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A deterministic quantum operation here is a CPTP map. By Stinespring dilation, there is an isometry $V:B\to B^{\prime }E$ with $\Lambda (\tau )={\mathrm{Tr}}_E(V\tau V^{†})$. Let ${\omega }_{A{B}^{\prime }E}=(I\otimes V){\rho }_{AB}(I\otimes V^{†})$. An isometry preserves all nonzero eigenvalues, soThe quantum mutual information is $I(A:B)=S(A)+S(B)-S(AB)$. The difference after discarding the environment iswhere the last inequality is Strong subadditivity of Von Neumann entropy, in the form $S(A{B}^{\prime })+S(B^{\prime }E)\ge S(B^{\prime })+S(A{B}^{\prime }E)$. Thus data processing for quantum mutual information givesSubsystem $A$ is unchanged; its prime simply labels the output. Keeping the environment would preserve the mutual information, while tracing it out cannot increase it.

Diagonalize the source average $\pi ={\sum }_j{\lambda }_j|j⟩⟨j|$, using its positive eigenvalues. For fixed $\delta >0$, let ${\Pi }_{n,\delta }$ project onto product eigenvectors whose eigenvalues obeyThe typical subspace theorem states that, for every $ϵ>0$ and all sufficiently large $n$,The probability statement is the weak law of large numbers applied to $-{\log }_2{\lambda }_j$; the dimension bounds follow by summing the typical eigenvalue bounds. This explains why the quantum typical subspace retains almost all probability using about $nS(\pi )$ qubits.

For $R>S(\pi )$ choose $0<\delta <R-S(\pi )$. Measure $\{{\Pi }_{n,\delta },I-{\Pi }_{n,\delta }\}$. On success, encode the projected state isometrically into a space of dimension $\dim {\mathcal{T}}_{n,\delta }$; on failure, output a separate fixed flag. Decode the successful sector by the inverse isometry, and map the flag to a fixed state $|{\phi }_0⟩$. The compressed dimension is at most $2^{n(S(\pi )+\delta )}+1$, hence fits within rate $R$ for all sufficiently large $n$. Both maps are trace-preserving quantum channels, not merely successful postselected operations.

Writing $\Pi ={\Pi }_{n,\delta }$, the composite channel isFor a pure source signal $|{\Psi }_k⟩$, set $a_k=⟨{\Psi }_k|\Pi |{\Psi }_k⟩$. Its squared quantum fidelity after the composite channel is at least $a_k^2$. Since the source is memoryless, its average state on $n$ uses is ${\pi }^{\otimes n}$. Convexity of the square givesThe Kraus formula for entanglement fidelity gives the same lower bound for $F_e({\pi }^{\otimes n},{\mathcal{N}}_n)$, because one Kraus operator is $\Pi$ and the other terms are nonnegative. Taking $ϵ\to 0$ proves reliable compression at every $R>S(\pi )$, in both average pure-signal fidelity and the stronger entanglement-fidelity sense. This is typical-subspace compression with a failure flag.

A Riemannian metric is a smooth field of positive-definite symmetric bilinear forms $g_x$ on the tangent spaces, equivalently a smooth section of ${\mathrm{Sym}}^2{T}^{\ast }M$ with that positivity. Its musical isomorphism isNondegeneracy makes every fiber map an isomorphism. In coordinates the map has matrix $g_{ij}$, and its inverse ${♯}_g$ has matrix $g^{ij}$. Smoothness of the inverse follows from the inverse-matrix formula and the nonvanishing determinant. Both maps cover the identity on $M$, giving a smooth vector bundle isomorphism.

For local flattening of a Riemannian metric, choose a manifold chart neighborhood $V$ around $p$ with closure contained in $U$, and let $h={\varphi }^{\ast }\delta$ on $V$. Choose a smooth bump function $0\le \chi \le 1$, with compact support in $V$ and equal to one on a smaller neighborhood $W$ of $p$. SetThe formulas glue smoothly because the modification is supported strictly inside $V$. A convex combination of positive-definite forms is positive definite. On $W$ the metric is exactly $h$, so $\varphi$ gives an isometry to a Euclidean open subset. Outside $U$ the original metric remains unchanged.

