious what the last sum is, but $P(x)$ must satisfy the following functional equation (since $F(x)$ is even and has period 1): 
  &=(x+1)^{2k-2}F\left(\frac{x}{x+1}\right)-x^{2k-2}F\left(\frac{x+1}{x}\right) \\
  &=x^{2k-2} P\left(\frac{1}{x}+1\right).

Definition 2 Let $\mathcal{P}_n$ be the space of polynomials of degree $\le n$ such that $$P(x+1)-P(x)=x^n P\left(\frac{1}{x}+1\right).$$

This is known as the space of period polynomials as they are related to period integrals of modular forms. One can produce polynomials in $\mathcal{P}_n$ by period integrals of modular forms.

Definition 3 Suppose $f\in \mathcal{M}_k$. We denote $\tilde f(z)=\sum_{n\ge1} \frac{a_n}{n^{2k-1}} q^n$ (the Eichler integral). Define $$r_f(z):=z^{2k-2}\tilde f\left(-\frac{1}{z}\right)-\tilde f(z).$$ Then one can use the definition of modular forms to check that the $(2k-1)$-st derivative of $f(z)$ is 0, hence $r_f(z)$ is a polynomial of degree $2k-2$. We define $r_f^+(z)$ to be even degree part of $r_f(z)$.

The fundamental theorem in the theory of modular forms is the following.

Theorem 4 (Eichler-Shimura) The map $f\mapsto r_f^+(z)$ induces is a canonical isomorphism $$\mathcal{M}_{2k}\cong \mathcal{P}_{2k-2}.$$
Example 7
  • When $k=2$, $\dim \mathcal{M}_4=1$, so $\mathcal{P}_{2}$ is 1-dimensional generated by $x^2-1$. As we have already seen this implies $A_D(x)=\alpha_D$ is a constant.
  • When $k=4$, $\dim \mathcal{M}_8=1$, so $\mathcal{P}_6$ is again 1-dimensional, generated by $x^6-1$. The same argument as in the case $k=2$ implies $B_D(x)=\beta_D$ is again a constant.
  • When $k=6$, $\dim \mathcal{M}_{12}=2$, and so there are two independent period polynomials in $\mathcal{P}_{10}$, which turns out to be $$x^{10}-1,\quad  x^2(x^2-1)^3.$$ Let $\Phi(x)=\sum_{n\ge1} \frac{\tau(n)}{n^{11}}\cos (2\pi nx)$. Then $r_\Delta^+(z)$ is given by (up to a transcendental period) $$x^{10}\Phi(1/x)-\Phi(x)=P(x)=x^{10}-1-\frac{691}{36}x^2(x^2-1)^3.$$ It is exactly the extra period polynomial causing $C_D(x)$ to be no longer a constant!
Remark 3 Why quadratic polynomials? Indeed Zagier's original motivation was to directly produce modular forms from cubic polynomials. It would be big news if one can succeed for such generalization (= a new appraoch to the modularity of elliptic curves!).
Last Update: 10/25/2017. Copyright © 2015 - 2017, Chao Li.


[1]Zagier, D., From quadratic functions to modular functions, Number theory in progress, Vol. 2 (Zakopane-Ko\'scielisko, 1997), de Gruyter, Berlin, 1999, 1147--1178.