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For $N\ge 2$, the effective principal congruence subgroup is torsion-free. Its modular curve cover of $X(1)$ has analytic ramification indices $2,3,N$ above the two elliptic orbits and the modular cusp. If $\mu$ is its projective index, the numbers of points above these three values are $\mu /2,\mu /3,\mu /N$. Applying the Riemann-Hurwitz formula gives $2g-2=\mu (1/6-1/N)$. Here $\mu =6$ for $N=2$ and $\mu =N^3{\prod }_{p\mid N}(1-p^{-2})/2$ for $N\ge 3$; level one is the identity cover.

The quotient $\mathbb{H}/\Gamma (2)$ is a genus-zero Riemann surface with three cusps and no elliptic Möbius transformations, hence is biholomorphic to $\mathbb{C}\setminus \{0,1\}$. The integer level-two principal congruence subgroup has projective index six; reduction modulo two gives three cusp classes.

The modular group is $S{L}_2(\mathbb{Z})$ acting on the complex upper half-plane byIts central element $-I$ acts trivially on points, so the effective group is $PS{L}_2(\mathbb{Z})$. It is generated by $S:\tau \mapsto -1/\tau$ and $T:\tau \mapsto \tau +1$: the Euclidean algorithm reduces the bottom row of a matrix using these operations until it is a translation. The relations $S^2=(ST{)}^3=1$ hold in the projective group.

A standard fundamental domain of the modular group isTo see that each orbit meets it, use $\mathrm{Im}(\gamma \tau )=\mathrm{Im}\tau /|c\tau +d{|}^2$. Choose a coprime integer pair $(c,d)$ minimizing the nonzero lattice modulus $|c\tau +d|$; a modular matrix with that bottom row maximizes the imaginary part. Translate to the central strip. If the result had modulus below one, inversion would further increase its imaginary part, a contradiction. Interior representatives are unique: for a point in the interior, $|c\tau +d|>1$ when $c\ne 0$. For $|c|\ge 2$ use $\mathrm{Im}\tau >\sqrt{3}/2$; for $|c|=1$ use the strip and the strict unit-circle bound. Thus such a matrix lowers the imaginary part, which would contradict the same fact applied to its inverse if both representatives were in the interior. Matrices with $c=0$ are translations, and cannot move one interior strip point to another.

The vertical sides are identified by $T$, and the circular halves by $S$. The exceptional points with elliptic stabilizers of the modular group are represented by $i$ and $\rho =e^{2\pi i/3}$, of projective stabilizer orders two and three. The rational boundary points together with infinity form one modular cusp orbit. In the hyperbolic metric the region has area ${\int }_{-1/2}^{1/2}(1-x^2{)}^{-1/2}dx=\pi /3$.

A useful subgroup is the level-two principal congruence subgroup,Reduction modulo two is onto $S{L}_2({\mathbb{F}}_2)$, which has six elements and permutes the three nonzero vectors. Hence its kernel has index six, and the quotient is $S_3$. In the projective action, $\Gamma (2)$ is torsion-free. Indeed finite-order nonidentity modular elements have trace zero or $\pm 1$; matrices in $\Gamma (2)$ have even trace, and trace zero is impossible because $a,d$ are odd while $b,c$ are even: $ad-bc=1$ with $d=-a$ would imply $bc=-a^2-1\equiv 2(\mathrm{mod}4)$. The subgroup has three modular cusp classes, represented by $\infty ,0,1$, each of width two. Parity of a primitive pair distinguishes these classes, and elementary level-two row operations reduce each class to its representative. Six translates of $\mathcal{F}$ supply a fundamental region. Its compactified level-two modular curve is obtained by adjoining these three modular cusp points.

Invariant holomorphic functions arise naturally from theta constants. In the radian convention of Question 4, put ${\Theta }_j(\tau )={\theta }_j(0,\tau )$, and defineHere ${\Theta }_2={\sum }_{n\in \mathbb{Z}}{q}^{(n+1/2{)}^2}$. The theta-constant inversion and translation laws, obtained by Gaussian Poisson summation and by reindexing, interchange ${\Theta }_2,{\Theta }_4$ under $S$ with the common factor $\sqrt{-i\tau }$, leave ${\Theta }_3$ fixed up to that factor, and under $T$ interchange ${\Theta }_3,{\Theta }_4$ while multiplying ${\Theta }_2$ by $e^{\pi i/4}$. Together with the Jacobi abstruse identity ${\Theta }_3^4={\Theta }_2^4+{\Theta }_4^4$, this givesThese transformations generate the six permutations of $0,1,\infty$. In particular $T^2$ and $S{T}^2{S}^{-1}$ fix $\lambda$; with $-I$, these generate $\Gamma (2)$. Thus the modular lambda function is invariant under this subgroup, but not under the whole modular group. Its even theta constants do not vanish on $\mathbb{H}$, so it is holomorphic there and avoids $0,1,\infty$. Near infinity, $\lambda =16q+O(q^2)$, with $q=e^{\pi i\tau }$, which is the width-two modular cusp coordinate. The two transformation laws give modular cusp values one at zero and infinity at one, with the same simple local orders. Thus $\lambda$ has exactly one simple pole on the compactified level-two quotient and no pole in the half-plane. A meromorphic map with one simple pole has degree one, so it identifies that quotient with the sphere and the open quotient with $\mathbb{C}\setminus \{0,1\}$. This proves the coordinate assertion and describes all three modular cusps.

