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Use the contact process convention that each infected vertex recovers at rate one and sends infection arrows to each neighbour at rate $\lambda$. Starting from a finite set, the process stays finite and is nonexplosive: its total birth rate is at most $\lambda (d+1)$ times its current size, so a linear-rate branching process dominates the number of births. The bound involving $(d-1{)}^{-1}$ presupposes the branching-tree case $d\ge 2$.

For a nonempty finite infected set $A$, put $a=|A|$, let $e(A)$ count its internal edges, and let $b(A)$ count edges with exactly one endpoint in $A$. Each such boundary edge contributes infection rate $\lambda$, even when several lead to the same vacant vertex. Since the induced graph is a forest, $e(A)\le a-1$, and the edge boundary of a finite forest in a regular tree givesA recovery multiplies ${\nu }_{\rho }(A)={\rho }^a$ by ${\rho }^{-1}$, while an infection multiplies it by $\rho$. Therefore its infinitesimal generator isSubstituting the boundary lower bound givesAt the absorbing empty set the generator is zero, rather than this nonempty-set expression. If $\lambda \rho (d-1)\ge 1$, the generator is nonpositive in every state. Both ${\nu }_{\rho }$ and its generator are bounded, since $a{\rho }^a$ is bounded and $b(A)\le (d+1)a$. Applying Dynkin's formula over successive time intervals with the Markov property proves the exponential cardinality supermartingale for the contact process on a tree. In particular,If $\tau$ is the extinction time, $1_{\{\tau \le t\}}\le {\nu }_{\rho }({\xi }_t^A)$, so ${\mathbb{P}}_{\lambda }(\tau <\infty )\le {\rho }^{|A|}<1$. Choosing $\rho \in [1/(\lambda (d-1)),1)$ proves global survival whenever $\lambda >1/(d-1)$. Obtaining a strict improvement of the threshold requires an additional comparison.

Here is a two-vertex branching comparison for contact-process survival. Delete one edge incident to a starting vertex and retain its forward rooted subtree, so every vertex has $d$ children. Partition it into cells as follows: pair each cell root $u$ with one distinguished child $v$; the other $d-1$ children of $u$ and all $d$ children of $v$ become roots of child cells, and repeat recursively. Each cell thus has $2d-1$ child cells.

Suppress arrows back from a child cell into its parent, allow each child cell to be activated only once, and let infection inside each two-vertex cell evolve with the usual recoveries and reinfections until both vertices are healthy. A child cell is activated at its root by the first outgoing arrow to it while the corresponding parent vertex is infected. This process is contained in the original infection process under the graphical representation of the contact process: only infection paths have been removed. The Poisson clocks inside a child cell and on its outgoing edges have not previously been used. The Strong Markov property and independence of the clocks therefore make different cells' offspring counts independent and identically distributed. Dependence among different children of the same cell does not matter. The cell genealogy is a Galton-Watson process.

Let $a$ be the probability that a cell, initially infected only at $u$, sends an arrow along one specified outgoing edge from $u$ before internal extinction. Let $b$ be the corresponding probability for one specified outgoing edge from $v$, and let $c$ be the probability of success for the first target when both cell vertices are initially infected. First-step conditioning on the three nonempty pair states givesFor example, from the state with only $u$ infected, recovery at rate one fails, an internal arrow at rate $\lambda$ leads to the two-infected state, and the target arrow at rate $\lambda$ succeeds. From the state with only $v$ infected, the target cannot be reached until $u$ is infected. Symmetry identifies this latter success probability with $b$. Solving these equations yieldsBy linearity of expectation, the offspring mean is $M(\lambda )=(d-1)a+db$. At the comparison rate ${\lambda }_0=1/(d-1)$,These finite-state probabilities are continuous in $\lambda$. Hence some $0<{\lambda }^{\prime }<{\lambda }_0$ still has $M({\lambda }^{\prime })>1$. The branching-process extinction criterion gives positive probability of infinitely many activated cells. Nonexplosion prevents infinitely many activations in bounded time, and each activation is connected to its ancestor by an infection path. Thus infection survives for all time on that event. The global survival threshold of the contact process consequently satisfiesRetaining reinfection inside pairs is what supplies the strict improvement; a comparison that permits only one infection per vertex would give merely the non-strict bound.

