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Fix the current sign convention by defining the localized variation as ${\delta }_{ϵ}S=-\int d^{d}x({\partial }_{\mu }{ϵ}^a)j_a^{\mu }$. For a first-derivative Lagrangian invariant without a boundary term under constant parameters, this means $j_a^{\mu }=-\partial \mathcal{L}/\partial ({\partial }_{\mu }\varphi )t_a\varphi$; include the usual improvement term if the constant variation is a total derivative. On solutions, arbitrary compactly supported parameters imply ${\partial }_{\mu }{j}_a^{\mu }=0$. The Noether charge is $Q_a=\int d^{d-1}x{j}_a^0$ and is conserved if the spatial flux vanishes. This sign convention matches the printed Ward identity; reversing the current also reverses the corresponding generator convention.

Assume an invariant regulated functional measure, invariant vacuum boundary conditions and no quantum anomaly. Changing variables in the normalized path integral gives $0=⟨{\delta }_{ϵ}X⟩+i⟨X{\delta }_{ϵ}S⟩$. Integration by parts then yieldsFor a product of scalar fields, the local Ward identity contact terms areAway from the insertions this is current conservation. Time-ordering or the distributional functional identity supplies the contact terms.

For the gauge theory write $[X,Y{]}^a=f^{abc}{X}^b{Y}^c$ and use the left-acting BRST differential $s$, with ${\delta }_{ϵ}=ϵs$. It obeys the graded Leibniz rule andAssume the structure constants satisfy the Jacobi identity and the dot product is invariant; antisymmetry alone would not be enough. Since $c$ and $D_{\mu }c$ are odd, their Lie brackets are symmetric in these two arguments. ConsequentlyAlso $s{F}_{\mu \nu }=D_{\mu }{D}_{\nu }c-D_{\nu }{D}_{\mu }c=[F_{\mu \nu },c]$. The graded Jacobi identity gives $s[c,c]=0$, so $s^{2}c=0$; the other three fields have zero second variation immediately. For independent odd parameters, ${\delta }_{{ϵ}^{\prime }}{\delta }_{ϵ}=-{ϵ}^{\prime }ϵs^2=0$. These are off-shell BRST nilpotence identities because the Nakanishi-Lautrup field $b$ is retained.

The Yang-Mills theory variation is proportional to $F^{\mu \nu }\cdot [F_{\mu \nu },c]=0$. The gauge-fixing variation is $({\partial }^{\mu }b)\cdot D_{\mu }c$, while the ghost variation is its negative; the $b^2$ term does not vary. Equivalently these terms are $s\Psi$ for the gauge-fixing fermion $\Psi =({\partial }^{\mu }\bar{c})\cdot A_{\mu }+\xi \bar{c}\cdot b/2$. BRST nilpotence makes this expression invariant. Ghost-number scaling also leaves every term invariant: the ghost and antighost factors carry opposite weights.

Here are the two explicit currents in the same sign convention as the Ward identity. Localizing the even ghost parameter gives coefficient $({\partial }_{\mu }\theta )[\bar{c}\cdot D^{\mu }c-({\partial }^{\mu }\bar{c})\cdot c]$. Localizing the odd parameter, keeping it on the left, gives coefficientThe last sign comes from moving the odd parameter through the odd antighost derivative. Since the current was defined as minus this coefficient,These are the ghost-number Noether current and the BRST current in derivative-b gauge fixing. If the opposite Noether sign is used, both displayed currents acquire an overall minus sign. Integrating the gauge-fixing term by parts changes the Noether representative by the associated boundary improvement; mixing the two Lagrangian conventions without that improvement gives incorrect signs.

