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Use separate moving orthonormal bases in the reference and current cylindrical coordinate systems. For pure torsion of an incompressible cylinder, the deformation map isThe unit vectors depend on the angle: ${\partial }_{\theta }{\mathbf{e}}_r={\mathbf{e}}_{\theta }$ and ${\partial }_{\theta }{\mathbf{e}}_{\theta }=-{\mathbf{e}}_r$. ConsequentlyIn particular, the last derivative includes the rotation of ${\mathbf{e}}_r$ along the cylinder. The material gradient has the reference coordinate scale factor $1/R$:The outer products specify the images of the three reference unit vectors. Thus, with current basis vectors indexing rows and reference basis vectors indexing columns,Its determinant is one, checking incompressibility. The radius is unchanged, so the current outer radius in part (d) is $a=A$. The formula at the coordinate axis is understood by continuity in Cartesian coordinates.

For an active clockwise rotation of column vectors in the usual oriented $xy$ plane, takeDirect matrix multiplication givesThus the rotation matrix is an orthogonal matrix, preserving the Euclidean inner product, with determinant $+1$.

For a passive one-dimensional Lorentz boost, use rapidity $\eta$, $\beta =\tanh \eta$:This is not $I$ unless $\eta =0$. Hence a nontrivial Lorentz boost is not an orthogonal matrix for the Euclidean inner product. Instead it preserves the Minkowski metric:by ${\cosh }^2\eta -{\sinh }^2\eta =1$. Thus it is pseudo-orthogonal. The zero-rapidity identity is the explicit exception to the assertion that a Lorentz boost is not orthogonal.

The supplied matrix obeysSeparate the even and odd powers in the matrix exponential to obtainIt fixes the $z$ component and is a counterclockwise active rotation viewed from $+z$. This sign differs from the clockwise convention in b, because the source fixes the sign of both $J_3$ and the exponential. Replacing $\theta$ by $-\theta$ yields the clockwise rotation.

With $(J_k{)}_{ij}=-i{ϵ}_{kij}$, the generators of rotations obeyThe commutator measures the leading difference between performing two infinitesimal rotations in opposite orders. For example, with $X=-i\alpha J_1$ and $Y=-i\beta J_2$,The failure to commute is itself an infinitesimal rotation about the perpendicular axis; the Lie algebra closes on the same three generators of rotations.

To pass from this local statement to the full group property, put $A_k=-i{J}_k$. These are real antisymmetric matrices, with $[A_i,A_j]={ϵ}_{ijk}{A}_k$. The Baker--Campbell--Hausdorff formula shows locally that the logarithm of a product of their exponentials remains a linear combination of the $A_k$, because all nested commutators do. Globally, each exponential has $R^{T}R=I$ and $\det R=e^{\mathrm{tr}A}=1$. If $R_1,R_2$ have these properties, thenEvery real three-dimensional orthogonal matrix of determinant $+1$ has an eigenvector with eigenvalue $+1$: its nonreal eigenvalues come in conjugate pairs, and the product of all three is $+1$. On the perpendicular plane it acts as a two-dimensional rotation. It therefore has axis-angle form and can be written $e^{-i\theta {\hat{n}}^k{J}_k}$. Thus the product is again a rotation. The identity is $\theta =0$, the inverse reverses $\theta$, and associativity comes from matrix multiplication. This proves that all these rotations form the special orthogonal group $SO(3)$. The commutator supplies local closure, while the orthogonality and axis-angle argument supplies the necessary global step.

Let $K_i$ generate Lorentz boosts with the same exponential convention as the rotation generators of rotations $J_i$. The Lie algebra of the Lorentz group isConsequently non-collinear Lorentz boosts do not close among themselves: the product contains a rotation, the Wigner rotation. For an explicit counterexample, in coordinates $(ct,x,y)$ write $c_a=\cosh a$, $s_a=\sinh a$ and similarly for $b$. ThenA pure rotationless Lorentz boost has a symmetric matrix, whereas this product has spatial entries $(xy)=0$ and $(yx)=s_a{s}_b\ne 0$ for $a,b\ne 0$. It is therefore a concrete product of two pure Lorentz boosts that is not a pure Lorentz boost. Collinear Lorentz boosts are the special case that do form a one-parameter group, with additive rapidities.