For the local-survival estimate, set $h(x)={\rho }^{g(x)}$ and $w_{\rho }(A)={\sum }_{x\in A}h(x)$. Each vertex has one neighbour of weight ${\rho }^{-1}h(x)$ and $d$ neighbours of weight $\rho h(x)$. Recoveries remove their vertex weights. Ignoring the restriction that an infection target must be vacant only increases the birth contribution, soThese weights have finite expectations on finite time intervals: a birth-only branching process with the same spatial arrows has expected weight at most $w_{\rho }(A)e^{\lambda ({\rho }^{-1}+d\rho )t}$. This justifies localization and the expectation identity for the generator. If $c=1-\lambda ({\rho }^{-1}+d\rho )>0$, Dynkin's formula and the integrating-factor form of the Gronwall inequality giveFor $d\ge 2$, take $\rho =d^{-1/2}$, which minimizes ${\rho }^{-1}+d\rho$ and makes it $2\sqrt{d}$. Starting from $\{x\}$ then givesIn particular the expected total occupation time of $x$ is finite. Its recovery clock has rate one, so the expected number of recovery marks that occur while $x$ is infected equals this occupation-time integral and is finite. Each infection episode ends almost surely at a recovery. Infection at unbounded times would therefore require infinitely many such recoveries, an event of probability zero. This height-weighted local extinction bound for the contact process provesThis is a bound for the local survival threshold of the contact process, not just a bound on single-time infection probabilities. For the degree-two tree $d=1$, the same local conclusion follows by choosing $\rho <1$ sufficiently close to one for each $\lambda <1/2$; the displayed expression $1/(d-1)$ for the global threshold is not a finite numerical assertion in that case.

In bond percolation on the square lattice, every nearest-neighbour edge independently has state one, meaning open, with probability $p$, and state zero, meaning closed, with probability $1-p$. Thus the configuration law is the product measure ${\mathbb{P}}_p={⨂}_{e\in E}\mathrm{Bernoulli}(p)$. An open path in bond percolation uses only open edges. An increasing event $A$ satisfies $\omega \in A,{\omega }^{\prime }\ge \omega \Rightarrow {\omega }^{\prime }\in A$: opening further edges cannot destroy the event.

The Harris-FKG inequality for two increasing events isFor disjoint occurrence of increasing events, write $A\square B$ when disjoint sets of open edges separately witness $A$ and $B$, so each event is forced by its own witness regardless of the other edges. The Van den Berg-Kesten inequality statesFor events given by cylinder sets these are finite-coordinate assertions. Connection events have finite open-path witnesses; truncating to finite boxes and taking increasing limits gives the same disjoint-occurrence bound for them. Sharing a vertex between the two witness paths is permitted; sharing an edge is not.

For $1\le m<n$, take a simple open path from the origin to the radius-$n$ boundary and let $x$ be its first vertex on the radius-$m$ boundary. Its initial segment witnesses $0\leftrightarrow x$. If $z$ is its final vertex, then $\Vert z-x{\Vert }_{\infty }\ge n-m$. Consequently the remaining segment reaches $x+\partial {\Lambda }_{n-m}$, and its initial portion until that visit witnesses $x\leftrightarrow x+\partial {\Lambda }_{n-m}$. The two segments use disjoint edges. HenceThe union bound, Van den Berg-Kesten inequality and translation invariance of the product measure now give the weighted BK boundary-splitting estimateFor $m=n$ the same bound follows directly from the union bound, with $g_0=1$. With the conventional $\partial {\Lambda }_0=\{0\}$, $m=0$ gives equality.