With no BRST anomaly, the conserved odd BRST charge has $Q_B^2=0$. The BRST cohomology identifies closed states $Q_B|\psi ⟩=0$ modulo exact states $Q_B|\chi ⟩$. Nilpotence puts every exact state in the closed space. In the usual indefinite gauge-fixed state space, a Hermitian BRST charge makes exact states orthogonal to closed states; the standard no-ghost/positivity assumptions then give a physical inner product on the quotient. The physical sector is its ghost-number-zero component,The ghost number assigns $+1$ to $c$, $-1$ to $\bar{c}$ and zero to gauge and auxiliary fields. Gauge-invariant observables and the chosen vacuum have ghost number zero; unphysical ghost excitations are removed in BRST pairs. Thus physical representatives are expected to satisfy $Q_G|\psi ⟩=0$. This zero-grading selection is part of the physical-state prescription, not a consequence of nilpotence alone.

For a Young tableau $t$, let $R_t$ permute the entries within its rows and $C_t$ within its columns. We use the Young symmetrizer conventionReversing the order gives another usual realization of the same irreducible polynomial module. On the tensor power $T=V^{\otimes n}$, the actions areThey commute because applying $g$ to every factor commutes with permuting the factors.

We state the permitted combinatorial input explicitly. A Young symmetrizer satisfies $h_t^2=H_{\lambda }{h}_t$, where $H_{\lambda }$ is the nonzero hook product of a partition. Thus $e_t=H_{\lambda }^{-1}{h}_t$ is a primitive idempotent. The standard-tableau decomposition of the right regular module is $\mathbb{C}{S}_n={⨁}_{\lambda ,t\mathrm{standard}}{e}_t\mathbb{C}{S}_n$. Tensor this right-module direct sum with the left module $T$. The map $e_t\mathbb{C}{S}_n{\otimes }_{\mathbb{C}{S}_n}T\to e_{t}T$ sending $e_{t}a\otimes v$ to $e_{t}av$ is an isomorphism, with inverse $w\mapsto e_t\otimes w$. Therefore the Young-symmetrizer tensor decomposition isThis is a direct sum of $GL(V)$-modules; individual summands need not be $S_n$-invariant.

We also state the allowed Schur algebra result, namely Schur–Weyl duality: the two actions are mutual commutants, andwhere the $D_{\lambda }(V)$ are pairwise nonisomorphic irreducible homogeneous polynomial representations of degree $n$. We also use the standard Schur algebra equivalence between its modules and homogeneous degree-$n$ polynomial representations, so these exhaust the irreducibles in that category. A primitive idempotent $e_t$ has one-dimensional image on $S^{\lambda }$ and zero image on the other simple factors, so $h_{t}T\cong D_{\lambda }(V)$. This identifies the requested irreducibles. They classify the polynomial degree-$n$ representations in this tensor power, not all rational representations of every degree.

The length bound for a Schur module is $D_{\lambda }(V)\ne 0$ if and only if $\ell (\lambda )\le m$. A column of length greater than $m$ antisymmetrizes more than $m$ vectors and gives zero. Conversely, for $\ell (\lambda )\le m$, fill every tensor position in row $i$ with the $i$th basis vector of $V$. Row symmetrization multiplies it by ${\prod }_i{\lambda }_i!$, and column antisymmetrization is nonzero because the vectors within each column are distinct basis vectors.

A rational representation of an algebraic group is a regular morphism into the general linear group of its representation space. For $G{L}_m$, its matrix entries belong to $\mathbb{C}[g_{ij},\det (g{)}^{-1}]$; rational here permits determinant denominators but not arbitrary poles on $G{L}_m$. A one-dimensional rational character of ${\mathbb{C}}^{\times }$ is a Laurent polynomial $p(z)$ with $p(zw)=p(z)p(w)$ and $p(1)=1$. Comparing Laurent coefficients shows that just one monomial occurs and its coefficient is $1$, hence $p(z)=z^r$ for $r\in \mathbb{Z}$.

Restrict a one-dimensional rational character of $G{L}_m$ to its diagonal torus. The same argument in several variables gives ${\prod }_i{x}_i^{r_i}$. Conjugation by permutation matrices makes all $r_i$ equal. On every diagonalizable invertible matrix it consequently agrees with $(\det g{)}^r$. The allowed Zariski-density statement, and equality of regular functions on a dense subset, giveThese are the one-dimensional rational characters of the general linear group.

