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There is an indexing issue in the printed topology. Its cumulative sums omit $a_0$, whereas the post-slot workload of a queue in part (a) includes $a_0$. To see the problem, take all $a_t=\lambda$ for $t\le -1$, with $\lambda <c$, and compare two inputs with $a_0=\lambda$ and $a_0=c+1$. Their distance according to the printed cumulative-sum expression is zero, yetThus the post-slot workload is not continuous in the literal printed topology. On two-sided sequences that expression is a seminorm, since it also ignores future coordinates; the fixed-mean input class is not a vector space (if signed inputs are allowed, it is an affine set).

The appropriate repair for the post-slot convention is to use one-sided past inputs including time zero and the weighted cumulative-input topology for a slotted queue:This is a finite metric on the fixed-mean class, since cumulative differences are $o(n)$; equality of all cumulative sums implies equality of all past coordinates. It is equivalent to augmenting the printed cumulative-difference distance by $|a_0-b_0|$: their cumulative sums differ only by this extra coordinate and a one-step shift.

Here is the full corrected continuity proof. Fix $a$ with $A_n(a)/n\to \lambda <c$, and let $\epsilon =(c-\lambda )/4$. Choose $N\ge 1$ such that $A_n(a)/n\le \lambda +\epsilon$ whenever $n\ge N$. If $d_{\#}(a,b)<\epsilon /2$, then, for all such $n$,Both queue workloads therefore maximize over the same finite collection $0\le n<N$, containing the value zero. For any two finite lists, the difference of their maxima is at most the maximum absolute difference of corresponding entries. Thusin this neighborhood. In particular the queue workload is locally Lipschitz continuous and hence continuous.

An alternative indexing repair retains the printed topology but interprets the time-zero function as the workload before slot $(-1,0)$, namely ${\sup }_{n\ge 0}\{{\sum }_{j=1}^n{a}_{-j}-cn\}$. The same finite-window argument proves its continuity. One must then shift the time labels consistently in the later recursions. The remaining answers use the post-slot convention and $d_{\#}$.

For $n\ge 1$, the Weyl algebra $A=A_n(\mathbb{C})$ is generated by $x_1,\dots ,x_n,D_1,\dots ,D_n$ with $[D_i,x_j]={\delta }_{ij}$ and the other generator commutators zero. Iterating the differential Ore extension construction gives the ordered basis $x^{\alpha }{D}^{\beta }$. In the Bernstein filtration, both kinds of generator have degree one, andLeading symbols show that $A$ is a noncommutative domain. The Hilbert basis theorem and homogeneous-generator lifting by decreasing degree make it both left and right Noetherian.

It is also a simple ring. In a nonzero two-sided ideal, choose a nonzero element and a monomial $x^{{\alpha }_0}{D}^{{\beta }_0}$ of maximal total degree with nonzero coefficient $c$. Applying $(\mathrm{ad}D{)}^{{\alpha }_0}(\mathrm{ad}x{)}^{{\beta }_0}$ eliminates every lower-degree term and every different term of that maximal degree. The result is the nonzero scalar $(-1{)}^{|{\beta }_0|}{\alpha }_0!{\beta }_0!c$. Thus the ideal contains $1$. Here the scalar extraction by iterated Weyl commutators uses characteristic zero. Despite simplicity, $A$ has no nonzero finite-dimensional modules: in such a representation $[D_1,x_1]=I$ would have trace $0=\dim M$, impossible over $\mathbb{C}$. Its interesting module theory is necessarily infinite-dimensional.

For a finitely generated left module $M$, choose a finite-dimensional generating subspace $M_0$ and set $M_j=F_{j}A{M}_0$. This is a good filtration of a module. Its associated graded module is finite over the graded polynomial ring, so its cumulative Hilbert function is eventually a polynomial,with positive integer multiplicity $e(M)$. These are the Bernstein growth dimension of a Weyl algebra module and its multiplicity of a filtered module. Two good filtrations bound each other after fixed shifts: lift finite homogeneous generators in one filtration, bound their degrees in the other, and use compatibility with $F_{j}A$; then reverse the roles. The growth degree and leading coefficient therefore do not depend on the good filtration of a module. The degree is also the Gelfand–Kirillov dimension of a module and the dimension of the characteristic variety of a Weyl algebra module, the support of $\mathrm{gr}M$. Since $M$ is a quotient of a finite sum of copies of $A$, $d(M)\le 2n$.

The essential lower bound is the Bernstein inequality for Weyl algebra modules:Here is a full proof. Multiplication gives a linear map $F_{j}A\to {\mathrm{Hom}}_{\mathbb{C}}(M_j,M_{2j})$. Prove it injective by induction on $j$. At $j=0$ a scalar killing $M_0\ne 0$ must vanish. If $a\in F_{j}A$ kills $M_j$, then for every generator $z$, its commutator $[a,z]\in F_{j-1}A$ kills $M_{j-1}$: both $azm$ and $zam$ are zero for $m\in M_{j-1}$. Induction gives $[a,z]=0$ for all generators. In the ordered basis, $[D_i,a]$ differentiates the $x_i$ exponents and $[x_i,a]$ differentiates the $D_i$ exponents with a minus sign. Vanishing of all these commutators in characteristic zero forces $a$ to be a scalar. It then vanishes because it kills $M_0$. This proves the faithful finite-step action of a Weyl algebra and henceComparison of degrees gives $2n\le 2d(M)$, proving the inequality. Equally, without invoking eventual polynomiality, this estimate directly bounds the Gelfand–Kirillov dimension of a module below by $n$. Right modules satisfy the same inequality through the filtration-preserving isomorphism $A\cong A^{\mathrm{op}}$, $x_i\mapsto x_i$, $D_i\mapsto -D_i$.

