If the similarity functions of a turbulent boundary layer are differentiable at zero and have finite positive neutral values, expanding $f=a+dz/\ell +O((z/\ell {)}^2)$ gives a shear proportional to $1/z-d/(a\ell )$. Integration yields a logarithmic term plus a linear correction. Dimensional reasoning alone does not fix the coefficients or guarantee this regular neutral limit. The gradient Richardson number is of order $z/\ell$, so $|z/\ell |\ll 1$ is a shear-dominated, nearly neutral regime on either side of stability.

A monad morphism $\alpha :S\Rightarrow T$ induces a functor ${\mathcal{C}}^T\to {\mathcal{C}}^S$ between Eilenberg-Moore categories, identical on underlying objects and arrows. Its unit equation gives the restricted algebra's unit law. Naturality and the multiplication equation give $(a{\alpha }_A)S(a{\alpha }_A)=aTa{\alpha }_{TA}S{\alpha }_A=a{\mu }_A^{T}T{\alpha }_A{\alpha }_{SA}=a{\alpha }_A{\mu }_A^S$. Naturality also preserves algebra morphisms.

Integrate field modes with momenta between a fixed microscopic ultraviolet cutoff ${\Lambda }_0$ and a lower cutoff $\Lambda$, retaining the modes below $\Lambda$ as backgrounds. The exact statistical Hamiltonian includes all generated interactions and field-independent constants. Integrating its retained modes reproduces the original partition function, so long-distance observables do not depend on the arbitrary intermediate cutoff. With sources retained, the corresponding coarse observables reproduce long-distance response and connected correlation functions. A local gradient expansion or a finite-coupling truncation is an additional approximation, rather than a consequence of cutoff independence. The momentum-shell construction is also explained in the Cambridge statistical field theory lecture notes.

An $R$-module homomorphism is an additive map $f:M\to N$ satisfying $f(rm)=rf(m)$.

A locally free module is one whose associated sheaf is locally a free module. In the finite constant-rank case, it is a finite projective module whose localization at a prime ideal has rank $r$ everywhere. Equivalently, it becomes $A_{f_i}^r$ on a finite cover by principal open subsets. Its associated sheaf is a locally free sheaf and defines an algebraic vector bundle.

A surjection onto such a module splits. If its source is also a finite projective module, the kernel is finite projective as a direct summand. A source hypothesis is essential: the projection $\mathbb{Z}\oplus \mathbb{Z}/2\mathbb{Z}\to \mathbb{Z}$ splits and has a free target, but its kernel $\mathbb{Z}/2\mathbb{Z}$ is not projective.

A module is finitely presented if it has finitely many generators and finitely many defining relations, equivalently an exact sequence of the displayed form with m and n finite. Over a commutative ring a relation matrix records the presentation, and its minors define the Fitting ideals.

// Target: knot-theory.bigb

A linear operator $T:V\to V$ on a vector space over a field $k$ makes $V$ a module over the polynomial ring $k[t]$: let $t$ act as $T$, let constants act as the original scalars, and define $p(t)\cdot v=p(T)v$. The identity $(pq)(T)=p(T)q(T)$ for polynomials $p,q$ verifies the multiplication axiom. The submodules are precisely the invariant subspaces of $T$. This construction is described in Keith Conrad's Introductory Notes on Modules, section 6.

For a computed example, take $k=\mathbb{Q}$, $V=k^2$, andThen $T{e}_2=e_1$, $T{e}_1=0$, and $T^2=0$. ConsequentlyEvery $(a,b)$ equals $(b+at)\cdot e_2$. The module homomorphism $k[t]\to V$, $p\mapsto p(T)e_2$, is therefore surjective, and its kernel is the ideal $(t^2)$. It induces a module isomorphism $V\cong k[t]/(t^2)$. This is a cyclic module and a torsion module over $k[t]$, since $t^{2}V=0$. Over $k$ it is a free module with basis $(e_1,e_2)$ and has no nonzero torsion elements of a module. Over $\mathbb{Q}$ its nonzero elements also have infinite additive order of a group element. Changing scalars changes the questions one asks about the same addition law.

Every abelian group has exactly one unital module structure over the integers. For $n>0$, define $nm$ by adding $m$ to itself $n$ times, put $0m=0$, and put $(-n)m=-(nm)$. The module axioms force this action, so no additional choice is involved. Moreover, every Abelian group homomorphism respects it and is a module homomorphism. See David McKinnon's Modules, the integer-scalar example.

For example, in the free module ${\mathbb{Z}}^2$,In the cyclic group $\mathbb{Z}/6\mathbb{Z}$, the same repeated-addition rule gives $4\overline{2}=\overline{8}=\overline{2}$ and $3\overline{2}=\overline{0}$. The scalar $3$ is nonzero in the scalar ring $\mathbb{Z}$, even though it kills the nonzero module element $\overline{2}$. Thus the scalar ring and the module must be kept distinct.

For a finite-index modular subgroup, a Manin symbol records a coset and its associated oriented unimodular edge. Over the rationals the presentation has the relations $[g]+[gS]=0$ and $[g]+[gST]+[g(ST{)}^2]=0$, for inversion $S$ and translation $T$. They express reversing an edge and the boundary of a triangle. Integral computation also removes torsion caused by elliptic stabilizers. The boundary map sends the symbol to its terminal cusp minus its initial cusp; its kernel is absolute homology.

