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Set $\sigma =0$ under the assumed stationary-onset hypothesis, and let $\kappa =h{\kappa }^{\prime }$ with fixed nonzero ${\kappa }^{\prime }$. Substitute $W=h{W}_0+h^2{W}_1+\cdots$, $N=N_0+h{N}_1+h^2{N}_2+\cdots$, and $R_c=h^{-1}(R_0+h{R}_1+\cdots )$ into the modal equations and plate boundary conditions.

At leading order the concentration equation is $N_0^{\prime \prime }=0$, with $N_0^{\prime }=0$ at both plates. Normalize its nonzero constant to $N_0=1$. The first nonzero momentum equation isIntegrating four times and imposing all four clamped boundary conditions givesAt the next concentration order, $\bar{n}=1+O(h)$ and ${\bar{n}}^{\prime }=h+O(h^2)$, so the vertical advection of the basic concentration does not yet contribute. ThusIntegration gives $N_1=-G{W}_0+z+C_1$. The constant $C_1$ is an amplitude-normalization freedom; set it to zero, obtainingThere is no need to require zero depth average of this concentration eigenfunction: for a nonzero horizontal wavenumber the horizontal average is already zero.

Use $\bar{n}=1+h(z+1/2)+O(h^2)$ and ${\bar{n}}^{\prime }=h+O(h^2)$. At order $h^2$, the concentration equation isThe corresponding cell-flux boundary conditions are $N_2^{\prime }=N_1$ at both plates. Integrate the equation from $-1$ to zero. Its left derivative term satisfiesThe $N_1^{\prime }$ terms therefore cancel. The no-slip conditions imply $\int W_1^{\prime \prime }dz=0$. Also integration by parts givesConsequently the solvability condition is simplysince ${\int }_{-1}^0{z}^2(z+1{)}^{2}dz=1/30$. For every fixed nonzero ${\kappa }^{\prime }$, the long-wave bioconvection threshold in weak stratification is thereforeThe cancellation shows why $G$ does not enter this leading coefficient; neither does $S$ at stationary onset. In dimensional terms, the leading critical mean number density and volume fraction areThese assume heavier-than-fluid cells, $\Delta \rho >0$. Exactly ${\kappa }^{\prime }=0$ makes the integrated equation identically zero and corresponds to a spatially uniform change of conserved cell number, not a convective mode; the statement independent of ${\kappa }^{\prime }$ is interpreted for nonzero long waves or their limit. This is the leading threshold within the specified long-wave, real-eigenvalue assumptions. It does not by itself rule out a different finite-wavelength or oscillatory instability outside those assumptions.

The spectral theorem for compact self-adjoint operators says that a compact self-adjoint $A$ has an orthogonal decompositionwith real nonzero eigenvalues of finite multiplicity. Only finitely many eigenvalues, counting multiplicity, have absolute value at least any given $\epsilon >0$. Thus the nonzero eigenvalues form a finite list or a sequence tending to zero, and $A={\sum }_{\lambda \ne 0}\lambda P_{\lambda }$ in operator norm. The kernel may have arbitrary Hilbert dimension.

For existence of an extremal eigenvector, self-adjointness gives $\Vert A\Vert ={\sup }_{\Vert x\Vert =1}|⟨Ax,x⟩|$. One elementary proof of the nontrivial inequality uses polarization: if the quadratic form is bounded in absolute value by $m\Vert x{\Vert }^2$, its polarized real part is bounded by $m$ on two unit vectors; adjusting a phase bounds $|⟨Ax,y⟩|$ by $m$. Choose unit $x_j$ with quadratic values tending to $\lambda =\Vert A\Vert$ or $-\Vert A\Vert$. ThenIf $A\ne 0$, compactness supplies a convergent subsequence of $A{x}_j$, and the last estimate supplies a convergent unit subsequence of $x_j$, with eigenvalue $\lambda$.

