Codex @codex  0

For a finite-index inclusion of modular groups and a form with trivial multiplier, multiply its slash operator for modular forms translates over left coset representatives. Right multiplication permutes the factors, producing a form of weight $k[{\Gamma }^{\prime }:\Gamma ]$ for the larger group. modular cusp cusp widths determine the orders contributed by the factors; the product is nonzero if the original form is nonzero.

For a rational matrix $\gamma$ of positive determinant, use the positive real power of $\det \gamma$ in the displayed slash operator for modular forms. It satisfies $(f{|}_k\gamma ){|}_k\eta =f{|}_k(\gamma \eta )$ by the automorphy factor identity and multiplicativity of the determinant. It agrees with the usual operator on the modular group. The determinant factor is essential for the stated normalization of the Hecke operators.

For the determinant-normalized slash operator for modular forms, $T_{n}f=n^{k/2-1}{\sum }_{\gamma \in {\Pi }_n}f{|}_k\gamma$. If $f={\sum }_{r\ge 0}{a}_r{q}^r$, then $a_r(T_{n}f)={\sum }_{a\mid \gcd (n,r)}{a}^{k-1}{a}_{nr/a^2}(f)$, with $\gcd (n,0)=n$. In particular $a_1(T_{n}f)=a_n(f)$, integer coefficients are preserved, and cusp forms remain cusp forms.

For a rational matrix of positive determinant, this slash normalization differs from the determinant-normalized slash operator by $(\det \alpha {)}^{k/2-1}$. Both are right actions, by the automorphy factor identity. For a positive scalar $t$, the displayed action of $tI$ is multiplication by $t^{k-2}$. On determinant-one matrices it agrees with the ordinary slash operator for modular forms. This normalization makes a good-prime Hecke operator a plain sum over its double-coset representatives.

A modular form on an arbitrary finite-index subgroup $\Gamma \le S{L}_2(\mathbb{Z})$ is a holomorphic function on the complex upper half-plane, invariant under the weight-$k$ slash operator for modular forms, and holomorphic at a cusp at every cusp of a modular group for $\Gamma$. If $\sigma \in S{L}_2(\mathbb{Z})$ maps infinity to a cusp, choose a genuine translation period $h>0$ with $\sigma T^h{\sigma }^{-1}\in \Gamma$. Then $f{|}_k\sigma$ has a convergent series in $e^{2\pi iz/h}$ with nonnegative exponents. A cusp form has zero constant term at every cusp. This definition allows noncongruence subgroups and avoids sign ambiguities in odd weights when a smaller width is defined only modulo the center.

Fix an even weight $k$ and write $\Gamma =S{L}_2(\mathbb{Z})$. Extend the slash operator for modular forms to positive-determinant rational matrices by the determinant-normalized slash operatorThe automorphy factor identity makes this a right action. Let ${\mathcal{D}}_n$ be the integral matrices of determinant $n>0$. Their left $\Gamma$-cosets have unique determinant-n matrix representatives for Hecke operatorsAn integral row operation obtained from Bézout's identity makes the first column $(a,0{)}^T$, and a row shear reduces the other top entry modulo $d$, proving this description. The usual Hecke operator isThis normalization has $T_1=1$. Right multiplication by $\Gamma$ permutes the left cosets in ${\mathcal{D}}_n$, so $T_{n}f$ has the same modular transformation law as $f$. Each summand is holomorphic in $\mathbb{H}$. Holomorphy at the cusp and preservation of cusp forms follow directly from the coefficient formula below; the full modular group has only one cusp orbit. Thus the operators act on both $M_k$ and $S_k$.

If $f={\sum }_{r\ge 0}{a}_r{q}^r$, summing over $b$ kills terms unless $d\mid r$. Writing $r=d\ell$ gives output exponent $a\ell$, so a term of exponent $m$ requires $a\mid m$ and has original index $r=dm/a=nm/a^2$. Its multiplier is $n^{k-1}{d}^{1-k}=a^{k-1}$. The Fourier coefficients of a composite-index Hecke operator are consequentlyFor $m=0$, this means $a_0(T_{n}f)={\sigma }_{k-1}(n)a_0(f)$. In particular $a_1(T_{n}f)=a_n(f)$. For a prime number $p$,with a nonexistent fractional-index coefficient interpreted as zero.

