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Write $g=u+iv$ and let its constant modulus be $C$. If $C=0$, then $g=0$. For $C>0$, differentiating $u^2+v^2=C^2$ with respect to $x$ and $y$, then applying the Cauchy-Riemann equations, givesThusThe coefficient matrix has determinant $-(u^2+v^2)=-C^2\ne 0$, forcing $u_x=v_x=0$. The Cauchy-Riemann equations then also give $u_y=v_y=0$. Therefore $u,v$ are constant on a connected domain andThis proves the constant-modulus holomorphic function result. If a domain is disconnected, the conclusion applies separately to each connected component.

Since $M^T=-M$, $(M^2{)}^T=(M^T{)}^2=M^2$, so $M^2$ is a real symmetric matrix. Moreover,proving that it is a negative semidefinite matrix.

For odd $n$, determinant invariance under transpose givesso $\det M=0$. Hence $M$ and $M^2$ have a zero eigenvalue. Finally, if $M^{2}w=0$, then $0=w^T{M}^{2}w=-\Vert Mw{\Vert }^2$, so $Mw=0$. Thuswhere equality of these kernels includes the reverse implication, which is immediate.

Write a matrix in the set as $M(a,b)$. The zero matrix $M(0,0)$ has no multiplicative inverse matrix, so $S$ is not a group. Its determinant iswhich vanishes only at that zero matrix. It is therefore the only element that needs to be removed.

For closure, direct matrix multiplication givesThe product of nonzero determinants is nonzero, so two nonzero elements never multiply to zero. The identity matrix is $M(1,0)$, associativity is inherited from matrix multiplication, andThus $\boxed{S\setminus \{0_{2\times 2}\}\text{ is a group}}$. It is the multiplicative group of nonzero quaternions in the complex matrix representation of quaternions.

Distinguish the quaternion group generators $\mathbf{i},\mathbf{j},\mathbf{k}$ from the complex scalar $i$. Denote their proposed image matrices byDirect multiplication givesFor example $AB=\mathrm{diag}(i,-i)=C$ and $ABC=C^2=-I_2$. Sending $I$ to $I_2$ and each negative quaternion element to the negative of its positive image now reproduces every quaternion product. The matrices all have determinant one and are invertible, so this is a group representation in $GL(2,\mathbb{C})$.

The eight matrices $\pm I_2,\pm A,\pm B,\pm C$ are distinct. Therefore the group homomorphism is injective, and the representation is faithful; equivalently its kernel is $\{I\}$.

The character of a representation is the trace of its representing matrix. ConsequentlyThese are constant on the five conjugacy classes $\{I\}$, $\{-I\}$, $\{\pm \mathbf{i}\}$, $\{\pm \mathbf{j}\}$, $\{\pm \mathbf{k}\}$.

The complete list of subgroups isTo see completeness, Lagrange's theorem allows only orders $1,2,4,8$. The only element of order two is $-I$, which fixes the order-two subgroup. Every noncentral element has order four and generates the corresponding listed cyclic subgroup. Any order-four subgroup contains such an element, because there are only two central elements; it must equal the cyclic subgroup generated by that element. This accounts for all possibilities.

The anticommutator condition and invertibility give the matrix similaritiesSince matrix trace is invariant under matrix similarity,Hence $\boxed{\mathrm{Tr}A=\mathrm{Tr}B=0}$. Taking determinants in the first matrix similarity also givesSince $A$ is an invertible matrix, $\det A\ne 0$, and $\boxed{n\text{ is even}}$. In terms of eigenvalues, the same matrix similarity pairs each nonzero eigenvalue $\lambda$ with $-\lambda$, with the same algebraic multiplicity, so an odd number of them is impossible.

The prescribed pole lets us choose $c=4$ and $d=-5-3i$. Write $f(z)=(Az+B)/(4z-5-3i)$. Fixing $1/2$ and $2$ givesso $A=5-3i$ and $B=-4$. ThusIts determinant is $-18\ne 0$, so it is a genuine Möbius transformation. Direct substitution gives $f(1/2)=1/2$, $f(2)=2$, and a pole at $(5+3i)/4$; the numerator at that pole is $9/2\ne 0$.