Geodesic completeness means that every geodesic with arbitrary initial point and tangent vector extends for every real affine parameter. On each connected component, the Hopf-Rinow theorem equates this with completeness of the Riemannian distance.

The Euclidean-end claim needs a compact-core interpretation that is absent from its literal hypotheses. Indeed, $M=\{x\in {\mathbb{R}}^n:|x|>1\}$ with $n\ge 2$, $X=\varnothing$ and $g=\delta$ satisfy those hypotheses. The geodesic $\gamma (t)=(2-t)e_1$ reaches the missing unit sphere at $t=1$ and cannot continue within $M$. Thus the printed assumptions alone do not prove completeness.

Here is the intended compact-core completeness for a Euclidean end. Write $\Phi :M\setminus X\to \{|x|>1\}$ for the end coordinates and assume additionally thatis compact for every sufficiently large $R$. Choose such an $R$ beyond the metric transition and large enough to include the initial point of a geodesic. Its speed $s$ is constant. On every segment outside $K_R$, the Euclidean radius changes at a rate at most $s$, so over a finite parameter interval it cannot exceed $R+sT$. The full segment therefore stays in the compact set $K_{R+sT}$. Its velocities also stay in a compact subset of $TM$, since their metric norm is fixed and the base set is compact. The smooth geodesic ordinary differential equation consequently extends past any supposed finite endpoint. The reversed-time argument is identical. This proves completeness under the compact-core condition, and explains exactly what the exterior-of-a-ball counterexample lacks.

For upward stability of Riemannian completeness, lengths of all curves satisfy $L_{\tilde{g}}\ge L_g$, hence $d_{\tilde{g}}\ge d_g$. A $d_{\tilde{g}}$-Cauchy sequence is therefore $d_g$-Cauchy and has a limit $p$ because $g$ is complete. Smooth positive-definite metrics induce the manifold topology. More explicitly, on a small coordinate ball around $p$, $\tilde{g}$ has a bounded largest matrix eigenvalue, so the coordinate straight segment gives $d_{\tilde{g}}(p,q)\le C|\varphi (q)-\varphi (p)|$. Thus convergence to $p$ is also in $d_{\tilde{g}}$. That distance is complete, and Hopf-Rinow yieldsFor disconnected $M$, apply this argument on each component.

Construct the Euclidean lattices in an orthogonal coordinate system whose squared axis lengths areThus an integer coordinate vector $z$ has squared norm ${\sum }_i{a}_i{z}_i^2$. These four numbers are linearly independent over $\mathbb{Q}$: the four sign changes of $\sqrt{2},\sqrt{3}$ in their field extension isolate the four coefficients of a rational relation. The independence will distinguish the two flat tori; equality of their length spectra will hold for every positive choice of axis weights.

LetHere and below these integer coordinates are interpreted in the weighted orthogonal axes. Direct multiplication gives $S^T=-S$, $S^2=-3I$, andIn particular both $L_{\pm }$ are full-rank Euclidean lattices. Denote their four generator columns by $v_i^{\pm }$. Modulo $3$ the two matrices have the same columns, and the nonzero words of the given ternary tetracode $T$ are exactlyThese are the reductions of $\pm v_i^{\pm }$, respectively. The first and fourth displayed words are the specified generators, and the middle two arise from their sum and difference, up to sign. Thus reduction modulo $3$ maps either Euclidean lattice onto precisely $T$.

To construct a length-preserving correspondence, find the kernels $M_{\pm }=L_{\pm }\cap 3{\mathbb{Z}}^4$ of this reduction. The inverse matrix is $A_{\pm }^{-1}=A_{\pm }^T/12$. Hence $3w\in L_{\pm }$ exactly when $A_{\pm }^{T}w\equiv 0(\mathrm{mod}4)$. These congruences are equivalent toFor clarity, the first $+$ congruence is $-(w_1+w_2+w_3+w_4)=0(\mathrm{mod}4)$; subtracting the others gives evenness of each required pair sum. For the $-$ matrix the first congruence is $2w_1-(w_1+w_2+w_3+w_4)=0(\mathrm{mod}4)$; the remaining congruences again give equality of the four parities. Substitution proves the converse in each case.

