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For a full-rank Euclidean lattice in ${\mathbb{R}}^n$, its Epstein zeta function is $E_{\Lambda }(s)={\sum }_{0\ne x\in \Lambda }|x{|}^{-2s}$, initially convergent for $\mathrm{Re}s>n/2$. A Mellin transform of its theta function gives its completed Epstein zeta function and a meromorphic continuation. Its sole pole has residue ${\pi }^{n/2}/[m(\Lambda )\Gamma (n/2)]$ at $s=n/2$.

For fixed $\mathrm{Im}\tau >0$, Gaussian decay of $|q{|}^{n^2}$ gives locally uniform convergence of the theta function series and all its $z$-derivatives. It is therefore entire. Shifting the series argument and reindexing givesSince ${\theta }_4(z)={\theta }_3(z+\pi /2)$, the corresponding second multiplier for ${\theta }_4$ has the opposite sign:Thus $h={\theta }_3/{\theta }_4$ obeys $h(z+\pi )=h(z)$ and $h(z+\pi \tau )=-h(z)$. These give candidate period lattices; to show they are exact, first locate every zero.

At $z_{\ast }=(\pi +\pi \tau )/2$,because the terms with indices $n$ and $-n-1$ cancel. The absolute convergence justifies that pairing. Quasi-periodicity propagates this zero to every $z_{\ast }+m\pi +n\pi \tau$.

For exhaustiveness and simplicity, let $v={\theta }_3^{\prime }/{\theta }_3$. It satisfies $v(z+\pi )=v(z)$ and $v(z+\pi \tau )=v(z)-2i$. Take a fundamental parallelogram for ${\Lambda }_0=\pi \mathbb{Z}+\pi \tau \mathbb{Z}$, translated so its boundary has no zero. The two sloping edges cancel by the first identity; the bottom and reversed top giveThe argument principle therefore counts one zero with multiplicity in every cell. Since a known zero class already exists, all zeros are simple and exactlyThe denominator's simple zeros are instead $\pi \tau /2+{\Lambda }_0$. These two cosets are disjoint: $\pi /2$ cannot lie in ${\Lambda }_0$ because $\mathrm{Im}\tau >0$. Thus there are no cancelled zeros or poles in the quotient.

Any period of $h$ or $h^2$ must translate its zero set onto itself, so must belong to ${\Lambda }_0$. A translation $m\pi +n\pi \tau$ multiplies $h$ by $(-1{)}^n$; the quotient is nonconstant, so an odd $n$ cannot be a period of $h$. Squaring removes precisely that sign. This proves the period lattice of theta3 over theta4 and its squared counterpart:The second lattice is for the squared function; no additional translation can preserve its zero set.

For a symmetric matrix $B$ with $Y=\mathrm{Im}B$ positive definite, the Riemann theta function isOn any compact set of $z$, the absolute value of the summand is bounded by $\exp (-c|n{|}^2+C|n|)$ for some $c>0$, so the series and all its derivatives converge uniformly there. It is therefore entire. Its constant Fourier coefficient in $\mathrm{Re}z$ is $1$, so it is not identically zero. Reindexing $n$ gives its parity and automorphy laws:Although the theta function is not itself a function on $J={\mathbb{C}}^g/\Lambda$, these factors define a holomorphic line bundle on $J$, and its zeros define a divisor $\Theta$ there.

For the period matrix of $X$, the Riemann vanishing theorem asserts that a fixed class $\kappa \in J$, the vector of Riemann constants in the convention used here, satisfiesWe prove both its support statement and its reduced divisor structure. This will make the multiplicity calculation in Solution 5 independent of any assumed multiplicity theorem.

The essential contour calculation is the theta pullback zero-divisor identity. Put $f_z(p)=\theta (u(p)-z,B)$, regarding this as a local expression for a section on $X$. If it is not identically zero, let $D_z$ be its zero divisor. We claimUse the cut polygon from Solution 1 and choose cuts avoiding the zeros. There ${\alpha }_z=d_p\log f_z$ is a meromorphic differential whose residues are the zero multiplicities. Under continuation by an $a_j$ period, $u$ changes by the integer vector $e_j$, so ${\alpha }_z$ is unchanged. Under a $b_j$ period, $u$ changes by $B{e}_j$, so automorphy gives ${\alpha }_z\mapsto {\alpha }_z-2\pi i{\omega }_j$. Pairing opposite boundary sides therefore givesFor example, the two $a_j$ sides, traversed with opposite orientations and separated by a $b_j$ translation, contribute ${\int }_{a_j}({\alpha }_z-({\alpha }_z-2\pi i{\omega }_j))$. This fixes the sign of the count.

