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For identical open-string tachyons, $s+t+u=-4/{\alpha }^{\prime }$. At ${\alpha }^{\prime }s=n-1$, the ordered Veneziano amplitude residues in the $s,t$ and $s,u$ terms differ by $(-1{)}^n$, so their symmetric sum cancels odd-level $s$-channel poles at generic $t$. In particular the massless vector does not couple to this identical-scalar singlet amplitude. It remains a state of the open-string spectrum and is visible in suitable ordered or color-dependent amplitudes. Pole absence due to an external-state selection rule must not be interpreted as absence of the exchanged state from the theory.

Vector normalization is not a linear map. With ${\mathbf{e}}_1=(1,0)$ and ${\mathbf{e}}_2=(0,1)$,Consequently, adding individually normalized directions does not generally give the direction of the original vector sum. Compute the intended vector sum first, then normalize it if it is nonzero. Cancellation is a further boundary case: $v+(-v)=0$ has no normalized direction even though both nonzero summands do. A numerical algorithm must check that case before dividing by the norm.

The local flow of a smooth vector field is the family of solutions of its ordinary differential equation, with ${\Phi }_0(p)=p$. Uniqueness gives ${\Phi }_{s+t}={\Phi }_s\circ {\Phi }_t$ wherever defined. A compactly supported smooth vector field on a manifold without boundary is complete, so its flow exists for all real times.

For an oriented closed polygon, triangulate with a common vertex $P$. Its total vector area is $\frac{1}{2}{\sum }_i(B_i-P)\times (B_{i+1}-P)$; the terms involving $P$ telescope, giving the displayed boundary-only formula. Its signed projected area along a unit vector $V$ is $V\cdot \mathbf{A}$. The maximum is $|\mathbf{A}|$, in direction $\mathbf{A}/|\mathbf{A}|$ if nonzero, and zero projections have directions perpendicular to $\mathbf{A}$. If $\mathbf{A}=0$, every direction has zero signed projection despite possible nonzero unsigned surface area.

A sheet with stress-free surfaces supports nearly uniform cross-thickness extensional motion. Its diverging long-wave mobility compensates the weak pressure gradient, leaving the nonzero rate limit in the creeping-flow model. The exactly uniform thickness change is excluded at fixed volume, and unsteady inertia or finite domain size can regularize the limiting prediction. Strong surfactant immobilization instead makes the rate vanish quadratically with wave number.

Let $U,V$ be disjoint, equally sized, nonempty sets, with $\min U<\min V$. Suppose an intersecting family is unchanged by all such UV-compressions of smaller size. Then its $(U,V)$-UV-compression is still an intersecting family. Indeed, a newly moved member $A^{\prime }=(A\setminus V)\cup U$ can only be disjoint from a retained old member $B$, since two newly moved members share $U$. Here $B\cap U=\varnothing$ and $V^{\prime }=B\cap V\ne \varnothing$. If $V^{\prime }=V$, retaining $B$ forces $(B\setminus V)\cup U$ to have been present, contradicting the old intersection property with $A$. Otherwise choose $U^{\prime }\subset U$ with $|U^{\prime }|=|V^{\prime }|$ and $\min U\in U^{\prime }$. Stability under the smaller $(U^{\prime },V^{\prime })$-UV-compression forces $(A\setminus V^{\prime })\cup U^{\prime }$ to be present; this member is disjoint from $B$, again a contradiction. Repeatedly applying a changing UV-compression of smallest size terminates, since the sum of lexicographic order positions decreases. A terminal set family is a lexicographic initial segment: a missing earlier member $X$ and present later member $Y$ would be moved by $U=X\setminus Y$, $V=Y\setminus X$.

