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For a finite-index subgroup of the modular group, take the union of translates of the standard fundamental domain of the modular group over left coset representatives. Boundary identifications are inherited from the subgroup action. The truncated region is compact, and its finitely many cusp ends have exponential parameters; this proves compactness of the compactified modular curve and boundedness of invariant cusp-form norms.

The closed standard fundamental domain of the modular group isIts vertical sides are paired by $T:z\mapsto z+1$, and its circular sides by $S:z\mapsto -1/z$. It is a fundamental region with boundary identifications. To obtain exactly one representative of each orbit, use $-1/2<x\le 1/2$ and, on $|z|=1$, retain the half with $x\ge 0$. We will include both vertices when discussing the closure.

To prove existence of a representative, fix $z\in \mathbb{H}$. Among primitive integer pairs $(c,d)$ choose one minimizing $|cz+d|$. The minimum exists: below any fixed bound, $|c|y\le |cz+d|$ bounds $c$, and then the real part bounds $d$, so only finitely many pairs occur. Extend this primitive pair to the bottom row of a determinant-one integer matrix $\gamma$. Sincethis makes the imaginary part maximal over the orbit. Translate by an integer to put the real part in $[-1/2,1/2]$. If the resulting point had modulus below one, inversion by $S$ would increase its imaginary part, a contradiction. Thus every orbit meets $\mathcal{F}$.

For uniqueness in the interior, note that $y\ge \sqrt{3}/2$ throughout $\mathcal{F}$ and that $|cz+d|\ge 1$ for every primitive pair. If $|c|\ge 2$, then $|cz+d|\ge |c|y\ge \sqrt{3}>1$. For $|c|=1$, the minimum over integer $d$ is attained at $d=0$ or, at an endpoint, at a neighboring integer: $|z|\ge 1$ and $|z\pm 1{|}^2=|z{|}^2\pm 2x+1\ge 1$. All larger $|d|$ give a strict inequality. If $c=0$, primitivity gives $d=\pm 1$.

If two points in $\mathcal{F}$ are equivalent, applying this inequality to the transformation and its inverse shows their imaginary parts are equal. For two interior points equality forces $c=0,d=\pm 1$, hence an integer translation. Their real parts lie in an interval of length one, so the translation is zero. The only matrices acting trivially are $\pm I$. On the boundary, the equality cases give precisely the stated vertical and circular identifications. This establishes the fundamental-domain claim, including its boundary convention.

An elliptic point of a modular curve means a fixed point of a nonidentity element of the effective modular group $PS{L}_2(\mathbb{Z})$; the central matrix $-I$ is not counted, since it acts trivially everywhere. The equality cases above show that the only such points in the closed region areTheir stabilizer subgroups in $S{L}_2(\mathbb{Z})$ arewhereIndeed $S^2=-I$, $(ST{)}^3=(TS{)}^3=-I$, and their fractional-linear fixed-point equations give the listed points. To see that these are the full stabilizer subgroups, equality in $|cz+d|\ge 1$ leaves only bottom rows $(0,\pm 1),(\pm 1,0)$, with the extra neighboring rows at the vertices. The fixed-point equation then fixes the top row. This yields exactly four matrices at $i$, six at each vertex, and just $\{\pm I\}$ elsewhere. These are the elliptic stabilizers of the modular group. After passing to $PS{L}_2(\mathbb{Z})$ their effective orders are two and three. The two vertices represent the same elliptic orbit.

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.

Put $q=e^{2\pi i\tau }$ and ${\sigma }_j(n)={\sum }_{d\mid n}{d}^j$. Separate the $m=0$ terms in the lattice Eisenstein series. Differentiating the cosecant partial-fraction identity $2k-2$ times gives, for $\Im z>0$,The initial identity is ${\sum }_n(z+n{)}^{-2}={\pi }^2{\csc }^2(\pi z)=-4{\pi }^2{\sum }_{r\ge 1}r{e}^{2\pi irz}$; differentiation accounts for the factorial and the sign. Pair $m$ and $-m$ and collect the products $mr$. Absolute convergence of the lattice sum for $k\ge 2$, and local uniform convergence of the resulting power series, justify these operations. Thus the requested Fourier expansions areHere $E_{2k}=G_{2k}/(2\zeta (2k))$, and the Bernoulli numbers have convention $B_2=1/6$. The second formula uses Euler's even-value identity for the Riemann zeta function, $2\zeta (2k)=(-1{)}^{k+1}{B}_{2k}(2\pi {)}^{2k}/(2k)!$. In particular $E_4=1+240\sum {\sigma }_3(n)q^n$ and $E_6=1-504\sum {\sigma }_5(n)q^n$.

