The quotient model of a lens space splits into two solid tori. In the boundary angle lattice ${\mathbb{Z}}^2+\mathbb{Z}(1/p,q/p)$, choose ${\mu }_1=(0,1)$ and ${\lambda }_1=(1/p,q/p)$. The other meridian is $(1,0)=-q{\mu }_1+p{\lambda }_1$. A second column completing this primitive slope to a determinant-one matrix specifies a gluing map after reversing one boundary-basis convention. Different completions differ by a meridional Dehn twist extending across the second solid torus, so they give the same three-manifold.

This second-order linear differential equation extends the Legendre differential equation by its order parameter $m$. Its solutions are associated Legendre functions. In a degree-one potential problem, with $x=\tanh \eta$, two explicit solutions are $e^{m\eta }(m-\tanh \eta )$ and $e^{-m\eta }(m+\tanh \eta )$. Their Wronskian with respect to $\eta$ is $2m(1-m^2)$, so the two formulas are not a fundamental pair at $m=0$ or $m=\pm 1$.

On a hemisphere of SU(2) as the three-sphere, write $A=u_{0}I+i{u}_i{\sigma }_i$, with real coefficients satisfying $u_0^2+|\mathbf{u}{|}^2=1$. Right multiplication by $I+i{v}_j{\sigma }_j$ gives $d{u}_i=(u_0{\delta }_{ij}+u_k{ϵ}_{jki})v_j$. This right multiplication flow defines left-invariant tangent fields. With the displayed factor $i$, they obey $T_{j}A=-A{\sigma }_j$, since ${\partial }_{u_i}{u}_0=-u_i/u_0$. Applying the commutator to $A$ and using the Pauli matrix commutator identity gives $[T_i,T_j]=-2i{ϵ}_{ijk}{T}_k$. The entries of $A$ contain every coordinate, so this determines the vector-field identity, not just its action in one isolated representation.

For $F⊣G$ inducing $T=GF$, the Eilenberg-Moore comparison functor has a left adjoint exactly when every algebra $(A,a)$ has a coequalizer $q_a:FA\to L(A,a)$ of $Fa,{\epsilon }_{FA}:FTA\rightrightarrows FA$. Transposition identifies maps equalizing this pair with monad algebra morphisms $(A,a)\to K(D)$, proving both directions of the criterion. The unit is $\lambda =G{q}_a{\eta }_A$ and satisfies $G{q}_a=\lambda a$. In the base category, $a:TA\to A$ is a split coequalizer of $Ta,{\mu }_A:T^{2}A\rightrightarrows TA$, split by ${\eta }_A$ and ${\eta }_{TA}$. Consequently $\lambda$ is invertible exactly when $G{q}_a$ is also a coequalizer. The unit is invertible everywhere precisely when $G$ preserves all these specified coequalizers; preservation of arbitrary coequalizers is not required.

With signed active covariate columns $X_A^s=(s_j{x}_j{)}_{j\in A}$ and their positive-definite Gram matrix $G_A=(X_A^s{)}^T{X}_A^s$, the displayed direction has unit norm and $(X_A^s{)}^{T}u=\alpha 1$. Thus it makes equal angles with unit-length signed active covariates, and movement along it reduces their signed residual correlations at the common rate $\alpha$. The regression coefficient direction is $d_A=\mathrm{diag}(s_A)w$, with inactive components zero.

For initially pristine gas, zero bulk mass return, a constant stellar yield $y$ and a well-mixed galactic outflow of mass-loading factor $\nu$, gas conservation gives $d{M}_g=-(1+\nu )d{M}_s$ and metal accounting gives $M_{g}dZ=yd{M}_s$. Consequently $M_g=M_{g,0}{e}^{-Z/y_e}$ with $y_e=y/(1+\nu )$. Since $d{M}_s/dZ=M_g/y$, complete depletion gives a normalized mass-weighted metallicity distribution function $p(Z)=e^{-Z/y_e}/y_e$ and mean stellar metallicity $y_e$. A time-independent initial mass function makes the distribution of surviving tracers proportional to this distribution; a sample selected by light need not have the same weights.

Let $F,G\ge 0$, $1<p<\infty$, and $G\in L^p$. Assume the displayed tail inequality and finite positive-threshold superlevel measures. First split $G$ at $a/2$ to obtain $a\mu \{F>2a\}\le {\int }_{\{G>a\}}G$. Layer cake representation and Tonelli theorem imply $F\in L^p$. Integrating the original inequality against $p{a}^{p-2}$ then gives $\Vert F{\Vert }_p^p\le q\int G{F}^{p-1}\le q\Vert G{\Vert }_p\Vert F{\Vert }_p^{p-1}$, with $q=p/(p-1)$. This is the analytic estimate behind the Doob Lp maximal inequality.

Write the Weierstrass elliptic function as $\wp (z)=z^{-2}+{\sum }_{n\ge 1}{c}_n{z}^{2n}$, with $c_1=3G_4$, $c_2=5G_6$ and $c_n=(2n+1)G_{2n+2}$. Differentiate the Weierstrass elliptic differential equation to obtain ${\wp }^{\prime \prime }=6{\wp }^2-g_2/2$. Comparing coefficients of $z^{2n-2}$ gives the displayed recurrence for $n\ge 3$. Every denominator is positive, so induction expresses every higher lattice Eisenstein series as a polynomial in $G_4,G_6$ with positive rational coefficients on its occurring monomials. In particular $G_8=3G_4^2/7$ and $G_{10}=5G_4{G}_6/11$.