Symmetrizing this subgroup invariant produces a full modular-invariant function:Substitution shows that this rational expression is unchanged by both $\lambda \mapsto 1-\lambda$ and $\lambda \mapsto \lambda /(\lambda -1)$, hence by $S$ and $T$. It is the Klein j-invariant. Its denominator does not vanish on $\mathbb{H}$, so it is holomorphic there; at the modular cusp it has $j\sim q^{-2}$, a simple pole in $Q=e^{2\pi i\tau }$. It therefore remains nonconstant without being a bounded holomorphic function on the compactified quotient. In fact a holomorphic invariant extending holomorphically to the compactified modular curve would be constant by the maximum principle.

This also connects invariance to elliptic curves: changing an oriented basis of the lattice $\mathbb{Z}+\tau \mathbb{Z}$ gives precisely the modular action after a rescaling. On the compact full modular quotient, $j$ has just the one simple modular cusp pole, so its map to the sphere has degree one. It therefore classifies the resulting homothety of complex lattices classes, while the modular lambda function retains a level-two labeling of the branch points in the Legendre model $y^2=x(x-1)(x-\lambda )$. Finally, a modular form of nonzero weight obeys a transformation law with a factor $(c\tau +d{)}^k$, not plain invariance; ratios of equal-weight forms instead give weight-zero modular functions. The distinction between forms, invariant holomorphic functions on the open half-plane, and their modular cusp behavior is essential to this construction.

Let $p:\mathbb{D}\to D$ be a universal covering map. On a small evenly covered neighborhood in $D$, choose a local inverse $s$ of $p$ and setAny two such lifts differ by a deck transformation of the disc. Every automorphism of the unit disk preserves the disc density: apply the derivative Schwarz-Pick lemma to the automorphism and its inverse to get equality. Hence the definition is independent of the local inverse. Uniqueness of the universal cover up to a covering isomorphism gives independence of the chosen cover as well. This is the Poincare metric on a Riemann surface, normalized to curvature $-1$.

For a piecewise continuously differentiable path $\gamma$, define its length by $L_D(\gamma )=\int {\lambda }_D(\gamma )|{\gamma }^{\prime }|$, and the distance by the infimum of lengths of joining paths. Path reversal and concatenation prove symmetry and the triangle inequality. The distance between distinct points is positive: choose a small closed Euclidean disc around one point not containing the other, on which the positive continuous density is bounded below by $c>0$. Every joining path must leave that disc and has length at least $c$ times its radius before doing so. Joining paths of finite length exist because a plane domain is path connected and the density is locally bounded. Thus this construction gives an actual metric, not just a nonnegative path-length expression.

For the punctured unit disc, use $p(w)=e^{iw}$ from the complex upper half-plane. The half-plane line element is $|dw|/\mathrm{Im}w$, while $|dz|=|z||dw|$ and $\mathrm{Im}w=\log (1/|z|)$. ConsequentlyThe domain called an annulus in the paper has inner radius zero, so it is this punctured-disc metric, not the metric of an annulus with positive inner radius.

We now prove the Great Picard theorem. In its holomorphic form, an essential singularity takes every complex value, with at most one exception, infinitely often in every punctured neighborhood. It suffices to prove that a holomorphic function on a punctured disc omitting $0$ and $1$ extends meromorphically through the puncture: if two values occurred only finitely often, a smaller punctured disc would omit both, and an affine change of target would reduce them to $0,1$.

Use the universal covering by the modular lambda function $\lambda :\mathbb{H}\to Y=\mathbb{C}\setminus \{0,1\}$, whose construction is given in the following solution. Its deck transformation group is the projective level-two principal congruence subgroup $\Gamma (2)$. Rescale the source punctured disc to unit radius. Lift $f(e^{2\pi iw})$ to a holomorphic function $L:\mathbb{H}\to \mathbb{H}$. Uniqueness of lifts givesfor a fixed deck transformation $T\in \Gamma (2)$. The Schwarz-Pick lemma givesas $\mathrm{Im}w\to \infty$. A nonidentity deck transformation has no interior fixed point, so it is either a hyperbolic Möbius transformation or a parabolic Möbius transformation. The hyperbolic case is conjugate to a dilation $w\mapsto aw$ with $a>0$, $aeq1$. Since $|d\log \mathrm{Im}w|\le d{s}_{\mathbb{H}}$, its displacement is at least $|\log a|>0$, contradicting this inequality.

