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The Lie derivative of a differential form satisfies ${\mathcal{L}}_X=d{\iota }_X+{\iota }_{X}d$. Both sides agree on functions. On a differential 1-form $\eta$, evaluation at a vector field $Y$ gives $X(\eta (Y))-\eta ([X,Y])$ on both sides, by the definition of the exterior derivative and the Lie bracket of vector fields. Both operators are degree-zero derivations of an algebra on the exterior algebra. Agreement on functions and coordinate differential 1-forms therefore proves agreement on every differential form. Since $d^2=0$, the formula also gives ${\mathcal{L}}_{X}d=d{\mathcal{L}}_X$, so the Lie derivative of a differential form commutes with the exterior derivative.

Since $T^r$ is a product of circles, the Künneth theorem identifies its cohomology ring with the exterior algebra on the coordinate degree-one classes. Inversion negates each integral degree-one class, so it acts as $(-1{)}^q$ in degree $q$. Even in characteristic two, each degree-one class has square zero.

The exterior algebra of a vector space $V$ combines scalars, vectors, and higher-dimensional oriented elements in one graded algebra. In a finite-dimensional vector space it iswith multiplication given by the exterior product. Degree zero is the scalar field, degree one is $V$, and degree two is the space of bivectors. Multiplication adds degrees: combining two vectors produces a bivector, rather than another ordinary vector.^[ref]

The essential rule is $v\wedge v=0$. Repeating the same direction cannot produce a nonzero oriented area. Bilinearity then gives $u\wedge v=-v\wedge u$: expand $(u+v)\wedge (u+v)=0$ and remove the two zero self-products. The alternating rule remains part of the definition in characteristic two, where sign reversal alone would not force a self-product to vanish.

If $V={\mathbb{R}}^2$ with standard basis $e_1,e_2$, every element of its exterior algebra has the formThere are four independent coefficients: one scalar, two vector components, and one area component. In ${\mathbb{R}}^3$, the respective degree dimensions are $1,3,3,1$, giving eight components in total. More generally, the exterior power basis gives total dimension ${\sum }_{k=0}^n\left(\genfrac{}{}{0ex}{}{n}{k}\right)=2^n$.

This language is useful when direction and orientation matter together. Determinants measure the action on top-degree exterior powers; differential forms use exterior powers of a dual space to describe quantities integrated along paths, surfaces, and higher-dimensional regions. The Grassmann algebra viewpoint also makes the sign changes in anticommuting generators systematic. A single multiplication rule connects geometry and algebra.

Exterior multiplication by a fixed differential form is the linear operator taking the wedge product of differential forms on the left. For a one-form $\omega$, it raises degree by one. Anticommutativity gives $e(\omega )e(\eta )+e(\eta )e(\omega )=0$ for one-forms. On the exterior algebra of an inner product space, pairing this operation with contraction produces Clifford multiplication $c(v)=e(v^{♭})-{\iota }_v$.

A Lie algebra representation on $V$ induces one on each exterior power by acting on every factor and summing. The tensor product of Lie algebra representations preserves the defining alternating relations, so this descends to the quotient in every characteristic of a field. In characteristic two define the exterior algebra by the relations $v\wedge v=0$, rather than by dividing a tensor antisymmetrizer by $2$.

A Gerstenhaber algebra has an associative product making it a graded commutative algebra and a degree-minus-one graded Lie bracket. In the left convention, $[u,vw]=[u,v]w+(-1{)}^{(|u|-1)|v|}v[u,w]$. The shifted degrees $|u|-1$ determine the signs of antisymmetry and the Jacobi identity. Hochschild cohomology and the exterior algebra of polynomial derivations are examples.

Getzler rescaling combines the change $x\mapsto \sqrt{t}x$ in geodesic normal coordinates with the Clifford algebra filtration, assigning degree one to a derivative and to a Clifford generator and degree minus one to a coordinate. The associated graded Clifford algebra is the exterior algebra. This is useful because the spinor supertrace annihilates every Clifford monomial except the top-degree one. Applied to the Lichnerowicz formula for a twisted Dirac operator, it retains a quadratic curvature model, whose heat kernel is computed by a Gaussian ansatz; all terms of smaller rescaling degree disappear from the local index limit. Thus the cancellations in a supersymmetric heat trace can be computed before solving the complete curved heat equation.

