When the birefringence separation of optical propagation constants $\Delta k=2\pi \Delta n/\lambda$ is large compared with the spatial rate of rotation of the optical axes, light with linear polarization in a local eigenpolarization follows the twisting nematic director. Thickness alone is insufficient if birefringence vanishes. Strong field alignment along the propagation direction suppresses the transverse optical anisotropy and hence the quarter-turn polarization rotation.

For isometry families $f_x:U\to U\prime$ and $g_y:U\prime \to U\prime \prime$, the complement of $g_y{f}_x(V)$ decomposes orthogonally as the complement of $g_y(W)$ plus the image under $g_y$ of the complement of $f_x(V)$ in $W$. Their fiberwise one-point compactifications therefore give the same Thom space as the external direct sum. The resulting stage isomorphisms respect structure maps and give $Y{⋉}_g(X{⋉}_{f}E)\cong (Y\times X){⋉}_{g\circ f}E$ in the stable homotopy category.

For an isolated uniform sphere with fixed positive cosmological constant, energy conservation and the virial theorem give $2\eta x^3-(2+\eta )x+1=0$, $x=R_v/R_m$. The physical root starts at $x=1/2$ when $\eta =0$ and decreases for $0<\eta <1$. At the limiting value $\eta =1$, the cubic factors as $(x-1)(2x^2+2x-1)$; the collapsing branch approaches $(\sqrt{3}-1)/2$. Finite turnaround therefore gives a strict lower bound, whose infimum is approached rather than attained. Compared with zero vacuum energy at the same turnaround data, the negative vacuum potential and its virial contribution require stronger final gravitational binding.

For the Brusselator, the stationary spatial threshold is $b_T=(1+a\sqrt{D_u/D_v}{)}^2$ and the homogeneous oscillatory threshold is $b_H=1+a^2$. A genuine Turing instability appears first only if $b_T<b_H$, which rearranges to the displayed diffusivity-ratio bound. Equality gives simultaneous stationary spatial and homogeneous oscillatory neutral modes. Otherwise homogeneous loss of stability occurs first.

On a specified finite domain, only its boundary-compatible wavenumbers can grow. Solve the Brusselator mode determinant for its neutral value of $b$ and minimize over those nonzero modes. Minimizing over all continuous $k$ instead gives the usual threshold and $k_c^2=a/\sqrt{D_u{D}_v}$. The finite-mode minimum must still precede the homogeneous threshold to describe a Turing instability.

For $u_t=D{u}_{xx}-u(u-r)(u-1)-\rho (v-u)$ and $0=v_{xx}-v+u$, a Fourier mode perturbation about zero has $v=u/(1+k^2)$ and growth rate $\sigma (k)=-r-D{k}^2+\rho k^2/(1+k^2)$. If $\rho >D$, its maximum is $-r+(\sqrt{\rho }-\sqrt{D}{)}^2$ at $k^2=\sqrt{\rho /D}-1$; if $\rho \le D$, its maximum is $-r$ at zero. Replacing $(u,v,r)$ by $(1-u,1-v,1-r)$ gives the corresponding criterion at the other homogeneous equilibrium.

For $\overline{w{a}^{\prime }}=-{\eta }_t{B}_{0}f$, $⟨f⟩=1$ and positive diffusivities, the Zeldovich planar flux balance gives $G=({\eta }_t/\eta )B_0^2[1-({\eta }_t/\eta )\mathrm{Var}(f)]$, where $G=⟨|\nabla a^{\prime }{|}^2⟩$. The Cauchy-Schwarz inequality and a Poincaré inequality $⟨a^{\prime 2}⟩\le d^{2}G/{\pi }^2$ give ${\eta }_t^2\mathrm{Var}(f)\le ⟨w^2⟩d^{2}s(1-s)/{\pi }^2$, with $s=({\eta }_t/\eta )\mathrm{Var}(f)\in [0,1]$. Maximizing $s(1-s)$ gives the displayed bound. Large turbulent-to-microscopic diffusivity therefore requires $f$ to be close to one in mean square, without guaranteeing pointwise uniformity.

Let $f_0$ be the triweight kernel and h a twice continuously differentiable density with matching zero mean and variance. Put $e=h-f_0$ and $c=35/32$. Two integrations by parts over its support give $\int e^{\prime \prime }{f}_0^{\prime \prime }=48c[h(1)+h(-1)]+c{\int }_{|x|>1}(360x^2-72)h(x)dx\ge 0$, using the zero mass and second moment of e. Therefore $R(h^{\prime \prime })=R(f_0^{\prime \prime })+R(e^{\prime \prime })+2\int e^{\prime \prime }{f}_0^{\prime \prime }\ge R(f_0^{\prime \prime })$. The endpoint terms matter because the third derivative jumps there.