In a complete market with a unit cash asset and normalized strictly positive state-price density $Z$, the optimal positive terminal portfolio wealth is $x/Z$ whenever it is replicable in the admissible class. The state-price budget constraint is an equality: $\mathbb{E}[Z(x/Z)]=x$. The utility conjugate inequality gives $\mathbb{E}\log H\le \log x-\mathbb{E}\log Z$ for every affordable positive claim. This value can be infinite, because $\mathbb{E}(\log Z{)}^{+}<\infty$ while $\mathbb{E}(\log Z{)}^{-}$ need not be finite. Even completeness for bounded claims suffices to approximate the value: set $b_n=\mathbb{E}[Z/(Z\vee n^{-1})]$ and $H_n=x/[b_n(Z\vee n^{-1})]$. Each claim is bounded and costs $x$, while its expected logarithmic utility tends to the displayed value by monotone convergence after an integrable lower bound is subtracted.

Let the increments define an upward skip-free random walk, and suppose $\varphi (\lambda )=\mathbb{E}{e}^{\lambda X_1}$ is finite for all $\lambda \ge 0$. If ${\lambda }_0>0$ satisfies $\varphi ({\lambda }_0)=1$ and ${\varphi }^{\prime }({\lambda }_0)>0$, the exponential tilting at ${\lambda }_0$ gives integrable increments of positive mean. The strong law of large numbers makes every nonnegative integer level attainable under this tilted law. Upward jumps are at most one, so the value at a first level hit is exactly that level. The stopped likelihood ratio under exponential tilting therefore gives hitting probability $e^{-{\lambda }_{0}k}$ for integer $k\ge 0$. Subtracting consecutive tail probabilities yields the displayed geometric distribution on $j=0,1,\dots$. For real thresholds use $\lceil k\rceil$ in place of $k$.

The least upper bound in a partially ordered set of a subset $A$, when it exists. It bounds every element of $A$ and is at most every other upper bound. For two elements it is written $x\vee y$; an empty join is a least element.

For an up-and-in claim in the dividend-free Black-Scholes model, set $\nu =(\rho -{\sigma }^2/2)/\sigma$, $\kappa =(S_0/b{)}^2$ and assume finite absolute payoff expectations. Its initial price equals the risk-neutral pricing value of the displayed European option. To prove this, express the log-price as $\sigma (W_t+\nu t)$ and apply the joint endpoint and maximum law for drifted Brownian motion. Above the logarithmic barrier $a=\log (b/S_0)/\sigma$, use the unrestricted endpoint density. Below it, the density of paths which hit the barrier is $e^{2a\nu }{\varphi }_T(y-2a-\nu T)$. Substitution $z=y-2a$ gives the scaled payoff, the threshold $\kappa b$, and the factor ${\kappa }^{-\nu /\sigma }$. This is an initial-price identity, not a pathwise equality of terminal payoffs.

For complex $S{L}_n$, choose a sequence taking an unstable vector to zero and apply singular value decomposition to each group element. Compactness gives a limiting right-unitary factor. Every nonzero torus-weight component of the transformed vector forces a strict linear inequality on the diagonal exponents. A rational solution of these finitely many strict inequalities scales to integral exponents with sum zero, producing an algebraic one-parameter subgroup that takes the vector to zero. The converse follows immediately from the subgroup limit.

A closed lambda term $E$ is easy if adjoining $E=M$ to the equations of beta equivalence remains nontrivial for every closed lambda term $M$. If $E$ is easy, then $ET$ is easy for every closed $T$: use the consistent equation $E=\lambda x.M$ to obtain $ET=M$. Since $I$ is not easy and solvability produces $I$ by applications, easy terms are unsolvable. The converse need not hold.

After a fixed even-depth prefix, Player I announces a move and a set $X$ of first exit positions from a closed set of infinite plays. Player II accepts, or challenges an exit $r\in X$. Acceptance follows the old game until the first exit, where I wins exactly when that exit belongs to $X$. A challenge forces play through $r$ and then resumes the old game. Infinite accepted plays remain in the closed payoff; infinite challenged plays are outside it. Thus the pulled-back payoff is clopen. For an I-strategy, transfer by initially accepting and switching to the challenged simulation at an exit in $X$. For an II-strategy, announce the set of exits that this strategy never challenges under any announcement; it must accept this set. An exit outside it permits a challenged simulation, while an exit inside it is a premature loss for II. These maps prove the game covering lifting condition, rather than merely continuity of the projection.