For complete reducibility of rational GL and SL representations, use the compact-group averaging argument. Average any positive definite Hermitian inner product over $SU(m)$ using normalized Haar measure. The orthogonal complement of an invariant subspace is then $SU(m)$-invariant. Differentiating makes it invariant under $\mathfrak{su}(m)$ and therefore under its complex span ${\mathfrak{sl}}_m(\mathbb{C})$. The elementary unipotent matrices $\exp (t{E}_{ij})$ generate $S{L}_m(\mathbb{C})$, so the complement is $S{L}_m$-invariant. This proves complete reducibility. Averaging over $U(m)$ similarly gives complete reducibility for rational $G{L}_m$ representations.

Here is an explicit rational extension from SL to GL. Decompose the representation space by the finite scalar center of $S{L}_m$ into subspaces $W_k$ on which $\zeta I$ acts as ${\zeta }^k$, with $0\le k<m$ and ${\zeta }^m=1$. These subspaces are invariant. For $g\in G{L}_m$, choose $a$ with $a^m=\det g$, set $s=a^{-1}g\in S{L}_m$, and defineChanging $a$ to $\zeta a$ changes $s$ to ${\zeta }^{-1}s$, so the two factors cancel. It is a homomorphism, since scalar roots multiply up to the same harmless factor, and it restricts to $\rho$ on $S{L}_m$.

It is rational as well. Every matrix coefficient of $\rho {|}_{W_k}$ has a polynomial representative on $S{L}_m$. Averaging that representative over the finite scalar center selects its homogeneous parts $P_d$ with $d\equiv k(\mathrm{mod}m)$. Substitution in the extension gives ${\sum }_d(\det g{)}^{(k-d)/m}{P}_d(g)$, a regular function on $G{L}_m$. Each $S{L}_m$-invariant subspace decomposes into its $W_k$ parts, and the extension acts on each part by a scalar times an $S{L}_m$ action. Thus these subspaces are also invariant under the chosen extension. Consequently $\rho$ is irreducible if and only if this ${\rho }^{\prime }$ is irreducible. For $m=1$, $S{L}_1$ is trivial and the trivial extension supplies the same conclusion.

Use the weighted Hilbert structure of velocity-reset relaxation, with inner product $⟨f,g⟩=\int f\overline{g}/M$ and norm $\Vert f{\Vert }_H$. The Cauchy-Schwarz inequality givesThus $\rho (f)$ is an absolutely convergent integral and a bounded linear functional. Since $\Vert M{\Vert }_H^2=\int M=1$, the normalized velocity-reset collision operator obeysThis establishes boundedness. The subsequent orthogonal decomposition improves the operator norm to exactly one.

The weighted inner product satisfies $⟨f,M⟩=\int f=\rho (f)$. ThereforeThe identity operator and $L$ are linear, bounded and self-adjoint, hence so is $\Pi$. Directly, $\Vert \Pi f{\Vert }_H=|\rho (f)|\le \Vert f{\Vert }_H$; equality holds at $f=M$. Thus $\Vert \Pi \Vert =1$.

Interpret $L$ as the velocity operator appearing in the equation and in the stated quadratic-form hypothesis. Let $E(t)=\int \int |f(t,x,v){|}^{2}dxdv$. Smoothness turns the distributional equation into the usual equation. Multiply it by $f$ and integrate; for complex functions use the real part of the inner product. Spatial integration by parts gives $\int v\cdot {\nabla }_x(f^2)=0$, while the coercivity hypothesis at each $x$ givesThus $e^{2\delta t}E(t)$ is nonincreasing. The coercive energy estimate for kinetic transport isNo division by $E$ is required, so the zero solution causes no exception. Compact spatial support, or sufficient decay to eliminate the boundary flux, is enough; compact support in time is unnecessary.