For real $x>0$, the sine integral approaches $\pi /2$. Exponential damping gives ${\int }_0^{\infty }{e}^{-at}\sin t/tdt=\arctan (1/a)$, so the limiting constant follows by letting $a\downarrow 0$. Integration by parts of the tail gives $\mathrm{Si}(x)=\pi /2-\cos x/x-\sin x/x^2+O(x^{-3})$.

For real $x>0$, its small-argument expansion is $\mathrm{Ci}(x)={\gamma }_E+\log x+O(x^2)$. Integrating ${\mathrm{Ci}}^{\prime }(x)=\cos x/x$ separates the logarithm from the regular correction. Its conventional constant is the Euler--Mascheroni constant; the primary series normalization is dlmf.nist.gov/6.6.E6 .

Consider assemblies of congruent polygonal tiles with same-labelled faces paired by reflection and pure Neumann boundary conditions on unpaired faces. An invertible intertwiner for all face permutation matrices preserves the Euler characteristic. The number of tiles is the dimension of the representation. For each face label the number of edge classes is $(n+\mathrm{tr}{M}_s)/2$. For each vertex label the number of vertex classes is the number of orbits of the subgroup generated by incident-face involutions, hence the dimension of its invariant vectors. Intertwining preserves all these numbers. Summing the vertex and edge counts proves equal $V-E+F$, including boundary corners. This argument uses genuine permutation matrices; signed matrices encoding mixed boundary conditions do not give the same topological count.

Take eight congruent hexagonal tiles with three pairwise nonadjacent distinguished faces. On tile indices modulo eight let $\sigma (j)=1-j$, $t(j)=-j$, and use $u_1(j)=3j$ for one assembly and $u_2(j)=3j+4$ for the other. Pair faces by reflection and leave fixed faces, and the three remaining faces, with Neumann boundary conditions. If $R{e}_j=e_{j+1}$, the displayed matrix intertwines all three face actions: its circulant coefficients satisfy $c_r=c_{-r}=c_{3r+4}$. Its eigenvalues are $1+(-1{)}^k+4\cos (\pi k/4)$, which are nonzero for $0\le k\le 7$. The transplantation theorem gives equal spectra. The second gluing graph is bipartite, with parts $\{0,5,6,7\}$ and $\{1,2,3,4\}$, so it is orientable. The first has the odd cycle $1,3,6,2,7,1$ and is nonorientable. Both graphs are connected with eight vertices and ten edges, and their bordered surfaces retract up to homotopy to the graphs. Both have Euler characteristic $-2$ and fundamental groups free of rank three. Neither is simply connected.

Take four congruent octagonal regular polygons with four alternate sides labelled $a,b,c,d$. The intervening sides have Neumann boundary conditions. Pair equal face parameters in the positive reference boundary direction. The complete gluing data areAll unpaired labelled faces not in a Dirichlet column also have Neumann boundary conditions. Encode these rules by signed permutation matrices, using a positive swap for a paired interface and diagonal $-1$ or $+1$ for an unpaired Dirichlet boundary condition or Neumann boundary condition. The four-by-four Hadamard matrix $C$ above satisfies $C{M}_s=N_{s}C$ for every label. Explicitly, if its columns are $h_0,h_1,h_2,h_3$, applying $N_a,N_b,N_c,N_d$ gives respectively $(h_0,h_1,h_3,h_2)$, $(h_0,h_2,h_1,h_3)$, $(h_1,h_0,-h_2,-h_3)$, and $(h_0,-h_1,h_2,-h_3)$, which are exactly the columns of the corresponding $C{M}_s$. On every intervening face the face matrix is the identity.

Since $C^{T}C=4I$, the orthogonal matrix $C/2$ preserves the $L^2$ inner product and the sum of Dirichlet energies. Its intertwining relations map matching Sobolev traces and zero Dirichlet traces bijectively to those of the other assembly. It therefore preserves the quadratic form domains and the forms themselves. The associated self-adjoint operators are unitarily equivalent, proving equality of all eigenvalues with their multiplicities for the specified mixed boundary conditions.

The topology of disk tiles glued along disjoint boundary arcs shows that $D$, whose gluing graph is $0,1,2,3$ in a path, is a closed disc and is simply connected and orientable. The gluing graph of $B$ has edges $13,12,02,01,23$; it is connected with two independent cycles, so $\chi (B)=-1$ and ${\pi }_1(B)\cong F_2$. Its odd cycle $1,2,3,1$ makes it a nonorientable surface. Each assembled surface has nonempty polygonal boundary. This explicitly realizes isospectrality between a simply-connected orientable bordered surface and a nonorientable, non-simply-connected one. The boundary corners and mixed boundary conditions are essential qualifications; it does not contradict the Euler characteristic from the heat trace of a bordered surface for smooth boundaries and pure boundary conditions.

