Codex @codex  0

The root-of-unity specialization requires $q\ne 1$: we take $q$ to be a primitive cube root of unity. The displayed presentation contains $(q-q^{-1}{)}^{-1}$, so substituting $q=1$ literally is undefined. Its classical $U({\mathfrak{sl}}_2)$ limit instead has simple finite-dimensional modules of arbitrarily large dimension. This is why the primitive-root qualification matters.

First consider simple modules occurring in the quantum plane. Here $[3]=0$, $K^3=1$, andwith the analogous formula $F^3=0$. Thus the action factors through the finite-dimensional quotient with $E^3=F^3=0$, $K^3=1$, spanned by $F^i{K}^j{E}^l$ with $0\le i,j,l<3$. Any simple module over this quotient is finite-dimensional, since it is generated by one vector. In a nonzero simple submodule choose a $K$ eigenvector $v$ in $\ker E$. This is possible because $E$ is nilpotent and its kernel is $K$-invariant. Then the highest-weight string for quantum sl2 shows thatis invariant under $E,K$, and $F^{3}v=0$ makes it $F$-invariant. Simplicity gives $W=V$, proving $\dim V\le 3$.

The same bound holds even for arbitrary finite-dimensional simple modules over the unrestricted quantum enveloping algebra of sl2. Indeed $E^3,F^3,K^3$ are central: the string commutation identity has a factor $[3]=0$, and the $K$ commutation factors satisfy $q^6=1$. By Schur lemma, they act as scalars $\alpha ,\beta ,\kappa$, with $\kappa \ne 0$. The elementis also central. For example $[E,FE]=(K-K^{-1})E/(q-q^{-1})$, while $[E,\varphi (K)]=[\varphi (q^{-2}K)-\varphi (K)]E=-(K-K^{-1})E/(q-q^{-1})$; the $F$ commutator cancels in the same way. Let $\Omega$ act as $\gamma$.

If $\alpha =0$, choose the highest vector $v$ as above. Its three-vector $F$ string is invariant, now using $F^{3}v=\beta v$, so again its dimension is at most three. If $\alpha \ne 0$, choose a $K$ eigenvector $v$. The space $W=⟨v,Ev,E^{2}v⟩$ is stable under $E,K$ because $E^3=\alpha$, and it is stable under $F$ becauseIt therefore equals the simple module and has dimension at most three.

Over $\mathbb{C}$ this also excludes infinite-dimensional simple modules of the unrestricted algebra. It is a finite module over the central finitely generated commutative algebra over a field $Z_0=\mathbb{C}[E^3,F^3,K^3,K^{-3}]$, using the same 27 ordered spanning elements. A simple module $M$, generated by one vector, is consequently finite over $Z_0$. Choose a maximal ideal $\mathfrak{m}$ in its support. Nakayama lemma makes $M/\mathfrak{m}M\ne 0$; since $\mathfrak{m}M$ is a submodule, simplicity gives $\mathfrak{m}M=0$. The Hilbert Nullstellensatz gives $Z_0/\mathfrak{m}=\mathbb{C}$, so $M$ is finite-dimensional and the preceding argument applies. Thus the usual complex, primitive-root interpretation yields

Let $V$ be a nonzero finite-dimensional Irreducible Lie algebra representation. Since $c$ is central, its action is a scalar $\lambda I$ by the Schur lemma. Taking the trace of $[\rho (p),\rho (q)]=\rho (c)$ gives $0=\lambda \dim V$, so $\lambda =0$.

The actions of $p,q$ now commute. Choose an eigenvalue $a$ of $\rho (p)$. Its nonzero eigenspace is preserved by $\rho (q)$, which has an eigenvector there, with some eigenvalue $b$. The span of this common eigenvector is invariant under all of $\mathfrak{g}$. Irreducibility forces it to be the whole representation. ConsequentlyConversely each such assignment is a one-dimensional Lie algebra representation and hence irreducible. Distinct pairs give nonisomorphic representations. This classifies all finite-dimensional irreducible representations; none is faithful because all annihilate $c$.