To identify the image of the unit disc, putThen $|\alpha {|}^2=8/17<1$, $|\lambda |=1$, andIn particular, its pole is outside the closed unit disc. The identityshows $f(D)\subseteq D$. For any $|w|<1$, solve $w=\lambda (z-\alpha )/(1-\overline{\alpha }z)$ to getIts denominator is nonzero, and the same modulus identity for this inverse gives $|z|<1$. Thus $f(D)=D$, and global injectivity of a Möbius transformation proves the required one-to-one correspondence.

Write $\Delta =ad-bc>0$. For real $a,b,c,d$ and $z\in H$, the denominator cannot vanish: if $c\ne 0$ its zero $-d/c$ is real, and if $c=0$ then $d\ne 0$. Taking the imaginary part after multiplying by $\overline{cz+d}$ givesThus $f(H)\subseteq H$. Alsoso it is conformal. Cross-multiplication shows that $f(z_1)=f(z_2)$ implies $\Delta (z_1-z_2)=0$, proving injectivity.

The inverse Möbius transformation isIts coefficients are real and its determinant is again $\Delta >0$, so the same imaginary part identity shows that it maps $H$ into $H$. Therefore every $w\in H$ has a preimage in $H$, proving surjectivity. Hence $f$ maps $H$ one-to-one and onto itself.

A sufficient condition is that $U$ is a connected space. Suppose, towards a contradiction, that $|f|$ is the same constant $R$ throughout $U$. If $R=0$, then $f=0$, contradicting $f(a)\ne f(b)$.

If $R>0$, differentiate $u^2+v^2=R^2$ in $x$ and $y$, and use the Cauchy-Riemann equations from part (a). They giveThe coefficient matrix for $(u_x,v_x)$ has determinant $-(u^2+v^2)=-R^2\ne 0$. Hence $u_x=v_x=0$, so $f^{\prime }=0$ everywhere. A holomorphic function with zero complex derivative is locally constant: within a small disc, differentiate along a straight segment. Local constancy and connectedness imply constancy on $U$, again a contradiction.

Consequently,This is the constant-modulus holomorphic function principle. The extra hypothesis matters: on two disjoint open discs, define $f=1$ on one and $f=-1$ on the other. It is holomorphic and nonconstant, yet $|f|=1$ everywhere.

Put $g=\gamma P_0$. For a temporally uniform envelope, $A_z=igA$ and $A_{TT}=0$, so the propagation equation reduces to $-gA+\gamma P_{0}A=0$. Thus the power stays $P_0$ and the accumulated nonlinear wave phase is $gz$. The optical Kerr effect gives an intensity-dependent refractive index, and hence self-phase modulation. For the usual power normalization $\gamma \simeq n_2{\omega }_0/(c{A}_{\mathrm{eff}})$, with effective guided-mode area $A_{\mathrm{eff}}$. Losslessness preserves the background power; it does not remove the self-phase modulation. The carrier wave phase has already been factored out.

Write $A=(\sqrt{P_0}+ϵ)e^{igz}$. To first order,Cancelling the background and its phase derivative leavesThe last term couples a perturbation to its complex conjugate: the pump can transfer energy into a pair of sidebands. It is not an independent wave phase of each Fourier component. This sideband coupling is the linear origin of fibre modulational instability.

For real $K,\Omega ,c_1,c_2$, put $d={\beta }_2{\Omega }^2/2$. Matching the two exponential coefficients givesA nonzero pair requires a zero determinant, henceFor unstable complex $K$, the real-coefficient assumption is too restrictive. The consistent pair is $ϵ=u{e}^{iKz-i\Omega T}+v^{\ast }{e}^{-i{K}^{\ast }z+i\Omega T}$, with complex $u,v$. Its equations give the same determinant and allow the growing and decaying branches. Merely inserting imaginary $K$ while treating the two original coefficients as real would not respect complex conjugation.

For $P_0>0$ and nonzero ${\beta }_2,\gamma$, exponential growth requires ${\beta }_2\gamma <0$ andFor ordinary positive Kerr nonlinearity this means anomalous group velocity dispersion, ${\beta }_2<0$. Define ${\Omega }_c=\sqrt{4g/|{\beta }_2|}$. The growing branch is $K=-iG$, with amplitude growth rateSideband power grows at rate $2G$, rather than $G$. The cutoff and zero-frequency modes have no exponential growth; in a degenerate zero-frequency sector a secular phase response can remain. Positive dispersion with positive $\gamma$ gives real $K$ and oscillatory perturbations.

