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Use the material-first nominal stress tensor convention $N_{\alpha i}$, whose reference-volume power is $N_{\alpha i}{\dot{A}}_{i\alpha }$. At a regular constraint point, $F_{,A}\ne 0$, all admissible deformation rates satisfy $F_{,A_{i\alpha }}{\dot{A}}_{i\alpha }=0$. Equality of mechanical power and the derivative of the strain energy density givesfor every vector in this eight-dimensional tangent hyperplane. Its orthogonal complement is the span of $F_{,A}$. Consequently the constraint reaction in hyperelastic stress isThe scalar Lagrange multiplier $q$ is not determined by the work identity: its contribution vanishes on every allowed rate. Regularity matters. For example, replacing a regular incompressibility constraint by $(\det A-1{)}^2=0$ would give zero gradient on the same admissible set and could not represent an arbitrary pressure reaction through that gradient.

For incompressibility use the regular function $F=J-1$, with $J=\det A$. Differentiating the alternating-tensor expression for $J$ givesThe first expression is the cofactor entry; contracting it with $A_{j\alpha }$ gives $J{\delta }_{ij}$, proving the inverse formula. On $J=1$ this yieldsWith this plus-sign convention $q$ is the negative of the usual pressure multiplier.

For an equibiaxially stretched sheet, incompressibility requires the principal stretches $(\lambda ,\lambda ,{\lambda }^{-2})$. The printed sheet expression's ${\lambda }^{-1/2}$ is inconsistent with this constraint and must be ${\lambda }^{-2}$; the balloon expression later on the same PDF page has the correct exponent. Let $W_j$ denote the derivative with respect to principal stretch ${\lambda }_j$, evaluated on this constrained path. With no applied thickness-direction traction,Isotropy gives $W_1=W_2$ at equal in-plane stretches. HenceThe equibiaxial nominal tension of an incompressible sheet is thereforeThe two equal in-plane nominal tensions do work $2N_{11}d\lambda$ per reference volume; the thickness-direction nominal stress does no work because it is zero. That is exactly the constrained energy differential, explaining energy conservation. As a check that the printed exponent is a genuine error, a neo-Hookean solid gives $N=\mu (\lambda -{\lambda }^{-5})$, zero at the undeformed state. The derivative with the erroneous exponent would instead give $3\mu /4$ at that state.

For the thin spherical balloon, use the leading reference shell volume $4\pi a_0^2{h}_0$. Its two tangential stretches are $\lambda =a/a_0$, while its thickness stretch is ${\lambda }^{-2}$ and its current thickness is $h=h_0{\lambda }^{-2}$. Define $\hat{W}(\lambda )=W(\lambda ,\lambda ,{\lambda }^{-2})$. Its total elastic energy at leading thin-shell order isDuring a quasistatic increment $da$, the enclosed volume changes by $d\mathcal{V}=4\pi a^{2}da$, and $d\lambda =da/a_0$. Equating energy change to work of the internal excess pressure givesThis is the inflation pressure of a thin incompressible elastic balloon. The approximation neglects through-thickness variation and curvature corrections beyond leading order, and requires $h/a\ll 1$. It also agrees with membrane force balance: $p=2h{\sigma }_t/a$, ${\sigma }_t=\lambda N$, and $N={\hat{W}}^{\prime }/2$. Here pressure is force per current area, while $N$ is force per reference area; confusing them would lose a factor of $\lambda$.

For a Hermitian operator $A$, every eigenvalue is real. Indeed, if $Av=\lambda v$ with $v\ne 0$, self-adjointness gives $\lambda ⟨v,v⟩=⟨Av,v⟩=⟨v,Av⟩=\overline{\lambda }⟨v,v⟩$. To prove the finite-dimensional spectral theorem, proceed by induction on the dimension. The fundamental theorem of algebra supplies a root of the characteristic polynomial, hence a unit eigenvector $v$. Its orthogonal complement is invariant: for $w\perp v$, $⟨Aw,v⟩=⟨w,Av⟩=0$. The restriction of $A$ to that complement is again Hermitian. Induction supplies an orthonormal basis of eigenvectors there; adjoining $v$ gives an orthonormal eigenbasis of the whole space. The zero-dimensional case is immediate. Consequently a Hermitian operator is diagonal in an orthonormal basis, with real diagonal entries.