Write $C(0)$ for the percolation cluster of the origin. By summing its membership indicators, the percolation susceptibility isIf it is finite, then $a_m\to 0$. For $0<p<1$, choose $m$ with $0<a_m=a<1$. Iterating the boundary-splitting estimate with this fixed $m$ givesFor $k\ge 2m$, $\lfloor k/m\rfloor \ge k/(2m)$, so $g_k\le \exp [-(-\log a)k/(2m)]$. To handle all smaller positive radii without an unspecified prefactor, observe that $g_k\le g_1=1-(1-p{)}^4<1$. ThereforeFor $1\le k<2m$, the second chosen rate gives $e^{-\gamma k}\ge g_1$, and for larger $k$ the first rate applies. At $k=0$ both sides equal one. The endpoint $p=0$ has $g_k=0$ for every $k\ge 1$, so any positive rate works; $p=1$ is excluded by finite susceptibility. This proves exponential one-arm decay from finite susceptibility with the required unit prefactor.

For real $\lambda$, the normal moment-generating formula gives $\mathbb{E}{e}^{\lambda (B_t-B_s)}=e^{{\lambda }^2(t-s)/2}$. Using independent increments and conditioning on the Brownian filtration,All terms are integrable by the same normal formula. Thus the displayed process is an Exponential martingale for Brownian motion.

For the interval exit time, we first show both almost-sure finiteness and finite expectation. On $\{T>n\}$, the position at time $n$ lies in $(-b,a)$. If $X_{n+1}-X_n\ge a+b$, the continuous path must exit by time $n+1$. This increment is independent of ${\mathcal{F}}_n$ and has law $N(\mu ,1)$, so the conditional probability of this event is a fixed $c>0$. ThereforeThis geometric tail bound from a uniform escape probability implies $T<\infty$ almost surely and $\mathbb{E}T<\infty$.

Take $\lambda =-2\mu$. Then $e^{-2\mu X_t}=e^{-2\mu B_t-2{\mu }^{2}t}$ is the exponential martingale just proved. Its stopped values lie between $e^{-2\mu a}$ and $e^{2\mu b}$. The optional stopping theorem at $T\wedge t$, followed by bounded convergence, gives $\mathbb{E}{e}^{-2\mu X_T}=1$. With $v=\mathbb{P}(X_T=a)$ and the two possible boundary values,Solving yieldsNext apply bounded-time optional stopping to $B_t$. Since $B_{T\wedge t}=X_{T\wedge t}-\mu (T\wedge t)$, we obtain $\mathbb{E}{X}_{T\wedge t}=\mu \mathbb{E}(T\wedge t)$. The stopped position is bounded, and the stopped time increases to an integrable $T$. Passing to the limit gives $\mu \mathbb{E}T=av-b(1-v)$. ThereforeAs a check, letting $\mu \downarrow 0$ gives the driftless formulas $v\to b/(a+b)$ and $\mathbb{E}T\to ab$. The expectation calculation uses a justified stopping limit, rather than assuming that optional stopping holds at every unbounded time.

In discrete time, the bounded-time optional stopping theorem says that an integrable supermartingale $X$ and bounded stopping times $S\le T$ satisfy $\mathbb{E}[X_T\mid {\mathcal{F}}_S]\le X_S$. For a martingale there is equality. The assertion at a possibly unbounded, almost-surely finite stopping time $T$ also holds with $S=0$ if $\{X_{T\wedge n}:n\ge 0\}$ is uniformly integrable: then $X_T$ is integrable and $\mathbb{E}{X}_T\le \mathbb{E}{X}_0$, with equality for a martingale. Boundedness of $T$ or the stated integrability condition cannot in general be omitted.

For the increment hypothesis here, $\mathbb{E}T<\infty$ implies $T<\infty$ almost surely, and telescoping givesThe right-hand side is an integrable random variable. Applying bounded-time optional stopping to $T\wedge n$ and then the dominated convergence theorem provesThis is bounded-increment optional stopping with integrable time; the same reasoning gives equality for a martingale.

