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If a domain of finite hyperbolic area $A$ contains a hyperbolic disc of radius $r$ in curvature $-1$, then $2\pi (\cosh r-1)\le A$, hence $r\le \mathrm{arcosh}(1+A/(2\pi ))$. For an interior point of a hyperbolic triangle, the open disc with radius equal to its distance from the boundary lies inside the triangle. Its area is at most $\pi$ by Gauss-Bonnet theorem, giving the elementary bound $r\le \mathrm{arcosh}(3/2)$; this need not be the sharp triangle inradius bound.

An ideal hyperbolic quadrilateral is the region bounded by four hyperbolic geodesics whose four consecutive endpoints are distinct points on the ideal boundary. The ideal vertices lie outside the hyperbolic plane. In the hyperbolic disk, the four vertices $1,i,-1,-i$ give a regular example: its sides are the interior arcs of the circles of radius one centered at $\pm 1\pm i$. Although the sides have infinite length, the quadrilateral has finite hyperbolic area $2\pi$ in curvature $-1$.

An ideal hyperbolic triangle has all three vertices on the boundary at infinity of the hyperbolic plane. All its angles are zero and its hyperbolic area in curvature $-1$ is $\pi$, by hyperbolic triangle area.

For a piecewise continuously differentiable curve $\gamma (t)=(x(t),y(t))$ in the upper half-plane model, the given Riemannian metric gives hyperbolic lengthThe metric matrix has determinant $y^{-4}$, so its area element is $y^{-2}dxdy$. Thus a measurable region $D$ has hyperbolic area ${\int }_D{y}^{-2}dxdy$. For the specified strip the integral isThe infinite Euclidean height is compatible with finite hyperbolic area because the area density decreases quadratically with height.

For curvature $-1$, the Poincare disc model has metric and area elementThe given minimizing-diameter property makes the distance from zero to Euclidean radius $R$ equal toRadial symmetry therefore identifies the whole hyperbolic circle in the Poincare disc. Integrating its enclosed hyperbolic area gives

For a geodesic hyperbolic triangle with angles $\alpha ,\beta ,\gamma$, the Gauss-Bonnet theorem states $A=\pi -(\alpha +\beta +\gamma )$, in particular $A\le \pi$, with strict inequality for ordinary finite vertices. Let $d$ be the distance of the interior point $P$ to the nearest side. The open disc of radius $d$ about $P$ is contained in the triangle: a path from $P$ to an exterior point must first cross the boundary at distance at least $d$. A hyperbolic isometry carries its center to zero, so its area is the area already calculated. ConsequentlyThis uses hyperbolic area bounds an inscribed disc and establishes the requested bound without assuming that $P$ is an incenter. A sharper universal triangle bound is possible, but is not needed here.

Use the curvature-minus-one Poincare disc model, whose Riemannian metric isIts geodesics are Euclidean diameters and arcs of circles meeting the boundary circle orthogonally. The hyperbolic area of a measurable region $D$ isAn isometry can move the center of a circle to the origin. Its hyperbolic radius $r$ and Euclidean radius $\rho$ then satisfy $r={\int }_0^{\rho }2du/(1-u^2)=2\mathrm{artanh}\rho$, so $\rho =\tanh (r/2)$. Integrating the area element of a surface in polar coordinates yieldsIntegrating the line element around the circle givesA geometric definition of Pi valid in this geometry isIndeed $\sinh r=r+O(r^3)$ and $\cosh r=1+r^2/2+O(r^4)$. The circumference-to-diameter ratio at a fixed positive radius is not constant, so the limiting qualification is necessary.

In the upper half-plane model $H=\{(x,y):y>0\}$ the hyperbolic metric isThe length of a differentiable curve is $\int \sqrt{{\dot{x}}^2+{\dot{y}}^2}dt/y$. The area element is the square root of the metric determinant times $dxdy$, so the hyperbolic area of a measurable region $S$ is ${\int }_S{y}^{-2}dxdy$.