A nonzero finitely generated module with $d(M)=n$ is a holonomic Weyl algebra module; the zero module is conventionally included. This minimal-growth class is stable under submodules, quotient modules and extensions. For $0\to N\to M\to Q\to 0$, take the induced and quotient filtrations. Their graded sequence is exact; Noetherianity of $\mathrm{gr}A$ makes the induced filtrations good. Thus the Hilbert functions add andif both nonzero terms have the top dimension, their multiplicities add. For a holonomic Weyl algebra module, every nonzero subquotient has dimension $n$, by the upper bound inherited from $M$ and Bernstein's lower bound. Each nonzero factor in a strict chain therefore consumes positive integral multiplicity. This proves the multiplicity bounds the length of a holonomic Weyl module:In particular a holonomic module is both Noetherian and Artinian and has finite length, although its vector-space dimension is infinite. This distinction separates finite-dimensional representation theory from finite-length Weyl module theory.

The polynomial module $\mathbb{C}[x_1,\dots ,x_n]$, with $x_i$ acting by multiplication and $D_i$ by differentiation, is generated by $1$. Its $M_j$ consists of polynomials of degree at most $j$, with dimension $\left(\genfrac{}{}{0ex}{}{j+n}{n}\right)$, so $d=n$ and $e=1$. It is simple: differentiating a maximal-degree monomial of a nonzero polynomial isolates a nonzero constant, and multiplication by polynomials then generates the whole module. A second basic example is the delta moduleUsing the reverse ordered basis $D^{\beta }{x}^{\alpha }$, its vector space is $\mathbb{C}[D_1,\dots ,D_n]$; $D_i$ acts by multiplication and $x_i$ by negative differentiation. The same argument proves simplicity and holonomicity. These two examples are exchanged by the automorphism $x_i\mapsto D_i$, $D_i\mapsto -x_i$. Free modules, in contrast, have growth dimension $2n$ and are not holonomic. Thus the central organizing invariants are good filtration of a module, growth dimension, characteristic support and multiplicity, with Bernstein's estimate explaining why dimension $n$ is the natural smallest nonzero size.

For $X\in \mathfrak{g}=T_{e}G$, let $X^L$ be the left-invariant vector field with value $X$ at the identity. Its integral curve through the identity is the unique one-parameter subgroup ${\gamma }_X$ with ${\gamma }_X^{\prime }(0)=X$. It exists for all real time: local existence followed by translation and the identity ${\gamma }_X(t+s)={\gamma }_X(t){\gamma }_X(s)$ extends it indefinitely. The Exponential map of a Lie group isIt is smooth and $d_0\exp ={\mathrm{id}}_{\mathfrak{g}}$, because differentiating $t\mapsto \exp (tX)$ at zero gives $X$. The inverse function theorem therefore makes it a local diffeomorphism at zero.

If $G$ is abelian, the product $t\mapsto \exp (tX)\exp (tY)$ is a one-parameter subgroup with initial tangent $X+Y$. Uniqueness gives $\exp (t(X+Y))=\exp (tX)\exp (tY)$, henceConversely, if this additive homomorphism identity holds, all $\exp (tX)$ and $\exp (sY)$ commute. Differentiating their adjoint action shows $[X,Y]=0$. Alternatively, their commuting images contain an identity neighbourhood; that neighbourhood generates the identity component of a Lie group, so the identity component is abelian. Conversely, if the Lie algebra is abelian, its left-invariant vector fields commute, so their flows commute. Hence all one-parameter subgroups commute; their product with initial tangent $X+Y$ is the one-parameter subgroup for $X+Y$. The Lie exponential map is therefore additive, and its image generates an abelian identity component. Thus the stated equivalence holds for connected $G$. Without connectedness the exact assertion is that the Lie exponential map is a homomorphism precisely when $G^{\circ }$ is abelian. The converse implication to all of $G$ is false: in $O(2)$ the Lie algebra consists of $tJ$, whereRotations satisfy $R_{t+s}=R_t{R}_s$, but a reflection $S$ satisfies $S{R}_t{S}^{-1}=R_{-t}$, so the full orthogonal group is not abelian.