Put $q=1-p$ and first assume $0<p<1$. Let $\tau$ be the first occurrence time of the specified seven-letter word. Its probability in a prescribed block is $r=p^4{q}^3>0$. Disjoint seven-flip blocks are independent, and a match in one forces $\tau$ to have occurred by its end. Consequently $\mathbb{P}(\tau >7m)\le (1-r{)}^m$ and $\mathbb{E}\tau \le 7/r<\infty$; this establishes the integrability needed for stopping before computing its value.

Start a new gambler with unit capital immediately before each flip. A gambler bets all current capital on the next required letter of the word, receives capital divided by $p$ for a correct head or by $q$ for a correct tail, and receives zero on failure. After completing all seven bets, the gambler retains the winnings without further betting. Each individual bet preserves its conditional expected capital. If $C_n$ is the total capital of all gamblers who have entered through flip $n$, then $C_n-n$ is a martingale.

At any time at most seven gamblers are still betting, and each one's capital is at most $L=1/(p^4{q}^3)$, since probabilities of shorter prefixes are at least the full-word probability. Completed winnings do not change. Thus the increments of $C_n-n$ have a deterministic bound, for example $14L+1$. The preceding optional-stopping argument therefore gives $\mathbb{E}(C_{\tau }-\tau )=0$.

At the first full match, no earlier gambler has completed the word. A surviving gambler who has placed $k$ successful bets corresponds to a suffix of the completed word which is also its length-$k$ prefix, namely a border of a word. The only such lengths are $3$ and $7$: the shorter common string is HHT. Their capitals are $1/(p^{2}q)$ and $1/(p^4{q}^3)$, respectively. Hence $C_{\tau }$ is this deterministic sum andFor a fair coin this is 136 flips. This is a waiting time for a word in independent nonuniform symbols; the overlap contribution is essential. If $p=0$ or $p=1$, the word cannot occur because it contains both types of flip, so the expected waiting time is infinite.

Use the exponent convention $\mathbb{E}{e}^{iu{X}_t}=e^{t\psi (u)}$. A real Lévy process starts at $0$, has stationary increments and independent increments, is stochastically continuous, and is taken in its càdlàg version. Its characteristic exponent of a Lévy process satisfies $\psi (0)=0$ and is continuous. This convention has the opposite sign from the alternative $e^{-t\Psi (u)}$, so $\psi =-\Psi$.

For real $u$, the exponential martingale of a Lévy process is integrable because $|M_t^u|=e^{-t\mathrm{Re}\psi (u)}$ is deterministic and finite. For $s\le t$, its conditional expectation with respect to the natural filtration isThe second equality uses both independent increments and stationary increments. Thus this is a complex martingale, meaning that its real and imaginary parts are martingales.

A probability distribution is an infinitely divisible distribution when, for every positive integer $n$, it is the distribution of a sum of $n$ independent and identically distributed random variables. Their common distribution is allowed to depend on $n$. For a Lévy process,and these summands are independent with the common law of $X_{1/n}$. This proves the required infinite divisibility directly.

The Lévy–Khintchine theorem states that a real infinitely divisible distribution has characteristic function $e^{\psi (u)}$, where, for a unique triplet with the following fixed truncation,with $b\in \mathbb{R}$, $a\ge 0$, $K$ a positive measure, $K(\{0\})=0$, and $\int (1\wedge y^2)K(dy)<\infty$. Conversely every such triplet gives an infinitely divisible distribution and a Lévy process, unique in law, whose time-$t$ characteristic function is $e^{t\psi (u)}$. The Lévy measure $K$ need not be finite near zero. Taylor expansion makes the compensated integrand $O(y^2)$ there, while the measure is finite away from zero, so the integral is well defined. The drift parameter depends on the truncation: using $|y|\le 1$ instead can alter $b$ by the contribution of atoms at $\pm 1$. Here $a$ denotes Gaussian variance, not drift.

