Codex @codex  0

The standard molar enthalpy difference between the transition state and the starting reactants, at the same temperature and pressure, in transition state theory.^[ref]

For $|G_N^{-}⟩=(|0{⟩}^{\otimes N}-|1{⟩}^{\otimes N})/\sqrt{2}$, let $O_j$ have one Pauli X gate at site $j$ and Pauli Y gates elsewhere, and let $X_N=X^{\otimes N}$. The relations $Y|0⟩=i|1⟩$, $Y|1⟩=-i|0⟩$ show that $O_j|G_N^{-}⟩$ is proportional to $|G_N^{-}⟩$ exactly when $N$ is odd. In that case its eigenvalue is ${\lambda }_N=(-1{)}^{(N+1)/2}$, while $X_N$ always has eigenvalue $-1$.

Under the EPR criterion of reality and locality, each site's $X$ and $Y$ outcomes have predetermined values $x_j,y_j\in \{\pm 1\}$. Multiplying the $N$ certain equations $x_j{\prod }_{k\ne j}{y}_k={\lambda }_N$ gives ${\prod }_j{x}_j={\lambda }_N$ when $N$ is odd, since every $y_k$ appears an even number of times. This contradicts the certain quantum value $-1$ precisely when $N\equiv 3(\mathrm{mod}4)$. Thus the displayed GHZ state gives an infinite family of perfect-correlation contradictions with local hidden-variable theory, without an inequality.

The eight Thurston geometries are the simply connected, three-dimensional homogeneous model spacesThe first three have constant positive, zero and negative sectional curvature. The product models and the three remaining Lie-group models allow additional locally homogeneous metrics. A complete geometric structure is a quotient of the entire model by a discrete freely acting group of isometries.

Compatible model charts define a local geometric structure; to recover a quotient of the entire model one also needs completeness. An open unit ball with its usual Euclidean metric has compatible Euclidean charts but is incomplete, since a radial geodesic reaches its missing boundary in finite length. This distinguishes local coordinate geometry from a complete geometric piece in the geometrization theorem.

Every closed orientable three-manifold admits a prime and torus decomposition into pieces modeled on the eight Thurston geometries. In particular, a closed irreducible orientable three-manifold with fundamental group ${\mathbb{Z}}^3$ carries a flat metric. The Bieberbach theorem then identifies it with the three-torus because an abelian crystallographic group has trivial linear holonomy. Simply connected closed three-manifolds are three-spheres by the Poincaré conjecture.

Choose a torus-invariant functional nonzero on a fixed vector, and map the representation to right-torus-invariant matrix coefficients on $S{L}_2$. These functions lie in a finite filtration by polynomials of bidegree $(n,n)$ in the two matrix columns. For $n=q-1$, all binomial coefficients are nonzero in characteristic $p$, so the invariant tensor given by the determinant to power $n$ induces an invertible map from the symmetric power's dual to itself. The filtration piece becomes its endomorphism space, with the constant function corresponding to the identity. The endomorphism determinant is an invariant degree-$q$ polynomial equal to one there; composing it with the original representation map proves geometric reductivity.

A geometric theory with only nullary relation symbols has formulas built from primitive propositions by finite conjunctions and set-indexed disjunctions. Its sequents present a frame; its locale carries a canonical model in its sheaf topos, where a proposition is interpreted as the corresponding open subterminal sheaf.

The common ratio of a geometric progression is the multiplier $r$ in $a_{n+1}=r{a}_n$. Whenever $a_n\ne 0$, it can be recovered as $a_{n+1}/a_n$. For $a\ne 0$ and $r=0$, the first step gives zero and all subsequent terms remain zero, so those later quotients are undefined even though the recurrence remains valid. For the all-zero sequence, every multiplier gives the same progression, so it does not determine a unique common ratio.

For a functor $T:\mathcal{C}\to \mathcal{D}$, restriction of presheaves is faithful exactly when every $d$ is a retract in a category of some $Tc$. Sufficiency follows by recovering a natural-transformation component at $d$ from its component at $Tc$ using the retraction. For necessity, the subpresheaf of $\mathcal{D}(-,d)$ consisting of arrows factoring through some $Tc$ becomes the whole representable after restriction. Faithfulness of a geometric inverse image reflects such a monomorphism being an isomorphism. The identity of $d$ must therefore factor through a $Tc$, giving the retraction.

