This course develops the classical theory of modular forms from the ground up, beginning with the action of the modular group on the upper half-plane and ending with modular forms as explicit, computable arithmetic objects. The central goal is to understand holomorphic functions with strong symmetry and growth conditions, and to see how those conditions force rich algebraic, analytic, and arithmetic structure. Along the way, the course introduces the basic examples, the Fourier or q-expansion viewpoint, and the special role played by Eisenstein series and cusp forms. The chapters are arranged to build a usable toolkit step by step. After the geometric and analytic foundations, the course turns to q-expansions and then uses the valence formula to control zeros, poles, and spaces of forms, leading to dimension formulas and concrete bases. Hecke operators and the Hecke algebra then provide the main arithmetic structure, with eigenforms serving as distinguished simultaneous eigenvectors. The final chapters connect this theory to orthogonality via the Petersson inner product, to [analytic continuation](/page/Analytic%20Continuation) and arithmetic information through L-functions, and to the synthesis of modular forms as objects that can be computed, decomposed, and applied in number theory. # 1. The Modular Group and the Upper Half-Plane This chapter introduces the geometric stage on which classical modular forms live. The main object is the complex upper half-plane, acted on by integer matrices through fractional linear transformations. We first study the action itself, then the special points and cusps that survive in the quotient, and finally the standard fundamental domain that turns the quotient into a concrete geometric object. The central obstruction is that a quotient by an infinite group is hard to see directly, so the chapter builds explicit representatives, boundary identifications, and stabilizer computations. ## Fractional Linear Transformations on the Upper Half-Plane How can an arithmetic group act on a complex domain in a way that preserves enough geometry for complex analysis? The answer is that matrices in $SL_2(\mathbb Z)$ act on the upper half-plane by Möbius transformations. This action is the source of the transformation law for modular forms in the next chapter. [definition: Upper Half-Plane] The complex upper half-plane is \begin{align*} \mathbb H := \{z \in \mathbb C : \operatorname{Im}(z) > 0\}. \end{align*} [/definition] The domain $\mathbb H$ is stable under the transformations we need because the imaginary part transforms by a positive factor. This is not automatic for arbitrary complex Möbius transformations: a real determinant condition is what prevents points from crossing the real axis. The group responsible for the arithmetic part of the theory is the integral determinant-one subgroup. [definition: Modular Group] The modular group is \begin{align*} SL_2(\mathbb Z) := \left\{\begin{pmatrix} a & b \\ c & d \end{pmatrix} : a,b,c,d \in \mathbb Z,\ ad-bc=1\right\}. \end{align*} [/definition] Since $-I$ acts in the same way as $I$ on $\mathbb H$, the effective group of transformations is often \begin{align*} PSL_2(\mathbb Z):=SL_2(\mathbb Z)/\{\pm I\}. \end{align*} We nevertheless keep $SL_2(\mathbb Z)$ in the notation because modular forms remember signs through their weights. Without this distinction, the geometry of the quotient would be correct but the later transformation law for odd weights would lose information. The next definition records the actual map whose kernel contains exactly this central sign ambiguity. [definition: Fractional Linear Action] The fractional linear action is the map \begin{align*} SL_2(\mathbb Z)\times \mathbb H &\longrightarrow \mathbb H,\\ (\gamma,z)&\longmapsto \gamma z, \end{align*} where, for $\gamma=\begin{pmatrix} a & b \\ c & d \end{pmatrix}\in SL_2(\mathbb Z)$, \begin{align*} \gamma z := \frac{az+b}{cz+d}. \end{align*} [/definition] The denominator never vanishes on $\mathbb H$: if $cz+d=0$ with $c,d\in\mathbb R$, then $z=-d/c$ is real when $c\ne0$, and if $c=0$ then $d\ne0$. For this formula to define an action on the upper half-plane, it must do more than avoid division by zero. We also need to know that applying such a matrix keeps the point in $\mathbb H$, and that composition of matrices agrees with composition of the resulting transformations. The next result supplies exactly these two checks, turning the formula into a genuine geometric action rather than just a formal expression. [quotetheorem:4213] [citeproof:4213] The formula also explains why the upper half-plane is natural: these transformations preserve orientation and hyperbolic height up to the factor forced by the denominator. The hypothesis $ad-bc=1$ is essential here; with negative determinant the imaginary part changes sign, and with non-real coefficients the real boundary need not be preserved. This is the first point where arithmetic, complex analysis, and geometry meet in the course: integral matrices give a discrete group, while the imaginary-part formula keeps the action inside a complex domain. [example: Translation and Inversion] Let \begin{align*} T=\begin{pmatrix}1&1\\0&1\end{pmatrix},\qquad S=\begin{pmatrix}0&-1\\1&0\end{pmatrix}. \end{align*} For $T$, the entries are $a=1$, $b=1$, $c=0$, and $d=1$, so the fractional linear action gives \begin{align*} Tz=\frac{1\cdot z+1}{0\cdot z+1} =\frac{z+1}{1} =z+1. \end{align*} Thus $T$ adds $1$ to the real part and leaves the imaginary part unchanged: \begin{align*} T(x+iy)=x+1+iy. \end{align*} So vertical strips are shifted horizontally by one unit. For $S$, the entries are $a=0$, $b=-1$, $c=1$, and $d=0$, hence \begin{align*} Sz=\frac{0\cdot z-1}{1\cdot z+0} =-\frac{1}{z}. \end{align*} If $z=re^{i\theta}$ with $r>0$ and $0<\theta<\pi$, then \begin{align*} Sz=-\frac{1}{re^{i\theta}} =-r^{-1}e^{-i\theta} =r^{-1}e^{i(\pi-\theta)}. \end{align*} Therefore $|Sz|=r^{-1}=1/|z|$, so points with $|z|<1$ are sent to points with $|Sz|>1$, and points with $|z|>1$ are sent to points with $|Sz|<1$. Also the argument changes from $\theta$ to $\pi-\theta$, which is reflection across the imaginary axis. These two transformations preview the horizontal translation and circle-inversion moves that build the quotient geometry. [/example] The preceding example also shows a problem that will reappear for functions: the denominator $cz+d$ is not a harmless by-product of the action. It controls both the change in height and the derivative of the transformation. To state modular transformation laws without rewriting this denominator every time, we isolate it as a separate function. [definition: Automorphy Factor] The automorphy factor is the map \begin{align*} j:SL_2(\mathbb Z)\times \mathbb H &\longrightarrow \mathbb C^*,\\ (\gamma,z)&\longmapsto j(\gamma,z), \end{align*} defined for $\gamma=\begin{pmatrix} a & b \\ c & d \end{pmatrix}\in SL_2(\mathbb Z)$ by \begin{align*} j(\gamma,z):=cz+d. \end{align*} [/definition] The automorphy factor satisfies a cocycle identity. This identity is the algebraic reason the slash operator for modular forms will define a [group action](/page/Group%20Action). Before defining modular forms, we need to know that the factor attached to a product of matrices can be computed from the factors attached to the two matrices separately. The next result supplies that compatibility, so that transformation laws can be imposed consistently on an entire group rather than checked as unrelated formulas. [quotetheorem:4214] [citeproof:4214] The cocycle identity is the compatibility condition that prevents transformation laws from depending on how a matrix is factored into simpler matrices. If $j$ did not satisfy this identity, applying two modular transformations successively would give a different factor from applying their product at once. This is why the denominator belongs in the theory as a structured automorphy factor rather than as an incidental term in a fraction. ## Stabilizers, Elliptic Points, Cusps, and the Quotient Which points of $\mathbb H$ have nontrivial symmetry under the modular group, and what happens at the boundary? Most points have no stabilizer beyond $\pm I$, but two exceptional orbits contain extra rotational symmetry. The boundary point at infinity is not in $\mathbb H$, yet it is essential because modular forms are tested there through Fourier expansions. [definition: Stabilizer] Let a group $G$ act on a set $X$. The stabilizer of $x\in X$ is \begin{align*} G_x:=\{g\in G:g\cdot x=x\}. \end{align*} [/definition] For the modular action, stabilizers are solved by a quadratic equation. If $\gamma z=z$, then \begin{align*} cz^2+(d-a)z-b=0. \end{align*} Thus fixed points in $\mathbb H$ occur only for elliptic transformations, and the standard representatives are $i$ and $\rho$. To record these exceptional quotient points without repeatedly solving stabilizer equations, we introduce a name for points whose effective modular symmetry is nontrivial. This terminology separates ordinary points of the quotient from the orbifold points that require special attention. [definition: Elliptic Points] A point $z\in\mathbb H$ is an elliptic point for $SL_2(\mathbb Z)$ if its stabilizer in $PSL_2(\mathbb Z)$ is nontrivial. [/definition] The two basic elliptic points are \begin{align*} i,\qquad \rho:=e^{2\pi i/3}=-\frac12+\frac{\sqrt3}{2}i. \end{align*} With this convention, $\rho$ lies on the left boundary of the standard fundamental domain. A definition by stabilizers leaves open whether many other interior points might also have nontrivial symmetry. For the quotient geometry, this matters because every such point would become an additional orbifold point that must be recorded separately. Before using the quotient as a geometric object, we need a classification that rules out hidden exceptional orbits and records the effective stabilizer orders after the central signs have been removed. This finite list is the input that later lets the compactified quotient be described without missing local symmetry cases. [quotetheorem:4215] [citeproof:4215] This theorem says that the quotient is not an ordinary surface at the images of $i$ and $\rho$; it has orbifold points of orders $2$ and $3$. The determinant-one hypothesis and the passage to $PSL_2(\mathbb Z)$ both matter: in $SL_2(\mathbb Z)$ the central signs remain, while in the effective geometric action they disappear. These exceptional stabilizers are the only interior symmetries that must be remembered when turning the orbit space into a geometric object. [example: Computing the Stabilizer of Infinity] Let \begin{align*} \gamma=\begin{pmatrix}a&b\\c&d\end{pmatrix}\in SL_2(\mathbb Z). \end{align*} Extend the fractional linear action to $\mathbb P^1(\mathbb R)=\mathbb R\cup\{\infty\}$ by \begin{align*} \gamma\infty= \begin{cases} a/c,&c\ne0,\\ \infty,&c=0. \end{cases} \end{align*} Thus $\gamma$ fixes $\infty$ exactly when $c=0$. Under this condition, \begin{align*} \gamma=\begin{pmatrix}a&b\\0&d\end{pmatrix}, \end{align*} and the determinant condition $ad-bc=1$ becomes \begin{align*} ad-b\cdot 0=1, \end{align*} so $ad=1$. Since $a,d\in\mathbb Z$, this forces \begin{align*} (a,d)=(1,1)\qquad\text{or}\qquad(a,d)=(-1,-1). \end{align*} If $a=d=1$, then \begin{align*} \gamma=\begin{pmatrix}1&b\\0&1\end{pmatrix}=T^b. \end{align*} If $a=d=-1$, then \begin{align*} \gamma=\begin{pmatrix}-1&b\\0&-1\end{pmatrix} =-\begin{pmatrix}1&-b\\0&1\end{pmatrix} =-T^{-b}. \end{align*} Therefore \begin{align*} (SL_2(\mathbb Z))_\infty =\{T^b:b\in\mathbb Z\}\cup\{-T^{-b}:b\in\mathbb Z\} =\{\pm T^n:n\in\mathbb Z\}. \end{align*} Passing to $PSL_2(\mathbb Z)$ identifies matrices that differ by the central sign $-I$, so the stabilizer of $\infty$ becomes the infinite cyclic group generated by the image of $T$. [/example] This computation is the boundary analogue of the elliptic stabilizer calculations. Interior stabilizers are finite rotations, whereas the stabilizer of $\infty$ is infinite cyclic and generated by translation. That difference is why the boundary point must be treated as a cusp rather than as another ordinary point of the quotient. The quotient is still incomplete if it only remembers points of $\mathbb H$: modular transformations also act on rational boundary directions, and those directions are where Fourier expansions will be measured. The next definition packages boundary directions by orbit, so that the compactification can add one point for each modularly distinct way to approach infinity. [definition: Cusp] A cusp for $SL_2(\mathbb Z)$ is an orbit of $\mathbb P^1(\mathbb Q)=\mathbb Q\cup\{\infty\}$ under the fractional linear action of $SL_2(\mathbb Z)$. [/definition] For the full modular group there is only one cusp, represented by $\infty$. Indeed, if $r=a/c\in\mathbb Q$ with $\gcd(a,c)=1$, choose $b,d\in\mathbb Z$ with $ad-bc=1$; then $\begin{pmatrix}a&b\\c&d\end{pmatrix}\in SL_2(\mathbb Z)$ sends $\infty$ to $a/c$. The point of adding cusps is that the quotient of $\mathbb H$ alone is not compact and misses the limiting directions in which Fourier expansions are taken. For smaller congruence subgroups this rational boundary may split into several inequivalent cusps, so the transitivity result below is a special feature of the full modular group. [quotetheorem:4216] [citeproof:4216] This theorem is special to $SL_2(\mathbb Z)$ and should not be read as a general statement about all modular curves. For congruence subgroups, the same boundary set $\mathbb P^1(\mathbb Q)$ usually decomposes into several orbits, and each orbit gives a different cusp. Here there is only one boundary orbit, so the compactified quotient will need only one added cusp point. [definition: Modular Quotient] The modular quotient is the orbit space \begin{align*} SL_2(\mathbb Z)\backslash\mathbb H:=\{SL_2(\mathbb Z)\cdot z:z\in\mathbb H\}. \end{align*} [/definition] The quotient becomes easier to understand after adding the cusp. Topologically, the compactified quotient has one cusp point and two elliptic orbifold points; analytically, the cusp is where Fourier expansions in the local parameter $q=e^{2\pi i z}$ will live. This connects the geometry of the quotient to number theory, because the coefficients of these expansions will become arithmetic data attached to modular forms. ## The Standard Fundamental Domain Can each orbit be represented by a point in a small, explicit region of the upper half-plane? Without such representatives, statements about functions on the quotient remain abstract because the same point has infinitely many images in $\mathbb H$. The standard fundamental domain answers this by combining horizontal translation with inversion in the unit circle. It is the geometric normal form behind many computations with modular forms. [definition: Standard Fundamental Domain] The standard fundamental domain for $SL_2(\mathbb Z)$ is \begin{align*} \mathcal F:=\left\{z\in\mathbb H: |z|\ge1,\ -\frac12\le \operatorname{Re}(z)\le \frac12\right\}. \end{align*} [/definition] The vertical sides are identified by $T:z\mapsto z+1$. The circular boundary arc $|z|=1$ is identified by $S:z\mapsto -1/z$, with the two special boundary points corresponding to the elliptic points. [illustration:standard-fundamental-domain] The picture suggests that every orbit should meet this region, but the useful statement must also say what kind of uniqueness remains after boundary identifications and stabilizers are taken into account. The following result turns the drawing into a precise reduction theorem for modular orbits. [quotetheorem:4217] [citeproof:4217] The theorem does not say that $\mathcal F$ is a fundamental domain in the naive sense of one point per orbit. Boundary points must be identified, and the elliptic points have extra stabilizers. The existence part depends on discreteness of the lattice $\mathbb Zz+\mathbb Z$, while the uniqueness part depends on being in the interior; both hypotheses fail at the boundary. This distinction is why later integrals over a fundamental domain must account for side pairings and elliptic points rather than treating $\mathcal F$ as an ordinary rectangle. [example: Reducing a Point to the Standard Fundamental Domain] Let $z=\frac35+2i$. Since $T=\begin{pmatrix}1&1\\0&1\end{pmatrix}$ acts by $Tz=z+1$, its inverse is \begin{align*} T^{-1}=\begin{pmatrix}1&-1\\0&1\end{pmatrix}, \end{align*} and the fractional linear action gives \begin{align*} T^{-1}z &=\frac{1\cdot z+(-1)}{0\cdot z+1}\\ &=z-1\\ &=\left(\frac35+2i\right)-1\\ &=-\frac25+2i. \end{align*} The real part now satisfies \begin{align*} -\frac12\le -\frac25\le \frac12, \end{align*} because $-\frac12=-\frac5{10}$ and $-\frac25=-\frac4{10}$. Its modulus squared is \begin{align*} \left|-\frac25+2i\right|^2 &=\left(-\frac25\right)^2+2^2\\ &=\frac4{25}+4\\ &=\frac4{25}+\frac{100}{25}\\ &=\frac{104}{25}>1. \end{align*} Therefore $-\frac25+2i$ satisfies both defining inequalities for $\mathcal F$: \begin{align*} -\frac12\le \operatorname{Re}\left(-\frac25+2i\right)\le \frac12, \qquad \left|-\frac25+2i\right|\ge 1. \end{align*} Thus $z=\frac35+2i$ is represented in the standard fundamental domain by $-\frac25+2i$, obtained by the reducing matrix \begin{align*} T^{-1}=\begin{pmatrix}1&-1\\0&1\end{pmatrix}. \end{align*} [/example] The reduction example illustrates the interior part of the theorem, where no side identifications are encountered. The next example records what changes when a representative lands on the boundary. These boundary cases are exactly where the [quotient topology](/page/Quotient%20Topology) is formed by gluing rather than by choosing a unique point. [example: Boundary Identifications] Let $y>0$. On the right vertical boundary, $z=\frac12+iy$, and \begin{align*} T^{-1}z &=\frac{1\cdot z+(-1)}{0\cdot z+1}\\ &=z-1\\ &=\left(\frac12+iy\right)-1\\ &=-\frac12+iy. \end{align*} Thus $T^{-1}$ identifies the right side $\operatorname{Re}(z)=\frac12$ with the left side $\operatorname{Re}(z)=-\frac12$ at the same height. Now let $z$ lie on the circular boundary arc, so $|z|=1$ and $\operatorname{Im}(z)>0$. Since \begin{align*} |z|^2=z\overline z=1, \end{align*} we have \begin{align*} \frac1z=\overline z. \end{align*} Therefore \begin{align*} Sz &=\frac{0\cdot z-1}{1\cdot z+0}\\ &=-\frac1z\\ &=-\overline z. \end{align*} Writing $z=x+iy$, this becomes \begin{align*} Sz=-(x-iy)=-x+iy, \end{align*} so $S$ reflects the circular arc across the imaginary axis. At the endpoint $i$, \begin{align*} Si=-\frac1i=\frac{-1}{i}\cdot\frac{i}{i}=\frac{-i}{-1}=i, \end{align*} so $i$ is fixed by $S$. For $\rho=-\frac12+\frac{\sqrt3}{2}i$, we have $\rho^2+\rho+1=0$, hence \begin{align*} \rho(\rho+1)=\rho^2+\rho=-1. \end{align*} Since \begin{align*} ST =\begin{pmatrix}0&-1\\1&1\end{pmatrix}, \end{align*} its action is \begin{align*} (ST)z=\frac{0\cdot z-1}{1\cdot z+1}=-\frac1{z+1}. \end{align*} Substituting $z=\rho$ gives \begin{align*} (ST)\rho &=-\frac1{\rho+1}\\ &=\rho, \end{align*} because $\rho(\rho+1)=-1$. Thus the boundary gluing fixes $i$ under $S$ and fixes $\rho$ under $ST$. [/example] ## Generators and Relations Why are the two geometric moves $S$ and $T$ enough to recover every matrix in $SL_2(\mathbb Z)$? The fundamental domain proof gives a reduction algorithm: repeatedly move points back into the strip and outside the unit circle. Translating this geometric reduction into matrices yields generators and the basic relations among them. [quotetheorem:4218] [citeproof:4218] This presentation is the bridge between geometry and algebra. Later, when checking modularity, it is often enough to verify the transformation law under $S$ and $T$ rather than under every matrix in $SL_2(\mathbb Z)$. [example: Checking the Relations] We verify the two relations by multiplying the matrices \begin{align*} S=\begin{pmatrix}0&-1\\1&0\end{pmatrix}, \qquad T=\begin{pmatrix}1&1\\0&1\end{pmatrix}. \end{align*} First, \begin{align*} S^2 &=\begin{pmatrix}0&-1\\1&0\end{pmatrix} \begin{pmatrix}0&-1\\1&0\end{pmatrix}\\ &=\begin{pmatrix} 0\cdot 0+(-1)\cdot 1 & 0\cdot(-1)+(-1)\cdot 0\\ 1\cdot 0+0\cdot 1 & 1\cdot(-1)+0\cdot 0 \end{pmatrix}\\ &=\begin{pmatrix}-1&0\\0&-1\end{pmatrix}\\ &=-I. \end{align*} Next, \begin{align*} ST &=\begin{pmatrix}0&-1\\1&0\end{pmatrix} \begin{pmatrix}1&1\\0&1\end{pmatrix}\\ &=\begin{pmatrix} 0\cdot 1+(-1)\cdot 0 & 0\cdot 1+(-1)\cdot 1\\ 1\cdot 1+0\cdot 0 & 1\cdot 1+0\cdot 1 \end{pmatrix}\\ &=\begin{pmatrix}0&-1\\1&1\end{pmatrix}. \end{align*} Squaring this matrix gives \begin{align*} (ST)^2 &=\begin{pmatrix}0&-1\\1&1\end{pmatrix} \begin{pmatrix}0&-1\\1&1\end{pmatrix}\\ &=\begin{pmatrix} 0\cdot 0+(-1)\cdot 1 & 0\cdot(-1)+(-1)\cdot 1\\ 1\cdot 0+1\cdot 1 & 1\cdot(-1)+1\cdot 1 \end{pmatrix}\\ &=\begin{pmatrix}-1&-1\\1&0\end{pmatrix}. \end{align*} Multiplying once more, \begin{align*} (ST)^3 &=(ST)^2(ST)\\ &=\begin{pmatrix}-1&-1\\1&0\end{pmatrix} \begin{pmatrix}0&-1\\1&1\end{pmatrix}\\ &=\begin{pmatrix} (-1)\cdot 0+(-1)\cdot 1 & (-1)\cdot(-1)+(-1)\cdot 1\\ 1\cdot 0+0\cdot 1 & 1\cdot(-1)+0\cdot 1 \end{pmatrix}\\ &=\begin{pmatrix}-1&0\\0&-1\end{pmatrix}\\ &=-I. \end{align*} In $PSL_2(\mathbb Z)$, the matrices $I$ and $-I$ define the same element, so these equalities become $S^2=I$ and $(ST)^3=I$. Thus the image of $S$ has order $2$, while the image of $ST$ has order $3$; geometrically, these are the rotational symmetries at the elliptic points $i$ and $\rho$ in the quotient. [/example] The first chapter leaves us with a concrete picture: the modular group folds the upper half-plane into a domain bounded by two vertical lines and a circular arc, with one cusp and two elliptic points. The next chapter puts holomorphic functions on this geometry. Modular forms will be functions on $\mathbb H$ whose transformation rule is controlled by the same automorphy factor $j(\gamma,z)=cz+d$ introduced here. At this point the abstract [symmetry condition](/theorems/1360) becomes concrete through $q$-expansions at infinity. The next chapter turns that expansion principle into the formal definition of modular forms and cusp forms, including the cusp condition and the first structural restrictions on possible weights. # 2. Modular Forms and q-Expansions Modular forms enter the course after the geometry of the upper half-plane because the definition mixes two different requirements: symmetry under the modular group and holomorphic behaviour at the missing boundary point called the