Let $R_i$ be the orthogonal reflection changing only coordinate $i$. If the coordinates of $w$ are even, this changes their sum by a multiple of $4$; if all are odd, it changes the sum by $2$ modulo $4$. ThereforeEach Euclidean lattice is the disjoint union of its kernel and the eight cosets $\pm v_i^{\pm }+M_{\pm }$, since $T$ has nine words. Define a map $\Phi :L_{+}\to L_{-}$ by using $R_1$ on $M_{+}$ and $R_i$ on each of $v_i^{+}+M_{+}$ and $-v_i^{+}+M_{+}$. Each piece maps bijectively onto the corresponding $-$ coset. Thus $\Phi$ is a bijection satisfyingIt also commutes with $z\mapsto -z$. The closed geodesics of a flat torus have lengths $\Vert z\Vert$ for nonzero translation vectors $z$ in its Euclidean lattice. Consequently the two length spectra agree, including the multiplicities of translation classes and with either consistent orientation convention. If only primitive closed geodesics are counted, the same conclusion follows by removing iterates in increasing order of length: the total multiplicity at $\ell$ is the sum of primitive multiplicities at $\ell /k$, and there are only finitely many such contributions below any fixed length. The correspondence need not itself preserve primitiveness to give this conclusion.

It remains to prove the two flat tori are not isometric. An isometry of flat tori lifts to an affine Euclidean isometry; its linear part of an affine map must carry $L_{+}$ onto $L_{-}$. In integer coordinates write this part as $B$. For every $z\in L_{+}$, both $z$ and $Bz$ have integer coordinates and have equal squared norms. Rational independence of the $a_i$ impliesApply this to $z=A_{+}n$ for all $n\in {\mathbb{Z}}^4$. Two squared linear forms agreeing on all integer points agree as polynomials; their difference factors, so one form is the other or its negative. Since $A_{+}$ is invertible, $B$ is consequently a diagonal sign change $D=\mathrm{diag}({\epsilon }_1,\dots ,{\epsilon }_4)$.

Modulo $3$, such a $D$ must preserve $T$. Each nonzero word of $T$ has exactly three nonzero coordinates, and for each choice of zero coordinate there are just two such words, differing by an overall sign. Preserving these words forces the three signs on each support to agree. The four overlapping supports then force ${\epsilon }_1={\epsilon }_2={\epsilon }_3={\epsilon }_4$. Thus only $D=I$ and $D=-I$ are possible. Neither changes $L_{+}$.

Finally $v_1^{+}\notin L_{-}$: it has the same ternary word as $v_1^{-}$, but their difference is $6e_1=3(2,0,0,0)$, which fails the congruence defining $M_{-}$. Hence $L_{+}\ne L_{-}$ and no such linear isometry exists. This completes the tetracode length-isospectral lattice construction.

In this context a nowhere locally homogeneous metric, also called a bumpy metric, is a smooth Riemannian metric for which no two distinct nonempty open subsets are isometric with their induced metrics. Equivalently every local isometry between open subsets is the identity wherever defined: a nonidentity local isometry sends some point $x$ to a distinct point, and restriction to sufficiently small disjoint neighborhoods would violate the first formulation. This is the local-isometry meaning of the terminology here; degeneracy of periodic geodesics is a different use of the word bumpy.

Sunada's local isometry lemma states that on a compact smooth manifold without boundary of dimension $m\ge 2$ the nowhere locally homogeneous metrics contain a residual set in the space of smooth Riemannian metrics with its $C^{\infty }$ topology. In particular they are dense, by the Baire category theorem. A residual set is a countable intersection of open dense sets. The dimension assumption matters: every one-dimensional Riemannian metric is locally $d{s}^2$ in arclength coordinates and has local translations.