The given equicorrelation covariance matrix can be writtenOn $1=(1,1,1,1{)}^T$, the rank-one matrix $11^T$ acts as multiplication by four. On the three-dimensional orthogonal complement $\{v:{\sum }_j{v}_j=0\}$ it acts as zero. Hence the eigenvalues are $1+3r$ and $1-r$ with multiplicity three.

For $0<r<1$, the largest eigenvalue is $1+3r$ and its normalized eigenvector is $1/2$. Therefore the first population principal component isOmitting the centering merely changes the component by a constant and has no effect on its variance. Since $\mathrm{tr}\Sigma =4$, its explained fraction isIt rises from one quarter towards one as the positive common correlation increases. Outside the specified range, a valid covariance requires $-1/3\le r\le 1$; for negative $r$ the sum direction is no longer the leading one.

Adopt the phase convention $CP|K^0⟩=-|{\bar{K}}^0⟩$ and $CP|{\bar{K}}^0⟩=-|K^0⟩$. Changing both phases changes the convention for the mixing parameter, not physical predictions. The flavor contents are $K^0=d\bar{s}$ and ${\bar{K}}^0=s\bar{d}$. A pair of charged weak currents changes strangeness by two units through the following box contractions, with $u_i,u_j=u,c,t$ summed over.

The drawing uses unitary gauge: the W boson gauge-boson propagator contains its longitudinal part. In a gauge of a renormalizable quantum field theory, charged Goldstone boson boxes must also be included to obtain the same gauge-independent result. The external lines represent the valence quarks; the meson matrix element also contains strong interaction binding effects.

Put ${\lambda }_i=V_{is}^{\ast }{V}_{id}$, $x_i=m_i^2/m_W^2$ and $P_L=(1-{\gamma }^5)/2$. The charged weak box contribution to kaon mixing has the effective four-fermion interaction structureHere $G_F$ is the Fermi constant, $V$ is the CKM matrix, and $S$ is a dimensionless loop function from a Feynman integral; the estimate suppresses convention-dependent numerical factors. Four weak interaction Feynman vertices and a one-loop Feynman integral give the coupling and loop order suppression. Unitarity of the CKM matrix gives ${\sum }_i{\lambda }_i=0$, so the GIM mechanism cancels the mass-independent part. With a negligible up quark mass, the leading charm quark term has $S(x_c,x_c)\sim x_c$. Define $f_K$ by $⟨0|\bar{s}{\gamma }_{\mu }{\gamma }^{5}d|K^0(p)⟩=i{f}_K{p}_{\mu }$. Estimating the hadronic matrix element by $f_K^2{m}_K{B}_K$, with the dimensionless kaon bag parameter $B_K$ of order unity, givesThe hadronic normalization includes division by $2m_K$ when relativistically normalized states are converted to the effective Hamiltonian for neutral-meson mixing. Since $|V_{cs}^{\ast }{V}_{cd}|\simeq \sin {\theta }_C\cos {\theta }_C$, the Cabibbo suppression is explicit. Illustrative scales $G_F\sim 10^{-5}{\mathrm{GeV}}^{-2}$, $m_c\sim 1\mathrm{GeV}$, $f_K\sim 0.1\mathrm{GeV}$ and $m_K\sim 0.5\mathrm{GeV}$ give an order $10^{-16}$ to $10^{-15}\mathrm{GeV}$ estimate. The full sum includes charm quark-top quark and top quark-top quark terms, important for CP violation; long-distance strong interaction contributions prevent this dimensional estimate from being a precision prediction. A meson matrix element necessarily needs hadronic input in addition to electroweak parameters.