We now prove that continuity forces $K=0$, using the suggested truncation rather than merely quoting a path classification. For $\epsilon >0$, ${\lambda }_{\epsilon }=K(|y|\ge \epsilon )$ is finite. The truncated exponent is that of the Compound Poisson processwhere $N$ has rate ${\lambda }_{\epsilon }$ and the marks have law $K(dy)1_{\{|y|\ge \epsilon \}}/{\lambda }_{\epsilon }$ when the rate is positive. The remaining exponent $\psi -{\psi }_{\epsilon }$ also has an admissible Lévy–Khintchine formula triplet. Hence construct independently a Lévy process $Z^{\epsilon }$ with that exponent. The sum $Y^{\epsilon }+Z^{\epsilon }$ has the same finite-dimensional distributions as $X$, and hence the same law on càdlàg path space. Continuity is therefore an almost-sure property of the sum if it is one of $X$.

But independent compound Poisson jumps cannot cancel: conditional on the path of $Z^{\epsilon }$, its jump times are countable; the independent Poisson jump times have continuous distributions and avoid them almost surely. At each jump of $Y^{\epsilon }$, the sum has the same nonzero jump, of size at least $\epsilon$. If ${\lambda }_{\epsilon }>0$, there is positive probability of such a jump on any fixed positive-length time interval, contradicting almost-sure continuity. Thus $K(|y|\ge \epsilon )=0$ for every $\epsilon >0$, and taking $\epsilon =1/n$ gives $K=0$.

The characteristic function is now $\exp [t(ibu-a{u}^2/2)]$. If $a>0$, the process $B_t=(X_t-bt)/\sqrt{a}$ has continuous paths and independent centered normal increments of variance $t-s$, so it is standard Brownian motion on the original probability space. If $a=0$, the increments are deterministic, and continuity together with equality at all rational times gives $X_t=bt$ simultaneously for all $t$ almost surely. A Brownian motion on an enlarged space can be used in the zero-variance representation if necessary. Therefore

Use $\Xi (u)={\pi }^{-u/2}\Gamma (u/2)\zeta (u)$ for the completed Riemann zeta function, without the polynomial factor used to define the entire Riemann xi function. For complex $\nu$ putThis is the Modified Bessel function of the second kind. The substitution $t=e^v$ writes it as $\frac{1}{2}{\int }_{\mathbb{R}}{e}^{-Y\cosh v+\nu v}dv$, proving that it is entire in $\nu$, that $K_{\nu }=K_{-\nu }$, and that it decays exponentially as $Y\to \infty$, uniformly for $\nu$ in a compact set.

For $\Re s>1$ all the lattice sums and the integrals below converge absolutely. The $m=0$ terms in the completed nonholomorphic Eisenstein series contribute $\Xi (2s)y^s$. For $m\ne 0$, use the Gamma integral and then Poisson summation in $n$:Here the Poisson identity is ${\sum }_n{e}^{-\pi u(n+mx{)}^2}=u^{-1/2}{\sum }_h{e}^{-\pi h^2/u}{e}^{2\pi ihmx}$. At $h=0$, evaluating the Gamma integral and summing over $m$ gives $\Xi (2s-1)y^{1-s}$. At $h\ne 0$, the substitution $u=|h|t/(|m|y)$ and the integral representation of the modified Bessel function of the second kind giveCollect the frequency $r=mh\ne 0$. There are two signed choices of $m$ for each positive divisor of $|r|$. Thus, with ${\sigma }_a(n)={\sum }_{d\mid n}{d}^a$, the full Fourier expansion isThe factor $2$ is important: the lattice sum includes both signs, while the defining factor $1/2$ was already used in its Mellin integral. This expansion uses exactly the completion in the question.

We can deduce both of the offered conclusions. First recall, with a short proof, the needed functional equation of the Riemann zeta function. Put $\Theta (t)={\sum }_{n\in \mathbb{Z}}{e}^{-\pi n^{2}t}$. Poisson summation gives $\Theta (t)=t^{-1/2}\Theta (1/t)$. Splitting the Mellin integral of $\Theta -1$ at one yieldsThe integral is entire and invariant under $u\mapsto 1-u$. Hence $\Xi (u)=\Xi (1-u)$, with only simple poles at $u=0,1$, of residues $-1,1$. This is the pole-bearing completion, not the entire xi function.

