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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.

Use units $c=\hslash =1$ and the Minkowski metric $g=\mathrm{diag}(1,-1,-1,-1)$. Composing two affine Poincaré group transformations, with the rightmost applied first, givesThe translations form a normal subgroup, so this is the semidirect product of the Lorentz group with ${\mathbb{R}}^{1,3}$. The transformations $(\mathrm{diag}(1,R),0)$, $R\in SO(3)$, fix time and the spatial origin. Their products and inverses correspond to $R{R}^{\prime }$ and $R^{-1}$, hence they form a subgroup isomorphic to SO(3).

A rotationless Lorentz boost changes the velocity of a frame without adding a spatial rotation. For $|\mathbf{v}|<1$ a convenient active convention iswith $B(0)=I$. It satisfies $B^{T}gB=g$, $\det B=1$, and $B(m,0)=(m\gamma ,m\gamma \mathbf{v})$. Collinear boosts form a one-parameter subgroup with additive rapidity, but general boosts do not. For nonzero boosts along $x$ and $y$, $(B_x{B}_y{)}_{02}={\gamma }_x{\gamma }_y{v}_y$ while $(B_x{B}_y{)}_{20}={\gamma }_y{v}_y$: their product is not symmetric, whereas every rotationless boost above is symmetric. The product contains a Wigner rotation. Infinitesimally, the Poincare algebra gives $[K_i,K_j]=-i{ϵ}_{ijk}{J}_k$ for Hermitian generators, so the boost generators do not close amongst themselves.

There is a group-versus-cover qualification in the spin range: an honest unitary representation of the proper orthochronous Poincaré group permits integer spin only. Half-integer spin requires its double cover $SL(2,\mathbb{C})⋉{\mathbb{R}}^{1,3}$, or a projective unitary representation of the ordinary group. The construction below gives the requested range on that double cover and descends for integer spin.

Fix $m>0$ and the rest momentum $k=(m,0)$. Its little group consists of rotations: if a proper orthochronous Lorentz transformation fixes $k$, its time column is $(1,0)$, the metric condition forces the corresponding time row, and the remaining block belongs to SO(3). On the Lorentz spinor double cover, this little group is SU(2). Choose its irreducible spin-$s$ representation, with rest-frame states $|k,s,\sigma ⟩$, $\sigma =-s,-s+1,\dots ,s$, and matrices $D^{(s)}$, the Wigner D-matrices.

For each $p$ on the forward mass shell, $p^2=m^2$, $p^0>0$, choose the standard rotationless Lorentz boost $L(p)=B(\mathbf{p}/p^0)$ and defineThe labels are canonical spin projections, defined by this choice of boosts. For any proper orthochronous $\Lambda$, factorThe last factor fixes $k$ and hence is a Wigner rotation. On the double cover, use its lift to SU(2). Taking the translation convention $U[I,a]|p,s,\sigma ⟩=e^{ip\cdot a}|p,s,\sigma ⟩$ yields the massive induced representation of the Poincare double cover:The argument of the translation phase is $\Lambda p$, since $U[\Lambda ,a]=U[I,a]U[\Lambda ,0]$. The Wigner rotation identityproves the Lorentz part of the representation law; the phases multiply according to the affine product already derived.

Normalize the generalized momentum states by $⟨p,s,\sigma |q,s,\tau ⟩=2p^0(2\pi {)}^3{\delta }^3(\mathbf{p}-\mathbf{q}){\delta }_{\sigma \tau }$. The invariant mass shell measure $d^{3}p/(2p^0)$ and unitarity of $D^{(s)}$ make the induced representation unitary on $L^2(d^{3}p/(2p^0))\otimes {\mathbb{C}}^{2s+1}$. This covariant normalization accounts for the absence of an energy square-root factor in the displayed transformation law. To see irreducibility, translations force an operator in the commutant to act within momentum fibres; the little-group action and Schur lemma make its action scalar on each irreducible spin fibre. Lorentz transitivity on the forward mass shell makes this scalar constant. The commutant is therefore scalar, so the unitary representation is irreducible. Conversely, the simultaneous translation spectrum and its little-group representation give the massive positive-energy sector of Wigner's classification.