For the first moment of the zeros, the residue theorem on the polygon gives, component by component,This is interpreted modulo the period lattice, with the lifts to the polygon fixed. Vary $z$ without letting zeros cross the cuts, and put $h=\delta \log f_z$. The coordinate $u_k$ does not vary, so integration by parts givesThere is no endpoint term: the closed polygon boundary returns both the Abel integral and $h$ to their starting values. Across an $a_j$ translation $h$ is unchanged, while across a $b_j$ translation it changes by $2\pi i\delta z_j$. Pairing the boundary edges consequently gives ${\int }_{\partial P}{\omega }_{k}h=-2\pi i{\sum }_j\delta z_j{\int }_{a_j}{\omega }_k=-2\pi i\delta z_k$. Thus $\delta u_g(D_z)=\delta z$, and $u_g(D_z)-z$ is constant. The calculation uses weighted zeros, so their collisions cause no problem. The parameters where $f_z$ is identically zero form a proper analytic subset: they are given by the vanishing of all its local Taylor coefficients, and parameters outside $\Theta$ are not in it. Its complement in the connected complex manifold $J$ is connected, so the constant is one class $\kappa$. This proves the pullback identity globally, with continuation of the zero divisors where necessary.

We must account for the parameters with identically zero pullback, rather than discard them. No irreducible component $C$ of $\Theta$ can consist entirely of such parameters. Otherwise parity would imply $C-u(p)\subset \Theta$ for every $p\in X$. The image of $C\times X$ under subtraction is irreducible, is contained in $\Theta$, and contains $C$ when $p=p_0$. Since $C$ is an irreducible component of the hypersurface, that image must be $C$. Hence $C-u(p)=C$ for all $p$. The surjectivity of the large-degree Abelian sum map, proved in Solution 3, says that these translations generate all of $J$. This would make the nonempty proper subset $C$ invariant under every translation, which is impossible.

A generic point $z$ of each component of $\Theta$ therefore has nonzero pullback. Since $\theta (-z)=\theta (z)=0$, its zero divisor contains $p_0$. Write $D_z=p_0+D^{\prime }$, with $D^{\prime }$ effective of degree $g-1$. The pullback identity gives $u_{g-1}(D^{\prime })=z+\kappa$. Each component of $\Theta$ is consequently contained in $W_{g-1}-\kappa$. By Solution 3 the latter is irreducible of dimension $g-1$, so any such component equals it. The zero locus is nonempty for $g>0$: any parameter with nonzero pullback has $g$ pullback zeros. We have proved equality of supports, in both directions.

Finally the automorphy factors make the First Chern class of the theta line bundle primitive. On the two-torus spanned by $a_i,b_j$, the logarithm of the $b_j$ factor changes by $-2\pi i{\delta }_{ij}$ after the $a_i$ translation; equivalently the transition for local frames has opposite winding. Thus $c_1(a_i,b_j)={\delta }_{ij}$, while $c_1(a_i,a_j)=c_1(b_i,b_j)=0$. In particular some integral pairing equals $1$. As its support is the single irreducible hypersurface $W_{g-1}-\kappa$, the zero divisor is $m(W_{g-1}-\kappa )$ for a positive integer $m$. Its integral First Chern class would then be divisible by $m$. Primitivity forces $m=1$, proving the displayed equality as reduced divisors.

The sign convention is $\kappa =u_g(D_z)-z$, so the translate is $W_{g-1}-\kappa$; a convention naming $-\kappa$ the Riemann constant reverses that displayed sign. For $g=1$, $\kappa =(1+B)/2$ modulo periods and the zero locus is the corresponding single point. For $g=0$, the empty-lattice theta series is $1$, its zero locus is empty, and the degree-$-1$ effective locus is empty; no negative symmetric product is required.

First alternative. The Euler product of the Riemann zeta function is absolutely convergent for $\sigma >1$. Its logarithm and the nonnegative trigonometric polynomialgiveConsequently $\zeta (\sigma {)}^3|\zeta (\sigma +it){|}^4|\zeta (\sigma +2it)|\ge 1$. Suppose $t\ne 0$ and the Riemann zeta function had a zero of order $m\ge 1$ at $1+it$. As $\sigma \downarrow 1$, its simple pole at one gives $\zeta (\sigma )=O((\sigma -1{)}^{-1})$, the alleged zero gives $\zeta (\sigma +it)=O((\sigma -1{)}^m)$, and $\zeta (\sigma +2it)$ remains bounded. The product would be $O((\sigma -1{)}^{4m-3})\to 0$, contradicting its lower bound. Thus there are no zeta zeros on $\Re s=1$; the point $s=1$ itself is a pole, not a zero.

