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Put $\delta =-A^2-A^{-2}$. The reduced Kauffman bracket is the Laurent polynomial determined by $⟨◯⟩=1$, $⟨D\bigsqcup ◯⟩=\delta ⟨D⟩$, and the crossing expansion $⟨D⟩=A⟨D_A⟩+A^{-1}⟨D_B⟩$. Fix the A smoothing so that a positive curl has multiplier $-A^3$; rotating the local crossing interchanges the smoothing descriptions. Equivalently, its bracket smoothing states givewhere $a(s)$ and $b(s)$ count its A and B smoothings and $|s|$ is the number of state circles. Resolving all crossings proves that the recursive rules define the same polynomial independently of the order of expansion.

Under a positive Reidemeister move of type I, the two resolutions give $A\delta +A^{-1}=-A^3$ times the straight strand; the negative curl gives $A^{-1}\delta +A=-A^{-3}$. Thus the bracket itself is not an unframed link invariant. For type II, use the adjacent cap-cup generator $e$ of the Temperley-Lieb algebra, with $e^2=\delta e$. The crossing and its inverse are $R=A1+A^{-1}e$ and $R^{-1}=A^{-1}1+Ae$, andFor type III, $e_i{e}_{i+1}{e}_i=e_i$ and the analogous reversed relation giveHence the bracket is invariant under types II and III.

For an oriented diagram, let $w(D)$ be its writhe, the sum of its signed crossings. Types II and III preserve writhe, while a positive or negative curl changes it by one or minus one. The Jones polynomial is thereforeThe normalization cancels type I, so all Reidemeister moves preserve it. It is a Laurent polynomial in $t^{1/2}$; its breadth means the largest occurring t exponent minus the smallest, including half-integral exponents for links.

Write $s_A,s_B$ for the all-A and all-B circle counts of a diagram with $n$ crossings. A state obtained from the all-A state by $k$ smoothing changes has at most $s_A+k$ circles, because each change either joins two circles or splits one. Its largest A exponent is consequently at most $n-2k+2(s_A+k-1)=n+2s_A-2$. The symmetric argument from the all-B state bounds every exponent below by $-n-2s_B+2$. Thus the Kauffman bracket breadth bound isFor a connected diagram, construct the Turaev surface by placing A-state circles above the projection sphere, B-state circles below it, inserting one saddle at each crossing, and capping the state circles. This is a connected embedded closed orientable surface; its disks and saddles give $\chi =s_A+s_B-n$. Since $\chi =2-2g_T\le 2$, $s_A+s_B\le n+2$. Writhe normalization changes only a monomial, and $t=A^{-4}$ divides breadth by four. Therefore

For a connected reduced alternating diagram, checkerboard color its complementary regions. Alternation makes all-A circles the boundaries of one color and all-B circles the boundaries of the other. Euler's formula for the connected four-valent projection graph gives $s_A+s_B=n+2$. Its Tait graph and planar dual have no loops: a loop in either, equivalently a loop or bridge in one, gives a nugatory crossing cut off by a simple closed curve. Reducedness excludes precisely such crossings. Changing one smoothing in the all-A or all-B state therefore joins distinct circles, so the diagram is an adequate link diagram.

This also proves that the extreme coefficients cannot cancel. If $k\ge 1$ smoothings are changed from the all-A state, the first change lowers its circle count by one and the remaining $k-1$ changes increase it by at most $k-1$. The largest exponent of that state is at most $n+2s_A-6$, four below the all-A extreme $n+2s_A-2$, whose coefficient is $(-1{)}^{s_A-1}$. The all-B extreme is uniquely attained for the same reason. Consequently

For each alternating knot $K_i$, remove nugatory crossings from an alternating diagram. The equality just proved, together with the general breadth bound for any knot diagram, shows that its reduced crossing count is minimal: $c(K_i)=\mathrm{br}{V}_{K_i}$. Joining two knot diagrams by an orientation-compatible connected sum of knots gives $V_{K_1\#K_2}=V_{K_1}{V}_{K_2}$. Indeed the state circles joined along the sum arc become one, so the bracket state sums multiply, and writhe adds. Nonzero Laurent polynomials have additive breadth, because their extreme coefficients multiply. Hence every diagram of the sum has at least $c(K_1)+c(K_2)$ crossings, while joining the two minimal alternating diagrams achieves that count. Thus