The jet bundle of maps $J^k(M,N)$ consists of equivalence classes $j_x^{k}f$ of smooth maps near $x$, where two maps agree to order $k$ at $x$ in coordinate charts. Its projection to $M\times N$ sends $j_x^{k}f$ to $(x,f(x))$. For a multi-index $\alpha$ in $m$ variables there are $\left(\genfrac{}{}{0ex}{}{m+r-1}{r}\right)$ derivatives of order $r$. Thus a coordinate chart consists of the source point, the target point, and $n$ coefficients for every derivative order $1$ through $k$. Summing the counts givesThis includes $J^0(M,N)=M\times N$. The fibre over $(x,y)$ is the space of truncated Taylor maps with fixed constant term $y$, locally modeled onFor $k=1$ its identification with $\mathrm{Hom}(T_{x}M,T_{y}N)$ is canonical. For higher $k$, changes of target coordinates mix derivatives of different orders, so this is a coordinate or connection-dependent description, not a canonical vector-bundle identification. The truncation $J^k\to J^{k-1}$ is an affine bundle modeled on the pullback of ${\mathrm{Sym}}^k(T^{\ast }M)\otimes TN$ over $M\times N$.

For the density assertion put $A={\overline{U}}_i$ and $B={\overline{U}}_j$. Their closures must be distinct. If $A=B$, the identity is an isometry for every Riemannian metric, and the requested complement is empty. Under the intended distinct-domain assumption, the smooth closed-ball hypothesis makes $A$ and $B$ regular closed domains, equal to the closures of their interiors. After interchanging them if necessary, there is a nonempty open set $W$ with compact closure in $\mathrm{int}(A)\setminus B$. Otherwise each interior would be contained in the other closure, forcing $A=B$.

Take any smooth Riemannian metric $g$. If ${\mathrm{vol}}_g(A)\ne {\mathrm{vol}}_g(B)$, it already lies outside ${\mathcal{S}}_{ij}$, because an isometry preserves the Riemannian volume form. If the volumes agree, choose a nonzero nonnegative smooth bump function $\eta$ supported in $W$ and setThese are positive definite Riemannian metrics, they agree with $g$ on $B$, and $g_t\to g$ in $C^{\infty }$ as $t\downarrow 0$, since every derivative of $g_t-g$ is $t$ times a fixed compactly supported smooth tensor. Their Riemannian volume forms satisfyHence ${\mathrm{vol}}_{g_t}(B)={\mathrm{vol}}_g(B)$ whereas ${\mathrm{vol}}_{g_t}(A)>{\mathrm{vol}}_g(A)$ for every $t>0$. In particularThe two domains cannot be isometric for $g_t$. Every $C^{\infty }$ neighborhood of $g$ therefore meets the complement of ${\mathcal{S}}_{ij}$, provingThis localized volume perturbation proves the requested density even when the two domains overlap, and works in every positive dimension. It does not by itself prove the stronger residual local-isometry statement: fixed isometric closures and arbitrary isometric open subsets are different conditions.

Use the unique curvature $-1$ hyperbolic metric on a closed Riemann surface of genus $g\ge 2$, supplied by the uniformization theorem. This is the implicit setting for the partition and spectral theorems; a genus-zero or genus-one surface does not have this hyperbolic metric.

A partition is a pants decomposition by disjoint essential simple closed geodesics, whose complement is a union of pairs of pants. A pair of pants is a sphere with three disks removed, furnished here with geodesic boundary. Its Euler characteristic is $-1$, so there must be $2g-2$ pants. Every pant has three boundaries and every cutting curve occurs twice, giving $3g-3$ cutting curves. The Bers pants decomposition theorem asserts that a constant $B_g$ depending only on $g$ bounds the lengths of all cuffs in some such pants decomposition of every closed genus-$g$ hyperbolic surface.

To see what geometry the cuffs determine, cut each pair of pants along its three perpendicular seams into two congruent right-angled hyperbolic hexagons. The three alternate sides are ${\ell }_1/2,{\ell }_2/2,{\ell }_3/2$, where the ${\ell }_i>0$ are its boundary lengths. The right-angled hyperbolic hexagon identity determines the seam $s_i$ opposite the half-cuff ${\ell }_i/2$ byWe use the standard existence and uniqueness theorem for a right-angled hyperbolic hexagon with prescribed positive alternating side lengths. Thus the three boundary lengths determine the pair of pants up to isometry.