Interpret $\theta$ as a continuous finite-dimensional complex group representation, so its values lie in ${\mathrm{GL}}_{\mathbb{C}}(V)$, and interpret approximation in the uniform norm. Assume the usual Hausdorff convention for the compact topological group. The following argument proves the required form of the Peter-Weyl theorem without first assuming that $G$ is a Lie group.

Start with normalized Haar measure. It is both left- and right-invariant on a compact group. Average any Hermitian inner product to make a finite-dimensional representation unitary. The left regular action $L_{a}h(x)=h(a^{-1}x)$ is already unitary on $L^2(G)$.

Choose a continuous nonnegative function $k$ supported in a small symmetric identity neighbourhood, with $k(z^{-1})=k(z)$ and $\int k=1$. Such functions are obtained from a nonzero continuous bump near the identity, symmetrizing and normalizing. DefineThe convolution operator $T_k$ commutes with left translations. Its continuous kernel is Hermitian, hence it is self-adjoint. It is compact on $L^2(G)$: approximate the continuous kernel uniformly by finite sums of separated functions of $x$ and $y$, using the Stone-Weierstrass theorem on the compact product. The corresponding operators have finite rank and converge in operator norm. It also maps into $C(G)$ and satisfiesby the Cauchy-Schwarz inequality; continuity follows from the uniform continuity of the translated kernel.

The spectral theorem for compact self-adjoint operators decomposes the orthogonal complement of its kernel into finite-dimensional eigenspaces with nonzero eigenvalues. Every such eigenspace is left-translation invariant, and all its elements are continuous, since $h=T_{k}h/\lambda$ there. If $W\subset C(G)$ is one of these invariant spaces, write $\rho (a)=L_a{|}_W$. Evaluation at the identity givesThus $h$ is a matrix coefficient of the dual representation. This identifies each finite spectral sum with finite-dimensional representation data.

To get uniform rather than just $L^2$ approximation, let $S_N$ project onto increasing finite sums of the nonzero eigenspaces. For $f\in C(G)$, $T_{k}f$ is orthogonal to the kernel, so $S_N{T}_{k}f\to T_{k}f$ in $L^2$. Applying the displayed smoothing bound givesMeanwhile, choosing the support of $k$ sufficiently small makes $T_{k}f$ uniformly close to $f$, by uniform continuity under small translations. Since $\Vert T_k{\Vert }_{C\to C}\le 1$, it also makes $T_k^{2}f$ uniformly close to $f$. Every $T_k{S}_N{T}_{k}f$ belongs to a finite sum of invariant eigenspaces, hence is a finite sum of matrix coefficients. This completes the convolution proof of uniform Peter-Weyl approximation.

A coefficient $\ell (\theta (g)v)$ equals $\mathrm{tr}(\alpha \theta (g))$ for the rank-one endomorphism $\alpha =v\otimes \ell$. Finite sums of coefficients can be combined using a direct-sum representation and a block-diagonal $\alpha$. Therefore the conclusion has exactly the requested form:

For a class function $f$, apply conjugation averaging on a compact group,It fixes $f$ and is a contraction in the uniform norm, so averaging an approximation preserves its error bound. A trace coefficient averages to $\mathrm{tr}(\overline{\alpha }\theta (g))$, whereThis endomorphism commutes with the representation. A finite-dimensional unitary representation splits into irreducibles by repeatedly taking invariant orthogonal complements. On each irreducible diagonal block, Schur lemma makes its average a scalar multiple of the identity, namely its trace divided by the block dimension. Off-diagonal blocks do not contribute to the total trace. Thus the averaged coefficient is a finite linear combination of irreducible characters. Consequently irreducible complex characters span a uniformly dense subspace of the continuous class functions.

Conjugating the coordinate reflection by the rotations supplies reflections in the other two coordinates. The quarter-turns, together with those reflections, also supply coordinate transpositions. Consequently the represented group consists of all signed permutation matrices, the full symmetry group of a cube of order $2^3\cdot 3!=48$.

Let $C$ be any real or complex matrix commuting with the group representation. Commutation with the three independent coordinate reflections forces all off-diagonal entries of $C$ to vanish. Commutation with the quarter-turn about the first axis equates its second and third diagonal entries, and the quarter-turn about the second axis equates its first and third entries. ThereforeFor completeness, if the complexified group representation had a proper invariant subspace, its orthogonal complement would be invariant too, since the generating matrices are unitary. The orthogonal projection onto that subspace would be a nonscalar commuting matrix, contradicting the calculation. Thus the complexification is irreducible, proving absolute irreducibility of the real group representation.