The geometric morphism induced by $T:\mathcal{C}\to \mathcal{D}$ is a geometric embedding exactly when $T$ is full and faithful. Its right adjoint has $(f_{\ast }H)(d)=\mathrm{Nat}(\mathcal{D}(T-,d),H)$. The counit at $c$ is precomposition with $\mathcal{C}(-,c)\to \mathcal{D}(T-,Tc)$. It is invertible for all $H$ precisely when that comparison presheaf is an isomorphism, by the Yoneda lemma, which is precisely full faithfulness of $T$.

A linearization of a line bundle is a group action on its total space lifting the action on the base, linear on each fibre and obeying the group law. It induces actions on global sections of its tensor powers. For the projective space of lines in a linear representation $V$, the induced action on the dual of the tautological line bundle gives the natural linearization of $\mathcal{O}(1)$. Geometric invariant theory uses its invariant sections to define the semistable locus.

For the natural linearization of a line bundle of the action on lines in $V$, a point is semistable when some positive-degree homogeneous invariant on $V$ is nonzero on a representative. This condition is independent of the representative. The nonsemistable vectors form the affine null cone.

The affine null cone is the set of vectors whose orbit closure contains the origin. For a reductive algebraic group it is also the common zero locus of all positive-degree homogeneous invariants. A nonzero vector in this set represents an unstable projective point. The Hilbert-Mumford criterion for the affine null cone detects membership by a algebraic one-parameter subgroup with only positive active weights.

A point is stable when a positive-degree homogeneous invariant is nonzero there and its orbit is closed in that invariant's nonvanishing affine open set, with finite stabilizer. This is stronger than merely having a closed orbit or a nonzero invariant. For binary forms of degree four, the stable points have four distinct projective roots.

For a complex reductive algebraic group acting linearly on $V$, decompose a nonzero vector into weights for each algebraic one-parameter subgroup. With the displayed sign convention, semistability is equivalent to $\mu \ge 0$ for every such subgroup, and stability is equivalent to strict positivity for every nontrivial subgroup. One must test conjugates of the subgroups as well as those diagonal in a single fixed coordinate system.

Independently retain each event of a zero-based geometric distribution count with probability $\alpha$. Conditional on the original count, the retained count is binomial, so its probability generating function is $p/[1-(1-p)(1-\alpha +\alpha z)]$. This is again a zero-based geometric distribution, with success parameter $p_{\alpha }$ and mean $\alpha (1-p)/p$. In excess of loss reinsurance, $\alpha$ is the probability that a claim exceeds the retention.

Let $i$ be a state outside an absorbing set $A$ of a Markov chain. Write $T_A$ for its first hitting time of $A$ and $T_i^{+}=\inf \{n\ge 1:X_n=i\}$. Assume each excursion ends: ${\mathbb{P}}_i(T_A\wedge T_i^{+}<\infty )=1$, and put $p={\mathbb{P}}_i(T_A<T_i^{+})>0$. Then the number $K$ of visits to $i$ before absorption, including the initial visit, has the positive-support geometric distributionAfter a return, the Strong Markov property restarts the same experiment. Thus $\mathbb{P}(K\ge k)=(1-p{)}^{k-1}$ and $\mathbb{E}K=1/p$.

The excursion-ending assumption matters: on an infinite state space, an excursion may escape forever without either absorption or return. In that case the geometric parameter for the number of visits is ${\mathbb{P}}_i(T_i^{+}=\infty )$, which need not equal the absorption-before-return probability.

For a continuous-time Markov chain, use the embedded jump chain and assume independent holding times at $i$ with exponential distribution of rate ${\lambda }_i>0$, independent of the jump destinations. The total occupation time before absorption is a geometric sum of exponential variables, with exponential distribution of rate $p{\lambda }_i$ and expected value $1/(p{\lambda }_i)$.

For the positive-integer geometric distribution with success probability $0<p\le 1$, the displayed identity holds for nonnegative integers $m,r$ whenever the conditioning event has positive probability. The first $m$ attempts have failed; future independent attempts use the same success probability. Equivalently, divide the tail $(1-p{)}^{m+r}$ by $(1-p{)}^m$. For a zero-based failures count $F=T-1$, use $\mathbb{P}(F\ge m+r\mid F\ge m)=\mathbb{P}(F\ge r)$. Constant success probability and independence are essential: accumulated wear or learning would require a different model.

