Invitation to Modular Forms: Representation of an integer into sum of eight squares

Created: August 23, 2025 8:54 PM

Last Updated: September 13, 2025

Note: This is an edited reupload which is intended to be hosted on my blog. The reason is that I've received a performance issue complain from the old site hosted at Notion, so here is a spare one. There are minor typographical fixes but no new content changes.

Introduction

We’re interested in the following computational problem: Given $n \in \mathbb{N}$, how many ways (i.e. number of tuples $(x_1, \dots, x_8) \in \mathbb{Z}^8$) are there such that $n$ can be written as $n = x_1^2 + \dots + x_8^2$.

Our basic computer programming knowledge tells us that we can design a naive algorithm as follows:

n = int(input())
ans = 0
for x1 in range(-n, n+1):
    for x2 in range(-n, n+1):
        for x3 in range(-n, n+1):
            for x4 in range(-n, n+1):
                for x5 in range(-n, n+1):
                    for x6 in range(-n, n+1):
                        for x7 in range(-n, n+1):
                            for x8 in range(-n, n+1):
                                if x1**2 + x2**2 + x3**2 + x4**2 + x5**2 + x6**2 + x7**2 + x8**2 == n:
                                    ans += 1
print(ans)

Of course, this takes a time complexity of $O(n^8)$, which is not very good. A basic obvious improvement is to observe that $x^2 \geq 0$ for all $x \in \mathbb{N}$, so in order for a tuple to work, no entry can be larger than $\sqrt{n}$ because otherwise (say, if $x_1 > \sqrt{n}$) we would have $x_1^2 + \dots + x_8^2 > n+x_2^2 + \dots + x_8^2 \geq n+1$. This allows us to decrease the loop to the size of $O(\sqrt{n})$.

n = int(input())
ans = 0
k = int(n**0.5)+1
for x1 in range(-k, k+1):
    for x2 in range(-k, k+1):
        for x3 in range(-k, k+1):
            for x4 in range(-k, k+1):
                for x5 in range(-k, k+1):
                    for x6 in range(-k, k+1):
                        for x7 in range(-k, k+1):
                            for x8 in range(-k, k+1):
                                if x1**2 + x2**2 + x3**2 + x4**2 + x5**2 + x6**2 + x7**2 + x8**2 == n:
                                    ans += 1
print(ans)

This takes time $O(n^4)$, a bit better, huh?

Obviously, from the eye of an algorithmist, we can do better. One direct way when we see something in this form is to do the meet-in-the-middle trick: instead of trying to solve for $x_1^2 + \dots + x_8^2 = n$, we solve for $x_1^2 + x_2^2 + x_3^2 + x_4^2 = m$ and $x_5^2 + x_6^2 + x_7^2 + x_8^2 = n-m$, count the number of solutions to this, and sum over the scenarios for each $m \in \{0, \dots, n\}$. Let us try to implement.

n = int(input())

def count_sol(m):
    k = int(m**0.5)+1
    local_ans = 0
    for a in range(-k, k+1):
        for b in range(-k, k+1):
            for c in range(-k, k+1):
                for d in range(-k, k+1):
                    if a**2 + b**2 + c**2 + d**2 == m:
                        local_ans += 1
    return local_ans

ans = 0
for m in range(n+1):
    ans += count_sol(m) * count_sol(n-m)
print(ans)

The count_sol subroutine runs in time $O(\sqrt{m}^4) = O(m^2)$. We repeat the process $2(n+1)$ times, each of $m \leq n$ so the total time is in $2(n+1) \cdot O(n^2) = O(n^3)$.

Can we do better? Still yes. Skilled algorithmists might even come up of this formulation immediately (note: I’m not that good though — I asked this problem to a bunch of friends and they immediately gave me the following formulation). We formulate this in terms of knapsack problem. Define $\mathrm{dp}_{i,j}$ to be the quantity $\#\{(x_1, \dots, x_i) \in \mathbb{Z}^i \colon \sum_{k=1}^i x_k^2 = j\}$, and look at what we can do about its recurrence.

In order to know the quantity $\mathrm{dp}_{i,j}$, we can enumerate all possible $x_i$’s and observe that

$$\begin{align*} &\{(x_1, \dots, x_i) \in \mathbb{Z}^i \colon \sum_{k=1}^i x_k^2 = j\} \\ &= \bigcup_{0 \leq x_i \leq \sqrt{j}} \{(x_1, \dots, x_{i-1}) \in \mathbb{Z}^{i-1} \colon \sum_{k=1}^{i} x_k^2 = j\} \times \{(x_i), (-x_i)\} \\ &= \bigcup_{0 \leq x_i \leq \sqrt{j}} \{(x_1, \dots, x_{i-1}) \in \mathbb{Z}^{i-1} \colon \sum_{k=1}^{i-1} x_k^2 = j-x_i^2\} \times \{(x_i), (-x_i)\}. \end{align*}$$