Take $A>0$ and use primes for spatial derivatives. Two integration by parts operations in the first variation of the elastic filament's bending energy giveThus the Euler-Lagrange equation is $A{h}^{\prime \prime \prime \prime }=0$, and the fluctuation differential operator is $K=A{d}^4/d{x}^4$. In the L2 space inner product, its boundary form isThe conjugate endpoint trace pairs are $(h,h^{\prime \prime \prime })$ and $(h^{\prime },h^{\prime \prime })$. Requiring one member of each pair to vanish gives the four standard self-adjoint endpoint conditions for filament bending, applied at both ends:

Each pair annihilates the boundary form for all $u,v$ in the domain. Conversely, the remaining two endpoint traces can be chosen freely: requiring the boundary form to vanish against every such $u$ forces an adjoint-domain function $v$ to satisfy the same two conditions. This proves self-adjointness, rather than just formal symmetry, on the corresponding fourth-order Sobolev space domain.

The count four concerns these elementary homogeneous choices. Identical-end boundary conditions do not restrict all self-adjoint operators to these four possibilities. For example, $h=0$, $h^{\prime \prime }=b{h}^{\prime }$ at both ends, with any fixed real $b\ne 0$, also annihilates the boundary form: the remaining expression is $-\overline{u^{\prime }}b{v}^{\prime }+b\overline{u^{\prime }}{v}^{\prime }=0$. This Robin boundary condition supplies a continuous family beyond the four listed pairs.

Use normalized Fourier coefficients and the inner productThe functions $e_n(t)=e^{int}$ are orthonormal. Hence the Fourier partial sum $S_{N}f={\sum }_{|n|\le N}\hat{f}(n)e_n$ is the orthogonal projection onto the trigonometric polynomials of degree at most $N$. For any such polynomial $P$, orthogonality givesso $\Vert f-S_{N}f{\Vert }_2\le \Vert f-P{\Vert }_2$.

Given $\epsilon >0$, the permitted density result supplies a trigonometric polynomial $P$ with $\Vert f-P{\Vert }_{\infty }<\epsilon$. Once $N$ includes its degree,Therefore the Fourier partial sums converge to $f$ in the normalized $L^2$ norm, giving exactly the stated mean-square limit.

A four-vector is a collection of four components $A^{\mu }=(A^0,\mathbf{A})$ that transforms by the same Lorentz transformation as $(ct,\mathbf{x})$. Use the metric signature $(-,+,+,+)$. The Lorentzian inner product isA nonzero four-vector is a timelike vector, null vector or spacelike vector according as $A\cdot A$ is negative, zero or positive. Reversing the metric signature reverses the signs used to name these three classes, without changing their geometric meaning.

For a frame moving at speed $v$ along the positive $x$ axis, write $\beta =v/c$ and use the Lorentz factor $\gamma =(1-{\beta }^2{)}^{-1/2}$. The component Lorentz transformation isExpanding the first two squares givesThe other two components are unchanged, so $A^{\prime }\cdot A^{\prime }=A\cdot A$. Consequently a timelike vector remains timelike under a Lorentz transformation.

For a nonzero null vector, $|\mathbf{A}|=|A^0|$ and $A^0\ne 0$. Set $a=A^0$ and $\hat{\mathbf{n}}=\mathbf{A}/a$. Then $|\hat{\mathbf{n}}|=1$ and $A=a(1,\hat{\mathbf{n}})$. If the zero four-vector is included among null vectors, take $a=0$ and any unit vector.

For two future-pointing null vectors, write $A=a(1,\hat{\mathbf{n}})$ and $B=b(1,\hat{\mathbf{m}})$ with $a,b>0$. The sum of future-pointing null vectors satisfiesbecause the ordinary inner product of two unit vectors is at most one. The sum is null if their spatial directions coincide, and timelike otherwise. Its positive time component makes the sum nonzero and future-pointing.