A graded algebra is a graded commutative algebra if $uv=(-1{)}^{pq}vu$ for homogeneous elements of a graded algebra of degrees $p,q$. In characteristic not equal to two, every odd-degree element then squares to zero. The exterior algebra and Hochschild cohomology with the Hochschild cup product are examples.

Let $Z={\mathbb{CP}}^5/{\mathbb{CP}}^2$. Its quotient cell structure has one cell in each of dimensions $0,6,8,10$. In particular $Z$ is five-connected. The relative cohomology sequence, or its cellular cochain complex, givesThe subscripts denote degrees. Under the quotient map back to ${\mathbb{CP}}^5$, the three positive classes pull back to $x^3,x^4,x^5$. All positive products vanish, since their degree is at least twelve. Thus the cohomology ring of a collapsed projective subspace in this case is the square-zero algebra $R$ on these three classes.

Now compute its Sullivan minimal model $(\Lambda V,d)$, using cohomological grading and the convention that even generators generate a polynomial ring and odd generators an exterior algebra. Five-connectivity gives $V^j=0$ for $j\le 5$. Since a minimal differential is decomposable, it cannot have a nonzero value below degree twelve. Matching cohomology through degree ten therefore forceswith no other generators in degrees at most ten. The first products must then be killed by generatorsThese follow by killing respectively the unwanted degree-twelve, degree-fourteen, and two degree-sixteen classes. Subsequent generators kill $bc,c^2$ and the resulting syzygies. For example $av-bu$ is closed of degree nineteen, so it requires a further generator of degree eighteen. Thus the full minimal model is not just the algebra on the first seven generators.

There is nevertheless a simple map from the full model to $R$: send $a,b,c$ to the three displayed classes and every other generator to zero. It is a chain map, since every generator differential is a sum of products of positive-degree elements, and all positive products in $R$ vanish. It induces an isomorphism in cohomology: the entire cohomology of $\Lambda V$ is concentrated in degrees $0,6,8,10$, and its indicated basis maps to the indicated basis of $R$. This proves formality of a topological space and identifies the full minimal model as a minimal resolution of $R$, rather than only matching the initial part of its cohomology.

An explicit description of that full resolution is useful. Take the free graded Lie algebraIts Chevalley–Eilenberg cochain algebra $C^{\ast }(L)$ has one generator of degree $|\ell |+1$ for every homogeneous Lie basis word $\ell$, and its quadratic differential is dual to the bracket, with the usual graded signs. Rescaling the bracket-word duals gives exactly the generators and differentials listed above. Positive Lie degrees make this algebra Sullivan minimal. Its cohomology is $R$: the universal enveloping algebra is $T(p,q,r)$, and the augmentation module has the exact free resolutionExactness is immediate because every nonempty tensor word has a unique last letter. Applying ${\mathrm{Hom}}_{T(p,q,r)}(-,\mathbb{Q})$ gives just the unit and the three dual classes in total degrees $6,8,10$; products of positive classes land in an absent Ext degree two and are zero. Chevalley–Eilenberg cochains of a graded Lie algebra compute this Ext, so this also verifies the full model's cohomology.

The wedge sum $W=S^6\vee S^8\vee S^{10}$ has the same connectivity, cohomology ring, and initial closed minimal generators. The same full-model-to-$R$ argument applies to $W$, so its minimal model is also $C^{\ast }(L)$, by uniqueness of Sullivan minimal models of quasi-isomorphic simply connected rational cochain algebras. The Sullivan minimal-model classification identifies isomorphic minimal models with the same rational homotopy type. ConsequentlyThe explicit infinite minimal resolution above is shared by both spaces.