Observe that the union in disjoint, so the cardinality of the whole is the sum of the cardinality of each case, so

$$\mathrm{dp}_{i,j} = \sum_{1 \leq a \leq \sqrt{j}} 2\mathrm{dp}_{i-1, j-a^2} + \mathrm{dp}_{i-1,j}.$$

With this formula, we can implement our dynamic programming algorithm as follows.

n = int(input())
dp = [[0 for j in range(n+1)] for i in range(9)]
dp[0][0] = 1
for i in range(1, 9):
    for j in range(0, n+1):
        dp[i][j] += dp[i-1][j]
        for a in range(1, int(j**0.5)+1):
            if j >= a**2:
                dp[i][j] += 2*dp[i-1][j-a**2]
print(dp[8][n])

This takes time $O(n^{3/2})$, which is already very much improved from before. Is this the end? Can we do better? An algorithmist might give up at this point, but I refuse to give up just yet! Can we use some mathematical magic to improve this even more? It turns out that we can!

In this article, I’m going to invite you to the fascinating (quite relatively deep) theory of modular forms, not as just an algorithmist, but more as an algorithmic number theorist.

The prerequisites may be quite demanding in order to fill in all the gaps, but I intend to make this suitable to most general audiences, so for deep results from mathematics, if you’re not used to it, please feel free to “just assume the claim is true, move on to see how things fit, and try to understand the proof later”. The true prerequisites are:

• Design and analysis of algorithms (and the fact that factoring an integer $n$ is currently still not known to be doable in polynomial time in $\log n$)
• Introductory, real, complex, and Fourier analysis (basic ordinary differential equation, standard topology in $\mathbb{R}$ and $\mathbb{C}$, differentiability, Lebesgue integration, dominated convergence theorem, Fourier transform, holomorphic function, contour integral, Laurent expansion)
• Linear algebra (vector space, basis, dimension) and abstract algebra (group, group action, field)

Acknowledgement

I’d like to dedicate this section to thank my professor, Emmanuel Kowalski, for teaching me the basics of number theory and modular forms. I want to also thank a friend of mine, Pitchayut (Mark) Saengrungkongka, for helping me through mathematical obstacles during my studies. Finally, after all the messes I’ve made in my life, I’d like to thank everyone who still believes in me, who still believes that I can become a mathematician too, despite my sluggish brain and the distractions that slowed me down.

Disclaimer. Any mistakes found in this article belong to me. I’m more than happy to receive feedback and report of errors.

Now let us begin.

Modular Forms

Note. The main reference for this whole section is the course Number Theory II: Introduction to Modular Forms given by Emmanuel Kowalski (see https://metaphor.ethz.ch/x/2025/fs/401-3112-23L/). Other small references will be given throughout the article.

It is quite surprising that modular form is a structure that is studied extensively and, even though initially it is of number theoretic nature, it now has applications in very different areas of mathematics. When I first see it, it looks very artificial, as if it comes out of nowhere. (Who on earth would have think up of this structure???) The following quote is taken from Martin Eichler.

There are five fundamental operations in mathematics: addition, subtraction, multiplication, division and modular forms.

So what is a modular form? In simple terms, it is a function $f \colon \mathbb{H} \to \mathbb{C}$ that satisfies the following property (here $\mathbb{H} := \{z \in \mathbb{C} \colon \operatorname{Im}(z) > 0\}$ is the Poincaré upper-half plane):

There exists an integer $k \geq 0$ such that $f(\frac{az+b}{cz+d}) = (cz+d)^kf(z)$ for all $z \in \mathbb{H}$ and for all $a, b, c, d \in \mathbb{Z}$ such that $ad-bc = 1$.

If furthermore, $f$ is holomorphic everywhere in $\mathbb{H}$, we say that it is a holomorphic modular form of weight $k$ and of level $1$. For meromorphic modular forms, we don’t require $f$ to be defined on the whole $\mathbb{H}$ — there can be isolated poles, only that the relation $f(\frac{az+b}{cz+d}) = (cz+d)^kf(z)$ holds for all $z$ such that $f(z)$ is defined (i.e. not a pole). We will generalize to other levels later.

Now we ramp up the machinery a bit for some spiciness. Define $\mathrm{GL}_n(K)$ as the set of $n \times n$ invertible matrices with entry in the field $K$. Equivalently, it is the set of $n \times n$ matrices with nonzero determinant. Since $\det(AB) = \det(A)\det(B)$ for square matrices $A, B$ of the same size, and by invertibility, we see clearly that $\mathrm{GL}_n(K)$ has a group structure defined by multiplication. Now, we define a subset $\mathrm{SL}_n(K) \subseteq \mathrm{GL}_n(K)$ as the subset of matrices with determinant one. This subset is also a group, so it’s a subgroup. We can generalize this a bit for the case that $K$ is not a field, for $\mathrm{GL}_n(K)$ it looks like we run into an immediate problem: it’s not a group anymore. However, fortunately, for $\mathrm{SL}_2(\mathbb{Z})$, there is no problem. If $\begin{pmatrix}a & b\\c & d\end{pmatrix} \in \mathrm{SL}_2(\mathbb{Z})$ is a matrix with determinant one, i.e. $ad-bc = 1$, then consider the matrix $\begin{pmatrix}-d & b\\c & -a\end{pmatrix}$. We have

$$\begin{pmatrix}a & b\\c & d\end{pmatrix}\begin{pmatrix}-d & b\\c & -a\end{pmatrix} = \begin{pmatrix}-ad+bc &ab-ba\\-cd +dc & bc -ad\end{pmatrix} = \begin{pmatrix}1 & 0\\0 & 1\end{pmatrix}.$$