Linearity of the integral in each argument gives a bilinear form, and $fg=gf$ gives symmetry. Also $(f,f)={\int }_{-1}^{1}f(x{)}^{2}dx\ge 0$. If a polynomial $f$ is nonzero somewhere on this interval, continuity makes $f^2$ strictly positive on a smaller interval, so the integral is positive. A polynomial vanishing on the whole interval is the zero polynomial. Thus the form is positive-definite and is an inner product.

Apply the Gram-Schmidt process to $1,x,x^2$. Odd functions have zero integral, so $x$ is already orthogonal to $1$, and the third orthogonal polynomial is $x^2-1/3$. Their squared norms areTherefore an orthonormal basis isThese are the first three normalized Legendre polynomials. Their distinct degrees also prove that they span the three-dimensional vector space.

A fiber metric on a real vector bundle $E\to M$ is a smoothly varying positive-definite inner product on each fiber. Choose a trivializing open cover $\{U_i\}$ and a smooth partition of unity $\{{\rho }_i\}$ subordinate to it. The partition theorem gives nonnegative functions summing to one, a locally finite family of subsets of supports, and $\mathrm{supp}{\rho }_i\subset U_i$; the usual Hausdorff second-countable smooth manifold hypotheses ensure this theorem applies. Transfer the Euclidean inner product to each local vector bundle trivialization, obtaining $h_i$, and setEach weighted term extends smoothly by zero outside $U_i$, and local finiteness makes the sum smooth in every vector bundle trivialization. At each $x$ some weight is positive, so $h_x(v,v)>0$ for every nonzero $v$. This proves existence of a fiber metric, with no orientability or triviality assumption.

A vector bundle morphism covering the identity is a smooth map $F:E^{\prime }\to E^{\prime \prime }$ that preserves base points and is a linear map on each fiber. Its induced map on the module of smooth sections is $s\mapsto F\circ s$, and is $C^{\infty }(M)$-linear. We prove the converse by constructing bundle morphisms from maps of smooth sections.

First the given map $\alpha$ is local. If a global section $s$ vanishes on a neighborhood of $x$, take a smooth bump function $\chi$ supported there with $\chi =1$ near $x$. Then $\chi s=0$, so $\chi \alpha (s)=\alpha (\chi s)=0$, and hence $\alpha (s)(x)=0$. Thus sections agreeing near $x$ have images agreeing at $x$.

Choose a local frame $e_1,\dots ,e_r$ on $U$, and a bump function equal to one on a smaller neighborhood $V$ of $x$ and supported in $U$. Multiplying the frame by that bump and extending by zero gives global smooth sections $t_j$ whose restrictions to $V$ are the frame. If $s(x)=0$, write $s=\sum a_j{e}_j$ on $V$. A second bump extends each $a_j$ to a global smooth function ${\tilde{a}}_j$ agreeing near $x$. By locality and $C^{\infty }(M)$-linearity,Every fiber vector $v\in E_x^{\prime }$ is the value of a global smooth section, by the same bumped-frame construction. Define $F_x(v)=\alpha (s)(x)$ for any such section. The just-proved vanishing statement makes this well-defined. The maps $F_x$ are linear maps, and locally their matrix columns are the smooth sections $\alpha (t_j)$ in a frame of $E^{\prime \prime }$. Thus $F$ is smooth, is a vector bundle morphism, and satisfies $\alpha (s)=F\circ s$. Fiberwise evaluation also proves uniqueness.

Finally apply the fiber metric construction to the tangent bundle. A Riemannian metric $g$ defines the musical isomorphismPositive definiteness makes it a fiberwise bijection; its inverse is smooth because inverse metric matrices vary smoothly. Hence $TM$ and $T^{\ast }M$ are isomorphic as real smooth vector bundles on every such manifold. The isomorphism depends on the chosen metric and is not canonical.

Use the given orientation throughout, and the usual convention that the manifold has no boundary. In an oriented manifold chart the metric volume form isIf $y$ is another oriented chart and $A=\partial x/\partial y$, then $g_y=A^t{g}_{x}A$, so $\sqrt{\det g_y}=(\det A)\sqrt{\det g_x}$ because $\det A>0$. This is exactly the transformation of the top exterior product. The formulas therefore agree on overlaps and define a smooth positive volume form. Equivalently it takes value one on any positively oriented orthonormal tangent frame.