Allowing all spatial directions and closure generates both the $K_i$ and, through their commutators, all $J_i$. The smallest connected group containing all these Lorentz boosts is the proper orthochronous Lorentz groupIt contains arbitrary compositions of spatial rotations and Lorentz boosts continuously connected to the identity. It has six continuous parameters. Here “smallest” refers to the subgroup generated by all continuous pure Lorentz boosts, not to an arbitrary collection already including discrete reversals.

The full Lorentz group $O(1,3)$ has four connected components, distinguished by the sign of the determinant and the direction of time. Representatives are the identity, parity $P=\mathrm{diag}(1,-1,-1,-1)$, time reversal $\mathsf{T}=\mathrm{diag}(-1,1,1,1)$, and $P\mathsf{T}=-I$. ThusMultiplication by these discrete transformations relates the components algebraically; no continuous path within the Lorentz group joins different components. Translations are not required for closure here; adding them gives the larger Poincaré group.

A locally invertible time-independent canonical transformation is canonical whenThe proposed coordinates are locally independent only for $q_1\ne 0$, since their configuration-space Jacobian has determinant $2q_1$. Work on one sign branch of $q_1$.

The mixed brackets first requireThe last bracket condition isExpressing $A,B$ in the $Q$ coordinates turns this into ${\partial }_{Q_2}A={\partial }_{Q_1}B$. Thus their shift one-form is closed and locally is the differential of an arbitrary smooth function $S(Q_1,Q_2)$. The canonical momenta for a squared coordinate therefore have the general local formConversely, any such $S$ satisfies all bracket conditions. This does not give one nonsingular transformation through $q_1=0$, or identify the two sign branches globally.

Choose $S=Q_2^3/3$. Then

Subtracting the two linearized equations eliminates their acceleration terms:For $g,k>0$, the assumed nontrivial simple harmonic motion therefore hasThe support displacement and pendulum angle oscillate in phase. Inserting this relation into either Euler-Lagrange equation yieldsso the one positive finite angular frequency isThe negative frequency describes the same real oscillation after a phase change. For a general phase, replace the sine in both coordinates by $A\cos (\omega t)+B\sin (\omega t)$.

There is no missing second finite branch. The quadratic Lagrangian iswhose kinetic matrix is $\left(\begin{array}{cc}1& 1\\ 1& 1\end{array}\right)$ and has rank one. Its characteristic determinant is ${\omega }_0^2{\omega }_1^2-({\omega }_0^2+{\omega }_1^2){\omega }^2$, not a quadratic polynomial in ${\omega }^2$. This singular kinetic matrix in small oscillations results from taking the support mass to be exactly zero. Adding a small support mass restores a second mode that runs to infinite frequency as that mass vanishes.

For a stiff spring,The hinge becomes fixed, recovering the simple pendulum frequency, as expected for the pendulum with a massless spring-mounted support. At finite $k$, horizontal support motion makes the frequency smaller.

For independent $\chi ,{\chi }^{\ast }$, the derivatives needed in the ${\chi }^{\ast }$ Euler-Lagrange field equation areConsequently the two conjugate field equations areWrite $\chi ={\chi }_1+i{\chi }_2$ with real components, and put $A={\hslash }^2/(2M)$. Separating real and imaginary parts givesFor each real field take the real part of a complex plane wave amplitude, ${\chi }_j=\mathrm{Re}[q_j{e}^{i(\mathbf{k}\cdot \mathbf{r}-\omega t)}]$. With ${ϵ}_k={\hslash }^2|\mathbf{k}{|}^2/(2M)$ the two equations becomeA nonzero normal mode requires their determinant to vanish, soThe positive-frequency Bogoliubov quasiparticle dispersion is thereforeThere is also the negative-frequency conjugate solution, as required for real component fields. Since $\mu >0$, ${\omega }^2\ge 0$, consistent with the stability of the chosen uniform ground state.

Let ${\delta }_0={\theta }_0-{\phi }_0$ be $0$ or $\pi$, and write the angular displacements as ${\eta }_1,{\eta }_2$. To quadratic order the restoring potential energy is $\kappa \cos {\delta }_0({\eta }_1-{\eta }_2{)}^2/2$. The linearized normal mode equation isThe determinant gives ${\omega }^2[{\omega }^2{I}_1{I}_2-\kappa \cos {\delta }_0(I_1+I_2)]=0$. Thus the normal modes of two spring-coupled coaxial rotors areFor the aligned family, the relative mode oscillates with positive frequencyFor the antipodal family, ${\omega }_{\mathrm{rel}}^2=-{\Omega }^2$, so that displacement has solutions $e^{\pm \Omega t}$ and the equilibrium is unstable.