Thus every $m>0$ and $s\in \{0,\frac{1}{2},1,\dots \}$ gives such an irreducible representation on the double cover. Its central element over a $2\pi$ spatial rotation acts as $(-1{)}^{2s}$. Since this rotation is the identity in the ordinary Poincaré group,

For the parity operator, put $\Pi =\mathrm{diag}(1,-1,-1,-1)$ and $\bar{p}=\Pi p=(p^0,-\mathbf{p})$. The extension obeys $PU[\Lambda ,a]P^{-1}=U[\Pi \Lambda \Pi ,\Pi a]$. At rest, $P$ commutes with the rotations; Schur lemma therefore gives $P|k,s,\sigma ⟩={\eta }_P|k,s,\sigma ⟩$ with a common phase $|{\eta }_P|=1$. Since $\Pi L(p)\Pi =L(\bar{p})$, the parity action on canonical massive spin states isFor the usual reflection convention $P^2=I$, ${\eta }_P=\pm 1$ is the intrinsic parity. Canonical spin is an axial quantity and its label $\sigma$ is unchanged; a helicity label instead changes sign because momentum is reversed. Either intrinsic parity extends a given massive spin representation.

The Born rule depends only on transition probabilitiesbetween rays. A physical transformation must preserve these probabilities. By Wigner theorem, such a ray transformation is induced by a unitary or antiunitary operator. For ordinary continuously connected transformations one chooses the unitary branch, so

The states $U(g_1)U(g_2)|\psi ⟩$ and $U(g_1{g}_2)|\psi ⟩$ represent the same ray. They may therefore differ by a state-independent quantum phase:Thus quantum transformations form a projective unitary representation.

If $g$ is a symmetry of the Hamiltonian operator $H$, it also obeysequivalently $[U(g),H]=0$. For a differentiable one-parameter subgroup, writewith a self-adjoint generator $Q$. Differentiating the symmetry equation at $s=0$ yieldsFor a time-independent $Q$, the Heisenberg picture equation isHence the generator is a conserved observable. Conversely, a self-adjoint conserved $Q$ commutes with $H$ and its unitary group generates symmetries. This is the conserved generator of a continuous quantum symmetry.

Such a continuous projective unitary representation admits strongly continuous rotations whose self-adjoint generator is bounded below and whose adjoint action rotates the loops. In the basic fermionic construction, integer one-particle modes are filled below zero. The rotation energy counts nonnegative particle energies and positive hole energies. It has finite-dimensional integer eigenspaces and implements circle rotations. Tensor products give higher nonnegative integer levels.

A projective representation is a group homomorphism into the projective linear group of a vector space. Choosing invertible linear representatives makes multiplication hold up to nonzero scalar factors. A projective unitary representation chooses unitary operators as representatives, whose scalar factors have modulus one.

The quotient identifies unitary operators differing by a scalar phase. A continuous projective unitary representation takes values in this quotient, with the quotient of the strong operator topology. A local continuous section fixes the phase by requiring a chosen nonzero matrix coefficient to be positive real.

Because state vectors differing by a quantum phase represent the same ray, a unitary realization of a group action need only satisfy

A projective unitary representation is a projective representation whose chosen representatives are unitary operators. The phase factors disappear on rays.

For the irreducible representation of the canonical anticommutation relations determined by $P$, a unitary one-particle map $u$ is unitarily implementable precisely when $P-uP{u}^{\ast }$ is Hilbert-Schmidt, equivalently when $[P,u]$ is Hilbert-Schmidt. The implementer is unique up to a scalar phase. The equivalent representation criterion for two polarizations is that their projections differ by a Hilbert-Schmidt operator. This yields a continuous projective unitary representation of the restricted unitary group.