Fix a topological pants decomposition, label its cuffs, and choose reference seam endpoints and orientations. Gluing two boundaries of the same length requires a translation along that boundary; its signed distance is a twist ${\tau }_i$. The gluing graph, the $3g-3$ positive cuff lengths, and the $3g-3$ twists therefore determine the unmarked hyperbolic surface up to isometry. A full boundary translation has period ${\ell }_i$, so for an unmarked gluing one may take $0\le {\tau }_i<{\ell }_i$, with endpoints identified. For a marked surface the whole real twist records the number of Dehn twists, and the Fenchel–Nielsen coordinates areThe topological gluing data are necessary when the decomposition is not fixed. Different markings or decomposition graphs can describe the same unmarked surface.

Teichmüller space ${\mathcal{T}}_g$ consists of pairs $(X,f)$, where $X$ is a closed genus-$g$ Riemann surface and $f:{\Sigma }_g\to X$ is an orientation-preserving marking, modulo the equivalence $(X,f)\sim (Y,h)$ when a conformal orientation-preserving map $a:X\to Y$ has $a\circ f$ homotopic to $h$. The Fenchel–Nielsen coordinates identify it with ${\mathbb{R}}_{>0}^{3g-3}\times {\mathbb{R}}^{3g-3}$, of real dimension $6g-6$. The moduli space of Riemann surfaces forgets the marking by quotienting by the mapping class group.

The Wolpert generic spectral rigidity theorem says that the locus of closed genus-$g$ hyperbolic surfaces possessing an isospectral but nonisometric partner lies in a locally real-analytic exceptional locus of lower dimension in Teichmüller space. Thus a generic surface is determined, up to isometry, by its unmarked length spectrum, equivalently by its Laplacian spectrum. Here an isometry may reverse orientation; the length spectrum cannot distinguish the two orientations of the same metric. The equivalence of the two spectra is the compact hyperbolic Selberg trace formula.

The simplifying result is the Buser finite length spectrum theorem: for every $g\ge 2$ and $\epsilon >0$ there is $L=L(g,\epsilon )$ such that two closed genus-$g$ hyperbolic surfaces with hyperbolic systoles at least $\epsilon$ have the same entire unmarked length spectrum if their length multisets up to $L$ agree. Multiplicities are included. We will count primitive unoriented closed geodesics; counting all iterates is equivalent by successively removing shorter iterates. The theorem is uniform on the thick part of moduli space, not merely a cutoff chosen separately for a particular pair.

Here is a proof using compactness and polynomial trace equations. We state the subsidiary facts and explain their role. First, Mumford's compactness theorem makes the genus-$g$ thick moduli space compact. It can also be seen from the Bers pants decomposition theorem: cuffs in a Bers decomposition lie in $[\epsilon ,B_g]$, there are finitely many pants decomposition graphs, and unmarked twists may be reduced modulo their cuff lengths. These data lie in finitely many compact boxes of Fenchel–Nielsen coordinates and cover the thick moduli space.

Second, these finite compact boxes give a compact family $K$ of marked hyperbolic structures on a fixed smooth surface ${\Sigma }_g$ representing every member of the thick moduli space. We use the standard smooth dependence of the glued metrics and their hyperbolic holonomy representations on Fenchel–Nielsen coordinates. Choose smooth representatives over finitely many parameter neighborhoods and conjugate the hyperbolic holonomy representation using a fixed lifted Riemannian orthonormal frame. Passing to finite compact subboxes gives a compact set of representations, together with a compact family of representative smooth metrics. Changing the markings between boxes causes no problem: all are markings from the same fixed ${\Sigma }_g$, and the union is finite.

Third, uniformization theorem identifies each marked hyperbolic surface with ${\mathbb{H}}^2/\rho ({\pi }_1{\Sigma }_g)$ for a faithful discrete hyperbolic holonomy representation into ${\mathrm{PSL}}_2(\mathbb{R})$. We use the standard lifting fact that a closed orientable hyperbolic holonomy representation admits a lift to ${\mathrm{SL}}_2(\mathbb{R})$, and lifts may be chosen continuously on small parameter neighborhoods. Choose lifts on finitely many such neighborhoods, allowing the finitely many possible sign choices. A representation is described by the matrices of $2g$ standard fundamental group generators. The polynomial encoding of hyperbolic geodesic lengths uses, for each fixed group word $w$, its squared matrix traceis a polynomial in their entries: products are polynomial, and the inverse of a determinant-one matrix is its polynomial adjugate. For a nontrivial hyperbolic word, the hyperbolic translation length obeysThus equality of the positive lengths is exactly equality of these trace polynomials, independent of lift signs.