cusp. The first chapter described how $SL_2(\mathbb Z)$ moves points of $\mathbb H$ and how the quotient is controlled by the generators $S:z\mapsto -1/z$ and $T:z\mapsto z+1$. This chapter turns that geometry into function theory: we define the transformation law, explain why periodicity produces [Fourier series](/page/Fourier%20Series) in $q=e^{2\pi iz}$, and isolate cusp forms as the functions whose first Fourier coefficient vanishes. ## The Transformation Law and the Slash Operator What kind of [holomorphic function](/page/Holomorphic%20Function) on $\mathbb H$ should count as a function on the quotient $SL_2(\mathbb Z)\backslash \mathbb H$? For weight $0$, invariance under $\gamma z$ is enough. In higher weight, the correct rule includes the automorphy factor $cz+d$, which records how fractional linear transformations distort tangent vectors and differential forms. Recall from Chapter 1 that the upper half-plane is $\mathbb H=\{z\in\mathbb C:\operatorname{Im}(z)>0\}$ and that $SL_2(\mathbb Z)$ acts on it by fractional linear transformations. If \begin{align*} \gamma=\begin{pmatrix}a&b\\ c&d\end{pmatrix}\in SL_2(\mathbb Z), \end{align*} then \begin{align*} \gamma z = \frac{az+b}{cz+d}, \end{align*} and the automorphy factor introduced there is $j(\gamma,z)=cz+d$. [definition: Slash Operator] Let $k\in \mathbb Z$ and let \begin{align*} \gamma=\begin{pmatrix}a&b\\ c&d\end{pmatrix}\in SL_2(\mathbb Z). \end{align*} The weight $k$ slash operator associated to $\gamma$ is the map \begin{align*} |_k\gamma:\operatorname{Map}(\mathbb H,\mathbb C)&\to \operatorname{Map}(\mathbb H,\mathbb C),\\ f&\mapsto f|_k\gamma, \end{align*} defined by \begin{align*} (f|_k\gamma)(z):=(cz+d)^{-k}f(\gamma z). \end{align*} [/definition] The condition $f|_k\gamma=f$ is the modular transformation law in operator form. Written without the slash operator, it is \begin{align*} f\left(\frac{az+b}{cz+d}\right)=(cz+d)^k f(z). \end{align*} This formulation is convenient because the slash operator is a right action: applying $\gamma$ and then $\delta$ corresponds to applying $\gamma\delta$. The slash notation would be misleading if it did not respect matrix multiplication. Invariance under one matrix and then another should imply invariance under their product, but that requires the automorphy factors from the two substitutions to combine exactly. The point to check is therefore not just notation: it is the compatibility that lets modular symmetry be treated as a group action. [quotetheorem:4219] [citeproof:4219] The right-action identity is what makes the compact generator test legitimate. Without it, invariance under two matrices would not automatically propagate to invariance under their products, because the automorphy factors could fail to multiply compatibly. The theorem uses the determinant-one fractional linear action and the specific factor $(cz+d)^{-k}$; changing either convention changes the order of multiplication or the power of the factor. It also says only that the slash operators compose correctly, not that a particular function is holomorphic or well behaved at the cusp. Since $SL_2(\mathbb Z)$ is generated by a small number of matrices, one would like to test modularity using only those generators. The obstruction is that the condition $f|_k\gamma=f$ must be stable under every word in the generators and their inverses. The following criterion packages the right-action identity into a finite invariance test. [quotetheorem:4220] [citeproof:4220] This criterion is algebraic: it reduces the number of transformation equations, but it does not reduce the analytic conditions in the definition below. Holomorphy on $\mathbb H$ and holomorphy at the cusp still have to be checked separately. Its usefulness is practical: in examples one verifies the $T$- and $S$-relations and then appeals to generation, instead of checking infinitely many matrices one by one. The result should therefore be read as a symmetry test, not as a replacement for the growth and regularity requirements that distinguish modular forms from merely weakly modular functions. [definition: Weakly Modular Function] Let $k\in\mathbb Z$. A weakly modular function of weight $k$ for $SL_2(\mathbb Z)$ is a [meromorphic function](/page/Meromorphic%20Function) $f:\mathbb H\to\mathbb C$ such that \begin{align*} f|_k\gamma=f \end{align*} for every $\gamma\in SL_2(\mathbb Z)$. [/definition] Weak modularity controls behaviour inside $\mathbb H$, but it says nothing yet about the cusp. Modular forms impose holomorphy both in the upper half-plane and at the cusp. This extra analytic condition is what rules out functions that satisfy the symmetry law but have uncontrolled growth or singular behavior at the missing boundary point. The definition below names the class that has both the modular transformation law and the regularity needed for Fourier expansion at infinity. [definition: Modular Form for $SL_2(\mathbb Z)$] Let $k\in\mathbb Z$. A modular form of weight $k$ for $SL_2(\mathbb Z)$ is a holomorphic function $f:\mathbb H\to\mathbb C$ such that \begin{align*} f|_k\gamma=f \end{align*} for every $\gamma\in SL_2(\mathbb Z)$, and $f$ is holomorphic at the cusp $\infty$. [/definition] Among modular forms, the ones that genuinely disappear at the added boundary point form the most important subspace. This vanishing condition is local at the cusp, and the next section translates it into a statement about the first coefficient of a [power series](/page/Power%20Series). [definition: Cusp Form for $SL_2(\mathbb Z)$] Let $k\in\mathbb Z$. A cusp form of weight $k$ for $SL_2(\mathbb Z)$ is a modular form $f$ of weight $k$ whose value at the cusp $\infty$ is $0$. [/definition] We write $M_k(SL_2(\mathbb Z))$, or simply $M_k$, for the complex [vector space](/page/Vector%20Space) of modular forms of weight $k$, and $S_k(SL_2(\mathbb Z))$, or simply $S_k$, for the subspace of cusp forms. The notation suppresses the group only in this course, where the full modular group is fixed. [example: Translation Invariance] Let $f:\mathbb H\to\mathbb C$ be a function and let \begin{align*} T=\begin{pmatrix}1&1\\0&1\end{pmatrix}. \end{align*} In the slash operator formula, the entries are $a=1$, $b=1$, $c=0$, and $d=1$, so \begin{align*} Tz&=\frac{1\cdot z+1}{0\cdot z+1}\\ &=\frac{z+1}{1}\\ &=z+1, \end{align*} and the automorphy factor is \begin{align*} cz+d&=0\cdot z+1\\ &=1. \end{align*} Therefore \begin{align*} (f|_kT)(z)&=(0\cdot z+1)^{-k}f(Tz)\\ &=1^{-k}f(z+1)\\ &=f(z+1). \end{align*} Thus the equation $f|_kT=f$ is exactly \begin{align*} f(z+1)=f(z) \end{align*} for every $z\in\mathbb H$, which means that $f$ is periodic of period $1$. This is the first bridge from modularity to Fourier series, because period-$1$ holomorphic functions can be studied through the coordinate $q=e^{2\pi iz}$. [/example] The second generator is not a translation, so it imposes a genuinely global symmetry on the upper half-plane. Comparing the two examples shows why modular forms are much more rigid than ordinary periodic holomorphic functions. [example: The Inversion Transformation] Let \begin{align*} S=\begin{pmatrix}0&-1\\1&0\end{pmatrix}. \end{align*} In the slash operator formula, the entries are $a=0$, $b=-1$, $c=1$, and $d=0$. Hence \begin{align*} Sz&=\frac{0\cdot z+(-1)}{1\cdot z+0}\\ &=\frac{-1}{z}\\ &=-\frac{1}{z}, \end{align*} and the automorphy factor is \begin{align*} cz+d&=1\cdot z+0\\ &=z. \end{align*} Therefore \begin{align*} (f|_kS)(z)&=(1\cdot z+0)^{-k}f(Sz)\\ &=z^{-k}f\left(-\frac{1}{z}\right). \end{align*} Thus the equation $f|_kS=f$ means \begin{align*} z^{-k}f\left(-\frac{1}{z}\right)&=f(z). \end{align*} Multiplying both sides by $z^k$ gives \begin{align*} f\left(-\frac{1}{z}\right)&=z^k f(z). \end{align*} This condition relates the value of $f$ at $z$ to its value at the inverted point $-1/z$, so it is a genuinely non-translation symmetry and is much more restrictive than ordinary period-$1$ invariance. [/example] ## Holomorphy at the Cusp and q-Expansions How can a function be holomorphic at a point that is not in $\mathbb H$? The cusp $\infty$ is approached by taking $\operatorname{Im}(z)\to\infty$, and translation invariance means that the coordinate measuring this approach should identify $z$ and $z+1$. The natural coordinate is \begin{align*} q=e^{2\pi iz}, \end{align*} because it maps the vertical strip modulo translation to the punctured unit disc. [definition: q-Coordinate] The $q$-coordinate is the map \begin{align*} \mathbb H/\langle z\mapsto z+1\rangle&\to \{q\in\mathbb C:0<|q|<1\},\\ [z]&\mapsto e^{2\pi iz}. \end{align*} [/definition] If $z=x+iy$, then $|q|=e^{-2\pi y}$. Thus $y\to\infty$ corresponds to $q\to0$. A periodic holomorphic function on $\mathbb H$ descends to a holomorphic function on the punctured disc $0<|q|<1$. The remaining question is whether the puncture at $q=0$ is removable for the descended function. This is the analytic meaning of being regular at the added cusp, and it is the condition that will distinguish modular forms from weakly modular functions with poles at infinity. [definition: Holomorphic at the Cusp] Let $f:\mathbb H\to\mathbb C$ be holomorphic and satisfy $f(z+1)=f(z)$. Define $F$ on $0<|q|<1$ by \begin{align*} F(e^{2\pi iz})=f(z). \end{align*} The function $f$ is holomorphic at the cusp $\infty$ if $F$ extends holomorphically to $q=0$. [/definition] After this definition, the value $f(\infty)$ means $F(0)$. The cusp form condition is $F(0)=0$. The definition is geometric, but computations need an explicit analytic test. Once the cusp has been changed into the point $q=0$, holomorphy there should be equivalent to having an ordinary power series with no negative powers. The theorem makes this equivalence precise and explains why Fourier coefficients are the natural coordinates for modular forms. [quotetheorem:4221] [citeproof:4221] The hypotheses in this criterion are both necessary. Periodicity is what makes $F(q)=f(z)$ independent of the chosen logarithm of $q$, while holomorphy at the cusp is what rules out negative powers such as $q^{-1}$. A non-periodic holomorphic function on $\mathbb H$ may have no single-valued expression in the $q$-coordinate, and a periodic holomorphic function may still have a pole or essential singularity at $q=0$. The criterion is local at the cusp: it does not by itself prove the $S$-transformation law or any other modular symmetry. In computations, however, it is the bridge from the geometric definition to arithmetic data, because the sequence $(a_n)_{n\ge0}$ is what later encodes divisor sums, congruences, and cusp-form information. The coefficients $a_n$ are called the Fourier coefficients or $q$-expansion coefficients of $f$. The notation \begin{align*} f(q)=\sum_{n=0}^{\infty}a_nq^n \end{align*} is common, but it abbreviates the function $z\mapsto f(z)$ after the change of variable $q=e^{2\pi iz}$. Once a modular form is written in the cusp coordinate $q$, the individual coefficients are only meaningful as part of the whole local power series. To compare forms, identify cusp forms, or compute arithmetic data, we need a name for this complete series rather than repeatedly referring to a holomorphic extension in the variable $q$. [definition: q-Expansion] Let $f\in M_k$. The $q$-expansion of $f$ at $\infty$ is the power series \begin{align*} f(z)=\sum_{n=0}^{\infty}a_nq^n, \qquad q=e^{2\pi iz}, \end{align*} obtained from the holomorphic extension to $q=0$. [/definition] The modular form condition can now be read in two layers. The equation for $T$ gives the possibility of a Fourier expansion, and holomorphy at $\infty$ removes negative powers of $q$. Cusp forms are exactly those modular forms whose expansion begins with $q^1$ or higher. [quotetheorem:4222] [citeproof:4222] The assumption $f\in M_k$ matters: a function can have zero constant term in a local $q$-series and still fail to be modular for the full group. Thus this theorem is not a test for modularity; it is a test for cuspidality after modularity and holomorphy at the cusp have already been established. In later computations, this is how Eisenstein series and cusp forms are separated: once the transformation law is known, the constant term of the $q$-expansion decides whether the form lies in $S_k$. [example: A Periodic Holomorphic Function That Is Not Holomorphic at the Cusp] Let $q=e^{2\pi iz}$ and define \begin{align*} f(z)=q^{-1}=e^{-2\pi iz}. \end{align*} The exponential function is holomorphic on $\mathbb C$, so $f$ is holomorphic on $\mathbb H$. For $z\in\mathbb H$, \begin{align*} f(z+1)&=e^{-2\pi i(z+1)}\\ &=e^{-2\pi iz-2\pi i}\\ &=e^{-2\pi iz}e^{-2\pi i}\\ &=e^{-2\pi iz}\cdot 1\\ &=f(z), \end{align*} so $f$ is $1$-periodic. In the $q$-coordinate, the descended function is \begin{align*} F(q)=q^{-1} \end{align*} on $0<|q|<1$. It cannot extend holomorphically to $q=0$, because \begin{align*} qF(q)&=q\cdot q^{-1}\\ &=1 \end{align*} for every $q\ne0$, so $F$ has a pole of order $1$ at $0$. Thus periodicity gives a well-defined function in the $q$-coordinate, but it does not by itself guarantee holomorphy at the cusp. [/example] The pole example shows that periodicity alone is too weak. A holomorphic function can also extend across the cusp and vanish there, but that still does not guarantee full modularity. [example: A Function Vanishing at the Cusp] Let $q=e^{2\pi iz}$ and define $f(z)=q=e^{2\pi iz}$. Since $z\mapsto 2\pi iz$ is holomorphic on $\mathbb H$ and the exponential function is holomorphic on $\mathbb C$, the composition $f$ is holomorphic on $\mathbb H$. For $z\in\mathbb H$, \begin{align*} f(z+1)&=e^{2\pi i(z+1)}\\ &=e^{2\pi iz+2\pi i}\\ &=e^{2\pi iz}e^{2\pi i}\\ &=e^{2\pi iz}\cdot 1\\ &=f(z), \end{align*} so $f$ is $1$-periodic. In the $q$-coordinate, the descended function is \begin{align*} F(q)=q. \end{align*} This extends holomorphically to $q=0$ by the same formula, and \begin{align*} F(0)=0. \end{align*} Equivalently, its $q$-expansion is \begin{align*} f(z)=q=0+1\cdot q, \end{align*} so the constant term is $0$ and the function vanishes at the cusp. This local cusp condition does not make $f$ a modular form for the full group. For \begin{align*} S=\begin{pmatrix}0&-1\\1&0\end{pmatrix}, \end{align*} the weight $k$ slash operator gives \begin{align*} (f|_kS)(z)&=z^{-k}f\left(-\frac{1}{z}\right)\\ &=z^{-k}e^{2\pi i(-1/z)}\\ &=z^{-k}e^{-2\pi i/z}. \end{align*} For instance, at $z=i$ this becomes \begin{align*} (f|_kS)(i)&=i^{-k}e^{-2\pi i/i}\\ &=i^{-k}e^{-2\pi}\\ &\ne e^{-2\pi}\\ &=f(i) \end{align*} unless $i^{-k}=1$, and even when that happens at this one point, the equality is not an identity in $z$. Thus $f(z)=q$ illustrates vanishing at the cusp, not full modularity. [/example] ## First Examples and Nonexamples Which modular forms appear before we construct Eisenstein series? The simplest tests already give useful information: constants give weight $0$ forms, the matrix $-I$ rules out odd weights, and naive periodic holomorphic functions usually fail the inversion symmetry. [quotetheorem:4223] [citeproof:4223] The holomorphy condition at the cusp is essential here. Without adding the cusp and requiring holomorphic extension there, the quotient is non-compact, and maximum-modulus arguments no longer force bounded holomorphic functions to be constant in the same way. The theorem also applies to holomorphic modular forms, not to meromorphic modular functions: the modular $j$-function is the standard later example of a nonconstant weight $0$ meromorphic modular function, with a pole at the cusp. Thus compactification is doing real analytic work, and it foreshadows the finite-dimensionality theorem for $M_k$ in higher weight. [example: Constants] Let $c\in\mathbb C$ and define $f:\mathbb H\to\mathbb C$ by $f(z)=c$ for every $z\in\mathbb H$. For weight $0$ and \begin{align*} \gamma=\begin{pmatrix}a&b\\ c_\gamma&d\end{pmatrix}\in SL_2(\mathbb Z), \end{align*} the slash operator gives \begin{align*} (f|_0\gamma)(z) &=(c_\gamma z+d)^0 f(\gamma z)\\ &=1\cdot f(\gamma z)\\ &=f(\gamma z). \end{align*} Since $f$ is constant, \begin{align*} f(\gamma z)&=c\\ &=f(z), \end{align*} so $f|_0\gamma=f$ for every $\gamma\in SL_2(\mathbb Z)$. The function is holomorphic on $\mathbb H$ because it is constant. It is also $1$-periodic, since \begin{align*} f(z+1)&=c\\ &=f(z). \end{align*} In the $q$-coordinate, the descended function is \begin{align*} F(q)=c, \end{align*} which extends holomorphically to $q=0$ by the same formula. Its power series is \begin{align*} F(q)=c=c+0\cdot q+0\cdot q^2+\cdots, \end{align*} so the constant term is $c$ and \begin{align*} F(0)=c. \end{align*} Thus every constant function is a weight $0$ modular form, and it is a cusp form exactly when \begin{align*} c=0. \end{align*} [/example] Constants therefore completely explain weight $0$, but they do not survive unchanged in higher weight. The next example isolates the obstruction caused by the automorphy factor. [example: Constants in Nonzero Weight] Let $f(z)=c_0$ with $c_0\ne0$, and suppose $k\ne0$. For \begin{align*} S=\begin{pmatrix}0&-1\\1&0\end{pmatrix}, \end{align*} the slash operator gives \begin{align*} (f|_kS)(z)&=z^{-k}f\left(-\frac{1}{z}\right)\\ &=z^{-k}c_0. \end{align*} If $f$ were modular of weight $k$, then $f|_kS=f$, so for every $z\in\mathbb H$, \begin{align*} z^{-k}c_0&=c_0. \end{align*} Multiplying both sides by $z^k$ gives \begin{align*} c_0&=z^k c_0, \end{align*} and since $c_0\ne0$, division by $c_0$ gives \begin{align*} 1&=z^k \end{align*} for every $z\in\mathbb H$. Taking $z=i$ and $z=2i$, both of which lie in $\mathbb H$, would give \begin{align*} i^k&=1,\\ (2i)^k&=1. \end{align*} Dividing the second equality by the first gives \begin{align*} \frac{(2i)^k}{i^k}&=\frac{1}{1},\\ 2^k&=1. \end{align*} But $2^k\ne1$ for every nonzero integer $k$. Thus a nonzero constant function cannot satisfy the weight $k$ transformation law when $k\ne0$; nonzero constants are modular only in weight $0$. [/example] There is an even stronger obstruction that affects every function, not just constants. It comes from the central matrix $-I$, whose action on points is invisible but whose automorphy factor still remembers the weight. [quotetheorem:4224] [citeproof:4224] The full modular group contains $-I$, so odd weights are killed before any analytic argument is needed. The hypothesis that the group is exactly $SL_2(\mathbb Z)$, or at least contains $-I$ acting with the usual integral-weight factor, is therefore essential. For example, on a subgroup not containing $-I$, the same contradiction is unavailable, and with multiplier systems the factor attached to $-I$ can be altered. The theorem also says nothing about even weights: evenness is necessary for nonzero full-level forms, but existence still requires constructions such as Eisenstein series or cusp forms. This distinction will matter later when the dimension formula separates weights with no forms, weights with Eisenstein series, and weights with genuine cusp forms. [example: A Nonexample from Odd Weight] Suppose $f:\mathbb H\to\mathbb C$ is holomorphic and satisfies the weight $1$ transformation law for every element of $SL_2(\mathbb Z)$. Consider \begin{align*} -I=\begin{pmatrix}-1&0\\0&-1\end{pmatrix}. \end{align*} Its entries are $a=-1$, $b=0$, $c=0$, and $d=-1$, so \begin{align*} (-I)z&=\frac{-z+0}{0\cdot z-1}\\ &=\frac{-z}{-1}\\ &=z, \end{align*} and its automorphy factor is \begin{align*} cz+d&=0\cdot z-1\\ &=-1. \end{align*} Therefore the weight $1$ slash operator gives \begin{align*} (f|_1(-I))(z)&=(0\cdot z-1)^{-1}f((-I)z)\\ &=(-1)^{-1}f(z)\\ &=-f(z). \end{align*} Since $f$ satisfies the weight $1$ transformation law, we also have \begin{align*} (f|_1(-I))(z)&=f(z). \end{align*} Combining the two equalities gives \begin{align*} f(z)&=-f(z),\\ 2f(z)&=0,\\ f(z)&=0 \end{align*} for every $z\in\mathbb H$. Thus any weight $1$ full-level modular form must be the zero function, so there are no nonzero weight $1$ modular forms for $SL_2(\mathbb Z)$. [/example] Eta-products are a useful motivation for later examples, even though the Dedekind eta function itself is not a modular form for $SL_2(\mathbb Z)$ with the definition used in this chapter. Its transformation law contains a multiplier, so it points toward modular forms with character and half-integral weight. [example: Eta Function as Motivation] The Dedekind eta function is formally introduced by \begin{align*} \eta(z)=q^{1/24}\prod_{n=1}^{\infty}(1-q^n), \qquad q=e^{2\pi iz}, \end{align*} where the fractional power is read on $\mathbb H$ as \begin{align*} q^{1/24}=e^{2\pi iz/24}=e^{\pi iz/12}. \end{align*} Under the translation $z\mapsto z+1$, the $q$-coordinate satisfies \begin{align*} q(z+1)&=e^{2\pi i(z+1)}\\ &=e^{2\pi iz+2\pi i}\\ &=e^{2\pi iz}e^{2\pi i}\\ &=q(z)\cdot 1\\ &=q(z). \end{align*} Thus each factor $1-q^n$ is unchanged by $z\mapsto z+1$. The initial fractional power changes by \begin{align*} q(z+1)^{1/24} &=e^{2\pi i(z+1)/24}\\ &=e^{2\pi iz/24}e^{2\pi i/24}\\ &=e^{2\pi i/24}q(z)^{1/24}. \end{align*} Therefore \begin{align*} \eta(z+1) &=q(z+1)^{1/24}\prod_{n=1}^{\infty}(1-q(z+1)^n)\\ &=e^{2\pi i/24}q(z)^{1/24}\prod_{n=1}^{\infty}(1-q(z)^n)\\ &=e^{2\pi i/24}\eta(z). \end{align*} Since $e^{2\pi i/24}\ne 1$, the eta function is not invariant under $T:z\mapsto z+1$. But for every integral weight $k$, the matrix $T=\begin{pmatrix}1&1\\0&1\end{pmatrix}$ has automorphy factor $0\cdot z+1=1$, so the weight $k$ transformation law for $T$ would require \begin{align*} \eta(z+1)=\eta(z). \end{align*} Thus $\eta$ is not a modular form for $SL_2(\mathbb Z)$ in the present integral-weight sense. The calculation shows exactly what goes wrong: the product part is periodic in $q$, while the fractional initial power contributes the nontrivial factor $e^{2\pi i/24}$. This is why eta-products naturally lead to multiplier systems and half-integral weight theories, which lie outside this course. [/example] This example marks the boundary of the present definitions. The course keeps to integral-weight full-level forms, using the basic $q$-expansion and cusp-form tests before turning to Eisenstein series, the discriminant, Hecke operators, and $L$-functions. [remark: What This Chapter Establishes] At this stage, a modular form of weight $k$ for $SL_2(\mathbb Z)$ is a holomorphic function on $\mathbb H$ satisfying the weight $k$ transformation law and admitting a power series at $q=0$. The generator $T$ produces the $q$-expansion, the cusp condition reads off its constant term, and the central element $-I$ removes all odd weights. The next chapter constructs nonconstant examples in even weights using Eisenstein series. [/remark] The Eisenstein series give the first explicit examples, but they also hint at a deeper pattern: modular forms are heavily constrained by their Fourier coefficients and vanishing at the cusp. Those constraints will now be turned into global structure, culminating in the valence formula and the description of the whole ring of level-1 modular forms. # 3. Eisenstein Series This chapter constructs the first large supply of holomorphic modular forms for the full modular group: the Eisenstein series. The previous chapter defined modular forms and explained why holomorphy at the cusp is encoded by a $q$-expansion. Here we build modular forms by summing over lattices, compute their Fourier expansions, and see the first signs that all modular forms for $SL_2(\mathbb Z)$ are generated by two special forms of weights $4$ and $6$. ## Constructing Modular Forms by Averaging over Lattices How can we produce a modular form without already knowing a holomorphic function satisfying the transformation law? The guiding idea is to average a simple homogeneous function over all non-zero lattice vectors in the lattice $\mathbb Z z + \mathbb Z$. Homogeneity gives the correct weight factor, while absolute convergence gives enough analytic control to prove holomorphy and rearrange sums. [definition: Eisenstein Series] Let $k \ge 4$ be an even integer. The Eisenstein series of weight $k$ for $SL_2(\mathbb Z)$ is the function $G_k:\mathbb H \to \mathbb C$ defined by \begin{align*} G_k(z) = \sum_{(m,n) \in \mathbb Z^2 \setminus \{(0,0)\}} \frac{1}{(mz+n)^k}. \end{align*} [/definition] The restriction $k \ge 4$ is the convergence threshold needed for the two-dimensional lattice sum. The parity condition is also forced for the full modular group: if $k$ is odd, the terms indexed by $(m,n)$ and $(-m,-n)$ cancel. Before this lattice sum can be used as a modular form, its analytic status has to be established: the infinite series must converge well enough to define a holomorphic function. The next theorem supplies that convergence and holomorphy input, separating the analytic foundation from the later modular-transformation argument. [quotetheorem:4225] [citeproof:4225] The threshold $k>2$ is not cosmetic: in two dimensions there are about $O(r)$ lattice points at lattice distance $r$, so the comparison series behaves like $\sum_r r^{1-k}$. Weight $2$ is therefore the borderline case, and the analogous series is not absolutely convergent; its regularized version is the source of the quasi-modular Eisenstein series $E_2$, not a modular form for $SL_2(\mathbb Z)$. Local [uniform convergence](/page/Uniform%20Convergence) on $\mathbb H$ gives holomorphy there; the later $q$-expansion will make the already asserted cusp holomorphy explicit and computable by showing the exact Fourier coefficients. [quotetheorem:4226] [citeproof:4226] The evenness hypothesis is essential for the full modular group because $-I\in SL_2(\mathbb Z)$ acts by the factor $(-1)^k$. A non-zero modular form of odd weight would have to satisfy $f(z)=(-1)^k f(z)$, forcing $f=0$. Absolute convergence is also essential in this proof: it justifies reindexing the lattice sum by the integral change of variables induced by $\gamma$. The theorem establishes that the Eisenstein series is a modular form; the $q$-expansion computed next is not needed to repair modularity, but to expose its arithmetic coefficients. [example: Pair Cancellation in Odd Weight] Suppose $k$ is odd and form the finite symmetric sum over a set $S\subset \mathbb Z^2\setminus\{(0,0)\}$ satisfying $(m,n)\in S$ if and only if $(-m,-n)\in S$. For each non-zero pair $(m,n)$, \begin{align*} \frac{1}{((-m)z+(-n))^k} &=\frac{1}{(-(mz+n))^k}\\ &=\frac{1}{(-1)^k(mz+n)^k}\\ &=-\frac{1}{(mz+n)^k}, \end{align*} because $k$ is odd, so $(-1)^k=-1$. Hence the two terms in each pair cancel: \begin{align*} \frac{1}{(mz+n)^k}+\frac{1}{((-m)z+(-n))^k} =\frac{1}{(mz+n)^k}-\frac{1}{(mz+n)^k} =0. \end{align*} Therefore every symmetric partial summation of the odd-weight lattice expression is zero. The same sign obstruction appears from the matrix $-I=\begin{pmatrix}-1&0\\0&-1\end{pmatrix}\in SL_2(\mathbb Z)$. If $f$ were a modular form of odd weight $k$ for the full modular group, then the transformation law would give \begin{align*} f(z) =f((-I)z) =(-1)^k f(z) =-f(z), \end{align*} so $2f(z)=0$ for every $z\in\mathbb H$, and hence $f=0$. Thus odd weights are excluded by the transformation law itself, not only by convergence considerations. [/example] To compare Eisenstein series across weights, it is convenient to normalize the constant term of the $q$-expansion to be $1$. For real $k>1$, the Riemann zeta value is \begin{align*} \zeta(k)=\sum_{n=1}^{\infty}\frac{1}{n^k}. \end{align*} Without this normalization, each weight carries the moving constant term $2\zeta(k)$, so products and comparisons of Fourier coefficients would be obscured by transcendental scaling factors. The constant term of $G_k$ comes from the vectors with $m=0$, namely $\sum_{n\ne0}n^{-k}=2\zeta(k)$. [definition: Normalized Eisenstein Series] Let $k \ge 4$ be even. The normalized Eisenstein series of weight $k$ is the function $E_k:\mathbb H\to\mathbb C$ defined by \begin{align*} E_k(z)=\frac{G_k(z)}{2\zeta(k)}. \end{align*} [/definition] Thus $E_k$ is a modular form of weight $k$ with constant term $1$. This normalization is the one compatible with Bernoulli numbers and divisor sums. ## Computing the Fourier Expansion What does the lattice definition look like from the viewpoint of the cusp? Since translation $z\mapsto z+1$ preserves the lattice sum, $G_k$ has a Fourier expansion in powers of $q=e^{2\pi iz}$. The aim of this section is to compute its coefficients explicitly. The computation separates the summation over $m=0$ from the summation over horizontal rows $m\ne0$. The analytic input is a classical partial fraction expansion, which turns the sum over $n\in\mathbb Z$ into an exponential series. [quotetheorem:4227] [citeproof:4227] The upper half-plane hypothesis is what converts the cotangent expansion into a positive-power exponential series in $e^{2\pi iz}$. Normal convergence on compact subsets is the justification for differentiating the exponential series term by term; differentiating only the symmetric partial fractions without this convergence control would not produce a usable Fourier expansion. The theorem is therefore a local analytic statement on $\mathbb H$, not a global identity at the poles of $\cot(\pi z)$. Its role is to convert each horizontal lattice row into a $q$-series whose coefficients can be collected arithmetically. The next obstruction is organizational rather than analytic: after every nonzero row has been expanded, a single coefficient of $q^n$ is no longer attached to one row. It receives contributions from every factorization $n=mr$, so the coefficient naturally asks for a weighted sum over the divisors of $n$. To state the Fourier expansion as a genuine arithmetic formula instead of a row-by-row double sum, we isolate that divisor-counting operation as notation. [definition: Divisor Sum] For $s\in\mathbb Z_{\ge0}$, the divisor-sum function is the map $\sigma_s:\mathbb N\to\mathbb Z$ defined by \begin{align*} \sigma_s(n)=\sum_{d\mid n} d^s. \end{align*} [/definition] The divisor sum $\sigma_{k-1}(n)$ measures the number of ways the lattice rows contribute to the coefficient of $q^n$, weighted by the appropriate power of the divisor. At this point the Eisenstein series has two descriptions: a lattice sum that shows modularity and a row-by-row analytic expansion that hints at arithmetic coefficients. To use $E_k$ as an actual modular form, we need the two descriptions to meet in a normalized $q$-expansion at the cusp. The theorem gives that expansion and records exactly how the divisor sums enter the Fourier coefficients. [quotetheorem:4228] [citeproof:4228] This theorem completes the verification that $E_k$ is holomorphic at infinity: its $q$-series has no negative powers. The assumptions remain visible in the formula: evenness is used to pair positive and negative rows, while $k\ge4$ is the convergence condition that permits the lattice sum to be rearranged. The Bernoulli-number normalization is not decorative; it is exactly what turns the analytic constant $2\zeta(k)$ into the arithmetic scalar $-2k/B_k$ after normalizing the constant term to $1$. The formula gives precise Fourier coefficients for Eisenstein series, but it does not by itself prove that all modular forms are built from them; that ring-structure statement requires later dimension and valence arguments. It also turns the abstract lattice average into an arithmetic generating function for divisor sums. [example: First Terms of E Four] For $k=4$, the Bernoulli number is $B_4=-1/30$. The scalar in the normalized Fourier expansion is \begin{align*} -\frac{2k}{B_k} &=-\frac{2\cdot 4}{-1/30}\\ &=-8\cdot(-30)\\ &=240. \end{align*} Therefore \begin{align*} E_4(z) &=1+240\sum_{n\ge1}\sigma_3(n)q^n. \end{align*} Now compute the first divisor sums from the definition $\sigma_3(n)=\sum_{d\mid n}d^3$: \begin{align*} \sigma_3(1)&=1^3=1,\\ \sigma_3(2)&=1^3+2^3=1+8=9,\\ \sigma_3(3)&=1^3+3^3=1+27=28. \end{align*} Substituting these values into the $q$-series gives \begin{align*} E_4(z) &=1+240\left(\sigma_3(1)q+\sigma_3(2)q^2+\sigma_3(3)q^3+\cdots\right)\\ &=1+240\left(q+9q^2+28q^3+\cdots\right)\\ &=1+240q+2160q^2+6720q^3+\cdots. \end{align*} Thus $E_4$ is a non-constant modular form for the full modular group, with its first non-constant coefficient equal to $240$. [/example] The weight-$4$ computation shows how divisor sums enter the first coefficients. Weight $6$ gives the companion series and introduces the sign change that later distinguishes the two elliptic points. [example: First Terms of E Six] For $k=6$, the Bernoulli number is $B_6=1/42$. The scalar in the normalized Fourier expansion is \begin{align*} -\frac{2k}{B_k} &=-\frac{2\cdot 6}{1/42}\\ &=-12\cdot 42\\ &=-504. \end{align*} Hence \begin{align*} E_6(z) &=1-504\sum_{n\ge1}\sigma_5(n)q^n. \end{align*} Now compute the first divisor sums from the definition $\sigma_5(n)=\sum_{d\mid n}d^5$: \begin{align*} \sigma_5(1)&=1^5=1,\\ \sigma_5(2)&=1^5+2^5=1+32=33,\\ \sigma_5(3)&=1^5+3^5=1+243=244. \end{align*} Substituting these values into the $q$-series gives \begin{align*} E_6(z) &=1-504\left(\sigma_5(1)q+\sigma_5(2)q^2+\sigma_5(3)q^3+\cdots\right)\\ &=1-504\left(q+33q^2+244q^3+\cdots\right)\\ &=1-504q-(504\cdot 33)q^2-(504\cdot 244)q^3+\cdots\\ &=1-504q-16632q^2-122976q^3+\cdots. \end{align*} The first non-constant coefficient is negative because $B_6>0$, so the scalar $-2k/B_k$ is negative in weight $6$. [/example] The Fourier coefficients grow arithmetically, but the modular transformation law came from geometry. This dual nature is why Eisenstein series are the basic bridge between modular forms and divisor sums. ## Products, Weight Twelve, and the First Structure Hint How much of the ring of modular forms can be seen from $E_4$ and $E_6$? Products of modular forms are again modular forms, with weights adding. Since $4$ and $6$ generate all sufficiently large even integers by addition, the forms $E_4$ and $E_6$ are natural candidates for algebraic generators. [quotetheorem:4229] [citeproof:4229] The cusp condition in the hypotheses matters: without holomorphy at infinity, multiplying two functions with the correct transformation law could still produce a meromorphic expression at the cusp. The theorem also has a limited conclusion: it explains how to build new forms from old ones, but it does not say that every modular form is obtainable as a polynomial in chosen generators. That stronger generation statement needs a dimension count or a vanishing argument, which has not yet been proved. The product construction is nevertheless the first mechanism that lets the low-weight Eisenstein series interact across weights. In weight $12$, two natural modular forms appear: $E_4^3$ and $E_6^2$. Both have constant term $1$, so their difference has zero constant term and is therefore a cusp form. [example: A Weight Twelve Cusp Form] Write $q=e^{2\pi iz}$. From the previous computations, \begin{align*} E_4(z)&=1+240q+2160q^2+O(q^3),\\ E_6(z)&=1-504q-16632q^2+O(q^3). \end{align*} For $E_4^3$, set $x=240q+2160q^2+O(q^3)$. Then \begin{align*} E_4^3 &=(1+x)^3\\ &=1+3x+3x^2+x^3\\ &=1+3(240q+2160q^2+O(q^3)) +3(240q+2160q^2+O(q^3))^2 +O(q^3). \end{align*} The square contributes, up to order $q^2$, \begin{align*} (240q+2160q^2+O(q^3))^2 &=(240q)^2+2(240q)(2160q^2)+(2160q^2)^2+O(q^3)\\ &=57600q^2+O(q^3), \end{align*} since the remaining displayed products have degree at least $3$ in $q$. Hence \begin{align*} E_4^3 &=1+720q+6480q^2+172800q^2+O(q^3)\\ &=1+720q+179280q^2+O(q^3). \end{align*} For $E_6^2$, set $y=-504q-16632q^2+O(q^3)$. Then \begin{align*} E_6^2 &=(1+y)^2\\ &=1+2y+y^2\\ &=1+2(-504q-16632q^2+O(q^3)) +(-504q-16632q^2+O(q^3))^2. \end{align*} Again keeping terms only through $q^2$, \begin{align*} (-504q-16632q^2+O(q^3))^2 &=(-504q)^2+2(-504q)(-16632q^2)+(-16632q^2)^2+O(q^3)\\ &=254016q^2+O(q^3). \end{align*} Therefore \begin{align*} E_6^2 &=1-1008q-33264q^2+254016q^2+O(q^3)\\ &=1-1008q+220752q^2+O(q^3). \end{align*} Subtracting the two expansions term by term gives \begin{align*} E_4^3-E_6^2 &=(1+720q+179280q^2+O(q^3)) -(1-1008q+220752q^2+O(q^3))\\ &=(720+1008)q+(179280-220752)q^2+O(q^3)\\ &=1728q-41472q^2+O(q^3). \end{align*} The product rule for modular forms gives $E_4^3,E_6^2\in M_{12}(SL_2(\mathbb Z))$, so their difference is also a weight $12$ modular form. Its constant term is $1-1=0$, so it is cuspidal, and it is non-zero because its $q$-coefficient is $1728$. [/example] This example shows that $E_4^3-E_6^2$ is the first explicit construction here that lands in a nonzero cusp-form space rather than merely producing Eisenstein series. To use this form consistently in later formulas, we need to fix its normalization rather than carry the coefficient $1728$ through every computation. Dividing by the first nonzero coefficient gives a canonical cusp form beginning with $q$, which will become the standard reference form of weight $12$. [definition: Discriminant Modular Form] The discriminant modular form is the function $\Delta:\mathbb H\to\mathbb C$ defined by \begin{align*} \Delta(z)=\frac{E_4(z)^3-E_6(z)^2}{1728}. \end{align*} It belongs to $S_{12}(SL_2(\mathbb Z))$. [/definition] With this normalization, $\Delta(z)=q+O(q^2)$, so it is a distinguished generator of the one-dimensional cusp space in weight $12$. The examples so far construct many forms, but they do not yet explain whether these constructions account for all modular forms. The natural structural question is whether every form can be expressed algebraically using the two basic Eisenstein series $E_4$ and $E_6$, with the discriminant marking the first cuspidal relation. The global structure theorem answers that question; the later chapters then develop the valence and dimension machinery that explains why the answer is forced. [quotetheorem:4230] [citeproof:4230] This structure theorem belongs conceptually with the valence formula of Chapter 4 and the dimension formula and monomial bases made explicit in Chapter 5. The present chapter has already shown why the statement is plausible: Eisenstein series provide canonical non-cuspidal forms in every even weight at least $4$, and products of $E_4$ and $E_6$ supply the basic algebraic building blocks. [remark: Why Weight Twelve Matters] Weight $12$ is the first weight in which two different monomials in $E_4$ and $E_6$ have the same weight, namely $E_4^3$ and $E_6^2$. Their difference cancels the constant term and produces a cusp form. In Chapter 4, the identity $E_4^3-E_6^2=1728\Delta$ becomes the basic relation connecting Eisenstein series, cusp forms, and the modular $j$-invariant. [/remark] The chapter therefore leaves us with three working objects: the normalized Eisenstein series $E_k$, the divisor-sum expansion of its coefficients, and the cusp form $\Delta$. These will be used repeatedly when the course turns from construction to structure: determining dimensions, proving generation, and studying Hecke operators through their action on Fourier coefficients. The valence formula explains how zeros, poles, and cusps control a modular form, and the ring structure shows that everything is generated by a small set of basic forms. With that structural picture established, we can move from existence and generation to counting, and ask for explicit bases in each weight. # 4. The Valence Formula and the Ring of Modular Forms Chapters 2 and 3 introduced modular forms through their transformation law and Fourier expansions, then constructed the Eisenstein series $E_k$ as the first large supply of examples. This chapter explains how much information is forced by the geometry of the quotient $SL_2(\mathbb Z)\backslash \mathbb H$. The central result is the valence formula, which counts zeros of a modular form with fractional weights at the elliptic points and at the cusp. We then use this count to identify the graded ring of modular forms as the [polynomial ring](/page/Polynomial%20Ring) generated by $E_4$ and $E_6$. ## Measuring Zeros on the Modular Curve A modular form is not an ordinary holomorphic function on $\mathbb H$: points related by $SL_2(\mathbb Z)$ represent the same point of the quotient, while the elliptic points have nontrivial stabilisers. The first question is therefore how to count zeros in a way that respects this quotient geometry. [definition: Order at an Ordinary Point] Let $f \in M_k$ and let $z_0 \in \mathbb H$. The order of vanishing of $f$ at $z_0$ is the integer $\operatorname{ord}_{z_0}(f) \ge 0$ such that \begin{align*} f(z) = (z-z_0)^{\operatorname{ord}_{z_0}(f)}g(z) \end{align*} near $z_0$, where $g$ is holomorphic and $g(z_0) \ne 0$. [/definition] This is the usual local order from complex analysis, but the quotient only sees orbits of points rather than individual representatives in $\mathbb H$. The order count will be meaningful on the modular curve only if changing representatives by an element of $SL_2(\mathbb Z)$ preserves the vanishing order. The modular transformation law suggests this should hold because the automorphy factor is nonzero on $\mathbb H$, and the next result makes that invariance precise. [quotetheorem:4231] [citeproof:4231] This invariance is the reason the later zero count can sum over orbits rather than over individual points of $\mathbb H$. The hypothesis that $f$ has a modular transformation law is essential: an arbitrary holomorphic function on $\mathbb H$ need not have the same vanishing order at equivalent points. The theorem also says only that orders are constant along ordinary orbits; it does not yet explain how to count the cusp or how to adjust for points with nontrivial stabiliser. Those two issues are handled next, first by the $q$-coordinate at infinity and then by fractional weights at elliptic points. The cusp is not a point of $\mathbb H$, so the ordinary local coordinate $z-z_0$ cannot measure vanishing there. To include the cusp in the same zero count as the interior points, we need a substitute local coordinate. The periodicity of a modular form supplies this coordinate through $q=e^{2\pi iz}$, so the first nonzero Fourier coefficient defines the missing notion of order at infinity. [definition: Order at the Cusp] Let $f \in M_k$ have Fourier expansion \begin{align*} f(z)=\sum_{n=0}^{\infty} a_n q^n, \qquad q=e^{2\pi iz}. \end{align*} The order of $f$ at the cusp is \begin{align*} \operatorname{ord}_{\infty}(f)=\min\{n\ge 0 : a_n\ne 0\}. \end{align*} For $f=0$, the order is not assigned in the valence formula. [/definition] Thus $f$ is a cusp form precisely when $\operatorname{ord}_{\infty}(f)\ge 1$. The cusp behaves like one additional point on the compactified quotient, but its local coordinate is $q$ rather than $z$. [example: Orders of Eisenstein Series at the Cusp] For $k\ge 4$ even, the normalized Eisenstein series has $q$-expansion \begin{align*} E_k(z) &=1-\frac{2k}{B_k}\sum_{n=1}^{\infty}\sigma_{k-1}(n)q^n\\ &=1+\left(-\frac{2k}{B_k}\sigma_{k-1}(1)\right)q+\left(-\frac{2k}{B_k}\sigma_{k-1}(2)\right)q^2+\cdots, \end{align*} where $q=e^{2\pi iz}$. Thus the coefficient of $q^0$ is $1$, and $1\ne 0$. By the definition of order at the cusp, \begin{align*} \operatorname{ord}_{\infty}(E_k) &=\min\{n\ge 0:\text{the coefficient of }q^n\text{ in }E_k\text{ is nonzero}\}\\ &=\min\{0,1,2,\ldots\}\\ &=0. \end{align*} By contrast, if $f\in S_k$ is a cusp form with expansion $f(z)=\sum_{n=0}^{\infty}a_nq^n$, then $a_0=0$. If $f$ is not the zero form and $m=\min\{n\ge 1:a_n\ne 0\}$, then \begin{align*} f(z)=a_mq^m+a_{m+1}q^{m+1}+\cdots, \end{align*} so $\operatorname{ord}_{\infty}(f)=m\ge 1$. The cusp order is therefore exactly the exponent of the first nonzero term in the $q$-expansion. [/example] The elliptic points require a correction. A small neighbourhood of $i$ in the quotient has local degree $2$, and a small neighbourhood of $\rho=e^{2\pi i/3}$ has local degree $3$. The formula will therefore count zeros at these points with weights $1/2$ and $1/3$. [definition: Elliptic Points] Let \begin{align*} i &= \sqrt{-1}, & \rho &= e^{2\pi i/3}=-\frac12+\frac{\sqrt{3}}{2}i. \end{align*} The points $i$ and $\rho$ are the elliptic points of the standard fundamental domain for $SL_2(\mathbb Z)$. [/definition] The stabiliser of $i$ is generated by $S:z\mapsto -1/z$ modulo $\{\pm I\}$, while the stabiliser of $\rho$ is generated by $ST:z\mapsto -1/(z+1)$ modulo $\{\pm I\}$. At a fixed point, the transformation law compares the value of $f$ with a root of unity times itself. If that root of unity is not $1$, the form must vanish there; thus the stabilisers impose weight-dependent restrictions that must be known before the global zero count can be trusted. At the elliptic points, the modular transformation law can force $f(p)$ to equal a nontrivial root of unity times itself. The resulting vanishing depends only on the weight modulo the stabiliser order, so the local question is to determine exactly which congruence classes of $k$ force zeros at $i$ and at $\rho$. Those