This means $\begin{pmatrix}-d & b\\c & -a\end{pmatrix}$ is the inverse of $\begin{pmatrix}a & b\\c & d\end{pmatrix}$! So $\mathrm{SL}_2(\mathbb{Z})$ forms a group.

Now, define a group action $|_k$ of $\mathrm{SL}_2(\mathbb{Z})$ acting from the right on the space of functions $\mathbb{H} \to \mathbb{C}$ by

$$f \Big\vert_k \begin{pmatrix}a & b\\c & d\end{pmatrix} := \left(z \mapsto f\left(\frac{az+b}{cz+d}\right)(cz+d)^{-k}\right).$$

It restricts to the space of holomorphic functions. Now, in fancier words, we can define modular forms of weight $k$ and of level $1$ to be functions $f \colon \mathbb{H} \to \mathbb{C}$ such that $f|_kg = f$ for all $g \in \mathrm{SL}_2(\mathbb{Z})$, i.e., it is the subset of $(\mathrm{SL}_2(\mathbb{Z}), |_k)$-invariant functions. In particular, it is enough to show that $f$ is invariant under a generating set of $\mathrm{SL}_2(\mathbb{Z})$ to conclude that it is invariant under the whole group. We will use this fact later.

Definition. (Modular form of weight $k$ and of level $1$) A function $f \colon \mathbb{H} \to \mathbb{C}$ is said to be a modular form of weight $k$ and of level $1$ if $f$ is invariant under the group action $|_k$ of the group $\operatorname{SL}_2(\mathbb{Z})$.

Now, consider the action of $\mathrm{SL}_2(\mathbb{Z})$ on $\mathbb{H}$ defined by $\begin{pmatrix}a & b\\c & d\end{pmatrix} \cdot z = \frac{az+b}{cz+d}$. This action sends point of $\mathbb{H}$ to another (or maybe the same) point of $\mathbb{H}$. We can start by just being randomly curious about this and ask interesting questions on this action. First of all, given a fixed $z \in \mathbb{H}$, what is the orbit of this point by the action? Well, we can pick $\begin{pmatrix}1 & 1\\0 & 1\end{pmatrix}$ from $\mathrm{SL}_2(\mathbb{Z})$ and observe that it sends $z$ to $\frac{1z+1}{0z+1} = z+1$. Therefore, $z+n$ is inside the orbit, for all $n \in \mathbb{Z}$. Of course, this is not the end. Let’s try something different. Try $\begin{pmatrix}0 & -1\\1 & 0\end{pmatrix} \in \mathrm{SL}_2(\mathbb{Z})$. This sends $z \mapsto \frac{-1}{z}$. It may be hard to visualize at first, but observe that if $|z| > 1$, it sends to $|\frac{-1}{z}| = \frac{1}{|z|} < 1$. This means, a point outside the unit disk can be sent into somewhere inside the unit disk. This allows us to conclude that “it is enough to just consider points inside a strip $[z_0, z_0+1]$ because the points outside this strip can be reached from one inside this strip by applying the group action” and also “it is enough to just consider points inside the unit disk because points outside the unit disk can be reached by the $z \mapsto -1/z$ transformation”. Turns out this is something that we’ll state it in more precise terms and use it repeatedly afterwards (but for reasons that will come later, instead of inside the disk, we’d rather use outside the disk).

The Fundamental Domain

Definition. The fundamental domain is defined as $\mathcal{D} := \{z \in \mathbb{H} \colon |z| \geq 1; |\operatorname{Re}(z)| \leq \frac{1}{2}\}$.

Theorem 1 [Serre, Ch. VII, §1, Th 1. and 2.].

1. For every $z \in \mathbb{H}$ there exists $g \in \mathrm{SL}_2(\mathbb{Z})$ such that $gz \in \mathcal{D}$.
2. Suppose $z, z' \in \mathcal{D}$ are two distinct points, such that there exists $g \in \mathrm{SL}_2(\mathbb{Z})$ with $z' = g \cdot z$, then either (a) $\operatorname{Re}(z) \in \{-\frac{1}{2}, \frac{1}{2}\}$ and $z = z' \pm 1$, or (b) $|z| = 1$ and $z' = \frac{-1}{z}$.
3. For all $z \in \mathcal{D}^\circ$, $\operatorname{Stab}_{\operatorname{SL}_2(\mathbb{Z})}(z) = \{\operatorname{id}\}.$
4. The group $\mathrm{SL}_2(\mathbb{Z})$ is generated by $S = \begin{pmatrix}0 & -1\\1 & 0\end{pmatrix}$ and $T = \begin{pmatrix}1 & 1\\0 & 1\end{pmatrix}$.

