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An inhomogeneous group $n$-cochain with values in a $G$-module $M$ is a function $G^n\to M$. The standard alternating group coboundary turns these groups into the cochain complex whose cohomology is $H^n(G,M)$.

Write a loop as $z^w{e}^{ih}$ with $h$ smooth real periodic and $h_0$ its mean. For the level-one representation with modes filled below zero and $\omega (h,k)=(2\pi {)}^{-1}{\int }_0^{2\pi }{h}^{\prime }k$, choose implementers $U(w,h)=S^w{e}^{iA(h)}$, where $S$ increases charge by one, $SQ{S}^{\ast }=Q-I$, and $[A(h),A(k)]=i\omega (h,k)I$. Their product differs from $U(w+v,h+k)$ by the displayed two-cocycle. Its phase is bilinear, which verifies the cocycle identity; integer $v$ makes it independent of the $2\pi$ ambiguity in $h_0$. Rephasing implementers changes it by a group coboundary.

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.

Galois cohomology records the obstruction to choosing Galois-invariant division points. Let $G_K=\mathrm{Gal}(\overline{K}/K)$ be the absolute Galois group and give its algebraic-point modules the discrete topology. A continuous $1$-cocycle with values in a Galois module $M$ is a function $c:G_K\to M$ satisfyingA group coboundary has the form $c_{\sigma }=\sigma a-a$. Quotienting cocycles by coboundaries defines $H^1(K,M)$. For the finite module $E[m]$ these cocycles have finite image and factor through finite data; continuity is essential because $G_K$ is a profinite group.

In characteristic zero, multiplication by $m\ge 2$ on the algebraic points of an elliptic curve is surjective, with kernel $E[m]\cong (\mathbb{Z}/m\mathbb{Z}{)}^2$. The short exact sequence of Galois modulesgives the Kummer exact sequence of an elliptic curveExplicitly, choose $Q$ with $mQ=P$ and set $\delta (P{)}_{\sigma }=\sigma Q-Q$. Changing $Q$ by an $m$-torsion point changes the cocycle by a coboundary. The class vanishes exactly when a suitable choice of $Q$ is Galois-fixed, that is, when $P\in mE(K)$. This proves the injection directly and identifies its arithmetic meaning.

The entire group $H^1(K,E[m])$ need not be finite. For example, Kummer theory gives $H^1(K,{\mu }_m)=K^{\ast }/K^{\ast m}$, which has classes supported on arbitrarily many different primes. The finite part needed for the Weak Mordell-Weil theorem comes from a ramification restriction on the image of $\delta$.

Choose a finite set $S$ of places containing the archimedean places, the primes dividing $m$ and all primes of bad reduction of an elliptic curve. At a finite place $v\notin S$, the curve has good reduction and $m$ is invertible in its valuation ring. The elliptic curve extends to a smooth proper group scheme, and $[m]$ on that model is finite etale. A point $P\in E(K_v)$ extends to an integral section by properness. Its division-point fibre is consequently finite etale over the valuation ring. Over the maximal unramified extension it has a point, so the Kummer cocycle restricts trivially to the inertia group. The module $E[m]$ itself is unramified there for the same reason. Thuswhere $H_S^1$ denotes classes unramified outside $S$.

Here is a proof that this restricted group is finite. Take a finite Galois extension $L/K$ containing all coordinates of $E[m]$ and the $m$th roots of unity, and enlarge $S$ by its ramified primes. Over $L$ the module is trivial and, after choosing a basis, is isomorphic to ${\mu }_m^2$. Hilbert theorem 90 and the multiplicative Kummer sequence therefore giveFor unramified Kummer classes with bounded prime support, a class unramified outside $S_L$ has valuations divisible by $m$ at every prime outside $S_L$: the valuation of an $m$th root in an unramified extension is integral. Consequently both coordinates lie in

To show this set finite, write the outside-$S_L$ divisor of $a$ as $mD$. The ideal class of $D$ belongs to the $m$-torsion of the ideal class group of the ring of $S_L$-integers. This gives the exact sequenceThe last map is onto: if $mD$ is principal in the $S_L$-ideal group, a generator represents a class with outside valuations divisible by $m$. The kernel consists exactly of S-units modulo $m$th powers. The Dirichlet unit theorem, with the finitely many inverted primes adjoined, makes the S-unit group finitely generated; the ideal class group is finite, and localization only quotients it. Both ends of the sequence are therefore finite.