For the first integral of the Chebyshev function, defineThe logarithmic derivative identity $-{\zeta }^{\prime }/\zeta (s)={\sum }_n\Lambda (n)n^{-s}$, valid for $\Re s>1$, leads toTo verify the kernel rather than assume an inversion formula, integrate $y^s/(s(s+1))$ on the vertical line $\Re s=c$. Closing to the left when $y>1$ gives the residues at zero and minus one, namely $1-y^{-1}$; closing to the right when $0<y<1$ gives zero. At $y=1$ the integral is zero as well. The horizontal integrals tend to zero because of the quadratic denominator, and the distant vertical integrals vanish in the indicated direction. Multiplying this kernel by $x$ with $y=x/n$ gives $(x-n{)}^{+}$. The absolute convergence of the Dirichlet series on $\Re s=c$, together with the integrable denominator, permits interchange with the integral and proves the formula.

Here is a quantitative contour route from this relation to the Prime number theorem. Question 3 establishes a zero-free region of the Riemann zeta function and the accompanying bounds for its logarithmic derivative, independently of the Prime number theorem. In a sufficiently narrow half-width region they implyon the left and horizontal sides of the rectangle with heights $\pm T$ and left edge ${\sigma }_L=1-b/(2\log (T+3))$, after making $b>0$ small enough also for bounded heights. The explanation of this bound is included at the end of Question 3. Take $c=1+1/\log x$ and $T=\exp (\sqrt{\log x})$. On the original line, the omitted tails aresince the Euler product bounds $|{\zeta }^{\prime }/\zeta (c+it)|$ by $-{\zeta }^{\prime }/\zeta (c)=O(\log x)$. Shift the truncated integral to the left edge. The only enclosed pole is at $s=1$, with residue $x^2/2$; the kernel poles at zero and minus one lie outside the rectangle. The left edge contributes $O(x^{2-b/(2\log (T+3))}{\log }^2(T+3))$, since $|s(s+1){|}^{-1}\ll (1+t^2{)}^{-1}$ there. The two horizontal edges contribute $O(x^2{\log }^2(T+3)/T^2)$. In particular, for some $a>0$,

It remains to remove the smoothing; differentiating an asymptotic formula would not justify this step. The monotonicity of the Second Chebyshev function gives, for $h=\epsilon x$ with $0<\epsilon <1$,First let $x\to \infty$ with $\epsilon$ fixed. After division by $x$, the lower and upper bounds tend to $1-\epsilon /2$ and $1+\epsilon /2$. Let $\epsilon \downarrow 0$ to obtain $\psi (x)\sim x$. Higher prime powers contribute $O(\sqrt{x}\log x)$ by the Chebyshev estimate, so $\theta (x)\sim x$. Finally partial summation givesThe integral is $O(x/{\log }^{2}x)$, by splitting at $\sqrt{x}$ and using $\theta (t)=O(t)$. This proves the Prime number theorem with $\pi (x)\sim x/\log x$.

Second alternative. We prove the functional equation of the Riemann zeta function from a theta function. Set $\Theta (u)={\sum }_{n\in \mathbb{Z}}{e}^{-\pi n^{2}u}$ for $u>0$. With Fourier transform convention $\hat{f}(\xi )={\int }_{\mathbb{R}}f(v)e^{-2\pi iv\xi }dv$, the Gaussian function $f(v)=e^{-\pi u{v}^2}$ hasIndeed, differentiation under the integral and integration by parts give ${\hat{f}}^{\prime }(\xi )=-2\pi \xi \hat{f}(\xi )/u$, while the Gaussian integral gives $\hat{f}(0)=u^{-1/2}$. Periodize $f$: the smooth period-one function ${\sum }_{n}f(v+n)$ has Fourier coefficients $\hat{f}(k)$, by integrating on $[0,1]$ and interchanging its absolutely convergent sum with the integral. Its absolutely convergent Fourier series, evaluated at zero, proves the Poisson summation formula in this case. HenceFor $\Re s>1$, termwise integration and the defining integral of the Gamma function give the Mellin representation of the completed Riemann zeta function:Split at one and substitute $u=1/v$ in the lower integral. The elementary part isand the remaining part becomes an integral over $[1,\infty )$. ThusThe integral is an entire function of $s$, because $\Theta (u)-1$ decays exponentially and uniform convergence holds on every compact subset of the complex plane. This proves meromorphic continuation of the completed Riemann zeta function, with only the displayed simple poles, and the right side is invariant under $s\mapsto 1-s$. ThereforeThe Gamma reflection formula and Gamma duplication formula give the equivalent form $\zeta (s)=2^s{\pi }^{s-1}\sin (\pi s/2)\Gamma (1-s)\zeta (1-s)$, understood by meromorphic continuation. Multiplication by $s(s-1)/2$ gives the entire Riemann xi function $\xi (s)=\xi (1-s)$.

