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The normalized Eisenstein series satisfy $E_4=1+240{\sum }_{m\ge 1}{\sigma }_3(m)q^m$ and $E_6=1-504{\sum }_{m\ge 1}{\sigma }_5(m)q^m$. Their weights are four and six. The valence formula for the modular group makes the normalized weight-twelve cusp form $\Delta$ have one simple cusp zero and no zero in $\mathbb{H}$. Comparing its first coefficient gives $\Delta =(E_4^3-E_6^2)/1728$.

Odd weights vanish because $-I$ acts by $(-1{)}^k$. Weight two vanishes by the valence formula for the modular group: the forced zero at $i$ alone contributes $1/2>2/12$. For every remaining nonnegative even weight, except two, choose $a,b\ge 0$ with $4a+6b=k$. Subtract the constant coefficient of $f$ times $E_4^a{E}_6^b$. The remainder is a cusp form; division by $\Delta$ produces a holomorphic modular form of weight $k-12$. The cusp and interior orders justify this division. Induction on the weight provesThe integral triangular basis of level-one modular forms is $F_j={\Delta }^j{E}_4^{a_j}{E}_6^{b_j}$, with $4a_j+6b_j=k-12j\ge 0$ and $k-12j\ne 2$. The induction just given shows these span, and their distinct leading terms $q^j$ show linear independence. Every coefficient is integral by the hypothesis on $\Delta$ and the displayed Eisenstein expansions. In $f={\sum }_j{c}_j{F}_j$, compare coefficients successively: $c_0=a_0$, then $c_1$ is $a_1$ minus an integer combination of already integral coefficients, and similarly for every $j$. The largest permitted $j$ is at most $\lfloor (k+1)/12\rfloor$, so the stipulated initial coefficients make every $c_j$ integral. Therefore all Fourier coefficients of $f$ are integers.

Hecke operators give a commuting family of arithmetic symmetries of modular forms. They preserve the weight and cusp forms, and diagonalizing them turns the Fourier coefficients of a modular form into multiplicative arithmetic data. Odd weight spaces are zero by the action of $-I$; in weight zero the operators act on constants by $T_{n}1={\sigma }_{-1}(n)$. For the substantive theory, fix an even positive weight $k$ and $\Gamma =S{L}_2(\mathbb{Z})$, and use the following normalization consistently.

Extend the slash operator for modular forms to $G{L}_2^{+}(\mathbb{Q})$ byIt is a right action and positive scalar matrices act trivially. Let ${\mathcal{D}}_n$ be the integral matrices of determinant $n>0$. The determinant-n matrix representatives for Hecke operators for $\Gamma \backslash {\mathcal{D}}_n$ areInteger row operations reduce the first column to $(a,0)$, after which a row shear reduces $b$ modulo $d$; uniqueness follows from these same conditions. The Hecke operator isRight multiplication by $\Gamma$ permutes the left cosets, so $T_{n}f$ is again invariant under the weight-$k$ slash operator for modular forms. Each summand is a holomorphic function on the complex upper half-plane, and the coefficient formula below proves holomorphy at a cusp. Thus $T_n$ acts on $M_k(\Gamma )$ and preserves $S_k(\Gamma )$. Geometrically, these operators sum over index-$n$ sublattices, or equivalently the associated finite-degree correspondences between complex tori; the factors above compensate for the weight's response to lattice scaling.

Write $f={\sum }_{r\ge 0}{a}_r{q}^r$. The sum over $b$ is zero unless $d\mid r$, in which case it is $d$. Substituting into the displayed definition proves the Fourier coefficients of a composite-index Hecke operator formulawith $\gcd (n,0)=n$. In particular, $a_0(T_{n}f)={\sigma }_{k-1}(n)a_0(f)$ and $a_1(T_{n}f)=a_n(f)$. At a prime this becomeswhere $a_{r/p}=0$ if $p\nmid r$.

The Hecke multiplication relations areThey can be checked by applying the coefficient formula twice and regrouping the divisors. More explicitly, coprime indices factor independently, so $T_m{T}_n=T_{mn}$ if $(m,n)=1$. At a prime, separating the terms in the divisor sum according to whether one more factor $p$ is available givesIteration gives $T_{p^a}{T}_{p^b}={\sum }_{j=0}^{\min (a,b)}{p}^{j(k-1)}{T}_{p^{a+b-2j}}$; combining the prime factors gives the full relation. Hence all Hecke operators commute, $T_1=1$, and the Hecke algebra of modular forms is generated by the $T_p$ together with scalars.