The metric induces an inner product on $p$-forms by making the increasing exterior products of an orthonormal coframe orthonormal. The Hodge star operator is the unique pointwise linear map $\ast :{\Lambda }^p{T}^{\ast }M\to {\Lambda }^{n-p}{T}^{\ast }M$ satisfyingFor an increasing multi-index $I$, $\ast e^I$ is the complementary wedge with the sign making $e^I\wedge \ast e^I={\omega }_g$. Swapping the blocks of $p$ and $n-p$ factors introduces $(-1{)}^{p(n-p)}$. ThereforeIn particular $\ast 1={\omega }_g$ and $\ast {\omega }_g=1$.

For compactly supported smooth $f$, the $(n-1)$-form $f\ast \alpha$ has compact support. Stokes theorem and the graded Leibniz rule giveA top form $\tau$ equals $(\ast \tau ){\omega }_g$. Consequently this Hodge integration by parts for one-forms becomes exactlyCompact support of $f$ is enough; $\alpha$ need not itself have compact support.

Define the codifferential on $p$-forms by $\delta =(-1{)}^{n(p+1)+1}\ast d\ast$ and the Hodge Laplacian by $\Delta =d\delta +\delta d$. A harmonic differential form is a smooth form in $\ker \Delta$. On functions this is the positive Laplace-Beltrami operator, $\Delta f=\delta df=-{\mathrm{div}}_g{\mathrm{grad}}_{g}f$. This sign convention is required by the product identity; it is the negative of the $\mathrm{div}\mathrm{grad}$ convention also commonly used for the Laplace-Beltrami operator.

For a $p$-form $\beta$, the square of the Hodge star operator and the codifferential formula giveApplying the second formula to $d\beta$ and the first to $\delta \beta$ givesThus Hodge star commutes with the Hodge Laplacian. Since $\ast$ is invertible, $\beta$ is harmonic if and only if $\ast \beta$ is harmonic. This holds without compactness; we have not used the generally false noncompact implication that a harmonic form must be closed and coclosed.

For a smooth function $h$ and one-form $\alpha$, the same graded Leibniz rule gives $\delta (h\alpha )=h\delta \alpha -\ast (dh\wedge \ast \alpha )=h\delta \alpha -⟨dh,\alpha {⟩}_g$. Apply this to $d(f_1{f}_2)=f_{2}d{f}_1+f_{1}d{f}_2$ to obtain the product rule for the positive Laplace-Beltrami operator

The Hodge decomposition theorem states that on a compact oriented boundaryless Riemannian manifold, smooth forms have the $L^2$-orthogonal decompositionand each de Rham cohomology class has a unique harmonic differential form representative. To spell out the last conclusion, a harmonic form is closed and coclosed because $⟨\Delta \eta ,\eta ⟩=\Vert d\eta {\Vert }^2+\Vert \delta \eta {\Vert }^2$. If a closed form decomposes as $h+du+\delta v$, then $d\delta v=0$; integration by parts gives $\Vert \delta v{\Vert }^2=⟨v,d\delta v⟩=0$. Thus it represents $h$. A harmonic exact form has zero norm by adjointness, proving uniqueness. The analytic existence of the decomposition is the stated Hodge decomposition theorem.

Now let $M$ be compact, connected and oriented. A harmonic function satisfies $0=⟨f,\Delta f⟩=\Vert df{\Vert }^2$, so it is constant. The Hodge star operator identifies ${\mathcal{H}}^0(M)$ with ${\mathcal{H}}^n(M)$, hence ${\mathcal{H}}^n(M)=\mathbb{R}{\omega }_g$. A Riemannian metric exists by Question 3 even if none was initially chosen. Using the harmonic representative of each class, we obtain the top de Rham cohomology of a compact connected oriented manifoldThe class is nonzero also directly from Stokes theorem, since ${\int }_M{\omega }_g>0$ while every exact top form has zero integral. Boundarylessness matters: a compact interval has zero first de Rham cohomology.