Now take a connected abelian Lie group of dimension $n$. Its Lie exponential map is a homomorphism whose image contains an identity neighbourhood, hence an open subgroup. A connected group has no proper open subgroup, so this map is onto. Its kernel $\Lambda$ is discrete, since it is injective on a neighbourhood of zero. The induced map is a bijective local diffeomorphism and therefore a Lie group isomorphismHere is the needed classification of $\Lambda$. Let $r$ be the dimension of its real span. Choose ${\lambda }_1,\dots ,{\lambda }_r\in \Lambda$ that form a real basis of that span. Their integer span ${\Lambda }_0$ has finite index in $\Lambda$: subtract integer multiples of the ${\lambda }_i$ to put each coset representative in a compact parallelepiped. Only finitely many points of $\Lambda$ lie there, since discreteness at zero gives a uniform positive separation between distinct points. Thus $\Lambda$ is finitely generated and torsion-free, and the structure theorem for finitely generated abelian groups makes it free abelian of rank $r$. Its integer basis is a real basis of the span. A linear change of coordinates consequently sends $\Lambda$ to ${\mathbb{Z}}^r\times \{0\}$ in ${\mathbb{R}}^n$.

This proves the classification of connected abelian Lie groups:The isomorphism depends on the choice of lattice basis and complementary vector space. Compactness is equivalent to $s=0$.

All cohomology below has rational coefficients. First describe the incidence space geometrically. The image of $g$ is $\mathbb{P}(W)\cong {\mathbb{CP}}^1$, inside $\mathrm{Gr}(1,4)\cong {\mathbb{CP}}^3$. For a line $U\subset W$, choosing a plane $V$ containing $U$ is equivalent to choosing the line $V/U$ in ${\mathbb{C}}^4/U$. If $\mathcal{L}$ is the tautological line bundle on $\mathbb{P}(W)$, thenThis is the projective bundle of a rank-three quotient bundle, with fibre ${\mathbb{CP}}^2$. Local frames of the quotient bundle give local product charts. In particular $F_W$ is an oriented real six-dimensional manifold.

For $V\in S_W\setminus \{W\}$, the intersection $V\cap W$ is exactly a line, so $f^{-1}(V)$ is a single point and the inverse map is $V\mapsto (V\cap W,V)$. The intersection line varies continuously, indeed holomorphically, wherever its dimension is constant: in local Grassmannian charts it is the kernel of a matrix of constant rank, whose kernel bases can be chosen locally. Thus $f$ is an isomorphism over the open stratum. At the remaining point,Set $K=R{f}_{\ast }{\mathbb{Q}}_{F_W}$, a derived direct image of sheaves. The proper base change for sheaves theorem identifies its derived stalks with the cohomology complexes of the fibres. Consequently $K$ is the constant rational sheaf in degree zero on $S_W\setminus \{W\}$, andThe cohomology sheaves are therefore finite-rank local systems on the two strata, with no other nonzero degrees. This proves that $K$ is a constructible complex of sheaves and verifies normalization. The open stratum has real dimension six, so the isolated point has codimension six. Its middle perversity is $\bar{m}(6)=2$, and the intersection-complex support and cosupport axioms require stalk vanishing in degrees above two. The fibre calculation verifies the support axiom, as well as vanishing in negative degrees.

For an oriented real six-dimensional manifold, Verdier duality gives $D_{F_W}{\mathbb{Q}}_{F_W}\cong {\mathbb{Q}}_{F_W}[6]$. For a proper map, Verdier duality commutes with derived direct image. Applying these statements here yieldsThe duality identity $i_W^{!}{D}_{S_W}K\cong D_{\{W\}}{i}_W^{\ast }K$ converts this intoHere $i_W^{!}K$ is the costalk, and dualization of rational complexes reverses degrees. Thus its only nonzero groups are $\mathbb{Q}$ in degrees four and six. In particular they vanish for $a\le 3$, which is precisely the cosupport axiom at this isolated six-dimensional singularity. The uniqueness characterization by constructibility, normalization, support and cosupport now identifies $K$ with the unshifted intersection complex of $S_W$. Taking hypercohomology, and using the composition of derived global sections with $R{f}_{\ast }$, givesFinally apply the fibration result allowed in the question to $g:F_W\to \mathbb{P}(W)$. Its base is the simply connected complex projective line, its fibre is the Complex projective plane, and the projective-bundle charts above verify the fibration hypothesis. Hence, as graded rational vector spaces,The resulting Betti numbers and Poincaré polynomial areThis uses $H^{\ast }({\mathbb{CP}}^r)=\mathbb{Q}[h]/(h^{r+1})$ with $\mathrm{deg}h=2$. No assertion that an arbitrary projective fibration has the product cohomology ring is needed.

Since the two-transitive group action is primitive, the orbits of any nontrivial normal subgroup form only the single block $\Omega$. Thus $N$ is transitive. If $N$ is regular, identify $\Omega$ with $N$ by $x\mapsto x\alpha$. Then $G=N{G}_{\alpha }$, $N\cap G_{\alpha }=1$, and conjugation by $G_{\alpha }$ gives automorphisms of $N$. The two-transitive group action makes these automorphisms transitive on $N\setminus \{1\}$. All nonidentity elements therefore have the same order, which must be a prime $p$: a suitable power of an element of composite order would have a smaller nontrivial order. Cauchy's theorem for finite groups now shows $|N|=p^d$.