Initially small ripples grow into pronounced intensity peaks and deplete the uniform background. Dispersion and self-phase modulation can then support localized bright pulses. A fundamental temporal bright soliton obeys $\gamma P_s{T}_0^2=|{\beta }_2|$, so its width is comparable to $1/{\Omega }_c$ when its peak power is comparable to the original power. The dominant ripple spacing is $2\pi /|\Omega {|}_{\max }$, not simply $1/{\Omega }_c$. Propagation distance $z$ is the evolution variable; $T$ measures time within the envelope. An ideal conservative fibre need not settle permanently into a unique pulse train: nonlinear compression, interactions and recurrence depend on the seed and boundary conditions.

Adjoin ${\mu }_{n+1}={\mu }_n$. At the origin the stable direction is the $x$ axis, with Floquet multiplier zero, while the $y$ and parameter directions have Floquet multiplier one. By the centre manifold theorem for a discrete dynamical system, write the extended centre manifold asOn it the scalar map is $Y=(1+\mu )y-\mu h(y,\mu )+y^3$. Its discrete centre-manifold invariance equation isPut $h=h_2+O({\rho }^3)$ with $\rho =|y|+|\mu |$. Since $Y=y+O({\rho }^2)$, the quadratic equation is simply $h_2(y,\mu )=\mu y$. ThusIn particular, setting the stable variable identically to zero would lose the leading parameter correction. The reduced map through total degree three isIn the natural pitchfork bifurcation scaling $\mu =O(y^2)$ this has leading form $y_{n+1}=y_n+\mu y_n+y_n^3$. Its nonzero fixed points satisfy $y^2\sim -\mu$, and their reduced fixed-point multipliers satisfy $1-2\mu +O({\mu }^2)>1$ for small $\mu <0$. The transverse Floquet multiplier remains inside the unit circle. Therefore the bifurcation at $\mu =0$ is a subcritical pitchfork bifurcation: the origin attracts for small negative $\mu$, each nonzero branch is a saddle fixed point of a map, and the origin is a saddle fixed point of a map for small positive $\mu$.

An exact check uses the full Jacobian matrix on either nonzero branch. Its trace is $T_{\ast }=1-2\mu +3{\mu }^2$ and its determinant is $D_{\ast }={\mu }^2$. Thuswhenever a nonzero branch exists. Also $p_{\ast }(0)={\mu }^2>0$, so one real Floquet multiplier lies in $(0,1)$ and the other in $(1,\infty )$. This proves that each nonzero branch is a saddle fixed point of a map throughout $\mu <0$ and $\mu >1$, in particular near the requested bifurcation.

For the $n$ distinct complex embeddings ${\sigma }_1,\dots ,{\sigma }_n$ of the number field, define the discriminant of elements of a number field byThe equality follows by multiplying the transpose of the embedding matrix by itself. Each field trace is rational: it is also the trace of the rational matrix of multiplication by $w_i{w}_j$ on $K$. Thus the determinant lies in $\mathbb{Q}$. If the $w_i$ are algebraic integers, the products and their conjugates are algebraic integers, and so are their sums. A rational algebraic integer is an integer, hence all trace entries and their determinant are integers.

Write $u_j={\sum }_i{a}_{ij}{v}_i$. The inclusion of the generated abelian groups makes $A=(a_{ij})$ an integer matrix. Its embedding matrix is the embedding matrix of the $v_i$ multiplied by $A$. Taking squared determinants givesFor independent algebraic integers $v_i$, choose an integral basis ${\omega }_i$ of ${\mathcal{O}}_K$. The same formula gives $D(v)=[{\mathcal{O}}_K:{\sum }_i\mathbb{Z}{v}_i{]}^{2}D(\omega )$. If $D(v)$ is nonzero and a square-free integer, the integer index must be one, proving that the $v_i$ form an integral basis.