We need the rational, rather than merely complex, version of the polynomial ring of level-one modular forms. Recall how the valence formula for the modular group supplies its proof. For a nonzero weight-$w$ modular form, with $\rho =e^{2\pi i/3}$,Integrate $f^{\prime }/f$ around a truncated standard fundamental domain of the modular group, indenting around boundary zeros. The paired vertical sides cancel by translation; pair the two halves of the circular edge using $f(-1/\tau )={\tau }^{w}f(\tau )$. The logarithmic derivative of ${\tau }^w$ supplies $w/12$, while the top edge supplies $v_{\infty }(f)$. The half-disc and third-disc at the two elliptic points account for the fractions $1/2,1/3$. The argument principle then gives the formula. All orders in it are nonnegative for a holomorphic modular form. It follows that negative weights have no forms, $M_0=\mathbb{C}$, and $M_2=0$: a weight-two form vanishes at $i$, contributing at least $1/2>2/12$.

Define ${\Delta }_0=(E_4^3-E_6^2)/1728$. It is a weight-twelve cusp form and its expansion starts with $q$. The valence formula for the modular group forces its order at infinity to be exactly one and leaves no interior zeros. Thus division by ${\Delta }_0$ identifies $S_w$ with $M_{w-12}$. Every nonnegative even $w\ne 2$ has a monomial $E_4^a{E}_6^b$ of weight $w$: solve $2a+3b=w/2$, taking $b=0$ when $w/2$ is even and $b=1$ when it is odd and at least three. Subtract the constant coefficient of a modular form times such a monomial and divide by ${\Delta }_0$. Induction on weight proves generation by $E_4,E_6$.

If the original Fourier coefficients are rational, every step preserves rationality: both generators and ${\Delta }_0$ have rational expansions, and formal division by ${\Delta }_0=q+O(q^2)$ does too. This proves the rational structure of level-one modular forms. Write $E_{2k}=\sum r_{a,b}{E}_4^a{E}_6^b$ with $r_{a,b}\in \mathbb{Q}$ and $4a+6b=2k$. Rescaling givesThe Bernoulli number formula above makes each zeta ratio rational: its powers of $\pi$ cancel precisely because the weights agree. This proves the required rational coefficients for the unnormalized lattice Eisenstein sums.

Finally $S_8=S_{10}=0$ because the quotient weights are negative, and $S_{14}=0$ because $M_2=0$. In each of these weights a form is determined by its constant coefficient. The indicated monomials have that coefficient equal to one, so

Use Euclidean worldsheet conventions with ${ϵ}^{12}=1$. Including a target metric, a Kalb–Ramond field $B_{\mu \nu }$ and the dilaton $\Phi$ givesThe imaginary coefficient of the antisymmetric term comes from Euclidean continuation; the Lorentzian action has the corresponding real coupling. For constant ${\Phi }_0$, the Gauss-Bonnet theorem gives $S_{\Phi }={\Phi }_0\chi (\Sigma )$, with Euler characteristic $\chi =2-2g$ on a connected closed oriented surface. Thus dilaton Euler-characteristic weighting givesAdding one handle multiplies a contribution by $g_s^2$, identifying $g_s$ as the string coupling. With canonically normalized external closed-string states, $n$ vertices add $g_s^n$, giving the string genus expansion $g_s^{2g-2+n}$.