First $E=\ker (A+I)$ is a linear subspace. To prove it is nonzero, use orthogonality and $\det A=-1$:Hence $\det (A+I)=0$. If $u\in E$ then $Au=-u$ also implies $A^{T}u=-u$. For $x\perp E$,so $E^{\perp }$ is invariant. If $E$ had dimension two, its one-dimensional orthogonal complement would be invariant and $A$ would act there as a scalar $\eta =\pm 1$. The determinant would be $(-1{)}^2\eta =-1$, so $\eta =-1$, forcing $A=-I$. Dimension three also forces $A=-I$. Both contradict the hypothesis. ConsequentlyChoose an orthonormal basis beginning with a unit vector in $E$ and followed by a basis of $\Pi =E^{\perp }$. In this basis the orthogonal matrix has block formThe first column of $R$ is some $(\cos \theta ,\sin \theta {)}^T$. Its perpendicular unit second column with positive orientation must be $(-\sin \theta ,\cos \theta {)}^T$. Thus the restriction is the plane rotationTrace invariance gives $\mathrm{tr}A=-1+2\cos \theta$, provingA Euclidean reflection in $\Pi$ fixes that plane pointwise and reverses $E$, so it occurs exactly when $R=I_2$. This is equivalent to $\cos \theta =1$, hence exactly to $\mathrm{tr}A=1$. Finally,This exhibits the three-dimensional improper orthogonal transformation as a rotation about the axis $E$ composed with Euclidean reflection in $\Pi$.

For a smooth function $f$ and the Levi-Civita connection $\nabla$, the Riemannian Hessian is the covariant two-tensorThe connection rules show that this is linear over smooth functions in both vector-field arguments, so it is a tensor. Torsion-freeness gives its symmetry:Its metric trace defines the paper's Laplace-Beltrami operator, $\Delta ={\mathrm{div}}_g{\mathrm{grad}}_g$; its eigenvalues on a closed manifold are nonpositive. We use $P=-\Delta$ when listing nonnegative eigenvalues or writing $e^{-tP}$.

At $p$, choose an orthonormal basis $e_1,\dots ,e_d$ of the tangent space and unit-speed geodesics ${\gamma }_i(t)={\exp }_p(t{e}_i)$. Differentiating twice along a curve givesThe covariant acceleration vanishes on a geodesic. Taking the trace at $t=0$ proves the geodesic trace formula for the Laplace-Beltrami operator:The last equality is Taylor's formula along the geodesics, and expresses the Laplacian entirely through nearby function values.

For stabilization, let $X_1,X_2$ be the assumed isospectral nonisometric four-dimensional tori, and let $k=n-4>0$. Take the same small flat torus $F_{\epsilon }={\mathbb{R}}^k/\epsilon {\mathbb{Z}}^k$ as an extra factor. On a Riemannian product,If ${\varphi }_j$ and ${\psi }_{\ell }$ are eigenfunctions with eigenvalues ${\lambda }_j$ and ${\mu }_{\ell }$, then ${\varphi }_j{\psi }_{\ell }$ has eigenvalue ${\lambda }_j+{\mu }_{\ell }$. Products of orthonormal eigenbases are complete in the product $L^2$ space, so this lists all eigenvalues with multiplicities, including coincidences between different sums. Consequently the two products are isospectral.

One must also prove that multiplying by a common factor has not accidentally made the two manifolds isometric. ChooseThe injectivity radii are positive because the factors are compact. A closed product geodesic of length less than $\delta$ projects to a closed geodesic of length less than $\delta$ in $X_i$. Such a projected geodesic must be constant: otherwise its initial vector over one period would be a nonzero vector of norm below the injectivity radius mapping back to its starting point under the exponential map. Thus all sufficiently short closed geodesics lie in the $F_{\epsilon }$ directions. Conversely the $k$ coordinate circles, of length $\epsilon$, pass through every point and span those directions.

It follows that the smooth distribution tangent to the small flat-torus factor is characterized intrinsically by the span of tangent vectors to closed geodesics of length below $\delta$. A hypothetical product isometry would preserve this smooth distribution and its orthogonal complement. The latter has leaves $X_i\times \{q\}$; the isometry and its inverse take whole leaves to whole leaves, thereby giving an isometry $X_1\cong X_2$, a contradiction. This proves isospectral stabilization by a small flat torus:This argument works even without assuming that the original four-dimensional torus metrics are flat.