The center of a group of prime-power order is nontrivial and is a characteristic subgroup. The transitive conjugation action on nonidentity elements therefore forces $Z(N)=N$. Hence $N$ is an elementary abelian group, which we identify with the additive vector space ${\mathbb{F}}_p^d$. Conjugation by $G_{\alpha }$ is linear on this vector space. It is faithful: an element fixing $\alpha$ and centralizing $N$ fixes every $x\alpha$, hence is the identity permutation. ConsequentlyHere the general affine group acts by translations followed by invertible linear maps.

Replace $F^n$ by a symplectic vector space $(V,B)$ of dimension $2m$, and elementary row operations by symplectic transvectionsThe cross terms in $B(Tx,Ty)$ cancel, and $B(v,v)=0$, so these maps preserve $B$. Their inverses are $T_{v,-c}$.

They generate $Sp(V)$ by the following symplectic basis elimination. If $B(u,v)\ne 0$, a single transvection with direction $v-u$ and parameter $1/B(u,v)$ sends $u$ to $v$. If $B(u,v)=0$ and $u\ne v$, choose $w$ with both $B(u,w)$ and $B(w,v)$ nonzero: the union of two proper hyperplanes does not exhaust a vector space, even over ${\mathbb{F}}_2$. Two transvections then suffice. Thus one can make a form-preserving linear map fix the first basis vector $e$. Next align its partner $f$, while keeping $e$ fixed. Two partners have difference in $e^{\perp }$, so the preceding single-transvection construction uses a direction in $e^{\perp }$ whenever their mutual pairing is nonzero. If it is zero, the intermediate partner $f+ae$, $a\ne 0$, has nonzero pairing with both. All these transvections fix $e$. Once both $e,f$ are fixed, recurse on their nondegenerate symplectic orthogonal complement. This proves generation over every field.

Use the action on the points of projective space, whose kernel is the scalar center $Z=\{\lambda I:{\lambda }^2=1\}$. For $m=1$ this is the $S{L}_2$ case. For $m\ge 2$ the action is primitive, although it is generally not two-transitive. The stabilizer of a line $L$ has two orbits on other lines: those orthogonal to $L$ and those not orthogonal to $L$. This follows by extending a hyperbolic pair, or an independent isotropic pair, to a symplectic basis. A block containing $L$ and a second line $M$ contains the entire corresponding orbit of the stabilizer of $L$. The stabilizer of $M$ then supplies a line of the other type: choose a vector orthogonal to one of $L,M$ and not to the other. Nondegeneracy and independence guarantee such a vector, also in dimension four. Hence the block contains both orbits and is the whole set.

The transvections along $L$ form an abelian normal subgroup of its stabilizer; their conjugates generate the group. Thus the Iwasawa simplicity lemma applies as soon as perfectness is established. For $|F|>3$, every transvection lies in an embedded $S{L}_2(F)$ acting on a hyperbolic plane, and 4(c) proves that embedded group perfect.

For the remaining fields, the perfectness of finite symplectic groups has a short direct proof. Over ${\mathbb{F}}_3$ with $m\ge 2$, all parameter-one transvections have the same class $t$ in the abelianization, and $3t=0$. In an isotropic plane with basis $u,v$, the four directions $u,v,u+v,u-v$ give commuting transvections whose product is $1$: their rank-one terms add to zero, sinceover ${\mathbb{F}}_3$. Thus $4t=0$, giving $t=0$. Parameter-two transvections are inverses, so all generators vanish in the abelianization. Over ${\mathbb{F}}_2$ with $m\ge 3$, use the seven nonzero vectors of an isotropic three-space. Their commuting transvections multiply to $1$, since each diagonal coordinate of ${\sum }_{0\ne v}v{v}^{\mathsf{T}}$ occurs four times and each off-diagonal coordinate twice. Their common abelianized class satisfies $2t=7t=0$, and again the group is perfect.

Consequently the projective symplectic group is simple, with the necessary exceptionsThe first two are the linear exceptions; the last is $S{p}_4(2)\cong S_6$, proved in 6(c). The word “projective” matters: the unquotiented symplectic group may have the nontrivial scalar center $\{I,-I\}$.

Let $L_A$ and $R_A$ be left and right multiplication by $A$ on the vector space $M_n(\mathbb{C})$. Then $\mathrm{ad}A=L_A-R_A$. On matrix units, or from $M_n(\mathbb{C})={\mathbb{C}}^n\otimes ({\mathbb{C}}^n{)}^{\ast }$, one obtainsFor example, the coefficient of $E_{ij}$ in $A{E}_{ij}B$ is $A_{ii}{B}_{jj}$; summing over $i,j$ proves the mixed-trace identity. Expanding the product of the two adjoint operators therefore gives the Killing form of the general linear Lie algebra:For $A,B$ in the special linear Lie algebra, both individual traces vanish. Furthermore $M_n(\mathbb{C})={\mathfrak{sl}}_n\oplus \mathbb{C}I$, and every adjoint operator kills the scalar summand, so its product has the same trace on $M_n$ and on ${\mathfrak{sl}}_n$. Consequently the Killing form of the special linear Lie algebra isNo uniqueness theorem for invariant bilinear forms on a Lie algebra is needed: this calculation establishes the constant directly. For the degenerate case $n=1$, the Lie algebra is zero, both forms vanish, and proportionality does not determine a unique constant.