Use cohomological degree, and write the Sullivan minimal models asandThe subscripts indicate degrees. The even generators generate a polynomial ring and the odd generators an exterior algebra. Their cohomology rings are respectively $\mathbb{Q}[x]/(x^{n+1})$ and $\mathbb{Q}[u]/(u^2)$, since each odd differential kills exactly the indicated non-zero-divisor polynomial relation. These are the usual projective-space and even-sphere minimal models.

Orient the top classes so that the degree-one condition is $f^{\ast }[u]=[x^n]$. A representative model map must send $u$ to $x^n$. It must send $v$ to an element of degree $4n-1$ with differential $x^{2n}$. The element $x^{n-1}y$ has exactly these properties. ThusThe differential check is $d(x^{n-1}y)=x^{n-1}{x}^{n+1}=x^{2n}=(f^{\ast }u{)}^2$. In these degrees there is no other available monomial: even powers of $x$ have even degree, and an odd monomial is a power of $x$ times $y$. Hence the coefficient is forced by the differential. For $n=1$ the two models coincide and the map is the identity model map.

Let ${\Phi }_t$ be the local flow of a vector field $V$. The Lie derivative of a differential form is ${\mathcal{L}}_V\eta ={\frac{d}{dt}|}_{t=0}{\Phi }_t^{\ast }\eta$. We prove Cartan's magic formulaBoth sides are degree-zero derivations of an algebra on the exterior algebra of differential forms. They agree on a function $f$, since ${\iota }_{V}df=V(f)$. They also agree on $df$: pullback commutes with the exterior derivative, so ${\mathcal{L}}_{V}df=d(Vf)$, while $(d{\iota }_V+{\iota }_{V}d)df=d(Vf)$. Locally every differential form is a sum of products of functions and coordinate differentials, proving the identity in every degree.

For a compactly supported Hamiltonian vector field $V_H$, take the convention ${\iota }_{V_H}\omega =-dH$. Since $d\omega =0$, Cartan's magic formula gives ${\mathcal{L}}_{V_H}\omega =0$, so its complete flow of a vector field consists of symplectomorphisms. Given $p$ and $v\in T_{p}M$, a bump function lets us choose a compactly supported $H$ with $d{H}_p=-{\iota }_v{\omega }_p$, and hence $V_H(p)=v$. Choose such fields for a basis of $T_{p}M$. Their successive small-time flows give a map with invertible derivative at the origin. The inverse function theorem makes the orbit of $p$ open. Every orbit is open; because $M$ is connected, there is only one orbit. Thus the compactly supported symplectomorphisms act transitively on points, by Hamiltonian transitivity on a connected symplectic manifold.

On the unit sphere, the standard symplectic area is $4\pi$. The equator has complementary disks of areas $2\pi ,2\pi$, whereas the latitude at height $1/2$ has complementary disks of areas $\pi ,3\pi$: the cap above height $h$ has area $2\pi (1-h)$. A symplectomorphism must preserve the unordered pair of complementary areas. Therefore no such symplectomorphism exists, by complementary areas obstruct symplectic equivalence of separating curves.

Use the complex inner product convention linear in its first argument. Give the exterior algebra its induced inner product, with an orthonormal wedge basis. Exterior multiplication is $e(v)\eta =v\wedge \eta$. Its adjoint is contraction:Applying this formula to $a\wedge \eta$ separates its first term from the remaining terms, giving $e(b{)}^{\ast }e(a)\eta =(a,b)\eta -e(a)e(b{)}^{\ast }\eta$. Hence the canonical anticommutation relations areFor irreducibility, choose an orthonormal basis $u_1,\dots ,u_d$. The occupancy projections are $N_j=e(u_j)e(u_j{)}^{\ast }$, and $P_0={\prod }_j(I-N_j)$ projects onto the vacuum $1\in {\Lambda }^{0}V$. For an increasing index set $J$, put $E_J=e(u_{j_1})\cdots e(u_{j_r})$. Then $E_I{P}_0{E}_J^{\ast }$ maps the wedge basis vector $u_J$ to $u_I$ and kills every other wedge basis vector. Thus these products are all matrix units and the generated star-algebra is $\mathrm{End}(\Lambda V)$. Any common invariant subspace for the creation and annihilation operators is therefore either zero or all of $W$.