Proof. Define $G' := \langle S, T \rangle \leq \mathrm{SL}_2(\mathbb{Z})$. We will eventually show that $G' = \mathrm{SL}_2(\mathbb{Z})$ later, but now let us show 1. Take $z \in \mathbb{H}$, and let $g = \begin{pmatrix}a & b\\c & d\end{pmatrix} \in G'$be arbitrary first (we will fix it later). We have

$$\begin{align*} \operatorname{Im}(gz) &= \operatorname{Im}\left(\frac{az+b}{cz+d}\right) \\ &= \operatorname{Im}\left(\frac{(az+b)\overline{(cz+d)}}{|cz+d|^2}\right) \\ &= \frac{1}{|cz+d|^2}\operatorname{Im}(a\overline{c}|z|^2+b\overline{cz} + a\overline{d}z + b\overline{d})\\ &= \frac{1}{|cz+d|^2} \operatorname{Im}(bc\overline{z}+adz)\\ &= \frac{1}{|cz+d|^2} (ad-bc) \operatorname{Im}(z)\\ &= \frac{\operatorname{Im}(z)}{|cz+d|^2}. \end{align*}$$

(We used the simple facts that $a, b, c, d \in \mathbb{Z}$, so $\overline{c} = c; \overline{d} = d$. We know that $ad-bc = 1$, and we know that $\operatorname{Im}(\overline{z}) = -\operatorname{Im}(z)$.) Now, the idea is that for a given $z$, we will try to send it as far away (from the origin) as possible, then slide back by repeatedly applying $T$ until it stays in $\mathcal{D}$. For any fixed $M > 0$, The set $S_{z, M} :=\left\{(c, d) \in \mathbb{Z}^2 \colon |cz+d| \leq M\right\}$ is finite. To see this, write $z = x+iy$ and observe that $|cz+d| \leq M$ if and only if $M^2 \geq |cz+d|^2 = |c(x+iy)+d|^2 = (cx+d)^2 + (cy)^2$. So if $(c, d)$ is in that set, then $|c| \leq \frac{M}{|y|}$ (finitely many choices), and for a fixed $c$, the integer $d$ must satisfy $|cx+d| \leq M$, that is, $d \in [-cx-M, -cx+M]$, which contains finitely many integers. This proves that the set $S_{z, M}$ is finite. If we fix $M = |z|$, then we see that $(1, 0) \in S_{z, M}$, so the set is not empty. Furthermore, we’re looking for such pair with smallest $|cz+d|$, so enlarging $M$ wouldn’t help. This means $\min_{(c,d) \in \mathbb{Z}^2} |cz+d|$ exists, with $(c, d) \in S_{z, |z|}$. However, we want to only look at pairs corresponding to elements of $G'$, so we consider the set

$$\tilde{S}_z := \left\{\begin{pmatrix}a & b\\c & d\end{pmatrix} \in G' \colon |cz+d| \leq |z|\right\}.$$

Well, $S \in \tilde{S}_z$ so it is not empty, and there are only finitely many choices of $(c, d)$ here, so, to minimize $|cz+d|$, we loop over all possible pairs in $S_{z,|z|}$, and check if there exists a pair $(a, b)$ making $\begin{pmatrix}a & b\\c & d\end{pmatrix} \in \tilde{S}_z$. We don’t need to loop over all possible $(a, b)$. Just that if one exists, we track the minimum over such $|cz+d|$. This shows that there exists $g \in \tilde{S}_z$ minimizing $|cz+d|$.

Now that we got $g \in G'$, we want to slide $gz$ back into the strip $[-\frac{1}{2},\frac{1}{2}]+i\mathbb{R}_{>0}$. Indeed, pick $n \in \mathbb{Z}$ such that $n+gz \in [-\frac{1}{2},\frac{1}{2}]$, and so $T^ngz \in [-\frac{1}{2},\frac{1}{2}]$. Furthermore, we have $|T^ngz| \geq 1$ because otherwise $|T^ngz| < 1$ and so when we consider $ST^ng \in G'$ we get

$$\operatorname{Im}(ST^ngz) = \frac{\operatorname{Im}(T^ngz)}{|T^ngz|} > \operatorname{Im}(T^ngz) = \operatorname{Im}(gz),$$

a contradiction of the maximality of $\operatorname{Im}(gz)$ for $g \in G'$. This shows that $T^ngz \in \mathcal{D}$.