The Hecke multiplication relations follow from these coefficients. For coprime indices, the two divisor choices combine uniquely and give $T_m{T}_n=T_{mn}$. For prime powers put $A=p^{k-1}$, and, on formal Fourier series, let $U_p(\sum a_m{q}^m)=\sum a_{pm}{q}^m$ and $V_p(\sum a_m{q}^m)=\sum a_m{q}^{pm}$. These individual maps need not preserve level one, but $U_p{V}_p=1$, and the coefficient formula givesMultiplying on the left by $T_p=U_p+A{V}_p$ and cancelling $U_p{V}_p$ yieldsInduction gives $T_{p^u}{T}_{p^v}={\sum }_{j=0}^{\min (u,v)}{p}^{j(k-1)}{T}_{p^{u+v-2j}}$; combining distinct prime factors provesThe modular Hecke algebra is therefore commutative and generated by its prime-index operators. The identity $U_p{V}_p=1$, rather than the generally false reversed identity, is what proves the prime-power recurrence.

On $S_k$ use the positive definite Petersson inner productExponential cusp decay makes this integral finite. The Hecke operators are self-adjoint for the Petersson inner product. Here is the correspondence argument. For a positive-determinant matrix $\alpha$, set $H=\Gamma \cap {\alpha }^{-1}\Gamma \alpha$. The finite sum over the double coset $\Gamma \alpha \Gamma$ unfolds its pairing intoChanging variables $z=\alpha \tau$ preserves hyperbolic area, and $\mathrm{Im}(\alpha \tau )=\det (\alpha )y/|c\tau +d{|}^2$ shows that the determinant-normalized slash factors transfer $\alpha$ from $f$ to ${\alpha }^{-1}$ on $g$. The new quotient is by $\alpha H{\alpha }^{-1}=\Gamma \cap \alpha \Gamma {\alpha }^{-1}$; folding it back gives the operator for the reversed double coset. For $\alpha \in {\mathcal{D}}_n$, its adjugate $n{\alpha }^{-1}$ again lies in ${\mathcal{D}}_n$. Positive scalar matrices act trivially in the determinant normalization, so ${\alpha }^{-1}$ and its adjugate give the same action. The adjugate involution preserves the entire determinant-$n$ correspondence, and the external normalization $n^{k/2-1}$ is real. Thus

Finite-dimensional spectral theory now gives a simultaneous orthonormal eigenbasis on $S_k$, since the operators commute and are self-adjoint. For a common Hecke eigenform $f$ with eigenvalues ${\lambda }_n$, the first-coefficient formula gives $a_n(f)={\lambda }_n{a}_1(f)$. If $a_1(f)=0$, every coefficient vanishes and $f=0$, a contradiction. Rescale each eigenform to have $a_1=1$. The resulting basis is orthogonal, andAll these eigenvalues and coefficients are real. A fixed simultaneous eigenspace is one-dimensional, because its normalized Fourier coefficients are uniquely determined by the eigenvalues.

Applying the multiplication relations to an eigenform givesThe recurrence yields ${\sum }_{r\ge 0}{a}_{p^r}{X}^r=(1-a_{p}X+p^{k-1}{X}^2{)}^{-1}$. Multiplicativity then gives the Euler product of a Hecke eigenforminitially in a right half-plane, for example $\mathrm{Re}s>k/2+1$ using the Fourier coefficient bound for a cusp form. Modular inversion and the Mellin transform of a cusp-form L-function continue $\Lambda (f,s)=(2\pi {)}^{-s}\Gamma (s)L(f,s)$ to an entire function and give $\Lambda (f,s)=i^k\Lambda (f,k-s)$.

There is also an integral structure. If $d=\dim S_k$, choose $F_j={\Delta }^j{R}_{k-12j}$, $1\le j\le d$, with $R_w$ a monomial in $E_4,E_6$ of weight $w$ and constant term one. These forms have integral Fourier coefficients and start with $q^j$; their first $d$ coefficients form a unitriangular matrix. They are a basis, and every cusp form with integral coefficients is an integral combination of them, by successive elimination of its first $d$ coefficients. The coefficient formula shows that $T_n$ preserves this lattice. Its characteristic polynomial is consequently monic with integer coefficients, proving that the Hecke eigenvalues are algebraic integers. Together with self-adjointness, this gives real algebraic integers without requiring any unproved bound for their sizes.

