Solution
= Solution
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 $r\geq0$. Use the <polarized fermionic Fock space>
$$
\mathcal F_P=\Lambda(PH)\,\widehat\otimes\,\Lambda(\overline{(I-P)H}),
$$
where the <exterior powers> are Hilbert-completed and the <tensor product> is graded. Its vacuum fills all modes $r<0$ and leaves $r\geq0$ empty; excitations are particles in the empty modes and holes in the filled modes. The <fermionic annihilation operators> $c_r$ and <fermionic creation operators> $c_r^*$ satisfy the <canonical anticommutation relations>
$$
\{c_r,c_s^*\}=\delta_{rs}I,\qquad\{c_r,c_s\}=\{c_r^*,c_s^*\}=0,
\qquad c_r\Omega=0\ (r\geq0),\quad c_r^*\Omega=0\ (r<0).
$$
Finite 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 $\Gamma(u)c^*(f)\Gamma(u)^*=c^*(uf)$, exactly when
$$
\boxed{[P,u]\text{ is Hilbert-Schmidt}.}
$$
Equivalently $P-uPu^*$ 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 https://arxiv.org/pdf/math/0409044[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 $g(z)=\sum_n g_nz^n\in LU(1)$, multiplication $M_g$ is a unitary one-particle operator. Its <matrix> entry from $e_s$ to $e_r$ is $g_{r-s}$. Therefore
$$
\|[P,M_g]\|_{\mathrm{HS}}^2
=\sum_{r,s}|\mathbf1_{r\geq0}-\mathbf1_{s\geq0}|^2|g_{r-s}|^2
=\sum_{n\in\mathbb Z}|n||g_n|^2.
$$
For 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 put
$$
E_{rs}=:c_r^*c_s:=c_r^*c_s-\delta_{rs}\mathbf1_{r<0}I,
\qquad Q=\sum_rE_{rr},\qquad L_0=\sum_r rE_{rr}.
$$
The sums are finite on each configuration. A particle at $r\geq0$ 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 $R_t=e^{itL_0}$ 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 <vectors>
$$
J_n=\sum_{r\in\mathbb Z}E_{r,r+n},\qquad J_n^*=J_{-n},\qquad J_0=Q.
$$
For fixed $n$ the sum has only finitely many nonzero actions on any configuration. Direct use of the <canonical anticommutation relations> gives
$$
[E_{rs},E_{tu}]=\delta_{st}E_{ru}-\delta_{ur}E_{ts}
+\delta_{st}\delta_{ur}(\mathbf1_{r<0}-\mathbf1_{s<0})I.
$$
On summing, the two noncentral sums cancel. If $m=-n$ the central sum is $\sum_r(\mathbf1_{r<0}-\mathbf1_{r+n<0})=n$: for $n>0$ the terms are precisely $r=-n,\ldots,-1$. Thus
$$
\boxed{[J_n,J_m]=n\delta_{n+m,0}I,\qquad[L_0,J_n]=-nJ_n.}
$$
In particular $J_n\Omega=0$ for $n>0$. The coefficient one in the central term is the level of this basic representation.
Let $h(\theta)=\sum_n h_ne^{in\theta}$ be smooth and real. The normal-ordered infinitesimal implementer of multiplication by $h$ is
$$
A(h)=\sum_n h_nJ_{-n}.
$$
The 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 $[A(h),c_r^*]=\sum_n h_nc_{r+n}^*$. 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.
Set
$$
\omega(h,k)=\frac1{2\pi}\int_0^{2\pi}h'(\theta)k(\theta)\,d\theta
=i\sum_n nh_nk_{-n}.
$$
It 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, yield
$$
e^{iA(h)}e^{iA(k)}=e^{-i\omega(h,k)/2}e^{iA(h+k)}.
$$
This 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 $z^we^{ih}$ 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> gives
$$
SQS^*=Q-I,\qquad SJ_nS^*=J_n\quad(n\ne0).
$$
Choose $U(w,h)=S^we^{iA(h)}$. Since $e^{iA(h)}S^v=S^ve^{i(A(h)+vh_0I)}$, one obtains the full <loop group two-cocycle for U(1)>
$$
\boxed{U(w,h)U(v,k)=c((w,h),(v,k))U(w+v,h+k),\quad
c((w,h),(v,k))=\exp\bigl(ivh_0-i\omega(h,k)/2\bigr).}
$$
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 $\mathcal F_P=\bigoplus_{q\in\mathbb Z}\mathcal F_q$ into charge <eigenspaces>. The charge-$q$ <ground state> $\Omega_q$ fills all integer modes $r<q$. Its energy is
$$
L_0\Omega_q=\frac{q(q-1)}2\Omega_q.
$$
For $q>0$, sum the energies of the added modes $0,\ldots,q-1$; for $q<0$, sum the hole energies of the removed modes $q,\ldots,-1$. 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 $b_n^*=J_{-n}/\sqrt n$, $n>0$. They obey $[b_n,b_m^*]=\delta_{nm}I$. Their occupation <vectors>
$$
\prod_{n>0}\frac{(J_{-n})^{m_n}}{\sqrt{n^{m_n}m_n!}}\Omega_q,
\qquad m_n\in\mathbb N_0\text{ with finite support},
$$
are <orthonormal> by commuting positive <fermionic currents on the circle> through negative <fermionic currents on the circle>. Their energies are $q(q-1)/2+\sum_{n>0}nm_n$. It remains to prove that they span, rather than assume the boson correspondence. Count configurations by formal charge $u$ and energy $t$. Empty modes $r\geq0$ can contain one extra particle, while filled modes $-r$, $r\geq1$, can contain one hole. Therefore the fermionic character is
$$
\prod_{r\geq0}(1+ut^r)\prod_{r\geq1}(1+u^{-1}t^r).
$$
The charge form of the <Jacobi triple product> proved in Question 4 identifies this with
$$
\sum_{q\in\mathbb Z}\frac{u^qt^{q(q-1)/2}}{\prod_{n\geq1}(1-t^n)}.
$$
The 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 identification
$$
\boxed{L_0=\frac{Q(Q-I)}2+\sum_{n>0}n b_n^*b_n.}
$$
The 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 $h_0Q$, nonzero modes of $h$ act as the square-summable oscillator field $\sum_{n>0}\sqrt n(h_n b_n^*+\overline{h_n}b_n)$, 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 $\|J_{-n}\Omega\|^2=kn$ 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.