The common zero mode has ${\eta }_1={\eta }_2=C+Dt$. It describes a neutral displacement or uniform common rotation, not a restoring oscillation. Thus the aligned family is stable with respect to relative motion and has orbital stability modulo common rotations. A particular static point is not strictly Lyapunov stable against unrestricted small velocity perturbations: arbitrarily small nonzero $J$ eventually moves the common angle far from that point. With $J=0$, the common angle is fixed and the aligned relative equilibrium point is stable. This distinguishes the symmetry-induced drift from the exponential instability of the antipodal family.

Measure intermediate time $u$ from $t^{\prime }$. The classical path with the fixed endpoints isIt satisfies $m{\ddot{x}}_c=-\alpha$. Direct integration of $m{\dot{x}}_c^2/2-\alpha x_c$ givesFor an arbitrary path write $x=x_c+\eta$ with $\eta (0)=\eta (T)=0$. Integrating the cross term by parts gives $-\int (m{\ddot{x}}_c+\alpha )\eta du=0$. Thus $S[x]=S_c+\int m{\dot{\eta }}^{2}du/2$ exactly. The fluctuation path integral is the free endpoint-fixed one, so its determinant supplies $\sqrt{m/(2\pi i\hslash T)}$. Multiplication by $\theta (T)e^{i{S}_c/\hslash }$ reproduces the previous kernel. This exact classical-path factorization of a quadratic path integral is possible because a linear potential introduces no additional quadratic fluctuation term.

The three-point density correlation of an ideal quantum gas follows first for a number-conserving Gaussian quantum state. There are six pairings, one for each permutation of three annihilation operators. The identity gives the product of densities; each transposition has sign $ϵ$; both three-cycles have positive sign. ThusThe last two terms are complex conjugates. They equal $2\mathrm{Re}(g_{12}{g}_{23}{g}_{31})$, not an additional signed pair term. This is the permanent of a matrix $(g_{ij})$ for bosons and its determinant for fermions. Wick's theorem supplies the six terms only with the state assumption stated in part (a).

For completeness an exact bosonic Fock-state alternative can be given without that assumption. Let $\mathbf{\ell }$ be a mode-occupation decrement with ${\sum }_{\alpha }{\ell }_{\alpha }=3$. Define $A_{\mathbf{\ell }}(1,2,3)$ as the sum of ${\prod }_{j=1}^3{\phi }_{{\alpha }_j}({\mathbf{r}}_j)$ over all distinct ordered triples having these mode multiplicities. Acting three annihilation fields on the number state leaves mutually orthogonal states for different $\mathbf{\ell }$. ThereforeHere $N^{\underline{\ell }}=N!/(N-\ell )!$, with zero when $\ell >N$. This expression includes repeated-mode occupation corrections absent from the Gaussian permanent; for a single mode it reduces to $N(N-1)(N-2){\prod }_j|\phi ({\mathbf{r}}_j){|}^2$.

At coincident points the Gaussian expression has $g_{ij}=n$. Its six contributions giveWhen all mutual separations become large and the density stays the same nonzero $n$, only $n_1{n}_2{n}_3=n^3$ remains. ConsequentlyThe bosonic value is $3!$ from Gaussian exchange pairings; the fermionic determinant vanishes when its rows coincide. Density-normalized correlations avoid the additional homogeneity assumption. For a specified bosonic number product, part (c)'s occupation sum must replace the six-pairing formula, so neither the universal factor six nor a universal printed ratio follows from that specification alone.

The literal printed difference is zero, because the same spin-index product appears twice, and zero is invariant under every transformation. The intended SU(2)-invariant antisymmetric spinor contraction uses opposite indices in the second term. With $\epsilon =\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)$, define $I={\varphi }^T\epsilon \chi$. For any two-by-two matrix,Thus for $U\in SU(2)$,Expanding the determinant identity proves the result without complex conjugating either spinor. This is the invariant alternating tensor behind the spin singlet state. If the two spinors are identical, the contraction vanishes. An antisymmetric internal two-boson state also needs an antisymmetric additional degree of freedom for overall bosonic symmetry; it does not lie in the fully symmetric fixed-two-mode sector discussed above.