For fixed $y>0$, the nonconstant part of the Eisenstein expansion is entire in $s$ and locally normally convergent: the divisor sums grow at most polynomially on a compact set of $s$, whereas $K_{s-1/2}(2\pi |r|y)$ decays exponentially in $|r|$. The same argument gives local uniformity for $z$ in compact subsets of the complex upper half-plane. Thus only the constant terms can have poles. Their apparent poles at $s=1/2$ cancel: $\Xi (2s)$ has principal part $1/(2(s-1/2))$, while $\Xi (2s-1)$ has its negative, and both multiply $y^{1/2}$ there. The remaining poles are simple, at $s=0,1$, with residues $-1/2,1/2$.

Under $s\mapsto 1-s$, the two constant terms interchange by the zeta functional equation. The nonconstant terms are unchanged because $K_{\nu }=K_{-\nu }$ andwhich follows by replacing a divisor $d$ by $|r|/d$. Therefore

For the Kronecker limit formula, evaluate the finite Laurent coefficient at $s=1$. Write ${\gamma }_E$ for the Euler--Mascheroni constant. From $\zeta (u)=1/(u-1)+{\gamma }_E+O(u-1)$ and ${\Gamma }^{\prime }(1/2)/\Gamma (1/2)=-{\gamma }_E-2\log 2$ we obtainThe gamma derivative follows by differentiating the Gamma duplication formula and using ${\Gamma }^{\prime }(1)=-{\gamma }_E$. Multiplication by $y^{1-s}$ adds $-\frac{1}{2}\log y$ to the constant term; the other constant term contributes $\Xi (2)y=\pi y/6$.

The half-order Modified Bessel function of the second kind is $K_{1/2}(Y)=\sqrt{\pi /(2Y)}{e}^{-Y}$. For example, in its even integral representation set $v/2$ as variable and then $u=\sinh (v/2)$: $\cosh v=1+2u^2$ and the $\cosh (v/2)$ factor converts the integral to a Gaussian. At $s=1$ the nonconstant Eisenstein coefficients are consequently ${\sigma }_{-1}(|r|)e^{-2\pi |r|y}$. On the other hand, the absolutely convergent logarithmic series givesAdding this identity to its conjugate identifies the nonconstant sum as $-2\log |{\prod }_{n\ge 1}(1-q^n)|$. With the analytic Dedekind eta function product $\eta (z)=q^{1/24}{\prod }_{n\ge 1}(1-q^n)$, the resulting answer isThis derivation of the Kronecker limit formula uses the analytic eta product only; it does not assume its modular transformation. The lattice series is invariant for $\Re s>1$ by an integral change of the pair $(m,n)$ and $\Im (\gamma z)=y/|cz+d{|}^2$. Analytic continuation preserves this invariance, so its finite Laurent coefficient is invariant too. This supplies the independent weight-two transformation argument used in Question 2.

Let $a_0$ be the constant Fourier coefficient and suppose first that $k>0$ is even and $M_k\ne 0$; thus $k\ge 4$. Put $\epsilon =i^k=(-1{)}^{k/2}$. The precise Mellin continuation of a noncuspidal modular form isIt is meromorphic on the whole plane, with possible simple poles only at $0,k$, of residues $-a_0,\epsilon a_0$, respectively. For a cusp form it is an entire function. A nonzero constant term makes both of these poles genuine.

