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Every finite coloring of a barrier in better-quasi-order theory has an infinite restriction on which all barrier members have the same color. It extends finite-dimensional Ramsey's theorem from fixed-length subsets to variable-length initial segments. A proof proceeds by well-founded induction on the initial-segment tree and a fusion construction selecting successively nested infinite tails. It implies that every finite set with the equality order is a better-quasi-ordering. The barrier/front partition proof is given in Section 3.2 of arxiv.org/pdf/1604.05866 .

For a balanced incomplete block design with $r>\lambda$, $N{N}^T$ has eigenvalues $r-\lambda >0$ on the zero-sum subspace and $r+(t-1)\lambda >0$ on the constant vector. Thus its rank is $t$. As $N$ is $t\times b$, $t=\mathrm{rank}(N{N}^T)\le \mathrm{rank}(N)\le b$.

A balanced incomplete block design is symmetric when its number of blocks equals its number of treatments. The identity $bk=tr$ then gives $r=k$. Its square incidence matrix of a set system $N$ obeys $N{N}^T=(r-\lambda )I+\lambda J$, with eigenvalues $r-\lambda$ of multiplicity $t-1$ and $r^2$ of multiplicity one.

For squared loss $E={\sum }_k(t_k-z_k{)}^2/2$, let $e_k=(t_k-z_k)g_k^{\prime }(x_k)$ and $e_j=g_j^{\prime }(x_j){\sum }_k{w}_{jk}{e}_k$. The chain rule gives $\partial E/\partial w_{jk}=-e_k{z}_j$ and $\partial E/\partial w_{ij}=-e_j{z}_i$. A gradient descent step of size $\eta$ uses $\Delta w_{jk}=\eta e_k{z}_j$ and $\Delta w_{ij}=\eta e_j{z}_i$, with every derivative evaluated at the same pre-update weights.

For ${\mathcal{L}}_{\mathrm{int}}=-g\varphi \bar{\psi }{\gamma }^5\psi$, the variational Dirac equation and adjoint Dirac equation are $(i/\partial -m-g\varphi {\gamma }^5)\psi =0$ and $i({\partial }_{\mu }\bar{\psi }){\gamma }^{\mu }+m\bar{\psi }+g\varphi \bar{\psi }{\gamma }^5=0$. Substitute these into the derivative of $j_5^{\mu }=\bar{\psi }{\gamma }^{\mu }{\gamma }^5\psi$. Anticommuting the chirality matrix past ${\gamma }^{\mu }$ gives two identical mass terms and two identical interaction terms, proving the displayed identity. These are classical field equations; the formula does not assert an unrenormalized quantum composite-operator identity. With Hermitian coupling $g=i{g}_P$, the second term is $-2g_P\varphi \bar{\psi }\psi$.

For $\mathcal{L}=\bar{\psi }(i/\partial -S+iP{\gamma }^5)\psi$, with real scalar backgrounds $S,P$, the Dirac equation is $(i/\partial -S+iP{\gamma }^5)\psi =0$ and the adjoint Dirac equation is $i({\partial }_{\mu }\bar{\psi }){\gamma }^{\mu }+S\bar{\psi }-iP\bar{\psi }{\gamma }^5=0$. Use $\{{\gamma }^5,{\gamma }^{\mu }\}=0$ and $({\gamma }^5{)}^2=1$ in the derivative of $j_5^{\mu }=\bar{\psi }{\gamma }^{\mu }{\gamma }^5\psi$ to obtain the displayed classical identity. Taking $S=m+g\varphi$ and $P=G\Phi$ includes both scalar and Hermitian pseudoscalar Yukawa couplings.

For an ordinary scalar field, its derivative is a polar covector, while the axial current is a pseudovector. Their contraction is invariant under the proper orthochronous Lorentz group but changes sign under parity. If the field is a pseudoscalar, the extra transformation sign makes this coupling parity even. In four spacetime dimensions its mass dimension is five, so Lorentz invariance alone does not imply power-counting renormalizability.

An axial charge is the spatial integral of the time component of an axial current. For a Dirac field and an internal generator $t$, it is $\int {\psi }^{†}{\gamma }_{5}t\psi d^{3}x$. Its canonical action rotates left- and right-handed fields oppositely. It is conserved only when the current divergence and flux at infinity vanish; Dirac masses and a chiral anomaly can obstruct conservation. The nonsinglet quark generators have no color anomaly and obey the nonsinglet axial charge algebra of quark densities.