Before gauge fixing, the Polyakov path integral for inserted string vertex operators is schematicallyThe quotient removes descriptions related by worldsheet diffeomorphisms and Weyl transformations. Gauge fixing has a nontrivial Jacobian, the Faddeev-Popov determinant. Anticommuting vector ghosts $c^a$ and symmetric trace-free antighosts $b_{ab}$ exponentiate this determinant as a local worldsheet ghost action. These worldsheet ghost fields implement the gauge quotient; they are not additional spacetime particles. Their zero modes must be treated separately rather than included in an invertible determinant.

Three kinds of geometry enter the scattering calculation. The worldsheet moduli describe genuinely different conformal shapes that remain after local gauge fixing. Each such shape must be integrated over. The mapping class group consists of orientation-preserving diffeomorphisms modulo those continuously deformable to the identity; it identifies different markings of the same conformal surface, so only one representative of each class is counted. The worldsheet conformal Killing group is the continuous residual conformal symmetry of a chosen surface. It moves insertion positions without changing the geometry and must be divided out, usually by fixing enough vertex positions.

For a torus, write $z\sim z+1\sim z+\tau$, with $\mathrm{Im}\tau >0$. Its shape modulus is $\tau$, its mapping class group acts through the modular group, and its connected worldsheet conformal Killing group acts by translations. Thus the shape integral runs over the standard fundamental domain of the modular group, while one insertion position can be fixed by translation. On the sphere, three positions can instead be fixed by Möbius transformations; for genus at least two the continuous conformal Killing group is trivial.

Now use the half-normalized worldsheet diffeomorphism ghost operator in the question. A vector field $v^a$ can preserve the representative metric after a compensating Weyl transformation precisely when $P_{1}v=0$. The trace of the metric variation sets $\delta \omega =\frac{1}{2}{\nabla }_a{v}^a$, leaving the Conformal Killing equationThe kernel of this operator gives the infinitesimal worldsheet conformal Killing group.

For a surviving infinitesimal worldsheet modulus, choose its metric variation $t_{ab}$ orthogonal to all gauge variations. Orthogonality to Weyl transformations requires $h^{ab}{t}_{ab}=0$. On the closed surface, the formal adjoint of the worldsheet conformal Killing operator follows by integration by parts:Orthogonality to every diffeomorphism variation is therefore the adjoint-kernel condition, givingIn a local conformal coordinate these imply ${\partial }_{\bar{z}}{t}_{zz}=0$ and its conjugate: the moduli variations are represented by holomorphic quadratic differentials. Multiplying the definition of $P_1$ by two changes its adjoint by two but changes neither kernel.

For four tachyon vertex operators on the torus, there is one complex shape modulus, four complex insertion positions and one complex translation to remove. The worldsheet integration count after conformal gauge fixing is thereforeFix the fourth insertion at $z_4=0$. A schematic gauge-fixed expression isHere the antighost pair supplies the modulus measure, and the ghost pair at the fixed vertex absorbs the translation zero modes. The three unfixed positions and $\tau$ are the four complex integrations. The punctured Riemann surface moduli dimension gives the same result, $3g-3+n=4$. Target-spacetime zero-mode integrals producing momentum conservation are separate from this requested count of worldsheet integrations.

Let $T=\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)$ and $S=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$. For $\gamma =\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ in the modular group, left multiplication by powers of $T$ and by $S$ performs subtraction and exchange on its first-column entries. The Euclidean algorithm reduces $(a,c)$ to $(\pm 1,0)$, since the determinant condition makes them coprime. The resulting matrix is $\pm T^m$, and $-I=S^2$ is already generated. Hence $S,T$ generate $S{L}_2(\mathbb{Z})$.

The standard fundamental domain of the modular group is $D=\{z\in \mathbb{H}:|\mathrm{Re}z|\le 1/2,|z|\ge 1\}$, with the vertical sides identified by $T$ and the circular sides by $S$. To prove every orbit meets it, choose a coprime integer pair $(c,d)$ minimizing $|cz+d|$. A minimum exists because only finitely many pairs have this quantity below any fixed bound: $|c|\mathrm{Im}z\le |cz+d|$ bounds $c$, then $d$. Complete the pair to a determinant-one matrix. Its image has maximal imaginary part in the orbit, since $\mathrm{Im}(\gamma z)=\mathrm{Im}z/|cz+d{|}^2$. Translate its real part into $[-1/2,1/2]$. If its modulus were below one, inversion would increase its imaginary part, a contradiction. This is the reduction to the standard modular region.