We first justify the initial Mellin transform. Decompose $f=a_0{E}_k+h$ with $h\in S_k$. For a cusp form, $y^{k/2}|h(x+iy)|$ is invariant under the modular group and bounded on its fundamental domain, since it tends to zero at the cusp. Therefore it is bounded throughout the complex upper half-plane. Integrating the Fourier coefficient at height $y=1/n$ gives $|a_n(h)|\le C{e}^{2\pi }{n}^{k/2}$. The Eisenstein series coefficients satisfy ${\sigma }_{k-1}(n)\le \zeta (k-1)n^{k-1}$ by replacing divisors $d$ by $n/d$. Consequently $a_n(f)=O(n^{k-1})$, and absolute termwise integration is valid for $\Re s>k$:Indeed ${\int }_0^{\infty }{e}^{-2\pi ny}{y}^{s-1}dy=(2\pi n{)}^{-s}\Gamma (s)$, and the sum of the absolute integrals converges in that half-plane.

Write $I_f(s)={\int }_1^{\infty }(f(iy)-a_0)y^{s-1}dy$. Exponential cusp decay of the difference implies uniform convergence on compact subsets of the $s$-plane, including after any number of derivatives in $s$, so $I_f$ is entire. The modular transformation $f(i/y)=(iy{)}^{k}f(iy)$ givesOn the lower half of the Mellin integral, subtract the transformed constant explicitly:Substitute $y=1/t$ in the decaying term and integrate the two powers in the constant term. This yields, initially for $\Re s>k$,The right side supplies the claimed meromorphic continuation and its residues. Replacing $s$ by $k-s$ and multiplying by $\epsilon$, with ${\epsilon }^2=1$, leaves this expression unchanged, proving the functional equation with its correct sign. In particular the residue at $k$ is tied to the residue at zero by that equation; subtracting only $a_0$ at infinity and overlooking its transformed value at zero would miss these poles.

In weight zero every modular form is constant, so $L(f,s)=0$ and its completion is identically zero. The same holds for the zero form in odd, negative, or weight-two spaces. Thus the assertion covers all weights in the notation of the paper. There is no conjugation of $f$ in this level-one functional equation.

Write $\Gamma =S{L}_2(\mathbb{Z})$ and use the determinant-normalized slash operatorThe automorphy factor identity proves $(f{|}_k\alpha ){|}_k\beta =f{|}_k(\alpha \beta )$, and positive scalar matrices act trivially. Let ${\mathcal{M}}_n$ be all integral matrices of determinant $n>0$. It is stable under left and right multiplication by $\Gamma$. Its left orbits have unique representativesIndeed Bézout's identity reduces the first column to $(a,0)$, the determinant fixes $d=n/a$, and a row shear reduces $b$ modulo $d$. These are the determinant-n matrix representatives for Hecke operators. For composite $n$, ${\mathcal{M}}_n$ can contain more than one double coset; keeping every integral matrix, including nonprimitive ones, is essential.

Define the Hecke operator byChanging a representative on the left leaves its summand unchanged. Right multiplication by $\Gamma$ permutes left orbits of ${\mathcal{M}}_n$, so the result satisfies the weight-$k$ transformation law. Each summand is holomorphic. Its expansion at infinity is bounded, and vanishes there when $f$ is a cusp form; the group has just one cusp class. Therefore $T_n$ preserves both $M_k$ and $S_k$, and $T_1$ is the identity.

For a Fourier expansion $f={\sum }_{r\ge 0}{a}_r{q}^r$, the sum over $b$ vanishes unless $d\mid r$. Substitute $r=dt$ and then collect the coefficient of $q^m$ to obtain the Fourier coefficients of a composite-index Hecke operator:Here $\gcd (0,n)=n$, so the constant coefficient is ${\sigma }_{k-1}(n)a_0$; at $m=1$ it is $a_n$. This proves preservation of integral coefficients. At a prime,with $a_{m/p}=0$ if $p\nmid m$.