An Archimedean ordered group with a two-sided invariant total order is commutative. If there is a least positive element, its cofinal powers exhaust the group. Otherwise each $c>1$ has $d>1$ with $d^2<c$. To see this, choose $1<s<r<c$ and use $r{s}^{-1}$ if its square is below $r$; if not, the inequality $(r{s}^{-1}{)}^2\ge r$ gives $s^2\le r<c$. If positive $g,h$ satisfied $gh<hg$, put $c=(gh{)}^{-1}hg$ and choose such a $d$. Cofinality gives $d^m\le g<d^{m+1}$ and $d^n\le h<d^{n+1}$. Two-sided order invariance then gives both $hg<d^{m+n+2}$ and $hg=ghc>d^{m+n+2}$, a contradiction. Replacing negative elements by their inverses reduces arbitrary commutativity to the positive case.

An analytic source monomial $g_{mn}{t}^m{h}^n$ in an additive renormalization-group transformation of free-energy density is resonant when its scaling degree equals the spatial dimension. The analytic background coefficient would require division by $1-b^{m{\lambda }_t+n{\lambda }_h-D}=0$. On a branch with $t\ne 0$ and ${\lambda }_t>0$, a particular nonanalytic background is $-g_{mn}{t}^m{h}^n\log |t|/({\lambda }_t\log b)$: substituting $t^{\prime }=b^{{\lambda }_t}t$, $h^{\prime }=b^{{\lambda }_h}h$ verifies the additive recursion exactly. Thus logarithms can be required even when the eliminated-mode source is analytic. This is distinct from marginal interaction running, which can create additional logarithmic corrections.

For an odd $n$-cycle $g$, an inverting permutation fixes one point and swaps $(n-1)/2$ pairs, so its sign is $(-1{)}^{(n-1)/2}$. Its other inverting permutations differ by elements of $C_{S_n}(g)=⟨g⟩$, all even. Thus inversion is possible inside $A_n$ exactly when that sign is positive. In particular $A_n$ is not an ambivalent group when $n\equiv 3(\mathrm{mod}4)$.

For independent bond percolation on a periodic plane lattice with three-fold rotation symmetry, the primal model at $p$ and the dual bond percolation model at $1-p$ cannot both percolate. To adapt Zhang's argument, divide a large boundary circle into six arcs, repeating the division in each $120$-degree sector. Vary the division point until the failure probabilities $a,b$ of the two alternating primal arc types are comparable within a factor $(1-p{)}^{-2}$. Moving past a single crossing edge changes a failure probability by at most a factor $(1-p{)}^{-1}$, by the FKG inequality. If the probability of no primal exterior arm is at most $\epsilon$, then $a^3{b}^3\le \epsilon$ and both failures are at most $(1-p{)}^{-1}{\epsilon }^{1/6}$. For the dual model, at least one arc type has failure at most ${\epsilon }^{1/6}$; its three rotations have the same bound. Four alternating primal/dual arms now occur with positive probability, contradicting the alternating exterior-arm separation lemma and uniqueness after a finite modification. This extends the four-fold argument to the martini lattice.

For constant coefficients and a transverse wavenumber $\ell \ne 0$, a planar mean-field dynamo retaining both alpha feedback and shear has dispersion $(\sigma +\eta p{)}^2={\alpha }^{2}p+i\alpha \Omega k$, $p=k^2+{\ell }^2$. Marginal stability gives ${\alpha }_m^2=4{\eta }^4{p}^4/(4{\eta }^2{p}^3+{\Omega }^2{k}^2)$. Differentiate with respect to $k^2$: an interior minimum solves $4{\eta }^2{p}^3+{\Omega }^2(3k^2-{\ell }^2)=0$. Its derivative with respect to $k^2$ is positive, so a unique positive minimizing $k^2$ exists only above the displayed shear threshold. Below it the minimum is at $k=0$. The sign of the marginal oscillation frequency is fixed by $2\eta p\omega =\alpha \Omega k$.

For a mean-field dynamo with uniform alpha, shear $\pi V/d$, diffusivity $\eta >0$ and fixed vertical wavenumber $\pi /d$, a horizontal Fourier mode obeys $(s+\eta (k^2+{\pi }^2/d^2){)}^2=i\alpha \pi Vk/d$. With $D=\alpha V{d}^2/({\pi }^2{\eta }^2)$ and $h=|k|d/\pi$, the growing branch has positive real part exactly when $|D|>2(1+h^2{)}^2/h$. Differentiation gives $3h^4+2h^2-1=0$, hence $h=1/\sqrt{3}$. Equality is a neutral oscillation, not exponential growth. The sign of $D$ reverses wave propagation without changing the growth threshold.