The Petersson inner product on the finite-dimensional cusp form space isThe integrand is invariant and the integral converges because cusp forms decay exponentially at the cusp. The Hecke operators are self-adjoint for the Petersson inner product at level one. To see the underlying adjoint calculation, unfold the finite correspondence for a determinant-$n$ matrix and change variable $\tau \mapsto \alpha \tau$. The invariant hyperbolic measure and the determinant-normalized slash factors transfer $\alpha$ from one side of the inner product to ${\alpha }^{-1}$. Scalar matrices act trivially, so one can replace ${\alpha }^{-1}$ by the integral adjugate $n{\alpha }^{-1}$, which again has determinant $n$. This reverses the correspondence and permutes the same collection of terms, with the same real prefactor $n^{k/2-1}$. Summing givesCommuting self-adjoint operators on a finite-dimensional inner product space admit a common orthonormal basis of eigenvectors. Thus $S_k(\Gamma )$ has a basis of simultaneous Hecke eigenforms, with real eigenvalues.

A nonzero simultaneous cuspidal Hecke eigenform has $a_1\ne 0$: if $T_{n}f={\lambda }_{n}f$ and $a_1=0$, comparison of the first coefficients gives $a_n={\lambda }_n{a}_1=0$ for every $n\ge 1$, so $f=0$. Normalize $a_1=1$. ThenConsequently the full Fourier series is determined by the eigenvalues, and a simultaneous eigenvalue system has a one-dimensional eigenspace. The Hecke multiplication relations translate intoIn particular the coefficients are multiplicative at coprime indices and are determined by the prime coefficients. This is the elementary multiplicity-one statement for normalized level-one Hecke eigenforms.

The Hecke eigenvalues are algebraic integers. One way to explain the arithmetic input is to use the rational structure supplied by the graded ring $M_{\ast }(\Gamma ,\mathbb{Q})=\mathbb{Q}[E_4,E_6]$. The rational cusp form space has a basis with integral Fourier coefficients, for example from appropriate monomials in $E_4,E_6,\Delta$. Its subgroup of forms whose entire expansions are integral is a full lattice: a finite initial coefficient map is injective by the valence formula for the modular group, so it embeds this subgroup into a finite-rank integer module. The coefficient formula shows that every $T_n$ preserves this lattice, and therefore has a monic integral characteristic polynomial. Its eigenvalues are algebraic integers. Simultaneous diagonalization also makes their number field finite over $\mathbb{Q}$; conjugating the rational coefficient data gives another eigenform, whose eigenvalues are again real by self-adjointness. Thus the coefficient field is totally real.

For a normalized cuspidal Hecke eigenform, define its L-function of a cusp form by $L(f,s)={\sum }_{n\ge 1}{a}_n{n}^{-s}$. The coefficient relations yield the Euler product of a Hecke eigenformIndeed the prime-power recurrence gives ${\sum }_{r\ge 0}{a}_{p^r}{X}^r=(1-a_{p}X+p^{k-1}{X}^2{)}^{-1}$, and multiplicativity assembles these factors. These identities hold analytically in an initial right half-plane, as well as formally. For instance boundedness of $y^{k/2}|f(\tau )|$ on the full complex upper half-plane gives $a_n=O(n^{k/2})$, so $\mathrm{Re}s>k/2+1$ suffices for absolute convergence.

The Mellin transform of a cusp-form L-function relates this arithmetic to the modular transformation:Splitting at one and using $f(i/y)=i^k{y}^{k}f(iy)$ givesExponential cusp decay makes the last integral entire in $s$. This proves the analytic continuation and functional equation of the L-function of a cusp form, complementing its Euler product. The root number is $i^k=(-1{)}^{k/2}$.

Two examples show the scope of the theory. For even $k\ge 4$, the Eisenstein series $E_k$ is an eigenform with $T_n{E}_k={\sigma }_{k-1}(n)E_k$. This follows from its divisor-sum expansion and the same multiplicativity relations; its associated positive-index Dirichlet series is proportional to $\zeta (s)\zeta (s-k+1)$. The constant coefficient gives $M_k(\Gamma )=\mathbb{C}{E}_k\oplus S_k(\Gamma )$, so an eigenbasis of the cusp space together with $E_k$ gives an eigenbasis of the whole modular-form space. The ordinary Petersson inner product above is only used on the cusp space, where it converges. In weight twelve the valence formula for the modular group gives $\dim S_{12}=1$: cancellation of the first coefficients of two independent cusp forms would give a nonzero form with order at infinity at least two, exceeding $12/12$. Thus the normalized modular discriminant is a Hecke eigenform; writing $\Delta ={\sum }_{n\ge 1}\tau (n)q^n$ gives the Ramanujan tau function relations $\tau (mn)=\tau (m)\tau (n)$ for coprime $m,n$ and $\tau (p^{r+1})=\tau (p)\tau (p^r)-p^{11}\tau (p^{r-1})$.