Write $f_x$ for the linear isometry specified at $x$. Choose a cofinal sequence $V_i$ from the indexing family $I$, and write the target part of the supplied subordinate flag for a family of linear isometries as $W_i$. After a cofinal enlargement it hasBoth flags exhaust their universe for spectra. Compactness of the parameter space ensures that a finite target stage can contain all images of each fixed source stage.

Let ${\xi }_i$ be the vector bundle on $X$ with fiberHere $\ominus$ is the orthogonal complement in the indicated ambient subspace. Its rank is $\dim W_i-\dim V_i$. Let $\mathrm{Th}({\xi }_i)$ be its Thom space, the disk bundle modulo its sphere bundle, so that every fiber is compactified and all its points at infinity become the single basepoint. For the zero bundle this convention gives $\mathrm{Th}(0_X)=X_{+}$. Define a target indexed prespectrum on the flag byThe smash product here is of based spaces, not yet the derived smash product of spectra.

To specify all structure maps, for $i\le j$ use the continuous fiberwise orthogonal isomorphismBoth sides identify with the complement of $f_x(V_i)$ in $W_j$. In particular this is the isometry obtained by orthogonally projecting onto $f_x(V_j\ominus V_i)$ and its complement; it is not an assertion that the displayed summands coincide separately. Pass to fiberwise one-point compactifications, identify the last summand with $V_j\ominus V_i$ using $f_x^{-1}$, and apply the source structure map. This definesMore explicitly, decompose $z+w\in W_j\ominus f_x(V_i)$, where $z\in W_j\ominus W_i$ and $w\in ({\xi }_i{)}_x$, as $w^{\prime }+f_x(v)$ with $w^{\prime }\in ({\xi }_j{)}_x$ and $v\in V_j\ominus V_i$. Send its pair with $e\in E(V_i)$ to the pair $(w^{\prime },{\sigma }_{V_i,V_j}(v,e))$ over $x$. The notation for finite vectors extends to the compactifications, sending infinities and basepoints to the basepoint. Orthogonal decomposition and associativity of the source structure maps prove the prespectrum identities.

Extend the indexed prespectrum from the cofinal flag to the target indexing family and apply spectrification $L$. The point-set definition isOn a map of topological spectra $u:E\to E^{\prime }$ it is induced at stage $i$ by $1_{\mathrm{Th}({\xi }_i)}\wedge u(V_i)$. Thus the prescribed isometry family, indexing family and flag have all been used. A common cofinal refinement identifies the resulting objects in the stable homotopy category; the point-set model retains the chosen flag. If $f$ is a constant standard inclusion, the complement bundles are trivial and this recovers the usual reindexed $X_{+}\wedge E$.

The associativity of twisted half-smash products follows from an actual isomorphism of the complement bundles. First work over compact parameter spaces, so that a common choice of flags can be made. Choose source, intermediate and target stages $V_i,W_i,Z_i$ satisfyingFor the two twisted half-smash products and their composite the relevant vector bundles have fibersPull the first two bundles back to $Y\times X$. Since $g_y$ is a linear isometry, the decomposition of $Z_i$ into $g_y(W_i)$ and its orthogonal complement gives the fiberwise vector bundle isomorphismIts inverse is given by the two orthogonal projections. The isomorphism varies continuously with $(y,x)$, and the ranks agree:The one-point compactification of an external orthogonal direct sum gives the smash product of its Thom spaces. Using $g_y$ to identify $g_y({\xi }_x)$ with ${\xi }_x$, the isomorphism therefore givesThe left side is a stage of $(Y\times X){⋉}_{h}E$ and the right side is the corresponding two-step prespectrum presentation of $Y{⋉}_g(X{⋉}_{f}E)$.

These stage comparisons commute with structure maps. Indeed, increasing a stage decomposes a suspension vector into its component in $g_y{f}_x(V_j\ominus V_i)$ and the two residual complement components. In either construction the same source component is pulled back to $V_j\ominus V_i$ and used in the same structure map of $E$; the residual components are left in the same final complement. Iterated orthogonal decomposition gives the same result whether the intermediate complement is separated first or last. They also commute with maps of $E$, since the comparison only rearranges the suspension coordinates.

