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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.

State the characteristic convention explicitly: for real $\alpha ,\beta$, define the theta function with characteristics byBinary subscripts ${\vartheta }_{ab}$ often denote this function with $(\alpha ,\beta )=(a/2,b/2)$, $a,b\in \{0,1\}$. On compact subsets of the $(z,\tau )$ domain, a summand and any derivative are bounded by a polynomial in $|n|$ times $e^{-c{n}^2+C|n|}$. Normal convergence therefore gives an entire function of $z$, holomorphic dependence on $\tau$, and legitimate termwise differentiation. In particular the heat equation for theta functions with characteristics isCharacteristics are defined modulo integers with a phase: $\theta [\alpha +1,\beta ]=\theta [\alpha ,\beta ]$ and $\theta [\alpha ,\beta +1]=e^{2\pi i\alpha }\theta [\alpha ,\beta ]$. Shifting the summation index proves the spatial quasi-periodsThese multipliers describe a degree-one line bundle on the complex torus with periods $1,\tau$, rather than an ordinary elliptic function. The logarithmic derivative is unchanged under one and decreases by $2\pi i$ under tau. Pairing opposite sides of a fundamental cell therefore makes its contour integral $2\pi i$. The argument principle counts exactly one zero in a cell, with multiplicity.

The characteristic function is a translate of the unshifted theta function:At $z=(1+\tau )/2$, the unshifted series cancels in pairs $n,-n-1$. Hence the zeros of theta functions with characteristics occur exactly and simply atReindexing also gives $\theta [\alpha ,\beta ](-z)=\theta [-\alpha ,-\beta ](z)$. For half-integer characteristics the parity is $e^{4\pi i\alpha \beta }$, so only $(1/2,1/2)$ is odd and vanishes at zero; the other three theta constants are nonzero.

The modular transformations of theta characteristics are obtained from Gaussian Poisson summation and elementary phase rearrangement. With the holomorphic branch of $\sqrt{-i\tau }$ positive at $\tau =i$,For the first formula, apply Poisson summation to $F(t)=e^{\pi i\tau (t+\alpha {)}^2+2\pi i(t+\alpha )(z+\beta )}$. Completing the square in its Fourier transform givesRearranging ${\sum }_{n}F(n)={\sum }_m\hat{F}(m)$ proves the stated inversion law. For translation by one, use $e^{\pi i{n}^2}=e^{\pi in}$ in the defining series; the remaining characteristic phase is precisely the displayed factor.

Write $Q=e^{\pi i\tau }$, so the modular-form variable of Question 1 is $q=Q^2$, and put ${\Theta }_2=\theta [1/2,0](0,\tau )$, ${\Theta }_3=\theta [0,0](0,\tau )$ and ${\Theta }_4=\theta [0,1/2](0,\tau )$. Inversion interchanges ${\Theta }_2,{\Theta }_4$ and fixes ${\Theta }_3$, each with factor $\sqrt{-i\tau }$. Translation interchanges ${\Theta }_3,{\Theta }_4$ and multiplies ${\Theta }_2$ by $e^{\pi i/4}$. The Jacobi triple product givesIt also gives the conventional odd theta function ${\theta }_1=-\theta [1/2,1/2]$ its product $2Q^{1/4}\sin (\pi z){\prod }_{m\ge 1}(1-Q^{2m})(1-2Q^{2m}\cos (2\pi z)+Q^{4m})$. Differentiate at zero and multiply the three even products to obtain the Jacobi derivative formulaIn the constant-product simplification, ${\prod }_m(1+Q^{2m})(1-Q^{4m-2})=1$, by separating even and odd positive exponents. The products prove nonvanishing of all three even constants as well.

The Jacobi abstruse identity can be proved directly, without assuming it from the products. Splitting pairs of summation indices according to whether their sum and difference are even or odd givesFor example, $u=(n+m)/2,v=(n-m)/2$ are either both integers or both half-integers, and $n^2+m^2=2u^2+2v^2$. The alternating signs give the second identity; shifting both original indices by one half gives the third. Squaring the first two and subtracting then proves ${\Theta }_3^4={\Theta }_2^4+{\Theta }_4^4$.

These identities connect theta theory to modular forms and elliptic curves. The fourth powers are weight-two forms on the principal level-two congruence subgroup. The modular lambda function $\lambda ={\Theta }_2^4/{\Theta }_3^4$ obeys $\lambda (-1/\tau )=1-\lambda (\tau )$ and $\lambda (\tau +1)=\lambda (\tau )/(\lambda (\tau )-1)$; the characteristic permutation kernel is $\Gamma (2)$. It parametrizes an elliptic curve with its ordered two-torsion in Legendre form $y^2=x(x-1)(x-\lambda )$. The corresponding Klein j-invariant is $256(1-\lambda +{\lambda }^2{)}^3/[{\lambda }^2(1-\lambda {)}^2]$. Finally the same products giveIndeed ${\Theta }_2{\Theta }_3{\Theta }_4=2Q^{1/4}{\prod }_m(1-Q^{2m}{)}^3$, whose eighth power divided by $256$ is $q{\prod }_m(1-q^m{)}^{24}$. This recovers the normalized modular discriminant of Question 1 and displays the bridge between Gaussian theta sums and the lattice discriminant of Question 2.

A theta function with characteristics evaluated at spatial argument zero. Its dependence on $\tau \in \mathbb{H}$ remains holomorphic. Among the four half-integer characteristic choices, the odd one vanishes identically, whereas the three even constants are nonzero on the half-plane. The zeros of theta functions with characteristics locate their spatial zero away from the origin. Ratios of equal powers of even constants yield the modular lambda function and other modular-invariant functions.

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.