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 exterior algebra of a vector space $V$ is ${\Lambda }^{\ast }{V}^{\ast }={⨁}_{r=0}^{\dim V}{\Lambda }^r{V}^{\ast }$, whose degree-$r$ elements are alternating $r$-linear forms. Its exterior product satisfies $\alpha \wedge \beta =(-1{)}^{rs}\beta \wedge \alpha$ for degrees $r,s$. For example, one-forms anticommute, so $d{x}^i\wedge d{x}^i=0$. A differential form of degree $r$ on a smooth manifold is a smooth section of ${\Lambda }^r{T}^{\ast }M$; in coordinates it is an alternating sum of products of $r$ coordinate one-forms. Its exterior derivative increases degree by one, satisfies $d^2=0$, and obeysThe coordinate definition and pullback of a differential form are compatible, making differential forms suitable for coordinate-independent integration.

An orientation and a nondegenerate metric tensor define the Hodge star operator by $\alpha \wedge \ast \beta =⟨\alpha ,\beta ⟩{\mathrm{vol}}_g$ for equal-degree forms. It maps degree $r$ to degree $d-r$. If the metric has $s$ negative directions, its square is ${\ast }^2=(-1{)}^{r(d-r)+s}$. Thus Hodge duality turns antisymmetric tensors into complementary-degree tensors, using the metric and orientation rather than merely the underlying vector space. The Generalized Stokes theorem says ${\int }_{M}d\eta ={\int }_{\partial M}\eta$, with the induced boundary orientation, for appropriate smooth forms and compactness or support assumptions. It generalizes the fundamental theorem of calculus, circulation version of Stokes theorem, and divergence theorem.

For the field calculation, choose signature $(-,+,+)$ and a fixed orientation on Minkowski spacetime. Write $a=\delta A$, so $\delta F=da$. Symmetry of the metric pairing and the Hodge star operator gives $\delta (F\wedge \ast F)=2da\wedge \ast F$. Applying the exterior derivative product rule,Since $F\wedge a=a\wedge F$, variation of the action isFix the boundary data or use compactly supported variations. Arbitrary interior one-forms $a$ then imply the bulk field equationWith $B=\ast F$, the Lorentzian Hodge star operator squares to $-1$ on both one- and two-forms, hence $F=-\ast B$. Thus the Maxwell–Chern–Simons one-form equation becomesThese two displayed equations are equivalent. Also $dF=0$ gives $d\ast B=0$. For $k\ne 0$ the latter follows from applying $d$ to the field equation; for $k=0$ it must be retained as the identity inherited from $F=dA$. Using the opposite metric-sign convention changes the corresponding ${\ast }^2$ sign; the signature-independent equation is $d\ast F+kF=0$.

Under the Abelian gauge transformation $A\mapsto A+d\Lambda$, the curvature $F=dA$ and the one-form $B$ are unchanged because $d^2\Lambda =0$. The field equation is therefore exactly gauge-invariant. The kinetic term is unchanged, while the other term changes byThe last equality uses $dF=0$ and the Generalized Stokes theorem. Thus the action density changes by a total derivative and the integrated action can change on a boundary. The source's statement that the action is not invariant needs this qualification: on a boundaryless spacetime, or when the boundary integral vanishes, the integrated action is invariant too. No such boundary restriction is needed for invariance of the bulk field equation.

For a left $\mathbb{Z}G$-module, group cohomology iswith $\mathbb{Z}$ the trivial module. Equivalently take a projective resolution $P_{\bullet }\to \mathbb{Z}$ and the cohomology of ${\mathrm{Hom}}_{\mathbb{Z}G}(P_{\bullet },M)$. This also describes the right-derived functors of the invariant-module functor $M\mapsto M^G$.