Finally suppose that the two-component link is split. Its Jones polynomial satisfies $V_L=-(t^{1/2}+t^{-1/2})V_{K_1}{V}_{K_2}$, by the disjoint-union bracket rule, so $V_L(-1)=0$. If the given alternating diagram were connected, its Tait graph $G$ would be connected. Evaluate the bracket at $A=e^{\pi i/4}$, where $\delta =0$ and $t=-1$. Only one-circle states survive. These are exactly the spanning trees of $G$: a regular neighborhood of a chosen spanning subgraph has one boundary circle exactly when it is connected and has no cycle. All spanning trees use the same number of black-joining smoothings, $|V(G)|-1$. Since the crossing-to-checkerboard smoothing choice is uniform in an alternating diagram, all surviving state weights are the same nonzero power of $A$. Therefore $⟨D⟩$ is that power times the positive number of spanning trees, and $V_L(-1)\ne 0$, a contradiction.

The diagram is thus disconnected. Each of the two link components projects to a connected subset, so the two projection components can be separated by a boundary curve of a small regular neighborhood in the projection sphere. Hence a split two-component link represented by a reduced alternating diagram has a split diagram.

For $a$ prime to $p$, putFor positive $a$, $b_a(X)=X^{1-a}{\sum }_{j=0}^{a-1}{X}^{2j}$ is a Laurent polynomial. For negative $a$, use $b_a=-b_{-a}$. Hence it is regular at $X=1$ with value $a$, and $f_a\in {\mathbb{Z}}_p[[T]{]}^{\times }$ has constant coefficient $a^{p-1}\equiv 1(\mathrm{mod}p)$. Evaluation at ${\zeta }_n$ shows that $c_n(a)$ is a unit and reduces to one in the residue field, so $c_n(a)\in U_n$.

Because $p$ is odd, ${\prod }_{\xi \in {\mu }_p}\xi =1$. Both maps $\xi \mapsto {\xi }^2$ and $\xi \mapsto {\xi }^{2a}$ permute ${\mu }_p$, and thereforeThis rational-function identity also holds for negative $a$ and extends across the removable point $X=1$. It implies $N{f}_a=f_a$. Part 1(b) now proves all the required norm compatibilities, and $c(a)\in U_{\infty }$, with Coleman series $f_a$.

Use the additive formal coordinate $z=\log (1+T)$, in which $D=d/dz$. ThenIts logarithm after normalization by $a^{p-1}$ is even, so all odd derivatives at zero vanish. For the even derivatives, the hyperbolic cotangent generating function for Bernoulli numbers givesHere the Bernoulli numbers use $B_1=-1/2$. Differentiating $k-1$ more times yields the answer for every index:These rational expressions lie in ${\mathbb{Z}}_p$ because they equal the integral logarithmic-derivative definition in part (a). In particular ${\delta }_1=0$; substituting $B_1$ into the even-index formula would give an incorrect answer. For $a=\pm 1$ the sequence is one and every derivative vanishes, as the formula also shows.

A cyclotomic Euler system must relate units at many auxiliary conductors, not just the successive layers of one p-power tower. A useful axiomatization is the root-of-unity Euler system. Fix a finite set of excluded primes $S$ containing $2$, and let $W_S$ be the roots of unity whose orders are prime to every prime in $S$. A system is a nonzero-valued function $\Phi :W_S\to {\overline{\mathbb{Q}}}^{\times }$ with three properties. It is even and Galois equivariant:It satisfies the distribution relation at every prime $\ell \notin S$:Finally, if $\ell \nmid \mathrm{ord}(\xi )$, it satisfies the local congruenceThe values in this congruence are required to be integral at the prime; in the unit example below they are local units. Evenness puts the values in the maximal real cyclotomic subfields. Equivariance identifies conjugates, distribution gives exact norm identities, and the congruence supplies information at newly introduced primes. These are the three distinct inputs used in the descent, rather than a single norm-coherence assumption.