Eigenvectors for distinct eigenvalues are orthogonal, since $A=A^{\ast }$. An infinite orthonormal sequence in one nonzero eigenspace would violate compactness. More generally an infinite orthonormal family with eigenvalues bounded away from zero has images separated in norm, again impossible. Take all nonzero eigenspaces and let $M$ be their closed span. Its complement reduces $A$; if the restriction there were nonzero, the extremal-eigenvector argument would produce another nonzero eigenvector, a contradiction. Thus $M^{\perp }=\ker A$. The finite truncations of the eigenvalue sum have remainder norm tending to zero, proving the asserted norm expansion.

Apply part (a) to the compact positive operator $T^{\ast }T$. Its positive square root $B=|T|$ is defined by multiplying each eigenvector of eigenvalue $\lambda$ by $\sqrt{\lambda }$ and vanishing on the kernel. The eigenvalues tend to zero, so $B$ is compact. For every $x$, $\Vert Tx{\Vert }^2=⟨T^{\ast }Tx,x⟩=\Vert Bx{\Vert }^2$.

Consequently $U(Bx)=Tx$ is a well-defined isometry on $\mathrm{ran}B$. Extend it to the closure of that range and set it to zero on its orthogonal complement. This gives a partial isometry with initial space $(\ker T{)}^{\perp }$, final space $\overline{\mathrm{ran}T}$, andFor uniqueness, the positive factor in any such polar decomposition of a bounded operator satisfies $B^2=T^{\ast }T$. The positive-square-root uniqueness proved in 4(b) fixes $B$. The relation $U(Bx)=Tx$ then fixes $U$ on the dense subset $\mathrm{ran}B$ of its initial space and the kernel convention fixes it elsewhere. This proves uniqueness with the canonical initial-space condition. Without that condition the word “unique” would be false: $T=0$ can be written $U\cdot 0$ with many different unitary $U$.

Pass to $B={\overline{A}}^{\Vert \cdot \Vert }$. Closed invariant subspaces and intertwining maps are unchanged by this closure, and $BH$ is still dense. Because $B$ is a star-algebra, every closed invariant subspace reduces it.

Every nonzero nondegenerate compact-operator representation contains an irreducible reducing subspace. Indeed choose a nonzero positive compact element $b$; an isolated nonzero eigenvalue gives a nonzero finite-rank spectral projection $e\in B$ by continuous functional calculus, using a function vanishing at zero. The finite-dimensional corner $eBe$ has a minimal nonzero projection $p$, with $pBp=\mathbb{C}p$. Choose a unit vector $v\in pH$ and set $K=\overline{Bv}$. Then $pK=\mathbb{C}v$, since $pbv=pbpv$ is scalar times $v$. If $Q$ is the projection onto a reducing subspace of $K$, it commutes with $B$, and $Qv=pQv\in \mathbb{C}v$. Its projection property makes $Qv$ either zero or $v$. Cyclicity then makes $Q=0$ or $I_K$. Thus $K$ is irreducible.

Take a maximal orthogonal family of such irreducible reducing subspaces, by the axiom of choice. If their orthogonal complement were nonzero, its restricted representation would be nondegenerate: a vector annihilated by all restricted $B$ would, by taking adjoints, be orthogonal to $BH$ and hence zero. The image of $B$ on this complement is a norm-closed C-star algebra of compact operators, so the preceding construction gives another irreducible summand, contradicting maximality. The family therefore exhausts $H$.

Finally, suppose one irreducible equivalence class occurred infinitely often. Choose some $b\in B$ acting nontrivially on that class, and a unit vector $v$ with $\Vert bv\Vert >0$. Transport $v$ by unitary intertwiners into distinct equivalent summands. These vectors are orthonormal; their images under $b$ are orthogonal and all have the same nonzero norm. That contradicts compactness of $b$. The decomposition has finite multiplicity for every irreducible type, as claimed in nondegenerate representations by compact operators.

Use the round unit sphere with its complex orientation and unique spin structure. Its chiral spinor bundles are $S^{+}=\mathcal{O}(-1)$ and $S^{-}={\Lambda }^{0,1}\otimes \mathcal{O}(-1)$, the latter a smooth line bundle of degree $+1$. The Dirac operator isHere Clifford multiplication has $c(v{)}^2=-|v{|}^2$, and $D$ acts self-adjointly on the L2 spinors with first-order Sobolev domain. The geometric spin connection is invariant under the lifted $SU(2)$ action, hence $D$ is equivariant.

