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A modular form is a holomorphic function on the complex upper half-plane that transforms with a fixed weight under a congruence subgroup and is holomorphic at every cusp.

The nonzero even theta constants and the Jacobi abstruse identity give $\lambda (-1/\tau )=1-\lambda (\tau )$ and $\lambda (\tau +1)=\lambda (\tau )/(\lambda (\tau )-1)$. It is invariant under the principal congruence subgroup of level two. This parameter puts the associated complex elliptic curve in Legendre form $y^2=x(x-1)(x-\lambda )$; its Klein j-invariant is $256(1-\lambda +{\lambda }^2{)}^3/[{\lambda }^2(1-\lambda {)}^2]$.

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.

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 an integer weight $k$, a cusp form for a congruence subgroup $\Gamma$ is a holomorphic function on $\mathbb{H}$ with $f{|}_k\gamma =f$ for every $\gamma \in \Gamma$, whose Fourier expansion of a modular form vanishes at every modular cusp. More explicitly, choose $\sigma \in S{L}_2(\mathbb{Z})$ representing a modular cusp $\sigma \infty$, and a genuine translation period $h>0$ with $\sigma T^h{\sigma }^{-1}\in \Gamma$. Then the requirement isconvergent for sufficiently large imaginary part. Thus the slash translate extends holomorphically at $q_h=0$ and has value zero. The vector space of these functions is $S_k(\Gamma )$.

A genuine period is convenient when $-I$ is absent and the weight is odd: the smaller effective cusp width may supply a negative rather than positive translation in $\Gamma$, but an actual period always exists, possibly twice that width. Using an actual period avoids imposing an incorrect periodicity. The zero-value condition is independent of the chosen representative or period. If $-I\in \Gamma$ and $k$ is odd, its slash relation forces $f=(-1{)}^{k}f=-f$, so the space is zero.

For positive integer $k$, the equivalence is the bounded invariant norm characterization of cusp forms. The three conditions below work together: meromorphicity and the bound remove interior poles; the transformation law produces a periodic local function at every modular cusp; the same bound makes that function tend to zero at the modular cusp. Conversely a cusp form has exponentially decaying Fourier expansions of a modular form, and finitely many compact core regions control its invariant norm everywhere. These implications are proved under the corresponding labelled items.

Without a positive-weight convention, weight zero is a genuine exception. The function $f=1$ satisfies all three printed conditions at $k=0$ for every congruence subgroup, but is not a cusp form. Indeed a weight-zero cusp form would descend to a holomorphic function on the compact curve $X(\Gamma )$ and hence be constant, and vanishing at a modular cusp would make it zero. Thus the literal unrestricted claim needs the qualification $k>0$.

If negative integer weights are included, both sides contain only the zero function. To check the bounded side, each modular cusp translate has at most power growth in $y$, hence at most a power of $\log (1/|q_h|)$. For a negative Laurent coefficient of order $m$, Cauchy's coefficient estimate bounds its absolute value by $C{r}^m(\log (1/r){)}^{|k|/2}\to 0$, so there is no negative coefficient and every modular cusp is holomorphic. If such a nonzero negative-weight form existed, the product of its slash translates over $\Gamma \backslash S{L}_2(\mathbb{Z})$ would be a nonzero holomorphic level-one modular form of negative weight. Right multiplication permutes the translates, and all modular cusp factors are bounded, so that product really is a level-one form. Its existence contradicts the valence formula. This also excludes negative-weight modular cusp forms. The nonzero counterexample is specifically weight zero.

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.

For an integer $k$ and a congruence subgroup $\Gamma \le S{L}_2(\mathbb{Z})$, the space $M_k(\Gamma )$ consists of holomorphic functions $f:\mathfrak{h}\to \mathbb{C}$ satisfyingfor every $\gamma =\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\in \Gamma$, and which are holomorphic at every cusp. This is the space of modular forms of weight $k$ and level $\Gamma$.

Let $\Gamma$ be a congruence subgroup. A modular form of integral weight $k$ and level $\Gamma$ is a holomorphic function $f:\mathfrak{h}\to \mathbb{C}$ such thatfor every $\gamma =\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\in \Gamma$, and such that $f$ is holomorphic at every cusp. In terms of the slash operator for modular forms, the first condition is $f{|}_k\gamma =f$; the second says that $f{|}_k\sigma$ has a Fourier expansion of a modular form with no negative powers in the local parameter whenever $\sigma \in S{L}_2(\mathbb{Z})$ sends infinity to a cusp.

For any finite-index subgroup $\Gamma \le S{L}_2(\mathbb{Z})$ and rational positive-determinant $\gamma$, the intersection $S{L}_2(\mathbb{Z})\cap {\gamma }^{-1}\Gamma \gamma$ has finite index. Multiply $\gamma$ by a positive rational scalar to obtain an integral matrix $A$ with positive integer determinant $D$. For $h=I+DB\in \Gamma (D)$, $Ah{A}^{-1}=I+AB\mathrm{adj}(A)$ is integral, so $\Gamma (D)$ lies in $S{L}_2(\mathbb{Z})\cap {\gamma }^{-1}S{L}_2(\mathbb{Z})\gamma$. This intersection has finite index. Its further intersection with ${\gamma }^{-1}\Gamma \gamma$ has relative index at most $[S{L}_2(\mathbb{Z}):\Gamma ]$. No assumption that $\Gamma$ itself is a congruence subgroup is needed.