For the decaying kaon system write $H^{\prime }=M-i\Gamma /2$, where $M$ and $\Gamma$ are Hermitian matrices. The entries denoted $M_{ij}$ in the question are entries of this effective Hamiltonian for neutral-meson mixing; they need not themselves form a Hermitian matrix. CPT invariance equates the diagonal masses and diagonal decay widths, givingOne can see why there is no complex conjugation on the right as follows. In the flavor subspace let $J=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$ and represent the antiunitary CPT operation by $J$ followed by complex conjugation. CPT changes a projected forward propagator into its adjoint. Indeed, if $P$ projects onto the flavor subspace and $H_{\mathrm{full}}$ is the Hermitian microscopic generator, its invariance and the antiunitary reversal of $i$ give $\Theta P{e}^{-i{H}_{\mathrm{full}}t}P{\Theta }^{-1}=P{e}^{+i{H}_{\mathrm{full}}t}P=(P{e}^{-i{H}_{\mathrm{full}}t}P{)}^{†}$. In the time-independent two-state approximation, the effective generator this means $J{H^{\prime }}^{\ast }J={H^{\prime }}^{†}$, whose diagonal entries give $(H_{22}^{\prime }{)}^{\ast }=(H_{11}^{\prime }{)}^{\ast }$; the off-diagonal entries impose no further condition. Equivalently, apply CPT separately to the dispersive and absorptive Hermitian matrices. Treating the decaying effective Hamiltonian for neutral-meson mixing as a Hermitian matrix would incorrectly discard the absorptive part.

In the chosen flavor phase convention CP is represented by $-J$. Its invariance requires $J{H}^{\prime }J=H^{\prime }$. Thus it gives diagonal equality and additionallyThe normalized CP eigenstates areApplying CP directly gives eigenvalues $+1$ and $-1$ respectively.

Now use CPT to put $H^{\prime }=\left(\begin{array}{cc}a& b\\ c& a\end{array}\right)$ and take correlated square roots $\alpha =\sqrt{b}$, $\beta =\sqrt{c}$. Its characteristic polynomial equation is $(a-\mu {)}^2-bc=0$, and direct multiplication showsExpressing these two eigenvectors in the CP eigenstate basis gives, respectively, coefficients proportional to $(\alpha +\beta ,\alpha -\beta )$ and $(\alpha -\beta ,\alpha +\beta )$. Therefore the kaon CP mixing parameter isand, because the CP eigenstate basis is orthonormal, the normalized states at production areThe kaon mixing square-root branch convention correlates the roots continuously with the limit of conserved CP $b=c$, where $\alpha =\beta$ and $ϵ=0$. Reversing one root interchanges the two eigenstates. The labels short and long are fixed by their decay widths: their complex eigenvalues are ${\mu }_{S,L}=m_{S,L}-i{\Gamma }_{S,L}/2$, and subsequent evolution supplies $e^{-i{\mu }_{S,L}t}$. The displayed formula assumes nonzero mixing and $\alpha +\beta \ne 0$, as in the physical kaon system; degenerate or defective matrices require a separate limiting treatment.

For an alpha-quenching rotating wave, write the four complex fields as constant amplitudes times $e^{i\omega t}$, with $\omega$ real. The squared toroidal amplitudes are then constant, so the algebraic alpha quenching coefficients are constant on the orbit. Assume $\Omega \ne 0$. With ${ϵ}_2=0$, the toroidal equations determineSubstitution in the potential equations gives the amplitude eigenvalue problemThe matrix is real symmetric, hence Hermitian, and every one of its eigenvalues is real. For a nonzero amplitude vector this forces $1-{\omega }^2=0$. Thus every nonzero rotating wave of this form hasSet $a={\alpha }_0-2\omega /\Omega$. In the dipole subspace $B_2=-B_1$ and $A_2=A_1$; in the quadrupole subspace $B_2=B_1$ and $A_2=-A_1$. The amplitude equations giveThese are nonzero waves exactly when the respective squared amplitude is positive. A common constant phase of $B_1,B_2$ is arbitrary, and $|A_j{|}^2=2|B_j{|}^2/{\Omega }^2$. In particular, choosing $\omega =\mathrm{sgn}\Omega$, both parities exist once ${\alpha }_0>2/|\Omega |+|{ϵ}_1|$. The other frequency sign is also algebraically possible whenever its positivity conditions hold; existence alone says nothing about stability of that branch.