Here is a concrete system extending the units in Q2. Fix a nonzero integer $a$ prime to $p$ and let $S$ contain $2$ and the prime divisors of $a$, but not $p$. PutEvery value is defined and nonzero: the excluded primes make $a$ invertible modulo the order of $\xi$, and that order is odd. The quotient is the same Laurent polynomial $b_a$ used in Q2. Replacing $\xi$ by ${\xi }^{-1}$ changes both numerator and denominator by a minus sign, so it is even; its rational coefficients give Galois equivariance.

For the distribution axiom, $\ell \notin S$ makes $\ell$ odd and prime to $a$. Both exponents $2$ and $2a$ permute ${\mu }_{\ell }$, and ${\prod }_{\rho \in {\mu }_{\ell }}\rho =1$. The calculation in Q2 therefore holds with $p$ replaced by $\ell$:Raising to $p-1$ proves the axiom. At removable points it holds by evaluating the regular Laurent polynomials, so it includes $\xi =1$ and ${\xi }^{\ell }=1$.

For the congruence axiom, a root $\rho$ of order $\ell$ reduces to one at every prime above $\ell$. The integral Laurent polynomial $b_a$ thus gives $b_a(\rho \xi )\equiv b_a(\xi )$. These values are local units when $\xi$ has order prime to $\ell$: its reduction retains that order, and, unless $\xi =1$, both $1-{\xi }^2$ and $1-{\xi }^{2a}$ remain nonzero; at $\xi =1$ the value is the unit $a$. The stated power preserves the congruence. Thus ${\Phi }_a$ satisfies all three Euler-system axioms, and its values at ${\zeta }_n$ are precisely $c_n(a)$. Products, Galois translates, suitable field norms and integral group-ring powers produce the other cyclotomic-unit systems needed in the theory.

The distribution relation explains the Euler factors. If $\ell \nmid m$, $\xi$ has order $m$ and $\rho$ is a primitive $\ell$th root, the conjugates over $\mathbb{Q}({\mu }_m)$ run through the nontrivial $\ell$th roots. Removing the single factor with $\rho =1$ from the distribution product givesWe use arithmetic Frobenius, so ${\mathrm{Fr}}_{\ell }(\xi )={\xi }^{\ell }$. This is the cyclotomic Euler norm factor at a new prime. At a higher power of a prime already dividing the conductor, all $\ell$ translates occur, and the ordinary successive-layer norm relation has no missing-factor quotient. A mere p-power compatible sequence therefore contains only part of the Euler-system information.

To explain the axiomatic method, fix a p-power modulus $P=p^r$ and auxiliary primes $\ell$ with $P\mid \ell -1$, split in the chosen real base field. For the cyclic auxiliary Galois group of order $\ell -1$, choose a generator ${\sigma }_{\ell }$ and defineSubtracting consecutive coefficients gives the concrete identityThis is the Kolyvagin derivative operator for a cyclic group. Apply products of these operators to Euler-system values at squarefree auxiliary conductors. The norm relations control the apparent failure of invariance modulo $P$; Kummer descent, using Hilbert 90 and retaining roots-of-unity corrections where present, turns the derivative data into classes over the base field. They have controlled valuations and are unramified away from the selected auxiliary primes and the excluded set. The local congruence relates the valuation component at a newly added prime to the finite residue-symbol component of the preceding class, with the sign and normalization determined by the chosen generator. This finite-to-singular relation is precisely the additional information not obtainable from norm compatibility alone.

The Chebotarev density theorem supplies auxiliary prime ideals in specified ideal classes while satisfying the splitting and congruence conditions. Reading the controlled valuations as principal-ideal relations successively bounds the p-primary elementary divisors of the ideal class group by the divisibility of the initial cyclotomic-unit class. On suitable nontrivial character components this gives the finite bound $\#A^{\chi }\le \#(E/C{)}^{\chi }$; passing through the norm tower yieldswhere divisibility is between characteristic generators, so the associated ideal containment is reversed. The Euler-system divisibility for real cyclotomic class modules requires the usual local conditions and control of finite errors; it is not a claim that arbitrary compatible units automatically annihilate every class group. The cyclotomic unit index formula, together with the global norm-defect comparison, gives the reverse equality. Q3's exact sequence then proves the main conjecture.