For clarity about the intervening spectrification, maps out of $LP$ into a genuine indexed spectrum $T$ are precisely compatible prespectrum maps $P\to T$. At stage $i$, adjunction from the Thom coordinate identifies such a map with a continuous familywhere the last identification is the genuine indexed spectrum structure of $T$. Compatibility between stages is exactly compatibility of this family with the source structure maps. Thus maps out of the twisted construction are continuous families of maps over the specified linear isometries. A family of such maps over $g$ out of $X{⋉}_{f}E$ is, by applying this universal property a second time, exactly a family of maps out of $E$ over $g_y{f}_x$. This is the same orthogonal-coordinate correspondence just constructed; it respects the identifications imposed by each spectrification. Consequently it descends from the prespectrum presentations to the associated spectra. Common cofinal refinements remove any incompatibility between the independently chosen flags.

The printed hypothesis does not require $Y$ to be compact. For an arbitrary compactly generated parameter space $Y$, use the same construction on compact test spaces mapping into $Y$. Each such family admits the common flags above, and the comparisons agree under restriction because their formula is the canonical orthogonal projection. They therefore assemble into the continuous global comparison in the compactly generated point-set construction. Equivalently, the preceding universal-property correspondence is valid for arbitrary $Y$ and does not impose a uniform finite target stage on all of it. Passing to cellular models for the derived construction gives the requested natural isomorphism

Let $\lambda =({\lambda }_1\ge {\lambda }_2\ge \cdots )$ be a partition of an integer $n$. Its Young diagram consists of cells $(i,j)$ with $1\le j\le {\lambda }_i$, and its conjugate partition has parts ${\lambda }_j^{\prime }=\#\{i:{\lambda }_i\ge j\}$. A numbered Young tableau fills these cells bijectively with $1,\dots ,n$. Its row and column stabilizers $R_t,C_t$ permute entries within each row and each column respectively. A tabloid $\{t\}$ remembers the sets in the ordered rows, forgetting their internal order. The Young subgroup $R_t\cong {\prod }_i{S}_{{\lambda }_i}$ is its stabilizer, so the Young permutation module isHere the group acts by permuting numbered entries and extending linearly. A submodule is an invariant linear subspace. Give the tabloid basis its orthonormal symmetric bilinear form, denoted $⟨,⟩$. This form is nondegenerate and invariant under the group. Define the column antisymmetrizer and polytabloid byThus $S^{\lambda }$ is the Specht module, generated by any one $e_t$, because the group is transitive on numbered Young tableaux. The orthogonal complement $(S^{\lambda }{)}^{\perp }$ consists of vectors pairing to zero with every polytabloid.

The key calculation isTo prove it, first take $v=\{u\}$. If two entries of one column of $t$ occur in one row of $u$, their transposition $s$ fixes $\{u\}$; pairing the terms indexed by $g$ and $gs$ cancels $b_t\{u\}$. This also works in characteristic two, where the paired equal terms have sum zero. Otherwise each row of $u$ contains at most one entry from each column of $t$. The first row has ${\lambda }_1$ entries and there are ${\lambda }_1$ columns, so it uses every column once. Removing those entries, the second row must use exactly the columns of remaining positive height, and induction does the same for all rows. Consequently there is a unique $h\in C_t$ with $\{u\}=\{ht\}$. Uniqueness follows from $R_t\cap C_t=1$. In that caseLinearity proves the identity.

Now let $U\le M^{\lambda }$ be a submodule. If $U$ is not contained in $(S^{\lambda }{)}^{\perp }$, choose $v\in U$ and $t$ with $⟨v,e_t⟩\ne 0$. The displayed identity puts $e_t$ in $U$, so its group translates put all of $S^{\lambda }$ in $U$. This proves the James submodule theorem without a semisimplicity assumption.

Put $R=S^{\lambda }\cap (S^{\lambda }{)}^{\perp }$. If $R\ne S^{\lambda }$, every proper submodule of $S^{\lambda }$ lies in $R$, by the same theorem. Hence $R$ is its unique maximal submodule, and $D^{\lambda }=S^{\lambda }/R$ is irreducible. If $R=S^{\lambda }$, the quotient is zero. This conclusion remains true after every field extension $E/F$: the integral tabloid construction becomes $M_E^{\lambda }=E{\otimes }_F{M}^{\lambda }$, and the radical of the restricted form is the kernel of its Gram matrix, so it commutes with extension of scalars. A nonzero Gram entry stays nonzero in $E$. Applying the theorem over $E$ makes $E{\otimes }_F{D}^{\lambda }$ irreducible. Thus a nonzero quotient is absolutely irreducible.