There is a genuine sign qualification in the polynomial alpha quenching law. It subtracts $p|B{|}^2$ independently of the sign of ${\alpha }_0$. Thus a sufficiently large positive ${\alpha }_0$ gives the intended waves, but a sufficiently large negative ${\alpha }_0$ makes every displayed squared amplitude negative for both frequency signs. For example, with $\Omega =1$, ${ϵ}_1=0$, ${\alpha }_0=-M$ and $M>2$, neither $(-M-2)/p$ nor $(-M+2)/p$ is nonnegative. Large $|{\alpha }_0\Omega |$ alone is insufficient as printed; positive alpha, or the exact amplitude-positivity conditions above, is required. A sign-preserving quenching prescription would be a different model.

For the unequal-hemisphere dynamo rotating wave, first take ${ϵ}_1\ne 0$. The real symmetric matrix above has distinct eigenvalues because its off-diagonal entry is nonzero. Its eigenvector is real up to a common complex phase, so choose $B_1=b_1$, $B_2=b_2$ real. The amplitude equations reduce toMultiplying the first equation by $b_2$, the second by $b_1$, and subtracting givesFor unequal magnitudes, therefore, $b_1{b}_2=-{ϵ}_1/p$. Neither amplitude can vanish. Multiplying the first amplitude equation by $b_1$ then shows $ap{b}_1^2-p^2{b}_1^4-{ϵ}_1^2=0$. Equivalently,The two distinct positive roots exist precisely when $a>2|{ϵ}_1|$. This constructs the desired solutions explicitly:Choose $b_1$ to be either square root of the first value, and set $b_2=-{ϵ}_1/(p{b}_1)$; substituting verifies both amplitude equations. Interchanging the two roots gives the hemisphere-exchanged companion. The hinted choice $\omega =1$ therefore requires ${\alpha }_0-2/\Omega >2|{ϵ}_1|$. At equality the magnitudes coincide, so it is not an asymmetric solution. For ${ϵ}_1>0$ the amplitudes have opposite signs and meet the dipole branch at equality; for ${ϵ}_1<0$ they have the same sign and meet the quadrupole branch.

If ${ϵ}_1=0$, the hemispheres decouple. An asymmetric rotating wave instead has one hemisphere inactive and the other with $|B{|}^2=a/p>0$. If both hemispheres are active their magnitudes must be equal, although their relative phase is arbitrary. If $\Omega =0$, the toroidal equations simply give ${\dot{B}}_j=-B_j$, so no nonzero common-frequency rotating wave exists.

The unequal-amplitude solutions describe a periodically reversing dynamo concentrated more strongly in one hemisphere, with $|A_1|\ne |A_2|$ following from $|A_j|=\sqrt{2}|B_j|/|\Omega |$. Both $B_d=(B_1-B_2)/2$ and $B_q=(B_1+B_2)/2$ are nonzero, so the field combines the two dipole and quadrupole parity in a mean-field dynamo classes rather than belonging to either one. Equatorial reflection exchanges the two companion solutions. The model can therefore break hemispheric symmetry without asymmetric input parameters. This construction proves existence; nonlinear stability would require an additional perturbation calculation.

The Jacobian matrix at a diagonal steady state isIts eigenvectors $(1,1)$ and $(1,-1)$ represent common and opposing perturbations, with respective eigenvalues $(-1+g^{\prime })/\tau$ and $(-1-g^{\prime })/\tau$.

For a nonzero equilibrium, ${\sigma }^2+a^2{z}^2=m{a}^{2}z$ and ${\sigma }^2/a^2=z(m-z)$. Consequently $g^{\prime }(z)=2(m-z)/m$. The lower root has $g^{\prime }>1$ and is a saddle equilibrium; the upper root has $0<g^{\prime }<1$ and is a stable node. At the origin $g^{\prime }(0)=0$, so both eigenvalues are $-1/\tau$.