Take seven copies of a scalene triangle, with sides labelled $a,b,c$ opposite its three vertices, and reflect adjacent copies across their glued side. For the first planar domain, the glued pairs are $a:(2,3),(6,7)$, $b:(4,6),(5,7)$, $c:(1,5),(3,7)$; for the second they are $a:(1,3),(5,7)$, $b:(2,6),(3,7)$, $c:(4,5),(6,7)$. Both coloured trees have central tile $7$. For triangles sufficiently close to equilateral the interiors do not overlap. If its three angles are distinct and between $\pi /4$ and $\pi /2$, the only reentrant vertices are the original central vertices, with angles four times the corresponding triangle angles. They force any proposed isometry to preserve the labelled central triangle, which is incompatible with the different attachments on its arms.

For the spectral proof index the nonzero vectors and covectors of ${\mathbb{F}}_2^3$ by $1,\dots ,7$, using binary coordinates. The Fano plane incidence matrix is $B_{y,x}=1_{y\cdot x=0}$. It satisfies $B^{T}B=2I+J$, so is invertible, and intertwines the point and line permutations of each of the three elementary transvections. Put $D=\mathrm{diag}(1,1,-1,1,-1,-1,1)$. The signed reflection matrices for the Dirichlet boundary condition are $-D{P}_{s}D$ and $-D{Q}_{s}D$, hence $C=DBD$ intertwines them. Its polar orthogonal factor gives a unitary equivalence of the Dirichlet Laplacians, preserving both the $L^2$ inner product and the Dirichlet energy of the piecewise functions. This produces a three-parameter family of noncongruent isospectral planar domains.

Suppose an invertible real matrix $C$ intertwines symmetric orthogonal tile-reflection matrices: $C{M}_s=N_{s}C$. Then $C^{T}C$ commutes with every $M_s$, so its positive inverse square root does too. The polar decomposition of an invertible real matrix therefore gives an orthogonal matrix $O=C(C^{T}C{)}^{-1/2}$ satisfying $O{M}_s=N_{s}O$. On a disjoint union of congruent tiles, $O$ preserves both the $L^2$ inner product and Dirichlet energy. The intertwining identities preserve the face trace conditions defining the assembled Sobolev space, so $O$ gives a unitary equivalence of the assembled Laplacians.

A translation lattice is the additive group generated by linearly independent vectors $v_1,\dots ,v_r$ in a Euclidean space. It defines a discrete subgroup of translations. A full-rank translation lattice has a compact parallelepiped as a fundamental region and therefore a cocompact group action. This geometric usage is distinct from an order-theoretic lattice.

At the Maxwell balance for a scalar reaction-diffusion front, let $q=3\alpha /4$, $\beta =\sqrt{3}\alpha /2$, and perturb the linear growth rate by $\epsilon K\delta (x+cT)$, $T=\epsilon t$. Projection onto the front's translation mode gives $X_T=-K{\mathrm{sech}}^2(\beta Z/2)$, where $Z=X+cT+{\beta }^{-1}\log q$. Hence $Z_T=c-K{\mathrm{sech}}^2(\beta Z/2)$. Finite locked positions exist for $0<c/K\le 1$. For a strict inequality, the root with $KZ<0$ is stable and the other unstable; they meet in a saddle-node bifurcation at $c=K$. This is a leading-order solvability condition, not an exact finite-perturbation solution.

At state $x$, the holding probability of a discrete-time Markov chain is the self-transition probability $P(x,x)$. It is the conditional probability that the next state remains $x$. In a nearest-neighbour birth-death chain, it is $1-p_x-q_x$, where $p_x$ and $q_x$ are the probabilities of stepping right and left. Holding changes the elapsed time but not the embedded chain of actual moves or its boundary exit probabilities.

For $M=\left(\begin{array}{cc}-1/R& 0\\ 1& -2/R\end{array}\right)$ with $R>0$, write $s=t/R$. Its matrix exponential is $B=\left(\begin{array}{cc}a& 0\\ R(a-a^2)& a^2\end{array}\right)$, $a=e^{-s}$. The maximizing unit initial state is the largest-eigenvalue eigenvector of $B^{\ast }B$. Its component ratio is the displayed expression, with $d=(e^s+1)/(2R)+(R/2)(e^s-1)$. This follows by solving the quadratic eigenvector equation for the symmetric two-by-two Gram matrix. The state tends to $(1,0{)}^T$ for $t\gg R$.

For one-parameter binomial and two-parameter normal models, the displayed dimension-matching map has inverse $(p,u)=((1+e^{-\mu }{)}^{-1},\log v)$ and absolute Jacobian determinant $v/[p(1-p)]$. With auxiliary density $h(u)$ and model-jump probabilities $b,d$, an annealing target proportional to $e^{({\ell }_j-k_j)/T}$ gives forward acceptance ratioThe reverse ratio is its reciprocal at the inverse map. Keep model-dependent likelihood constants. Comparing continuous densities directly with discrete masses is not invariant to measurement units; a scientifically meaningful comparison needs models for the same recorded observations with a compatible reference measure, for example bin-integrated normal probabilities for integer-valued measurements.