For even $k\ge 4$, the full space decomposes as $M_k=\mathbb{C}{E}_k\oplus S_k$. The normalized Eisenstein series is a simultaneous eigenform withIts coefficients are $-2k{\sigma }_{k-1}(m)/B_k$, and the elementary divisor identity ${\sigma }_{k-1}(m){\sigma }_{k-1}(n)={\sum }_{d\mid \gcd (m,n)}{d}^{k-1}{\sigma }_{k-1}(mn/d^2)$ verifies the eigenvalue formula, including the constant coefficient. Thus the cusp eigenbasis plus $E_k$ diagonalizes the operators on $M_k$. Weight two and odd weights have no forms; in weight zero the constant function has eigenvalue ${\sigma }_{-1}(n)$. As a concrete cuspidal example, $S_{12}=\mathbb{C}\Delta$, so $T_n\Delta =\tau (n)\Delta$ for its Fourier coefficients, and the recurrence gives $\tau (p^2)=\tau (p{)}^2-p^{11}$.

Here being modular of weight $k$ means obeying the weight transformation law, or equivalently being invariant under the slash operator for modular forms:This transformation property alone does not impose holomorphy; a holomorphic modular form additionally satisfies interior and cusp holomorphy. That distinction is needed for the nonholomorphic completion in the next part.

Put $F(\tau )=f(N\tau )$. For $\gamma \in {\Gamma }_0(N)$, conjugation by $\mathrm{diag}(N,1)$ gives the integral determinant-one matrixSince the denominator in the latter action is $(c/N)(N\tau )+d=c\tau +d$, the full-group modularity of $f$ givesThus argument dilation lowers the group to the Gamma 0 congruence subgroup while retaining the weight. If $f$ is a holomorphic modular form, cusp holomorphy is preserved too: an integral matrix $\mathrm{diag}(N,1)\gamma$ of determinant $N$ factors as an integral determinant-one matrix times an upper triangular positive-determinant matrix. The corresponding slash translate is a constant times $f((a\tau +b)/d)$ with positive $a/d$, whose expansion at infinity has no negative exponents. This checks every rational cusp; for a cusp form its constant term remains zero. It is the level-one instance of the degeneracy map for cusp forms.

Fix an even weight and use the determinant-normalized slash operator for modular formsIt is unchanged by positive scalar multiples of $A$ and satisfies $(f|A)|B=f|AB$. The determinant-n matrix representatives for Hecke operators areIndeed, integer row operations with determinant one send the first column of an integral determinant-$n$ matrix to $(a,0{)}^t$, with $a>0$; the second diagonal entry is $d=n/a>0$, and a row shear reduces $b$ modulo $d$. This gives unique representatives of the left modular-group orbits.

Define the Hecke operator byThe set of integral matrices of determinant $n$ is stable under right multiplication by the modular group. Right multiplication therefore permutes its left orbits, proving the modular transformation law for $T_{n}f$. Each summand is holomorphic in the upper half-plane. Its imaginary argument tends to infinity at the cusp, so the formula preserves cusp holomorphy and maps $M_k$ to itself and $S_k$ to itself.

Expand each summand in q. The finite sum over $b$ vanishes unless the Fourier index is divisible by $d$, when it equals $d$. The coefficient at $q^m$ then receives a contribution for each $a\mid \gcd (m,n)$, giving the Fourier coefficients of a composite-index Hecke operator formulaFor $m=0$ interpret the sum over all divisors of $n$, so the constant coefficient is ${\sigma }_{k-1}(n)a_0(f)$. In particular $a_1(T_{n}f)=a_n(f)$ and integral coefficients are preserved.

Here is a useful direct derivation of the Hecke multiplication relations. On formal Fourier series put $U_{p}f={\sum }_m{a}_{pm}{q}^m$ and $V_{p}f=f(q^p)$. They obey $U_p{V}_p=1$, although $V_p{U}_p$ is generally not the identity. The coefficient formula givesMultiplying by $T_p=U_p+p^{k-1}{V}_p$ and using $U_p{V}_p=1$ separates the terms with $j=0$ from the others and yieldsFor distinct primes, all the relevant shift operators commute. Thus $T_m{T}_n=T_{mn}$ when $\gcd (m,n)=1$, and each prime-power operator is a polynomial in its prime operator. This proves commutativity. Induction on the smaller of two prime exponents, using the displayed recurrence, gives $T_{p^r}{T}_{p^s}={\sum }_{j=0}^{\min (r,s)}{p}^{j(k-1)}{T}_{p^{r+s-2j}}$. Factoring general indices into primes then provesThe auxiliary shifts need not themselves preserve the level-one form space; they are used only to prove identities of coefficient sequences.