Phew. That was long. Now let us move to the second point (which is even longer, because I want to make it very clear to myself, and not leave the casework to the reader). We want to now show that if two points inside the fundamental domain are congruent modulo $G'$, then they must be in a few special cases only. Suppose $z, z' \in \mathcal{D}$ are two distinct points and $\begin{pmatrix}a & b\\c & d\end{pmatrix} =g \in G'$ such that $z' = gz$. Assume without loss of generality that $\operatorname{Im}(z) \leq \operatorname{Im}(z')$ (otherwise just swap $z$ and $z'$). And so $\operatorname{Im}(z) \leq \operatorname{Im}(gz) = \frac{\operatorname{Im}(z)}{|cz+d|^2}$, so $|cz+d| \leq 1$. Write now $z = x+iy$ with $|x| \leq \frac{1}{2}$ and $y > \frac{\sqrt{3}}{2}$. If $|c| \geq 2$ , then

$$\begin{align*} |cz+d|^2 &= |cx+icy+d|^2 = (cx+d)^2 + (cy)^2 \\ &\geq c^2y^2 \geq 4 \frac{\sqrt{3}^2}{2^2} = 3, \end{align*}$$

a contradiction. So $|c| < 2$, i.e. $c \in \{-1, 0, 1\}$. If $c = 0$ then $|cz+d| \leq 1$ implies $d \in \{-1, 1\}$ (remember that $c = d = 0$ is not possible since this means $\det g = 0$, impossible in $G' \subseteq \mathrm{SL}_2(\mathbb{Z})$). Furthermore, in $\mathrm{SL}_2(\mathbb{Z})$, we have $ad - bc = 1$ so for $c = 0$, $ad = 1$ means $a = d$. This means $z' = gz = \frac{az+b}{d} = z \pm (\text{integer})$, but $z, z' \in \mathcal{D}$ so such integer must be $\pm 1$ only. And in this case, $|\operatorname{Re}(z)-\operatorname{Re}(z')| = 1$, so they must lie on somewhere inside the vertical boundaries of $\mathcal{D}$. This concludes that in case $c = 0$, we have $\operatorname{Re}(z) = \pm \frac{1}{2}$ and $z' = z \pm 1$, i.e., case (a) in the theorem.

Now, for the case that $c \in \{-1, 1\}$, we have $1 \geq |cz+d| = |cx+icy+d| = (cx+d)^2 + y^2 \geq (cx+d)^2 + (\frac{\sqrt{3}}{2})^2 = (cx+d)^2 + \frac{3}{4}.$ This means $|cx+d| \leq \frac{1}{2}$. This means $d$ cannot be too far away, say, $|d| > 1$ is not possible, because then $cx \in [-d-\frac{1}{2}, -d+\frac{1}{2}]$ is impossible for $|cx| = |x| \leq \frac{1}{2}$. For $d = -1$, the only possible place for $cx$ to be is $\frac{1}{2}$, and in this case, $1 \geq |cz+d|^2 = (cx+d)^2 + y^2 = \frac{1}{4} + y^2$, so $\frac{\sqrt{3}}{2} \leq y \leq \frac{\sqrt{3}}{2}$, i.e., $y = \frac{\sqrt{3}}{2}$, and for $d = 1$, $cx = -\frac{1}{2}$, and $1 \geq |cz+d|^2 = \frac{1}{4}+y^2$ so $y = \frac{\sqrt{3}}{2}$ as well. In these two subcases, we have $z = \pm \frac{1}{2} + i\frac{\sqrt{3}}{2}$, and so $|z| = 1$.

In both subcases, we have $d+2cx = 0$. Consider the fact that $gz = z' \in \mathcal{D}$, so $|\operatorname{Re}(gz)| \leq \frac{1}{2}$, i.e. that

$$\begin{align*} \frac{1}{2} &\geq \left|\operatorname{Re}\left(\frac{az+b}{cz+d}\right)\right| \\ &= \left|\operatorname{Re}\left(\frac{(az+b)\overline{(cz+d)}}{(az+b)\overline{(cz+d)}}\right)\right| \\ &= \frac{1}{|cz+d|^2} |\operatorname{Re}((a(x+iy)+b)(c(x-iy)+d))| \\ &= \frac{1}{|cz+d|^2} |(ax+b)(cx+d) + acy^2|. \end{align*}$$

But we knew that $|cz+d| \leq 1$, so $\frac{1}{2} \geq |acx^2+bcx+adx+bd+acy^2|$. In both subcases, $1 = |z|^2 = x^2+y^2$, so we have $\frac{1}{2} \geq |ac + (bc+ad)x + bd| = |ac + (2bc+1)x + bd| = |b(d+2cx) + ac+x| = |ac+x|$. If $|a| \geq 2$, then $2 \leq |ac+x-x| \leq |ac+x| + |x| = |ac+x| + \frac{1}{2}$, a contradiction, so $a \in \{-1, 0, 1\}$.