With $J=\mathrm{diag}(1,-1)$ these matrices obey $M^{†}JM=J$ and $\det M=1$, identifying the group as SU(1,1). Products preserve the displayed conjugate-entry pattern and determinant; the identity belongs, andalso belongs. Multiplication and inversion are smooth. The real constraint $|\alpha {|}^2-|\beta {|}^2=1$ has nonzero gradient on its level set, so the matrices form a smooth real three-dimensional manifold. Associativity is inherited from matrix multiplication. These facts establish a Lie group, which is noncompact because $|\beta |$ is unbounded.

The defining condition isThe entries are complex and the matrices are invertible; products and inverses retain determinant one. There is no unitarity condition $S^{†}S=I$ in general. Four complex entries subject to one complex determinant equation give three complex, or six real, local group parameters.

The association between four real coordinates and Hermitian two-by-two matrices is a real-linear bijection. Congruence by $S$ preserves Hermiticity, so $X^{\prime }=SX{S}^{†}$ defines another real four-vector depending linearly on the original. Its determinant isThus the induced invertible linear map preserves the Minkowski quadratic form:The maps obtained continuously from the identity are proper and orthochronous. For example, a future timelike vector gives a positive-definite $X$, which remains positive definite under invertible congruence; its time direction is preserved. The matrices $S$ and $-S$ induce the same four-vector transformation, illustrating the Lorentz spinor double cover.

The source uses $x+iy$ in the upper-right entry, corresponding to $X=ctI+x{\sigma }_x-y{\sigma }_y+z{\sigma }_z$ for the standard Pauli matrices. This is a permissible basis convention, but its minus sign in the $y$ basis must also be retained in the spinor identification.

Assume the composite pure state has unit norm. Put its amplitudes into the matrixThe partial trace contracts the same $B$ index between ket and bra:ThusIt is Hermitian, positive semidefinite and has trace one. Its determinant is particularly useful:This follows from multiplicativity of the determinant, or by expanding the two-by-two expression. If the supplied amplitudes were not normalized, the physical reduced state would be $C{C}^{†}/\mathrm{tr}(C{C}^{†})$ instead.

At equilibrium $v=u/(u+K)$ and adding $u^{\prime }+ϵv^{\prime }$ gives $0=\eta u-\lambda v$. The positive equilibrium isThis condition also gives $\eta <1$. The nullclines are $v=u/(u+K)$ and $v=(1-\eta )u/(u+K-\lambda )$. The second lies above the first for $0<u<u_e$ and below it for $u>u_e$. Below the first nullcline $v^{\prime }>0$; above it $v^{\prime }<0$. Below the second nullcline $u^{\prime }<0$; above it $u^{\prime }>0$.

At the origin the Jacobian has trace $\eta -1-K/ϵ<0$ and determinant $(\lambda -\eta K)/ϵ>0$. Its off-diagonal entries are positive, making the eigenvalues real, so this is an attracting node. At the positive equilibrium the slow reduced derivative is $\eta -\lambda K/(u_e+K{)}^2=\eta v_e>0$, while the fast derivative is $-(u_e+K)/ϵ<0$; equivalently the full Jacobian determinant is negative. Thus the positive point is a saddle equilibrium. Its stable manifold separates extinction from unbounded substrate growth.

The reaction Jacobian isIts trace is $-1/2$ and determinant is $1$, so both eigenvalues have negative real parts. Explicitly they are $(-1\pm i\sqrt{15})/4$. Therefore the homogeneous reaction equilibrium is a stable focus in the absence of diffusion.

Put $q=k_n^2$. The mode matrix has trace $-1/2-(1+D)q<0$, so a growing eigenvalue occurs exactly when its determinant is negative:This upward quadratic has two positive roots only if $D<1/2$ and $D^2-5D+1/4>0$. Within $0<D<1$ this isFor such $D$, defineThe unstable modes satisfy $\boxed{\sqrt{q_{-}}<k_n<\sqrt{q_{+}}}$. These are strict inequalities: at either endpoint the leading eigenvalue is zero rather than positive.

For the modal matrix $M(s)$, $s\ge 0$, its trace is always negative:Its determinant isThe proposed inequality is exactly $\beta \kappa \eta <\alpha \kappa +1$. It makes the coefficient of $s$ positive, while the leading and constant coefficients are already positive. Therefore $D(s)>0$ for every $s\ge 0$, independent of which discrete wave numbers the domain allows.

A real two-by-two matrix with negative trace and positive determinant has both eigenvalues with negative real part: if they are real they have positive product and negative sum, and if nonreal their common real part is half the trace. Every mode therefore decays, proving linear stability for all domain sizes under the stated sufficient condition.