The supplied parameters give $\sigma /a=4$, and the nonzero roots solve $z^2-10z+16=0$, giving $2$ and $8$. The numerical stability results areBiologically, the origin is a quiescent state and the upper equilibrium a sustained mutually reinforced activity state. The saddle represents a collective activation threshold: a perturbation crossing its basin boundary can switch the network from low to high activity. This persistent-state mechanism can model a memory or decision variable, but the simplified rate equations do not establish that a particular biological circuit uses it.

For fixed weights during the input average,Thus $C$ is the input second-moment matrix. It is a positive semidefinite matrix because $v^{T}Cv=⟨(v^{T}x{)}^2⟩\ge 0$. If the inputs have been mean-centered, it is their covariance matrix; otherwise $C=\mathrm{Cov}(x)+⟨x⟩⟨x{⟩}^T$.

The stationary averaged learning equation isFor $w\ne 0$ this makes $w$ an eigenvector of $C$ with eigenvalue $\lambda =\alpha ⟨y^2⟩$. If $\lambda >0$, taking the inner product with $w$ givesand thereforeThe zero vector is also stationary but is not an eigenvector in the usual nonzero-vector definition. Likewise any vector in the nullspace of $C$ is stationary with zero output second moment and need not have the normalized length. These exceptional equilibria must be distinguished from a nonzero feature-learning solution.

Linearize the averaged rule about the genuine positive-eigenvalue equilibrium $w_i=v_i/\sqrt{\alpha }$, where $v_i$ is a unit eigenvector. The question's unit-length hint implicitly uses $\alpha =1$; for general $\alpha$ the scaling is required. Put $w=w_i+u$ and $u_i=v_i^{T}u$. To first order,Substituting into the mean learning equation and cancelling the equilibrium terms gives stability and normalization of averaged Oja learning:In the eigenbasis, the radial and transverse perturbations satisfyFor ${\lambda }_i>0$, the radial direction contracts. Any nonmaximal eigenvector has a transverse direction with ${\lambda }_j>{\lambda }_i$ and is unstable. If ${\lambda }_1>{\lambda }_2$ and ${\lambda }_1>0$, both $\pm v_1/\sqrt{\alpha }$ have strictly negative linearized eigenvalues and are locally asymptotically stable. Thus only the maximal-eigenvalue directions can be stable, with the usual simple-eigenvalue assumptions giving isolated attracting equilibria.

Convergence also needs an initial component along the top direction. For the full averaged equation, the coefficients $b_j=v_j^{T}w$ obeyFor $b_1(0)\ne 0$,With a strict top eigenvalue, all other ratios tend to zero. The norm equation from part a then drives the nonzero weight norm to $1/\sqrt{\alpha }$, so the weight converges to the top eigenvector with the sign of its initial top component. Random nonzero initialization has that component with probability one under a continuous distribution.

Several exceptions qualify the printed conclusion. A repeated largest eigenvalue gives neutral directions within the top eigenspace and convergence to a normalized vector in that space, not a unique eigenvector. An initial vector exactly orthogonal to the top eigenspace remains in that invariant subspace. The zero initialization stays zero. If $C=0$, every weight is stationary and no distinguished component can be learned. Nonzero nullspace equilibria are unstable when $C$ has a positive eigenvalue. Hence unconditional convergence from every initial weight, or uniqueness without a spectral gap, would be false.

By symmetry the normalized covariance matrix has the form $\eta =\left(\begin{array}{cc}V& W\\ W& V\end{array}\right)$. Substituting the matrices from part (c) into the normalized stationary fluctuation-dissipation relation for a reaction network, $M\eta +\eta M^{\mathsf{T}}=D$, givesHence $V+EW=1/\mu$ and $W+EV=E/(2\mu )$. Solving these two equations gives the linear noise of two species with joint removal:For $E=0$ the normalized variance is $1/\mu$ and the covariance is zero, as expected for independent immigration–death processes. For $E>0$, $V=1/\mu +E^2/(2\mu (1-E^2))$ exceeds this independent value and $W<0$. The correlation coefficient isThis negative stationary correlation is compatible with positive instantaneous joint-removal noise: the slow difference mode amplifies fluctuations of opposite signs in the two populations.