For an explicit model, the homogeneous bar free resolution has $P_n=\mathbb{Z}[G^{n+1}]$, with diagonal $G$-action and boundary the alternating sum of coordinate deletions. It is free over $\mathbb{Z}G$: each orbit has a unique representative with first coordinate $1$. The augmented complex is exact because inserting $1$ as a first coordinate supplies a contracting homotopy of underlying abelian groups; its lack of $G$-equivariance does not affect exactness. The associated inhomogeneous group cochains are functions $f:G^n\to M$, withFor group cochains of degree zero, $(dm)(g)=gm-m$. Adjacent deletions cancel in pairs, giving $d^2=0$ and hence $H^n=\ker d/\mathrm{im}d$.

In degree zero this givesIt is also ${\mathrm{Hom}}_{\mathbb{Z}G}(\mathbb{Z},M)$, since a map is determined by the invariant image of $1$.

In degree one the cocycle condition is $f(gh)=f(g)+gf(h)$, soThese are crossed homomorphisms modulo principal crossed homomorphisms. For trivial action they reduce to ordinary group homomorphisms, so $H^1(G,M)=\mathrm{Hom}(G_{\mathrm{ab}},M)$.

There are two further useful descriptions. If $I=\ker (\mathbb{Z}G\to \mathbb{Z})$ is the augmentation ideal, a crossed homomorphism gives an equivariant map $I\to M$ by $g-1\mapsto f(g)$, because $gh-1=(g-1)+g(h-1)$. Applying $\mathrm{Hom}(-,M)$ to $0\to I\to \mathbb{Z}G\to \mathbb{Z}\to 0$ therefore givesAlso a section of $M⋊G\to G$ has the form $g\mapsto (f(g),g)$ and is a group homomorphism exactly when $f$ is a crossed homomorphism. Conjugation of sections by an element of $M$ changes $f$ by a principal crossed homomorphism. Thus $H^1$ classifies the $M$-conjugacy classes of such sections, or equivalently of complements to $M$ in the semidirect product.

In degree two, a normalized two-cocycle $c:G\times G\to M$ satisfiesDefine multiplication on $M\times G$ byThe displayed cocycle identity is precisely associativity; normalization gives the identity and inverses exist. This produces a group extension $1\to M\to E_c\to G\to 1$ with the prescribed action on $M$.

Conversely a set-theoretic section $s:G\to E$, $s(1)=1$, gives $c(g,h)=s(g)s(h)s(gh{)}^{-1}$ in the additive kernel of a group homomorphism. Associativity gives the cocycle condition. Replacing $s(g)$ by $b(g)s(g)$ changes $c$ byThe resulting extensions are equivalent through a change of coordinates in $M\times G$. Conversely any extension equivalence preserving both kernel of a group homomorphism and quotient yields such a change of section. HenceThe zero class corresponds exactly to a split extension. With trivial action these are central group extensions; addition of classes corresponds to the Baer sum. This proves the second group cohomology classifies group extensions interpretation.

For $G={\mathbb{Z}}^2$ and trivial coefficients $\mathbb{Z}$, put $R=\mathbb{Z}[x^{\pm 1},y^{\pm 1}]$. The Koszul resolution for a rank-two free abelian group iswhere $d_1(a,b)=(x-1)a+(y-1)b$ and $d_2(c)=(-(y-1)c,(x-1)c)$. To check exactness at $R^2$, reduce $d_1(a,b)=0$ modulo $x-1$. Since $y-1$ is not a zero divisor in $\mathbb{Z}[y^{\pm 1}]$, write $b=(x-1)c$, and cancellation gives $a=-(y-1)c$. Injectivity of $d_2$ follows because $R$ is a domain. The augmentation kernel is generated by $x-1,y-1$, proving the other exactness statement.

Applying ${\mathrm{Hom}}_R(-,\mathbb{Z})$ makes both differentials zero, since $x,y$ act trivially. ThusDegree-one classes assign arbitrary integers to the two generators. Geometrically these are also the cohomology groups of the torus, a classifying space for ${\mathbb{Z}}^2$; its cohomology ring is the exterior algebra on the two degree-one classes.