The central matrices $\pm I$ act identically on the complex upper half-plane, so the effective group is ${\mathrm{PSL}}_2(\mathbb{Z})$. Away from points with nontrivial effective stabilizer, properly discontinuous action supplies ordinary quotient-disc charts. At a fixed point $z_0$, the coordinate $u=(\tau -z_0)/(\tau -{\overline{z}}_0)$ identifies the stabilizer action with a rotation; its invariant coordinate is $u^e$, where $e$ is the effective stabilizer order. These charts give the quotient its Riemann surface structure.

The elliptic stabilizers of the modular group occur only in the orbits of $i$ and $\omega =e^{2\pi i/3}$. They have effective orders two and three. Consequently the analytic ramification indices of the map from the half-plane areThe stabilizers in ${\mathrm{SL}}_2$ have orders four and six, but the central factor does not double these indices.

All rational boundary points, including infinity, lie in one cusp of a modular group: a primitive column $(a,c)$ can be completed to a determinant-one integral matrix taking infinity to $a/c$. The stabilizer of infinity is generated effectively by $T:\tau \mapsto \tau +1$, and the coordinate $q=e^{2\pi i\tau }$ identifies its high horodisc quotient with a punctured disc. Adding $q=0$ fills that disc. This constructs the compactified modular curve from the extended half-plane; it does not use the ordinary subspace topology on the rational boundary.

For compactness use the standard fundamental domain of the modular group. Its part below a fixed height $Y>1$ is compact, since its imaginary part is at least $\sqrt{3}/2$. The part above $Y$, modulo translation and with the modular cusp added, is a closed disc in the $q$-coordinate. Their images cover the quotient, so it is compact. The same argument with finitely many translates proves compactness for every finite-index subgroup.

A weight-$k$ modular form for $\Gamma =S{L}_2(\mathbb{Z})$ is a holomorphic function $f$ on the complex upper half-plane satisfyingand holomorphic at a cusp. Since the full modular group has one cusp class, this last condition means that the period-one function has a convergent expansion $f(\tau )={\sum }_{n\ge 0}{a}_n{q}^n$, $q=e^{2\pi i\tau }$, near infinity. A cusp form has $a_0=0$. The element $-I$ makes a nonzero weight odd form impossible.

For weight zero, $f$ descends to a holomorphic function on the quotient of the complex upper half-plane by the modular group. At an elliptic fixed point, invariance under its finite stabilizer makes the local Taylor series a series in the quotient coordinate. At the cusp, the $q$ expansion extends it over $q=0$. The standard fundamental domain of the modular group, with its boundary identified and its cusp added, is a compact Riemann surface, the compactified modular curve $X(1)$. The maximum modulus principle now provesHolomorphy at the cusp is essential to this conclusion.

For positive even $k\ge 4$, define the unnormalized Eisenstein seriesThe real-linear map $(m,n)\mapsto m\tau +n$ is invertible, and on a compact subset of the complex upper half-plane its norm is uniformly comparable to $\sqrt{m^2+n^2}$. Thus the sum converges absolutely and locally uniformly for $k>2$, proving holomorphy. For $\gamma =\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$,The map $(m,n)\mapsto (ma+nc,mb+nd)$ is a bijection on the nonzero integer pairs, giving $G_k(\gamma \tau )=(c\tau +d{)}^k{G}_k(\tau )$.

For the expansion, the cotangent partial-fraction Fourier kernel follows by differentiating the partial-fraction expansion of $\pi \cot \pi z$ and its geometric-series expansion $-\pi i(1+2{\sum }_{r\ge 1}{e}^{2\pi irz})$, valid for $\mathrm{Im}z>0$. It gives, for even $k\ge 2$,The $m=0$ part of $G_k$ is $2\zeta (k)$, where $\zeta$ is the Riemann zeta function; positive and negative $m$ contribute equally. Apply the kernel at $z=m\tau$ for $m>0$, and regroup the absolutely convergent sum according to $N=mr$. With the divisor sum ${\sigma }_{k-1}(N)={\sum }_{r\mid N}{r}^{k-1}$, this provesThe expansion has no negative powers and converges for $|q|<1$, so it also proves holomorphy at a cusp and completes the proof of modularity.