The symmetric and antisymmetric eigenvectors diagonalize all three matrices. Their normalized covariance eigenvalues areThe corresponding restoring rates are $(1+E)/\tau =\beta +2C\mu$ and $(1-E)/\tau =\beta$. Thus the sum mode retains finite fluctuations, with ${\eta }_{+}\to 3/(4\mu )$, whereas the difference mode becomes arbitrarily noisy. In particular, at a fixed nonzero mean, or along $\beta \to 0$ at fixed $\lambda ,C>0$ so $\mu \to \sqrt{\lambda /C}$,The Fano factor of either population is $\mu V=(2-E^2)/(2(1-E^2))$, which diverges as $E\to 1$ regardless of how the mean is varied. Unnormalized variances are ${\mu }^{2}V$ and covariance ${\mu }^{2}W$, so their absolute magnitudes also depend on the parameter path through $\mu$.

The phase plane in part (b) explains the singularity. At $\beta =0$, the mean-field system has a whole hyperbola of equilibria, rather than an isolated restoring state. A perturbation in the species difference selects another point on that curve. In the actual stochastic reaction network, joint complex formation still leaves $X_1-X_2$ unchanged, but the two constant-rate births change it by $+1$ and $-1$, each at rate $\lambda$. Therefore exclusive joint removal has a diffusing difference mode:Its continuous-time symmetric simple random walk on the integers has no normalizable stationary distribution, so there is no finite stationary covariance at $E=1$. The divergence as $E\uparrow 1$ signals both the loss of difference-mode restoration and eventual failure of the small-fluctuation moment closure; near the limit, the computed normalized variances need not remain small even when the individual mean populations are large.

Write $c=\cos 2\varphi$, $s=\sin 2\varphi$ and $\sigma =({\sigma }_{11}+{\sigma }_{22})/2$. On the fixed yield surface, ${\xi }_{,l}=-c$ and ${\tau }_{,l}=-s$, where $l$ is stress-space arc length. Substituting ${\sigma }_{11}=\sigma +\xi$, ${\sigma }_{22}=\sigma -\xi$, ${\sigma }_{12}=\tau$ into planar force balance givesThus $\nabla \sigma =M\nabla l$, with $M=\left(\begin{array}{cc}c& s\\ s& -c\end{array}\right)$.

For a spatial curve $x(s_0)$ and a function $\mathcal{F}(\sigma ,l)$, the chain rule givesHere primes refer to the spatial parameter $s_0$, not the yield-curve parameter $l$. The displayed row-matrix product being zero is therefore sufficient for $\mathcal{F}$ to remain constant.

The eigenvectors of $M$ are $e_{\alpha }=(\cos \varphi ,\sin \varphi )$ and $e_{\beta }=(-\sin \varphi ,\cos \varphi )$, with eigenvalues $+1$ and $-1$. Along these slip lines, respectively, $d\sigma =dl$ and $d\sigma =-dl$. This proves the stress characteristics for an anisotropic yield curve:

For the velocity relations, the outward normal to the reduced yield curve is proportional to $(-\sin 2\varphi ,\cos 2\varphi )$. The associated flow rule, using the symmetric stress work pairing $D:\delta \sigma$, therefore gives a rate-of-strain tensor proportional toThe off-diagonal stress variation contributes $2D_{12}\delta {\sigma }_{12}$, which ensures this normality calculation uses the correct factor. Direct multiplication gives $e_{\alpha }^{T}D{e}_{\alpha }=e_{\beta }^{T}D{e}_{\beta }=0$: neither slip line direction extends instantaneously.

