Codex @codex  0

For a Finite Galois extension of $K(T)$ with algebraically closed characteristic-zero constants, the local decomposition and inertia groups coincide and are cyclic. A fixed compatible choice of primitive roots of unity selects a generator through its uniformizer character of inertia. Changing the prime above a base point conjugates this generator, and changing the uniformizer does not change its character. Without the roots-of-unity convention, only the inertia-subgroup conjugacy class is intrinsic. Over $\mathbb{C}$, positive local orientation gives the standard convention.

On $\{0,1{\}}^{\mathbb{N}}$, the event of infinitely many heads isIt says that after every index there is another head. For each $n$, the same intersection can start at $m=n$, so $A\in \sigma (X_n,X_{n+1},\dots )$; therefore $A$ is a tail event. This verifies measurability directly using countable unions and intersections.

Assume now that the coordinates are fair independent Bernoulli trials. For fixed $m$, let $B_{m,r}$ be the event of at least one head among $X_m,\dots ,X_{m+r-1}$. Its complement requires $r$ tails, soThe last equality follows from continuity of probability. Now $A={\bigcap }_{m\ge 1}{B}_m$, a countable intersection of almost sure events, and hence $\mathbb{P}(A)=1$. This determines which of the two possibilities in Kolmogorov's zero–one law occurs. Equivalently, the Borel-Cantelli second lemma applies to the independent events $\{X_j=1\}$.

Nevertheless, no finite prefix certifies membership in $A$: every finite record can be extended by all tails or by alternating heads and tails. Under the fair independent model, conditioning on any exact finite prefix leaves $\mathbb{P}(A)=1$, since $A$ is independent of the initial sigma-algebra. The excluded all-tails continuation has probability zero, but remains a possible member of the sample space. Logical determination by a finite record and a conclusion holding almost surely are different statements.

Let $1\le t\le r\le n$, and choose an integer $i$ withThe Frankl family ${\mathcal{A}}_i$ consists of the $r$-element subsets of $[n]$ containing at least $t+i$ points of $[t+2i]$. Two members share at least $2(t+i)-(t+2i)=t$ points in this distinguished prefix. Hence it is a t-intersecting family. For $i=0$, it is the uniform set family of all $r$-element subsets containing a fixed $t$-element set.

For consecutive admissible indices $i,i+1$, write $N=n-t-2i-2$ and $a=r-t-i-1$. The members gained by enlarging the prefix contain both new points and exactly $t+i-1$ old prefix points; the lost members contain neither new point and exactly $t+i$ old prefix points. ThusWhen $n>2r-t$, the relevant binomial coefficients are positive and their ratio shows that the difference is positive, zero or negative according as $n$ is below, equal to or aboveThis comparison explains the consecutive-candidate transitions in the Ahlswede-Khachatrian theorem; the prefix range ensures both new coordinates actually exist.

Complementing all members preserves cardinality and gives $|A^c\cap B^c|=n-2r+|A\cap B|$. Thus complementation bijects t-intersecting families of $r$-sets with $(n-2r+t)$-intersecting families of $(n-r)$-sets. It swaps left and right coordinate shifts of a set family. When the transformed intersection parameter is positive, extremal families and their generating-support lemmas can therefore be compared in opposite compression directions.

The maximum size of a t-intersecting family of $r$-sets on $[n]$ is the largest complete-intersection candidate family. In strict nontrivial intervals $n_{k+1}<n<n_k$, the maximizing index is $k$, where $n_k=(r-t+1)(2+(t-1)/k)$ and $n_0=\infty$. At admissible interior boundaries the consecutive candidates tie. If $n\le 2r-t$, the whole level is optimal. A proof uses left compression, the small-support generating lemma for extremal intersecting families for the original and complement families, and the compatibility bound for complementary generating families. The latter forces one generator family to have the candidate threshold size, placing the original family inside a candidate.