Observe that in both subcases, $2cxd = -1$. If $b \ne 0$, then $ad = 1+bc$ means $ad-1$ also cannot be zero, i.e. $ad \ne 1$. If $a = -d$ then $\frac{1}{2} \geq|ac+x| = |-dc+x| = |\frac{-cdx+x^2}{x}| = |\frac{\frac{1}{2}+\frac{1}{4}}{\pm\frac{1}{2}}| = \frac{3}{2}$, a contradiction, so the only option when $b \ne 0$ is that $a = 0$. In such scenario, $1 = ad-bc = -bc$ so $b = -c$. Now for $b = 0$, we have $1 = ad - bc = ad$ so $a = d$.

To summarize, for the case (b.1) that $c \in \{-1, 1\}$ and $d \ne 0$, we have deduced the candidates for $g$:

$$g \in \left\{\begin{pmatrix}0 & -c\\c & d\end{pmatrix} \colon cx = -\frac{d}{2}\right\} \cup \left\{\begin{pmatrix}d & 0\\c & d\end{pmatrix} \colon cx = -\frac{d}{2}\right\}.$$

If $g$ is in the first set, $gz = \frac{-c}{cz+d} = \frac{-c}{c(x+iy)+d} = \frac{-c}{\frac{d}{2}+ic\frac{\sqrt{3}}{2}} = \frac{-1}{\frac{d}{2c}+i\frac{\sqrt{3}}{2}} = \frac{-1}{-x+iy} = \frac{1}{x-iy} = \frac{1}{\overline{z}} = \frac{z}{|z|^2} = z$, a contradiction with the fact that $z' \ne z$. If $g$ is in the second set, $gz = \frac{dz}{cz+d} = \frac{dz}{cz+d} \frac{\overline{cz+d}}{\overline{cz+d}} = \frac{(dz)(\overline{cz+d})}{|cz+d|^2} = dc|z|^2 + d^2z = cd+z = -2x+x+iy = -x+iy = -\bar{z} = -\frac{z\bar{z}}{z} = -\frac{1}{z}$, and now it is clear that $g = \begin{pmatrix}1 & 0\\-2x & 1\end{pmatrix}$ or $g = \begin{pmatrix}-1 & 0\\2x & -1\end{pmatrix}$ for the case (b.1).

Now we’re left with the case that $c \in \{-1, 1\}$ and $d = 0$ (b.2). In this case, $1=ad-bc=-bc$ so $b = -c$. We’ve assumed that $1 \geq |cz+d|$, but for $d = 0$ and $|c| = 1$, this means $|z| = 1$. Recall that $|\operatorname{Re}(gz)| \leq \frac{1}{2}$ implies that $\frac{1}{2} \geq |acx^2+bcx+adx+bd+acy^2|$. Plug in what we have to get $\frac{1}{2} \geq |acx^2 -c^2x +acy^2| = |ac(x^2+y^2) - x|.$ Suppose $|a| \geq 2$, then, we have, by triangle inequality, that $|ac(x^2+y^2)| \leq |ac(x^2+y^2)-x| +|x|$, and so

$$\begin{align*} \frac{1}{2} &\geq |ac(x^2+y^2)-x| \\ &\geq |ac(x^2+y^2)|-|x| \\ &= |a| |c| |z|^2 -|x| \\ &\geq 2 \cdot 1 \cdot 1 - \frac{1}{2} = \frac{3}{2}, \end{align*}$$

a contradiction. So, $a \in \{-1, 0, 1\}$. If $a \ne 0$, then $a = \pm 1$, so $ac = \pm 1$, and $\frac{1}{2} \geq |\operatorname{Re}(gz)| = \frac{1}{|cz+d|^2} |(ax+b)(cx+d) + acy^2| = |(ax-c)(cx) + acy^2| = |acx^2 + acy^2 - x| = |ac|z|^2 - x| = |ac-x|$ would mean $x$ must be near $ac = \pm 1$ with distance no more than $\frac{1}{2}$. The only case is back to when $|x| = \frac{1}{2}$, and this shows that $y = \frac{\sqrt{3}}{2}$ and $|\operatorname{Re}(gz)| = \frac{1}{2}$ also in this case. Furthermore, if $ac = -1$ then $x = -\frac{1}{2}$ and if $ac = 1$ then $x = \frac{1}{2}$. We can summarize this as $2acx=1$ (or equivalently $ac = 2x$). This shows that $|gz|^2 = \frac{|az-c|^2}{|cz|^2} = (ax-c)^2 + (ay)^2 = 2-2acx = 1$, so $|gz|=1$ also. Now, we have

$$\begin{align*} gz &= \frac{az-c}{cz} = \frac{a}{c}-\frac{1}{z} = ac-\frac{1}{z} = 2x-\frac{1}{z} \\ &= z+\bar{z}+\frac{1}{z} = \frac{z^2+|z|^2-1}{z} = \frac{z^2}{z} = z, \end{align*}$$

a contradiction. This rules out $a = \pm 1$. Now let us check $a = 0$. We have

$$gz = \frac{-c}{cz} = \frac{-1}{z}.$$

This shows that for $c \ne 0$, $|z| = 1$ and $z' = -\frac{1}{z}$, and completes the case (b).