The Gauss-Bonnet theorem for a geodesic hyperbolic triangle of angles $\alpha ,\beta ,\gamma$ in curvature $-1$ gives area $\pi -\alpha -\beta -\gamma$; an ideal vertex has angle zero. The horizontal edge in the present region is not a geodesic, so it is simpler to use the area element directly:The open boundary does not affect the integral because its area is zero.

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.

Use the primitive real-analytic Eisenstein series with exactly the normalization in the PDF. Its coset expression uses ${\Gamma }_{\infty }=\{\pm \left(\begin{array}{cc}1& n\\ 0& 1\end{array}\right):n\in \mathbb{Z}\}$, so primitive bottom rows differing by sign represent the same coset. The factor $1/2$ in the primitive-pair sum accounts for this sign identification. Unlike the sum over all nonzero integer pairs, this Eisenstein series has no additional factor $2\zeta (2w)$. Write $w$ for the Rankin–Selberg integral parameter and $s$ for the Dirichlet series parameter to keep their shift visible.

First establish an initial region where all rearrangements are absolute. The invariant norm of a modular form $y^{k/2}|f(\tau )|$ is bounded: it is $\Gamma$-invariant, continuous on a truncated fundamental domain, and tends to zero at the cusp. Integrating one period of the Fourier series givesWith $y=1/n$, this proves the Fourier coefficient bound for a cusp form $a_n=O(n^{k/2})$, and similarly $b_n=O(n^{k/2})$. Hence $D(f,g,s)$ is initially absolutely convergent for $\mathrm{Re}s>k+1$.

For $\mathrm{Re}w>1$, the primitive real-analytic Eisenstein series is locally uniformly convergent and holomorphic in $w$. On the standard fundamental domain it has at most polynomial growth in $y$, locally uniformly in this half-plane. These are the analytic properties of the Eisenstein series needed here; they can also be checked directly from its defining sum. For example, for real $\sigma >1$ and $y\ge 1$, majorize the primitive sum by the sum over all nonzero pairs and useThis bound follows by comparing a shifted one-dimensional sum with its integral and a largest summand. Summing over $m\ne 0$ gives $E(\tau ,\sigma )=O(y^{\sigma }+y^{1-\sigma })$. For complex $w$, $|E(\tau ,w)|\le E(\tau ,\mathrm{Re}w)$, and the estimates are uniform on compact parameter sets. Since $f$ and $g$ decay exponentially on the cusp end, the integral $I(f,g,w)$ is therefore holomorphic throughout $\mathrm{Re}w>1$ by dominated integration.

The product $f(\tau )\overline{g(\tau )}{y}^k$ is $\Gamma$-invariant: the two modular factors cancel the transformation $\mathrm{Im}(\gamma \tau )=y/|c\tau +d{|}^2$. With the hyperbolic area $d\mu =dxdy/y^2$, Rankin–Selberg unfolding gives, initially for $\mathrm{Re}w>2$,Absolute unfolding is justified by replacing $f\overline{g}$ by $|fg|$ and using the same finite fundamental-domain integral with $E(\tau ,\mathrm{Re}w)$. The subsequent termwise integrations can be justified particularly economically by Cauchy-Schwarz inequality: on each horizontal line,The coefficient bounds make the right side $O(y^{-k-1})$ as $y\downarrow 0$, and it decays exponentially as $y\to \infty$. Multiplication by $y^{\mathrm{Re}w+k-2}$ is integrable at zero when $\mathrm{Re}w>2$. Thus the integral of the inner Fourier series may be evaluated first. Orthogonality of the exponentials givesThe Mellin transform of each exponential isIn the same initial region the sum of the absolute values of these integrals is finite by $a_n\overline{b_n}=O(n^k)$. Consequently the exact Rankin–Selberg integral identity isIn particular the two parameters are not equal: the Dirichlet series argument is shifted by $k-1$.

The holomorphy of $I$ on $\mathrm{Re}w>1$ now gives the required continuation:The right side agrees with the original Dirichlet series on $\mathrm{Re}s>k+1$ and is holomorphic on the larger half-plane. No continuation of the Eisenstein series beyond $\mathrm{Re}w>1$ is needed.

