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In charge $q$, let ${\Omega }_q$ fill all integer modes below $q$. Its rotation energy is $q(q-1)/2$. The negative Heisenberg current algebra of a fermion on the circle modes create mutually commuting oscillators $b_n^{\ast }=J_{-n}/\sqrt{n}$ with $[b_n,b_m^{\ast }]={\delta }_{nm}$. Their normalized occupation monomials applied to ${\Omega }_q$ are orthonormal. The Jacobi triple product identifies the fermionic charge-energy generating function with ${\sum }_q{u}^q{t}^{q(q-1)/2}{\prod }_{n>0}(1-t^n{)}^{-1}$, proving completeness in each finite-dimensional energy space. Thus the rotation Hamiltonian is $Q(Q-1)/2+{\sum }_{n>0}n{b}_n^{\ast }{b}_n$.

Use period-one spatial normalization and $Q=e^{\pi i\tau }$. The conventional odd function is ${\theta }_1=-\theta [1/2,1/2]$ with the characteristic convention of theta function with characteristics. Differentiating its Jacobi triple product gives ${\theta }_1^{\prime }(0)=2\pi Q^{1/4}{\prod }_{m\ge 1}(1-Q^{2m}{)}^3$. The three even constant products have product $2Q^{1/4}{\prod }_{m\ge 1}(1-Q^{2m}{)}^3$, proving the identity. The sign changes if one instead calls the characteristic series itself theta1.

The Jacobi triple product givesEvery factor is nonzero for $|q|<1$, and the product converges to a nonzero limit, so $\theta$ has no zero in the upper half-plane.

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.

One form of the Jacobi triple product is, for $0<|q|<1$ and $z\ne 0$,Both sides converge locally uniformly on the punctured $z$ plane. Here is a finite-product proof which also determines the multiplicative constant.

Put $Q=q^2$ and $(Q;Q{)}_m={\prod }_{j=1}^m(1-Q^j)$, with value one at $m=0$. Define the Gaussian binomial coefficient ${\left[\genfrac{}{}{0ex}{}{m}{k}\right]}_Q=(Q;Q{)}_m/[(Q;Q{)}_k(Q;Q{)}_{m-k}]$. The recurrence ${\left[\genfrac{}{}{0ex}{}{m}{k}\right]}_Q={\left[\genfrac{}{}{0ex}{}{m-1}{k}\right]}_Q+Q^{m-k}{\left[\genfrac{}{}{0ex}{}{m-1}{k-1}\right]}_Q$ proves by induction thatRewrite the two finite factors asThe finite identity, with $n=k-N$, makes this ${\sum }_{n=-N}^N{z}^n{q}^{n^2}{\left[\genfrac{}{}{0ex}{}{2N}{N+n}\right]}_Q$. Multiply by $(Q;Q{)}_N$. For each fixed integer $n$, the resulting coefficient tends to one, because $(Q;Q{)}_m$ tends to a nonzero limit and ${\left[\genfrac{}{}{0ex}{}{2N}{N+n}\right]}_Q\to 1/(Q;Q{)}_{\infty }$.

All finite $(Q;Q{)}_m$ are uniformly bounded above and away from zero. Thus these coefficients have one uniform bound, and ${\sum }_{n\in \mathbb{Z}}|q{|}^{n^2}|z{|}^n$ is uniformly summable on compact annuli. Dominated convergence passes the limit through the finite sums and proves the boxed formula. At $q=0$ the formula follows by its limiting value one. The frequently used alternative form isIt follows by the substitution $q^2=t$, $z=u/q$, and is the version used for the charge-counting identity in Question 8.

The loop group here is $LU(1)=C^{\infty }(S^1,U(1))$, with pointwise multiplication. Although it is abelian, its nontrivial positive-energy quantum representations are projective: implementers can commute only up to a phase. A positive energy representation of the loop group of U(1) has rotations $R_t$ whose self-adjoint generator is bounded below and which implement $g(z)\mapsto g(e^{it}z)$ in the projective unitary group. The fermionic construction gives both that phase and an explicit bosonic description.

Choose the one-particle Hilbert space $H=L^2(S^1)$ with modes $e_r(z)=z^r$, $r\in \mathbb{Z}$, and projection $P$ onto $r\ge 0$. Use the polarized fermionic Fock spacewhere the exterior powers are Hilbert-completed and the tensor product is graded. Its vacuum fills all modes $r<0$ and leaves $r\ge 0$ empty; excitations are particles in the empty modes and holes in the filled modes. The fermionic annihilation operators $c_r$ and fermionic creation operators $c_r^{\ast }$ satisfy the canonical anticommutation relationsFinite particle-hole configurations form an orthonormal basis. As in the finite-dimensional exterior algebra construction, their creation and annihilation operators act irreducibly: an operator in the commutant sends the vacuum to a scalar vacuum, since the joint kernel of its vacuum annihilators is one-dimensional, and then acts by that scalar on the dense configuration basis.