This account separates the theorem inputs from the explicit calculations: the distribution and congruence identities for cyclotomic units are verified above, and the derivative operator is computed directly; the Chebotarev/Kummer descent bounds and the global index comparison are the deeper arithmetic steps. Their usefulness is that the descent uses the axioms rather than a special closed formula for each individual auxiliary unit.

There are no numerical continuity bounds determined by $w$ and $a$ alone: the subdivision mask coefficients matter. First check constant reproduction, ${\sum }_j{M}_{ϵ-aj}=1$ for every residue $ϵ$, and convergence. For the usual stationary scalar scheme, two useful calculations are a uniform contraction test and a spectral obstruction test. Higher smoothness is tested with derived difference subdivision masks.

Write $M(z)={\sum }_j{M}_j{z}^j$ and $s_a(z)=1+z+\cdots +z^{a-1}$. Using backward differences $\Delta P_i=P_i-P_{i-1}$, the generating-function identityproves the subdivision difference scheme relation $\Delta S_M=S_{M/s_a}\Delta$. Iterating gives ${\Delta }^r{S}_M=S_{M/s_a^r}{\Delta }^r$. To test a candidate $C^r$ limit, suppose the requisite sum rules hold, so $M/s_a^{r+1}$ is a finite Laurent polynomial. The scaled derivative controls $q_i^{\ell }=a^{r\ell }{\Delta }^r{P}_i^{\ell }$ are refined by the subdivision maskwhose residue sums are one. Their first differences are refined byThe powers of $a$ are essential because each refinement shrinks parameter intervals by $a$.

A lower continuity bound comes from proving uniform contraction. The infinity norm of one difference-refinement step is bounded byIf $q<1$, adjacent derivative controls contract geometrically at every location. More generally, form the $p$-step subdivision maskand test ${\max }_{ϵ\mathrm{mod}{a}^p}{\sum }_j|(B_r^{[p]}{)}_{ϵ-a^{p}j}|<1$. This proves convergence of the constant-preserving derivative scheme to a continuous limit. One justification is to compare consecutive piecewise-linear interpolants: their difference is bounded by a constant times the maximum adjacent-control difference, and the geometrically decreasing bound is summable. To identify the limit as $P^{(r)}$, use fixed-degree Cardinal B-spline interpolants of the original refined controls. Their $r$th derivatives are sums of the scaled finite differences; convergence of both functions and derivatives gives $P\in C^r$. Testing successive $r$ supplies a guaranteed lower bound.

An upper continuity bound comes from an observable noncontracting shape mode. Build finite local subdivision matrices for $B_r$ on a refinement-invariant control neighborhood. For a digit $ϵ$ their entries are $(A_{ϵ}{)}_{ij}=(B_r{)}_{ϵ+i-aj}$, since the new index is $ak+ϵ+i$. Restrict to the realizable difference space and discard modes that do not affect the limit. A word ${ϵ}_1,\dots ,{ϵ}_p$ corresponds to a nested parameter location. If its product has an observable eigenvalue $\lambda$ with $|\lambda |>1$, repetition of this word makes some scaled derivative differences grow rather than tend to zero. For a stable limit representation, $C^r$ regularity would force these differences to vanish; hence general-position data cannot have $C^r$ continuity. A modulus-one mode requires examination of whether it actually persists, rather than an automatic rounding rule.

The two calculations can be organized through the joint spectral radius $\hat{\rho }$ of these restricted matrices:These norm and spectral bounds for subdivision regularity make the test computable from the subdivision mask. The right bound certifies all-word contraction when it is less than one; the left bound detects a repeated-word obstruction when it exceeds one. Longer products improve the tests. A single matrix's spectrum is not a substitute for controlling arbitrary products. The product estimates also appear in Charina's analysis of subdivision regularity; the strict contraction and observable-mode arguments above explain their respective roles.

For $G=U(n)$ take the diagonal maximal torusThe spectral theorem for unitary matrices says that every element is conjugate to an element of $T$. A diagonal element with pairwise distinct eigenvalues has centralizer exactly $T$. Its conjugates therefore determine its eigenvalues up to permutation, and the normalizer quotient is $W=N_G(T)/T\cong S_n$. The character lattice of a torus $T$ is ${\mathbb{Z}}^n$, with $e^{\lambda }(z)=z_1^{{\lambda }_1}\cdots z_n^{{\lambda }_n}$. Conjugation on the matrix unit $E_{ij}$ has weight $e_i-e_j$, so these are the roots. Choose $e_i-e_j$ positive for $i<j$; thenDominant integral weights are precisely the integer tuples ${\lambda }_1\ge \cdots \ge {\lambda }_n$. Negative entries are allowed: they correspond to determinant twists, so polynomial representations alone do not exhaust the irreducible representations of $U(n)$.