Finally, the inflation-restriction exact sequence bounds the kernel of restriction from $H^1(K,E[m])$ to $H^1(L,E[m])$ by the finite group $H^1(\mathrm{Gal}(L/K),E[m])$. The unramified subgroup has finite image, contained in $L(S_L,m{)}^2$, and finite kernel, so it is finite. The Kummer injection now provesThis is the Weak Mordell-Weil theorem, proved without first assuming finite generation of $E(K)$.

For computation one refines the unramified group by local solvability. The Selmer group of an elliptic curve consists of classes whose restriction at every completion lies in the corresponding local Kummer image. It is finite and sits inwhere the Tate–Shafarevich group $\mathrm{Sha}$ measures classes in $H^1(K,E)$ that become trivial at every completion. Thus locally soluble descent equations can give an upper bound without every class coming from a rational point. No finiteness assumption on the whole Tate-Shafarevich group is needed for the weak theorem. Combining the finite quotient with the height descent lemma from the height essay yields the full Mordell-Weil theorem. Replacing multiplication by $m$ with a smaller isogeny gives the same cohomological framework for two-isogeny descent.

The inhomogeneous group cochain group iswith $C^0(G,M)=M$. Its group coboundary $d:C^r(G,M)\to C^{r+1}(G,M)$ isOne checks that $d^2=0$, and group cohomology is $H^r(G,M)=\ker d/\mathrm{im}d$.

A group extension of $G$ by the $\mathbb{Z}G$-module $M$ is an exact sequencewhose conjugation action on $i(M)$ agrees with the prescribed action of $G$ on $M$. It is a split group extension when $\pi$ has a group-homomorphic section $s:G\to E$. Two such extensions are equivalent group extensions when an isomorphism of their middle groups is the identity on $M$ and induces the identity on $G$. Transporting a section through that isomorphism proves that every extension equivalent to a split extension is split.

Choose a set-theoretic section $s$ with $s(1)=1$. Its failure to preserve multiplication is the normalized two-cocycleAssociativity gives the two-cocycle identity, and replacing $s$ changes $c$ by a group coboundary. The resulting class $[c]\in H^2(G,M)$ is therefore intrinsic to the extension, as expressed by second group cohomology classifies group extensions.

Now write $G=⟨x,y⟩\cong {\mathbb{Z}}^2$ and let $I$ be the augmentation ideal of $\mathbb{Z}G$. The Koszul resolution for a rank-two free abelian group gives, after applying ${\mathrm{Hom}}_{\mathbb{Z}G}(-,M)$, the last coboundaryIts image is $IM$. For $M=\mathbb{Z}G/I^2$ this yields the second cohomology of a rank-two free abelian group with truncated group-ring coefficients calculationThe canonical map $M\to \mathbb{Z}$ induces the identity on these final quotients, so $H^2(G,M)\to H^2(G,\mathbb{Z})$ is surjective; indeed it is an isomorphism.

Let $F_2=⟨x,y⟩$ and let ${\gamma }_i(F_2)$ be its lower central series. The class-two quotient $F_2/{\gamma }_3(F_2)$ is the Integer Heisenberg group. In the class-three free nilpotent group $N=F_2/{\gamma }_4(F_2)$, the module ${\gamma }_2(F_2)/{\gamma }_4(F_2)$ is cyclic over $\mathbb{Z}G$ on $[x,y]$ and is isomorphic to $\mathbb{Z}G/I^2$. Quotienting it by $I/I^2$ gives the central kernel $⟨[x,y]⟩\cong \mathbb{Z}$ of the Heisenberg group. The kernel ofis ${\gamma }_3(F_2)/{\gamma }_4(F_2)$, freely generated by $[[x,y],x]$ and $[[x,y],y]$, and is central. Thus it is ${\mathbb{Z}}^2$. This is the central nonsplit extension of the integer Heisenberg group by ${\mathbb{Z}}^2$. If it split, centrality would give $N\cong H(\mathbb{Z})\times {\mathbb{Z}}^2$, whose abelianization has rank four; but $N$ has abelianization ${\mathbb{Z}}^2$. Hence the extension is nonsplit.

For $a,b\in I$, the square-zero ideal condition givesThus the square-zero unit subgroup $1+I$ is abelian, andis a group isomorphism from the additive group of $I$.