If $T$ is the identity, compose $L$ with the Cayley transform between the half-plane and disk. Its periodicity lets the resulting bounded function descend to the punctured disc and extend by the Riemann removable singularity theorem. A nonconstant extension takes its central value inside the disc by the maximum modulus principle. The covering map then gives an extension of $f$; a constant lift gives a constant $f$.

In the parabolic case, its fixed point is a rational modular cusp. Indeed, a parabolic integral matrix of trace $\pm 2$ has a rational eigenvector for its repeated eigenvalue. Conjugate by an element of the modular group sending this cusp to infinity. The level-two subgroup is normal, so the conjugated transformation is $\zeta \mapsto \zeta +2n$ for a nonzero integer $n$. The conjugated lift $P$ consequently satisfiesThe modular transformations permute the three target cusp values $0,1,\infty$, as seen from the lambda transformation formulas below. It is therefore enough to examine $\lambda (P(w))$.

To fix the sign of $n$, the function $\mathrm{Im}P$ is periodic in the real direction. Its mean $M(y)={\int }_0^1\mathrm{Im}P(x+iy)dx$ satisfies $M^{\prime }(y)=2n$ by the Cauchy-Riemann equations and the increment of $\mathrm{Re}P$. Positivity for all $y>0$ forces $n>0$. The functionis well-defined, nonvanishing and bounded by one on the punctured disc, so it extends holomorphically to zero. Its winding number along a circle around the puncture is $n$, because a unit increment in $w$ increases $\pi iP$ by $2\pi in$. The argument principle therefore shows that $Q$ has a zero of order $n$ at the origin.

At the infinite cusp, the modular lambda function has the convergent expansion $\lambda (\zeta )=16q+O(q^2)$ in $q=e^{\pi i\zeta }$. Thus $\lambda (P(w))=16Q(z)+O(Q(z{)}^2)$ extends and tends to zero. Undoing the target permutation gives a removable singularity or a pole for $f$. In every case the puncture is not essential, which proves the holomorphic Great Picard theorem. For a meromorphic function with an essential singularity, if three spherical values occurred only finitely often, a Möbius transformation would make the omitted values $0,1,\infty$. The same proof applies because omission of infinity makes the transformed function holomorphic. Hence the meromorphic version has at most two exceptional spherical values.

Take the Jacobi theta functions at zero, with $q=e^{\pi i\tau }$:Define the modular lambda function byThis is an explicit holomorphic expression on the complex upper half-plane. The product formsshow that all three are nonzero there. The theta identity ${\theta }_3^4={\theta }_2^4+{\theta }_4^4$ gives $1-\lambda ={\theta }_4^4/{\theta }_3^4$, so $\lambda$ omits $0,1$ and infinity on the half-plane.

The theta-constant inversion and translation laws, obtained from Poisson summation and changing indices in the defining sums, giveThese formulas imply invariance under the projective groupgenerated by $\tau \mapsto \tau +2$ and $\tau \mapsto \tau /(2\tau +1)$. This is the effective level-two principal congruence subgroup.

Its action is properly discontinuous because it is a discrete subgroup of $PS{L}_2(\mathbb{R})$. It is also free. An elliptic integral matrix has trace $0$ or $\pm 1$. The odd traces are excluded because $a+d$ is even. Trace zero would give $d=-a$ and $bc=-a^2-1\equiv 2(\mathrm{mod}4)$, impossible when $b,c$ are even. Thus the only elements acting trivially are the already quotiented central matrices $\pm I$.

A fundamental domain is the ideal quadrilateralwith vertices $-1,0,1,\infty$. The vertical sides are paired by translation by two, and the semicircular sides by $\tau \mapsto \tau /(2\tau +1)$. Reduction using these pairings gives the domain; equivalently it is a union of six translates of the standard modular fundamental domain, consistent with the six elements of $PS{L}_2({\mathbb{F}}_2)$. On adjoining the cusp points, there are three vertex classes, two paired edges and one face. The resulting orientable compact surface has Euler characteristic $3-2+1=2$, hence is a sphere. This is the compactification of the level-two modular curve.