The integer $n$ in degree two has an explicit extension representative. On triples $(a,b,m)\in {\mathbb{Z}}^3$, setThe bilinear term is a two-cocycle. With generators $x=(1,0,0)$, $y=(0,1,0)$ and central $z=(0,0,1)$,Every central extension of ${\mathbb{Z}}^2$ by $\mathbb{Z}$ has this form: choose lifts of the two generators and the fixed generator of the kernel of a group homomorphism, then their group commutator determines $n$ and every element has a unique form $z^m{y}^b{x}^a$. Changing the lifts by central elements leaves that commutator unchanged. Thus distinct integers give inequivalent extensions, and the cocycles $nad$ add with $n$. The case $n=0$ is ${\mathbb{Z}}^3$; $n=1$ is the integer Heisenberg group. This realizes the integral central extensions of a rank-two free abelian group concretely.

Let $M$ be a closed manifold of even dimension $n=2m$, with a Riemannian metric and a spin structure. Let $S=S^{+}\oplus S^{-}$ be its spinor bundle, and let $W$ be a Hermitian vector bundle with a unitary connection of vector-bundle curvature $F^W=({\nabla }^W{)}^2$. We use Clifford multiplication satisfying $c(v)c(w)+c(w)c(v)=-2⟨v,w⟩$, so the Twisted Dirac operatoris self-adjoint and $D_W^{-}=(D_W^{+}{)}^{\ast }$. Here $(e_i)$ is an oriented local orthonormal basis and $(e^i)$ its dual. The Atiyah-Singer index theorem in this setting isThe second factor is the Chern character; the first is the A-hat formThe square root has constant term one. All functions of the vector-bundle curvature matrices mean power series truncated above the dimension of $M$. The brackets select the top-degree differential form. The Bianchi identity shows that these forms are closed, and Chern-Weil connection transgression shows that their de Rham cohomology classes do not depend on the chosen connections on a vector bundle.

The global step is supersymmetric quantum mechanics. On the Hilbert space $L^2(S\otimes W)$, take the grading operator $\Gamma$, the odd supercharge $D_W$, and the nonnegative Hamiltonian $H=D_W^2$. They obey $\Gamma D_W=-D_W\Gamma$ and $D_W^2=H$. The principal symbol of $D_W$ is invertible away from the zero covector, so elliptic regularity and Rellich-Kondrachov compactness theorem give a smooth discrete orthonormal eigenbasis for $H$, finite-dimensional eigenspaces, and a trace-class operator $e^{-tH}$ for $t>0$. Define its supertrace by $\mathrm{Str}A=\mathrm{Tr}(\Gamma A)$.

For every positive eigenvalue $\lambda$ of $H$, the map ${\lambda }^{-1/2}{D}_W$ is an isomorphism between its positive and negative grading eigenspaces, with inverse itself. Their contributions to the supertrace cancel. The zero eigenspaces are $\ker D_W^{+}$ and $\ker D_W^{-}$. ConsequentlyThis also proves independence of $t$ without differentiating an unbounded operator inside a trace. If $K_t$ is the heat kernel, the heat kernel trace formula givesThe remaining task is to compute this local integrand as $t\downarrow 0$.

Fix $p\in M$, use geodesic normal coordinates $x$, and trivialize $S\otimes W$ by radial parallel transport. Expanding $D_W^2$ at $p$ gives the Lichnerowicz formula for a twisted Dirac operator:Indeed the terms with identical indices give the rough Laplacian. The other terms are ${\sum }_{i<j}c(e^i)c(e^j)[{\nabla }_i,{\nabla }_j]$. Decompose this vector-bundle curvature into spin curvature and $F^W$; substituting the spin curvature into the Clifford algebra relations and the algebraic Bianchi identity leaves $\mathrm{Scal}/4$. Thus the calculation retains the curvature of the twisting bundle explicitly.