The supplied zeta values yield the Fourier expansion of a normalized Eisenstein seriesBoth $E_4^3$ and $E_6^2$ are modular forms of weight twelve. Their constant terms cancel, so their difference divided by $1728$ is a cusp form. Integrality requires an argument: dividing an integral series by $1728$ alone would not suffice. Expanding directly gives the integrality identity for the modular discriminantFor every integer $d$, $5d^3+7d^5$ is divisible by twelve. Modulo three, $d^3\equiv d^5\equiv d$; modulo four, an even $d$ makes both powers divisible by four, while an odd $d$ has $d^3\equiv d^5\equiv d(\mathrm{mod}4)$. Thus each coefficient $(5{\sigma }_3(n)+7{\sigma }_5(n))/12$ is integral. The other three series visibly have integral coefficients, provingThis proves integrality without assuming a product expansion for the modular discriminant.

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

For a weight-$k$ cusp form, the invariant norm of a modular form $y^{k/2}|f(z)|$, with $y=\mathrm{Im}z$, is unchanged by the modular group: $\mathrm{Im}(\gamma z)=y/|cz+d{|}^2$ and $|f(\gamma z)|=|cz+d{|}^k|f(z)|$ cancel exactly.

Every orbit meets the standard fundamental domain of the modular group $\mathcal{F}=\{z:|\mathrm{Re}z|\le 1/2,|z|\ge 1\}$, where $y\ge \sqrt{3}/2$. The Fourier expansion of a modular form at the cusp has $f(q)=qh(q)$, with $h$ holomorphic near zero, since $f$ is a cusp form. Thus $|f(x+iy)|=O(e^{-2\pi y})$ uniformly in $x$ as $y\to \infty$, and $y^{k/2}|f|$ tends to zero there. On the remaining compact part of $\mathcal{F}$ it is bounded by continuity. Invariance therefore gives

For every $y>0$, Fourier coefficient extraction givesFor $n\ge 1$, minimize the last factor by taking $y=k/(4\pi n)$. This proves the Fourier coefficient bound for a cusp formThe assertion concerns positive indices; the constant coefficient vanishes because $f$ is a cusp form.

Let $S=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$ and $T=\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)$. The modular group acts on the complex upper half-plane by fractional linear transformations, andFor fixed $z=x+iy$, the values $|cz+d|$ with primitive integer pairs $(c,d)$ have a positive minimum: the conditions $|cz+d|\le R$ bound both $|c|\le R/y$ and $|d|\le R+|cx|$, leaving finitely many pairs. By Bezout identity, a minimizing pair is the bottom row of a matrix in $S{L}_2(\mathbb{Z})$. Applying that matrix gives a point of maximal imaginary part in its orbit.

Translate it by a power of $T$ into $-1/2<x\le 1/2$. It must have modulus at least one, since otherwise $S$ would increase its imaginary part. If it is on the unit circle with negative real part, apply $S$, which there changes $x$ to $-x$. This gives an element of the specified half-open standard fundamental domain of the modular group. It proves existence by reduction to the standard modular region.

For uniqueness, if $z$ lies in the closed standard region then $y\ge \sqrt{3}/2$ andfor every primitive pair. If $c=0$, this follows from $d=\pm 1$; if $|c|\ge 2$, it follows from $|c|y\ge \sqrt{3}$. If $|c|=1$, then $d=0$ uses $|z|\ge 1$, while $|d|\ge 1$ givesIf both $z$ and $z^{\prime }=\gamma z$ lie in the region, applying this inequality to $\gamma$ and ${\gamma }^{-1}$ forces their imaginary parts to agree, hence $|cz+d|=1$.