A function on the complex upper half-plane is modular invariant if it is unchanged by the fractional-linear action of the modular group. This does not imply holomorphicity: a nonholomorphic Eisenstein series is modular invariant, whereas a modular function additionally satisfies meromorphicity and cusp conditions.

A Dirichlet character $\psi$ modulo $N$ specifying the displayed transformation law is the nebentypus of a modular form. The action of $-I$ forces $\psi (-1)=(-1{)}^k$ for a nonzero form. It describes a character eigenspace of the diamond action rather than an additional analytic condition at modular cusps.

The integral representation of the modified Bessel function of the second kind gives $K_0(x)={\int }_0^{\infty }{e}^{-x\cosh t}dt$ for $x>0$. Taking the Fourier transform of the even continuation and setting $u=\sinh t$ gives $2{\int }_0^{\infty }(1+k^2+u^2{)}^{-1}du=\pi /\sqrt{1+k^2}$. The large-wavenumber terms $\pi /|k|-\pi /(2|k{|}^3)$ are also obtained from the local logarithmic singularity $-\log |x|-(x^2/4)\log |x|$ using the Fourier transform of the absolute logarithm.

The positive occupation operators are $N_e={\sum }_s\int d^{3}p{a}_{\mathbf{p}}^{s†}{a}_{\mathbf{p}}^s/(2\pi {)}^3$ and $N_{\bar{e}}={\sum }_s\int d^{3}p{b}_{\mathbf{p}}^{s†}{b}_{\mathbf{p}}^s/(2\pi {)}^3$. Both commute with the free Hamiltonian. The normal-ordered local density integral $\int :{\psi }^{†}\psi :d^{3}x$ is their difference, not their sum. It generates the global phase symmetry. For electron charge $-e$, electric charge is $-eQ$. Pair creation can change total occupation while preserving this net charge.

For any continuous map between the displayed Real projective spaces, the induced map on cohomology sends the degree-one generator to $\alpha x+\beta y$. Its $2n$th power is $\left(\genfrac{}{}{0ex}{}{2n}{n}\right){\alpha }^n{\beta }^n{x}^n{y}^n$, since all other terms exceed the truncation exponent of one factor. The coefficient is zero modulo two because $\left(\genfrac{}{}{0ex}{}{2n}{n}\right)=2\left(\genfrac{}{}{0ex}{}{2n-1}{n-1}\right)$. Naturality of the evaluation pairing then makes the top induced map on homology zero. This obstruction does not apply to $n=0$, when both spaces are points.

The unit sphere bundle of the tautological line is the double cover $S^n\to {\mathbb{RP}}^n$. In its Gysin sequence, multiplication by the mod-two Euler class $a$ is an isomorphism in successive interior degrees, since the sphere has no interior cohomology. Connectedness in degree zero makes multiplication into degree one injective. At the top, vanishing above dimension $n$ makes the transfer from $H^n(S^n;{\mathbb{F}}_2)$ surjective; together with that injection it makes the top group one-dimensional. Thus every $a^j$, $0\le j\le n$, is the nonzero generator and the next power vanishes.

A Möbius transformation with two distinct fixed points is conjugate to $z\mapsto az$. It is loxodromic when $|a|\ne 1$; on three-dimensional hyperbolic space it translates along the axis joining its fixed points and rotates around that axis by $\arg a$. The positive-real multiplier case has no rotation and is also called hyperbolic. When listing elliptic, parabolic, hyperbolic and loxodromic as separate types, loxodromic refers to the remaining case with nonzero rotation. A determinant-one matrix with nonreal trace has reciprocal eigenvalues that are neither real nor of modulus one, so it supplies this latter case.

The Möbius transformation $M(z)=(3z-1)/(3-z)$ fixes the antipodal pair $1,-1$ under stereographic projection. But $M^{\prime }(1)=2$, so it multiplies infinitesimal spherical lengths at that fixed point by two, as follows from the round line element $2|dz|/(1+|z{|}^2)$. It is not an isometry and hence not a sphere rotation. Fixing an antipodal pair is therefore necessary for a nontrivial rotational Möbius transformation, but is not sufficient.

For $|w|<R$, this Möbius transformation maps the radius-$R$ disc to the unit disc, vanishes at $w$ and has modulus one on the boundary. Expanding gives $|R^2-\overline{w}z{|}^2=R^2|z-w{|}^2$ when $|z|=R$; its pole lies outside the disc. Dividing a holomorphic function by factors corresponding to its interior zeros removes those zeros without changing its boundary modulus. This gives a direct factorization proof of Jensen's formula.

Let $T_b(z)=z+b$ and $I(z)=1/z$ on the Riemann sphere. For $s\ne 0$, direct composition gives the displayed identity. Every nonzero complex scaling $z\mapsto \lambda z$ is obtained by choosing $s^2=-\lambda$. Since translations, scalings and reciprocal inversion generate the Möbius group, translations and reciprocal inversion alone also generate it.

A nonconstant Möbius transformation mapping the real line into the unit circle has the displayed form with $|\lambda |=1$ and $k$ nonreal. Indeed $|ax+b{|}^2=|cx+d{|}^2$ for real $x$ yields $|a|=|c|\ne 0$, $|b/a|=|d/c|$ and $\mathrm{Re}(b/a)=\mathrm{Re}(d/c)$. Thus $d/c$ is $b/a$ or its complex conjugate; the first possibility makes the map constant, so the second is necessary. Conjugate numerator and denominator have equal modulus on the real line, proving the converse as well.