forced zeros are part of the input for the global count. [quotetheorem:4232] [citeproof:4232] This theorem depends on the full level-one group, because the stabilisers of $i$ and $\rho$ are what impose the congruence restrictions on the weight. It detects only whether the value must vanish at the elliptic point; it does not determine the exact order of vanishing, nor does it rule out additional zeros elsewhere on the quotient. The statement is also asymmetric in the two elliptic points because their stabilisers have different orders. The valence formula will combine these forced zeros with the cusp order and the ordinary zeros, turning the local fixed-point restrictions into a global count. ## The Valence Formula The main counting problem is to relate the analytic zeros of $f$ to its weight. Since multiplying by a modular form adds weights and adds orders of zeros, we expect a linear formula. The answer is that a nonzero form of weight $k$ has total weighted number of zeros equal to $k/12$. [illustration:truncated-fundamental-domain] [quotetheorem:4233] [citeproof:4233] The formula applies only to nonzero modular forms for the full modular group; the zero form has infinite vanishing order everywhere, so no finite counting identity is possible. The hypotheses also matter because different congruence subgroups have different fundamental domains, cusp widths, and elliptic stabilisers, hence different valence formulas. For level one, the identity says that the compactified quotient has degree $1/12$ with respect to weight. It immediately gives strong restrictions: for example, if $0<k<12$ then a nonzero form has very few possible zeros, and these restrictions will drive both the construction of $\Delta$ and the proof that $E_4$ and $E_6$ generate the whole ring. [example: The Zeros of $E_4$] The normalized Eisenstein series $E_4$ has $q$-expansion beginning \begin{align*} E_4(q)=1+240q+2160q^2+\cdots. \end{align*} The coefficient of $q^0$ is $1\ne 0$, so by the definition of order at the cusp, \begin{align*} \operatorname{ord}_{\infty}(E_4)=0. \end{align*} Since $4\not\equiv 0\pmod 6$, the elliptic point condition at $\rho$ gives $E_4(\rho)=0$, hence \begin{align*} \operatorname{ord}_{\rho}(E_4)\ge 1. \end{align*} Applying the valence formula with $k=4$ gives \begin{align*} 0+\frac{1}{2}\operatorname{ord}_{i}(E_4)+\frac{1}{3}\operatorname{ord}_{\rho}(E_4)+\sum_z\operatorname{ord}_z(E_4) &=\frac{4}{12}\\ &=\frac{1}{3}. \end{align*} Reordering the nonnegative terms, \begin{align*} \frac{1}{3}\operatorname{ord}_{\rho}(E_4)+\frac{1}{2}\operatorname{ord}_{i}(E_4)+\sum_z\operatorname{ord}_z(E_4) =\frac{1}{3}. \end{align*} Because $\operatorname{ord}_{\rho}(E_4)\ge 1$, \begin{align*} \frac{1}{3}\operatorname{ord}_{\rho}(E_4)\ge \frac{1}{3}. \end{align*} All other terms in the valence formula are nonnegative vanishing orders, so equality with $\frac{1}{3}$ forces \begin{align*} \operatorname{ord}_{\rho}(E_4)=1,\qquad \operatorname{ord}_{i}(E_4)=0,\qquad \sum_z\operatorname{ord}_z(E_4)=0. \end{align*} Thus $E_4$ has a simple zero at $\rho$ and no other zeros on the quotient. [/example] The companion computation for $E_6$ shows the same mechanism at the other elliptic point. Together, $E_4$ and $E_6$ provide the basic test cases for how congruence restrictions and the valence formula interact. [example: The Zeros of $E_6$] The normalized Eisenstein series $E_6$ has $q$-expansion beginning \begin{align*} E_6(q)=1-504q-16632q^2+\cdots. \end{align*} The coefficient of $q^0$ is $1\ne 0$, so by the definition of order at the cusp, \begin{align*} \operatorname{ord}_{\infty}(E_6) &=\min\{n\ge 0:\text{the coefficient of }q^n\text{ in }E_6\text{ is nonzero}\}\\ &=0. \end{align*} The weight is $6$, and \begin{align*} 6\equiv 2 \pmod 4, \end{align*} so the elliptic point condition at $i$ gives $E_6(i)=0$. Hence \begin{align*} \operatorname{ord}_{i}(E_6)\ge 1. \end{align*} Applying the valence formula with $k=6$ gives \begin{align*} \operatorname{ord}_{\infty}(E_6) +\frac{1}{2}\operatorname{ord}_{i}(E_6) +\frac{1}{3}\operatorname{ord}_{\rho}(E_6) +\sum_z\operatorname{ord}_z(E_6) &=\frac{6}{12}. \end{align*} Substituting $\operatorname{ord}_{\infty}(E_6)=0$ and reducing the right-hand side, \begin{align*} \frac{1}{2}\operatorname{ord}_{i}(E_6) +\frac{1}{3}\operatorname{ord}_{\rho}(E_6) +\sum_z\operatorname{ord}_z(E_6) &=\frac{1}{2}. \end{align*} Since $\operatorname{ord}_{i}(E_6)\ge 1$, \begin{align*} \frac{1}{2}\operatorname{ord}_{i}(E_6)\ge \frac{1}{2}. \end{align*} All remaining terms are nonnegative vanishing orders, so equality with $\frac{1}{2}$ forces \begin{align*} \operatorname{ord}_{i}(E_6)=1,\qquad \operatorname{ord}_{\rho}(E_6)=0,\qquad \sum_z\operatorname{ord}_z(E_6)=0. \end{align*} Thus $E_6$ has a simple zero at $i$ and no other zeros on the quotient. [/example] These examples also show how a small weighted total can force uniqueness of the zero pattern. In weight $2$, the same principle becomes even more restrictive and rules out nonzero forms entirely. [remark: No Weight Two Modular Form] If $f\in M_2$ were nonzero, the valence formula would give a weighted zero count equal to $1/6$. The allowed contributions from ordinary points, the cusp, $i$, and $\rho$ cannot add to $1/6$ subject to the elliptic divisibility constraints. Hence $M_2=0$. [/remark] The same reasoning is a dimension-counting machine. Once a modular form is forced to have more zeros than $k/12$ permits, it must be the zero form. ## The Discriminant and the $j$-Invariant The valence formula becomes most effective after constructing a cusp form of weight $12$ with a simple zero at infinity and no zeros in $\mathbb H$. The difference $E_4^3-E_6^2$ has exactly the right weight and vanishes at the cusp because both Eisenstein series have constant term $1$. [definition: Modular Discriminant] The modular discriminant is the function \begin{align*} \Delta: \mathbb H &\longrightarrow \mathbb C, & z &\longmapsto \frac{E_4(z)^3-E_6(z)^2}{1728}. \end{align*} [/definition] The normalising factor is chosen so that the first nonzero $q$-coefficient is $1$. The next theorem shows in particular that $\Delta \in S_{12}$, so the formula defines not just a holomorphic function on $\mathbb H$ but a level-one cusp form of weight $12$. In terms of the Dedekind eta function, this form is also given by $\Delta(q)=q\prod_{n=1}^{\infty}(1-q^n)^{24}$, a product formula that will be useful later when Hecke eigenforms enter the course. [quotetheorem:4234] [citeproof:4234] The nonvanishing of $\Delta$ on $\mathbb H$ is the structural point: it means division by $\Delta$ lowers the weight by $12$ without creating any poles in the upper half-plane. The theorem does not say that $\Delta$ is the only cusp form of weight $12$ until a dimension argument is supplied; it only identifies its divisor on the compactified quotient. Its simple zero at the cusp will make multiplication by $\Delta$ an isomorphism between lower-weight modular forms and cusp forms in weights at least $12$. This is the bridge from the valence formula to the ring structure theorem. The quotient of two weight $12$ modular forms is a modular function. Since $\Delta$ has no zeros on $\mathbb H$, it can be used as a denominator without introducing poles away from the cusp. This suggests looking for a distinguished modular function that has exactly one pole at infinity and no other poles. Such a function can serve as a coordinate on the compactified quotient, turning the abstract orbit space into something that can be studied through one meromorphic function. The natural candidate must have weight $0$, so the weights in the numerator and denominator must match. Cubing $E_4$ produces weight $12$, and division by the nonvanishing cusp form $\Delta$ gives a modular function with its only possible pole at the cusp. We now need to name this function because it becomes the basic coordinate for modular functions on the full modular group. The definition packages the preceding weight and divisor considerations into a single object whose pole behavior at infinity can be tracked through its $q$-expansion. [definition: Modular $j$-Invariant] The modular $j$-invariant is the function \begin{align*} j: \mathbb H &\longrightarrow \mathbb C, & z &\longmapsto \frac{E_4(z)^3}{\Delta(z)}. \end{align*} [/definition] It is invariant under $SL_2(\mathbb Z)$ and holomorphic on $\mathbb H$, with a simple pole at the cusp. Thus it descends to a meromorphic function on the compactified quotient $SL_2(\mathbb Z)\backslash\mathbb H\cup\{\infty\}$. This function gives a coordinate on the compactified quotient curve, and in later courses it classifies complex elliptic curves up to isomorphism. [example: First Terms of the $j$-Expansion] Using \begin{align*} E_4(q)&=1+240q+2160q^2+6720q^3+\cdots,\\ \Delta(q)&=q-24q^2+252q^3-1472q^4+\cdots, \end{align*} we first expand the numerator in $j=E_4^3/\Delta$. Up to order $q^3$, \begin{align*} E_4(q)^3 &=(1+240q+2160q^2+6720q^3+\cdots)^3\\ &=1+3(240q)+\bigl(3(2160q^2)+3(240q)^2\bigr)\\ &\qquad+\bigl(3(6720q^3)+6(240q)(2160q^2)+(240q)^3\bigr)+\cdots\\ &=1+720q+(6480+172800)q^2\\ &\qquad+(20160+3110400+13824000)q^3+\cdots\\ &=1+720q+179280q^2+16954560q^3+\cdots. \end{align*} Since \begin{align*} \Delta(q)=q(1-24q+252q^2-1472q^3+\cdots), \end{align*} write \begin{align*} \frac{1}{1-24q+252q^2-1472q^3+\cdots} =1+Aq+Bq^2+Cq^3+\cdots. \end{align*} Multiplying and matching coefficients with $1$ gives \begin{align*} (1-24q+252q^2-1472q^3)(1+Aq+Bq^2+Cq^3) &=1+\cdots,\\ A-24&=0,\\ B-24A+252&=0,\\ C-24B+252A-1472&=0. \end{align*} Thus \begin{align*} A&=24,\\ B&=24A-252=576-252=324,\\ C&=24B-252A+1472=7776-6048+1472=3200. \end{align*} Therefore \begin{align*} j(q) &=\frac{E_4(q)^3}{\Delta(q)}\\ &=q^{-1}(1+720q+179280q^2+16954560q^3+\cdots)(1+24q+324q^2+3200q^3+\cdots)\\ &=q^{-1}\bigl(1+(720+24)q+(179280+720\cdot 24+324)q^2\\ &\qquad +(16954560+179280\cdot 24+720\cdot 324+3200)q^3+\cdots\bigr)\\ &=q^{-1}\bigl(1+744q+196884q^2+21493760q^3+\cdots\bigr)\\ &=q^{-1}+744+196884q+21493760q^2+\cdots. \end{align*} The leading term $q^{-1}$ records that $j$ has a simple pole at the cusp, while the large positive coefficients are the first hint of the later connection with moonshine, although that topic lies outside this course. [/example] This calculation also points to a convention issue: different normalisations move scalar factors between the operator and its eigenvalues. [remark: Alternative Normalisation] Some authors write $J=j/1728$. With our convention, $j=E_4^3/\Delta$ has principal part $q^{-1}$ at infinity and constant term $744$. [/remark] ## The Ring Generated by $E_4$ and $E_6$ The final question of the chapter is whether all modular forms for $SL_2(\mathbb Z)$ can be built from the Eisenstein series already constructed. The valence formula gives enough control to prove that no other generators are needed. [definition: Graded Ring of Modular Forms] The graded ring of modular forms for $SL_2(\mathbb Z)$ is \begin{align*} M_* = \bigoplus_{k\ge 0} M_k, \end{align*} with multiplication given by pointwise multiplication of modular forms. [/definition] Since $E_4\in M_4$ and $E_6\in M_6$, every polynomial $P(E_4,E_6)$ is a modular form whose weight is determined by assigning degrees $4$ and $6$ to the variables. The remaining obstruction is surjectivity and independence: there could be modular forms not reached by these polynomials, or there could be hidden algebraic relations among the two generators. The valence formula gives the control needed to rule out both possibilities and identify the whole graded ring. [quotetheorem:4235] [citeproof:4235] The level-one hypothesis is essential: for congruence subgroups, extra cusps and different elliptic data usually produce additional generators. The theorem is also a graded statement, so the weights of $E_4$ and $E_6$ are part of the conclusion rather than decorative information. Its main structural consequence is that cusp forms are obtained by multiplying lower-weight forms by the discriminant. If $S_k$ denotes the subspace of cusp forms in $M_k$, then the induction proves \begin{align*} S_k = \Delta M_{k-12} \end{align*} for $k\ge 12$, with $S_k=0$ for $k<12$. This reduces many questions about cusp forms to questions about lower weights and prepares the finite-dimensional spaces on which Hecke operators will act. [example: Low-Weight Spaces] The structure theorem identifies $M_k$ with the span of all monomials $E_4^aE_6^b$ whose weights satisfy \begin{align*} 4a+6b=k,\qquad a,b\ge 0. \end{align*} For the low weights, the possible pairs $(a,b)$ are found by solving this equation: \begin{align*} k=0:&\qquad 4a+6b=0 \quad\Longrightarrow\quad (a,b)=(0,0),\\ k=2:&\qquad 4a+6b=2 \quad\Longrightarrow\quad \text{no nonnegative integer solution},\\ k=4:&\qquad 4a+6b=4 \quad\Longrightarrow\quad (a,b)=(1,0),\\ k=6:&\qquad 4a+6b=6 \quad\Longrightarrow\quad (a,b)=(0,1),\\ k=8:&\qquad 4a+6b=8 \quad\Longrightarrow\quad (a,b)=(2,0),\\ k=10:&\qquad 4a+6b=10 \quad\Longrightarrow\quad (a,b)=(1,1). \end{align*} Thus \begin{align*} M_0&=\mathbb C, & M_2&=0, & M_4&=\mathbb C E_4,\\ M_6&=\mathbb C E_6, & M_8&=\mathbb C E_4^2, & M_{10}&=\mathbb C E_4E_6. \end{align*} In weight $12$, the equation is \begin{align*} 4a+6b=12. \end{align*} The nonnegative integer solutions are \begin{align*} (a,b)=(3,0)\qquad\text{and}\qquad (a,b)=(0,2), \end{align*} so \begin{align*} M_{12}=\mathbb C E_4^3\oplus \mathbb C E_6^2. \end{align*} By the definition of the discriminant, \begin{align*} \Delta=\frac{E_4^3-E_6^2}{1728}, \end{align*} and multiplying both sides by $1728$ gives \begin{align*} E_4^3-E_6^2=1728\Delta. \end{align*} Thus weight $12$ is the first weight in this list where both an Eisenstein-type form and a nonzero cusp form appear. [/example] The low-weight table illustrates that the whole dimension problem has become a lattice-point count in the generators of weights $4$ and $6$, with cusp forms obtained by multiplying by $\Delta$. The remaining problem is to turn that count into a formula that works uniformly in every weight, including the parity and exceptional congruence cases. The theorem below packages that bookkeeping into the closed dimension formula used throughout the theory. [quotetheorem:4236] [citeproof:4236] This dimension formula turns the ring theorem into a practical computational tool, but it is specific to level one. The parity clause comes from the element $-I$ in the full modular group; for groups where $-I$ is absent or acts differently, odd weights need not vanish for the same reason. The congruence class $k\equiv 2\pmod{12}$ is also not a universal phenomenon, but the result of counting monomials in generators of weights $4$ and $6$. In later chapters, Hecke operators will act on each finite-dimensional space $M_k$ and especially on $S_k=\Delta M_{k-12}$, where the first genuinely new cusp form is $\Delta$ itself. Once dimensions are known, the next problem is to choose natural bases and write forms explicitly in them. That leads directly to Hecke operators, which organize each finite-dimensional space arithmetically rather than just as a vector space of forms. # 5. Dimension Formulas and Explicit Bases Chapters 3 and 4 constructed Eisenstein series and the discriminant form $\Delta$, giving us explicit modular forms of weights $4$, $6$, and $12$. This chapter turns those examples into a counting machine. The central questions are: how many modular forms of each weight are there, how many of them vanish at the cusp, and how can we write down concrete bases without solving functional equations from scratch? ## Multiplication by the Discriminant Form How can a cusp condition be converted into an ordinary holomorphic modular form condition? The point of $\Delta$ is that it vanishes at the cusp to order exactly one and has no zeros on the upper half-plane. Dividing a cusp form by $\Delta$ therefore removes precisely the forced vanishing at infinity while preserving holomorphy away from infinity. [definition: Order At Infinity] For each integer $k$, the order at infinity is the function \begin{align*} \operatorname{ord}_{\infty}:M_k(SL_2(\mathbb Z)) \longrightarrow \mathbb Z_{\ge 0}\cup\{\infty\} \end{align*} defined as follows. If $f \in M_k(SL_2(\mathbb Z))$ has $q$-expansion \begin{align*} f(z) = \sum_{n=0}^{\infty} a_n q^n, \qquad q=e^{2\pi iz}, \end{align*} then \begin{align*} \operatorname{ord}_{\infty}(f) = \min\{n \ge 0 : a_n \ne 0\}, \end{align*} with $\operatorname{ord}_{\infty}(0)=\infty$. [/definition] A modular form is a cusp form exactly when its order at infinity is at least $1$. For the discriminant form, \begin{align*} \Delta(z)=q\prod_{n=1}^{\infty}(1-q^n)^{24}=q-24q^2+252q^3-1472q^4+\cdots, \end{align*} so $\operatorname{ord}_{\infty}(\Delta)=1$. [quotetheorem:4237] [citeproof:4237] The hypotheses on $\Delta$ are doing two separate jobs. Its exact order-one vanishing at infinity removes precisely the cusp condition; if the chosen divisor vanished to order $2$, then dividing a cusp form with first term $q$ would produce a pole at infinity. Its absence of zeros on the upper half-plane prevents new singularities; division by a modular form with an interior zero would only work for cusp forms vanishing to at least the same order at that point. The theorem therefore does not say that every cusp condition can be divided away by an arbitrary cusp form. It is the bridge between cusp forms and all modular forms, and once the dimensions of the spaces $M_k$ are known, the dimensions of $S_k$ follow by subtracting $12$ from the weight. [example: Cusp Forms Of Weight Twelve] Taking $k=12$ in the discriminant division isomorphism gives \begin{align*} S_{12}(SL_2(\mathbb Z)) &=\{\Delta g : g\in M_{0}(SL_2(\mathbb Z))\}\\ &=\Delta M_0(SL_2(\mathbb Z)). \end{align*} Every element of $M_0(SL_2(\mathbb Z))$ is constant, so \begin{align*} M_0(SL_2(\mathbb Z))=\mathbb C\cdot 1. \end{align*} Therefore \begin{align*} S_{12}(SL_2(\mathbb Z)) &=\Delta(\mathbb C\cdot 1)\\ &=\{\Delta(c\cdot 1):c\in\mathbb C\}\\ &=\{c\Delta:c\in\mathbb C\}\\ &=\mathbb C\Delta. \end{align*} Thus $S_{12}(SL_2(\mathbb Z))$ is one-dimensional with basis $\{\Delta\}$. Since \begin{align*} \Delta(z)=q-24q^2+252q^3-\cdots, \end{align*} its first Fourier coefficient is $1$, so $\Delta$ is normalized. If $f\in S_{12}(SL_2(\mathbb Z))$ is normalized, then $f=c\Delta$ for some $c\in\mathbb C$, and comparing first Fourier coefficients gives $1=c\cdot 1$, hence $c=1$. Therefore the unique normalized cusp form of weight $12$ is $\Delta$. [/example] ## Counting Modular Forms For The Full Modular Group What numerical restrictions does modularity impose before we write down any formula? Odd weights vanish for the full modular group because $-I \in SL_2(\mathbb Z)$ acts by $(-1)^k$. In weight $2$ there is also no holomorphic modular form: the natural Eisenstein series $E_2$ is not modular, and the valence formula rules out a nonzero element of $M_2$. [quotetheorem:4238] [citeproof:4238] The assumptions $k\in \mathbb Z$ and full level are both essential. Odd weights vanish because $-I$ belongs to $SL_2(\mathbb Z)$; for groups not containing $-I$ this argument would not apply in the same form. The exceptional residue class is already visible at weight $2$, where the natural candidate $E_2$ is holomorphic but not modular, so the theorem is not a statement about all holomorphic $q$-series with plausible transformation behaviour. The formula says that the ring of full-level modular forms grows by one dimension roughly every twelve weights, and it sets up the next task: finding explicit forms whose number matches this count. [quotetheorem:4239] [citeproof:4239] The lower bound $k\ge 12$ is forced by the first possible cusp form, namely $\Delta$ in weight $12$. For example, a putative nonzero cusp form of weight $10$ would have at least one zero at infinity, leaving too little total zero count under the valence formula. The theorem does not identify individual cusp forms; it only counts them by reducing to ordinary modular forms twelve weights lower. These dimension formulas turn qualitative structure into computable linear algebra, and in the rest of the chapter we use them to certify that explicit lists of monomials are not merely spanning candidates but bases. ## Monomial Bases In Eisenstein Series How do we construct actual modular forms once the dimension is known? The weights of $E_4$ and $E_6$ suggest looking at monomials $E_4^aE_6^b$. Their weights are $4a+6b$, so a basis in weight $k$ should be built from the nonnegative integer solutions of this equation. [definition: Eisenstein Monomial Of Weight K] Let $k$ be an even integer. An Eisenstein monomial of weight $k$ is a product \begin{align*} E_4^aE_6^b \end{align*} with $a,b \in \mathbb Z_{\ge 0}$ and $4a+6b=k$. [/definition] The equation $4a+6b=k$ has no solution when $k$ is odd, and it has fewer solutions than a generic linear equation because both weights are even. The dimension formula says that, except for the same weight-$2$ obstruction, the number of such monomials is exactly the dimension of $M_k$. [quotetheorem:4240] [citeproof:4240] The theorem depends on the dimension formula as an external count; it is not using the graded ring theorem as a black box. The weight equation $4a+6b=k$ is essential: a product such as $E_4E_6$ belongs to weight $10$, not to a neighbouring even weight. The theorem also does not claim uniqueness for arbitrary polynomial expressions in $E_4$ and $E_6$ across different weights; it asserts uniqueness after the total weight has been fixed. This is the practical form of the structure theorem $M_*(SL_2(\mathbb Z))=\mathbb C[E_4,E_6]$, and it gives explicit bases that can now be transferred to cusp spaces. [example: Bases In Weights Twelve Sixteen And Eighteen] For weight $12$, the monomial equation is \begin{align*} 4a+6b=12. \end{align*} Dividing by $2$ gives $2a+3b=6$. Since $a,b\in \mathbb Z_{\ge 0}$, we check the possible values of $b$: \begin{align*} b=0 &\implies 2a=6 \implies a=3,\\ b=1 &\implies 2a=3 \implies a=\frac{3}{2}\notin \mathbb Z,\\ b=2 &\implies 2a=0 \implies a=0, \end{align*} and $b\ge 3$ gives $2a=6-3b<0$. Thus the only solutions are $(a,b)=(3,0)$ and $(0,2)$. By the *Monomial Basis Theorem*, \begin{align*} M_{12}(SL_2(\mathbb Z)) = \operatorname{span}_{\mathbb C}\{E_4^3,E_6^2\}. \end{align*} The cusp dimension formula gives $\dim S_{12}(SL_2(\mathbb