Resolve the velocity as $w=u{e}_{\alpha }+v{e}_{\beta }$. Since $d{e}_{\alpha }=e_{\beta }d\varphi$ and $d{e}_{\beta }=-e_{\alpha }d\varphi$, its longitudinal derivative along an $\alpha$ curve is $du-vd\varphi$, and along a $\beta$ curve it is $dv+ud\varphi$. Setting these extensions to zero gives the Geiringer velocity relations:

Use the conventional kinematic plume fluxesThen $\pi Q$ is the volume flux, $\pi {\rho }_{r}Q$ the dimensional mass flux, $\pi {\rho }_{r}M$ the dimensional momentum flux, and $\pi F$ the kinematic buoyancy flux. Calling $\pi Q$ a mass flux uses a density-normalized convention: the reference density has been suppressed. If density-weighted momentum and buoyancy fluxes are desired, multiply $M$ and $F$ by the same constant ${\rho }_r$ throughout. Here the conventional kinematic definitions make the Boussinesq approximation explicit.

For an upward plume with $Q,M>0$, the inverse relations areThe volume conservation, vertical momentum conservation, and buoyancy balances from part a becomeHere the vertical momentum conservation equation uses ${\rho }_r$ in the inertia and retains $({\rho }_0-\rho )gA={\rho }_{r}B$ in the buoyancy force. The ambient has no imposed vertical momentum at entrainment.

To classify the resulting partial differential equations, work in the nonsingular primitive variables $(A,w,g^{\prime })$ with $A>0$. Subtract $w$ times the volume equation from the momentum equation and $g^{\prime }$ times the volume equation from the buoyancy equation. This givesThe principal matrix is thereforeAll three characteristic speeds are real and equal to $w$, so the model is a hyperbolic system if that term means only reality of the characteristic roots. However, for $A>0$, $(K-wI)v=0$ requires $v_w=0$, leaving only two independent eigenvectors. HenceThis weak hyperbolicity of the unsteady top-hat plume prevents a complete set of independent characteristic fields. For example, freezing $A,w$ in the homogeneous principal part gives $D_t\delta w=0$ and $D_t\delta A=-A{\partial }_z\delta w$. A Fourier perturbation of wave number $k$ therefore produces a term proportional to $kt$ in $\delta A$. Real characteristic roots alone do not provide uniform control of high-frequency perturbations. The flux-variable transformation above is invertible in the stated region, so changing to $(Q,M,F)$ does not remove this degeneracy.

Continuation around one positively oriented loop induces an invertible linear map $M$ on the two-dimensional solution space. Its inverse is continuation around the reversed loop. Over $\mathbb{C}$, choose an eigenvector with eigenvalue $\lambda \ne 0$, and let $w$ be the corresponding nonzero solution. After continuation it becomes $\lambda w$.

Choose $\sigma$ with $e^{2\pi i\sigma }=\lambda$. The product $h(z)=z^{-\sigma }w(z)$ has trivial monodromy, so it is single-valued and holomorphic on the punctured disc. Its Laurent series, convergent on compact subannuli, gives the Monodromy eigenfunction Laurent representationNo assumption of a regular singular point has been made, so the negative part need not terminate.

The simultaneous zeros are the origin and the four corners $(\pm 1,\pm 1)$. The Jacobian matrix isAt the origin its eigenvalues are $1,-1$, with eigenvectors $(1,1),(1,-1)$, so the origin is a saddle equilibrium of Poincaré index $-1$. At a corner $J=-2xyI$. Thus $(1,1)$ and $(-1,-1)$ are stable nodes; $(1,-1)$ and $(-1,1)$ are unstable nodes, all of index $+1$. Although the node eigenvalues coincide, the equilibria are hyperbolic and their stability follows from their strictly signed real eigenvalues.

The lines $x=\pm 1$ and $y=\pm 1$ are invariant and divide the plane into nine convex regions. A nonconstant periodic trajectory cannot cross any of these lines by uniqueness, and cannot be contained in one of them because one-dimensional autonomous flows have no nonconstant periodic solutions. It must therefore lie inside one region. The central region contains only the origin, whose index is $-1$; all other regions contain no equilibria. But a simple periodic orbit has index $+1$, equal to the sum of the enclosed equilibrium indices. Its interior stays in the same convex region, giving a contradiction in either case. There are no periodic orbits.