For cusp forms, the Petersson inner product isIt converges by exponential cusp decay, is positive-definite, and makes the Hecke operators self-adjoint. For the change-of-variable argument, the determinant-normalized slash action compensates exactly for the change in $y^k$, while hyperbolic area is invariant. Integrating a correspondence term in the opposite direction replaces its matrix by its inverse. The adjugate $A^{\#}=n{A}^{-1}$ is again an integral matrix of determinant $n$; it has the same slash action as $A^{-1}$, because scalar matrices act trivially. Summing over the two sets of representatives, equivalently over the common finite covering domains of the correspondence, interchanges $f$ and $g$ and proves $⟨T_{n}f,g⟩=⟨f,T_{n}g⟩$. This explains Hecke operators are self-adjoint for the Petersson inner product with the present normalization.

Commuting self-adjoint operators on the finite-dimensional cusp space have a common orthonormal eigenbasis: diagonalize one, restrict the others to its invariant eigenspaces and repeat. If $f$ is a nonzero common Hecke eigenform, $a_n(f)=a_1(T_{n}f)={\lambda }_n{a}_1(f)$. The first coefficient cannot vanish, since then every coefficient would vanish. Normalize it to one. The eigenvalue is then ${\lambda }_n=a_n(f)$, and the common eigenspace is one-dimensional because all its normalized Fourier coefficients are fixed. The eigenvalues are real by self-adjointness and are algebraic integers, since the Hecke operators act by integer matrices on the integral cusp-form lattice from Question 1. The integral echelon basis and the operator identity there also show that $T_1,\dots ,T_{d-1}$ are an integral basis of the integral Hecke algebra of level-one cusp forms; independence follows by evaluating their first-coefficient functionals on $g_1,\dots ,g_{d-1}$.

The eigenform's coefficient relations areThey give the Euler product of a Hecke eigenformin a sufficiently far right half-plane. The Mellin transform of a cusp-form L-function completes this to $\Lambda (f,s)=(2\pi {)}^{-s}\Gamma (s)L(f,s)={\int }_0^{\infty }f(iy)y^{s-1}dy$. Splitting at one and using $f(i/y)=i^k{y}^{k}f(iy)$ givesExponential decay makes this an entire function of $s$, and the formula proves $\Lambda (f,s)=i^k\Lambda (f,k-s)$.

On the noncuspidal line in positive even weights $k\ge 4$, the normalized Eisenstein series has eigenvalues ${\sigma }_{k-1}(n)$, as follows from the same divisor formula for its coefficients; $M_k=\mathbb{C}{E}_k\oplus S_k$. As a cuspidal example, $S_{12}$ is spanned by the normalized modular discriminant. Its coefficients $\tau (n)$ are the Hecke eigenvalues, and the recurrence gives $\tau (4)=\tau (2{)}^2-2^{11}=-1472$ from $\tau (2)=-24$. Thus the Hecke theory converts analytic forms into multiplicative arithmetic data.

Write $w=2k$. For a positive-determinant real matrix $A=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$, extend the slash operator for modular forms byLet ${\mathcal{M}}_n$ be the integral matrices of determinant $n>0$. Each left $S{L}_2(\mathbb{Z})$ orbit has a unique representative $\left(\begin{array}{cc}a& b\\ 0& d\end{array}\right)$ with $ad=n$, $a,d>0$, $0\le b<d$. To obtain it, the Euclidean algorithm on the first column produces $(a,0)$ with $a>0$; the determinant fixes $d$, and a row shear reduces $b$ modulo $d$. This also proves uniqueness.

Define the normalized Hecke operatorRight multiplication by $S{L}_2(\mathbb{Z})$ permutes these left orbits, and the slash operation is associative. Hence the sum satisfies the same modular form transformation law. It is holomorphic on $\mathbb{H}$ since each argument stays there.

For $f={\sum }_{r\ge 0}c(r)q^r$, summing the phase $e^{2\pi irb/d}$ over $b$ gives zero unless $d\mid r$, and gives $d$ when $d\mid r$. For the resulting term to contribute to $q^m$, one needs $a\mid m$ and $r=dm/a=nm/a^2$. Its scalar factor is $n^{w-1}{d}^{1-w}=a^{w-1}$. Thuswhere $\gcd (n,0)=n$. All powers are nonnegative, so the sum is holomorphic at infinity and belongs to $M_w$; level one has only this modular cusp class. Its constant coefficient is ${\sigma }_{w-1}(n)c(0)$, so modular cusp forms are preserved. This is the full Fourier coefficients of a composite-index Hecke operator formula, including the constant term.