Segal's quantization criterion for fermions says that a unitary one-particle operator $u$ has a unitary implementer $\Gamma (u)$ on this Fock space, satisfying $\Gamma (u)c^{\ast }(f)\Gamma (u{)}^{\ast }=c^{\ast }(uf)$, exactly whenEquivalently $P-uP{u}^{\ast }$ is Hilbert-Schmidt. Implementers are unique up to phase, hence form a projective unitary representation of the restricted unitary group. This is the fermionic, polarized criterion; it is not a condition that the entire operator $u-I$ be Hilbert-Schmidt. The criterion and the corresponding continuity topology are stated in Theorem III.2.1.1 of the fermionic construction of loop-group representations. For the compact principal-angle part of a pair of polarizations, the threshold has a concrete interpretation. Its two-dimensional blocks are rotations through angles ${\theta }_j$; finitely many unmatched modes can be handled separately. Each angle ${\theta }_j$ rotates the vacuum into $\cos {\theta }_j$ times its original component plus $\sin {\theta }_j$ times a particle-hole pair. After discarding finitely many exceptional blocks, the successive tensor products are Cauchy precisely when ${\prod }_j\cos {\theta }_j$ converges to a nonzero value, equivalently ${\sum }_j{\sin }^2{\theta }_j<\infty$. In this block model the sum is exactly half the squared Hilbert-Schmidt norm of the projection difference; finitely many unmatched modes account for a change of charge. Infinitely many unmatched modes already violate the criterion. This explains why infinitely many arbitrarily small mixings can fail to be implementable.

For $g(z)={\sum }_n{g}_n{z}^n\in LU(1)$, multiplication $M_g$ is a unitary one-particle operator. Its matrix entry from $e_s$ to $e_r$ is $g_{r-s}$. ThereforeFor a fixed difference $n$, exactly $|n|$ pairs cross the polarization boundary. Smoothness makes the sum finite, so Segal's quantization criterion for fermions constructs a continuous projective unitary representation of every smooth loop. Smooth convergence gives operator-norm convergence of $M_g$ and Hilbert-Schmidt convergence of its commutator, which is sufficient for the strong projective continuity of the implementers.

Normal ordering fixes the charge and rotation energy. On the dense finite-configuration domain putThe sums are finite on each configuration. A particle at $r\ge 0$ adds charge one and energy $r$; a hole at $r<0$ adds charge minus one and energy $-r$. Consequently $L_0$ is a nonnegative self-adjoint diagonal operator with integer spectrum and finite-dimensional eigenspaces. The unitaries $R_t=e^{it{L}_0}$ are strongly continuous, have period $2\pi$, and implement $e_r\mapsto e^{itr}{e}_r$. Conjugating multiplication by this one-particle rotation gives $M_{g(e^{it}\cdot )}$, so their Fock implementers rotate the loop representation projectively. This is an actual positive-energy circle action.

The scalar anomaly is visible in the Heisenberg current algebra of a fermion on the circle. Define on finite-energy vectorsFor fixed $n$ the sum has only finitely many nonzero actions on any configuration. Direct use of the canonical anticommutation relations givesOn summing, the two noncentral sums cancel. If $m=-n$ the central sum is ${\sum }_r(1_{r<0}-1_{r+n<0})=n$: for $n>0$ the terms are precisely $r=-n,\dots ,-1$. ThusIn particular $J_n\Omega =0$ for $n>0$. The coefficient one in the central term is the level of this basic representation.

Let $h(\theta )={\sum }_n{h}_n{e}^{in\theta }$ be smooth and real. The normal-ordered infinitesimal implementer of multiplication by $h$ isThe sign of the current index matters: multiplication by $z^n$ raises the one-particle mode by $n$, so it corresponds to $J_{-n}$. It follows either directly from the bilinear commutator or from the fermionic currents on the circle that $[A(h),c_r^{\ast }]={\sum }_n{h}_n{c}_{r+n}^{\ast }$. For trigonometric $h$ these are oscillator fields on the finite-energy domain; their self-adjoint closures exponentiate to the implementers of $e^{ih}$. For general smooth $h$, their extensions are obtained by the oscillator-field construction below: ${\sum }_{n>0}n|h_n{|}^2<\infty$ guarantees the square-summable field coefficients, and Fourier truncation converges in the corresponding Weyl relations implementers. Thus no unjustified infinite operator product is required.