Write $L\alpha =\omega \wedge \alpha$, the Lefschetz operator of a Kähler manifold. The metric and volume form define the $L^2$ inner product on smooth complex forms, and $\Lambda =L^{\ast }$ is its formal adjoint. Similarly, ${\bar{\partial }}^{\ast }$ is the formal adjoint of the Dolbeault operator, characterized by $⟨\bar{\partial }\beta ,\alpha ⟩=⟨\beta ,{\bar{\partial }}^{\ast }\alpha ⟩$. The Dolbeault Laplacian isOn the compact manifold without boundary, integration by parts givesIf the Laplacian vanishes, both terms are zero. Conversely, if both operators annihilate $\alpha$, the defining formula annihilates it. Thus harmonicity is equivalent to being both $\bar{\partial }$-closed and ${\bar{\partial }}^{\ast }$-closed.

Because $\omega$ is closed and has type $(1,1)$, $\bar{\partial }\omega =0$ and $[L,\bar{\partial }]=0$. The supplied identity from the Kähler identities gives $[L,{\bar{\partial }}^{\ast }]=-i\partial$. With ordinary commutators for the even-degree operator $L$,The last identity follows from $d^2=0$. This also proves that the Lefschetz operator preserves harmonic forms.

For the cohomology map one can work directly with forms: $\bar{\partial }({\omega }^k\wedge \alpha )={\omega }^k\wedge \bar{\partial }\alpha$, since $\omega$ has even degree. It takes closed forms to closed forms and exact forms to exact forms. Therefore the $k$th power of $L$ inducesThe bidegree is $(p+k,q+k)$, including the zero groups outside the dimension range. No isomorphism claim is needed here.

Finally put $\theta =-i\partial f$, so $\bar{\partial }\theta =i\partial \bar{\partial }f={\omega }^{\prime }-\omega$. LetBoth Kähler forms are $\bar{\partial }$-closed, so $\bar{\partial }{T}_k=0$. For a closed representative $\alpha$,This primitive has type $(p+k,q+k-1)$; the degree-one sign produces no additional term because $T_k\wedge \alpha$ is closed. HenceThis is the dependence of Lefschetz maps on the Dolbeault class.

Multiplying the matrices gives the Heisenberg group lawDifferentiating at the identity gives all strictly upper-triangular matrices. With $X=E_{12}$, $Y=E_{23}$ and $Z=E_{13}$, the commutator givesThis is the Heisenberg Lie algebra.

The right Maurer-Cartan form isThus the right-invariant coframe of the real Heisenberg group is ${\sigma }^1=dx$, ${\sigma }^2=dy$, ${\sigma }^3=dz-ydx$. Under a right translation,we have $d{x}^{\prime }=dx$, $d{y}^{\prime }=dy$, and $d{z}^{\prime }-y^{\prime }d{x}^{\prime }=dz-ydx$. All three right-invariant differential forms are preserved.

Every right-invariant Riemannian metric is determined by an arbitrary inner product at the identity. In this coframe its most general expression isEquivalently,Products here are symmetric products. If a pseudo-Riemannian metric tensor is intended, replace positive definiteness by nondegeneracy. Since every right translation preserves the coframe, it is an isometry. The action is faithful, and $a\mapsto R_{a^{-1}}$ is a group homomorphism, embedding a copy of $G$ into the isometry group.

The one-parameter right translations produce the Killing frame for a right-invariant Heisenberg metric:Their flows are respectively $(x+t,y,z)$, $(x,y+t,z+xt)$ and $(x,y,z+t)$. Each preserves $h$, and their Lie brackets of vector fields are $[K_X,K_Y]=K_Z$, with the other two zero. This explicitly realizes the Heisenberg Lie algebra. These Killing vector fields are left-invariant vector fields; the vector fields dual to the right-invariant coframe are instead ${\partial }_x+y{\partial }_z,{\partial }_y,{\partial }_z$, and should not be substituted for these generators.