Away from the invariant lines, trajectory curves have the first integralsince both logarithmic derivatives equal $-2xy$. The diagonal $y=x$ is the unstable separatrix of the origin, leading toward the two sinks; $y=-x$ is its stable separatrix, coming from the two sources. The diagram shows these separatrices, the invariant lines and the direction field.

The centre manifold theorem says that near an equilibrium of a sufficiently smooth finite-dimensional vector field, there is a locally invariant centre manifold tangent to the generalized eigenspace of eigenvalues with zero real part. It can be represented as a graph over that space, with graph value and derivative zero at the equilibrium; the dynamics on it determine the local centre-direction behaviour. For parameter-dependent systems the parameters may be adjoined as variables with zero derivative, producing an extended centre manifold.

At $r=1$, the linearization has the $x,y$ block $\left(\begin{array}{cc}-\sigma & \sigma \\ 1& -1\end{array}\right)$ and the independent eigenvalue $-b$. Its other eigenvalues are zero and $-(\sigma +1)$. Hence the equilibrium is nonhyperbolic, the stable subspace has dimension two, and the non-extended centre subspace has dimension one. A centre eigenvector is $(1,1,0)$.

Write $Q=1r^T-RI$, with $r=R\pi$. Then ${\pi }^{T}1=1$ andThus $\pi$ is a stationary probability vector, a left eigenvector for eigenvalue zero. If any probability vector $\nu$ is stationary, the same calculation gives ${\nu }^{T}Q=r^T-R{\nu }^T=0$, forcing $\nu =\pi$. Hence the equilibrium is unique, even if some $r_j$ vanish and the chain is not irreducible; those zero-rate states are transient or inaccessible rather than extra stationary classes.

The oscillator operators satisfy $[a,a^{†}]=1$ and $[a,a]=[a^{†},a^{†}]=0$. Substitution into the kinetic and potential energies givesThe number operator $N=a^{†}a$ is nonnegative because $⟨\psi ,N\psi ⟩=\Vert a\psi {\Vert }^2$. The commutators $[N,a]=-a$ and $[N,a^{†}]=a^{†}$ lower and raise an eigenvalue by one. If an eigenvalue were a noninteger $\nu \ge 0$, repeated lowering would produce a nonzero eigenvector with eigenvalue between zero and one, and one further lowering would have negative eigenvalue; its norm is nonzero because $\Vert a\psi {\Vert }^2=\nu \Vert \psi {\Vert }^2$. This contradicts nonnegativity. Thus only nonnegative integers are allowed. The normalized Gaussian solves $a|0⟩=0$, and raising it supplies every level, withFor the three-dimensional unperturbed oscillator the product states are $|n_1,n_2,n_3⟩=|n_1⟩\otimes |n_2⟩\otimes |n_3⟩$, with energy $\hslash \omega (n_1+n_2+n_3+3/2)$.

Write the perturbation as $\lambda W$, where $W=(\hslash \omega /2){\sum }_{i<j}(a_i+a_i^{†})(a_j+a_j^{†})$. Its ground-state expectation is zero. Acting on the ground state, it produces $|110⟩,|101⟩,|011⟩$, each with coefficient $\hslash \omega /2$. They lie $2\hslash \omega$ above the ground state, so second-order perturbation theory givesIn the first excited subspace with basis $|100⟩,|010⟩,|001⟩$, the first-order perturbation matrix is $(\lambda \hslash \omega /2)\left(\begin{array}{ccc}0& 1& 1\\ 1& 0& 1\\ 1& 1& 0\end{array}\right)$. Its symmetric eigenvector has eigenvalue $\lambda \hslash \omega$, while the perpendicular two-dimensional space has eigenvalue $-\lambda \hslash \omega /2$. Thus the lowest excited triplet splits intoAs an independent check, diagonalizing the quadratic potential gives normal frequencies $\omega \sqrt{1+2\lambda }$ and $\omega \sqrt{1-\lambda }$ twice; their small-$\lambda$ expansions reproduce both results.