If every nonempty finite intersection of members of an open cover has vanishing higher sheaf cohomology for $\mathcal{F}$, then the cover's Čech cohomology computes $H^p(X,\mathcal{F})$. An affine open cover of a separated scheme satisfies this condition for a quasi-coherent sheaf because its finite intersections are affine.

On $X={\mathbb{A}}_k^1$, let $K=k(t)$ and let $\mathcal{K}$ be its constant sheaf. For the two closed rational points $0,1$, take skyscraper sheaves $S_0,S_1$ with value $K$, and put $\mathcal{F}=\ker (\mathcal{K}\to S_0\oplus S_1)$. The maps on the two stalks are identities, so this is a sheaf surjection. All three displayed outer sheaves are flasque. The induced global map is the diagonal $K\to K^2$, so $H^1(X,\mathcal{F})=K^2/\Delta K\simeq K$. The one-member affine cover has zero positive Čech cochains. This even gives an ${\mathcal{O}}_X$-module example: let regular functions act on all copies of $K$ by multiplication as rational functions, not by evaluation at the closed point. The kernel is not quasi-coherent.

On the standard two-affine cover of the projective line, with coordinate $t$, a frame for the twisting sheaf on projective space $\mathcal{O}(n)$ on the second chart equals $t^n$ times the frame on the first. Its Čech cochain complex is $k[t]\oplus k[t^{-1}]\to k[t,t^{-1}]$, with map $(a,b)\mapsto t^{n}b-a$. The kernel has basis $1,t,\dots ,t^n$ for $n\ge 0$. The cokernel has basis $t^{n+1},\dots ,t^{-1}$ for $n\le -2$. The two-chart complex also proves vanishing in degrees at least two.

The two affine opens $D(x),D(y)$ cover the punctured affine plane. Their intersection has ring $k[x^{\pm 1},y^{\pm 1}]$. The Čech cohomology in degree one is the quotient by $k[x^{\pm 1},y]+k[x,y^{\pm 1}]$. Its basis consists precisely of the Laurent monomials whose two exponents are negative, so this vector space is infinite-dimensional. Global functions, by contrast, form $k[x,y]$.

For $r\ge 1$, a twisting sheaf on projective space has $H^0({\mathbf{P}}_k^r,\mathcal{O}(n))=k[X_0,\dots ,X_r{]}_n$ when $n\ge 0$, and top cohomology with basis the Laurent monomials of total degree $n$ whose exponents are all negative. All intermediate sheaf cohomology vanishes. Their dimensions are $\left(\genfrac{}{}{0ex}{}{n+r}{r}\right)$ and $\left(\genfrac{}{}{0ex}{}{-n-1}{r}\right)$ respectively, with the top group nonzero only for $n\le -r-1$. Decompose the standard Čech cochain complex by exponent vector: no negative exponents gives a simplex with cohomology in degree zero; a nonempty proper negative-support set admits a contracting homotopy inserting a vertex outside that set; all negative exponents give only the top term. For $r=0$, every twist is trivial on $\mathrm{Spec}k$ and has $H^0=k$.

During radiation domination, $a\propto \tau$ and $8\pi G{\rho }_R{a}^2=3/{\tau }^2$. The adiabatic initial conditions reduce the CDM density equation in a matter-radiation universe to ${\delta }_C^{\prime \prime }+{\delta }_C^{\prime }/\tau -4{\delta }_C/{\tau }^2=0$. Its Euler-Cauchy equation powers are two and minus two. The growing solution is the displayed one, in the stated CDM-comoving synchronous gauge.

A J-holomorphic curve is regular when the linearized Cauchy–Riemann operator is surjective. Regularity only for simple curves is weaker than regularity for all maps: multiple covers of an exceptional sphere can fail regularity even when that simple sphere is regular.

A real-linear Cauchy–Riemann operator on a complex rank-$n$ vector bundle over a closed genus-$g$ Riemann surface is a Fredholm operator between the usual Sobolev spaces and has the displayed index. Zero-order real-linear terms do not change this index, since they form a continuous deformation of Fredholm operators. For the bundle pulled back by a sphere, subtracting the six-dimensional reparametrization group gives the dimension formula for regular J-holomorphic spheres.

Normalize the second-order equation as $y^{\prime \prime }+P(z)y^{\prime }+Q(z)y=0$. If every point other than zero and infinity is ordinary and both exceptions are at worst regular singular points, then $zP$ and $z^{2}Q$ extend holomorphically across zero and stay bounded at infinity. Liouville's theorem makes them constants, yielding the displayed Euler-Cauchy equation. If both exceptions must be genuinely singular, exclude $(A,B)=(0,0)$ and $(2,0)$, which make zero and infinity respectively ordinary.

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.