Normalize Haar measure on $G$ and $T$ to have mass one, and putThe Weyl integration formula for U(n) for a continuous class function isFor a general continuous function, replace $f(t)$ in the torus integral by ${\int }_{G/T}f(gt{g}^{-1})d(gT)$, using the normalized invariant quotient measure.

To prove the formula, consider the conjugation map $q:G/T\times T\to G$, $q(gT,t)=gt{g}^{-1}$. On regular torus elements its fibres have exactly $n!$ points: the possible orderings of the distinct eigenvalues. Equip the group with an invariant inner product on its Lie algebra and split $\mathfrak{u}(n)=\mathfrak{t}\oplus {\mathfrak{t}}^{\perp }$. After translating the derivative back to the identity, its torus directions contribute the identity, while its transverse directions contribute ${\mathrm{Ad}}_{t^{-1}}-I$. Each pair $i<j$ supplies a real two-dimensional off-diagonal plane. On that plane ${\mathrm{Ad}}_t$ is multiplication by $z_i/z_j$, regarded as a plane rotation, so the absolute real determinant of the transverse derivative isThe change-of-variables theorem, divided by the covering multiplicity, now gives the desired weighted integral up to the constant relating the normalized invariant measures. The nonregular locus has measure zero: repeated eigenvalues are the zero-set of the nonzero discriminant, and the torus coincidence sets likewise have measure zero. Thus its omission causes no change to the integral. To determine the normalization, expand $\Delta$ as a sum of $n!$ distinct torus characters. Fourier orthogonality gives ${\int }_T|\Delta {|}^{2}dt=n!$. Applying the formula to $f=1$ fixes the coefficient to $1/n!$. This proves both the class-function and conjugacy-averaged forms.

We can now derive the character formula from integration, rather than merely quote it. Use the basic highest-weight theorem: a finite-dimensional irreducible representation has a unique highest weight $\lambda$ with a one-dimensional highest-weight space; all its other weights are $\lambda$ minus a nonnegative sum of positive roots. Its character is Weyl invariant. Also, character orthogonality for compact groups gives ${\int }_G|{\chi }_{\lambda }{|}^2=1$. This orthogonality follows by averaging the representation on $\mathrm{End}(V_{\lambda })$: the integral projects onto intertwining operators, whose dimension is one by the Schur lemma.

Let $\delta =(n-1,n-2,\dots ,0)$ and, for a strictly decreasing integer tuple $\eta$, put $A_{\eta }(z)=\det (z_i^{{\eta }_j})$. The finite Laurent polynomial $\Delta {\chi }_{\lambda }$ is alternating. Every alternating Laurent polynomial is a linear combination of the $A_{\eta }$: collect its monomials by permutation orbits, and observe that an orbit with repeated exponents has zero coefficient by antisymmetry. Different $A_{\eta }$ are orthogonal on $T$ and each has squared norm $n!$.

The coefficient of $z^{\lambda +\delta }$ in $\Delta {\chi }_{\lambda }$ is exactly one. Indeed, a contribution from the term $z^{w\delta }$ would require a weight $\lambda +\delta -w\delta$ in $V_{\lambda }$. For $w\ne 1$, the vector $\delta -w\delta$ is a nonzero nonnegative combination of the positive simple roots, so that weight lies above $\lambda$ and is impossible. The identity permutation supplies the highest-weight coefficient one. ConsequentlyApply the integration formula to $|{\chi }_{\lambda }{|}^2$. Torus orthogonality givesAll the other coefficients vanish, provingThe quotient extends across repeated eigenvalues because its product with the denominator is the already defined continuous character. Equivalently, alternant divisibility makes it a Laurent polynomial invariant under permutations; multiply by a determinant power if negative exponents must first be removed.