Fourth, the length spectrum of a closed hyperbolic surface is locally finite with finite multiplicities. We need the uniform finiteness of short geodesic classes across the compact marked family. More uniformly, for each finite $b$, only finitely many unoriented primitive conjugacy classes can have length at most $b$ in any metric of $K$. To justify uniformity, compactness of the smooth metrics gives a common bilipschitz equivalence comparison with one fixed metric $h_0$ on ${\Sigma }_g$. If ${\ell }_h(\alpha )\le b$, the minimizing geodesic has $h_0$ length at most $Cb$, so ${\ell }_{h_0}(\alpha )\le Cb$. Local finiteness for $h_0$ then gives a finite list. Local finiteness itself follows from proper discontinuity of the cocompact group of the hyperbolic holonomy representation: conjugate a geodesic axis to meet a fixed compact fundamental set, and a bounded-length axis gives a group element moving that compact set a bounded distance, of which there are only finitely many.

Finally, the Hilbert basis theorem says that a polynomial ring in finitely many real variables is Noetherian. Consequently a decreasing sequence of sets cut out by polynomial equations eventually stabilizes. Indeed, their vanishing ideals form an increasing chain and stabilize. Finite unions of such sets are still algebraic: one can take all products of one defining polynomial from each component. This finiteness fact is the engine that turns arbitrarily many spectral comparisons into finitely many.

Enumerate the primitive unoriented conjugacy classes as ${\alpha }_1,{\alpha }_2,\dots$. For each $n$ letand let $F_n$ be the finite set of primitive classes that can have length at most $b_n$ anywhere in $K$. Define $Z_n$ in the finite-dimensional space of pairs of generator matrices as follows: it consists of pairs $(\rho ,\sigma )$ for which there exists an injectionand also an injection satisfying the analogous equations with $\rho$ and $\sigma$ interchanged. There are only finitely many injections. Each choice imposes finitely many polynomial equations; taking their finite unions and then intersecting the two directions proves that $Z_n$ is a real affine algebraic set. On determinant-one representations these equations are exactly the desired matching of lengths. Defining the polynomials on the ambient matrix-entry space is harmless; the argument will only apply them to representations in $K$.

Put $W_n=Z_1\cap \cdots \cap Z_n$. The sets $W_n$ form a decreasing chain of affine algebraic sets, hence for some $N$,For representations in $K$, membership in this intersection is equivalent to equality of the full length spectra. One direction follows by matching equal lengths, including their multiplicities. For the other, fix a length $\ell$. All primitive classes of length $\ell$ on the first surface occur in some finite initial segment. Its injective matching shows that the second surface has at least that multiplicity at $\ell$; the reverse injection gives the reverse inequality. Local finiteness makes both multiplicities finite. Doing this for every $\ell$ proves multiset equality.

TakeIf two surfaces in $K$ have equal length multisets up to $L$, each initial segment through $n\le N$ can be matched injectively on the other surface. Its matching classes lie in $F_n$, since the matched lengths are at most $b_n$. The reverse matching is available as well. Thus the pair belongs to $W_N$, hence to every $W_n$, and the preceding paragraph yieldsAll choices of the compact family were made using only $g$ and $\epsilon$, so the resulting cutoff has the asserted dependence. If the thick family is empty the assertion is vacuous. This proves Buser's finite length spectrum theorem without claiming that equal full length spectra always force isometry; exceptional isospectral pairs are compatible with the theorem and are precisely why Wolpert's theorem is a generic statement.

Use the natural empty-sum convention ${\mu }_1(0)=0$ for cumulative coherence. This is needed for the printed case $s=1$. For $x=0$ the conclusion is immediate. Otherwise let $S$ be its support of a vector, with $k=|S|\le s$. The Gram matrix $G=A_S^{\ast }{A}_S$ has diagonal entries one because the columns have unit Euclidean norm. Each off-diagonal row sum is bounded byThe last inequality uses monotonicity of cumulative coherence: enlarging an index set only adds nonnegative summands. There are enough indices to enlarge it because $s\le N$.