It is important to supply a further argument for absolute convergence: analytic continuation alone does not imply it. For real $w>1$, both $E(\tau ,w)$ and $|f(\tau ){|}^2$ are nonnegative, and their fundamental-domain integral is finite by the cusp estimates above. Tonelli theorem therefore unfolds $I(f,f,w)$ even before knowing convergence of the Dirichlet series. On each fixed horizontal line, the Fourier series converges absolutely and uniformly, so integration in $x$ givesA second use of Tonelli theorem, now in $y$, proves the positive unfolding of a cusp-form square:Since every real $s>k$ has $s=w+k-1$ with $w>1$, we have absolute convergence of $D(f,f,s)$ for $\mathrm{Re}s>k$. Absolute convergence for complex $s$ follows by taking its real part. In fact Cauchy-Schwarz inequality also gives absolute convergence of $D(f,g,s)$ there from the two corresponding square sums.

Finally let $\sigma >(k+1)/2$ and choose $k<\beta <2\sigma -1$. The preceding square-coefficient bound and Cauchy-Schwarz inequality giveThis proves absolute convergence of cusp-form L-series from square coefficients, and henceEquivalently, the suggested elementary inequality works by choosing $k-\sigma <\alpha <\sigma -1$: after multiplication by $n^{-\sigma }$ its two bounds sum to a convergent $\sum n^{-(\sigma -\alpha )}$ and a convergent $\sum |a_n{|}^2{n}^{-(\sigma +\alpha )}$. The existence of this interval is precisely $2\sigma >k+1$.

Interpret the printed congruence entrywise in the integer lattice:It then defines the usual level-two principal congruence subgroup of ${\mathrm{SL}}_2(\mathbb{Z})$, despite the PDF's ambient ${\mathrm{SL}}_2(\mathbb{R})$. A literal ideal congruence inside $\mathbb{R}$ would be vacuous because $2\mathbb{R}=\mathbb{R}$, and would make the asserted conclusion false. The integer-lattice interpretation is essential.

Pass to $\overline{\Gamma (2)}=\Gamma (2)/\{\pm I\}$, which has exactly the same action. Reduction modulo two maps the modular group onto ${\mathrm{SL}}_2({\mathbb{F}}_2)$, a group of order six: the reductions of $S$ and $T$ generate it. Its kernel is $\overline{\Gamma (2)}$, so the index of a subgroup is six. The standard fundamental domain of the modular group has hyperbolic area $\pi /3$, and hence the quotient has hyperbolic area $2\pi$.

There are no nonidentity elliptic Möbius transformations in $\overline{\Gamma (2)}$. An integral matrix representing an elliptic Möbius transformation has trace $0$ or $\pm 1$. Here the trace is even, excluding $\pm 1$; trace zero would give $d=-a$ and $1=-a^2-bc\equiv -1(\mathrm{mod}4)$, impossible. Thus the effective action is a free properly discontinuous group action, and the quotient is a Riemann surface.

A cusp of a modular group is represented by a rational boundary point. Their orbits correspond towhich has $6/2=3$ elements. They are represented by $\infty ,0,1$, or by the three nonzero parity vectors of a primitive numerator-denominator pair. Each width of a cusp is two. A union of six copies of the standard fundamental domain of the modular group gives a fundamental region for this subgroup. Removing small horocycle neighbourhoods of its cusps leaves a compact core. Adding one point at each cusp of a modular group, using the local parameter $e^{\pi iz}$ after moving that cusp to infinity, gives a compact Riemann surface $\overline{X}$.

For a finite-area hyperbolic surface of genus $g$ with $r$ cusps, the Gauss-Bonnet theorem gives area $2\pi (2g-2+r)$. Thus $2\pi =2\pi (2g-2+3)$ and $g=0$. A compact genus-zero Riemann surface is the Riemann sphere, by the uniformization theorem. A Möbius transformation sends the three added points to $0,1,\infty$. Restricting it gives