Finally, $\delta -\rho =\frac{n-1}{2}(1,\dots ,1)$ is fixed by the Weyl group, so its common factor cancels in the quotient. Since $A_{\rho }=e^{\rho }{\prod }_{\alpha >0}(1-e^{-\alpha })$, the determinant expression is exactlyUsing $\delta$ in the derivation avoids treating a possibly half-integral $\rho$ as an actual character of the torus. At the identity, taking the leading Vandermonde term of the two determinants gives the accompanying dimension formula

The dense algebraic torus has coordinate ring $k[M]$, a Laurent polynomial ring and hence a unique factorization domain. Its divisor class group is therefore zero. Restrict an arbitrary Weil divisor $D$ to the algebraic torus and choose a rational function on an algebraic variety $f$ whose divisor there is $D{|}_T$. Then $D-\mathrm{div}(f)$ is supported on the algebraic torus boundary. The prime divisors in that boundary are precisely the $F_j$ corresponding to the toric rays, by the torus orbit-cone correspondence. ConsequentlySince the variety is smooth, every Weil divisor is Cartier. This gives invariant representatives for all divisor classes, without assuming that the original divisor was invariant.

Now use such a representative. The associated vector space isThe principal divisor on a toric variety formula gives $\mathrm{div}({\chi }^m)={\sum }_j⟨m,e_j⟩F_j$. Define the lattice polytope of a toric divisorFor a maximal toric cone $\sigma$, its basic toric ray vectors are a basis of $N$, so there is a unique integral $m_{\sigma }$ with $⟨m_{\sigma },e_j⟩=-a_j$ on its toric rays. The algebraic torus character ${\chi }^{m_{\sigma }}$ is a local frame for $\mathcal{O}(D)$ on $U_{\sigma }$. Every global section restricts on the algebraic torus to a finite Laurent polynomial ${\sum }_m{c}_m{\chi }^m$. Relative to this frame, regularity on $U_{\sigma }$ means ${\sum }_m{c}_m{\chi }^{m-m_{\sigma }}\in k[{\sigma }^{\vee }\cap M]$. Since distinct algebraic torus characters are independent, this holds exactly when each exponent with nonzero coefficient satisfies the inequalities for the toric rays of $\sigma$. Taking all charts provesCompleteness makes $P_D$ bounded when nonempty: a nonzero recession vector would be nonnegative on every toric ray of a complete toric fan, and therefore on all of $N_{\mathbb{R}}$, which is impossible. The displayed space is consequently finite-dimensional.

At the torus-fixed point $p_{\sigma }$ in the smooth chart $U_{\sigma }\cong {\mathbb{A}}^n$, a section ${\chi }^m$ has local monomial ${\chi }^{m-m_{\sigma }}$. Its value is nonzero exactly when this monomial is constant, namely when $m=m_{\sigma }$. Thus if the complete linear system of a divisor has no base point, some section is nonzero at $p_{\sigma }$, forcing $m_{\sigma }\in P_D$. Conversely, if $m_{\sigma }\in P_D$ for every maximal toric cone, its algebraic torus character is a global section whose local expression on $U_{\sigma }$ is one. It therefore vanishes nowhere on that whole chart, and these charts cover $X$. This proves the toric basepoint-free criterion:The analogous toric ampleness criterion, stated without proof, replaces the inequalities off $\sigma$ by strict ones:Equivalently, the normal fan of a polytope $P_D$ is exactly $\Sigma$.

For the final fan subdivision, take $n\ge 2$ so that $\tilde{e}$ is a new toric ray, as the construction requires. The vectors $e_1,\dots ,e_n$ are a lattice basis, so their sum is primitive. Replacing any one basis vector by this sum gives another basis. Thus the star subdivision gives a smooth toric blowup at a torus-fixed point $\pi :X_{{\Sigma }^{\prime }}\to X_{\Sigma }$, with algebraic exceptional divisor $E=\tilde{F}$. On the old chart, a local equation of $D$ is ${\chi }^{-m_{\sigma }}$, whose order on the new toric ray is $-⟨m_{\sigma },\tilde{e}⟩={\sum }_{i=1}^n{a}_i=a$. All old toric ray coefficients are unchanged. ThereforeWe prove ampleness of $D_c=c{D}^{\prime }-E$ directly by inequalities, giving the ample divisor after blowing up a torus-fixed point result.