The Gershgorin circle theorem places every eigenvalue of $G$ within distance ${\mu }_1(s-1)$ of one. Since $G$ is a Hermitian matrix, its eigenvalues are real, and the finite-dimensional spectral theorem givesThis is the cumulative coherence bound for restricted isometry. The lower bound remains valid when ${\mu }_1(s-1)>1$, although it is then nonpositive. No assumption that the matrix is already a near isometry is required. The distortion is bounded by ${\mu }_1(s-1)$ on every order-$s$ sparse vector.

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.

Let $E=L^1({\mathbb{R}}_x^d\times {\mathbb{R}}_v^d)$, and denote the free-transport semigroup by $(U_{t}g)(x,v)=g(x-tv,v)$. For fixed $v$, the spatial translation has unit Jacobian determinant, so the Tonelli theorem givesThus the given free term is an isometry on $E$. The operators form a strongly continuous semigroup: continuity first holds for smooth compactly supported functions by dominated convergence, and density plus the isometry extends it to every $L^1$ function.

The Fubini's theorem and integral triangle inequality giveDefine the normalized velocity-reset collision operator using the normalized velocity-reset projection $Pg(x,v)=\rho (g)(x)M(v)$ and $B=P-I$. Since $M\ge 0$ and $\int M=1$,Also $P^2=P$; the collision gain replaces the velocity distribution by $M$ while preserving the spatial mass. In Bochner integral notation the printed, undamped source operator isUsing the transport isometry and the integral triangle inequality provesThese estimates hold for measurable, locally time-bounded $E$-valued functions. If the displayed supremum is infinite, the numerical bound is interpreted in the extended sense; the construction below works in a space where it is finite.

A spacetime is a smooth four-dimensional manifold equipped with a smooth Lorentzian metric, conventionally of metric signature $(-,+,+,+)$, and a choice of time orientation. The usual manifold assumptions include the Hausdorff space and second-countable space conditions. A physical model also specifies matter fields and requires the Einstein field equations and matter equations.

Diffeomorphism invariance of general relativity means that relabelling events by a smooth invertible map, while transforming the metric tensor and every matter field together, preserves the form of the equations. Passively, a coordinate change gives new components for the same geometric fields. Actively, pulling all fields back by a diffeomorphism gives another representative of the same physical geometry, subject to any prescribed boundary conditions or boundary symmetries.

The geometry and matter together carry the physical content; coordinate labels do not. A diffeomorphism need not be an isometry of a fixed metric, and unrelated metrics are not automatically physically equivalent.

All vector spaces in this solution are complex, while convex combinations use real coefficients. An extreme point $x$ of a convex set $C$ is one for which $x=ty+(1-t)z$, $y,z\in C$ and $0<t<1$, forces $y=z=x$. Equivalently, $x$ is not the midpoint of two distinct points of $C$.

The Krein-Milman theorem says that every nonempty compact convex set $K$ in a Hausdorff locally convex space is the closed convex hull of its extreme points. Here is a proof. A face of a convex set $F\subseteq K$ is a convex set such that whenever an interior point of a segment in $K$ belongs to $F$, both endpoints belong to $F$. Consider nonempty compact faces, ordered by reverse inclusion. A chain has nonempty intersection by compactness and the finite intersection property; that intersection is again a compact face. The Zorn lemma therefore gives a minimal nonempty compact face $F$.

If $F$ contains distinct $x,y$, a continuous real linear functional $\ell$ on the underlying real locally convex space distinguishes them. Such a linear functional exists because the space is Hausdorff and locally convex, by the Hahn-Banach theorem. The maximizers of $\ell$ on $F$ form a nonempty proper compact face of $F$. A face of a face is a face of $K$, contradicting minimality. Therefore $F$ is a singleton, yielding an extreme point. The same argument inside any nonempty compact face of $K$ supplies an extreme point of $K$ lying in that face.

Let $D$ be the closed convex hull of the extreme points of $K$. It is a nonempty closed subset of compact $K$, hence compact. If $x_0\in K\setminus D$, the Hahn-Banach separation theorem provides a continuous real linear functional $\ell$ with $\ell (x_0)>{\sup }_D\ell$. Its maximizer set on $K$ is a nonempty compact face, which contains an extreme point $e$ of $K$. But $e\in D$ and $\ell (e)\ge \ell (x_0)>{\sup }_D\ell$, a contradiction. Thus $\boxed{K=\overline{\mathrm{co}}(\mathrm{ext}K)}$, proving Krein-Milman theorem.