Apply the Serre derivative in weight twelve to the modular discriminant. The preceding result makes $D_{12}\Delta =D\Delta -E_2\Delta$ a weight-fourteen cusp form. The vanishing of weight-fourteen level-one cusp forms follows directly from the valence formula for the modular group: a nonzero such form has order at infinity at least one, and $f(i)=i^{14}f(i)=-f(i)$ forces order at $i$ at least one. Thus its weighted number of zeros would be at least $1+1/2>14/12$, a contradiction. Therefore $D\Delta =E_2\Delta$.

Compare the Fourier coefficients at positive indices using $\Delta ={\sum }_{n\ge 1}\tau (n)q^n$ and $E_2=1-24{\sum }_{r\ge 1}{\sigma }_1(r)q^r$. This gives $n\tau (n)=\tau (n)-24{\sum }_{r=1}^{n-1}{\sigma }_1(r)\tau (n-r)$. For the Ramanujan tau function, the concise recurrence isAt $n=1$ the sum is empty and both sides are zero; normalization supplies $\tau (1)=1$. For instance the next coefficient is $\tau (2)=-24$.

Use the valence formula for the modular group: for a nonzero modular form of weight $k$ on $S{L}_2(\mathbb{Z})$,The sum contains one representative of each other modular group orbit of zeros. The orders are nonnegative because $f$ is holomorphic on the complex upper half-plane and at infinity. The half and third weights account for the elliptic stabilizers of the modular group.

In weight two, the transformation under $S=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$ at its fixed point gives $f(i)=i^{2}f(i)=-f(i)$. Thus any nonzero $f$ would have $v_i(f)\ge 1$. The valence formula for the modular group would give a left side at least $1/2$ and a right side $1/6$, which is impossible. Therefore the unheaded request is answered by vanishing of weight-two level-one modular forms:

A modular form of integer weight $k$ for a finite-index subgroup $\Gamma \le S{L}_2(\mathbb{Z})$ is a holomorphic function on the complex upper half-plane satisfying $f{|}_k\gamma =f$ for $\gamma \in \Gamma$ and holomorphic at every cusp of a modular group. Explicitly, for each $\sigma \in S{L}_2(\mathbb{Z})$ choose a positive integer $h_{\sigma }$ with $\sigma T^{h_{\sigma }}{\sigma }^{-1}\in \Gamma$. Then $f{|}_k\sigma$ is periodic with period $h_{\sigma }$, and its expansion in $q_{h_{\sigma }}=e^{2\pi iz/h_{\sigma }}$ must have no negative powers. Equivalently it must be bounded as $\mathrm{Im}z\to \infty$, uniformly in one period. This definition applies to noncongruence subgroups too. If $-I\in \Gamma$, nonzero forms require even weight.

Put $d=[S{L}_2(\mathbb{Z}):\Gamma ]$ and let $f\ne 0$. Form the product over left cosetsChanging a representative does not affect a factor. Right multiplication permutes the cosets, so the product is a nonzero level-one modular form of weight $kd$. It is holomorphic on the upper half-plane and at infinity, since every factor is bounded there by cusp holomorphy.

Let $h$ be the least positive integer with $T^h\in \Gamma$, and write $f={\sum }_{n\ge 0}{b}_n{q}_h^n$. If $r$ is its first nonzero index, the $h$ distinct cosets $\Gamma T^j$, $0\le j<h$, give the factors $f(z+j)$. Their product has leading term a nonzero constant times $e^{2\pi irz}=q^r$. All remaining factors are bounded at infinity. Hence $F(z)=O(e^{-2\pi r\mathrm{Im}z})$, so its level-one order satisfies $v_{\infty }(F)\ge r$. Notice that using just the single factor $f$ would only give $r/h$; the entire translation orbit is needed.