Now the third part follows by the same argument as in the previous part: Take $z \in \mathcal{D}^\circ$. Suppose $g = \begin{pmatrix}a & b\\c & d\end{pmatrix} \in \operatorname{Stab}_{\operatorname{SL}_2(\mathbb{Z})}(z)$, then $gz = z$. We have $\operatorname{Im}(gz) = \frac{\operatorname{Im}(z)}{|cz+d|^2} = \operatorname{Im}(z)$, so $|cz+d|=1$. By the same argument as before, it is impossible for $|c| \geq 2$, so we know that $c \in \{-1, 0, 1\}$. If $c \ne 0$ then we follow the same argument as before to deduce that $z = \pm \frac{1}{2} + \frac{\sqrt{3}}{2}i$, a contradiction since we’ve assumed that $z$ is in the interior, so we conclude that $c = 0; d = 1$, and $ad-bc = 1$ implies $a = 1$. This means $z = gz = \frac{az+b}{cz+d} = z+b$, so $b = 0$ and this tells us that $\operatorname{Stab}_{\operatorname{SL}_2(\mathbb{Z})}(z) \subseteq \{\operatorname{id}\}$.

Now let us show the fourth part, that $G' = \mathrm{SL}_2(\mathbb{Z})$. Take $g \in \mathrm{SL}_2(\mathbb{Z})$ and let us show that $g \in G'$. Choose $z_0 := 2i \in \mathcal{D}^\circ$, and let $z = gz_0 \in \mathbb{H}$. We’ve seen that by 1., there exists $g' \in G'$ such that $g'z \in \mathcal{D}$. The points $z_0$ and $g'z = g'gz_0$ are both in $\mathcal{D}$ and are congruent modulo $G'$. But $z_0$ is not on the boundary, so both points fail to be distinct. Therefore, $g'gz_0 = z_0 \in \mathcal{D}^\circ$. So $g'g \in \operatorname{Stab}_{\operatorname{SL}_2(\mathbb{Z})}(z_0) = \{\operatorname{id}\}$, i.e. $g = g'^{-1} \in G'$. This shows that $\operatorname{SL}_2(\mathbb{Z}) \subseteq G'$ and hence completes the proof. $\square$

We now turn to the more restricted notion: holomorphic modular forms which is also defined and holomorphic at infinity.

At Infinity

The real line $\mathbb{R}$ is not compact. The sequence $1, 2, 3, \dots$ has no accumulation point in $\mathbb{R}$. However, we can compactify it by adding a point at infinity, say, define $\hat{\mathbb{R}} := \mathbb{R} \cup \{\infty\}$. In the same way, the complex plane is not compact, but we can think of it as a sphere with one punctured hole corresponding to $\infty$.

Riemann Sphere, GKFXtalk, de:User:Bjoern_klipp, CC BY-SA 3.0 http://creativecommons.org/licenses/by-sa/3.0/, via Wikimedia Commons

Now, with this picture in mind, we go back to modular forms and try to complete them at infinity.

Proposition 2. Let $f$ be a meromorphic modular form of weight $k$, then there exists a meromorphic function $\tilde{f} \colon \mathbb{D}^\times \setminus \{\text{isolated poles}\}\to \mathbb{C}$ such that $f(z) = \tilde{f}(e^{2\pi i z})$ for all $z \in\mathbb{H} \setminus \{\text{isolated poles}\}$, where $\mathbb{D}^\times := \{z \in \mathbb{C} \colon 0 < |z| < 1\}$.

Proof. First, for any $a \in \mathbb{R}$, there is a bijection

$$\Phi_a \colon \begin{cases}\text{Strip}_a &\to \mathbb{D}^\times \setminus \{re^{2\pi i a} \colon 0 < r < 1\}\\ z &\mapsto e^{2\pi i z}\end{cases}$$