Z))=1$, and \begin{align*} \Delta=\frac{E_4^3-E_6^2}{1728} \end{align*} is a nonzero cusp form of weight $12$, so \begin{align*} S_{12}(SL_2(\mathbb Z))=\mathbb C\Delta. \end{align*} For weight $16$, the equation $4a+6b=16$ becomes $2a+3b=8$. The nonnegative integer possibilities are \begin{align*} b=0 &\implies 2a=8 \implies a=4,\\ b=1 &\implies 2a=5 \implies a=\frac{5}{2}\notin \mathbb Z,\\ b=2 &\implies 2a=2 \implies a=1, \end{align*} and $b\ge 3$ gives $2a=8-3b<0$. Hence the solutions are $(4,0)$ and $(1,2)$, so the monomial basis is \begin{align*} \{E_4^4,E_4E_6^2\}. \end{align*} For weight $18$, the equation $4a+6b=18$ becomes $2a+3b=9$. Now \begin{align*} b=0 &\implies 2a=9 \implies a=\frac{9}{2}\notin \mathbb Z,\\ b=1 &\implies 2a=6 \implies a=3,\\ b=2 &\implies 2a=3 \implies a=\frac{3}{2}\notin \mathbb Z,\\ b=3 &\implies 2a=0 \implies a=0, \end{align*} and $b\ge 4$ gives $2a=9-3b<0$. Thus the solutions are $(3,1)$ and $(0,3)$, giving the basis \begin{align*} \{E_4^3E_6,E_6^3\}. \end{align*} These three weights illustrate the same rule: solving $4a+6b=k$ lists the Eisenstein monomials, and the monomial basis theorem turns that list into an actual basis of $M_k(SL_2(\mathbb Z))$. [/example] ## Cusp Bases By Multiplying With Delta Once the bases for $M_k$ are known, how should we choose bases for cusp forms? The discriminant division isomorphism gives the answer with no additional computation: take a basis for $M_{k-12}$ and multiply every element by $\Delta$. [quotetheorem:4241] [citeproof:4241] The condition $k\ge 12$ is necessary because the construction begins with the weight-$12$ form $\Delta$. If $k=10$, the displayed equation would ask for monomials of weight $-2$, and the corresponding cusp space is zero. The theorem does not say that these basis elements are eigenforms for Hecke operators; they are convenient linear algebra coordinates before the Hecke action is diagonalised. This basis is often the most convenient one for computations with cusp forms, because it separates the forced vanishing at the cusp, carried by $\Delta$, from the freely chosen modular form of lower weight. [example: A Basis For S Twenty Four] For $S_{24}(SL_2(\mathbb Z))$, the *Delta Times Monomial Basis For Cusp Forms* says that a basis is obtained by taking all products \begin{align*} \Delta E_4^aE_6^b \end{align*} with $a,b\in \mathbb Z_{\ge 0}$ and \begin{align*} 4a+6b=24-12=12. \end{align*} Solving this equation, divide by $2$ to get \begin{align*} 2a+3b=6. \end{align*} Since $b\ge 0$ and $2a=6-3b\ge 0$, we must have $b\le 2$. Checking these possibilities gives \begin{align*} b=0 &\implies 2a=6 \implies a=3,\\ b=1 &\implies 2a=3 \implies a=\frac{3}{2}\notin \mathbb Z,\\ b=2 &\implies 2a=0 \implies a=0. \end{align*} Thus the only nonnegative integer solutions are $(a,b)=(3,0)$ and $(a,b)=(0,2)$. Therefore the corresponding cusp basis is \begin{align*} \{\Delta E_4^3,\Delta E_6^2\}. \end{align*} Equivalently, division by $\Delta$ sends these two forms to the basis elements $E_4^3$ and $E_6^2$ of $M_{12}(SL_2(\mathbb Z))$, so the two displayed products account for all of $S_{24}(SL_2(\mathbb Z))$. [/example] This example also shows why a basis is not the same thing as a preferred basis. The two displayed elements are natural from the dimension formula, but later arithmetic questions will ask for combinations that interact well with Hecke operators. [remark: Normalized Forms And Later Hecke Theory] A nonzero cusp form $f=\sum_{n\ge 1}a_nq^n$ is normalized when $a_1=1$. In weight $12$, the cusp space is one-dimensional and $\Delta=q-24q^2+\cdots$, so $\Delta$ is the unique normalized cusp form of weight $12$. In higher-dimensional cusp spaces, normalization alone does not choose a unique form; Hecke eigenform conditions will supply the extra structure in later chapters. [/remark] The dimension formulas identified the spaces in which the key objects live, but they do not distinguish one form from another. Hecke operators provide the extra arithmetic structure needed to pick out eigenforms and to understand how modular forms decompose under correspondences. # 6. Hecke Operators: Definitions and First Properties Hecke operators are the first point in the course where modular forms are organized by arithmetic correspondences rather than only by transformation laws and Fourier expansions. The guiding question is this: how can one average a modular form over finite-index modifications of lattices in a way that preserves modularity and produces concrete arithmetic information in the coefficients? This chapter introduces the classical operators $T_n$, proves their action on $q$-expansions, and checks that they preserve both modular forms and cusp forms. ## Double Cosets and the Classical Formula The problem is to define an averaging operator that is intrinsic to the modular group but still computable from Fourier series. A matrix of determinant $n$ sends a lattice to a sublattice of index $n$, and the Hecke operator should sum over the finitely many such possibilities modulo change of basis. For the full modular group this finite correspondence is encoded by determinant-$n$ matrices modulo the change of basis on the target lattice. A naive sum over all matrices in $M_2(\mathbb Z)$ of determinant $n$ would be infinite and would count the same sublattice many times, because left multiplication by $SL_2(\mathbb Z)$ only changes the chosen basis. The first step is therefore to replace matrices by left-coset representatives inside the determinant-$n$ double-coset correspondence. [definition: Integral Hecke Coset Representatives] Let $n \in \mathbb N$ and let \begin{align*} \Gamma &= SL_2(\mathbb Z),\\ \Delta_n &= \{\alpha \in M_2(\mathbb Z) : \det \alpha = n\}. \end{align*} The determinant-$n$ integral Hecke correspondence is represented by the left-coset space $\Gamma \backslash \Delta_n$ for the left action of $\Gamma$ on $\Delta_n$ by multiplication. [/definition] The averaging operator will be useful only if the finite correspondence can be written in coordinates that interact well with Fourier expansions. Arbitrary determinant-$n$ matrices obscure the powers of $q$ that appear after substitution. What is needed is a normal form that separates scaling from translation. Triangular representatives do exactly this, and the finite sum over translations is what will later produce the divisibility condition in the coefficient formula. A Hecke average over determinant-$n$ matrices is only well defined after choosing one representative for each relevant left coset. If a representative is missed, the correspondence is incomplete; if one is counted twice, the averaging operator has the wrong multiplicities. The triangular normal form is therefore the finite coordinate system needed before any formula for the operator can be trusted. [quotetheorem:4242] [citeproof:4242] The determinant hypothesis is essential: if determinant is not fixed, there is no finite list of triples $(a,b,d)$, and the resulting sum would not define a finite averaging operator. The theorem does not say that the displayed matrices form a subgroup, nor that right multiplication by $\Gamma$ preserves this particular triangular list; it only gives one finite set of representatives for left cosets. Its usefulness is computational, because the variable $b$ will later be summed over a complete residue system modulo $d$. The coset representatives turn the correspondence into an explicit finite sum of pullbacks $z\mapsto (az+b)/d$. To make this sum act on weight-$k$ functions with the conventional scaling, one must include a normalization depending on $n$ and $d$; otherwise the later Fourier-coefficient formula would have nonstandard weights. This gives the classical operator attached to the determinant-$n$ correspondence. [definition: Classical Hecke Operator] Let $\mathbb H=\{z\in\mathbb C:\operatorname{Im}(z)>0\}$, let $k \in \mathbb Z$, and let $n \in \mathbb N$. The classical Hecke operator of weight $k$ and index $n$ is the map \begin{align*} T_n : \operatorname{Hol}(\mathbb H) \to \operatorname{Hol}(\mathbb H) \end{align*} defined by \begin{align*} (T_n f)(z) = n^{k-1}\sum_{\substack{ad=n\\0\le b<d}} d^{-k} f\left(\frac{az+b}{d}\right). \end{align*} [/definition] With $ad=n$, the same formula can be written as \begin{align*} (T_n f)(z)=\sum_{\substack{ad=n\\0\le b<d}} \frac{a^{k-1}}{d} f\left(\frac{az+b}{d}\right). \end{align*} This is the convention used in these notes. Some books place different powers of $n$ in the definition; those choices shift eigenvalues by predictable scalar factors, and in particular may prevent normalized eigenvalues from being literally equal to normalized Fourier coefficients without rescaling. To connect this geometric average with arithmetic data, we must know how it transforms the coefficients of a $q$-expansion. The triangular representatives make the calculation finite and force only those terms whose exponents satisfy a divisibility condition to survive the sum over translations. The important test is whether the analytic definition of $T_n$ can be read purely from Fourier coefficients. Establishing that formula is what turns Hecke operators from geometric averages into arithmetic operators on sequences. This coefficient formula is needed before Hecke operators can be used in computations: without it, $T_n$ is still defined by evaluating $f$ at many transformed points, but its effect on the arithmetic data $a_m(f)$ remains opaque. The theorem isolates exactly which old coefficients contribute to each new coefficient and with what weights. [quotetheorem:4243] [citeproof:4243] This formula is the computational heart of Hecke theory. The hypothesis that $f$ has a Fourier expansion with no negative powers matters: if negative powers were present, the same calculation would not prove holomorphy at the cusp for $T_n f$. The theorem does not assert multiplicativity of the operators $T_n$ themselves; it only gives their action on coefficients, and the algebra relations require a separate argument in the next chapter. What it gives now is the bridge from geometric averaging to arithmetic recursions among Fourier coefficients. [example: Prime Hecke Coefficient Formula] Let $p$ be prime and write $f(z)=\sum_{r=0}^{\infty}a_r(f)q^r$. By the *Hecke Formula on Coefficients* with $n=p$, \begin{align*} a_m(T_p f) &=\sum_{d\mid (m,p)} d^{k-1}a_{mp/d^2}(f). \end{align*} Since $p$ is prime, the positive divisors of $p$ are $1$ and $p$. If $p\nmid m$, then $(m,p)=1$, so only $d=1$ contributes: \begin{align*} a_m(T_p f) &=1^{k-1}a_{mp/1^2}(f)\\ &=a_{pm}(f). \end{align*} If $p\mid m$, then the common divisors are $1$ and $p$, so \begin{align*} a_m(T_p f) &=1^{k-1}a_{mp/1^2}(f)+p^{k-1}a_{mp/p^2}(f)\\ &=a_{pm}(f)+p^{k-1}a_{m/p}(f). \end{align*} Combining the two cases gives \begin{align*} a_m(T_p f)=a_{pm}(f)+p^{k-1}a_{m/p}(f), \end{align*} with $a_{m/p}(f)=0$ when $p\nmid m$. Thus $T_p$ combines the coefficient at index $pm$ with the coefficient at index $m/p$, and the backward-index term appears exactly for indices divisible by $p$. [/example] ## Preservation of Modular Forms and Cusp Forms The next question is structural: after averaging over these determinant-$n$ matrices, do we remain inside the finite-dimensional spaces $M_k$ and $S_k$? This is not automatic, because the summands are translates by matrices that usually do not lie in $SL_2(\mathbb Z)$, so a single summand need not be modular for the original group. The point of the double-coset construction is that the failure of one summand to be invariant is repaired by permutation of the whole finite family. [quotetheorem:4244] [citeproof:4244] The theorem has two parts hidden inside it: invariance under the modular group and regularity at the cusp. The full modular group hypothesis is being used through the chosen coset decomposition; for congruence subgroups the statement remains true only after using the corresponding level-compatible Hecke correspondences. The theorem does not say that $T_n$ preserves every normalization of a form, since constants and first coefficients may be rescaled. It does guarantee that the finite-dimensional space $M_k(SL_2(\mathbb Z))$ is a legitimate linear-algebra setting for Hecke theory. For cusp forms there is one further boundary condition. A modular form can be holomorphic at the cusp without vanishing there, so preservation of $M_k$ alone is not enough for $S_k$. The coefficient formula isolates this issue in the constant term. The point is to check that the Hecke averaging process does not create a new constant term from positive Fourier coefficients. If the first coefficient at the cusp remains zero after applying $T_n$, then the boundary vanishing condition is preserved along with modularity. This is the extra input needed before the cusp space can be treated as an invariant space for Hecke operators. [quotetheorem:4245] [citeproof:4245] This preservation result is what allows the later spectral theory: the vector spaces $M_k$ and $S_k$ carry families of linear operators $T_n$. The hypothesis that $f$ is cuspidal is used exactly through $a_0(f)=0$; for instance an Eisenstein series has nonzero constant term and is not made cuspidal by applying $T_n$. The theorem does not identify eigenvectors or eigenvalues, but it creates the invariant subspace on which such questions make sense. In later chapters we will exploit commutation relations and diagonalize these operators using the Petersson inner product. [example: Hecke Operator on an Eisenstein Series] Let $k\ge 4$ be even and put $s=k-1$. Write \begin{align*} E_k(z)=1-\frac{2k}{B_k}\sum_{m=1}^{\infty}\sigma_s(m)q^m, \qquad \sigma_s(m)=\sum_{d\mid m}d^s. \end{align*} Thus $a_0(E_k)=1$ and, for $m\ge 1$, \begin{align*} a_m(E_k)=-\frac{2k}{B_k}\sigma_s(m). \end{align*} For a prime $p$, the prime coefficient formula gives \begin{align*} a_m(T_pE_k)=a_{pm}(E_k)+p^s a_{m/p}(E_k), \end{align*} where $a_{m/p}(E_k)=0$ if $p\nmid m$. First consider the constant term. Since $a_0(E_k)=1$, the coefficient formula at $m=0$ gives \begin{align*} a_0(T_pE_k) &=1^s a_0(E_k)+p^s a_0(E_k)\\ &=1+p^s\\ &=(1+p^s)a_0(E_k). \end{align*} Now let $m\ge 1$. If $p\nmid m$, then every divisor of $pm$ is either a divisor $d$ of $m$ or $pd$ with $d\mid m$, so \begin{align*} \sigma_s(pm) &=\sum_{d\mid m}d^s+\sum_{d\mid m}(pd)^s\\ &=\sum_{d\mid m}d^s+p^s\sum_{d\mid m}d^s\\ &=(1+p^s)\sigma_s(m). \end{align*} Since $a_{m/p}(E_k)=0$ in this case, \begin{align*} a_m(T_pE_k) &=-\frac{2k}{B_k}\sigma_s(pm)\\ &=-\frac{2k}{B_k}(1+p^s)\sigma_s(m)\\ &=(1+p^s)a_m(E_k). \end{align*} If $p\mid m$, write $m=p^u r$ with $u\ge 1$ and $p\nmid r$. Then \begin{align*} \sigma_s(m)&=\left(\sum_{i=0}^{u}p^{is}\right)\sigma_s(r),\\ \sigma_s(pm)&=\left(\sum_{i=0}^{u+1}p^{is}\right)\sigma_s(r),\\ \sigma_s(m/p)&=\left(\sum_{i=0}^{u-1}p^{is}\right)\sigma_s(r). \end{align*} Therefore \begin{align*} \sigma_s(pm)+p^s\sigma_s(m/p) &=\left(\sum_{i=0}^{u+1}p^{is}\right)\sigma_s(r) +p^s\left(\sum_{i=0}^{u-1}p^{is}\right)\sigma_s(r)\\ &=\left(\sum_{i=0}^{u+1}p^{is}+\sum_{i=1}^{u}p^{is}\right)\sigma_s(r)\\ &=\left(1+2\sum_{i=1}^{u}p^{is}+p^{(u+1)s}\right)\sigma_s(r), \end{align*} while \begin{align*} (1+p^s)\sigma_s(m) &=(1+p^s)\left(\sum_{i=0}^{u}p^{is}\right)\sigma_s(r)\\ &=\left(\sum_{i=0}^{u}p^{is}+\sum_{i=1}^{u+1}p^{is}\right)\sigma_s(r)\\ &=\left(1+2\sum_{i=1}^{u}p^{is}+p^{(u+1)s}\right)\sigma_s(r). \end{align*} Hence $\sigma_s(pm)+p^s\sigma_s(m/p)=(1+p^s)\sigma_s(m)$, and so \begin{align*} a_m(T_pE_k) &=-\frac{2k}{B_k}\sigma_s(pm) +p^s\left(-\frac{2k}{B_k}\sigma_s(m/p)\right)\\ &=-\frac{2k}{B_k}\left(\sigma_s(pm)+p^s\sigma_s(m/p)\right)\\ &=-\frac{2k}{B_k}(1+p^s)\sigma_s(m)\\ &=(1+p^s)a_m(E_k). \end{align*} All Fourier coefficients of $T_pE_k$ are therefore $(1+p^{k-1})$ times the corresponding coefficients of $E_k$, so \begin{align*} T_pE_k=(1+p^{k-1})E_k. \end{align*} Thus the Eisenstein series $E_k$ is a Hecke eigenform for $T_p$ with eigenvalue $1+p^{k-1}$ under this normalization. [/example] ## Normalization Conventions and Prime Operators The final question in this chapter is how the abstract definition should be normalized for arithmetic use. If the operator is scaled differently, the same eigenspaces may appear, but the eigenvalues no longer coincide with the normalized Fourier coefficients without an additional factor. The convention chosen above makes the formula for prime operators especially transparent and prepares the eigenform normalization used in the construction of $L$-functions. [definition: Normalized Hecke Eigenform] A nonzero modular form $f\in M_k(SL_2(\mathbb Z))$ is a normalized Hecke eigenform if \begin{align*} T_n f=\lambda_n f \quad \text{for all } n\in\mathbb N, \qquad a_1(f)=1. \end{align*} [/definition] For a normalized eigenform, the coefficient formula at $m=1$ gives \begin{align*} \lambda_n a_1(f)=a_1(T_nf)=a_n(f). \end{align*} Thus $\lambda_n=a_n(f)$. This is why the normalization $a_1(f)=1$ is so effective: the Hecke eigenvalues become the Fourier coefficients themselves. For computations, the general coefficient formula is still too broad because it sums over all common divisors of $m$ and $n$. The most important local case is when the index is a prime: then the divisor sum has only two possible contributions, giving a simple recurrence that can be iterated in examples and later assembled into Euler products. We therefore need the prime-index specialization as a standalone tool. It is the local rule that will be used repeatedly when computing examples and when translating Hecke eigenvalue relations into Euler factors. [quotetheorem:4246] [citeproof:4246] This prime formula is the one most often used in calculations. The primality hypothesis is essential for the two-term shape: if $n$ has several prime factors, all common divisors of $m$ and $n$ contribute, so the formula has more than two local terms. The theorem does not say that $T_p$ has eigenvectors in every space by itself, only how it acts once a Fourier expansion is known. It also foreshadows the Euler product: the coefficient at $pm$ and the coefficient at $m/p$ are the two local directions at the prime $p$. [example: Computing the First Terms of T Two Delta] Let \begin{align*} \Delta(z)=\sum_{m=1}^{\infty}\tau(m)q^m =q-24q^2+252q^3-1472q^4+4830q^5-6048q^6+\cdots \end{align*} be the discriminant cusp form of weight $12$, so $\tau(m)=a_m(\Delta)$ and $a_1(\Delta)=1$. For $p=2$, the *Prime Hecke Operator Formula* gives \begin{align*} a_m(T_2\Delta)=a_{2m}(\Delta)+2^{12-1}a_{m/2}(\Delta) =a_{2m}(\Delta)+2^{11}a_{m/2}(\Delta), \end{align*} with $a_{m/2}(\Delta)=0$ when $2\nmid m$. For $m=1$, since $2\nmid 1$, the second term is zero: \begin{align*} a_1(T_2\Delta) &=a_2(\Delta)+2^{11}a_{1/2}(\Delta)\\ &=-24+2^{11}\cdot 0\\ &=-24. \end{align*} For $m=2$, the index $m/2=1$ is integral, so \begin{align*} a_2(T_2\Delta) &=a_4(\Delta)+2^{11}a_1(\Delta)\\ &=-1472+2048\cdot 1\\ &=576. \end{align*} For $m=3$, since $2\nmid 3$, the second term is again zero: \begin{align*} a_3(T_2\Delta) &=a_6(\Delta)+2^{11}a_{3/2}(\Delta)\\ &=-6048+2^{11}\cdot 0\\ &=-6048. \end{align*} Thus \begin{align*} T_2\Delta=-24q+576q^2-6048q^3+\cdots. \end{align*} By *[Hecke Operators Preserve Cusp Forms](/theorems/4245)*, $T_2\Delta\in S_{12}(SL_2(\mathbb Z))$. Since $S_{12}(SL_2(\mathbb Z))$ is one-dimensional and $\Delta$ spans it, there is a scalar $c$ such that $T_2\Delta=c\Delta$. Comparing the coefficient of $q$ gives \begin{align*} c\,a_1(\Delta)=a_1(T_2\Delta), \end{align*} so \begin{align*} c\cdot 1=-24, \qquad c=-24. \end{align*} Therefore \begin{align*} T_2\Delta=-24\Delta. \end{align*} The eigenvalue is the second Fourier coefficient of the normalized form, namely $\tau(2)=a_2(\Delta)=-24$. [/example] The recurrence also explains why prime-index operators contain the essential local information; composite indices can later be reconstructed from the Hecke relations. [remark: Why Prime Operators Are Enough Later] Chapter 7 proves that the full Hecke algebra is generated by the operators $T_p$ together with their powers, subject to the multiplicative relations among the $T_n$. For this reason the prime formula is not merely a special case; it is the local building block for the Euler product of a Hecke eigenform. [/remark] The chapter has established the basic mechanism: determinant-$n$ correspondences produce endomorphisms of $M_k$ and $S_k$, and their effect on $q$-expansions is explicit. From here the course turns to the algebra generated by the $T_n$, where commutativity and multiplicativity turn these operators into arithmetic structure rather than isolated linear maps. The algebra of Hecke operators turns the spaces of modular forms into a system governed by commuting endomorphisms. With eigenforms in hand, the next step is to put the operators into a Hilbert-space framework so that orthogonality and self-adjointness can be used to isolate the spectral theory. # 7. Hecke Algebra and Eigenforms This chapter turns the abstract construction of Hecke operators into an algebra acting on spaces of modular forms. The guiding question is how the operators $T_n$ interact with each other, and why their algebraic relations are strong enough to organize cusp forms into eigenforms. The main payoff is that, after normalizing an eigenform, its Fourier coefficients are not auxiliary data: they are the Hecke eigenvalues themselves. This chapter therefore moves from geometric correspondences to linear algebra, and then back to arithmetic information in Fourier coefficients. ## Multiplication of Hecke Operators The first problem is to understand whether applying two Hecke correspondences in succession has a predictable effect. Since $T_n$ was defined by summing over representatives of integral matrices of determinant $n$, composing $T_m$ and $T_n$ amounts to counting the same lattices, or equivalently the same double cosets, in more than one way. [quotetheorem:4247] [citeproof:4247] This formula is the structural identity behind all later coefficient relations. The hypothesis $k\ge 2$ keeps us in the standard holomorphic modular-form setting used throughout the course, and the level-one group $\operatorname{SL}_2(\mathbb Z)$ is what makes the double-coset representatives and the displayed normalization of $T_n$ match the coefficient formula used below. The theorem does not say that $T_mT_n$ is always a single Hecke operator: common divisors of $m$ and $n$ produce correction terms, so non-coprime indices remember extra overlap between the two correspondences. In the special case $\gcd(m,n)=1$, only $d=1$ occurs, so the relation becomes $T_mT_n=T_{mn}$. [example: Coprime Hecke Operators] Let $m,n\in\mathbb N$ with $\gcd(m,n)=1$. In the [Hecke multiplicativity formula](/theorems/4247), \begin{align*} T_mT_n=\sum_{d\mid \gcd(m,n)}d^{k-1}T_{mn/d^2}, \end{align*} the only positive divisor of $\gcd(m,n)=1$ is $d=1$. Hence \begin{align*} T_mT_n &=1^{k-1}T_{mn/1^2}\\ &=T_{mn}. \end{align*} Now let $f$ be a simultaneous Hecke eigenform, so $T_r f=\lambda_r f$ for every $r\ge 1$. Applying the operator identity to $f$ gives \begin{align*} T_mT_n f&=T_{mn}f. \end{align*} The left-hand side is \begin{align*} T_mT_n f &=T_m(\lambda_n f)\\ &=\lambda_n T_m f\\ &=\lambda_n\lambda_m f\\ &=\lambda_m\lambda_n f, \end{align*} using linearity of $T_m$ and the eigenvalue equations for $T_n$ and $T_m$. The right-hand side is \begin{align*} T_{mn}f=\lambda_{mn}f. \end{align*} Therefore \begin{align*} \lambda_m\lambda_n f=\lambda_{mn}f. \end{align*} Since an eigenform is nonzero, this implies \begin{align*} \lambda_m\lambda_n=\lambda_{mn}. \end{align*} Thus coprime indices multiply at the level of Hecke eigenvalues, which is the operator-theoretic source of multiplicativity for normalized Fourier coefficients. [/example] The prime-power part of the same formula is just as important. Taking $m=p$ and $n=p^r$ gives the two divisors $d=1,p$ when $r\ge 1$. Coprime multiplicativity does not determine the eigenvalues at powers of a single prime. Those coefficients require a local relation, because multiplying by $T_p$ can either raise the prime power or lower it through the correction term in the Hecke product formula. The recurrence below isolates this prime-power behavior and is the finite algebraic shadow of an Euler factor. [quotetheorem:4248] [citeproof:4248] This recurrence turns the sequence of eigenvalues at powers of $p$ into a second-order recursion. The primality of $p$ is essential for this two-term shape: the divisor structure of $p$ and $p^r$ has only the common divisors $1$ and $p$, whereas composite indices would introduce more correction terms. The power $p^{k-1}$ records the weight of the modular form and cannot be removed without changing the normalization of the Hecke operators. This is the local calculation that later becomes the Euler factor of the $L$-function of an eigenform. ## The Hecke Algebra and Simultaneous Eigenforms The next question is whether the operators $T_n$ can be studied as one compatible algebra rather than as unrelated endomorphisms. The multiplication formula first shows that they commute. The analytic diagonalization theorem will come later, after the Petersson inner product has been introduced; for now the algebra explains why simultaneous eigenforms are the right objects to look for. [definition: Hecke Algebra] Let $k\ge 2$. The Hecke algebra $\mathbb T_k$ is the subalgebra of $\operatorname{End}_{\mathbb C}(M_k(\operatorname{SL}_2(\mathbb Z)))$ generated by the operators $T_n$ for $n\ge 1$. Its cuspidal Hecke algebra is the image of the same algebra in $\operatorname{End}_{\mathbb C}(S_k(\operatorname{SL}_2(\mathbb Z)))$. [/definition] The identity $T_1=I$ makes $\mathbb T_k$ a unital algebra. Since every product of generators can be rewritten as a finite linear combination of the same generators, this algebra is controlled by the multiplication formula rather than by arbitrary compositions. To justify studying simultaneous eigenforms, we need more than closure under multiplication: the operators must be compatible enough that eigenspaces for one operator are not disrupted by another. Commutativity is the algebraic condition that makes this compatibility possible before any analytic diagonalization theorem is introduced. [quotetheorem:4249] [citeproof:4249] This commutativity is the first reason the Hecke operators behave like a family of compatible measurements rather than unrelated endomorphisms. Without it, an eigenvector for one operator need not remain inside an eigenspace after applying another, so simultaneous eigenforms would not be a natural expectation. The theorem only gives an algebraic compatibility statement; it does not by itself provide diagonalizability or orthogonality. For that stronger conclusion, the cuspidal space also uses the Petersson inner product and the self-adjointness of the Hecke operators. [example: The One-Dimensional Cusp Space in Weight Twelve] The cusp space $S_{12}(\operatorname{SL}_2(\mathbb Z))$ is one-dimensional, with basis vector the discriminant modular form \begin{align*} \Delta(z) &=q\prod_{n=1}^{\infty}(1-q^n)^{24}\\ &=\sum_{n=1}^{\infty}\tau(n)q^n. \end{align*} The first term of the product expansion is $q$, so the coefficient of $q$ is $\tau(1)=1$; in particular $\Delta\ne 0$. Fix $m\ge 1$. The Hecke operator $T_m$ acts on $S_{12}(\operatorname{SL}_2(\mathbb Z))$, so \begin{align*} T_m\Delta\in S_{12}(\operatorname{SL}_2(\mathbb Z)). \end{align*} Because $S_{12}(\operatorname{SL}_2(\mathbb Z))=\mathbb C\Delta$, there is a scalar $\lambda_m\in\mathbb C$ such that \begin{align*} T_m\Delta=\lambda_m\Delta. \end{align*} This holds for every $m\ge 1$, so $\Delta$ is an eigenvector for every Hecke operator $T_m$ at once. Thus $\Delta$ is a simultaneous Hecke eigenform, and the eigenvalue of $T_m$ is exactly the scalar $\lambda_m$ defined by $T_m\Delta=\lambda_m\Delta$. [/example] This example explains why the Ramanujan tau function is the first natural test case for the theory. Its arithmetic properties are forced by the general Hecke algebra identities. ## Normalized Eigenforms and Fourier Coefficients The final question in this chapter is how to read Hecke eigenvalues from the $q$-expansion. For cusp forms, normalization fixes the first nonzero coefficient, and the explicit action of $T_n$ on Fourier expansions then identifies all coefficients with eigenvalues. [definition: Normalized Eigenform] A normalized eigenform in $S_k(\operatorname{SL}_2(\mathbb Z))$ is a simultaneous Hecke eigenform \begin{align*} f(z)=\sum_{r=1}^{\infty}a_r(f)q^r \end{align*} with $a_1(f)=1$. [/definition] The normalization removes scalar ambiguity. If $f$ is an eigenform and $cf$ is a nonzero scalar multiple, the two forms have the same Hecke eigenvalues, so requiring $a_1(f)=1$ chooses a preferred representative whenever $a_1(f)\ne 0$. After this scalar choice, the remaining question is whether the eigenvalues are visible directly in the Fourier expansion. The coefficient formula for $T_n$ shows that the first coefficient of $T_nf$ reads off the $n$th coefficient of $f$, so a normalized simultaneous eigenform should encode all Hecke eigenvalues in its $q$-series. [quotetheorem:4251] [citeproof:4251] The theorem changes the meaning of the Fourier expansion. The normalization $a_1(f)=1$ is essential: if $g=cf$ with $c\ne 0$, then $g$ has the same Hecke eigenvalues as $f$, but its coefficients are multiplied by $c$, so $a_n(g)$ equals $c\lambda_n$ rather than $\lambda_n$. The restriction to cusp forms matches the course's use of the Petersson inner product and avoids the extra Eisenstein contribution that appears in the full modular-form space. For normalized eigenforms, the analytic expansion at the cusp is at the same time a list of simultaneous eigenvalues for the Hecke algebra. Once coefficients are identified with eigenvalues, the algebraic relations among the operators should become arithmetic relations among Fourier coefficients. This is the main payoff of normalization: the Hecke product formula can be read directly in the $q$-expansion. The next statement translates the coprime product and prime-power recurrence into coefficient identities. [quotetheorem:4252] [citeproof:4252] These relations give a practical recipe for recovering many coefficients from a much smaller set of data. First compute the coefficients at prime powers using the recurrence, then multiply across coprime prime-power factors using the first formula. The coprimality hypothesis is essential in the multiplicative step: when the same prime divides both indices, the correction term $p^{k-1}a_{p^{r-1}}(f)$ appears and ordinary complete multiplicativity fails. [example: Multiplicativity of the Ramanujan Tau Function] For the discriminant modular form \begin{align*} \Delta(z) &=q\prod_{r=1}^{\infty}(1-q^r)^{24}\\ &=\sum_{n\ge 1}\tau(n)q^n, \end{align*} the first displayed factor is $q$, so the coefficient of $q$ is $\tau(1)=1$. Thus $\Delta$ satisfies the normalization condition $a_1(\Delta)=1$. Since $S_{12}(\operatorname{SL}_2(\mathbb Z))$ is one-dimensional and spanned by the nonzero cusp form $\Delta$, every Hecke operator $T_\ell$ sends $\Delta$ to a scalar multiple of itself: \begin{align*} T_\ell\Delta=\lambda_\ell\Delta \end{align*} for some $\lambda_\ell\in\mathbb C$. Hence $\Delta$ is a normalized simultaneous Hecke eigenform. Let $m,n\in\mathbb N$ with $\gcd(m,n)=1$. By *Fourier Coefficients as Hecke Eigenvalues*, applied to the normalized eigenform $\Delta$, we have \begin{align*} \tau(r)=a_r(\Delta)=\lambda_r \end{align*} for every $r\ge 1$. By *Multiplicativity of Normalized Eigenform Coefficients*, coprime indices satisfy \begin{align*} a_{mn}(\Delta)=a_m(\Delta)a_n(\Delta). \end{align*} Substituting $a_r(\Delta)=\tau(r)$ into this identity gives \begin{align*} \tau(mn) &=a_{mn}(\Delta)\\ &=a_m(\Delta)a_n(\Delta)\\ &=\tau(m)\tau(n). \end{align*} Thus the coprime multiplicativity of $\tau$ follows from the Hecke algebra and the normalization of $\Delta$, rather than from expanding the infinite product defining $\Delta$. [/example] The coprime formula is only half of multiplicativity. Powers of a single prime require the Hecke recurrence, and the tau function gives the first concrete test. [example: Prime-Power Recurrence for the Tau Function] Again take $\Delta\in S_{12}(\operatorname{SL}_2(\mathbb Z))$, so its weight is $k=12$, and write \begin{align*} \Delta(z)=\sum_{n\ge 1}\tau(n)q^n. \end{align*} Since $\Delta$ is a normalized Hecke eigenform, the prime-power recurrence from *Multiplicativity of Normalized Eigenform Coefficients* applies to its coefficients: \begin{align*} a_{p^{r+1}}(\Delta) &=a_p(\Delta)a_{p^r}(\Delta)-p^{k-1}a_{p^{r-1}}(\Delta). \end{align*} Substituting $k=12$ gives $k-1=11$, so for every prime $p$ and every $r\ge 1$, \begin{align*} \tau(p^{r+1}) &=a_{p^{r+1}}(\Delta)\\ &=a_p(\Delta)a_{p^r}(\Delta)-p^{11}a_{p^{r-1}}(\Delta)\\ &=\tau(p)\tau(p^r)-p^{11}\tau(p^{r-1}). \end{align*} In the special case $r=1$, this becomes \begin{align*} \tau(p^2) &=\tau(p)\tau(p^1)-p^{11}\tau(p^0)\\ &=\tau(p)^2-p^{11}\tau(1). \end{align*} The normalization of $\Delta$ gives $\tau(1)=1$, hence \begin{align*} \tau(p^2)=\tau(p)^2-p^{11}. \end{align*} Thus the coefficient at $p^2$ is forced by the coefficient at $p$ together with the weight-dependent correction term $p^{11}$. [/example] The chapter has reduced several arithmetic identities to linear algebra plus the double-coset multiplication formula. The next stage of the course uses these identities to factor Dirichlet series into Euler products and to attach analytic $L$-functions to eigenforms. The Hecke algebra has supplied commuting operators and coefficient identities. The next chapter adds the Petersson inner product, which explains when these algebraic eigendirections can be chosen orthogonally and how projections onto them should be computed. # 8. Petersson Inner Product and Orthogonality This chapter puts a Hilbert-space structure on the finite-dimensional space of cusp forms and uses it to explain why Hecke eigenforms behave like an orthogonal basis. In Chapters 6 and 7, Hecke operators were introduced algebraically through double cosets, Fourier coefficients, and their multiplicative relations. The new question is analytic: can we choose eigenforms in a way that separates their Fourier expansions and makes the Hecke action diagonal? The answer is the Petersson inner product. Its definition integrates modular-invariant data over a fundamental domain, so the geometry of $SL_2(\mathbb Z)\backslash \mathbb H$ enters directly. The rapid decay of cusp forms at infinity is exactly what makes the integral converge. ## Measuring Cusp Forms on the Modular Surface What should the inner product of two modular forms be? A first attempt, integrating $f(z)\overline{g(z)}$ over a fundamental domain, is not invariant under change of representatives because modular forms transform with weight. The correction is to insert a power of $y=\operatorname{Im}(z)$ and use the invariant hyperbolic measure. [definition: Petersson Inner Product] Let $k \ge 0$ be an integer, and let $f,g \in S_k(SL_2(\mathbb Z))$. The Petersson inner product of $f$ and $g$ is \begin{align*} (f,g)_{\mathrm{Pet}} = \int_{\mathcal F} f(z)\overline{g(z)}y^k\,\frac{dx\,dy}{y^2}, \end{align*} where $z=x+iy$ and $\mathcal F$ is any fundamental domain for the action of $SL_2(\mathbb Z)$ on $\mathbb H$. [/definition] The factor $y^k$ cancels the weight-$k$ transformation law, while $dx\,dy/y^2$ is invariant under fractional linear transformations. Thus the integrand descends to a well-defined function times measure on the quotient modular surface. Even with the factor $y^k$, the definition would fail if changing variables by a modular transformation changed the weighted measure. The obstruction is simultaneous: the modular factors from $f$ and $g$, the transformation of $y$, and the Jacobian of the fractional [linear map](/page/Linear%20Map) must cancel in one expression. This invariance is what lets the integral descend from a chosen region in $\mathbb H$ to the quotient. [quotetheorem:4253] [citeproof:4253] This calculation explains why the definition is independent of the chosen fundamental domain, provided the integral converges. The convergence issue is concentrated at the cusp, because the usual standard fundamental domain has finite hyperbolic area away from infinity. [example: Convergence Near the Cusp] Let $f,g \in S_k(SL_2(\mathbb Z))$ have Fourier expansions \begin{align*} f(z)=\sum_{n\ge 1}a_n q^n, \qquad g(z)=\sum_{n\ge 1}b_n q^n, \end{align*} where $q=e^{2\pi i z}=e^{2\pi i x}e^{-2\pi y}$. Since $f$ and $g$ are cusp forms, their $q$-expansions have no constant term. Thus, in the local coordinate $q$ at the cusp, we may write \begin{align*} f(z)=qF(q), \qquad g(z)=qG(q), \end{align*} where $F$ and $G$ are holomorphic near $q=0$. Choose $Y_0$ large enough that $|q|\le e^{-2\pi Y_0}$ lies inside this coordinate neighbourhood, and set \begin{align*} M_f=\sup_{|q|\le e^{-2\pi Y_0}}|F(q)|,\qquad M_g=\sup_{|q|\le e^{-2\pi Y_0}}|G(q)|. \end{align*} These suprema are finite because $F$ and $G$ are continuous on a compact disk. For $|x|\le 1/2$ and $y\ge Y_0$, we have $|q|=e^{-2\pi y}$, so \begin{align*} |f(z)|=|qF(q)|\le M_f e^{-2\pi y},\qquad |g(z)|=|qG(q)|\le M_g e^{-2\pi y}. \end{align*} Therefore \begin{align*} |f(z)\overline{g(z)}| = |f(z)|\,|g(z)| \le M_fM_g e^{-4\pi y}. \end{align*} The contribution of the cusp region $|x|\le 1/2$, $y\ge Y_0$ to the Petersson integral is bounded by \begin{align*} \int_{Y_0}^{\infty}\int_{-1/2}^{1/2} |f(z)\overline{g(z)}|y^k\,\frac{dx\,dy}{y^2} &\le \int_{Y_0}^{\infty}\int_{-1/2}^{1/2} M_fM_g e^{-4\pi y}y^{k-2}\,dx\,dy\\ &= M_fM_g\int_{Y_0}^{\infty}e^{-4\pi y}y^{k-2}\,dy, \end{align*} because the interval $[-1/2,1/2]$ has length $1$. For fixed $k$, the factor $y^{k-2}$ grows at most polynomially, while $e^{-4\pi y}$ decays exponentially; explicitly, after increasing $Y_0$ if necessary, $y^{\max(k-2,0)}\le e^{2\pi y}$ for all $y\ge Y_0$, hence \begin{align*} e^{-4\pi y}y^{k-2}\le e^{-2\pi y}. \end{align*} Since \begin{align*} \int_{Y_0}^{\infty}e^{-2\pi y}\,dy = \frac{e^{-2\pi Y_0}}{2\pi}, \end{align*} the cusp contribution converges for every fixed weight $k$. [/example] The convergence calculation shows that the integral defining the Petersson pairing is meaningful on cusp forms, but an inner product also needs positivity and compatibility with the quotient by the modular group. These points are what allow analytic integration over a fundamental domain to become linear algebra on $S_k(SL_2(\mathbb Z))$. The theorem records that the pairing has exactly the properties needed for the spectral theory of Hecke operators. [quotetheorem:4254] [citeproof:4254] This turns $S_k(SL_2(\mathbb Z))$ into a finite-dimensional [Hilbert space](/page/Hilbert%20Space). [Finite dimensionality](/theorems/1534) is important: once the Hecke operators are self-adjoint, ordinary linear algebra gives orthogonal eigenspace decompositions. The construction is specific to cusp forms in this discussion because their exponential decay makes the cusp contribution converge without regularization. Conceptually, the Petersson product is the device that turns modular forms from holomorphic functions with symmetries into vectors whose orthogonality and projections can be studied. That perspective is essential for choosing eigenbases and for comparing Fourier coefficients through [self-adjoint operators](/page/Self-Adjoint%20Operators). ## Hecke Operators as Self-Adjoint Operators Why should Hecke operators be compatible with this inner product? The guiding principle is that $T_n$ averages $f$ over finitely many modular correspondences, and the same correspondence can be traversed in the opposite direction when paired against $g$. [definition: Hecke Self-Adjointness] Let $V$ be a complex [inner product space](/page/Inner%20Product%20Space) and let $A:V\to V$ be linear. The operator $A$ is self-adjoint with respect to the inner product $(\cdot,\cdot)$ if \begin{align*} (Au,v)=(u,Av) \end{align*} for all $u,v\in V$. [/definition] For modular forms, the operator $T_n$ is defined by summing over representatives of a double coset. For the full modular group, the relevant correspondence is symmetric enough that the adjoint operator is again $T_n$. The key analytic question is whether this symmetry is visible inside the Petersson pairing. If it is, Hecke operators become spectral operators on cusp forms rather than merely coefficient transformations. [quotetheorem:4255] [citeproof:4255] The theorem is the point where the analytic and algebraic parts of the course meet. Algebra defines $T_n$ and gives relations among the Hecke operators; the inner product shows that these operators are normal, indeed self-adjoint, so their eigenspaces are orthogonal. [remark: Normalisation Conventions] Some books use a normalised Hecke operator differing from $T_n$ by a scalar depending on $n$ and $k$. Self-adjointness is unchanged when the scalar is real. In these notes, $T_n$ denotes the convention used earlier in the course, so the Fourier coefficient formula and the inner product statement refer to the same operator. [/remark] Self-adjointness becomes useful only after we ask how different eigenvectors sit inside the Petersson inner product space. If two cusp forms are eigenvectors for the same Hecke operator with distinct eigenvalues, then the operator cannot mix their directions: symmetry of the inner product forces the two directions to be perpendicular. This is the mechanism that turns Hecke eigenvalues into an orthogonal coordinate system. [quotetheorem:4256] [citeproof:4256] This is the basic orthogonality argument used throughout the rest of the course. To separate two simultaneous Hecke eigenforms, it is enough to find one Hecke operator for which their eigenvalues differ. The hypothesis that the forms are eigenvectors for the same self-adjoint operator is essential: arbitrary cusp forms need not be orthogonal just because their Fourier expansions look different. In practice the theorem lets us use arithmetic data, namely Hecke eigenvalues, to make a geometric conclusion about the Petersson inner product. ## Orthogonal Eigenbases and Fourier Expansions How does orthogonality affect the actual Fourier expansions of cusp forms? The Hecke operators act directly on coefficients, so diagonalising them gives distinguished forms whose coefficients encode the full Hecke system. [definition: Normalized Hecke Eigenform] A normalized Hecke eigenform in $S_k(SL_2(\mathbb Z))$ is a nonzero cusp form \begin{align*} f(z)=\sum_{n\ge 1}a_n q^n \end{align*} which is an eigenvector for every $T_m$ and satisfies $a_1=1$. [/definition] The normalization $a_1=1$ removes scalar ambiguity. For the standard Hecke convention on level $1$, the eigenvalue of $T_m$ on a normalized eigenform is its $m$-th Fourier coefficient. Having defined the preferred eigenvectors, we still need to know that there are enough of them to describe every cusp form. The spectral theorem for the commuting self-adjoint Hecke operators supplies exactly this missing basis statement. [quotetheorem:4257] [citeproof:4257] The basis is not merely convenient; it is the coordinate system in which the arithmetic of modular forms becomes visible. In that basis, the Hecke action is diagonal and Fourier coefficients become eigenvalue data. The normalization matters because an eigenline has no preferred vector until the first coefficient is fixed; after imposing $a_1=1$, the same object can be compared across all Hecke operators. This also explains why cusp forms are the right finite-dimensional setting here: the Petersson inner product and the commuting Hecke action together produce a basis whose elements have intrinsic arithmetic meaning. Once an eigenbasis has been chosen, the next question is how an arbitrary cusp form is expressed in that basis. Orthogonality supplies projections onto the individual eigendirections, while the normalization of the basis elements makes their Fourier coefficients directly readable. The resulting expansion is the bridge from an abstract spectral decomposition to concrete $q$-series calculations. [quotetheorem:4258] [citeproof:4258] This result is the practical payoff of Petersson orthogonality. Once a form is projected onto the eigenbasis, all of its coefficients are assembled from the coefficient systems of normalized eigenforms. [example: Conceptual Diagonalisation of S Twenty Four] The dimension formula gives $\dim S_{24}(SL_2(\mathbb Z))=2$, so two linearly independent elements of $S_{24}$ form a basis. Since $\Delta$ has weight $12$, $E_{12}$ has weight $12$, and $E_4^3$ has weight $12$, both $\Delta E_{12}$ and $\Delta E_4^3$ lie in $S_{24}(SL_2(\mathbb Z))$. To see that