Use the specified ring isomorphism $R/I\cong \mathbb{Z}G$. For $x\in \mathbb{Z}G$, choose a lift $r\in R$ and define $x\cdot a=ra$ for $a\in I$. Two lifts differ by an element of $I$, whose product with $a$ vanishes, so this is well defined. Right multiplication is handled identically. The two actions commute by associativity, making $I$ a bimodule. If $u\in R$ lifts $g\in G$, then $u$ is a unit: a lift $v$ of $g^{-1}$ makes both $uv$ and $vu$ elements of $1+I$, hence units, and a ring element with both a left and a right inverse is invertible. Conjugation therefore definesChanging $u$ by an element of $I$ does not change this expression because $I^2=0$. Moreover,so $a\mapsto 1+a$ is an isomorphism of $\mathbb{Z}G$-modules for these conjugation actions.

Let $R^{\times }\to (R/I{)}^{\times }$ be reduction on unit groups, and define $U$ as the inverse image of the distinguished subgroup $G\subseteq (\mathbb{Z}G{)}^{\times }$. Every $g\in G$ has a unit lift by the preceding argument, and the kernel consists exactly of the units congruent to $1$, namely $1+I$. Multiplication in $R$ therefore gives the group extensionChoose a set-theoretic section $s:G\to U$ with $s(1)=1$. Its extension cocyclesatisfies the two-cocycle identity by associativity. A different section changes $c$ by a group coboundary, so second group cohomology classifies group extensions gives a well-defined classThe same construction for $(R_1,I_1)$ gives $U_1$, an extension cocycle $c_1$, and $x_1=[c_1]\in H^2(G,1+I_1)$.

The answer to the final question is no. An abstract ring isomorphism $R_1\cong R$ need not carry the distinguished ideal $I_1$ to $I$, need not induce the identity under the two chosen identifications of the quotient rings with $\mathbb{Z}G$, and need not induce the prescribed $\mathbb{Z}G$-module isomorphism $\theta$. Hence it need not give an isomorphism of the two displayed group extensions, so it imposes no equality $\varphi (x_1)=x$. That equality does hold if the ring isomorphism has all three compatibility properties, because it then carries one extension cocycle to the other up to a group coboundary.

Put $Q=G/H$ and write $M^H$ for the invariant submodule. The five-term exact sequence in group cohomology associated with the Lyndon–Hochschild–Serre spectral sequence isThe inflation map in group cohomology composes a cocycle on $Q$ with the quotient homomorphism $G\to Q$. The restriction map in group cohomology restricts a cocycle from $G$ to $H$. The quotient action on the middle term is, for $q=gH$ and a one-cocycle $f$,This is independent of the lift and of the representative at the level of cohomology. Finally, the transgression in group cohomology extends a $Q$-invariant class on $H$ to a one-cochain on $G$; its coboundary is $H$-basic and descends to the two-cocycle on $Q$ representing $d_2[f]$. Changing the extension changes that cocycle by a group coboundary.

For the application, choose free generators $x_1,\dots ,x_n$ of $F$ and normal generators $r_1,\dots ,r_n$ of $R$. Since a finite nonabelian simple group $K$ is a perfect group, its abelianization is zero. The five-term sequence for $1\to R\to F\to K\to 1$ with trivial coefficients containsThe first term is zero because $K$ is finite, and the last term is zero because a free group has cohomological dimension one. It remains to prove that restriction is surjective.

The invariant submodule of homomorphisms $R\to \mathbb{Z}$ is exactlyThe images of the relators $r_1,\dots ,r_n$ generate $R/[F,R]$, so such a homomorphism $\lambda$ is determined by the integer vector $b=(\lambda (r_1),\dots ,\lambda (r_n){)}^T$. Let $A$ be the relator exponent-sum matrix. This square integer matrix presents $K^{\mathrm{ab}}$, which is zero, so $A$ is a unimodular matrix. There is therefore an integer vector $v$ satisfying $Av=b$. Define $\psi :F\to \mathbb{Z}$ by assigning to $x_i$ the $i$th entry of $v$. The definition of $A$ gives $\psi (r_j)=\lambda (r_j)$ for every $j$. Since the relator images generate $R/[F,R]$, the restriction of $\psi$ to $R$ equals $\lambda$. Restriction is surjective, exactness now gives

For $m\in M$, the map $g\mapsto g\cdot m-m$ is a principal crossed homomorphism. It is the group coboundary of the zero-cochain $m$, so first group cohomology is the quotient of crossed homomorphisms by principal crossed homomorphisms.

Equivalence classes of group extensions of $G$ by an abelian $G$-module $A$ correspond to $H^2(G,A)$. A section $s:G\to E$ produces the extension cocycleand changing the section changes $c$ by a group coboundary.