All three cusp widths are two, so their local parameters are obtained from $q=e^{\pi i\tau }$ after a modular change of coordinate. The defining products giveat infinity: there is a simple zero. At the cusp zero, inversion gives $1-\lambda =16q_0+O(q_0^2)$, so the value is $1$. At the cusp one, combine inversion and translation; the value is infinity and the pole is simple. The three cusp classes are precisely the parity classes of primitive rational pairs $(p,q)$: odd-even, even-odd and odd-odd. A Bézout identity, with parity adjusted by adding multiples of $(p,q)$, places any such cusp in the orbit of $\infty$, $0$ or $1$, respectively.

The descended meromorphic function on the compactified quotient thus has exactly one simple pole. Its degree is one, since the degree equals its total pole multiplicity; it is therefore a biholomorphic coordinate on that sphere. Removing the three cusps identifies the quotient with $\mathbb{C}\setminus \{0,1\}$. Consequentlyis a covering map with deck group $\overline{\Gamma (2)}$. The source half-plane is simply connected, so it is the required universal covering map. This proves the universal covering by the modular lambda function, including the absence of interior branching. Three arbitrary distinct punctures are handled by a target Möbius transformation.

Suppose $K=\ker \phi$ for a group homomorphism $\phi :G\to H$. Its identity belongs to $K$, and for $k,l\in K$ we have $\phi (k{l}^{-1})=1$, so it is a subgroup. Moreover, for $g\in G$ and $k\in K$,Hence $gK{g}^{-1}\subseteq K$; applying this also to $g^{-1}$ gives equality, so $K$ is a normal subgroup.

Conversely, if $K$ is a normal subgroup, multiplication of cosets in the quotient group $G/K$ is well-defined. The map $g\mapsto gK$ is a group homomorphism and has kernel of a group homomorphism exactly $K$. This proves both directions without requiring $\phi$ to be injective or surjective.

For the matrix application, reduce all entries modulo $p$:Reduction respects matrix multiplication and the determinant, so this is a group homomorphism. Its kernel of a group homomorphism consists exactly of matrices congruent to the identity modulo $p$, which are the given matrices. Therefore $K=\ker \rho$ is a normal subgroup, the level-$p$ principal congruence subgroup. Surjectivity of reduction is not needed for this argument.

Write $Y={\mathbb{P}}^1\setminus \{0,1,\infty \}$. One concrete uniformization uses the image in ${\mathrm{PSL}}_2(\mathbb{R})$ of the level-two principal congruence subgroupIt acts on the complex upper half-plane by Möbius transformations. Integer matrix entries make the group discrete, and the action is properly discontinuous. There are no nonidentity elliptic stabilizers: an elliptic element would have integer even trace with absolute value less than two, hence trace zero. But then $d=-a$ gives $bc=-(a^2+1)\equiv 2(\mathrm{mod}4)$, contradicting the evenness of $b,c$. Thus the quotient is an ordinary Riemann surface, with no orbifold points.

A fundamental region is the ideal quadrilateralIts vertical sides are identified by $T(\tau )=\tau +2$, and its semicircular sides by $U(\tau )=\tau /(1-2\tau )$. The ideal vertices are $-1,0,1,\infty$; $-1$ and $1$ represent one puncture, while zero and infinity give two others. Filling these three punctures gives genus zero: the identified quadrilateral has three vertices, two paired edges and one face, so its compactification has Euler characteristic $3-2+1=2$. Its complex compactification is a sphere, and a Möbius transformation puts the punctures at $0,1,\infty$. HenceThis is the thrice-punctured sphere as a modular quotient. Equivalently, the uniformization theorem applied to its universal covering map produces the upper-half-plane model: the sphere cannot cover this noncompact surface, and an entire covering from $\mathbb{C}$ would contradict Little Picard theorem by omitting both zero and one.

The curvature-minus-one metric $d{s}_{\mathbb{H}}=|d\tau |/\mathrm{Im}\tau$ is invariant under every real Möbius transformation preserving $\mathbb{H}$. Indeed, for determinant-one coefficients,Thus it descends to the hyperbolic metric on $Y$. This defines its distance by the infimum of path lengths; a choice of local lift does not affect lengths.

For the Schwarz-Pick lemma, first prove the Schwarz lemma. If $h:\mathbb{D}\to \mathbb{D}$ is a holomorphic function and $h(0)=0$, the function $h(z)/z$ extends as a holomorphic function across zero. On $|z|=r$ its modulus is at most $1/r$. The maximum modulus principle and $r\uparrow 1$ yield $|h(z)|\le |z|$ and $|h^{\prime }(0)|\le 1$. Equality at a nonzero point or in the derivative makes $h(z)/z$ a constant of modulus one, again by the maximum modulus principle, so $h$ is a rotation.