these two forms really give a basis, compare their first two Fourier coefficients. Using \begin{align*} \Delta &= q-24q^2+O(q^3),\\ E_{12} &= 1+\frac{65520}{691}q+O(q^2),\\ E_4 &= 1+240q+O(q^2), \end{align*} we get \begin{align*} \Delta E_{12} &=(q-24q^2+O(q^3))\left(1+\frac{65520}{691}q+O(q^2)\right)\\ &=q+\left(\frac{65520}{691}-24\right)q^2+O(q^3)\\ &=q+\frac{65520-16584}{691}q^2+O(q^3)\\ &=q+\frac{48936}{691}q^2+O(q^3). \end{align*} Also, \begin{align*} E_4^3 &=(1+240q+O(q^2))^3\\ &=1+3\cdot 240q+O(q^2)\\ &=1+720q+O(q^2), \end{align*} so \begin{align*} \Delta E_4^3 &=(q-24q^2+O(q^3))(1+720q+O(q^2))\\ &=q+(720-24)q^2+O(q^3)\\ &=q+696q^2+O(q^3). \end{align*} If $\Delta E_{12}=c\,\Delta E_4^3$, then comparing the coefficient of $q$ gives $1=c$. Comparing the coefficient of $q^2$ would then give \begin{align*} \frac{48936}{691}=696. \end{align*} But \begin{align*} 696\cdot 691=480936\ne 48936, \end{align*} so the two forms are not scalar multiples. Hence $\Delta E_{12}$ and $\Delta E_4^3$ are linearly independent, and therefore form a basis of the two-dimensional space $S_{24}$. The Petersson inner product gives a different, more canonical basis. The Hecke operators $T_n$ commute with one another and are self-adjoint for the Petersson inner product. Therefore, on the two-dimensional inner product space $S_{24}$, if one operator $T_p$ has two distinct eigenvalues $\lambda_1\ne\lambda_2$, its two eigenspaces are one-dimensional and orthogonal. Because every $T_n$ commutes with $T_p$, each $T_n$ preserves these eigenspaces, so the two eigenlines are simultaneous Hecke eigenlines. After scaling the two eigenvectors so that their first Fourier coefficient is $1$, the resulting basis consists of two normalized linear combinations of $\Delta E_{12}$ and $\Delta E_4^3$, and the $n$-th Fourier coefficient of each basis element is the corresponding $T_n$-eigenvalue. [/example] This computation shows how a non-eigen basis is converted into eigenlines. Orthogonality then explains why distinct normalized eigenforms separate cleanly in the Petersson geometry. [remark: Why Distinct Normalized Eigenforms Are Orthogonal] If $f$ and $g$ are distinct normalized simultaneous eigenforms, then their Hecke eigenvalue systems differ for some $T_n$ in the level-$1$ setting. Since the eigenvalue of $T_n$ is the $n$-th Fourier coefficient, this says $a_n(f)\ne a_n(g)$ for some $n$. The orthogonality theorem for distinct eigenvalues then gives $(f,g)_{\mathrm{Pet}}=0$. [/remark] ## Consequences for the Rest of the Course What has changed after introducing the Petersson inner product? Before this chapter, Hecke operators were explicit transformations of Fourier series. After this chapter, they are also self-adjoint operators on a finite-dimensional Hilbert space, and that extra structure forces a strong decomposition theory. [explanation: Three Roles of Petersson Orthogonality] First, it gives a canonical way to separate independent eigenforms. Orthogonality means that different Hecke systems do not interfere under projection. Second, it justifies treating normalized eigenforms as the basic arithmetic objects rather than arbitrary bases of $S_k$. In later chapters, Euler products and $L$-functions are attached to eigenforms because their Fourier coefficients multiply according to Hecke relations. Third, it gives analytic control over coefficient expansions. Inner products compute projections, projections compute coordinates in the eigenbasis, and those coordinates determine Fourier coefficients through finite sums. [/explanation] The main lesson is that cusp forms are the right domain for the Petersson inner product because their vanishing at the cusp makes the integral finite. On this domain, the Hecke operators are self-adjoint and commuting, so the spectral theorem produces an orthogonal basis of normalized eigenforms. This is the bridge from classical modular-form calculations to the eigenform-centered viewpoint used in the theory of modular $L$-functions. The Petersson theory supplies the analytic setting in which eigenforms separate cleanly, and the Hecke relations supply the multiplicative coefficients needed for Euler products. The final chapter uses that combination to view modular forms not as isolated examples, but as computable objects with geometry, algebra, and analysis all working together. # 9. L-Functions of Modular Forms The preceding chapters attached Hecke operators to modular forms and showed that simultaneous eigenforms have Fourier coefficients with strong multiplicative structure. This chapter turns that algebraic structure into analytic objects. The guiding question is: how much of a modular form can be recovered from the Dirichlet series formed from its Fourier coefficients, and how does modularity force analytic continuation, a functional equation, and an Euler product? ## Dirichlet Series Attached to Modular Forms Given a modular form $f(z)=\sum_{n\geq 0} a_n q^n$, the first attempt at an analytic invariant is to place the coefficients into a Dirichlet series. The constant term must be treated separately, since $n^{-s}$ begins at $n=1$, and the cleanest theory is obtained for cusp forms. [definition: L-Function Of A Cusp Form] Let $f\in S_k(SL_2(\mathbb Z))$ have Fourier expansion \begin{align*} f(z)=\sum_{n=1}^{\infty} a_n q^n, \qquad q=e^{2\pi i z}. \end{align*} Let \begin{align*} H_f=\left\{s\in\mathbb C: \sum_{n=1}^{\infty} a_n n^{-s} \text{ converges}\right\}. \end{align*} The $L$-function of $f$ is the map \begin{align*} L(f,-):H_f&\to\mathbb C, & s&\mapsto \sum_{n=1}^{\infty} \frac{a_n}{n^s}. \end{align*} [/definition] The notation hides two different variables: $z$ belongs to the upper half-plane, while $s$ is the complex variable of the Dirichlet series. The cusp condition removes $a_0$ and makes $f(iy)$ decay rapidly as $y\to\infty$, which is the analytic input behind the Mellin transform in the next section. If $f$ is not cuspidal, then $f(iy)$ usually tends to its constant term as $y\to\infty$, so the Mellin integral of $f(iy)$ itself diverges at infinity. Thus the cusp hypothesis is not only aesthetic: it is what lets the Fourier expansion be converted directly into an integral without first subtracting a constant term. [example: First Terms Of The Delta L-Function] The discriminant cusp form \begin{align*} \Delta(z)=q\prod_{n=1}^{\infty}(1-q^n)^{24} \end{align*} lies in $S_{12}(SL_2(\mathbb Z))$. To see the first coefficients, terms with $n\geq 5$ contribute only powers $q^5,q^6,\ldots$ to the product before the leading factor $q$, so up to $q^4$ we compute \begin{align*} (1-q)^{24}&=1-24q+276q^2-2024q^3+10626q^4+\cdots,\\ (1-q^2)^{24}&=1-24q^2+276q^4+\cdots,\\ (1-q^3)^{24}&=1-24q^3+\cdots,\\ (1-q^4)^{24}&=1-24q^4+\cdots . \end{align*} Multiplying these truncated factors gives \begin{align*} &(1-24q+276q^2-2024q^3+10626q^4) (1-24q^2+276q^4)\\ &\qquad=1-24q+(276-24)q^2+(-2024+576)q^3 +(10626+276-6624)q^4+\cdots\\ &\qquad=1-24q+252q^2-1448q^3+4278q^4+\cdots, \end{align*} then \begin{align*} &(1-24q+252q^2-1448q^3+4278q^4)(1-24q^3)\\ &\qquad=1-24q+252q^2+(-1448-24)q^3+(4278+576)q^4+\cdots\\ &\qquad=1-24q+252q^2-1472q^3+4854q^4+\cdots, \end{align*} and finally \begin{align*} &(1-24q+252q^2-1472q^3+4854q^4)(1-24q^4)\\ &\qquad=1-24q+252q^2-1472q^3+(4854-24)q^4+\cdots\\ &\qquad=1-24q+252q^2-1472q^3+4830q^4+\cdots. \end{align*} Thus \begin{align*} \Delta(z)=q-24q^2+252q^3-1472q^4+4830q^5+\cdots, \end{align*} so $\tau(1)=1$, $\tau(2)=-24$, $\tau(3)=252$, $\tau(4)=-1472$, and $\tau(5)=4830$. Its associated Dirichlet series is therefore \begin{align*} L(\Delta,s) =\sum_{n=1}^{\infty}\frac{\tau(n)}{n^s} =1-\frac{24}{2^s}+\frac{252}{3^s}-\frac{1472}{4^s}+\frac{4830}{5^s}+\cdots . \end{align*} This shows that the $L$-function records the same Fourier coefficients as $\Delta$, but places the coefficient of $q^n$ on the multiplicative scale $n^{-s}$. [/example] For a general modular form $f\in M_k(SL_2(\mathbb Z))$ with nonzero constant term, the same formal series $\sum_{n\geq 1} a_n n^{-s}$ ignores $a_0$. This is useful for Eisenstein series, where the nonconstant coefficients are divisor sums and the resulting Dirichlet series factors into zeta-functions. [example: Eisenstein Coefficients And Zeta Products] For even $k\geq 4$, take the normalized Eisenstein series \begin{align*} E_k(z)=1-\frac{2k}{B_k}\sum_{n=1}^{\infty}\sigma_{k-1}(n)q^n, \qquad \sigma_{k-1}(n)=\sum_{d\mid n}d^{k-1}, \end{align*} where $B_k$ is the $k$th Bernoulli number. We compute the Dirichlet series attached to the nonconstant coefficients $\sigma_{k-1}(n)$. For $\operatorname{Re}(s)>k$, both zeta series below converge absolutely, so their product may be expanded and rearranged by absolute convergence: \begin{align*} \zeta(s)\zeta(s-k+1) &=\left(\sum_{m=1}^{\infty}\frac{1}{m^s}\right) \left(\sum_{d=1}^{\infty}\frac{1}{d^{s-k+1}}\right)\\ &=\sum_{m=1}^{\infty}\sum_{d=1}^{\infty}\frac{1}{m^s d^{s-k+1}}\\ &=\sum_{m=1}^{\infty}\sum_{d=1}^{\infty}\frac{d^{k-1}}{m^s d^s}\\ &=\sum_{m=1}^{\infty}\sum_{d=1}^{\infty}\frac{d^{k-1}}{(md)^s}. \end{align*} Now group the double sum according to the value $n=md$. For a fixed positive integer $n$, the pairs $(m,d)$ with $md=n$ are exactly the pairs $(n/d,d)$ with $d\mid n$, so the coefficient of $n^{-s}$ is \begin{align*} \sum_{\substack{m,d\geq 1\\ md=n}}d^{k-1} =\sum_{d\mid n}d^{k-1} =\sigma_{k-1}(n). \end{align*} Therefore, in the initial half-plane $\operatorname{Re}(s)>k$, \begin{align*} \sum_{n=1}^{\infty}\frac{\sigma_{k-1}(n)}{n^s} =\zeta(s)\zeta(s-k+1). \end{align*} Thus the nonconstant Fourier coefficients of $E_k$ produce no new primitive $L$-function here; their Dirichlet series is already a product of two classical zeta-functions. [/example] This comparison explains why cusp forms, isolated in Chapter 2 by the vanishing of the constant term and organized in Chapters 7 and 8 by Hecke eigenbases, are the main focus here. Eisenstein series already contain products of classical zeta-functions, while cusp forms produce new analytic objects whose properties are forced by modularity and Hecke symmetry. ## The Mellin Transform Of A Cusp Form The main analytic question is how to continue $L(f,s)$ beyond its first half-plane of convergence. The route is to evaluate the cusp form on the imaginary axis and take its Mellin transform; the exponential decay of the Fourier expansion turns the integral into the Dirichlet series. [definition: Completed L-Function] Let $f\in S_k(SL_2(\mathbb Z))$, and let $H_f$ be the half-plane of convergence of $L(f,s)$. The completed $L$-function of $f$ is the map \begin{align*} \Lambda(f,-):H_f&\to\mathbb C, & s&\mapsto (2\pi)^{-s}\Gamma(s)L(f,s). \end{align*} [/definition] The gamma factor is not a cosmetic addition. It is exactly the factor produced by integrating $e^{-2\pi n y}y^{s-1}$ over $y>0$, and it is also the factor that gives the functional equation its symmetric form. To use modularity for analytic continuation, the completed function must first be expressed as an integral attached directly to $f(iy)$. The Mellin transform identity provides that bridge from the Fourier series to the analytic function $\Lambda(f,s)$. [quotetheorem:4259] [citeproof:4259] The formula says that the completed $L$-function is a Mellin transform of the function $y\mapsto f(iy)$. The cusp hypothesis is used at both ends of the integral: as $y\to\infty$ it gives exponential decay, and after the substitution $y=1/t$ it also controls the behaviour near $0$ through modularity. For a non-cusp Eisenstein series, the constant term would contribute an integral comparable to $\int_1^\infty y^{s-1}\,dy$, so the same formula would need correction terms. The theorem is initially a statement only in a right half-plane where the Dirichlet series and the interchange of sum and integral are justified; analytic continuation comes from the next modularity argument, not from the Dirichlet series alone. It turns modularity under $z\mapsto -1/z$ into the symmetry $s\leftrightarrow k-s$. [quotetheorem:4260] [citeproof:4260] For level $1$, the factor $i^k$ is the sign $(-1)^{k/2}$. This sign is the root number in this simplest setting. The proof depends on the specific generator $S:z\mapsto -1/z$ of $SL_2(\mathbb Z)$; for higher level groups the same idea survives, but the functional equation involves the level, character, and an Atkin-Lehner type sign. Cuspidality is again essential for the completed function to be entire: non-cusp forms typically have poles coming from Eisenstein or zeta-function terms. The functional equation gives a symmetry of values, but it does not by itself locate zeros, prove non-vanishing at the centre, or determine special values arithmetically. [example: Central Symmetry For The Delta Function] For $\Delta\in S_{12}(SL_2(\mathbb Z))$, the completed $L$-function is \begin{align*} \Lambda(\Delta,s)=(2\pi)^{-s}\Gamma(s)L(\Delta,s). \end{align*} Applying the *Functional Equation For Level One Cusp Forms* with $k=12$ gives \begin{align*} \Lambda(\Delta,s)=i^{12}\Lambda(\Delta,12-s). \end{align*} Since \begin{align*} i^{12}=(i^4)^3=1^3=1, \end{align*} the functional equation becomes \begin{align*} \Lambda(\Delta,s)=\Lambda(\Delta,12-s). \end{align*} The fixed point of the reflection $s\mapsto 12-s$ is found by solving \begin{align*} s&=12-s,\\ 2s&=12,\\ s&=6. \end{align*} Equivalently, for any complex number $t$, \begin{align*} \Lambda(\Delta,6+t) =\Lambda(\Delta,12-(6+t)) =\Lambda(\Delta,6-t). \end{align*} Thus the completed function is symmetric about the central point $s=6$: knowing its values at $6+t$ determines its values at the reflected points $6-t$. [/example] This is the first point in the course where the modular transformation law becomes a statement about an analytic function of a new complex variable. The next step is to add the Hecke eigenform hypothesis, which changes the Dirichlet series from a sum into an Euler product. ## Euler Products For Hecke Eigenforms The guiding question is when the coefficients $a_n$ behave like the coefficients of an arithmetic object with local factors at primes. The answer is supplied by normalized Hecke eigenforms: the Hecke relations give exactly the multiplicativity needed to factor the Dirichlet series prime by prime. [definition: Normalized Hecke Eigenform] A cusp form $f\in S_k(SL_2(\mathbb Z))$ with Fourier expansion $f(z)=\sum_{n=1}^{\infty}a_nq^n$ is a normalized Hecke eigenform if $a_1=1$ and $f$ is an eigenvector for every Hecke operator $T_n$. [/definition] For such a form, the eigenvalue of $T_n$ is $a_n$. The normalization $a_1=1$ fixes the scaling ambiguity: multiplying an eigenform by a nonzero constant changes all Fourier coefficients but should not change the underlying eigensystem. If $a_1\neq 1$, the eigenvalue of $T_n$ is $a_n/a_1$ rather than $a_n$, so the coefficient formulas below would have extra normalization factors. The multiplicative relations from the Hecke algebra are therefore relations among the Fourier coefficients themselves. [quotetheorem:4261] [citeproof:4261] The prime-power recurrence is the local calculation behind the Euler product. Normalization matters because it makes the constant term of the local generating series equal to $a_1=1$, which is what allows an inverse quadratic factor rather than a rescaled one. Simultaneous eigenform status is also essential: for an arbitrary cusp form, $T_n f$ need not be a scalar multiple of $f$, so the Hecke algebra identities give relations among forms rather than multiplicative identities among the numbers $a_n$. Thus a general cusp form has an $L$-series, but its coefficients do not usually satisfy the prime-power recursion. It says that the generating function for the sequence $a_{p^r}$ has a quadratic denominator. [quotetheorem:4262] [citeproof:4262] This theorem is the analytic shadow of the Hecke algebra. Non-eigenforms do not generally have Euler products, because a sum of eigenforms usually mixes different local factors and destroys coefficient multiplicativity. The product is also initially an identity only in a half-plane of absolute convergence; outside that region the analytically continued $L$-function is studied using the completed function and its functional equation, not by multiplying the infinite product term by term. The two terms in the local quadratic encode the eigenvalue $a_p$ and the weight factor $p^{k-1}$. Practically, the factor at $p$ lets us recover all coefficients $a_{p^r}$ from $a_p$ by the recurrence, while coprime multiplicativity then reconstructs $a_n$ from the prime-power data. [example: Euler Product Of The Delta Function] The form $\Delta$ is a normalized Hecke eigenform of weight $12$, so by *[Euler Product For Normalized Hecke Eigenforms](/theorems/4262)* its $L$-function has the Euler product \begin{align*} L(\Delta,s) &=\prod_p\left(1-\tau(p)p^{-s}+p^{12-1-2s}\right)^{-1}\\ &=\prod_p\left(1-\tau(p)p^{-s}+p^{11-2s}\right)^{-1}. \end{align*} For $p=2$, the coefficient computed earlier is $\tau(2)=-24$, hence \begin{align*} 1-\tau(2)2^{-s}+2^{11-2s} &=1-(-24)2^{-s}+2^{11-2s}\\ &=1+24\cdot 2^{-s}+2^{11-2s}. \end{align*} Thus the $2$-local factor is \begin{align*} \left(1+24\cdot 2^{-s}+2^{11-2s}\right)^{-1}. \end{align*} To see how this factor encodes the powers of $2$, put $X=2^{-s}$. The prime-power recurrence from *Hecke Recursion For Fourier Coefficients* gives \begin{align*} \tau(2^{r+1}) &=\tau(2)\tau(2^r)-2^{11}\tau(2^{r-1})\\ &=-24\tau(2^r)-2^{11}\tau(2^{r-1}) \end{align*} for $r\geq 1$, with $\tau(1)=1$ and $\tau(2)=-24$. For example, \begin{align*} \tau(4)&=-24\tau(2)-2^{11}\tau(1)\\ &=(-24)(-24)-2048\cdot 1\\ &=576-2048\\ &=-1472, \end{align*} and \begin{align*} \tau(8)&=-24\tau(4)-2^{11}\tau(2)\\ &=(-24)(-1472)-2048(-24)\\ &=35328+49152\\ &=84480. \end{align*} Therefore the beginning of the $2$-power series is \begin{align*} \sum_{r=0}^{\infty}\tau(2^r)X^r =1-24X-1472X^2+84480X^3+\cdots, \end{align*} and this is exactly the expansion of \begin{align*} (1+24X+2^{11}X^2)^{-1}. \end{align*} The local factor at $2$ therefore packages all coefficients $\tau(2^r)$ into one quadratic denominator, while the full Euler product combines the analogous data over all primes. [/example] The Euler product brings modular forms into the same analytic family as the Riemann zeta function and Dirichlet $L$-functions, but with degree two local factors. In later courses these quadratic factors are reinterpreted through Galois representations; in this course, they arise from the Hecke action on classical modular forms. The course has now moved from the geometry of the modular curve to explicit Fourier expansions, Hecke symmetries, and $L$-functions. The synthesis chapter ties these strands together by showing how level-1 modular forms can be computed from a small amount of structural data and a few basic generators. # 10. Synthesis: Classical Modular Forms as Computable Objects This closing chapter collects the two main threads of the course into a single computational picture. The geometry of $SL_2(\mathbb Z)\backslash \mathbb H$ explains why level $1$ modular forms are finite-dimensional and controlled by a few special points, while the $q$-expansion turns those geometric objects into explicit arithmetic sequences. The Hecke operators then add enough structure that many spaces admit bases whose Fourier coefficients can be computed, multiplied, and packaged into $L$-functions. ## From Geometry to $q$-Expansions The first question is how much global information about a modular form is already visible from a finite amount of local data. Earlier chapters answered this through the standard fundamental domain, the cusp at infinity, and the valence formula: a holomorphic modular form cannot have arbitrary zeros, because the quotient has finite hyperbolic area and only two elliptic points. [quotetheorem:4263] [citeproof:4263] The formula is the bridge from geometry to computation. It says that the number of zeros, counted with orbifold weights, is prescribed by $k/12$, so a modular form is strongly constrained once its first few Fourier coefficients are known. The nonzero hypothesis is essential because the zero form has infinite order of vanishing everywhere and cannot be assigned a finite zero count. Holomorphy is also essential: meromorphic modular functions satisfy a related divisor-degree statement with poles subtracted, while non-holomorphic objects are not controlled by this holomorphic zero-counting argument. The special coefficients $1/2$ and $1/3$ belong to level $1$, where the only elliptic points are represented by $i$ and $\rho$; at higher level the elliptic and cusp contributions change with the subgroup. Thus the theorem does not classify modular forms by itself, but it gives the finite numerical constraint that makes the later dimension and ring computations possible. [example: Zeros of the Basic Generators] The normalized Eisenstein series satisfy $E_4(q)=1+240q+O(q^2)$ and $E_6(q)=1-504q+O(q^2)$, so $v_\infty(E_4)=v_\infty(E_6)=0$. Applying *Valence Formula for Level One* gives \begin{align*} \frac{1}{2}v_i(E_4)+\frac{1}{3}v_\rho(E_4)+\sum_{p\ne i,\rho}v_p(E_4)&=\frac{4}{12}=\frac{1}{3},\\ \frac{1}{2}v_i(E_6)+\frac{1}{3}v_\rho(E_6)+\sum_{p\ne i,\rho}v_p(E_6)&=\frac{6}{12}=\frac{1}{2}. \end{align*} For $E_4$, the matrix $\begin{pmatrix}0&-1\\1&1\end{pmatrix}$ fixes $\rho$, since \begin{align*} -\frac{1}{\rho+1}=\rho \end{align*} using $\rho^2+\rho+1=0$. The weight $4$ transformation law gives \begin{align*} E_4(\rho)=(\rho+1)^4E_4(\rho). \end{align*} Since $\rho+1=e^{\pi i/3}$, we have $(\rho+1)^4=e^{4\pi i/3}\ne 1$, hence $E_4(\rho)=0$ and $v_\rho(E_4)\ge 1$. Substituting this into the valence identity, \begin{align*} \frac{1}{3}v_\rho(E_4)\le \frac{1}{3}, \end{align*} so $v_\rho(E_4)=1$, while $v_i(E_4)=0$ and every other interior order is $0$. For $E_6$, the matrix $\begin{pmatrix}0&-1\\1&0\end{pmatrix}$ fixes $i$, and the weight $6$ transformation law gives \begin{align*} E_6(i)=i^6E_6(i)=-E_6(i). \end{align*} Thus $2E_6(i)=0$, so $E_6(i)=0$ and $v_i(E_6)\ge 1$. The valence identity then forces \begin{align*} \frac{1}{2}v_i(E_6)\le \frac{1}{2}, \end{align*} so $v_i(E_6)=1$, while $v_\rho(E_6)=0$ and there are no other interior zeros. Finally, $\Delta(q)=q-24q^2+O(q^3)$ gives $v_\infty(\Delta)=1$, and the weight $12$ valence identity becomes \begin{align*} 1+\frac{1}{2}v_i(\Delta)+\frac{1}{3}v_\rho(\Delta)+\sum_{p\ne i,\rho}v_p(\Delta)=\frac{12}{12}=1. \end{align*} All