There is an especially strong cancellation in the spinor supertrace. With $\Gamma =i^{m}c(e^1)\cdots c(e^{2m})$, every ordered product of distinct Clifford generators of degree less than $2m$ has zero supertrace, whileFor an even product of lower degree, conjugate $\Gamma$ times that product by a missing Clifford generator: it changes sign, so its matrix trace is zero. An odd product exchanges the grading summands and also has zero supertrace. For the full product, $(c(e^1)\cdots c(e^{2m}){)}^2=(-1{)}^m$, and $\dim S=2^m$, giving the displayed constant.

To exploit this, identify the Clifford algebra as a filtered vector space with the exterior algebra by sending an ordered Clifford product to the corresponding wedge product of differential forms. Implement Getzler rescaling: put $x=\epsilon y$, $t={\epsilon }^{2}s$, multiply the degree-$k$ exterior component of the kernel by ${\epsilon }^{-k}$, and multiply the whole kernel by ${\epsilon }^{2m}$. A derivative and a Clifford generator have rescaling degree one; a coordinate has degree minus one. Clifford contractions lower the degree by two relative to the wedge product of differential forms, so the leading product becomes exterior multiplication.

In the radially parallel frame, the linear Taylor term of the connection is determined by its vector-bundle curvature: if $A_i$ is its matrix, then $A_i(x)=\frac{1}{2}{x}^j{F}_{ji}(p)+O(|x{|}^2)$. To check the coefficient, radial gauge means $x^i{A}_i=0$, and integration of $x^j{F}_{ji}$ along the radial segment gives $A_i(x)={\int }_0^{1}s{x}^j{F}_{ji}(sx)ds$. Combining this formula with the spin representation, the symmetries of Riemann curvature tensor, and the Lichnerowicz formula for a twisted Dirac operator, the rescaled square has leading modelHere $\mathcal{R}$ is the skew matrix of tangent-bundle curvature two-forms at $p$; changing the curvature-matrix sign convention changes the sign inside the covariant derivatives but not the determinant below. The coefficients act in $\Lambda T_p^{\ast }M\otimes \mathrm{End}(W_p)$. The scalar-curvature term has degree zero and disappears after multiplication by ${\epsilon }^2$; the linear spin-connection term and the Clifford action of $F^W$ have degree two and survive. Higher Taylor terms have smaller degree and disappear.

For clarity, the analytic meaning of this limiting calculation is the following. Construct the local heat kernel expansion to any fixed order and insert the Taylor expansion of the rescaled coefficients into its transport equations. At each exterior degree, the leading equation is the heat equation of ${\mathcal{H}}_p$; the remaining terms acquire positive powers of $\epsilon$. Equivalently, successive substitutions ingive the same expansion. Polynomial spatial factors are integrable against the local Gaussian kernels, and only finitely many exterior degrees exist. A cutoff outside a fixed normal neighbourhood contributes $O(e^{-c/t})$. Choosing the initial expansion order arbitrarily large bounds its differentiated remainder to any required order. This proves convergence of the rescaled diagonal kernel to the model diagonal kernel, uniformly in $p$ on the compact manifold. Its degree-$2m$ component has no net power of $\epsilon$, precisely the component detected by the spinor supertrace.

We can compute the model kernel rather than invoke a local index formula. Set $F=F_p^W$ and seek its kernel from the origin in the formThe scalar even exterior coefficients of $\mathcal{R}$ commute. On this ansatz the rotational first-order term vanishes because $Q$ is a function of ${\mathcal{R}}^2$. Expanding the model operator gives $-\sum {\partial }_i^2+\frac{1}{2}(\mathcal{R}y)\cdot \nabla +\frac{1}{16}{y}^T{\mathcal{R}}^{2}y+F$, so substitution into the heat equation givesThe Euclidean initial condition $Q(s)\sim s^{-1}I$, $a(s)\sim (4\pi s{)}^{-m}$ selectsThese expressions have well-defined power series at $\mathcal{R}=0$. This is the curvature version of the Gaussian calculation underlying the Mehler formula for Hermite functions.