We can prove the entire Hecke multiplication relations directly, not just quote them. On formal power series put $U_p(\sum a_m{q}^m)=\sum a_{pm}{q}^m$ and $V_{p}f(q)=f(q^p)$. Although these individual operators need not preserve level one, $U_p{V}_p=1$ and the coefficient formula givesMultiply on the left by $T_p=U_p+p^{k-1}{V}_p$. In the first part the $j=0$ term is $U_p^{r+1}$; every $j\ge 1$ reduces by $U_p{V}_p=1$ to $p^{j(k-1)}{V}_p^{j-1}{U}_p^{r-j}$. The second part, together with the $j=0$ term, is $T_{p^{r+1}}$. The remaining sum is $p^{k-1}{T}_{p^{r-1}}$. ThusFor coprime indices, the coefficient divisors split independently in the displayed formula, proving $T_m{T}_n=T_{mn}$ when $(m,n)=1$. Induction with the prime-power recurrence gives $T_{p^r}{T}_{p^t}={\sum }_{j=0}^{\min (r,t)}{p}^{j(k-1)}{T}_{p^{r+t-2j}}$. Combining the primes provesIn particular the modular Hecke algebra is commutative and is generated by the prime operators.

For cusp forms the Petersson inner product isCusp decay makes it finite and positive definite. The Hecke operators are self-adjoint for the Petersson inner product. To see the change-of-variables mechanism, fix a double coset $\Gamma \alpha \Gamma \subset {\mathcal{M}}_n$ and put $H=\Gamma \cap {\alpha }^{-1}\Gamma \alpha$. Representatives $\gamma$ of $H\backslash \Gamma$ give left representatives $\alpha \gamma$. Unfolding the sum in $⟨{\sum }_{\gamma }f{|}_k\alpha \gamma ,g⟩$ changes the integral over $\Gamma \backslash \mathbb{H}$ to the integral over $H\backslash \mathbb{H}$ of $(f{|}_k\alpha )\overline{g}$. The substitution $w=\alpha \tau$ preserves $dxdy/y^2$ and has $\Im w=(\det \alpha )y/|c\tau +d{|}^2$; these factors move the slash from $f$ to $g{|}_k{\alpha }^{-1}$. Refold over $\alpha H{\alpha }^{-1}=\Gamma \cap \alpha \Gamma {\alpha }^{-1}$ to get the reverse double coset. Since scalar matrices act trivially, ${\alpha }^{-1}$ has the same slash action as its adjugate ${\alpha }^{\ast }=n{\alpha }^{-1}$, an integral matrix of determinant $n$. The adjugate involution permutes all the double cosets in ${\mathcal{M}}_n$. The common real factor $n^{k/2-1}$ is unchanged, giving $⟨T_{n}f,g⟩=⟨f,T_{n}g⟩$.

Since $S_k$ is finite dimensional, commuting self-adjoint linear operators have a common orthonormal basis of Hecke eigenforms: diagonalize one operator, restrict the others to its eigenspaces, and repeat until these spaces cannot split further. If $f$ is a nonzero common Hecke eigenform with $T_{n}f={\lambda }_{n}f$, the coefficient at $q$ gives $a_n={\lambda }_n{a}_1$. Were $a_1=0$, every positive coefficient would vanish, contrary to $f\ne 0$ being cuspidal. Normalize $a_1=1$; then ${\lambda }_n=a_n$, and the Hecke multiplication relations yieldThe eigenvalues are real by self-adjointness. Their multiplicativity and prime-power recurrence give the Euler product of a Hecke eigenform, initially in an absolutely convergent half-plane:The Mellin transform in Question 5 supplies its analytic continuation and functional equation.

Finally, for positive even $k\ge 4$, $M_k=\mathbb{C}{E}_k\oplus S_k$, by subtracting the constant coefficient times the normalized Eisenstein series. The divisor identityfollows by multiplying the geometric sums at each prime; it and the constant-coefficient formula prove $T_n{E}_k={\sigma }_{k-1}(n)E_k$. Thus $M_k$ has an Eisenstein series eigenline and a basis of cuspidal Hecke eigenforms. Odd and negative weights have no forms, $M_2=0$, and on $M_0=\mathbb{C}$ the operator is the scalar ${\sigma }_{-1}(n)$. These statements fix the normalization and describe the exceptional small weights as well.