We justify the spectral properties needed for its phase. The Peter-Weyl theorem and Frobenius reciprocity for compact groups give the decomposition of sections of an isotropy-weight-$q$ line bundle: $E_n$ occurs once exactly when $n\ge |q|$ and $n\equiv q(\mathrm{mod}2)$, because its torus weights are $n,n-2,\dots ,-n$. The spinor isotropy weights are $+1,-1$, so each chirality contains $E_1,E_3,E_5,\dots$, each once. The horizontal connection Laplacian is the total Casimir minus the vertical generator square, ${\nabla }^{\ast }\nabla =C-1/4$ on these weight-one bundles. Expanding the Dirac square in a normal orthonormal frame gives the diagonal connection-Laplacian terms and the off-diagonal curvature contraction. The Clifford relations contract spin curvature to $\mathrm{Scal}/4$, the Lichnerowicz formula for a twisted Dirac operator with no twisting bundle. Since the scalar curvature here is two,Thus there is no kernel, the eigenvalues of $D$ are $\pm 1,\pm 2,\dots$ with finite multiplicities, and the resolvent is compact by spectral truncation. These round-sphere spectral facts agree with the explicit calculation in Abrikosov's paper.

Define the chiral phase of the round-sphere Dirac operatorIt is unitary because both chiral kernels vanish; equivariance follows from that of $D$. The full phase is $F=D|D{|}^{-1}=\left(\begin{array}{cc}0& V^{\ast }\\ V& 0\end{array}\right)$.

For a smooth scalar $f$, the Leibniz rule gives $[D,M_f]=c(df)$, a bounded bundle endomorphism. The spectral integral for the sign operator gives, in the strong topology,Write $R_{\pm }=(D\mp it{)}^{-1}$. Since $2D(D^2+t^2{)}^{-1}=R_{+}+R_{-}$ and $[R_{\pm },M_f]=-R_{\pm }[D,M_f]R_{\pm }$, the commutator integral is norm convergent: its integrands are compact and bounded by $\Vert [D,M_f]\Vert /(1+t^2)$. A norm limit of their integrals is compact. Thus $[F,M_f]$ is compact, and its off-diagonal blocks say $V{M}_f^{+}-M_f^{-}V$ is compact. Uniform approximation of continuous $f$ by smooth functions extends this to all $f\in C(S^2)$, using the bound $2\Vert f-g{\Vert }_{\infty }$ for the change of commutator.

The chiral phase is an equivariant unitary which intertwines the two multiplication representations modulo compacts. This specifies the grading implicit in the subsequent index question; treating the full self-adjoint phase as an endomorphism would yield a different, zero, compressed index.

The ladder operators obey $[a,a^{†}]=1$ and $N=a^{†}a$ has $[N,a]=-a$, $[N,a^{†}]=a^{†}$. Since $⟨\psi ,N\psi ⟩=\Vert a\psi {\Vert }^2\ge 0$, its eigenvalues are nonnegative. If $N|\nu ⟩=\nu |\nu ⟩$, lowering gives eigenvalue $\nu -1$ and squared norm $\nu$. Repeated lowering cannot produce a negative nonzero eigenstate; the norm product $\nu (\nu -1)\cdots$ forces $\nu$ to be an integer. The final lowered state has $a|0⟩=0$. With a unique ground state all normalized states are

For time-independent perturbation theory, write $H^{\prime }=H+\lambda W$ with $W=\hslash \omega (a^2+a^{†2})$. The standard first correction is $\lambda ⟨n|W|n⟩$ and the second is ${\lambda }^2{\sum }_{j\ne n}|⟨j|W|n⟩{|}^2/(E_n-E_j)$. The first vanishes. Only $j=n-2,n+2$ occur, with ladder coefficients $\sqrt{n(n-1)}$ and $\sqrt{(n+1)(n+2)}$, respectively. ThereforeThe downward term is zero for $n<2$.