A topological group is a group with a topology for which $(x,y)\mapsto xy$ and $x\mapsto x^{-1}$ are continuous; we use the Hausdorff convention. Translations and inversion are then homeomorphisms.

For the final generator-extraction assertion, use the profinite Frattini subgroupFor a finitely generated pro-p group, the Burnside basis theorem says that $G/\Phi (G)$ is a finite-dimensional vector space over ${\mathbb{F}}_p$ and a subset topologically generates $G$ precisely when its image spans that vector space. In particular its dimension is the minimal number $d(G)$ of topological generators. A given $d$-element generating set shows $d(G)\le d$. The image of any other generating set spans the same finite-dimensional vector space, so it contains a basis of $d(G)$ elements. Choose their preimages in that generating set and apply the Burnside basis theorem again. Every generating set contains a generating subset of size $d(G)\le d$. This argument also applies to an infinite generating set.

The characteristic equation of a linear recurrence isFor $0<\theta <\pi$, these characteristic roots are distinct. Two real solutions of the second-order difference equation are $\cos (n\theta )$ and $\sin (n\theta )$. Their values at $n=0,1$ form a matrix with determinant $\sin \theta \ne 0$, proving linear independence. Consequently they span the two-dimensional vector space of recurrence solutions.

At $\theta =0$ the two roots coalesce at one. The linear recurrence becomes $X_{n+2}-X_{n+1}=X_{n+1}-X_n$, so its successive differences are constant. A basis is therefore $1,n$, and every solution is $A+Bn$.

For the specified initial conditions, the zero-angle solution is $X_n(0)=n$. For nonzero angle, it is convenient to center the basis at $n=1$:This gives $X_1=1$ and $X_2=\cos \theta +2-\cos \theta =2$, and each summand obeys the linear recurrence. For every fixed integer $n\ge 1$, the elementary limit $\sin z/z\to 1$ givesThe convergence here is for fixed $n$; it does not assert uniform convergence over all indices.

The minimal polynomial $m_{\beta }(t)$ of the linear map $\beta$ is the unique monic polynomial of smallest degree satisfying $m_{\beta }(\beta )=0$. An annihilating polynomial exists because the powers $I,\beta ,{\beta }^2,\dots$ lie in the finite-dimensional vector space $\mathrm{End}(V)$ and are therefore linearly dependent. If two monic annihilating polynomials had the same least degree, their difference would be an annihilating polynomial of smaller degree, so they coincide.

Write $m_{\beta }(t)=a_0+tq(t)$. If $a_0\ne 0$, evaluation at the linear map gives $\beta q(\beta )=-a_{0}I$. Since $\beta$ commutes with every polynomial in itself, this is a two-sided inverse:Conversely, if $\beta$ is an invertible linear map and $a_0=0$, then $m_{\beta }(t)=tq(t)$ and multiplication of $\beta q(\beta )=0$ by ${\beta }^{-1}$ gives $q(\beta )=0$, contradicting the defining least degree. Thus invertibility is equivalent to a nonzero constant term of the minimal polynomial. In the zero-dimensional case the minimal polynomial is $1$ and the same conclusion holds.

Let $\dim V=n$ and $\dim W=k$. Extend a basis $w_1,\dots ,w_k$ of the vector subspace $W$ to $w_1,\dots ,w_n$ of $V$, and let $f_1,\dots ,f_n$ be its dual basis. Every linear functional is uniquely ${\sum }_i{c}_i{f}_i$. It vanishes on $W$ exactly when $c_1=\cdots =c_k=0$. Thus the annihilator of a vector subspace has basis $f_{k+1},\dots ,f_n$, and

To apply this to a matrix, regard its rows as elements of $({\mathbb{R}}^n{)}^{\ast }$ and let $R$ be their linear span. Elementary invertible row and column operations reduce a rank-$r$ matrix to a block identity $\mathrm{diag}(I_r,0)$, showing that $\dim R=r$. The equations require that every functional in $R$ vanish on $x$. The canonical identification ${\mathbb{R}}^n\cong (({\mathbb{R}}^n{)}^{\ast }{)}^{\ast }$ sends $x$ to evaluation $f\mapsto f(x)$: it is injective because coordinate functionals separate points, and is bijective by equal dimensions. Consequently the solution vector space is the annihilator of $R$ under this identification. The dimension formula just proved givesA basis of this kernel consists of exactly $n-r$ linearly independent solutions, and every solution is their linear combination.

Use the standard boundaryless-manifold convention, so $M$ is a closed manifold. Its orientation and Riemannian metric determine a unique positive Riemannian volume form ${\mu }_g$ taking value one on every positively oriented orthonormal frame. In an oriented atlas it isFor the coordinate invariance of the Riemannian volume form, on an overlap put $J=\partial y/\partial x$. Then $g_x=J^T{g}_{y}J$ and $d{y}^1\wedge \cdots \wedge d{y}^n=(\det J)d{x}^1\wedge \cdots \wedge d{x}^n$. The oriented transition has $\det J>0$, so the square root of the metric determinant and the coordinate volume transform together. The local forms therefore agree and define ${\mu }_g$ globally.