For a general holomorphic function $F:\mathbb{D}\to \mathbb{D}$, conjugate by the automorphisms of the unit disk $A_a(z)=(z-a)/(1-\overline{a}z)$, putting $a$ and $F(a)$ at zero. Applying the proved Schwarz lemma givesThe disk hyperbolic metric is $2|dz|/(1-|z{|}^2)$, and its distance is twice the inverse hyperbolic tangent of the displayed pseudohyperbolic distance. Both forms of Schwarz-Pick lemma therefore express metric contraction. Equality in the derivative at one point makes $F$ a disk automorphism by the equality case already proved.

For $X=\mathbb{D}\setminus \{0\}$ and a holomorphic function $f:X\to Y$, compose with a universal covering map of $X$ and lift to a universal covering map of $Y$. The simply connected source of the lift is a disk or half-plane. Schwarz-Pick lemma there bounds the lifted metric derivative, and the covering maps are local isometries, soThe second inequality follows by applying the length inequality to every joining path and taking the infimum. The hyperbolic metric on the punctured disk is its own intrinsic metric,obtained from the covering $z=e^{i\tau }$. It is not merely the ordinary disk metric restricted to the punctured disk.

Yes, equality can occur for distinct points. The cyclic subgroup $⟨T⟩$ lies in $\Gamma (2)$. Its quotient is conformally the punctured disk, using $z=e^{\pi i\tau }$, because this identifies precisely $\tau \sim \tau +2$. The subgroup inclusion induces a covering map that is a holomorphic functionIt is a local isometry. Choose a sufficiently small geodesically convex ball around any target point that is evenly covered. Two distinct sufficiently close points in one lift of that ball are joined by the lift of its minimizing target geodesic. The lift has the same length, giving $d_X\le d_Y$ for this pair, while contraction gives the reverse inequality. Thus equality holds.

For context, if the lifted map is not an automorphism, the derivative inequality is strict at every point; integrating along a minimizing geodesic in the complete punctured-disk metric gives a strict distance inequality for every distinct pair. Equality forces the covering-map case, but even a covering map need not preserve all distances globally because shorter paths can pass through other sheets. This is the distinction captured by equality in hyperbolic Schwarz-Pick contraction.

Write $A,B$ for the two given congruence subgroups, to distinguish them from the notation ${\Gamma }_1(N)$. Their double coset has a finite decompositionFiniteness follows because $B\cap {\alpha }^{-1}A\alpha$ has finite index in $B$. To check this without assuming integrality of $\alpha$, clear its denominators: a sufficiently deep principal congruence subgroup lies both in $B$ and, after conjugation by $\alpha$, in $A$. Representatives for $(B\cap {\alpha }^{-1}A\alpha )\backslash B$ give the displayed decomposition.

Using the Hecke-normalized rational slash operator from the PDF, define the double-coset operator on modular forms byReplacing ${\alpha }_j$ by $\gamma {\alpha }_j$ for $\gamma \in A$ changes no term because $f[\gamma {]}_k=f$. Right multiplication by $\beta \in B$ permutes these left cosets, so $([A\alpha B]f)[\beta {]}_k=[A\alpha B]f$. Each summand is holomorphic in the complex upper half-plane. At any rational modular cusp, factor ${\alpha }_j\rho =\sigma C$ with $\sigma \in S{L}_2(\mathbb{Z})$ and $C$ upper triangular of positive slope, as in 2(i). The cusp expansion of $f[\sigma {]}_k$ has no negative terms, and substitution by $C$ preserves boundedness. The sum is periodic in a cusp coordinate for $B$, so it has a removable singularity there. This provesIf $f$ vanishes at all its cusps, the same argument gives vanishing for the image too.

Now set $\Gamma ={\Gamma }_1(N)$. For an integer $d$ prime to $N$, choose ${\sigma }_d\in {\Gamma }_0(N)$ with lower-right entry congruent to $d$ modulo $N$. Such a lift exists by choosing $a$ with $ad\equiv 1(\mathrm{mod}N)$ and using $\left(\begin{array}{cc}a& (ad-1)/N\\ N& d\end{array}\right)$. The subgroup $\Gamma$ is normal in ${\Gamma }_0(N)$, and its quotient is $(\mathbb{Z}/N\mathbb{Z}{)}^{\times }$ through the lower-right entry. Define the diamond operator byDifferent lifts differ by a member of $\Gamma$, so this is well defined. The diamond operators satisfy $⟨d⟩⟨e⟩=⟨de⟩$.