Now put $\beta =3\alpha +4$. Since $\mathbb{Q}(\beta )=\mathbb{Q}(\alpha )$ has degree three, the monic polynomialis its minimal polynomial of an algebraic element. The field norm is the negative of its constant term:For the discriminant calculation, $f^{\prime }(\alpha )=3{\alpha }^2+4\alpha$. In the basis $1,\alpha ,{\alpha }^2$, using ${\alpha }^3=-2{\alpha }^2-1$, multiplication by this element has matrixThus the given derivative-norm formula gives $D(1,\alpha ,{\alpha }^2)=-59$. Since $\alpha$ satisfies a monic polynomial with integer coefficients, $1,\alpha ,{\alpha }^2$ are algebraic integers; the fact that a squarefree discriminant certifies an integral basis proves

For the basis $w_1,\dots ,w_d$, its discriminant isEquivalently it is the square of the determinant of the matrix of the $d$ embeddings applied to the basis vectors. It is nonzero because number fields are separable. An integral basis is a basis of the ring of integers as a free abelian group; in particular it is a rational basis of the field and expresses every algebraic integer uniquely with integer coefficients.

To prove existence, start with any rational basis and multiply each vector by a suitable nonzero integer to make it integral. This is possible by clearing coefficients in its monic rational minimal polynomial of an algebraic element and rescaling the variable. Let $M$ be the lattice spanned over $\mathbb{Z}$ by these integral vectors, and let $D\ne 0$ be its discriminant. For any algebraic integer $\alpha ={\sum }_j{c}_j{w}_j$, each trace $\mathrm{Tr}(\alpha w_i)$ is an integer. The resulting linear system has integral coefficient matrix with determinant $D$. Its adjugate matrix shows $c_j\in D^{-1}\mathbb{Z}$. ThusA subgroup of a finitely generated free abelian group is free and finitely generated. Here its rank is $d$, since it contains $M$ and is contained in a rank-$d$ lattice. Its free generators are the required integral basis.

For squarefree $d\ne 1$, the quadratic integral basis isThe source allows squarefree nonzero integers, so the degenerate value $d=1$ gives $K=\mathbb{Q}$ and basis $\{1\}$ instead of a two-element basis.

In the biquadratic field define the three automorphisms by changing respectively the sign of $\sqrt{r}$, the sign of $\sqrt{2}$, and both signs. Their actions areHence the three sums are $2a+2b\sqrt{2}$, $2a+2c\sqrt{r}$ and $2a+2d\sqrt{2r}$. If $\mu$ is integral, each sum is integral and lies in its displayed quadratic subfield. Their radicands are squarefree and respectively congruent to two, three and two modulo four. The quadratic basis formula therefore forces all their two coefficients to be integers, proving

The determinant is invariant under transpose. Therefore the characteristic polynomials coincide:Their polynomial roots, including their algebraic multiplicities, are the same, so $A$ and $A^T$ have the same eigenvalues. This allows column sums as well as row sums to supply Gershgorin discs for the same spectrum.

The four row Gershgorin discs have centres $9,3,-4,-5$ and radii $1,2,2,1$, respectively. Their real-axis intersections areNone of the Gershgorin discs contains zero, so the Gershgorin circle theorem givesThe discs centred at $9$ and $3$ are each isolated. By Gershgorin eigenvalue counting, they contain exactly one eigenvalue apiece, and both eigenvalues are simple. Each must be real: the characteristic polynomial is real, so a nonreal root would bring a distinct conjugate root into the same disc. Denote these eigenvalues by $\beta \in [8,10]$ and $\alpha \in [1,5]$. The remaining two eigenvalues lie in the union of the negative-centred discs.

For the shift $4$, the distances satisfyfor either of the other eigenvalues $\gamma$, since its real part is at most $-2$. Thus $\alpha$ is uniquely closest to $4$. Once $4$ is checked not to be an eigenvalue, the natural full target-disc interval enclosure and a conservative asymptotic direction factor areAs always, inverse iteration requires a nonzero initial target component. The following calculation verifies the shift and gives sharper information.