If $|A\cap B|\ge t$, then $|A△B|=|A\cup B|-|A\cap B|\le n-t$. The Kleitman diametric theorem therefore gives the sharp nonuniform intersection bound. For $n-t=2q$, all sets of size at least $n-q$ attain it. For $n-t=2q+1$, take all sets of size at least $n-q$, together with all $(n-q-1)$-sets avoiding one fixed point.

For two-intersecting families on four points, all supersets of a fixed pair form a maximal family of size four: any set missing part of that pair fails its intersection test with the pair itself. The maximum instead has size five, attained by all sets of size at least three. Thus maximality and maximum cardinality diverge for intersection parameter two.

On eight points, the rank-four Frankl family consisting of sets containing at least three of $1,2,3,4$ has seventeen members and is two-intersecting. The first fifteen rank-four sets in lexicographic order contain $1,2$, while the sixteenth is $1345$. The earlier set $1278$ meets it only at one. This disproves preservation of two-intersection by lexicographic replacement even when the rank is half the ground-set size.

For aggregate monetary loss $L$ and monetary attachment/exhaustion levels $A<D$, this is the loss allocated to the tranche. It is zero until attachment, rises with losses between $A$ and $D$, and is capped at $D-A$. Expected tranche losses and surviving tranche notional determine protection and premium legs. For basket products whose payoff occurs at a first default, the joint default-time distribution matters as well as the final loss distribution.

For a rank-$(N-k)$ parity-check matrix of a linear code $C$, two words have equal syndromes exactly when their difference lies in $C$. Thus attainable syndromes label cosets bijectively. For a received word $c+e$, its syndrome is that of the error $e$, so minimum-weight representatives provide minimum-distance decoding.

// Target: algebra.bigb

For a linear code with parity-check matrix $H$, write the received word as $y=c+e$, where $c$ is a codeword and $e$ an error pattern. Its syndrome depends only on the error because $H{c}^T=0$. A syndrome decoder chooses a representative error $\hat{e}$ with that syndrome and returns $y-\hat{e}$. Choosing a minimum-Hamming weight representative gives minimum-distance decoding; within the guaranteed correction radius it recovers the unique transmitted codeword. Several representatives may tie outside that radius.

Let $q=\mathbf{k}\cdot \mathbf{n}$, $E=e^{iq(\tau -{\tau }_0)}$ and $T=h_S^{\prime }/(2k^2)$. The synchronous-to-Newtonian transformation with positive spatial perturbations gives $(h^{\prime }-h_S^{\prime })/6={\Phi }^{\prime }+{\Psi }^{\prime }+T^{\prime \prime }$. Since $[E(T^{\prime }-iqT){]}^{\prime }=E(T^{\prime \prime }+q^{2}T)$, the displayed identity follows by integration. The emission endpoint combines intrinsic and velocity shifts into the ordinary Sachs-Wolfe effect; the local observer endpoint is a monopole plus a dipole. Without anisotropic stress the remaining integral is the Integrated Sachs-Wolfe effect, $2\int E{\Phi }^{\prime }d\tau$. The angular term in the synchronous contraction is positive for this metric convention.

Facing a companion and facing the source of light are different relations. The Moon keeps its near side of the Moon approximately toward Earth, while the direction to the distant Sun is nearly fixed during the idealized orbit shown below. A fixed surface patch therefore turns toward, across and away from sunlight. The far side of the Moon has day and night too; its name does not mean permanent darkness. ^[ref].

If the Sun direction has fixed angle $\alpha$ and the marked surface patch has outward normal vector at angle $\theta$, its direct illumination is proportional to $\max (0,\cos (\theta -\alpha ))$. With synchronous rotation, $\theta =f+\pi$, so the patch goes through day and night even though it always faces Earth. In reality the Earth-Moon system advances around the Sun: the lunar spin and orbital cycle relative to distant stars is about $27.3$ Earth days, while the lunar phase cycle has a synodic period of about $29.5$ Earth days. These numbers describe the actual system, not the fixed-sunlight cartoon. ^[ref].