For a prime $p\nmid N$, put $A_p=\mathrm{diag}(1,p)$ and define the Hecke operatorChoose $a,b\in \mathbb{Z}$ with $ap-bN=1$ and set ${\sigma }_p=\left(\begin{array}{cc}a& b\\ N& p\end{array}\right)$. A complete set of left-coset representatives isHere $A_j=A_p{T}^j$ and $B_p=A_p\left(\begin{array}{cc}ap& b\\ N& 1\end{array}\right)$, with both right factors in $\Gamma$. To prove completeness, $H=\Gamma \cap A_p^{-1}\Gamma A_p$ consists of the members of $\Gamma$ whose upper-right entry is divisible by $p$. Reduction of $\Gamma$ modulo $p$ is onto $S{L}_2({\mathbb{F}}_p)$: use the Chinese remainder theorem to impose identity modulo $N$ and any chosen determinant-one matrix modulo $p$, then lift to $S{L}_2(\mathbb{Z})$ by elementary matrices. Reduction of $H$ is the lower triangular subgroup. Its left cosets correspond to the $p+1$ lines of top rows in ${\mathbb{F}}_p^2$. The right factors $T^j$ give lines $[1:j]$, while the last factor gives $[0:b]$ with $b\not\equiv 0(\mathrm{mod}p)$. They are all the lines, proving the decomposition.

The printed slash normalization now gives $f[A_j{]}_k=p^{-1}f((\tau +j)/p)$ andThus

To prove commutation, use the marked-lattice model of a modular form. A pair $(L,t)$ consists of a lattice and a point of exact order $N$ modulo $L$, and a weight-$k$ function satisfies $F(uL,ut)=u^{-k}F(L,t)$. In a basis $L=\mathbb{Z}\tau +\mathbb{Z}$ with $t=1/N$, it represents $f(\tau )$. The diamond operator multiplies $t$ by $d$. The same Hecke operator isEvery marked point retains exact order $N$: multiplication by $p$ is invertible on its cyclic group, since $p\nmid N$. This lattice sum agrees with the displayed slash sum. Its $p$ overlattices with basis $((\tau +j)/p,1)$ give the first terms, and the remaining one with basis $(\tau ,1/p)$ has normalized marked point $p/N$, giving $p^{k-1}(⟨p⟩f)(p\tau )$. Multiplying the mark by $d$ commutes with passage to each overlattice and does not change the set of overlattices. Hence

Finally write $f={\sum }_{m\ge 0}{a}_m(f)q^m$. The first term of $T_p$ contains the finite averageIt therefore contributes $a_{np}(f)$ to the coefficient of $q^n$. The second term contributes $p^{k-1}{a}_{n/p}(⟨p⟩f)$ if $p\mid n$, and zero otherwise. This proves the good-prime and bad-prime Hecke coefficient formula at a good prime, without assuming that $f$ has a particular character:where a nonintegral coefficient index means zero. For $n=0$ both terms are retained. On a diamond-character eigenspace the second term becomes $\chi (p)p^{k-1}{a}_{n/p}(f)$.

For either coefficient system $\mathbb{Z}$ or ${\mathbb{F}}_p$, multiplying two matrices keeps all entries in that system, and multiplicativity of the determinant preserves determinant one. The identity matrix has determinant one, and for a determinant-one matrix the matrix inverse isIts entries again belong to the same coefficient system and its determinant is one. Together with associativity of matrix multiplication, these facts establish all the group axioms. Thus both sets are special linear groups.

Reduce each entry modulo five to defineReduction respects sums and products, so $\overline{MN}=\overline{M}\overline{N}$. It also respects the determinant, ensuring that the image has determinant one. Hence $\pi$ is a group homomorphism. Its kernel of a group homomorphism consists exactly of determinant-one integer matrices congruent to the identity modulo five:The determinant-one condition remains part of this set; the four integers cannot be chosen independently without it. A kernel of a group homomorphism is a normal subgroup, since $\pi (gk{g}^{-1})=\pi (g)e\pi (g{)}^{-1}=e$ for $k\in \ker \pi$. This is the principal congruence subgroup of level five.

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

Use the Gaussian curvature $-1$ normalizationof the Poincaré half-plane model. This is a complete conformal metric, invariant under real Möbius transformations. Its quotient by the free effective principal congruence subgroup is again complete: a geodesic lifts to the complete covering plane and extends there for all time. Transporting the quotient metric through the biholomorphism supplies the required complete conformal metric.

A nonconstant closed geodesic corresponds to a hyperbolic Möbius transformation $A$ of $\Gamma (2)$, and its hyperbolic translation length is $2\mathrm{arcosh}(|\mathrm{tr}A|/2)$. Indeed its eigenvalues have magnitudes $\lambda ,{\lambda }^{-1}$ with $\lambda >1$, and its axis quotient has length $2\log \lambda$. For $A=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ in $\Gamma (2)$, $bc\equiv 0(\mathrm{mod}4)$ implies $ad\equiv 1(\mathrm{mod}4)$. The odd numbers $a,d$ therefore have the same residue modulo four, andA hyperbolic Möbius transformation must have $|\mathrm{tr}A|>2$, so its smallest possible absolute trace is six. It is attained byConsequentlyThis element is primitive: a proper power would have a shorter root represented by a hyperbolic Möbius transformation, contradicting the trace bound. No essential simple closed geodesic exists on the three-punctured sphere, since every essential simple loop is peripheral and corresponds to a parabolic Möbius transformation. The minimizing closed geodesic is therefore nonsimple. Finally, the PDF leaves the numerical Gaussian curvature unspecified: the displayed length uses Gaussian curvature $-1$; for Gaussian curvature $-{\kappa }^2$, it is divided by $\kappa$.