The precise nonvanishing criterion isA regular partition has fewer than $p$ rows of each positive length. Here is the integral pairing argument behind this condition. Write $m_j$ for the number of rows of length $j$. Expanding two polytabloids into tabloids, simultaneous permutations of equal-length rows partition of an integer all contributions to their pairing into free orbits of size ${\prod }_j{m}_j!$; the signs agree within each orbit because the same row permutation is applied in every relevant column on both sides. Therefore every pairing is divisible by ${\prod }_j{m}_j!$. If some $m_j\ge p$, the entire restricted form is zero modulo $p$. Conversely, reverse the order of the entries in each row of a tableau to obtain $t^{\ast }$. Counting the matching column permutations in the two expansions gives, up to the harmless overall sign,Indeed the $m_j$ equal rows of length $j$ can be matched independently in each of their $j$ positions; after these choices the decreasing row lengths force every other matching. For a regular partition this integer is nonzero modulo $p$, so the form and its radical quotient are nonzero. In characteristic zero it is nonzero for every shape. This supplies the requested justification of the criterion.

Apply the Schmidt decomposition to the bipartite pure state. Local unitary operators put it in the form $|{\psi }_{\theta }⟩=\cos \theta |00⟩+\sin \theta |11⟩$, with $0<\theta \le \pi /4$ after interchanging basis vectors. Both Schmidt coefficients are nonzero because the state is entangled. Put $s=\sin 2\theta >0$.

For this state, direct application of the Pauli operators givesFor example ${\sigma }_x\otimes {\sigma }_x$ interchanges $|00⟩$ and $|11⟩$, giving $2\cos \theta \sin \theta$; the crossed products take these two kets into their orthogonal complement.

Choose the CHSH axes for an entangled pure two-qubit state:Their associated Bloch vectors have unit length, so each is a valid two-outcome projective measurement with projectors $(I\pm A_i)/2$ or $(I\pm B_j)/2$. The correlation combination isA local hidden-variable assignment with outcomes $\pm 1$ gives absolute value at most two: exactly one of $b_0+b_1$ and $b_0-b_1$ vanishes, while the other is $\pm 2$. Averaging these assignments preserves that bound, which is the CHSH inequality.

Therefore $\boxed{\mathcal{S}=2\sqrt{1+{\sin }^{2}2\theta }>2}$. For the original state $(U\otimes V)|{\psi }_{\theta }⟩$, conjugate Alice's observables by $U$ and Bob's by $V$. These conjugations rotate their measurement axes and preserve all the displayed correlations. Thus every entangled pure two-qubit state has the requested violating axes. The maximally entangled case gives $2\sqrt{2}$, while the unentangled limit merely saturates two.

Sylvester's law of inertia says that any real quadratic form has coordinates in which it is a sum of $p$ positive squares minus $m$ negative squares, with $z$ unused coordinates, and that the numbers $(p,m,z)$ are invariant under every invertible linear change of coordinates. Its rank is $p+m$; its signature of a quadratic form is commonly $p-m$, with $(p,m)$ also called its signature pair.

The symmetric matrix here is $J-I$, where $J=11^T$ and $1=(1,\dots ,1{)}^T$. On the line spanned by $1$, $J-I$ has eigenvalue $n-1$. If ${\sum }_i{x}_i=0$, then $Jx=0$, so that the orthogonal complement has eigenvalue $-1$ with multiplicity $n-1$. Thus the off-diagonal all-ones quadratic form hasFor $n=1$ the form vanishes, so rank and scalar signature are both zero, with inertia $(0,0,1)$. Including this case is necessary because the distinguished eigenvalue then becomes zero.

Use a strictly positive bank account as numéraire, and write all prices and payoffs in discounted units. There are $d$ risky assets with deterministic initial price vector $S_0$, terminal price vector $S_1\in L^2(P)$, and gains $Y=S_1-S_0$. A self-financing portfolio with initial capital $x$ and constant risky holdings $\theta$ has discounted terminal wealth $x+{\theta }^{T}Y$. The initial amount in the bank account is $x-{\theta }^T{S}_0$, so $x$ denotes total initial capital, not just the cash holding. Assume the usual absence of arbitrage, and hence existence of an equivalent martingale measure; wherever a general probability space is used below, explicitly assume that such a measure exists and makes the asset payoffs integrable.

