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For $T:{\mathcal{C}}^{\mathrm{op}}\times \mathcal{C}\to \mathcal{D}$, an end is a universal object $E$ with maps $e_C:E\to T(C,C)$ satisfying $T(1_C,f)e_C=T(f,1_D)e_D$ for every $f:C\to D$. Every other such family factors uniquely through $E$. For small $\mathcal{C}$ in a complete category, it is the equalizer of the two maps ${\prod }_{C}T(C,C)\rightrightarrows {\prod }_{f:C\to D}T(C,D)$.

For a Lie algebra representation with generators referred to an orthonormal invariant form, the sum ${\sum }_a{t}_a{t}_a$ commutes with every generator. Indeed, the commutator is proportional to a contraction of antisymmetric structure constants of a Lie algebra with the symmetric product $t_a{t}_b+t_b{t}_a$. The Schur lemma makes it scalar in an irreducible complex representation; that scalar is its quadratic Casimir eigenvalue. With trace normalization $\mathrm{tr}(t_a{t}_b)=C(R){\delta }_{ab}$, taking the trace gives $C(R)\dim \mathfrak{g}=C_2(R)\dim R$.

A magnetic island in a planar magnetic field is a region of closed magnetic field lines surrounding an extremum of the Cartesian magnetic flux function. A surrounding separatrix can pass through a saddle of that function and separate the island from other flux regions. In ideal incompressible flow, the flux-function value carried by each fluid particle and the area enclosed by each material contour are conserved.

Sampling consecutive order-six Cardinal B-splines at their support midpoints gives the displayed finite B-spline collocation matrix. No exterior columns are retained. Its row strict diagonal dominance margin is at least $(66-2\cdot 26-2)/120=1/10$, so the inverse infinity-norm bound from diagonal dominance gives $\Vert A^{-1}{\Vert }_{{\ell }^{\infty }}\le 10$. The uniform knot samples are $N_{0,6}(0),\dots ,N_{0,6}(6)=(0,1,26,66,26,1,0)/120$, obtained recursively from the Cox-de Boor recurrence.

For a Gaussian forward-rate field with deterministic volatility, put $F=P(t,U)/P(t,T)$, $K_b=(1+\delta K{)}^{-1}$, and $v^2={\int }_t^T\Vert \Sigma (s,U)-\Sigma (s,T){\Vert }^{2}ds$. Under the T-forward measure, $F$ is lognormal with that integrated variance. With $d_1=\log (F/K_b)/v+v/2$ and $d_2=d_1-v$, the displayed expression is the caplet price. At $v=0$ use the discounted intrinsic bond-put value. Gaussian instantaneous rates do not make the reset floating rate Gaussian; the exponential bond relation is what yields this bond-option formula.

For a proper plane domain $D$, choose $a\in D$ and $w\notin D$ attaining $\delta (a)=\mathrm{dist}(a,\mathbb{C}\setminus D)$. The holomorphic functionhas exactly one zero in $D$, a simple zero at $a$. For $|z-w|>2\delta (a)$ its zero-free holomorphic logarithm satisfiesThus products of such factors can realize prescribed zeros approaching the boundary. If the zero has multiplicity $m$, choose the logarithmic error smaller by a factor $m$ before taking the $m$th power. On a compact subset of $D$, eventual uniform summability of these logarithms gives a nonvanishing product tail and a holomorphic function with precisely the requested zeros.

For nonzero prescribed zeros $a_n$ with $|a_n|\to 1$, the Weierstrass elementary factors of increasing order give the displayed holomorphic function on the unit disc; $m$ handles finitely many zeros at zero. On any compact disc, eventually $|z/a_n|\le q<1$, and $|\log E_n(z/a_n)|\le q^{n+1}/[(n+1)(1-q)]$. The infinite product convergence from logarithmic tails gives a nonvanishing tail. The finite initial factors supply exactly the prescribed zeros and multiplicities. No Blaschke condition is required unless boundedness is imposed.

Let $S_L$ have a Poisson distribution with mean $L\lambda$, where $\lambda >0$, and put $u_L=L\lambda +C{L}^{(1+\beta )/2}$ with $C>0$. Assume $0<\alpha <\beta <1$. The upper restriction is essential: this is a moderate deviation, with a threshold shift larger than $\sqrt{L}$ but smaller than $L$.

For an input rate $\lambda +\delta$, the Poisson rate function isTaking $\delta =C{L}^{(\beta -1)/2}$ givesThe expansion follows directly from the Poisson weights and Stirling's formula. Because $\beta >\alpha$, clipping has superexponentially small probability at speed $L^{\alpha }$. In particular, for every $\epsilon >0$,This is the relevant exponential equivalence. More directly, every threshold $r_L<u_L$ gives identical events $\{\min (S_L,u_L)>r_L\}=\{S_L>r_L\}$. At $\beta =1$, the full Poisson rate function replaces its quadratic approximation.

For $\lambda ,C>0$ and $0<\beta <1$, clip Poisson input of mean $L\lambda$ at $L\lambda +C{L}^{(1+\beta )/2}$. The Poisson moderate deviation principle and contraction principle for large deviations give the displayed output rate at the clipping scale. At smaller exponents the output retains the original rate by moderate deviations below a bufferless queue cap. At larger positive-deviation scales every overflow event is eventually impossible because the requested threshold exceeds the cap.

The relative position of Borel subgroups $B_1,B_2$ is the Weyl group index of their simultaneous conjugacy orbit. After moving the first subgroup to a fixed $B$, the second has the form $gB{g}^{-1}$ and its index is the double coset $BgB$. This identifies the diagonal action on pairs of flags with the Bruhat decomposition.