where $\text{Strip}_a :=\{z \in \mathbb{C} \colon \operatorname{Re}(z) \in (a, a+1); \operatorname{Im}(z) \in (0, +\infty)\}$, which is very nice in our use case. Furthermore, $\Phi$ is holomorphic everywhere in its strip, and we can compute the derivative as $\Phi'(z) = 2\pi i e^{2\pi i z}$ which never vanishes in the strip. This shows that $\Phi$ is a conformal equivalence. Now, we can define $\tilde{f}$ by patching up two charts. The first chart is $\Phi_{-\frac{1}{2}}$sending our favorite strip to $\mathbb{D}^\times \setminus (-1, 0)$, and the second chart is $\Phi_{\frac{1}{2}}$ sending an offset strip to $\mathbb{D}^\times \setminus (0, 1)$. We define $\tilde{f}(q)$ as $f(\Phi_{-\frac{1}{2}}^{-1}(q))$ whenever $q \in \mathbb{D}^\times \setminus (-1, 0)$ and define as $f(\Phi_0^{-1}(q))$ whenever $q \in \mathbb{D}^\times \setminus (0, 1)$. Observe that this is well-defined (except for some not too many points that are not defined even on $f$ since it’s meromorphic) since the values agree on the intersection. Indeed, if $q \in \mathbb{D}^\times \setminus (-1, 1)$ then $z_{-\frac{1}{2}} := \Phi_{-\frac{1}{2}}^{-1}(q)$ and $z_0 := \Phi_0^{-1}(q)$ satisfies $e^{2\pi iz_{-1/2}} = e^{2\pi i z_0}$, so $e^{2 \pi i(z_{-1/2}-z_0)} = 1$. This means $n := z_{-1/2}-z_0 \in \mathbb{Z}$, and so $f(z_{-1/2}) = f(z_0 + n) = f\left(\begin{pmatrix}1 & n\\0 & 1\end{pmatrix} z_0\right) = (0z_0+1)^k f(z_0) = f(z_0)$ by the modularity of $f$ (if it is defined there). The two charts extend each other to $\tilde{f} \colon \mathbb{D}^\times \to \mathbb{C}$, which is meromorphically defined, such that $f(z) = \tilde{f}(e^{2\pi iz})$ for all $z \in \mathbb{H}$ wherever defined. (One can check this by sliding $z$ back to our favorite strip or the offset strip, and note that $f$ is periodic due to modularity, and $z \mapsto e^{2\pi iz}$ is already periodic by default) $\square$

Note. Inspecting the proof carefully, we haven’t actually used the (rather strong) fact that $f$ is modular. We’ve only used the fact that $f$ is periodic. We will recall this construction for “just periodic, but not necessarily modular” meromorphic functions later on.

So now the disk $\mathbb{D}^\times$ is our model for $\mathbb{H}$. Think of it as the upper hemisphere of the Riemann sphere, with the center removed. Do you see the point of doing this now? If we can complete $\tilde{f}$ at the hole, we would complete $f$ at infinity! If there is a meromorphic continuation of $\tilde{f}$ to all of $\mathbb{D} := \mathbb{D}^\times \cup \{0\}$, then we’re done.

Definition. A meromorphic modular form $f$ is said to be meromorphic at infinity if there exists a meromorphic continuation of $\tilde{f}$ to $\mathbb{D}$. It is holomorphic at infinity if $\tilde{f}$ is holomorphic at $0$. If $\tilde{f}$ is defined at $0$, we may write $f(\infty)$ to denote $\tilde{f}(0)$.

Fourier Expansion

For meromorphic modular form that is also meromorphic at infinity, by complex analysis, we can expand $\tilde{f}$ into a Laurent series $\tilde{f}(q) = \sum_{n \geq k} a_n q^n$ for some $k \in \mathbb{Z}$ (maybe negative) and for some sequence $(a_n)_{n \geq k}$ of complex numbers with $a_k \ne 0$, such that equality holds for all $q$ in a neighborhood of $0$ (except maybe zero itself if $k < 0$). Such expansion is unique, i.e., for a given $f$, the pair $(k, (a_n)_{n \geq k})$ is uniquely defined (even if we pick different neighborhoods, the pair coincides). Equivalently, we have $f(z) = \sum_{n \geq k} a_ne^{2 \pi i n z}$ for all $z$ with large enough $\operatorname{Im}(z)$, with an extra value of $f(\infty) = \sum_{n \geq k} a_n$. To simplify things, we would rather denote $(a_n)_{n \in \mathbb{Z}}$ but keep in mind that it is always zero when the index is negative enough.

Definition. For a meromorphic modular form $f$ that is also meromorphic at infinity, there exists a unique sequence $(a_n)_{n \in \mathbb{Z}}$ of complex numbers, which takes value zero when the index is negative enough, such that $\tilde{f}(q) = \sum_{n \in \mathbb{Z}} a_nq^n$ in a neighborhood of zero but maybe without zero itself. The sequence is called the Fourier expansion of $f$ at infinity, and the sequence is called the Fourier coefficients.

Definition. We write $\mathcal{M}_k$ for the set of holomorphic modular forms of weight $k$, which is also holomorphic at infinity, and we write $\mathcal{S}_k$ for the subset of those forms that take value $0$ at infinity.

Also observe that the Fourier coefficients starts at $n \geq 0$ for forms in $\mathcal{M}_k$ and starts at $n \geq 1$ for forms in $\mathcal{S}_k$.

Now that we’ve defined $\mathcal{M}_k$ and $\mathcal{S}_k$, what do you see? This is where things get interesting.

Proposition 3. For $f, h \in \mathcal{M}_k$ and $\lambda \in \mathbb{C}$, we have $f+h \in \mathcal{M}_k$ and $\lambda f \in \mathcal{M}_k$. The result is still true if we replace $\mathcal{M}_k$ by $\mathcal{S}_k$.