The attainable claims are precisely the linear subspaceThe finite-dimensional subspace is closed, so every square-integrable contingent claim $H$ has a unique orthogonal projection onto $\mathcal{A}$. This is its optimal one-period least-squares hedging payoff. Holdings are unique after redundant assets have been removed. Putand suppose the remaining covariance matrix $C$ is invertible. Expanding the squared error givesFirst minimize over $x$, then complete the square in $\theta$. ThusThe residual $L=H-x_{\ast }-{\theta }_{\ast }^{T}Y$ satisfiesIndeed, every other squared error equals this minimum plusThis proves optimality, and also shows that zero error is equivalent to attainability of a European contingent claim. If initial capital $x$ is prescribed instead, the normal equations for linear least squares give $\mathbb{E}[Y{Y}^T]\theta =\mathbb{E}[Y(H-x)]$; one must not use the free-capital formula without this adjustment.

A dominated martingale measure is a probability measure $Q\ll P$ such that ${\mathbb{E}}_{Q}Y=0$. Its Radon-Nikodym derivative $Z$ satisfies $Z\ge 0$, ${\mathbb{E}}_{P}Z=1$, and ${\mathbb{E}}_P[ZY]=0$. An equivalent martingale measure additionally requires $Z>0$ almost surely. Every integrable attainable claim $H=x+{\theta }^{T}Y$ then has price $x={\mathbb{E}}_{Q}H$, independent of which such measure is chosen. For an unattainable contingent claim, different dominated martingale measures can give different prices; least-squares approximation chooses a projection criterion rather than exact replication.

Write $Y=m+M$, where $M=Y-m$ is the centered martingale part of the one-period gains. The minimal martingale measure in a one-period market changes the means of the traded gains to zero while preserving all square-integrable mean-zero directions $L$ orthogonal to $M$. Its candidate density isDirect computation gives ${\mathbb{E}}_P{Z}_{\ast }=1$ and ${\mathbb{E}}_P[Z_{\ast }Y]=m-C{C}^{-1}m=0$. If ${\mathbb{E}}_{P}L=0$ and ${\mathbb{E}}_P[ML]=0$, then ${\mathbb{E}}_P[Z_{\ast }L]=0$, as required. Conversely, an $L^2$ density with this preservation property is orthogonal to the orthogonal complement of $\mathrm{span}\{1,M_1,\dots ,M_d\}$, so it lies in that span. Its normalization and martingale constraints force exactly $Z_{\ast }$. Moreover,Every other square-integrable signed pricing density has the form $Z_{\ast }+D$, with $D$ orthogonal to $1$ and $Y$, hence to $Z_{\ast }$. Thereforeso the minimal density is also the minimum-norm one-period pricing weight. This coincidence concerns one period; it is not an unrestricted assertion about dynamic hedging models.

Positivity needs checking. If $Z_{\ast }>0$ it defines an equivalent martingale measure; if merely $Z_{\ast }\ge 0$, it defines a dominated martingale measure. Otherwise it defines only a signed martingale measure. For example, take $S_0=2$ and $Y=-1,1,3$ with physical probabilities $1/100,89/100,10/100$. Then $m=59/50$, $C=1019/2500$, and $Z_{\ast }=-4350/1019$ in the third state. There is nonetheless an equivalent martingale measure assigning probabilities $3/5,3/10,1/10$, whose mean gain is zero. Thus absence of arbitrage does not guarantee positivity of the least-squares density, and a least-squares initial cost need not be an arbitrage-free price for an unattainable payoff.

Finally prove completeness and uniqueness of dominated martingale measures. Suppose first that every bounded contingent claim is attainable. For any event $A$, replicate $1_A=x_A+{\theta }_A^{T}Y$. Under any dominated martingale measure, $Q(A)=x_A$. Thus all such measures agree on every event. Existence follows from our no-arbitrage assumption, giving uniqueness.

Conversely, fix an equivalent martingale measure $Q_0$, and let $k=\dim \mathcal{A}$. If the probability space admits $k+1$ disjoint events $A_1,\dots ,A_{k+1}$ of positive probability, choose a basis $a_1,\dots ,a_k$ of $\mathcal{A}$, including $1$. The $k\times (k+1)$ matrix with entries ${\mathbb{E}}_{Q_0}[a_i{1}_{A_j}]$ has a nonzero null vector $(c_j)$. Consequently the bounded, nonzero random variablesatisfies ${\mathbb{E}}_{Q_0}h=0$ and ${\mathbb{E}}_{Q_0}[hY]=0$. For $0<|\epsilon |<1/\Vert h{\Vert }_{\infty }$, defineThese are distinct equivalent martingale measures, because their densities are positive and their gain moments remain zero. Thus uniqueness rules out such a partition. A probability space with no $k+1$ disjoint positive-probability events consists, modulo null sets, of at most $k$ atoms: repeatedly split any non-atomic positive event to obtain a larger finite partition. On $N$ atoms the payoff space has dimension $N$, while $\mathcal{A}$ has dimension $k\le N$. Hence $N=k$ and $\mathcal{A}$ is the whole payoff space. Every contingent claim is attainable. This provesHere completeness means all bounded claims, or all square-integrable claims in the $L^2$ formulation. The existence/no-arbitrage hypothesis matters: on a one-state space, cash and a stock with a sure positive discounted gain span every payoff, but no dominated martingale measure can make that gain have zero mean.