For a nonzero null momentum and a nontrivial BRST charge on the gauge/ghost one-particle sector, closed vector states obey $k\cdot \epsilon =0$ and exact vector states have polarization proportional to $k$. The Faddeev-Popov ghost field state is exact and the antighost field state is not closed. With mostly-plus Minkowski metric, choose $k=(\omega ,0,0,\omega )$; closure sets ${\epsilon }^3={\epsilon }^0$, so the norm is $|{\epsilon }^1{|}^2+|{\epsilon }^2{|}^2$. Quotienting by the null longitudinal polarization leaves two positive-norm transverse polarizations per group index. Self-adjointness and nilpotence alone do not establish positivity on an arbitrary state space: the nonzero momentum, nontrivial charge and stated indefinite inner products are essential here.

On a positive-definite Hilbert space, these two identities imply $\Vert Q\psi {\Vert }^2=⟨\psi ,Q^2\psi ⟩=0$, hence $Q=0$ on its invariant domain. A nontrivial covariant BRST charge therefore acts before reduction on an indefinite-inner-product gauge/ghost space. Exact vectors are null and orthogonal to closed vectors, and BRST cohomology quotients them out. Positivity of the reduced physical space requires the additional gauge-theory structure; it does not follow from abstract nilpotence alone.

For Brownian motion started at zero in dimension three, let ${\sigma }_R$ be its first hit of the sphere of radius $R>r$. The optional stopping theorem applied to $|B_t{|}^2-3t$ proves that ${\sigma }_R$ is finite almost surely. The Strong Markov property and the Brownian sphere-hitting probability in dimension three then give the displayed return probability. Choose radii $R_m=2^{m}r$. The return events after ${\sigma }_{R_m}$ decrease and their probabilities tend to zero. Almost surely one of these exit times is followed by no visit to the closed ball of radius $r$. Apply this to all positive integer $r$ to obtain $|B_t|\to \infty$ almost surely. This gives full transience of Brownian motion in dimension at least three, rather than only a positive escape probability.

Let $f$ be positive, continuous and nondecreasing, with ${\int }_0^{1}f(t)dt/t<\infty$. The small-value probability for the Brownian running maximum is at most a constant times the value. Brownian scaling and dyadic integral comparison make the exceptional probabilities $\mathbb{P}(S_{2^{-n-1}}<2^{-n/2}f(2^{-n}))$ summable. The Borel-Cantelli first lemma and monotonicity interpolate between dyadic times. Apply this argument to $mf$ for every positive integer $m$ on a countable intersection of probability-one events to obtain the infinite lower limit.

For $a>0$ and $t>0$, the distribution function is $\Phi ((a-\nu t)/\sqrt{t})-e^{2a\nu }\Phi ((-a-\nu t)/\sqrt{t})$. The Brownian reflection principle gives the zero-drift terminal density below the barrier as ${\varphi }_t(y)-{\varphi }_t(2a-y)$ for $y<a$. Weight it by $e^{\nu y-{\nu }^{2}t/2}$ using the Girsanov theorem and complete squares in the two Gaussian integrals.

For Brownian motion, write the later maximum as $B_b+Z+H$, where $Z=B_c-B_b$ has a nondegenerate normal distribution and $H={\sup }_{0\le u\le d-c}(B_{c+u}-B_c)$. The independent increments make $Z$ independent of the earlier history and of $H$. Conditional on those two objects, equality of the two maxima prescribes one value of $Z$, an event of probability zero. This argument uses the positive gap $c-b$ and does not assume independence of the two absolute maxima.

Stop the Exponential martingale for Brownian motion $e^{\theta B_s-{\theta }^{2}s/2}$ when $B$ first reaches $\delta$ or at $t$. On the hitting event its value is at least $e^{\theta \delta -{\theta }^{2}t/2}$, so that event has probability at most $e^{-\theta \delta +{\theta }^{2}t/2}$. Minimize over $\theta >0$ and apply the same bound to $-B$. The union bound gives the displayed inequality for $t>0$; at zero the probability is zero.

For $d\ge 3$, fixed $x\ne 0$ and $t>0$, apply optional stopping to $u(B_{t\wedge T_{\epsilon }})$ with $u(z)=|z{|}^{2-d}$. The resulting identity is ${\epsilon }^{2-d}{\mathbb{P}}_x(T_{\epsilon }\le t)=u(x)-\mathbb{E}u(B_t)+\mathbb{E}[u(B_t);T_{\epsilon }\le t]$. The last term tends to zero by integrability of $u(B_t)$ and the vanishing total hitting probability. The Newtonian potential of the Brownian heat kernel identifies the remaining difference.

The Brownian first-passage subordinator has a countable dense set of jump levels almost surely, and is continuous at every other level. Its paths are finite nondecreasing functions, so on a bounded level interval there can be only finitely many jumps larger than any fixed positive size; a countable union proves countability. For density, suppose its path were continuous throughout an open level interval. It is strictly increasing, since the original continuous Brownian path takes different values at different passage times. Its continuous inverse on that interval would parametrize a positive-length time interval on which the Brownian path increases monotonically. This contradicts nowhere monotonicity of Brownian motion. Thus every level interval contains a jump. In particular dense discontinuities do not mean that the path is nowhere continuous: it is continuous at every level outside a countable random set.

Successive exits from intervals of radius $a$ around the current position embed a simple symmetric random walk into Brownian motion. The Strong Markov property makes the durations independent with common law $a^2{\sigma }_1$; the displacements divided by $a$ are independent symmetric signs. The displacement sequence is independent across exits; no independence between a given displacement and its duration is needed for these conclusions.