The equality cases describe every possible boundary identification. For $c=0$ the map is an integer translation; the half-open vertical strip retains one representative. For $|c|=1,d=0$, equality requires $|z|=1$, and the map is an inversion followed by an integer translation. Inversion exchanges the two halves of the circle, so retaining only the right half removes the duplication; the only translation that returns the exchanged endpoint is the one identifying the two corners. The remaining equality cases have $|c|=|d|=1$, $|z|=1$ and $x=\pm 1/2$, and give only those same corner identifications. Our region retains the right corner alone. Therefore $z^{\prime }=z$, proving

The equality analysis also gives the elliptic stabilizers of the modular group. The central matrices $\pm I$ fix every point. The point $i$ is fixed by $S$, with $S^2=-I$, and the retained corner ${\rho }_{+}=e^{i\pi /3}$ is fixed by $R=TS$, with $R^3=-I$. There are no other noncentral stabilizers, and the possible bottom rows in the equality cases give exactlyThe corresponding stabilizer orders in $PS{L}_2(\mathbb{Z})$ are two, three and one; the requested group is $S{L}_2(\mathbb{Z})$, so the center must be included.

The closed string already contains the particle required for quantum gravity. For bosonic string theory in its critical dimension of string theory, the mass-shell condition and closed-string level matching areAt $N=\tilde{N}=1$, the states ${ϵ}_{mn}{\alpha }_{-1}^m{\tilde{\alpha }}_{-1}^n|0;k⟩$ are massless. After imposing the Virasoro constraints and quotienting null string states, the transverse polarization tensor decomposes into symmetric trace-free, antisymmetric and scalar pieces under $SO(D-2)$. They describe a graviton, the Kalb–Ramond field and the dilaton. In particular, the symmetric trace-free sector is a massless spin-two graviton. Its number of degrees of freedom is $D(D-3)/2$.

The state–operator correspondence associates each physical string state with a string vertex operator. For example, the matter part of a massless closed-string vertex operator isIts conformal weights are $(1,1)$, so its integrated form $\int d^{2}z{V}_{ϵ,k}$ is invariant under changes of string worldsheet coordinates. At fixed insertion positions it is accompanied by $c\bar{c}$ from the worldsheet ghost fields. Physical vertices represent BRST cohomology classes; longitudinal changes of the polarization tensor are BRST-exact operators and decouple from physical scattering amplitudes. For the graviton, this string-state gauge redundancy becomes the linearized target-space diffeomorphism $h_{mn}\mapsto h_{mn}+{\partial }_m{\xi }_n+{\partial }_n{\xi }_m$.

The Polyakov path integral computes a closed-string scattering amplitude by inserting the external string vertex operators, integrating their unfixed positions and the inequivalent worldsheet moduli, and including the worldsheet ghost fields. On a Riemann sphere three positions are fixed by the residual conformal group. For example, contractions of tachyon vertex operators produce the Koba-Nielsen factor, whose position integral gives the Virasoro–Shapiro amplitude.

The string dual resonance property means that one crossing-symmetric scattering amplitude has equivalent expansions in the different channels: its poles exhibit intermediate string states in each channel, rather than separate channel contributions being added again. The scattering-amplitude factorization at a pole identifies the intermediate particle and its couplings. Set $8\pi T=1$, so ${\alpha }^{\prime }=4$ and the closed-string tachyon has $M^2=-1$. For four such external states the Mandelstam variables obey $s+t+u=-4$. The Virasoro–Shapiro amplitude has generic $s$-channel poles at $s=-1,0,1,\dots$, exactly the bosonic string mass spectrum in these units.

The massless pole gives an especially direct check. With overall normalization $C$ and $u=-4-s-t$, the Gamma function recurrence givesThusThe quadratic residue in $t$ contains a spin-two exchange. A scalar exchange alone could not give this angular dependence. Combined with the known massless spectrum and scattering-amplitude factorization, it identifies exchange of the graviton, with possible scalar contributions from the dilaton; the antisymmetric Kalb–Ramond field does not couple to two identical scalar tachyons here. Consequently the graviton participates in interactions, rather than being an isolated free state.