Let $u_1,\dots ,u_n$ be the basis of $M$ dual to $e_1,\dots ,e_n$. On the new maximal toric cone ${\sigma }_i$, which omits $e_i$, the required local algebraic torus character isIndeed its pairings on $e_j$, $j\ne i$, are $-c{a}_j$, while its pairing on $\tilde{e}$ is $-ca+1$, exactly the negative of the new coefficient $ca-1$. Its inequality on the omitted old toric ray has strict gap one:For any toric ray $e_j$ outside the original toric cone, put ${\delta }_j=⟨m_{\sigma },e_j⟩+a_j$. The original ampleness gives positive integers ${\delta }_j$, and the new strict gaps arefor all sufficiently large $c$.

For any unchanged maximal toric cone $\eta \ne \sigma$, the local algebraic torus character is $c{m}_{\eta }$. All strict inequalities on old toric rays follow from those for $D$. Its inequality on the new toric ray has gapEach summand is nonnegative, and at least one is positive because $\eta \ne \sigma$ cannot contain all its toric rays. Hence ${ϵ}_{\eta }$ is a positive integer, making this gap positive for $c\ge 2$. There are only finitely many remaining inequalities. For example, any integer satisfyingworks, with the inner maximum omitted when its index set is empty. The toric ampleness criterion now givesThe new-ray assumption is essential to the printed construction. In dimension one, $\tilde{e}=e_1$ and the fan subdivision is unchanged, so the two purported distinct divisors coincide. If the displayed definition of $D^{\prime }$ is applied literally on ${\mathbb{P}}^1$ with coefficients $a_1=-2$, $a_2=3$, then $D$ has degree one but $D^{\prime }$ has degree $-1$, contradicting the conclusion. The argument above applies to the intended blowup in dimension at least two.

Assume $0\le k\le n$ with $n,k$ integers, as required for the quantum binomial expression. Let $L_{n-1}$ be the $n$-dimensional irreducible sl2 Lie algebra representation, with one-dimensional weight spaces of weights $n-1-2j$, $0\le j<n$. The exterior-power Lie algebra representation on ${\Lambda }^k{L}_{n-1}$ has characterThis is already a Laurent polynomial with nonnegative integer coefficients.

To identify it with the requested quotient, introduce $t=q^{-2}$ and the Gaussian binomial coefficientSplitting the subsets according to whether $n-1$ is included provesThe product ${\prod }_{j=1}^k(1-t^{n-k+j})/(1-t^j)$ satisfies the same recurrence: after factoring common numerator and denominator terms, the needed identity is $(1-t^{n-k})+t^{n-k}(1-t^k)=1-t^n$. Induction on $n$ therefore proves the product identity, including its polynomiality.

Since the quantum integer is $[a{]}_q=q^{a-1}(1-q^{-2a})/(1-q^{-2})$, taking the quotient of the products givesThus this is the symmetric quantum binomial coefficient. All its weights have parity $k(n-k)$, because the exponent in the exterior-power sum differs from $k(n-k)$ by an even integer. Applying part i to this finite-dimensional representation proves symmetric unimodality in steps of two. Equivalently, the ordinary polynomial $G_{n,k}(t)$ has a unimodal coefficient sequence at consecutive powers of $t$.

Again, the full integer-exponent coefficient sequence in $q$, including the absent parity, need not be unimodal: for $n=2,k=1$ the answer is $q+q^{-1}$. This counterexample makes precise the convention under which the claimed quantum unimodality holds. The cases $k=0$ or $k=n$ give the constant one.

The overlap of the two charts is obtained by inverting $c=t^3$. Since $c^2=b^3$, this also inverts $b=t^2$; then $t=c/b$ and $t^{-1}=b/c$. Thus its coordinate ring is $R[t,t^{-1}]$. The other chart has coordinate $s=t^{-1}$. A global section is therefore an element in the intersectioninside $R[t,t^{-1}]$: a Laurent polynomial with both only nonnegative and only nonpositive exponents is constant.