remaining terms are nonnegative, so each is $0$; $\Delta$ has its single zero at the cusp and no zeros in $\mathbb H$. [/example] The course then used these constraints to identify the whole graded ring of level $1$ modular forms. This is the structural result that makes the theory computable. The zero computations show that $E_4$, $E_6$, and $\Delta$ have the right special vanishing behavior, but they do not by themselves show that there are no other independent level-one modular forms. The remaining obstruction is generation: a form of arbitrary weight might, a priori, contain information not expressible through the two Eisenstein series. Resolving this turns the analytic classification problem into algebra in explicit generators. [quotetheorem:4264] [citeproof:4264] This theorem turns a question about functions on $\mathbb H$ into algebra in two generators. Since $E_4$ and $E_6$ have explicit Fourier expansions, every level $1$ modular form has an effectively computable $q$-expansion. The level $1$ assumption is doing real work: for general congruence subgroups the graded ring may require more generators and relations, and cusp structure must be tracked separately. Holomorphy is also part of the statement, since allowing poles leads instead to modular functions generated by ratios such as the $j$-invariant. The roles of $E_4$, $E_6$, and $\Delta$ are complementary: $E_4$ and $E_6$ generate the holomorphic ring, while $\Delta$ isolates the cusp condition by contributing exactly one zero at infinity and none in $\mathbb H$. [example: A Basis in Weight Twelve] In weight $12$, the only monomials in $E_4$ and $E_6$ of total weight $12$ are $E_4^3$ and $E_6^2$, since $4a+6b=12$ has the nonnegative integer solutions $(a,b)=(3,0)$ and $(0,2)$. Thus the level one structure theorem gives \begin{align*} M_{12}=\mathbb C E_4^3+\mathbb C E_6^2. \end{align*} The two forms are linearly independent: their difference has nonzero $q$-coefficient, as the expansions below show. Using \begin{align*} E_4&=1+240q+2160q^2+6720q^3+17520q^4+O(q^5),\\ E_6&=1-504q-16632q^2-122976q^3-532728q^4+O(q^5), \end{align*} we compute \begin{align*} E_4^3 &=1+3(240)q+\left(3(2160)+3(240)^2\right)q^2\\ &\quad+\left(3(6720)+6(240)(2160)+(240)^3\right)q^3\\ &\quad+\left(3(17520)+6(240)(6720)+3(2160)^2+3(240)^2(2160)\right)q^4+O(q^5)\\ &=1+720q+179280q^2+16954560q^3+396974160q^4+O(q^5), \end{align*} and \begin{align*} E_6^2 &=1+2(-504)q+\left((-504)^2+2(-16632)\right)q^2\\ &\quad+\left(2(-122976)+2(-504)(-16632)\right)q^3\\ &\quad+\left(2(-532728)+2(-504)(-122976)+(-16632)^2\right)q^4+O(q^5)\\ &=1-1008q+220752q^2+16519104q^3+399517776q^4+O(q^5). \end{align*} Therefore \begin{align*} E_4^3-E_6^2 &=1728q-41472q^2+435456q^3-2543616q^4+O(q^5)\\ &=1728\left(q-24q^2+252q^3-1472q^4+O(q^5)\right). \end{align*} The constant term cancels, so $E_4^3-E_6^2$ is a cusp form, and the standard normalization is \begin{align*} \Delta=\frac{E_4^3-E_6^2}{1728}=q-24q^2+252q^3-1472q^4+O(q^5). \end{align*} Thus $S_{12}=\mathbb C\Delta$, and choosing the normalized Eisenstein series $E_{12}$ for the complementary Eisenstein line gives \begin{align*} M_{12}=\mathbb C E_{12}\oplus \mathbb C\Delta. \end{align*} The calculation shows concretely how the two algebraic generators split weight $12$ into its Eisenstein part and its one-dimensional cusp part. [/example] ## From $q$-Expansions to Arithmetic Once a modular form is written as $f(z)=\sum_{n\ge 0}a_nq^n$, the next question is which arithmetic patterns in the sequence $(a_n)$ are forced by modularity. The answer comes from the Hecke operators: they are geometric correspondences on lattices, but on $q$-expansions they become explicit coefficient operators. [definition: Hecke Operator on $q$-Expansions] Let $k\ge 0$ and let $n\in\mathbb N$. The Hecke operator $T_n:M_k(SL_2(\mathbb Z))\to M_k(SL_2(\mathbb Z))$ sends a modular form $f(z)=\sum_{m\ge 0}a_mq^m$ to the modular form $T_nf$ whose Fourier expansion is \begin{align*} (T_nf)(z)=\sum_{m\ge 0}\left(\sum_{d\mid \gcd(m,n)} d^{k-1}a_{mn/d^2}\right)q^m. \end{align*} [/definition] This formula is the computational form of a double-coset construction. In particular, for a prime $p$ it becomes \begin{align*} (T_pf)(z)=\sum_{m\ge 0}(a_{pm}+p^{k-1}a_{m/p})q^m, \end{align*} where $a_{m/p}=0$ if $p\nmid m$. The coefficient recipe is useful only if it stays inside the spaces of modular forms and cusp forms already under study. The next result verifies that applying $T_n$ to a genuine level-one form produces another genuine level-one form, with cuspidality preserved when the constant term vanishes. [quotetheorem:4245] [citeproof:4245] The preservation statement is not merely formal: it says that the coefficient formula is still computing genuine modular forms, not just new formal power series. Holomorphy at the cusp matters because the absence of negative powers of $q$ is part of being a holomorphic modular form on the compactified quotient, and the condition $a_0=0$ is exactly what distinguishes cusp forms at infinity in level $1$. For an arbitrary $q$-series, the same coefficient recipe need not produce anything modular, because the transformation law under $SL_2(\mathbb Z)$ is missing. The Hecke operators do more than preserve the relevant spaces: they commute with each other and interact well with the Petersson inner product, so they can be simultaneously diagonalised on cusp forms. [quotetheorem:4266] [citeproof:4266] Finite-dimensionality is essential here because the argument is ordinary spectral theory on $S_k$, not a convergence theorem for an infinite operator family. The restriction to cusp forms supplies the Petersson inner product without Eisenstein boundary terms, and the normality and commutativity of the Hecke operators are what allow [simultaneous diagonalisation](/theorems/2403) rather than unrelated eigenbases for different $T_n$. In a higher-dimensional eigenspace for a single operator, one operator alone would not pick out canonical forms; the full commuting Hecke algebra is what separates the arithmetic directions in the space. For a normalized eigenform, the Fourier coefficients are therefore not merely coordinates: they are the Hecke eigenvalues. The next issue is what relations those eigenvalues satisfy. The operators $T_m$ and $T_n$ do not form an arbitrary commuting family; their multiplication law reflects divisibility relations among integers. Translating that operator algebra through a normalized eigenform should impose arithmetic identities on the coefficients $a_n$, and those identities are exactly what make Euler products possible later. [quotetheorem:4252] [citeproof:4252] The normalization $a_1=1$ is what turns eigenvalues into coefficients without an extra scalar factor; without it, the same eigenline would give coefficients scaled by an arbitrary nonzero constant. The eigenform hypothesis is also essential: a general cusp form can be written as a linear combination of eigenforms, but its coefficients need not be multiplicative. These identities do not say that the coefficients are periodic or determined by finitely many primes; they say that all coefficients are assembled from the prime-power data in exactly the way required for an Euler product. This is the arithmetic mechanism behind the $L$-functions introduced below. [example: The Ramanujan Delta Function as an Eigenform] Since $S_{12}=\mathbb C\Delta$ and each $T_n$ preserves $S_{12}$ by *Hecke Operators Preserve Cusp Forms*, the one-dimensional space $S_{12}$ is stable under every $T_n$. Hence for each $n$ there is a scalar $\lambda_n$ such that $T_n\Delta=\lambda_n\Delta$. The normalization \begin{align*} \Delta(q)=q-24q^2+252q^3-1472q^4+O(q^5) \end{align*} gives \begin{align*} \Delta(z)=\sum_{n\ge 1}\tau(n)q^n, \qquad \tau(1)=1,\quad \tau(2)=-24,\quad \tau(3)=252,\quad \tau(4)=-1472. \end{align*} Because the coefficient of $q$ in $T_n\Delta$ is $\tau(n)$, the eigenvalue is $\lambda_n=\tau(n)$ after comparing the $q$-coefficient in \begin{align*} T_n\Delta=\lambda_n\Delta. \end{align*} Applying *Multiplicativity of Normalized Eigenform Coefficients* with $p=2$, $r=1$, and $k=12$ gives \begin{align*} \tau(2^2)&=\tau(2)\tau(2)-2^{12-1}\tau(1)\\ &=(-24)(-24)-2^{11}\cdot 1\\ &=576-2048\\ &=-1472. \end{align*} Thus the recurrence gives $\tau(4)=-1472$, exactly the coefficient of $q^4$ in the displayed expansion of $\Delta$. For a normalized weight $12$ eigenform, the prime-power recurrence is encoded by the local Euler factor \begin{align*} \left(1-\tau(p)p^{-s}+p^{11-2s}\right)^{-1}, \end{align*} so the associated $L$-function has Euler product \begin{align*} L(\Delta,s)=\prod_p \left(1-\tau(p)p^{-s}+p^{11-2s}\right)^{-1}. \end{align*} The calculation at $p=2$ shows how the Fourier coefficients of $\Delta$ already satisfy the local recurrence required by this Euler product. [/example] ## Building the Computational Package The practical problem is how to build the modular forms of a fixed weight and extract their arithmetic data. The course's answer is a finite procedure: construct a basis using $E_4$ and $E_6$, compute the matrices of enough Hecke operators from the $q$-expansion formula, diagonalise on the cusp subspace, and normalize the resulting eigenforms. [example: A Normalized Eigenbasis in Weight Sixteen] By *[Structure Theorem for Level One Modular Forms](/theorems/4264)*, weight $16$ forms are spanned by monomials $E_4^aE_6^b$ with $4a+6b=16$. The nonnegative integer solutions are $(a,b)=(4,0)$ and $(1,2)$, so \begin{align*} M_{16}(SL_2(\mathbb Z)) =\mathbb C E_4^4\oplus \mathbb C E_4E_6^2. \end{align*} The same theorem gives \begin{align*} S_{16}(SL_2(\mathbb Z)) =\Delta M_4(SL_2(\mathbb Z)). \end{align*} Since $4a+6b=4$ has the only nonnegative solution $(a,b)=(1,0)$, we have $M_4=\mathbb C E_4$, and therefore \begin{align*} S_{16}=\mathbb C(\Delta E_4). \end{align*} Using \begin{align*} \Delta&=q-24q^2+252q^3-1472q^4+O(q^5),\\ E_4&=1+240q+2160q^2+6720q^3+O(q^4), \end{align*} we multiply term by term: \begin{align*} \Delta E_4 &=(q-24q^2+252q^3-1472q^4+O(q^5))(1+240q+2160q^2+6720q^3+O(q^4))\\ &=q+(240-24)q^2+(2160-24\cdot 240+252)q^3\\ &\quad +(6720-24\cdot 2160+252\cdot 240-1472)q^4+O(q^5)\\ &=q+216q^2+(2160-5760+252)q^3\\ &\quad +(6720-51840+60480-1472)q^4+O(q^5)\\ &=q+216q^2-3348q^3+13888q^4+O(q^5). \end{align*} Thus the normalized generator of $S_{16}$ is \begin{align*} f_{16}=\Delta E_4=\sum_{n\ge 1}a_nq^n, \end{align*} with \begin{align*} a_1=1,\qquad a_2=216,\qquad a_3=-3348,\qquad a_4=13888. \end{align*} Because $S_{16}$ is one-dimensional and each $T_n$ preserves cusp forms by *Hecke Operators Preserve Cusp Forms*, every $T_n$ acts on $S_{16}$ by a scalar. Since $f_{16}$ is normalized, *Multiplicativity of Normalized Eigenform Coefficients* identifies that scalar with $a_n$. For $p=2$, $r=1$, and $k=16$, the prime-power recurrence gives \begin{align*} a_{2^{1+1}}&=a_2a_{2^1}-2^{16-1}a_{2^{0}}\\ a_4&=a_2^2-2^{15}a_1\\ &=216^2-2^{15}\cdot 1\\ &=46656-32768\\ &=13888. \end{align*} This matches the $q^4$ coefficient computed above. The Euler factor at $2$ is therefore \begin{align*} (1-a_2 2^{-s}+2^{16-1-2s})^{-1} =(1-216\cdot 2^{-s}+2^{15-2s})^{-1}. \end{align*} The example shows that in weight $16$, the cusp space has a single normalized Hecke eigenform, and its first coefficients already satisfy the prime-power relation required by the Euler product. [/example] Weight $16$ is one-dimensional, so no diagonalisation choice is visible. Weight $24$ is the first useful place to see an actual Hecke matrix split a cusp space into eigenlines. [example: Diagonalising a Higher-Dimensional Cusp Space] In weight $24$, the level-one structure theorem gives $S_{24}=\Delta M_{12}$, and the weight $12$ computation gives $M_{12}=\mathbb C E_4^3\oplus \mathbb C E_6^2$. Set \begin{align*} h_1=\Delta E_4^3,\qquad h_2=\Delta E_6^2. \end{align*} Using the previously computed expansions \begin{align*} E_4^3&=1+720q+179280q^2+16954560q^3+396974160q^4+4632858720q^5+O(q^6),\\ E_6^2&=1-1008q+220752q^2+16519104q^3+399517776q^4+4624512480q^5+O(q^6),\\ \Delta&=q-24q^2+252q^3-1472q^4+4830q^5-6048q^6+O(q^7), \end{align*} we multiply term by term: \begin{align*} h_1 &=q+(720-24)q^2+(179280-24\cdot720+252)q^3\\ &\quad +(16954560-24\cdot179280+252\cdot720-1472)q^4\\ &\quad +(396974160-24\cdot16954560+252\cdot179280-1472\cdot720+4830)q^5\\ &\quad +(4632858720-24\cdot396974160+252\cdot16954560-1472\cdot179280+4830\cdot720-6048)q^6+O(q^7)\\ &=q+696q^2+162252q^3+12831808q^4+34188270q^5-882400608q^6+O(q^7), \end{align*} and \begin{align*} h_2 &=q+(-1008-24)q^2+(220752+24\cdot1008+252)q^3\\ &\quad +(16519104-24\cdot220752-252\cdot1008-1472)q^4\\ &\quad +(399517776-24\cdot16519104+252\cdot220752+1472\cdot1008+4830)q^5\\ &\quad +(4624512480-24\cdot399517776+252\cdot16519104-1472\cdot220752-4830\cdot1008-6048)q^6+O(q^7)\\ &=q-1032q^2+245196q^3+10965568q^4+60177390q^5-1130921568q^6+O(q^7). \end{align*} The two forms are linearly independent because their $q$-coefficients are both $1$ but their $q^2$-coefficients are $696$ and $-1032$. For a weight $24$ cusp form $f=\sum a_mq^m$, the prime Hecke formula gives \begin{align*} T_2f=\sum_{m\ge 1}\left(a_{2m}+2^{23}a_{m/2}\right)q^m, \end{align*} with $a_{m/2}=0$ when $2\nmid m$. Therefore \begin{align*} T_2h_1&=696q+(12831808+2^{23})q^2+O(q^3)\\ &=696q+21220416q^2+O(q^3), \end{align*} so if $T_2h_1=\alpha h_1+\beta h_2$, then \begin{align*} \alpha+\beta&=696,\\ 696\alpha-1032\beta&=21220416. \end{align*} Substituting $\beta=696-\alpha$ gives \begin{align*} 696\alpha-1032(696-\alpha)&=21220416,\\ 1728\alpha-718272&=21220416,\\ 1728\alpha&=21938688,\\ \alpha&=12696, \end{align*} and hence $\beta=696-12696=-12000$. Similarly, \begin{align*} T_2h_2&=-1032q+(10965568+2^{23})q^2+O(q^3)\\ &=-1032q+19354176q^2+O(q^3). \end{align*} Writing $T_2h_2=\alpha h_1+\beta h_2$ gives \begin{align*} \alpha+\beta&=-1032,\\ 696\alpha-1032\beta&=19354176, \end{align*} so \begin{align*} 696\alpha-1032(-1032-\alpha)&=19354176,\\ 1728\alpha+1065024&=19354176,\\ 1728\alpha&=18289152,\\ \alpha&=10584, \end{align*} and $\beta=-1032-10584=-11616$. Thus, in the ordered basis $(h_1,h_2)$, \begin{align*} [T_2]= \begin{pmatrix} 12696&10584\\ -12000&-11616 \end{pmatrix}. \end{align*} Its characteristic polynomial is \begin{align*} \lambda^2-(12696-11616)\lambda+\left(12696(-11616)-10584(-12000)\right) &=\lambda^2-1080\lambda-20468736. \end{align*} The discriminant is \begin{align*} 1080^2+4\cdot20468736 =1166400+81874944 =83041344 =24^2\cdot144169, \end{align*} so the two $T_2$-eigenvalues are \begin{align*} \lambda_\pm=540\pm 12\sqrt{144169}. \end{align*} If $f_\pm=x_\pm h_1+y_\pm h_2$ is normalized by requiring its $q$-coefficient to be $1$, then $x_\pm+y_\pm=1$ and its $q^2$-coefficient is \begin{align*} 696x_\pm-1032y_\pm=\lambda_\pm. \end{align*} Using $y_\pm=1-x_\pm$ gives \begin{align*} 696x_\pm-1032(1-x_\pm)&=\lambda_\pm,\\ 1728x_\pm&=\lambda_\pm+1032,\\ x_\pm&=\frac{540\pm12\sqrt{144169}+1032}{1728} =\frac{131\pm\sqrt{144169}}{144}, \end{align*} and therefore \begin{align*} y_\pm =1-x_\pm =\frac{13\mp\sqrt{144169}}{144}. \end{align*} Thus \begin{align*} f_\pm =\frac{131\pm\sqrt{144169}}{144}h_1 +\frac{13\mp\sqrt{144169}}{144}h_2 \end{align*} are the two normalized $T_2$-eigenforms in $S_{24}$. For $T_3$, the weight $24$ formula is \begin{align*} T_3f=\sum_{m\ge 1}\left(a_{3m}+3^{23}a_{m/3}\right)q^m. \end{align*} The first two coefficients have no $3^{23}$ contribution for $m=1,2$, so \begin{align*} T_3h_1&=162252q-882400608q^2+O(q^3),\\ T_3h_2&=245196q-1130921568q^2+O(q^3). \end{align*} Solving as before, \begin{align*} T_3h_1&=-413748h_1+576000h_2,\\ T_3h_2&=-508032h_1+753228h_2, \end{align*} so \begin{align*} [T_3]= \begin{pmatrix} -413748&-508032\\ 576000&753228 \end{pmatrix}. \end{align*} The $q^3$-coefficient of $f_\pm$ is \begin{align*} a_3(f_\pm) &=162252\frac{131\pm\sqrt{144169}}{144} +245196\frac{13\mp\sqrt{144169}}{144}\\ &=\frac{162252\cdot131+245196\cdot13}{144} \pm\frac{162252-245196}{144}\sqrt{144169}\\ &=169740\mp576\sqrt{144169}. \end{align*} Applying the displayed matrix for $T_3$ to the same vectors gives the same eigenvalues, so $T_2$ and $T_3$ are diagonal on the normalized basis $f_+,f_-$. Consequently the first two nontrivial Euler factors are \begin{align*} L_2(f_\pm,s) &=\left(1-(540\pm12\sqrt{144169})2^{-s}+2^{23-2s}\right)^{-1},\\ L_3(f_\pm,s) &=\left(1-(169740\mp576\sqrt{144169})3^{-s}+3^{23-2s}\right)^{-1}. \end{align*} The calculation shows explicitly how diagonalising Hecke matrices turns the basis $\Delta E_4^3,\Delta E_6^2$ into normalized eigenforms whose Fourier coefficients supply the local Euler factors. [/example] The same package also explains why modular forms are unusually well suited for computation. A finite-dimensional vector space is represented by finitely many monomials in $E_4$ and $E_6$, the Hecke action is represented by explicit matrices on Fourier coefficients, and normalized eigenforms convert those matrices into multiplicative arithmetic functions. [remark: Level One Computational Package] For the level $1$ holomorphic theory developed here, computation has a finite workflow: use the ring structure to choose a basis in a fixed weight, compute enough Fourier coefficients to determine the Hecke action, diagonalize the commuting Hecke operators on cusp forms, and then attach Euler products to the normalized eigenforms. This package is specific to the setting of the course. At higher level one must replace the two-generator ring by a more complicated finite computation involving the chosen congruence subgroup, its cusps, and its Hecke algebra. The Euler product conclusion depends on working with normalized cusp eigenforms; arbitrary modular forms and non-eigen cusp forms still have $q$-expansions, but their coefficient sequences do not generally satisfy the same multiplicative relations. [/remark] ## The Associated $L$-Function The final analytic question is how the Fourier expansion remembers the modular transformation $z\mapsto -1/z$. The Mellin transform converts the $q$-series along the positive imaginary axis into a Dirichlet series, and modularity becomes a functional equation. [quotetheorem:4269] [citeproof:4269] The cusp condition is what makes this Mellin transform clean at both ends of the integral. As $y\to\infty$, cusp decay follows from the absence of a constant term; as $y\to 0$, modularity converts the question back to decay at infinity. For Eisenstein series or other non-cusp modular forms, constant terms create divergent contributions unless correction terms are subtracted or the integral is regularized. The formula therefore identifies the completed Dirichlet series in its initial half-plane of convergence, but analytic continuation and the functional equation require the modular transformation argument that comes next. The completed $L$-function is usually written \begin{align*} \Lambda(f,s)=(2\pi)^{-s}\Gamma(s)L(f,s). \end{align*} The transformation law for $f$ relates the integral over $(0,1)$ to the integral over $(1,\infty)$ and yields the functional equation. At this point the remaining question is global rather than coefficientwise: how does the symmetry $z\mapsto -1/z$ of level $1$ modular forms appear in the analytic function built from the coefficients? The gamma factor in $\Lambda(f,s)$ is chosen so that the Mellin transform changes cleanly under this substitution. The theorem records the resulting symmetry of the completed $L$-function and explains why the center of the functional equation is $s=k/2$. [quotetheorem:4260] [citeproof:4260] This functional equation is the analytic endpoint of the course, but its exact form depends on the hypotheses. Level $1$ gives the simple involution $z\mapsto -1/z$ and hence the factor $i^k$ in this normalization; at higher level the Fricke involution, nebentypus character, and conductor change the completed function and the sign. The cusp condition again removes boundary terms, so the Mellin integral represents an entire completed $L$-function rather than a meromorphic object with correction terms. Thus the same Fourier coefficients that diagonalise Hecke operators also define Dirichlet series with Euler products and functional equations, provided they come from normalized cusp eigenforms in this setting. ## Boundaries of the Course The natural question at the end is what has been deliberately excluded. The course has stayed at level $1$, over $\mathbb C$, and within the classical analytic theory; this makes the main ideas visible without the additional bookkeeping of level structure, congruence subgroups, and coefficient fields. [remark: Congruences Between Modular Forms] Congruences such as Ramanujan's congruence \begin{align*} \tau(n)\equiv \sigma_{11}(n) \pmod{691} \end{align*} belong to the arithmetic theory of integral and $p$-adic modular forms. They require keeping track of coefficient rings rather than working only over $\mathbb C$. [/remark] This first boundary concerns coefficients: the course has used complex vector spaces because that is the cleanest setting for dimension, diagonalisation, and analytic continuation. [remark: Higher Level and Modular Curves] For congruence subgroups such as $\Gamma_0(N)$, the quotient has more cusps, more possible elliptic behaviour, and a richer Hecke algebra including operators at primes dividing $N$. The geometric object is a modular curve with level structure, and its compactification carries arithmetic information not visible in the level $1$ quotient alone. [/remark] The next boundary concerns geometry: changing the subgroup changes the quotient, and therefore changes both the computational basis problem and the Hecke operators available. [remark: Galois Representations] The sequel attaches Galois representations to normalized eigenforms, relating Fourier coefficients $a_p$ to traces of Frobenius elements. This direction explains why Hecke eigenvalues are not only analytic data but also encode arithmetic of field extensions and algebraic varieties. [/remark] The central lesson is that classical level $1$ modular forms are computable because geometry, algebra, analysis, and arithmetic all describe the same finite-dimensional objects. The fundamental domain controls zeros and dimensions, the ring theorem constructs the forms, the Hecke operators identify canonical bases, and the Mellin transform turns eigenvalue data into $L$-functions.