SetIt is real and antisymmetric for real $h,k$. The Heisenberg current algebra of a fermion on the circle relations give $[A(h),A(k)]=i\omega (h,k)I$. The Weyl relations, or the finite-mode Baker-Campbell-Hausdorff computation followed by the square-summable limit, yieldThis is an explicit two-cocycle on the identity component of the loop group. A different phase choice changes it by a group coboundary, but its antisymmetric commutator remains $e^{-i\omega (h,k)}$, and hence cannot be removed.

The winding components also matter. Every smooth loop is $z^w{e}^{ih}$ with $w\in \mathbb{Z}$, real periodic $h$, and mean $h_0$ determined modulo $2\pi$. Let $S$ implement multiplication by $z$. It moves the filled boundary one place upwards, so it raises charge by one; phases of charge ground states can be chosen so that $S{\Omega }_q={\Omega }_{q+1}$. Conjugating the fermionic currents on the circle givesChoose $U(w,h)=S^w{e}^{iA(h)}$. Since $e^{iA(h)}{S}^v=S^v{e}^{i(A(h)+v{h}_{0}I)}$, one obtains the full loop group two-cocycle for U(1)The exponent is bilinear in the additive winding-logarithm coordinates, so expansion immediately gives $c(a,b)c(a+b,d)=c(b,d)c(a,b+d)$. Replacing $h$ by $h+2\pi \ell$ changes $A(h)$ by $2\pi \ell Q$, whose exponential is identity since charges are integers; the scalar cocycle also stays unchanged since $v\in \mathbb{Z}$. Thus the expression is well-defined on loops, not just on chosen logarithms. It records both the derivative anomaly and the commutator of winding with constant loops.

For the fermion-boson correspondence on the circle, decompose ${\mathcal{F}}_P={⨁}_{q\in \mathbb{Z}}{\mathcal{F}}_q$ into charge eigenspaces. The charge-$q$ ground state ${\Omega }_q$ fills all integer modes $r<q$. Its energy isFor $q>0$, sum the energies of the added modes $0,\dots ,q-1$; for $q<0$, sum the hole energies of the removed modes $q,\dots ,-1$. Every other charge-$q$ configuration is obtained by moving finitely many occupied levels upwards, so this is the lowest energy. Since $J_n$ preserves charge and lowers energy by $n>0$, $J_n{\Omega }_q=0$ for $n>0$.

Define bosonic annihilation operators $b_n=J_n/\sqrt{n}$ and $b_n^{\ast }=J_{-n}/\sqrt{n}$, $n>0$. They obey $[b_n,b_m^{\ast }]={\delta }_{nm}I$. Their occupation vectorsare orthonormal by commuting positive fermionic currents on the circle through negative fermionic currents on the circle. Their energies are $q(q-1)/2+{\sum }_{n>0}n{m}_n$. It remains to prove that they span, rather than assume the boson correspondence. Count configurations by formal charge $u$ and energy $t$. Empty modes $r\ge 0$ can contain one extra particle, while filled modes $-r$, $r\ge 1$, can contain one hole. Therefore the fermionic character isThe charge form of the Jacobi triple product proved in Question 4 identifies this withThe reciprocal product counts bosonic occupations, or partitions of an integer. For fixed charge and energy, the finite-dimensional fermion space has exactly as many basis configurations as the independent fermionic current on the circle monomials already constructed. Hence those monomials exhaust every such space, proving their dense completeness. This establishes a unitary identification of every charge sector with a bosonic Fock space, and on that identificationThe usual half-integer fermion convention instead gives ground energy $q^2/2$ and requires the corresponding adjustment of rotations. Here integer modes were deliberately chosen to obtain a genuine circle action with integral nonnegative energies. The two ground states of energy zero, at charges zero and one, are therefore consistent, not a negative-energy defect.

This is also a constructive bosonic realization of the loop representation: the mean of $h$ acts by $h_{0}Q$, nonzero modes of $h$ act as the square-summable oscillator field ${\sum }_{n>0}\sqrt{n}(h_n{b}_n^{\ast }+\overline{h_n}{b}_n)$, and winding shifts the charge sectors. Taking $k$ tensor copies adds the fermionic current on the circle central terms and gives level $k\in \mathbb{N}$. Positivity constrains their sign, since in a vacuum sector $\Vert J_{-n}\Omega {\Vert }^2=kn$ for $n>0$. The level-one construction, its explicit loop cocycle, and its complete bosonic occupation basis show how an abelian classical loop group acquires noncommuting quantum currents while retaining positive rotation energy.

The specialization of the Jacobi triple product to the Jacobi theta function isSince $\tau \in \mathfrak{h}$, one has $|q|<1$, so every displayed factor is nonzero. MoreoverThe standard convergence criterion for infinite products therefore shows that the product converges to a nonzero value. This proves the nonvanishing of the Jacobi theta function throughout $\mathfrak{h}$.