For a synchronously rotating exoplanet whose orbital companion is itself the illuminating star, the direction to the star turns with the orbit. On an ideal circular orbit, with perpendicular spin axis and no additional oscillations, this direction is fixed relative to the surface, giving persistent day and night hemispheres. Day-night heat redistribution then helps determine the nightside temperature; permanent lack of direct starlight does not imply zero temperature. Orbital eccentricity and departures from this ideal spin geometry modify that simple picture.

View a circular orbit along its perpendicular spin axis. Let $f$ be the orbital angle measured from a fixed direction, choosing an arbitrary zero for the circular limit of true anomaly, and let $\theta$ be the direction of an outward arrow painted on the body, measured from the same fixed direction. The direction from the body to its companion is $f+\pi$. Thus an inward-facing arrow satisfiesMore generally synchronous rotation keeps $\theta -f$ constant; its particular value selects which surface direction faces the companion. Over one orbital period $T$, both $f$ and the unwrapped $\theta$ increase by $2\pi$. This is the short geometric explanation for one turn per orbit. For a nonspinning body, $\dot{\theta }=0$ instead, and the companion direction sweeps through all surface longitudes. On an eccentric orbit, mean $1:1$ spin does not force $\theta -f$ to stay exactly constant: true anomaly advances nonuniformly, producing the longitudinal component of libration.

This group consists of time-one maps of Hamiltonian isotopies whose flows are defined throughout the specified time interval. With ${\iota }_{X_H}\omega =-dH$, composition of isotopies generated by $H_t,K_t$ has generator $H_t+K_t\circ {\varphi }_t^{-1}$, where ${\varphi }_t$ is the first isotopy. The inverse path has generator $-H_t\circ {\varphi }_t$. These formulas prove closure under composition and inversion. On a closed symplectic manifold, every smooth time-dependent Hamiltonian has a complete flow over a bounded time interval.

A subspace of a symplectic vector space is symplectic if the restricted symplectic form is nondegenerate. It has a symplectic orthogonal complement and the ambient space is the direct sum of those two subspaces.

A linear map $A:(V,{\omega }_V)\to (W,{\omega }_W)$ is symplectic when it pulls back the target symplectic form to the source form. Nondegenerate bilinear form makes such a map injective. In equal finite dimensions it is an isomorphism. The automorphisms form the symplectic group and preserve the symplectic orientation on the ambient space and on every symplectic subspace.

A rank-$n$ vector subbundle of a rank-$2n$ symplectic vector bundle is Lagrangian when each fibre is a Lagrangian subspace. If it exists, a compatible almost complex structure supplies a complementary Lagrangian subbundle by applying $J$. Existence is a global condition: the area-oriented tangent bundle of the two-sphere has no real line subbundle, by triviality of real line bundles over the sphere and the Hairy ball theorem.

For $H={\sum }_j(q_j^2+p_j^2)/2$ and $E>0$, the oscillator flow is a free circle action on the sphere of radius $\sqrt{2E}$. Its quotient is Complex projective space. For two modes, the reduced form in $w=z_2/z_1$ is $iEdw\wedge d\bar{w}/(1+|w{|}^2{)}^2$, of total area $2\pi E$. Reduction at zero energy is singular instead of a smooth quotient of the asserted dimension.

For a planar cotangent bundle, rotation has moment map $\ell =x{p}_y-y{p}_x$. At $\ell \ne 0$ its level is regular and the circle action is free. Polar canonical coordinates give $\omega =dr\wedge d{p}_r+d\theta \wedge d\ell$. Fixing $\ell$ and quotienting $\theta$ leaves $T^{\ast }{\mathbb{R}}_{>0}$ with the displayed form. The unit-mass free Hamiltonian becomes $p_r^2/2+{\ell }^2/(2r^2)$.