First establish rational conjugation of finite-index modular subgroups without assuming that $\Gamma$ is a congruence subgroup. Multiply $\gamma$ by a positive integer to obtain an integral matrix $A$, and let $D=\det A>0$. Conjugation is unchanged by this scalar. If $h=I+DB\in \Gamma (D)$, thenThus the principal congruence subgroup $\Gamma (D)$ is contained in $H=S{L}_2(\mathbb{Z})\cap {\gamma }^{-1}S{L}_2(\mathbb{Z})\gamma$. It has finite index because reduction modulo $D$ has finite image. Inside $H$, pullback under conjugation of $\Gamma$ has relative index at most $[S{L}_2(\mathbb{Z}):\Gamma ]$. ConsequentlyThis argument does not assert that an arbitrary finite-index subgroup contains a principal congruence subgroup.

Use the determinant-normalized slash operatorThe positive real power of the determinant is used; on $S{L}_2(\mathbb{Z})$ this reduces to the usual slash operator for modular forms. The automorphy factor identity gives the right-action rule $(f{|}_k\gamma ){|}_k\eta =f{|}_k(\gamma \eta )$.

A modular form on a finite-index subgroup of integer weight $k$ is a holomorphic function on the complex upper half-plane, invariant under this weight-$k$ action of $\Gamma$, and holomorphic at a cusp at each of its cusps. A cusp of a modular group is a $\Gamma$ orbit in ${\mathbb{P}}^1(\mathbb{Q})$. If $\sigma \in S{L}_2(\mathbb{Z})$ carries infinity to its representative, choose a positive integer $h$ with $\sigma T^h{\sigma }^{-1}\in \Gamma$. Such $h$ exists by finite index. Then $f{|}_k\sigma$ is periodic and has a convergent expansion in $q_h=e^{2\pi iz/h}$ near zero; holomorphy means no negative exponents, and being a cusp form means zero constant term. Using an actual translation period avoids possible signs if a smaller width of a cusp is defined only modulo the center, particularly in odd weights.

For cusp holomorphy under rational slash operators, choose $\sigma \in S{L}_2(\mathbb{Z})$ with $\sigma \infty =\gamma \infty$, possible by completing a primitive integer pair to a determinant-one matrix. Then ${\sigma }^{-1}\gamma =\left(\begin{array}{cc}a& b\\ 0& d\end{array}\right)$, with $a/d>0$. Up to a nonzero constant factor,The imaginary part of the argument tends to infinity with that of $z$, so this remains bounded by the cusp expansion of $f{|}_k\sigma$. It tends to zero if $f$ is a cusp form. Moreover $f{|}_k\gamma$ is invariant under ${\Gamma }^{\prime }$: for $\eta \in {\Gamma }^{\prime }$, $\gamma \eta {\gamma }^{-1}\in \Gamma$ and the right-action rule applies. Finite index gives a translation period for ${\Gamma }^{\prime }$, so boundedness is a removable singularity at zero in that periodic parameter. This proves holomorphy at infinity. For every other cusp, apply the same argument to the rational matrix $\gamma \rho$, with $\rho \in S{L}_2(\mathbb{Z})$. Thus all cusp conditions hold, and

For the character twist by rational translations of a cusp form, put $B_j=\left(\begin{array}{cc}1& j/N\\ 0& 1\end{array}\right)$ and $K={\Gamma }_1(N)\cap {\Gamma }_0(N^2)$. For $\eta =\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\in K$, direct conjugation givesIndeed $N^2\mid c$ and $a\equiv d\equiv 1(\mathrm{mod}N)$. Every $f{|}_k{B}_j=f(z+j/N)$ is therefore $K$-invariant and vanishes at all its cusps by the preceding rational-translate argument. Their finite weighted sum is a cusp form, for every Dirichlet character:

There is, however, a missing primitivity hypothesis in the printed final expansion claim. The exact Fourier expansion of a modular form is alwaysValues of a Dirichlet character on units have modulus one, so $\chi (j{)}^{-1}=\overline{\chi (j)}$. For unit $n$, substitution gives $S_{\overline{\chi }}(n)=\chi (n)g(\overline{\chi })$, where $g$ is the Gauss sum of a Dirichlet character. For nonunit $n$, this vanishing formula requires a primitive Dirichlet character.