For real $r$-forms the Hodge star operator is defined byThe pointwise wedge pairing is nondegenerate, so this uniquely defines a smooth map $\ast :{\Omega }^r(M)\to {\Omega }^{n-r}(M)$. On an oriented orthonormal coframe, it sends a basis wedge to the signed complementary wedge. Applying it twice gives

For Hodge star eigenvalues on graded differential forms, extend $\ast$ complex-linearly to the direct sum of all form degrees. This interpretation matters: on a fixed degree it is an endomorphism only in middle degree. In positive odd dimension, $r(n-r)$ is even for every $r$, so ${\ast }^2=I$ and only $\pm 1$ can occur. Both occur, with eigenforms $1+{\mu }_g$ and $1-{\mu }_g$.

In positive even dimension, ${\ast }^2$ is $+I$ on even-degree forms and $-I$ on odd-degree forms. Hence ${\ast }^4=I$ and the only possible eigenvalues are $\pm 1,\pm i$. The first two occur as above. For the latter, choose a smooth local orthonormal frame and multiply its first dual differential one-form by a nonzero smooth bump function supported in the chart. This gives a form $\eta$ for which $\eta$ and $\ast \eta$ are independent. For $\lambda =\pm i$,Thus all four occur. Consequently, for $n\ge 1$,On degree $m$ in dimension $2m$, the Hodge star has eigenvalues $\pm 1$ for even $m$ and $\pm i$ for odd $m$; signed complementary middle-degree wedges show both occur for $m\ge 1$. In zero dimensions it is multiplication by the signed unit volume at each point, so only the signs actually present occur.

For Hodge integration by parts in arbitrary degree, let $\alpha \in {\Omega }^{r-1}(M)$ and $\beta \in {\Omega }^r(M)$. The graded Leibniz rule and Stokes theorem on the closed manifold giveUsing the given codifferential and the square of the Hodge star,Thereforewhich proves that $\delta$ is the formal adjoint of $d$. The complex version inserts conjugation in the usual Hermitian inner product.

The Hodge decomposition theorem gives an $L^2$-orthogonal decompositionwhere ${\mathcal{H}}^k(M)=\ker \Delta$ is finite dimensional and consists of the harmonic differential forms. Each de Rham cohomology class has a unique harmonic representative. Equivalently, with harmonic projection $P_{\mathcal{H}}$ there is a smooth Green operator of the Hodge Laplacian $G$ satisfying $\Delta G=G\Delta =I-P_{\mathcal{H}}$.

The solvability condition for the Hodge Poisson equation isNecessity follows from $⟨\Delta \alpha ,h⟩=⟨\alpha ,\Delta h⟩=0$. Conversely the orthogonality condition gives $P_{\mathcal{H}}\beta =0$, so ${\alpha }_0=G\beta$ solves the Poisson equation for differential forms.

The nonnegativity of the Hodge Laplacian identityidentifies its kernel with closed and coclosed forms. Thus the affine space of solutions of the Hodge Poisson equation isFor completeness, if a closed form is decomposed as $\omega =h+d\xi +\delta \zeta$, then $d\delta \zeta =0$, and $\Vert \delta \zeta {\Vert }^2=⟨\zeta ,d\delta \zeta ⟩=0$. Hence its cohomology class is represented by $h$. An exact harmonic form $h=d\xi$ must vanish, since $\Vert h{\Vert }^2=⟨\delta h,\xi ⟩=0$. This proves the canonical isomorphism ${\mathcal{H}}^k(M)\cong H_{\mathrm{dR}}^k(M)$, so the solution set is an affine space whose translation vector space is isomorphic to the requested de Rham cohomology.

For a set of primes $\pi$, a Hall $\pi$-subgroup $H$ of a finite group is a subgroup whose order uses only primes in $\pi$ and whose subgroup index uses only primes outside $\pi$. Equivalently, $|H|$ is the full $\pi$-part of $|G|$. A Hall subgroup is a Hall $\pi$-subgroup for some $\pi$. The theorem of P. Hall for finite soluble groups states: Hall $\pi$-subgroups exist for every $\pi$, any two are conjugate, and every $\pi$-subgroup is contained in a Hall $\pi$-subgroup.

For the existence proof, first give the explicit coprime splitting calculation needed in the induction. Let $N$ be elementary abelian of field characteristic $r$, normal in $E$, with $r\nmid |E/N|$. Write $Q=E/N$ and use additive notation on $N$. A normalized section $s:Q\to E$ gives an action $x\cdot n=s(x)ns(x{)}^{-1}$ independent of its choice, since $N$ is abelian. Write $s(x)s(y)=f(x,y)s(xy)$. Associativity givesSum over $z\in Q$, put $S(x)={\sum }_{z}f(x,z)$, and divide by $|Q|$ in the vector space $N$. This is legitimate because $r\nmid |Q|$. With $b(x)=|Q{|}^{-1}S(x)$ one obtainsThe new section $s^{\prime }(x)=(-b(x))s(x)$ therefore has zero multiplication defect and is a group homomorphism. Its image is a complement to $N$.