For the tridiagonal matrix, let $p_j(t)$ be the determinant of the leading $j\times j$ block of $tI-A$. Expansion along the last row givesIn particular, $\det A=577$, $p_4(4)=-239$ and $p_4(5)=-583$. Both shifted linear systems are nonsingular. Further exact evaluations areThe intermediate value theorem gives a real root in each of $(-6,-5)$, $(-4,-3)$, $(3,4)$ and $(8,9)$. As $p_4$ has polynomial degree four, these are all its roots and each is simple. In particular, $A$ is a diagonalizable matrix, $\alpha \in (3,4)$ and $\beta \in (8,9)$. The closest competitor to $\alpha$ at the shift $4$ is $\beta$, so this refinement yieldsFor the shift $5$, the original Gershgorin discs only give $|\alpha -5|\le 4$ and $|\beta -5|\ge 3$; those bounds do not decide which is closer. The refined interval enclosures do:for the negative eigenvalues. Therefore the answer for this particular matrix is yes:Both shifts select the same eigenvector line and alternate signs because $\alpha <\mu$. The factor bounds concern asymptotic directional convergence, rather than a universal error reduction from the first iterate for a nonsymmetric matrix. The wider interval $[1,5]$ remains a valid full-disc enclosure for both shifts; $[3,4]$ records the stronger information obtained from the characteristic polynomial.

Clearing the denominators in the rational interpolation conditions givesThese are $n+1$ linear equations for $(m+1)+\ell =n+1$ unknowns. The $i$th row of the coefficient matrix, in the order $a_0,\dots ,a_m,b_0,\dots ,b_{\ell -1}$, isThis rational interpolation linearization is equivalent to the original fraction conditions only if $r(x_i)\ne 0$ at every interpolation node. If $r(x_i)=0$, the cleared equation merely forces $q(x_i)=0$, leaving the fraction undefined.

For $f_i=x_i^{m+1}$, the coefficient row becomesThis is the Vandermonde matrix, whose determinant is ${\prod }_{0\le i<j\le n}(x_j-x_i)\ne 0$. Equivalently, a vector in its null space would define a degree-at-most-$n$ polynomial vanishing at all $n+1$ distinct interpolation nodes, which is zero by the preceding uniqueness proof. Thus the cleared linear system has exactly one solution.

There is also a constructive description of that solution. Define the monic polynomialThe polynomial $x^{m+1}r(x)-q(x)$ is monic of degree $n+1$ and vanishes at every interpolation node. It must therefore equal $W(x)$. By polynomial division, write uniquelyThe quotient $R$ is monic of degree $\ell$, andis the unique coefficient solution. This does not assert that $R$ is nonzero at every interpolation node.

For a counterexample with $m=0$, $\ell =1$, choose $x_0=0,x_1=1$ and the specified data $f_i=x_i$, so $f_0=0,f_1=1$. The cleared equations reduce togiving the unique solution $q(x)=0$, $r(x)=x-1$. But at $x_1=1$ the quotient is $0/0$. Indeed, any genuine rational interpolant of this prescribed form would need $r(0)\ne 0$; then $q(0)/r(0)=0$ forces the constant numerator to be zero. At a nonzero denominator its value at $1$ would also be zero, never one. Cancelling the common factor in the cleared candidate would merely give the identically zero rational function, which still fails the data. Hence

Applying the implicit midpoint rule to the simple harmonic oscillator system givesFactor the change in the quadratic form $y^2+z^2$ and substitute these equations:Thus every step preserves the first integralFor completeness, the implicit equations are solvable for every real $h$: putting $s=h/2$, their coefficient matrix has determinant $1+s^2>0$ and the update isThis is an orthogonal matrix, as follows from $(1-s^2{)}^2+4s^2=(1+s^2{)}^2$. It therefore preserves the Euclidean norm, explaining geometrically the exact energy conservation. This is a concrete instance of Runge-Kutta conservation of quadratic invariants. Preservation of the circle does not assert that the numerical and exact solutions have the same phase at a fixed nonzero step size.

There are errors in the original PDF at this point, not merely in the TeX transcription. The second exponential must denote the eigenvalue ${\lambda }_{-}(z)=e^{-{\gamma }^{-}z}$, not the propagation coefficient itself. More substantially, the printed prefactor $\mathrm{sech}(\delta z)$ is inconsistent with the stated exponential eigenvalues. It must be $\cosh (\delta z)$. The original's primed Jones-matrix label distinguishes the distributed component from the earlier static matrix; below write $\mathcal{J}(z)$ to avoid confusing that prime with a derivative.