The permitted valence formula for the modular group bounds $v_{\infty }(F)\le kd/12$, because all its other weighted zero orders are nonnegative. Thus $r\le kd/12$. With $R=\lfloor kd/12\rfloor$, the map $f\mapsto (b_0,\dots ,b_R)$ is injective: a nonzero form in its kernel would have $r\ge R+1$. This proves the dimension bound for modular forms on a finite-index subgroup:

For the dilation assertion, the original PDF states $N_{1}D\mid N$, a product-divisibility condition, rather than the TeX's chain $N_1\mid D\mid N$. Use the PDF condition. Put $A=\mathrm{diag}(D,1)$. If $\gamma =\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\in {\Gamma }_0(N)$, thensince $N_{1}D\mid c$. Therefore, for $g(z)=f(Dz)$, either direct substitution or the slash operator for modular forms gives $g{|}_k\gamma =g$. It is holomorphic on the upper half-plane.

To verify every cusp, take $\sigma \in S{L}_2(\mathbb{Z})$ and choose $\tau \in S{L}_2(\mathbb{Z})$ taking infinity to the rational cusp $A\sigma \infty$. Then $B={\tau }^{-1}A\sigma$ is rational upper triangular, with positive ratio of its two diagonal entries. Consequentlyis a nonzero constant factor times $f{|}_k\tau$ evaluated at an affine map $z\mapsto uz+v$ with $u>0$. The imaginary part tends to infinity, where $f{|}_k\tau$ is bounded. Thus $g{|}_k\sigma$ is bounded too. Together with its periodicity this proves cusp holomorphy by the Riemann removable singularity theorem. Hence the oldform by argument dilation satisfies

Finally ${\Gamma }_0(3)$ has index four: reduction modulo three gives a transitive $S{L}_2({\mathbb{F}}_3)$ action on the four points of ${\mathbb{P}}^1({\mathbb{F}}_3)$, and ${\Gamma }_0(3)$ is the inverse image of the stabilizer of infinity. Reduction is surjective, since elementary determinant-one matrices generate $S{L}_2({\mathbb{F}}_3)$ and lift integrally. The dimension bound gives $\dim M_4({\Gamma }_0(3))\le 2$.

The normalized Eisenstein series $E_4(z)$ and the oldform $E_4(3z)$ both belong to this space. Their Fourier expansion of a normalized Eisenstein series starts asA linear relation between them has zero constant term only if its two scalars sum to zero; its $q$ coefficient then forces both scalars to vanish. They are independent, proving the weight-four modular forms on Gamma 0 3 description

Let $N+1=\dim M_k(S{L}_2(\mathbb{Z}))$. For each $0\le i\le N$, the dimension formula, equivalently the valence formula for the modular group, ensures thatfor some nonnegative integers $a_i,b_i$. DefineThe modular discriminant, $E_4$, and $E_6$ have integral Fourier coefficients and leading terms $q$, $1$, and $1$, respectively. HenceTheir distinct orders of vanishing make the $h_i$ linearly independent, so they form a basis.

Starting with $f_N=h_N$, define $f_i$ downwards by subtracting from $h_i$ the integral multiples of $f_{i+1},\dots ,f_N$ needed to kill the coefficients of $q^{i+1},\dots ,q^N$. This integer Gaussian elimination preserves all integral coefficients and givesThis is the integral echelon basis of level-one modular forms.

Let $\rho =e^{2\pi i/3}$. For a nonzero meromorphic modular form $f$ of weight $k$, the valence formula for the modular group isHere $v_z(f)$ is positive for a zero and negative for a pole, and the factors $1/2$ and $1/3$ account for the stabilizers of the two elliptic points. Equivalently, the weighted number of zeros minus poles in $\Gamma (1)\backslash \mathfrak{h}$ is $k/12-v_{\infty }(f)$.

A level-one modular form with rational Fourier coefficients is a weighted homogeneous polynomial with rational coefficients in the normalized Eisenstein series $E_4,E_6$. Subtract its constant coefficient times a same-weight monomial. The remainder is a cusp form; divide it by ${\Delta }_0=(E_4^3-E_6^2)/1728=q+O(q^2)$. The valence formula for the modular group shows that ${\Delta }_0$ has no interior zeros, so this quotient is holomorphic and has rational coefficients. Induction on weight proves the assertion. Euler's Bernoulli number formula for even zeta values then gives the corresponding rational weighted polynomial in the unnormalized lattice Eisenstein sums $G_4,G_6$.

A nonzero weight-fourteen cusp form would have order at infinity at least one and order at $i$ at least one, since $i^{14}=-1$. The valence formula for the modular group would then give $1+1/2\le 14/12$, which is impossible.

A weight-two modular form obeys $f(i)=i^{2}f(i)=-f(i)$, so it vanishes at $i$. The valence formula for the modular group would then have left side at least $1/2$, but right side $2/12=1/6$. All orders are nonnegative by holomorphy, giving a contradiction for nonzero $f$.