For a nonempty compact Hausdorff space $K$, the extreme points of the dual unit ball of C(K) areThis is the permitted description without proof, with $C(K)$ denoting the complex space of continuous functions on a compact space equipped with the supremum norm. If $K$ is empty, $C(K)=\{0\}$ and the closed unit ball of its continuous dual space has the single extreme point $0$ instead.

The complex Banach–Stone theorem states that a surjective complex-linear isometric isomorphism of normed spaces $T:C(K)\to C(L)$ has the formwhere $u\in C(L)$ and $\phi :L\to K$ is a homeomorphism. Conversely, every such map is a surjective complex-linear isometric isomorphism of normed spaces for the supremum norm.

To prove this, suppose first that $K,L$ are nonempty. The Banach-space adjoint $T^{\ast }$ is a bijective isometric isomorphism of normed spaces on the continuous dual spaces, so it bijects their closed unit balls and preserves extreme points. The displayed extreme point description gives uniquelyUniqueness follows by evaluating at the constant function $1$, and then using that continuous functions separate points of a compact Hausdorff space. Evaluation gives the required formula for $Tf$, and $u=T1$ is continuous. Surjectivity of $T^{\ast }$ on extreme points proves surjectivity of $\phi$: the preimage of any ${\delta }_x$ is $\alpha {\delta }_y$ for some $y$. If $\phi (y_1)=\phi (y_2)$, every function in the range of $T$ has equal values at these two points after division by $u$; surjectivity of $T$ and separation of points force $y_1=y_2$. Thus $\phi$ is bijective.

For every $f\in C(K)$, $f\circ \phi =(Tf)/u$ is continuous. The evaluation map $K\to {\mathbb{C}}^{C(K)}$, $x\mapsto (f(x){)}_f$, is a continuous injection of a compact Hausdorff space into a Hausdorff product topology, hence a homeomorphism onto its image. Continuity of every coordinate $f\circ \phi$ proves continuity of $\phi$. A continuous bijection between compact Hausdorff spaces is a homeomorphism. Conversely the weighted-composition formula plainly preserves the supremum norm, and its inverse isIf one compact space is empty, a surjective isometric isomorphism of normed spaces forces the other to be empty, and the empty homeomorphism gives the corresponding trivial case. This completes Banach–Stone theorem.

Neither the space of sequences converging to zero $c_0$ nor $L^1[0,1]$ can be isometrically a continuous dual space of a Banach space. Indeed, every nonzero continuous dual space has a nonempty weak-star compact closed unit ball by Banach-Alaoglu theorem, and Krein-Milman theorem then guarantees an extreme point. A bijective linear isometry preserves extreme points of closed unit balls.

The closed unit ball of $c_0$ has no extreme points. Given $x\in c_0$ with $\Vert x{\Vert }_{\infty }\le 1$, choose $n$ with $|x_n|<1$ and $0<\epsilon <1-|x_n|$. The two distinct elements $x\pm \epsilon e_n$ remain in that closed unit ball and have midpoint $x$.

The closed unit ball of the complex Lp space $L^1[0,1]$ also has no extreme points. An element of norm less than one can be perturbed by a sufficiently small nonzero Lp space element in both directions. If $\Vert f{\Vert }_1=1$, the non-atomic measure $|f(t)|dt$ admits a measurable set $A$ of mass $1/2$; for example the continuous function $s\mapsto {\int }_0^s|f(t)|dt$ attains $1/2$. Put $h=f(1_A-1_{A^c})$. For $0<\epsilon <1$, the distinct functions $f\pm \epsilon h$ both have Lp normwith the two coefficients interchanged for the minus sign. Their midpoint is $f$. Thus neither proposed space is isometrically a Banach dual.

Finally, $[0,1]$ is a connected space, whereas $[0,1]\cup [2,3]$ is disconnected. They cannot be homeomorphic. By Banach–Stone theorem, $C[0,1]$ and $C([0,1]\cup [2,3])$ are not isometrically isomorphic as complex Banach spaces.