Let $L/K$ be a finite unramified extension of complete fields with a Non-Archimedean absolute value, with separable residue extension and $[L:K]=[k_L:k_K]=n$. If $x$ is integral and its residue generates $k_L/k_K$, then $1,x,\dots ,x^{n-1}$ form a $K$-basis. Divide any nonzero combination by a coefficient of largest absolute value. Its remaining coefficients are integral, at least one is a unit, and the residue-basis independence makes the reduction nonzero. The displayed norm formula follows. In particular, the combination is integral exactly when every coefficient is integral, proving ${\mathcal{O}}_L={\mathcal{O}}_K[x]$ without requiring a discrete valuation.

Let $K={\mathbb{Q}}_3$, $L=K(i)$ with $i^2=-1$, and ${\alpha }^2=3(1+i)$. The extension $L/K$ is unramified quadratic, whereas $N=L(\alpha )$ is totally ramified quadratic over $L$, by the Eisenstein polynomial $T^2-3(1+i)$. Also $i={\alpha }^2/3-1$, so $N=K(\alpha )$. The Hensel lemma gives $r\in K$ with $r^2=-2$. Put $\beta =(1+i)/r$; then ${\beta }^2=-i$ and $\beta \overline{\beta }=-1$. Consequently $i\mapsto -i$, $\alpha \mapsto \beta \alpha$ defines an automorphism whose square fixes $i$ and negates $\alpha$. It generates the cyclic Galois group of order four. Its unique quadratic subfield is $L$, which is unramified. A decomposition as an unramified field and a totally ramified field would require a totally ramified quadratic subfield, and therefore cannot exist.

A universal-existential sentence has a universal block followed by an existential block and a quantifier-free formula. Such axioms are preserved under unions of chains of structures: the universally chosen tuple occurs in one stage, whose existential witnesses remain in the union. An inductive first-order theory has an axiomatization by these sentences.

Label the vertices of a universal countable uniform hypergraph by the natural numbers and remove the largest vertex of each hyperedge. The resulting $(r-1)$-uniform hypergraph is complete. For any prescribed $(r-1)$-set $F$, apply the hypergraph extension property to a finite initial segment containing $F$, prescribing that $F$ extend to an edge. The resulting vertex is larger than every member of $F$, so this deletion produces $F$. This supplies a useful example of a projection that destroys the nonedge information of a random structure.

A Hamiltonian operator has perfect transfer between basis states $|a⟩,|b⟩$ at time $t_{\ast }$ if $e^{-iH{t}_{\ast }}|a⟩=e^{i\chi }|b⟩$ for a known global phase $\chi$. In an excitation-conserving qubit network whose vacuum is fixed, the initial product state with $\alpha |0⟩+\beta |1⟩$ at site $a$ and zeros elsewhere evolves to $\alpha |\mathrm{vac}⟩+\beta e^{i\chi }|b⟩$. A local phase gate at $b$ removing $\chi$ restores the unknown qubit there. This explains why transfer of the single-excitation basis state suffices for qubit transfer when the vacuum phase is also accounted for.

Two operators on Hilbert spaces are unitarily equivalent if a unitary operator maps their operator domains onto one another and intertwines their actions. Their spectra and eigenvalue multiplicities then agree. For nonnegative self-adjoint operators it suffices that the unitary maps their closed quadratic form domains onto one another and preserves the forms, since the associated operator is determined by the form.

For a real orthogonal gauge group representation, the tangent vectors to the vacuum orbit are $v_a=T_a{\varphi }_0$. Choose an independent basis of these vectors. Under an infinitesimal gauge transformation, the derivative of the displayed slice conditions with respect to the broken gauge parameters is the Gram matrix $v_a^T{v}_b$. It is invertible on the independent tangent directions, so the implicit function theorem selects this unitary gauge locally near the vacuum. The remaining scalar fluctuations are orthogonal to the gauge orbit, and their derivative has no quadratic mixing with the massive gauge fields. The statement is local and does not assert a global gauge slice through zeros of the scalar field.