Use the metric $\eta =\mathrm{diag}(1,-1,-1,-1)$ and first consider the connected Poincaré group, ${\mathbb{R}}^{1,3}⋊S{O}^{+}(1,3)$. Physical half-integer spins require unitary representations of its double cover ${\mathbb{R}}^{1,3}⋊SL(2,\mathbb{C})$; genuine representations of the group without this cover permit only integer spin. Write $P^{\mu }=d(P^{\mu })$ and $M^{\mu \nu }=d(M^{\mu \nu })$. Strong continuity gives self-adjoint generators, with commutators understood on a common invariant dense domain.

The commuting translation generators have a joint spectral resolution. Lorentz transformations carry the spectral four-momentum $p$ to $\Lambda p$, so an irreducible representation has its momentum spectral measure concentrated on a single Lorentz orbit. For positive-energy particles the orbits of interest are the future mass shells $p^2=m^2>0$, or $p^2=0$ with $p\ne 0$ and $p^0>0$. The zero-momentum orbit and spacelike or negative-energy orbits are outside these two particle cases.

Choose ${ϵ}^{0123}=1$ and define the Pauli-Lubanski pseudovector with the overall sign conventionThis sign makes the spatial rest-frame vector equal to $m\mathbf{J}$. The generators' Hermiticity and the antisymmetric epsilon contraction make $W^{\mu }$ Hermitian: the ordering correction from commuting $P$ through $M$ contracts metric tensors into epsilon and vanishes. The same antisymmetry proves Pauli-Lubanski orthogonality, $W^{\mu }{P}_{\mu }=0$.

The assumed commutators say that $P^{\mu }$ and $W^{\mu }$ transform as four-vectors and that $W$ commutes with all translations. Consequentlycommute with the entire Poincare algebra. For instance the two terms in the Lorentz commutator with a contracted vector square cancel. The Casimir operators therefore have fixed spectral values in a unitary irreducible representation, by the Schur lemma applied to their spectral projections. A second ingredient, the little group, distinguishes the internal states.

Fix a standard momentum $k$ on the orbit. Its little group is the subgroup of Lorentz transformations leaving $k$ fixed. Choose $L(p)$ with $L(p)k=p$. A Lorentz transformation acts on the internal fibre by the Wigner rotationIf $d_K$ is a unitary irreducible representation of this stabilizer, the induced representation acts on wavefunctions asThe measure $d^{3}p/(2p^0)$ on either future mass shell is Lorentz invariant, so this action is unitary. The identityproves the group multiplication law. Conversely, covariance of the translation spectral resolution supplies precisely these fibres and their stabilizer action. A commuting operator acts fibrewise; transitivity relates its fibre values, and irreducibility of $d_K$ makes it scalar. This explains why the construction gives, and classifies, the unitary irreducible representations on each orbit.

For a timelike momentum choose $k=(m,0,0,0)$. A Lorentz transformation fixing it acts only as a rotation of its spatial orthogonal complement, so the massive particle little group is $SO(3)$, or $SU(2)$ in the double cover. PutThe Poincare algebra gives $[J_i,J_j]=i{ϵ}_{ijk}{J}_k$. The unitary irreducible representations of $SU(2)$ are labelled by $j=0,\frac{1}{2},1,\dots$, with ${\mathbf{J}}^2=j(j+1)$ and $J_3$ eigenvalues $-j,-j+1,\dots ,j$. Evaluation of the Pauli-Lubanski pseudovector at $k$ gives $W^0=0$, $\mathbf{W}=m\mathbf{J}$. Thus the massive Pauli-Lubanski Casimir and particle labels areThe full massive induced representation of the Poincare double cover is infinite-dimensional because momentum ranges continuously over the orbit; its finite internal multiplicity is $2j+1$.