We will also need a single conjugation carrying one complement to the other. If two homomorphic sections differ by $d(x)$, their difference satisfies $d(xy)=d(x)+x\cdot d(y)$. Summing over $y$ gives $d(x)=c-x\cdot c$, where $c=|Q{|}^{-1}{\sum }_{y}d(y)$. Thus the second section is $cs(x)c^{-1}$. This proves, by explicit averaging, both coprime splitting over an elementary abelian normal subgroup and the existence of such a conjugation by an element of $N$.

Now induct on $|G|$ for Hall subgroup existence in soluble groups. A minimal normal subgroup $N$ of a finite soluble group is elementary abelian of some field characteristic $r$. Indeed, $N^{\prime }$ is a characteristic subgroup of $N$ and normal in $G$; it cannot equal $N$ because $N$ is soluble, so it is trivial. In this abelian group, a nontrivial primary component is a characteristic subgroup, forcing a single prime, and the characteristic subgroup annihilated by that prime forces exponent $r$.

By induction $G/N$ has a Hall $\pi$-subgroup $\overline{H}$, with preimage $M$. If $M<G$, induction in $M$ supplies a Hall $\pi$-subgroup of $M$, which is also Hall in $G$ because $[G:M]$ is prime to all primes in $\pi$. If $M=G$, the quotient is a $\pi$-group. When $r\in \pi$, $G$ itself is a $\pi$-group. When $r\notin \pi$, the averaging construction splits off $N$, and its complement has order $|G/N|$ and is the required Hall $\pi$-subgroup. This completes the existence proof, including the identity group and empty prime set.

A Sylow basis is a family $\{P_p:p\mid |G|\}$ containing one Sylow subgroup for each prime, with $P_p{P}_q=P_q{P}_p$ for each pair. This is permutability of subgroups, not commutation of all their elements. The product of any subfamily is a subgroup: commuting the factors as sets proves closure and closure under inversion. Inductively its order is the product of the prime-power orders, since the newly adjoined factor has trivial intersection with the earlier product. It is thus the corresponding Hall subgroup.

Prove existence of a Sylow basis by induction on $|G|$, again using elementary abelian $N$ of field characteristic $r$. Choose a basis $\{{\overline{P}}_p\}$ of $G/N$, taking ${\overline{P}}_r=1$ if $r$ does not divide that quotient's order. The product $\overline{H}={\prod }_{p\ne r}{\overline{P}}_p$ is its Hall $r^{\prime }$-subgroup. In its full preimage $T$, the averaging lemma supplies a complement $H$ to $N$, mapping isomorphically to $\overline{H}$. For $p\ne r$, let $P_p$ be the inverse image of ${\overline{P}}_p$ under this isomorphism within $H$. Let $P_r$ be the full preimage of ${\overline{P}}_r$ in $G$. These are Sylow subgroups of $G$. The other-prime pairs permute because they do so in $\overline{H}$. For $P_r$ and $P_p$, their product is contained in the full preimage of the subgroup ${\overline{P}}_r{\overline{P}}_p$ and has exactly that preimage's order $|N||{\overline{P}}_r||{\overline{P}}_p|$. Equality follows, so the product is a subgroup and these two factors permute as well. This constructs a basis.

To obtain a simultaneous conjugation of two bases, take two bases. Their images in $G/N$ are bases: $N$ lies in every Sylow $r$-subgroup, and the other-prime Sylow subgroups project injectively. By induction, conjugate one basis so that all corresponding images agree. The products of their non-$r$ factors are then two complements to $N$ in the same preimage $T$ of their common Hall $r^{\prime }$-subgroup. The averaging lemma for complements supplies a single element of $N$ conjugating these complements. In the resulting common complement each corresponding Sylow subgroup is the unique lift of its agreed quotient subgroup, so all non-$r$ factors agree. The $r$-factors already equal the full preimage of their common quotient factor and are preserved by conjugation from $N$. Thus all Sylow bases of a finite soluble group are conjugate.

In $S_4$ there are three Sylow $2$-subgroups of order eight: the action on the three pair-partitions of four points has kernel $V_4$, and these subgroups are the preimages of the three order-two subgroups of $S_3$. There are four Sylow $3$-subgroups, one for each three-point support. Every choice $P_2,P_3$ has trivial intersection and product of cardinality $8\cdot 3=24$, so $P_2{P}_3=S_4=P_3{P}_2$. ThereforeFor $E=V⋊S_4$, $V={\mathbb{F}}_5^4$, the Sylow $5$-subgroup is uniquely $V$. In a Sylow basis the product $P_2{P}_3$ is a Hall $5^{\prime }$-subgroup, hence a complement to $V$. Conversely, a basis of any complement together with $V$ is a basis of $E$, because $V$ is normal. The averaging lemma makes all complements conjugate by $V$. The stabilizer of the standard complement $S_4$ under this conjugation is $C_V(S_4)$: if a vector normalizes it, its group commutator with each complement element lies in both $V$ and that complement, hence is trivial. The fixed vectors under the natural coordinate permutation action are exactly $(a,a,a,a)$, so $|C_V(S_4)|=5$. There are consequently $5^4/5=125$ complements, and each has twelve bases. The complement is uniquely recovered from its basis as $P_2{P}_3$, so no basis is counted twice. The counting Sylow bases in a coprime elementary abelian extension formula yields

Work in the homotopy category of simply connected CW complexes. This is one of the permitted assumptions on the fundamental group; it makes every coefficient system below untwisted. A rational space is a simply connected space whose homotopy groups are vector spaces over $\mathbb{Q}$. A rationalization of a topological space is a map $r:X\to X_{\mathbb{Q}}$ to such a space which inducesIts purpose includes a universal mapping property, which will be described after the construction.