Let $N(T)$ count positive ordinates of Nontrivial zeros of the Riemann zeta function up to height $T$, including multiplicity. The Riemann–von Mangoldt formula isMore precisely, away from ordinates of zeros, the constant term is $7/8$, the additional term is $S(T)={\pi }^{-1}\arg \zeta (1/2+iT)$, and the remaining error is $O(T^{-1})$. Here the argument is continued along $2\to 2+iT\to 1/2+iT$, and $S(T)=O(\log T)$. The coarse formula displayed above holds at all large heights with the usual right-continuous counting convention. Subtract it at $T+1$ and $T$: the main term changes by $O(\log T)$ and each error is $O(\log T)$. Hence

For infinitely many reciprocal-logarithmic gaps between zeta zeros, fix any $0<c<2\pi$, for example $c=\pi$. Suppose all sufficiently late gaps satisfied ${\gamma }_{n+1}-{\gamma }_n\le c/\log n$. Summing and usinggives ${\gamma }_n\le (c+o(1))n/\log n$. The last elementary estimate follows by integral comparison with $1/\log t$ and one integration by parts. The Riemann–von Mangoldt formula would implya contradiction. This reasoning is valid whether repeated ordinates are listed according to multiplicity or only distinct ordinates are listed, since in either case $n\le N({\gamma }_n)$. Consequently ${\gamma }_{n+1}-{\gamma }_n>c/\log n$ infinitely often, for any fixed $0<c<2\pi$.

Extend $\chi$ by zero to integers not coprime to $N$, and write $g(\chi )={\sum }_{a\mathrm{mod}N}\chi (a)e^{2\pi ia/N}$ for its Gauss sum of a Dirichlet character. We use the analytic branch of $\sqrt{-i\tau }$ that is positive when $\tau$ is positive imaginary. With the convention $\hat{F}(t)={\int }_{\mathbb{R}}F(x)e^{-2\pi ixt}dx$, the Gaussian Fourier transform and its differentiated version areThe first formula follows by completing the square, first on the imaginary axis and then by analytic continuation; the second follows by differentiating the Fourier transform in $t$. All the functions are Schwartz functions on the real axis.

For completeness, the finite Fourier transform of a primitive Dirichlet character satisfiesFor a unit $r$ this is substitution by its inverse. If $p\mid (r,N)$, primitivity supplies a unit $u\equiv 1(\mathrm{mod}N/p)$ with $\chi (u)\ne 1$: otherwise the character would factor through modulus $N/p$. Multiplication by $u$ fixes the additive exponential but multiplies the sum by $\chi (u)$, so the sum vanishes. This proves the identity also on nonunits. Finite Parseval identity now gives $|g(\chi ){|}^2=N$, since exactly $\phi (N)$ frequencies have nonzero character values. In particular this Gauss sum of a Dirichlet character is nonzero.

Apply Poisson summation to each residue class: ${\sum }_{j}F(a+Nj)=N^{-1}{\sum }_r{e}^{2\pi ira/N}\hat{F}(r/N)$. Multiply by $\chi (a)$ and sum over $a$. The formulas above give the two character theta transformationsandIn the odd case the prefactor originally reads $g(\chi )/(N^2\tau \sqrt{-i\tau })$; using $\tau =i(-i\tau )$ explains the factor $i$. Opposite parity makes the corresponding series identically zero by pairing $n,-n$. This is why the weight-three-halves series is needed for odd Dirichlet characters.

To deduce the products from these transformations, use two specific primitive Dirichlet characters. The even character ${\chi }_{12}$ takes values $1,-1,-1,1$ on $1,5,7,11$ modulo twelve. It is primitive: its values do not factor through six or four, hence through no proper divisor of twelve. Its Gauss sum of a Dirichlet character is $2\sqrt{3}=\sqrt{12}$. PutThe even transformation gives $A(-1/\tau )=\sqrt{-i\tau }A(\tau )$; also $A(\tau +1)=e^{\pi i/12}A(\tau )$, since each nonzero term has $n^2\equiv 1(\mathrm{mod}24)$. Thus $A^{24}$ transforms as a weight-twelve modular form under the generators $S,T$ of the modular group. It is holomorphic and starts with $q$, so it is a normalized cusp form. The proof in Question 1 shows $S_{12}=\mathbb{C}{\Delta }_0$, giving $A^{24}={\Delta }_0$. In particular $A$ has no interior zeros, by the valence formula for the modular group.