Here is its proof in that case. Choose a prime $p\mid (n,N)$. Primitivity supplies a unit $u\equiv 1(\mathrm{mod}N/p)$ with $\chi (u)\ne 1$: otherwise the character would factor through the surjective reduction to units modulo $N/p$. Surjectivity follows by lifting a unit and, if needed, adjusting the lift to avoid the additional prime $p$, using the Chinese remainder theorem. Multiplication by $u$ fixes $e^{2\pi inj/N}$ because $p\mid n$, but multiplies the character factor by a nontrivial constant. Hence $S_{\overline{\chi }}(n)=0$. The finite Fourier transform of a primitive Dirichlet character now gives the corrected formulaThe constant is nonzero: finite exponential orthogonality gives ${\sum }_{n\mathrm{mod}N}|S_{\overline{\chi }}(n){|}^2=N\phi (N)$, whereas the proved formula makes this $\phi (N)|g(\overline{\chi }){|}^2$. Thus $|g(\overline{\chi })|=\sqrt{N}$.

For a concrete counterexample to the printed unrestricted claim, take $N=2$, the principal Dirichlet character, and $f=\Delta$. The translation sum is $\Delta (z+1/2)={\sum }_n(-1{)}^n\tau (n)q^n$, whose $q^2$ coefficient is $-24$. Any constant multiple of the proposed odd-index-only series has $q^2$ coefficient zero. Thus the general modularity conclusion is proved, while the claimed simplification is false without the stated extra hypothesis.

Consider the principal congruence subgroupIt has finite index because $S{L}_2({\mathbb{F}}_3)$ is a finite group. It is torsion-free: by part (b), a finite-order element is conjugate to $A^i$ or $B^j$, while the reductions of $A$ and $B$ modulo $3$ still have orders $6$ and $4$, respectively. Hence no nonidentity power in either vertex group reduces to the identity.

It follows that $\Gamma (3)$ intersects every conjugate of the two vertex stabilizers trivially, so its action on the Bass-Serre tree is free. A group with a free group action on a tree is a free group. Consequently $S{L}_2(\mathbb{Z})$ contains the free subgroup $\Gamma (3)$ of finite index; equivalently, it is a virtually free group.

For $N\ge 3$, the principal congruence subgroup has projective index $\mu =\frac{1}{2}{N}^3{\prod }_{p\mid N}(1-p^{-2})$, no effective elliptic stabilizers, and all cusp widths equal to $N$. Hence the number of cusps is $\mu /N$, and the genus formula for a modular curve gives the displayed formula. The exceptional levels one and two both have genus zero; the full genus-zero list is $N=1,2,3,4,5$. Level six has genus one and every larger level has positive genus.

In curvature $-1$, its length is $2\mathrm{arcosh}3$. The level-two principal congruence subgroup has traces congruent to two modulo four, so the smallest hyperbolic absolute trace is six, achieved by $\left(\begin{array}{cc}5& 2\\ 2& 1\end{array}\right)$. The hyperbolic translation length formula gives the result; the shortest closed geodesic is nonsimple.

A determinant-one matrix over ${\mathbb{F}}_2$ permutes the three nonzero vectors of ${\mathbb{F}}_2^2$. This group action is faithful because fixing two basis vectors forces the matrix to be the identity. There are $(4-1)(4-2)=6$ invertible matrices, all of determinant one since the only nonzero field element is one, so the action identifies the group with the symmetric group $S_3$. Reduction of integer determinant-one matrices modulo two is onto: $\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$ and $\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)$ induce two distinct transpositions, which generate $S_3$. Its kernel consists of matrices congruent to the identity modulo two, the level-two principal congruence subgroup.

The level-two principal congruence subgroup $\Gamma (2)\subset {\mathrm{PSL}}_2(\mathbb{R})$ consists of integer determinant-one matrices congruent to the identity modulo two, modulo the scalar sign. It acts freely and properly discontinuously on the complex upper half-plane. An ideal quadrilateral with vertices $-1,0,1,\infty$ has its paired sides identified by $\tau \mapsto \tau +2$ and $\tau \mapsto \tau /(1-2\tau )$. The quotient has genus zero and three punctures, so it is conformally the sphere with three points removed. The invariant metric $|d\tau |/\mathrm{Im}\tau$ descends to its curvature-minus-one hyperbolic metric.

// Destination: analysis.bigb

The modular lambda function is invariant under the projective level-two principal congruence subgroup. That group acts freely, and its quotient is the level-two modular curve. Adding its three modular cusps gives a sphere, and $\lambda$ has cusp values $0,1,\infty$, with simple orders in exponential cusp coordinates. Therefore it identifies the quotient with the sphere minus those three points. Since the complex upper half-plane is simply connected, this is a universal covering map, with the projective level-two group as its deck transformation group.