For the diagonal case, putThe Kaluza-Klein decomposition along a Killing field uses $V=h(K_X,K_X)=\alpha +\chi y^2$. Completing the square in $dx$ gives the Heisenberg metric Kaluza-Klein reduction:These depend only on the quotient coordinates $(y,z)$ and satisfy $h=V(dx+A{)}^2+\gamma$. For a diagonal indefinite metric tensor, the same expressions hold wherever $V\ne 0$; a null Killing vector field cannot be treated with this completed-square decomposition.

A separable Hilbert space has a countable subset dense in its norm topology.

The printed assertion about all locally square-integrable functions is false. The proposed average is not even finite for every such function: for $f(x)=x$,It also fails positive definiteness. The nonzero function $f=1_{[0,1]}$ hasConsequently this formula cannot define an inner product, much less a Hilbert space, on $L_{\mathrm{loc}}^2(\mathbb{R})$.

A precise version of the intended nonseparability argument uses the mean-square completion of trigonometric polynomials. Start with the real vector space $V$ of finite linear combinations of $1$, $\cos (\lambda x)$, and $\sin (\lambda x)$, with arbitrary $\lambda >0$. Product-to-sum identities show that all the proposed cross averages exist. Distinct frequencies are orthogonal, each sine and cosine has squared norm one, and the constant function has squared norm two. Thus, after collecting equal frequencies,This is positive definite on $V$. Its Hilbert space completion $H$ contains the uncountable orthonormal set $\{\cos (\lambda x):\lambda >0\}$. The distance between two distinct members is $\sqrt{2}$. Their open balls of radius $1/2$ are pairwise disjoint, and a dense subset must meet each one. A countable dense subset is therefore impossible: this corrected completed space is nonseparable. Completion is an essential additional construction; it does not validate the printed claim about all of $L_{\mathrm{loc}}^2$.

The Riesz representation theorem states that every bounded linear functional $L$ on a real or complex Hilbert space $H$ is represented by a unique $h\in H$:For the complex case take the inner product to be linear in its first argument.

If $L=0$, choose $h=0$. Otherwise its kernel $K$ is a closed linear subspace. Choose $x$ with $L(x)\ne 0$ and let $z=x-P_{K}x$, using the orthogonal projection. Then $z\ne 0$, $z\perp K$, and $L(z)=L(x)\ne 0$. For every $v$,Therefore take $h=L(z)z/\Vert z{\Vert }^2$ in the real case, and $h=\overline{L(z)}z/\Vert z{\Vert }^2$ in the complex case. The conjugate in the latter formula compensates for conjugate linearity in the second argument.

The Cauchy-Schwarz inequality gives $|L(v)|\le \Vert v\Vert \Vert h\Vert$, and evaluation at $v=h/\Vert h\Vert$ when $h\ne 0$ gives equality of the norms. If two vectors represent $L$, their difference is orthogonal to every vector, including itself, hence zero. This proves all assertions of the Riesz representation theorem.

Work in the mean-zero Sobolev spaceThe integral is a continuous functional on $H^1(U)$, so $H$ is closed. The Neumann-Poincare inequality shows that $\Vert v{\Vert }_a=\Vert \nabla v{\Vert }_2$ is equivalent to the usual $H^1$ norm on $H$, making $a$ a complete inner product there.

For $f\in L^2(U)$ the functional $L(v)={\int }_{U}fv$ satisfiesThe Riesz representation theorem, or the Lax-Milgram theorem, gives a unique $u\in H$ with $a(u,v)=L(v)$ for all $v\in H$. To recover every $H^1(U)$ test, write $v=(v-v_U)+v_U$. The constant contributes zero to $a$ and contributes $v_U{\int }_{U}f=0$ to $L$. Thus the same equality holds for all $v\in H^1(U)$.