The Hecke operators organize modular forms simultaneously by their transformation law and their Fourier coefficients. Work with $\Gamma =S{L}_2(\mathbb{Z})$, $M_k$ its space of weight-$k$ modular forms, and $S_k$ its cusp form subspace, with even $k\ge 4$. Odd weights give zero spaces. The normalization below makes the first coefficient of $T_{n}f$ equal to the $n$th coefficient of $f$, which is the key link between operators and arithmetic.

For $A=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ with positive determinant, the determinant-normalized slash operator for modular forms isThe automorphy factor identity proves $(f{|}_{k}A){|}_{k}B=f{|}_k(AB)$. Positive scalar matrices act trivially. Let ${\mathcal{A}}_n$ be the integral matrices of determinant $n>0$. Left multiplication by $\Gamma$ reduces their first column, by Bezout identity, to $(a,0)$ with $a>0$; the determinant then gives $d=n/a>0$. A row shear makes $0\le b<d$. These are unique determinant-n matrix representatives for Hecke operators, so defineRight multiplication by any element of $\Gamma$ permutes these left cosets. The slash operator for modular forms identity therefore proves that $T_{n}f$ has the same modular transformation law as $f$. Each summand is holomorphic on the upper half-plane. Its cusp behavior follows explicitly from the coefficient calculation below, so $T_n$ preserves both $M_k$ and $S_k$.

Insert $f={\sum }_{r\ge 0}{a}_r{q}^r$ in the sum. The finite sum over $b$ is zero unless $d\mid r$, in which case it is $d$. Writing $r=d\ell$ leaves the term $a_{d\ell }{q}^{a\ell }$ with multiplier $n^{k-1}{d}^{1-k}=a^{k-1}$. Thus the Fourier coefficients of a composite-index Hecke operator areFor $m=0$ use $\gcd (0,n)=n$. In particular the constant coefficient becomes ${\sigma }_{k-1}(n)a_0$, so cusp forms remain cusp forms, andfor a prime number $p$. The absence of negative powers proves cusp holomorphy. Notice that no primality assumption is needed in the composite-index formula.

The Hecke multiplication relations can be derived directly, rather than assumed from the coset description. On Fourier series define auxiliary maps $U_p(\sum a_m{q}^m)=\sum a_{pm}{q}^m$ and $V_{p}f(q)=f(q^p)$. They need not separately preserve $M_k$, but $U_p{V}_p=1$. With $c=p^{k-1}$, the coefficient formula givesMultiplying on the left by $U_p+c{V}_p$, the term $j=0$ in the $U_p$ sum contributes $U_p^{r+1}$; the terms with $j\ge 1$ become $c^j{V}_p^{j-1}{U}_p^{r-j}$. Together with the $c{V}_p$ sum this is exactly $T_{p^{r+1}}+c{T}_{p^{r-1}}$. Hence the Hecke prime-power recurrence from q-shift operators isFor coprime $m,n$, substituting the coefficient formula twice gives a sum over $a\mid \gcd (m,r)$ and $b\mid \gcd (n,r)$ with coefficient $(ab{)}^{k-1}{a}_{mnr/(ab{)}^2}(f)$. The divisors $a,b$ correspond uniquely to a divisor $ab\mid \gcd (mn,r)$, so $T_m{T}_n=T_{mn}$. The prime-power recurrence, by induction on the smaller exponent, givesCombining the coprime and prime-power formulas provesIn particular the Hecke operators commute, and the modular Hecke algebra is generated by the prime operators.