The Serre derivative argument in Question 2 gives $D{\Delta }_0=E_2{\Delta }_0$. Hence $DA/A=E_2/24$. The analytic product $P(\tau )=q^{1/24}{\prod }_{n\ge 1}(1-q^n)$ is nonzero and has the same logarithmic derivative, since $DP/P=1/24-\sum {\sigma }_1(n)q^n$. Therefore $A/P$ is constant on the connected complex upper half-plane; its limit at infinity is one. This proves $A=P$, without assuming either of the product identities being deduced. Pair the terms $n,-n$ in $A$, choosing representatives $n=6r+1$. Since ${\chi }_{12}(6r+1)=(-1{)}^r$, we getThis is the Euler pentagonal identity, derived by the character theta proof of the eta product.

For the second product use the odd primitive character ${\chi }_4(1)=1$, ${\chi }_4(3)=-1$, with Gauss sum of a Dirichlet character $2i$. SetThe odd transformation gives $B(-1/\tau )=(-i\tau {)}^{3/2}B(\tau )$ and the series gives $B(\tau +1)=e^{\pi i/4}B(\tau )$. Thus $B^8$ is another normalized weight-twelve cusp form, and $B^8={\Delta }_0=A^{24}$. Because $A$ is nonzero, $B/A^3$ has eighth power one, so it is constant; its leading coefficient fixes it to one. Pairing positive and negative odd integers now givesSince $B=A^3=q^{1/8}{\prod }_{n\ge 1}(1-q^n{)}^3$, the Jacobi cubic product identity follows:All pairings and differentiations take place in locally uniformly convergent theta functions or products on $|q|<1$.

Use the Fourier transform convention $\hat{g}(\xi )={\int }_{\mathbb{R}}g(x)e^{-2\pi ix\xi }dx$. For $t>0$, completing the square in the Gaussian integral givesBoth functions are rapidly decreasing, so the Poisson summation formula applies:In terms of the Jacobi theta function, this is $\vartheta (i/t)=\sqrt{t}\vartheta (it)$. The theta series converges locally uniformly in the complex upper half-plane, defining a holomorphic function. The branch of $\sqrt{-i\tau }$ with positive value at $\tau =it$ is holomorphic there, because $-i\tau$ lies in the right half-plane. Both sides of the proposed inversion identity are holomorphic; agreement on the positive imaginary axis and the identity theorem giveEquivalently, the complex Gaussian has transform $(-i\tau {)}^{-1/2}\exp (-\pi i{\xi }^2/\tau )$, which produces the same identity directly by Poisson summation.

Put $\psi (t)=(\vartheta (it)-1)/2={\sum }_{n\ge 1}{e}^{-\pi n^{2}t}$. For $\mathrm{Re}s>1$, integrate term by term in its Mellin transform:The theta function inversion givesSplit the integral at one. In the term containing $\psi (1/t)$ put $t=1/u$. The elementary remainder integral isThus the pole-subtracted theta integral for the completed zeta function isThe integral is entire by exponential decay. This supplies a meromorphic continuation with simple poles at zero and one, of residues $-1$ and $1$. The expression is unchanged by $s\mapsto 1-s$, proving the functional equation of the Riemann zeta functionThe reflection and duplication identities $\Gamma (z)\Gamma (1-z)=\pi /\sin \pi z$ and $\Gamma (z)\Gamma (z+1/2)=2^{1-2z}\sqrt{\pi }\Gamma (2z)$ convert it toMultiplying $Z$ by $s(s-1)/2$ removes its two poles and yields the Riemann xi function, which is an entire function, satisfying $\xi (s)=\xi (1-s)$.

For a symmetric matrix $B$ with positive definite imaginary part, this theta function converges normally on compact subsets of ${\mathbb{C}}^g$. It obeys $\theta (z+m+Bn,B)=e^{-\pi i{n}^{t}Bn-2\pi i{n}^{t}z}\theta (z,B)$, so its zeros define a divisor on the associated complex torus. If $B$ is a curve's normalized period matrix, its zero divisor is a translate of the effective degree-$g-1$ locus, by the Riemann vanishing theorem. A general symmetric positive period matrix need not come from a curve.