An Eilenberg–MacLane space $K(A,n)$ has exactly one nonzero homotopy group, ${\pi }_n=A$, where $A$ is abelian when $n\ge 2$. The representability theorem says $[B,K(A,n)]\cong H^n(B;A)$, naturally in $B$ and $A$. More generally, a principal fibration is a homotopy pullback of a path-space fibration $PC\to C$ along a map from its base to $C$; the fibre $\Omega C$ acts on the fibre paths by concatenation. For the Postnikov tower construction take $C=K(A,n+1)$. A principal Eilenberg–MacLane fibration with fibre $K(A,n)$ is the homotopy pullback of the path fibration of $K(A,n+1)$. Equivalently, its total space is the homotopy fibre of a map $k:B\to K(A,n+1)$. It is classified by the cohomology class $[k]\in H^{n+1}(B;A)$; there is no nontrivial action of the base fundamental group on the fibre coefficients in this principal description.

The Postnikov tower theorem supplies spaces $P_{n}X$ and compatible mapswith ${\pi }_i(P_{n}X)={\pi }_i(X)$ for $i\le n$ and zero for $i>n$. For a simply connected $X$, the fibre of $P_{n}X\to P_{n-1}X$ is $K({\pi }_{n}X,n)$, and the fibration is principal. Its Postnikov invariant isrepresented by $P_{n-1}X\to K({\pi }_{n}X,n+1)$, whose homotopy fibre is $P_{n}X$. Thus the homotopy groups alone are not the tower data: the classes $k_n$ specify how each new group is attached. For instance a zero class gives a product with $K({\pi }_{n}X,n)$.

State also the standard rational comparison results used in constructing the tower: $K(A,n)\to K(A\otimes \mathbb{Q},n)$ is a rational homology equivalence; a map between simply connected spaces is a rational homology equivalence if and only if it induces isomorphisms on all rational homotopy groups; and a rational homology equivalence induces isomorphisms in cohomology with any constant rational vector-space coefficients. The last assertion uses the universal coefficient theorem over the field $\mathbb{Q}$. These results are used here without proof, as allowed for this essay.

Construct rational stages inductively, with $P_1^{\mathbb{Q}}=\ast$. Suppose $r_{n-1}:P_{n-1}X\to P_{n-1}^{\mathbb{Q}}$ has already been constructed and is a rational homology equivalence. Put $V_n={\pi }_{n}X\otimes \mathbb{Q}$ and change coefficients in $k_n$ to obtain ${\overline{k}}_n\in H^{n+1}(P_{n-1}X;V_n)$. The comparison result gives an isomorphismLet $k_n^{\mathbb{Q}}$ be its unique inverse image and defineEquality of the pulled-back classes gives a homotopy-commuting square of classifying maps, and hence a map $r_n:P_{n}X\to P_n^{\mathbb{Q}}$ over $r_{n-1}$. On the fibre it is the map induced by ${\pi }_{n}X\to V_n$. The homotopy exact sequences therefore show that $r_n$ tensors all homotopy groups through degree $n$ with $\mathbb{Q}$, while the higher groups of both stages vanish. The rational comparison theorem makes $r_n$ a rational homology equivalence, allowing the induction to continue.

Take the homotopy inverse limit of the rational tower:The convergence theorem for a Postnikov tower gives $X\simeq {\mathrm{holim}}_n{P}_{n}X$ and the compatible stage maps induce $r:X\to X_{\mathbb{Q}}$. To see what convergence says here, fix a homotopy degree $i$. From stage $n\ge i$ onwards the group and its transition maps are constant; the inverse-limit group is ${\pi }_{i}X\otimes \mathbb{Q}$ and the associated derived-limit obstruction is zero. Thus this construction has exactly the claimed rational homotopy groups. Taking a homotopy limit, rather than an arbitrary set-theoretic inverse limit, is part of the construction.

The rational localization theorem says that this map has the universal propertyIn other words every map from $X$ to a rational space factors through $r$, and the factorization is unique up to homotopy. This concerns homotopy classes, not strict uniqueness of point-set maps. The theorem can be formulated as localization of the simply connected homotopy category at rational homotopy equivalences; rational spaces are its local objects. It also implies uniqueness of rationalization of a topological space: apply the property to two candidate rational spaces to obtain inverse equivalences under $X$. In particular the construction forgets torsion in homotopy while retaining precisely the information needed for maps to rational targets.