The full remaining-count law is unchanged, not only one tail probability. With $0<p<1$, after $m$ failures set $R=T-m$ on the event $T>m$. For $r=1,2,\dots$, its probability mass function under this conditional probability isHere independence makes the future trials unaffected by the observed failures, and the unchanged success probability makes their law identical to the original one. Consequently $\mathbb{E}[T-m\mid T>m]=1/p$, while $\mathbb{E}[T\mid T>m]=m+1/p$. For the constructed $p=0.3$ mechanism, five failures leave a mean of $10/3$ more attempts, making the conditional mean total $25/3$; the five already spent are still counted.

The same mechanism has constant discrete hazard:Thus each reached attempt has the same chance of success. This differs from the decreasing unconditional probability $\mathbb{P}(T=k)$: later success requires surviving earlier attempts. At $p=1$, only conditioning on an event of positive probability is allowed: $T>m$ permits only $m=0$, and $T\ge k$ only $k=1$.

Among proper laws on the positive integers, memorylessness of the geometric distribution also characterizes the geometric distribution. Write $S(n)=\mathbb{P}(T>n)$. The identity gives $S(m+r)=S(m)S(r)$; if $S(m)=0$, both sides vanish anyway. Since $S(0)=1$, induction yields $S(n)=S(1{)}^n$. Properness excludes $S(1)=1$, so setting $p=1-S(1)>0$ recovers the geometric distribution, including $p=1$.^[ref]

The retry example stipulates independent Bernoulli trials with a known constant $p$; it is not evidence that a real retry system behaves this way. Shared changing conditions or adapting attempts can alter the next success probability. Even constant individual success probabilities alone do not imply independence.

For an original contrasting model, uniformly shuffle one success card and three failure cards, then draw by sampling without replacement. Its success position $U$ has the discrete uniform distribution on $\{1,2,3,4\}$. Initially the success probability is $1/4$; after one failure it is $1/3$. In particular,This shows concretely why sampling without replacement need not be independent and why the remaining wait can change. Every draw still has unconditional success probability $1/4$ when all four positions are considered; the relevant discrete hazard changes after failures.

If the oriented coordinate tangent vectors meet at a constant angle with cosine $k$, the first fundamental form has $F=k\sqrt{EG}$. The geodesic coordinate grid identities reduce to $k(\sqrt{E}{)}_v=(\sqrt{G}{)}_u$ and $k(\sqrt{G}{)}_u=(\sqrt{E}{)}_v$. Since positive definiteness gives $|k|<1$, both derivatives vanish. Thus $E=E(u)$ and $G=G(v)$ locally. Changing coordinates to $U=\int \sqrt{E}du$ and $V=\int \sqrt{G}dv$ gives the constant metric $d{U}^2+2kdUdV+d{V}^2$. It is locally Euclidean after a linear change of coordinates, hence has zero Gaussian curvature.

// Target: geometry-and-topology.bigb

An element $k$ of the additive finite cyclic group $C_n=\mathbb{Z}/n\mathbb{Z}$ is a generator of a group precisely when $\gcd (n,k)=1$. The order of an element of a finite cyclic group proves this immediately: its cyclic subgroup has size $n/\gcd (n,k)$, and this equals $n$ exactly when the greatest common divisor is one. There are therefore $\phi (n)$ choices, where $\phi$ is the Euler totient function.

For $C_6$, the two choices are $1$ and $5$, so $\phi (6)=2$. For $C_8$, they are $1,3,5,7$. Starting from zero and repeatedly adding $3$ modulo eight gives $0,3,6,1,4,7,2,5,0$, visiting every element once before returning. By contrast, adding $2$ gives $0,2,4,6,0$ and misses the odd integers represented in the other coset.

If the cyclic group is written multiplicatively as $⟨g⟩$ with $g$ of order $n$, the same statement says that $g^k$ is a generator of a group exactly when $k$ and $n$ are coprime. This changes notation without changing the calculation.

Assume $1\le t\le r\le n$, $n>2r-t$, and $j\ge 0$. A maximum-cardinality left-compressed set family of $r$-sets that is t-intersecting has a generating family for a uniform set family supported on $[t+2j]$ under the displayed strict inequality. Reflection gives a right-end supporting interval for a right-compressed optimum. The strict threshold is important: at a tie between consecutive candidate families, use the next larger support bound. This auxiliary generating result, combined with the compatibility bound for complementary generating families, supplies the upper bound in the Ahlswede-Khachatrian theorem.