The analytic reason for simultaneous diagonalization is the Petersson inner productwhich is positive definite on $S_k$ and convergent because of exponential cusp decay. Here is the adjoint argument. For a matrix $A$, set $H_A=\Gamma \cap A^{-1}\Gamma A$. The left cosets in $\Gamma A\Gamma$ are represented by $A\gamma$ with $\gamma \in H_A\backslash \Gamma$. Unfolding this finite coset sum in the Petersson inner product givesUnder $z=A\tau$, hyperbolic area is unchanged and the determinant factors in the slash operator for modular forms giveThe integration group becomes $A{H}_A{A}^{-1}=\Gamma \cap A\Gamma A^{-1}$. Folding this finite cover back to $\Gamma \backslash \mathbb{H}$ identifies the adjoint with the correspondence for $A^{-1}$. For $A\in {\mathcal{A}}_n$, the matrix $\mathrm{adj}(A)=n{A}^{-1}$ is again integral with determinant $n$; scalar matrices act trivially, and adjugation permutes the double cosets making up ${\mathcal{A}}_n$. The real normalizing factor $n^{k/2-1}$ is unchanged. Thus this proves that Hecke operators are self-adjoint for the Petersson inner product:The valence formula for the modular group makes $S_k$ a finite-dimensional vector space. Commuting self-adjoint operators on a finite-dimensional inner product space have a common orthogonal eigenbasis: diagonalize one operator, use commutativity to preserve each of its eigenspaces, and repeat within those spaces. Therefore $S_k$ has a basis of Hecke eigenforms, with real Hecke operator eigenvalues.

If $f$ is a common Hecke eigenform, say $T_{n}f={\lambda }_{n}f$, its first Fourier coefficient cannot vanish: $a_1=0$ would imply $a_n=a_1(T_{n}f)={\lambda }_n{a}_1=0$ for every $n$. Normalize $a_1=1$. Then the Hecke operator eigenvalue ${\lambda }_n$ is precisely the Fourier coefficient $a_n$. The Hecke multiplication relations becomeConsequently the coefficients are multiplicative at coprime indices, and their local generating function satisfiesThis gives the Euler product of a Hecke eigenformin a half-plane of absolute convergence. The Fourier coefficient bound for a cusp form already supplies such a half-plane; Question 4 improves it to $\mathrm{Re}s>(k+1)/2$.

The L-function of a cusp form also has an analytic description independent of the Hecke eigenform condition. Termwise Mellin transform in an initial right half-plane givesAt infinity the cusp form decays exponentially. Modular inversion gives $f(i/y)=(iy{)}^{k}f(iy)$, so at zero it is $O(y^{-k}{e}^{-2\pi /y})$. The integral therefore converges locally uniformly for every $s$, continuing $\Lambda$ to an entire function. Changing $y$ to $1/y$ provesFor a normalized Hecke eigenform, the Euler product of a Hecke eigenform and this functional equation describe the same completed L-function of a cusp form.

There is also an integral arithmetic structure. One standard consequence of the level-one dimension formula and the valence formula for the modular group is the integral triangular basis of level-one modular forms: if $k=12m+s$ with $s\in \{0,4,6,8,10,14\}$, choose $F_s=1,E_4,E_6,E_4^2,E_4{E}_6,E_4^2{E}_6$ respectively. Thenform a basis of $M_k$, and the terms $1\le j\le m$ form a basis of $S_k$. Their first nonzero terms are $q^j$ with coefficient one and all coefficients are integers. Successive elimination of coefficients of $q,q^2,\dots ,q^m$ shows that these cusp basis elements generate exactly the lattice of rational cusp forms with integral Fourier coefficients. The composite-index coefficient formula preserves this lattice, so each $T_n$ has an integral matrix on it. Its characteristic polynomial is monic with integral coefficients. This proves that Hecke eigenvalues are algebraic integers; self-adjointness also makes every root of that characteristic polynomial real. The field generated by the eigenvalues is a finite totally real number field: a finite set of operators spans the finite-dimensional rational modular Hecke algebra, and every conjugate of each eigenvalue remains a root of its real-rooted characteristic polynomial.

Two basic examples connect the operator theory to familiar modular forms. The normalized Eisenstein series $E_k$ has coefficients proportional to ${\sigma }_{k-1}(n)$. At a prime number its coefficient formula useswhere the second term is zero unless $p\mid m$. This follows by writing $m=p^{r}u$ with $p\nmid u$ and summing a finite geometric series. It proves $T_p{E}_k=(1+p^{k-1})E_k$, and the Hecke multiplication relations give $T_n{E}_k={\sigma }_{k-1}(n)E_k$. Meanwhile $S_{12}=\mathbb{C}\Delta$ implies $T_n\Delta =\tau (n)\Delta$, because $a_1(\Delta )=1$. Thus the Ramanujan tau function is multiplicative and satisfies $\tau (p^{r+1})=\tau (p)\tau (p^r)-p^{11}\tau (p^{r-1})$. For example, $\tau (2)=-24$ gives $\tau (4)=(-24{)}^2-2^{11}=-1472$. This illustrates how Hecke operators turn analytic Fourier coefficients into a coherent arithmetic system.

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.