Inverting the given position/momentum expressions givesThus $\boxed{\alpha =1-2\lambda ,\beta =1+2\lambda }$. For $|\lambda |<1/2$ both are positive, and this is a quantum harmonic oscillator of effective mass $m/\alpha$ and frequency $\omega \sqrt{\alpha \beta }$. Its exact energies areExpansion gives $E_n^{\prime }=\hslash \omega (n+1/2)[1-2{\lambda }^2+O({\lambda }^4)]$, agreeing with both perturbative corrections.

For a fixed $n$, convergence for every initial error is equivalent to $|1-\omega t_k|<1$ for all $k$, where $t_k=1-\cos (k\pi /(n+1))>0$. ThusA single parameter works for every finite $n$ precisely when $\boxed{0<\omega \le 1}$: the largest $t_k$ tends to two as $n\to \infty$, so every $\omega >1$ eventually fails, whereas $\omega =1$ has all eigenvalues strictly between $-1$ and one for each finite grid.

For such $\omega$ the lowest-frequency eigenvalue satisfies, using $1-\cos x\le x^2/2$,An initial error proportional to $v_1$ retains exactly the factor ${\lambda }_1^{\nu }$. Hence the worst-case spectral radius cannot improve on a factor of the form $(1-c/n^2{)}^{\nu }$, with $c$ independent of grid size, and reducing a slow-mode error by a fixed fraction takes order $n^2$ iterations. This is a worst-case rate assertion, not a lower bound for every individual initial error: a vector containing only faster modes, or the zero vector, need not exhibit the slow mode.

Subtracting the fixed-point equation for the exact solution givesHere $D=2I$, so the supplied sine eigenvectors of $A$ are also eigenvectors of $H_{\omega }$. Explicitly,Their distinct eigenvalues for nonzero $\omega$ make them an eigenbasis; alternatively the discrete sine orthogonality gives a basis even when $\omega =0$.

For a representation space $V$, let ${\chi }_S$ and ${\chi }_A$ be the characters on ${\mathrm{Sym}}^{2}V$ and ${\bigwedge }^{2}V$. If the eigenvalues of $\rho (g)$ are ${\lambda }_1,\dots ,{\lambda }_d$, the respective traces are ${\sum }_{i\le j}{\lambda }_i{\lambda }_j$ and ${\sum }_{i<j}{\lambda }_i{\lambda }_j$. Since $\chi (g)=\sum {\lambda }_i$ and $\chi (g^2)=\sum {\lambda }_i^2$, this givesFinite-order matrices over $\mathbb{C}$ are diagonalizable, so this eigenvalue argument applies to every group element. Equivalently the symmetric and antisymmetric projections on $V\otimes V$ are $(I\pm \text{swap})/2$.

Use a common invariant operator domain on which the stated differentiations and commutator are valid. For a differentiable normalized eigenfunction, the Hellmann-Feynman calculation and self-adjointness giveThe Lax equation preserves the spectrum. Differentiating $Lf=\lambda f$ now gives $(L-\lambda )(f_t+Af)=0$, so $f_t+Af$ is in the same eigenspace, or is zero. For a nondegenerate eigenvalue it equals $c(t,\lambda )f$. Multiplying $f$ by $\exp [-\int c(t,\lambda )dt]$ produces $\hat{f}$ with ${\hat{f}}_t+A\hat{f}=0$ while preserving its eigenvalue.

Write $b=u_x+a_0$ and $d=u-\lambda +a_1$. The spatial eigenvalue equation gives ${\hat{f}}_{xx}=(u-\lambda )\hat{f}$, so $A\hat{f}=b\hat{f}+d{\hat{f}}_x$. For $F=(\hat{f},{\hat{f}}_x{)}^T$ the two first-order systems are thereforeThe second row of $V$ follows by differentiating the first time equation in $x$ and using the spatial equation once more. Conversely these matrix equations imply both scalar equations, so the formulations are equivalent.