Decoupling longitudinal graviton polarizations forces a universal coupling to the conserved stress-energy tensor. Consistency of this massless spin-two gauge invariance extends the linearized coupling to the nonlinear dynamics of general relativity. At distances large compared with $\sqrt{{\alpha }^{\prime }}$, the gravitational sector of the effective action begins with the Einstein-Hilbert action, alongside dilaton and Kalb–Ramond field terms and higher-derivative string corrections. The infinite tower of string states supplies the short-distance completion of this quantum gravity expansion.

Interactions are organized by string worldsheet topology. A constant dilaton gives the string coupling $g_s=e^{{\Phi }_0}$, and a connected oriented surface of genus $g$ has Euler characteristic $\chi =2-2g$. Its weight is $g_s^{-\chi }$; with $n$ normalized external vertices,The sphere is tree order, the torus is one loop, and each additional handle adds a factor $g_s^2$. Each coefficient is an integral over worldsheet moduli, so this is string perturbation theory in $g_s$, with an independent low-energy expansion in ${\alpha }^{\prime }$.

For the one-loop vacuum contribution, torus modular invariance identifies $\tau$ with $(a\tau +b)/(c\tau +d)$, $ad-bc=1$. Integration is over the standard fundamental domain of the modular group,The potential short-proper-time region ${\tau }_2\to 0$, responsible for a point-particle ultraviolet divergence, is absent. It would count metrics already represented elsewhere in $\mathcal{F}$. Torus modular invariance is therefore the geometric reason for one-loop ultraviolet finiteness. More explicitly, in $D=26$ the vacuum integrand is proportional to $d^2\tau {\tau }_2^{-14}|\eta (\tau ){|}^{-48}$, with $\eta$ the Dedekind eta function; the complete expression is invariant under the modular group.

Ultraviolet finiteness does not make the bosonic one-loop vacuum energy finite. The remaining long-tube region ${\tau }_2\to \infty$ is an infrared divergence from the tachyon: $|\eta (\tau ){|}^{-48}\sim e^{4\pi {\tau }_2}$. It signals instability of the bosonic vacuum. Tachyon-free consistent backgrounds remove this particular obstruction, though other infrared effects must still be treated. The ultraviolet improvement comes from the extended string and the complete spectrum together with the worldsheet gauge identifications; the genus expansion remains a perturbative description of quantum gravity.

For fixed $\tau =x+iy$, the set $\{|c\tau +d|:(c,d)\in {\mathbb{Z}}^2\setminus \{0\}\}$ has a positive minimum. Choose a primitive pair $(c,d)$ attaining it and complete it to a matrix $\gamma \in S{L}_2(\mathbb{Z})$. Sincethis point has maximal imaginary part in the orbit. Translate by a power of $T:\tau \mapsto \tau +1$ to arrange $|\mathrm{Re}\tau |\le 1/2$. If now $|\tau |<1$, applying $S:\tau \mapsto -1/\tau$ strictly increases the imaginary part, a contradiction. Thus $|\tau |\ge 1$, proving that every orbit meets the standard fundamental domain of the modular group $\mathcal{F}$.

For $\tau =x+iy\in \mathfrak{h}$ and $\gamma =\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\in \Gamma (1)$,The set $\{c\tau +d:(c,d)\in {\mathbb{Z}}^2\setminus \{0\}\}$ is a lattice in $\mathbb{C}$, so it has a shortest nonzero vector. Dividing $(c,d)$ by their greatest common divisor can only shorten it; hence a minimizing pair may be chosen coprime and completed to the bottom row of some $\gamma \in S{L}_2(\mathbb{Z})$. Consequently the orbit contains a point ${\tau }_0$ of maximal imaginary part.

Apply a power of $T$ so that $-1/2\le \mathrm{Re}{\tau }_0\le 1/2$. If $|{\tau }_0|<1$, thencontradicting maximality. Thus $|{\tau }_0|\ge 1$, and ${\tau }_0$ lies in the standard fundamental domain of the modular group.

The torus worldsheet path integral is invariant under the modular group acting on its complex parameter $\tau$. Integrating over the standard fundamental domain of the modular group counts each inequivalent torus once and excludes the short-proper-time ultraviolet region.