Let $k=R/(\pi )$. Reducing the presentation of $B$ givesAs a vector space and as a ring extension, this iswhere $a^2=0$ and $a$ is annihilated by $b,c$. In particular $a\ne 0$: the free basis of $B$ in part ii remains a basis after tensoring with $k$. Although $a$ mapped to $\pi t$ before reduction, reducing the subring is not the same operation as taking its image inside $k[t]$; that distinction is exactly where the extra section arises.

On the overlap, $c$ is invertible and $ac=0$, so $a$ restricts to zero. Therefore the section $a$ on the first chart glues to zero on the second chart and defines a nonzero global nilpotent section $\eta$ on $X_s$. More precisely the reduced first-chart ring is $k[t^2,t^3]$, andin $k[t,t^{-1}]$. The only extra compatible sections are multiples of $a$. HenceThe map printed in the question reduces $R$ modulo $\pi$ and includes its image as constants. Its kernel is $\pi R$ and its cokernel is $k\eta$. More strongly, even the usual base-change map after tensoring,is not surjective. This is global-section base-change failure in a flat projective family, despite the properness and flatness proved above.

Choose an ordered affine open cover $\mathcal{U}=(U_i)$ of the separated scheme. For a quasi-coherent sheaf $\mathcal{F}$, the Čech cochain groups and differential areTerms corresponding to empty intersections are zero. Restriction commutes with restriction, so the terms in $d^2$ cancel in pairs. DefineHere an affine cover computes the Čech groups independently of the chosen affine cover. The reason is that finite intersections of affine opens in a separated scheme are affine: the intersection of two is the inverse image of the closed diagonal inside their affine product. Repeating gives the finite-intersection assertion. Quasi-coherent sheaves have no higher cohomology on these intersections. Apply an injective resolution of the sheaf and form the double complex of its Čech cochains. In one direction the higher cohomology vanishes on every affine intersection, leaving precisely the displayed Čech complex. In the other direction the augmented Čech complex of the flasque resolution terms is exact, leaving the global-section resolution computing sheaf cohomology. Thus the two cohomologies agree, and refinement maps between affine-cover complexes induce the same groups. No such affine-intersection claim is being made for arbitrary nonseparated schemes.

For an exact sequence of quasi-coherent sheaves, sections on an affine intersection give an exact sequence of modules, including surjectivity: this follows from the equivalence between quasi-coherent sheaves and modules on an affine scheme. Taking products yields a short exact sequence of Čech cochain complexes. To describe the connecting map, lift a degree-$p$ cocycle in $C^p({\mathcal{F}}^{\prime \prime })$ to a cochain in $C^p(\mathcal{F})$. Its differential maps to zero in $C^{p+1}({\mathcal{F}}^{\prime \prime })$ and therefore is a cocycle in $C^{p+1}({\mathcal{F}}^{\prime })$. Changing the lift adds a coboundary there; changing the original representative does the same. This defines the connecting homomorphism. The same lift-and-differentiate argument shows that a class is in its kernel exactly when it lifts to a cocycle in the middle complex, and verifies exactness at the other terms. Thus the resulting sequence isThe sequence must be in quasi-coherent sheaves for this affine-sections argument; that is the intended category here.

For the punctured affine plane, use $U=D(x)$ and $V=D(y)$. Their rings are $k[x^{\pm 1},y]$ and $k[x,y^{\pm 1}]$, and the intersection ring is $k[x^{\pm 1},y^{\pm 1}]$. The two-open Čech complex isConsequentlyThe denominator contains precisely the span of Laurent monomials for which at least one exponent is nonnegative. Every Laurent polynomial has a unique monomial expansion, so the remaining monomials are linearly independent in the quotient and span it. Thereforea countably infinite-dimensional vector space. The brackets here denote cohomology classes. This calculates the Čech cohomology of the punctured affine plane, rather than merely observing that some nonzero cohomology exists.

Use the generators $e,f,h$ of the sl2 Lie algebra with $[h,e]=2e$, $[h,f]=-2f$, and $[e,f]=h$. For a finite-dimensional Lie algebra representation, $h$ is diagonalizable with integral weights. Its formal character of a weight module isHere $q$ is a formal variable; equivalently the expression is ${\mathrm{tr}}_V(q^h)$, or the formal sum ${\sum }_m(\dim V_m)e^m$ with $e^m=q^m$. This records every weight multiplicity, rather than only the dimension.