Complex eigenvalues have no intrinsic largest/smallest ordering, so $+$ and $-$ are modal labels. Keep the circular eigenvectors fixed, put $\gamma =({\gamma }^{+}+{\gamma }^{-})/2$ and $\delta =({\gamma }^{+}-{\gamma }^{-})/2$, and use the spectral decomposition from part (c):Its diagonal entries are $(e^{-{\gamma }^{+}z}+e^{-{\gamma }^{-}z})/2=e^{-\gamma z}\cosh (\delta z)$, and its upper off-diagonal entry is $i(e^{-{\gamma }^{+}z}-e^{-{\gamma }^{-}z})/2=-i{e}^{-\gamma z}\sinh (\delta z)$. ThereforeWhere $\cosh (\delta z)\ne 0$, this can be factored asThe unfactored expression is valid even at zeros of the cosine-like factor when $\delta$ is imaginary. The printed expression is ${\mathrm{sech}}^2(\delta z)$ times this correct matrix; its eigenvalues are consequently $e^{-{\gamma }^{\pm }z}{\mathrm{sech}}^2(\delta z)$ rather than those specified. A second check is $\det \mathcal{J}=e^{-2\gamma z}$, whereas the printed version has determinant $e^{-2\gamma z}{\mathrm{sech}}^4(\delta z)$. Thus the requested printed identity cannot be proved except in the degenerate case $\delta =0$; the following parts use the consistent propagator.

Use coordinates $(ct,x,y,z)$ and put $\beta =v/c$, $\gamma =(1-{\beta }^2{)}^{-1/2}$. For an observer moving at $+v\hat{\mathbf{x}}$, the coordinate Lorentz boost isThe given electromagnetic field tensor transforms as $F^{\prime }=\Lambda F{\Lambda }^{\mathsf{T}}$. For example,whileThe $01$ component is unchanged because its boost determinant is ${\gamma }^2(1-{\beta }^2)=1$. The $23$ component is unchanged because neither index lies in the boost plane. Reading off all six components givesThe signs follow the stated positive observer velocity and the supplied tensor convention. In vector form for a boost in any direction, ${\mathbf{E}}_{\parallel }^{\prime }={\mathbf{E}}_{\parallel }$, ${\mathbf{E}}_{\perp }^{\prime }=\gamma ({\mathbf{E}}_{\perp }+\mathbf{v}\times \mathbf{B})$, with analogous ${\mathbf{B}}_{\perp }^{\prime }=\gamma ({\mathbf{B}}_{\perp }-\mathbf{v}\times \mathbf{E}/c^2)$.

Use the right-action convention printed in the paper: $\omega 1=\omega$ and $(\omega g)h=\omega (gh)$. The orbit of a group action and stabilizer areThe identity belongs to $H$, products of elements of $H$ fix $\omega$, and $\omega g=\omega$ implies $\omega g^{-1}=\omega$. Thus $H$ is a subgroup. For $k\in G$,proving $\mathrm{Stab}(\omega g)=g^{-1}Hg$.

The rotational action on the eight vertices of a cube is transitive. Model the vertices as $(\pm 1,\pm 1,\pm 1)$. For any desired sign triple, choose a permutation whose parity makes the corresponding signed permutation matrix have determinant one; it maps $(1,1,1)$ to that vertex and preserves the cube. A rotation fixing a vertex fixes the axis through that vertex, the centre and the opposite vertex. It cyclically permutes the three incident edges, giving precisely rotations by $0,2\pi /3,4\pi /3$. Thus the stabilizer is cyclic of order three.

The unit disk has area $\pi$, so its joint density is $f_{X,Y}(x,y)={\pi }^{-1}{1}_{\{x^2+y^2\le 1\}}$. The polar-coordinate Jacobian has absolute determinant $r$. Thereforeand is zero outside this rectangle, up to irrelevant boundary choices. Integrating givesThe joint density equals the product of the two marginal densities, so $R$ and $\Theta$ are independent. The angle at the centre can be chosen arbitrarily since that point has probability zero.