A positive bump of the real line is an order automorphism whose nonempty support of a permutation is one open interval and which moves every point of that interval to the right. It fixes the complement. Iterating any interior point tends to the right endpoint in positive time and to the left endpoint in negative time.

An ordered permutation group is order n-transitive if it can send any increasing $n$-tuple to any other increasing $n$-tuple. This compares tuples with the same relative order, unlike unrestricted n-transitive group actions. In the usual transitive chain setting, order two-transitivity implies every finite order-transitivity for a lattice-ordered permutation group.

Every lattice-ordered group embeds in a lattice-ordered permutation group on one totally ordered set. Choose a value in a lattice-ordered group for each nonidentity element, act on its ordered right cosets, and combine these chains in an order sum. The original element moves its own value's identity coset, making the action faithful. The coset lattice formulas make the embedding preserve both join and meet.

A transitive lattice-ordered permutation group is order-primitive when its only invariant equivalence relations with convex classes are equality and the universal relation. This concerns convex blocks in a chain, rather than arbitrary blocks of a primitive group action.

Let $Z_n$ clip the minimum of $k$ independent exponential sample means into $[a,b]$, with $0<a<1/\lambda <b$. The rate function under scalar clipping is the minimum cost on this interval. Apply Varadhan's lemma to $\log z$. On the lower side its objective is increasing; on the upper side its derivative is $(k+1)/z-k\lambda$. Thus the optimum is $z_{\ast }=(k+1)/(k\lambda )$ once $k\ge (\lambda b-1{)}^{-1}$, giving the displayed limit. The expectation is of the high power, rather than a power of the mean; the latter has limit $-\log \lambda$.

In the self-coupling-dominated approximation $d{\alpha }_H/d\ln {\mu }^2=b{\alpha }_H^2$, $b>0$, the running coupling reaches a perturbative Landau pole at the displayed scale. With the conventional Higgs potential, $m_H^2=2{\lambda }_H{v}^2$ and one can normalize ${\alpha }_H={\lambda }_H/(4\pi )$. A larger Higgs boson mass therefore decreases the logarithmic interval before the self-coupling becomes strong. Requiring perturbative validity up to a fixed high cutoff gives an upper Higgs-mass bound in this approximation. The full Standard Model beta function also contains gauge and Yukawa couplings; its sign is not universally positive. A perturbative pole is not a theorem excluding every nonperturbative ultraviolet completion.

The covariant Landau gauge propagator is the transverse limit of the inversion of the gauge-fixed Maxwell kinetic operator. Contracting with $k^{\mu }$ gives zero away from the pole. The Lorenz gauge condition still has residual transformations satisfying $\square \chi =0$, so boundary conditions and the pole prescription complete the choice.

Let $A$ code application and $C$ code numeral quotation. For a lambda term $F$, choose a fresh $x$ and put $H=\lambda x.F(Ax(Cx))$ and $Q=Hd(H)$. Then $Ad(H)(Cd(H)){\equiv }_{\beta }d(Q)$, so $Q{\equiv }_{\beta }Fd(Q)$. This diagonal argument uses the syntax of the actual term $Q$, including any free variables, not an equivalence-class code.

A leaf of a Lagrangian foliation is an immersed Lagrangian submanifold tangent to its integrable distribution. A prequantum line bundle restricts to a flat connection along such a leaf, because its curvature is proportional to the symplectic form. Nonzero global parallel sections require trivial holonomy along every leaf loop.

At distinct interpolation nodes, differentiating Lagrange interpolation gives $p^{(k)}(x)={\sum }_{i}p(t_i){\ell }_i^{(k)}(x)$. The triangle inequality gives the displayed upper bound, attained by prescribing each nodal value to be the sign of the corresponding derivative weight. For increasingly ordered real nodes, the Markov interlacing lemma and monotonicity of polynomial critical points in their roots imply that $(-1{)}^i{\ell }_i^{(k)}(x)$ has at most one sign change, ignoring zero weights. Thus an extremizer has alternating nodal signs with at most one reversal of that alternating pattern. This reduces a choice among $2^{n+1}$ sign patterns to $n+1$ candidate polynomials.

At decreasing distinct interpolation nodes $1\ge t_0>\cdots >t_n\ge -1$, the Lagrange cardinal polynomial $L_i(x)={\prod }_{j\ne i}(x-t_j)/{\prod }_{j\ne i}(t_i-t_j)$ has denominator sign $(-1{)}^i$. For $1\le k\le n$, its numerator's $k$th derivative at one is $k!$ times a sum of products of $n-k$ nonnegative factors $1-t_j$. At most one factor vanishes, and at least one product avoids it, so that sum is strictly positive. The same alternating nodal signs therefore maximize every endpoint derivative simultaneously.

For a cusp form $f=\sum c_n{q}^n$ and coprime integers $a,N$, the additive twist inserts the additive character $e^{2\pi ian/N}$. It is the Dirichlet series associated with the rational translate $f(a/N+\tau )$. Its completed Mellin transform uses the scale $N^s$ and the Gamma function.