For isotropic monochromatic forcing, the longitudinal coefficient in a mean-field electromotive force is proportional to ${\int }_0^1{\mu }^2(1+h^2{\mu }^2{)}^{-2}d\mu =[\arctan h-h/(1+h^2)]/(2h^3)$, where $h$ is proportional to mean-field strength. This gives generic $|B{|}^{-3}$ quenching. The natural angular alpha tensor is anisotropic in a nonzero mean field: its transverse coefficient instead contains ${\int }_0^1(1-{\mu }^2)(1+h^2{\mu }^2{)}^{-2}d\mu$ and scales as $|B{|}^{-1}$. Only the longitudinal contraction enters $\mathcal{E}=\mathbf{\alpha }\mathbf{B}$. An effective scalar representation may therefore have every diagonal entry equal to the longitudinal coefficient without being the unique natural response tensor.

In a Parker dynamo wave model, an alpha effect of the form $\alpha ={ϵ}^{-1}{\alpha }_0\sin (\omega t/ϵ)$ has zero time average but can generate a finite averaged coupling. The method of multiple scales gives $\tilde{A}=-({\alpha }_0/\omega )\cos (\omega \tau )\overline{B}$ and $\tilde{B}=-(\Omega {\alpha }_0/{\omega }^2)\sin (\omega \tau ){\partial }_x\overline{B}$, where $\tau =t/ϵ$. Consequently $C=\Omega {\alpha }_0^2/(2{\omega }^2)$. With diffusion operator ${\partial }_x^2-{\ell }^2$, the averaged Fourier mode growth exponents are $-k^2-{\ell }^2\pm |k|\sqrt{\Omega C}$. For freely adjustable real $k$, growth requires $\Omega C>4{\ell }^2$. The result is an asymptotic averaged model for fixed nonzero $\omega$, not an exact replacement at finite $ϵ$.

Let $M_n={\sum }_{k\le n}{a}_k{Z}_k$, and stop on first reaching a fixed positive threshold. If the coefficient squares are summable, uniform integrability gives the stopped mean zero, with the terminal limit used on $T=\infty$. If the coefficient squares diverge, $\mathbb{P}(M_n\ge R)\to 1/2$ for each fixed $R$. Thus $\mathbb{P}({\sup }_n{M}_n=\infty )\ge 1/2$. This event is unaffected by changing finitely many of the independent variables, so the Kolmogorov zero-one law makes its probability one. Hence $T<\infty$ and $M_T$ is at least the positive threshold; its expectation, finite or infinite, cannot be zero.

A rotating wave in a complex dynamo quenching model has constant amplitude and a common phase factor $e^{i\omega t}$. In the two-hemisphere model with toroidal coupling parameter ${ϵ}_2=0$, eliminating the poloidal amplitudes givesThe matrix is Hermitian, so its eigenvalue is real and a nonzero rotating wave with $\Omega \ne 0$ has $\omega =\pm 1$. Writing $a={\alpha }_0-2\omega /\Omega$, pure dipole and quadrupole waves have $|B_j{|}^2=(a-{ϵ}_1)/p$ and $(a+{ϵ}_1)/p$, respectively, when those quantities are positive.

The displayed entire function has exactly the periods $2\pi i\mathbb{Z}$: a period $c$ would make $e^z(e^c-1)$ a constant integer multiple of $2\pi i$, forcing $e^c=1$. If a polynomial addition relation existed, specialize its middle argument at a point with $e^c=\alpha >0$ irrational, avoiding the finitely many specializations making the entire polynomial vanish. With $t=e^z$, the relation becomes a nonzero finite sum of exponentials $e^{(m+\alpha n)t}$. These real exponents are distinct, so the largest one dominates as $t\to +\infty$, a contradiction. This demonstrates that an algebraic addition theorem is stronger than simple periodicity.

The Conway polynomial is normalized by ${\nabla }_U=1$ for the unknot, zero for a split unlink with more than one component, and the skein relation ${\nabla }_{L_{+}}-{\nabla }_{L_{-}}=z{\nabla }_{L_0}$. For a knot, the symmetric Alexander polynomial with value one at $t=1$ is ${\Delta }_K(t)={\nabla }_K(t^{1/2}-t^{-1/2})$. This fixes the Laurent-unit ambiguity of the Seifert-matrix presentation of the Alexander module.