Although $1/r$ is harmonic away from the origin in three dimensions, it is not a regular harmonic function on a ball containing the origin. Its gradient is $-\hat{r}/r^2$, so the displayed improper volume integral follows by canceling the $r^2$ in the spherical volume element. On the annulus $\epsilon <r<a$, Green's first identity gives outer boundary contribution $-4\pi$ and inner contribution $4\pi \epsilon /a$, recovering the same limit. Its global Laplacian contains a Dirac delta distribution, so the regular-harmonic gradient-pairing theorem is inapplicable.

A super vector space is a vector space graded by the two-element group. Degree-preserving maps and the usual graded tensor product form a monoidal category. Over a field of characteristic different from two, this category has two different symmetries: ordinary interchange of tensor factors and interchange multiplied by $(-1{)}^{pq}$ on homogeneous factors of degrees $p,q$. The latter is the Koszul sign rule.

For a positively graded vector space $V$, the free graded Lie algebra $\mathbb{L}(V)$ receives a linear map from $V$ and extends every graded linear map $V\to L$ uniquely to a Lie homomorphism. Over $\mathbb{Q}$, it is the Lie subalgebra generated by $V$ inside the tensor algebra $T(V)$, using $[u,v]=uv-(-1{)}^{|u||v|}vu$. Its universal enveloping algebra is $T(V)$: both have the same universal property for maps from $V$ into associative algebras. The primitives of the tensor Hopf algebra with primitive generators are exactly $\mathbb{L}(V)$.

For a positively graded Lie algebra $L$ over $\mathbb{Q}$, finite-dimensional in each degree, its cochain algebra is free graded commutative on the graded dual of $sL$, with differential dual to the Lie bracket and its Koszul signs. The Jacobi identity gives $d^2=0$. A Lie element of degree $r$ contributes a cochain generator of degree $r+1$. Its cohomology computes ${\mathrm{Ext}}_{U(L)}(\mathbb{Q},\mathbb{Q})$ with total grading; the standard free enveloping-algebra resolution gives this identification. For a free graded Lie algebra, the tensor-algebra augmentation has a length-one free resolution, explaining the square-zero cohomology of a rational wedge sum of spheres.

The tensor product of graded algebras uses the displayed multiplication, incorporating the sign for interchanging homogeneous factors. This is the algebra structure in the cohomological Künneth theorem, and explains the signs of mixed intersection products on product manifolds.

For an augmented graded algebra $A$, its augmentation ideal $A^{+}=\ker \epsilon$ has indecomposable quotient $Q(A)=A^{+}/(A^{+}{)}^2$. For a Sullivan minimal model $(\Lambda V,d)$, this quotient is $V$ with zero differential because every $dv$ is decomposable. The map $H^{+}(\Lambda V,d)\to V$ induced by projection is dual to the rational Hurewicz homomorphism under the finite-type model identification.

For a good filtration of a module over a fixed finite-dimensional filtration of a ring whose associated graded algebra is commutative and finitely generated, the cumulative Hilbert function is eventually a quasipolynomial by the Hilbert-Serre theorem. Its leading asymptotic degree is the dimension of the nonzero filtered module. Monotonicity forces the leading coefficients in its residue classes to agree. If the associated graded algebra is a standard graded algebra, the Hilbert function is eventually polynomial. Two good module filtrations bound one another after fixed shifts, so they have the same growth degree. Set $d(0)=-\infty$. For the Bernstein filtration this is the Bernstein growth dimension of a Weyl algebra module.

Suppose $R$ is a right Noetherian ring and every essential right ideal contains a regular element of a ring. Preimages of essential regular principal right ideals under left multiplication are essential, so they contain regular elements and prove the right Ore condition. In the classical right ring of quotients $Q$, every right ideal is generated by its contraction to $R$, making $Q$ right Noetherian. An essential right ideal of $Q$ contracts to an essential right ideal of $R$, hence contains a regular element and is all of $Q$. A maximal disjoint complement to any right ideal has essential sum with it, so every right ideal is complemented. The right regular module is therefore a finite direct sum of simple modules, making $Q$ semisimple Artinian.

A nonzero two-sided ideal of the classical right ring of quotients contains a nonzero element of the original ring: multiply a nonzero fraction by its regular right denominator. If the original ring is prime, the contractions of any two such ideals have nonzero product, so the quotient ring is prime too. The Goldie theorem makes it semisimple Artinian, and the Artin–Wedderburn theorem expresses it as a finite product of matrix rings over division rings. Primeness excludes more than one factor.

For an effective Gödel numbering $M\mapsto \#M$, the quotation $d(M)=c_{\#M}$ is a Church numeral coding the syntax of $M$. It is not a code of a beta equivalence class. The operations taking syntax codes to the code of an application, or to the code of the numeral itself, are total computable functions and hence lambda-definable functions.

Adding sodium hydroxide to acidified glycine successively removes its two titratable protons. Two broad buffer solution regions appear around the two pKa values, separated by a steep pH rise near the isoelectric point. This is useful evidence for two separated acid-base chemistry steps even though the changing chemical species are invisible. In the approximation that added hydroxide only changes glycine protonation, $b=1-\bar{z}$ is the added base amount of substance per initial amount of substance of glycine. Then $b\approx 0.5$, $1$ and $1.5$ correspond to $\mathrm{p}{K}_{a1}$, $\mathrm{pI}$ and $\mathrm{p}{K}_{a2}$; endpoints are not exactly pure chemical species. ^[ref]

For the solid-curve model, start with $n_G=0.00200\mathrm{mol}$ of glycine in $V_0=0.0200\mathrm{L}$ and exactly $n_G$ of fully dissociated hydrochloric acid; this fixes the chloride inventory, but does not force every glycine molecule to remain $G^{+}$. Let the sodium hydroxide concentration be $C_B=0.100\mathrm{mol}{\mathrm{L}}^{-1}$ and $b=n_{\mathrm{NaOH}}/n_G$. Assume volume is additive and no other ions are present; stipulate $K_w=10^{-14}$ for self-ionization of water at a temperature of $25{}^{\circ }\mathrm{C}$. ThenBalancing all positive and negative electric charges requiresSolving this equation gives the solid curve. It remains finite at $b=0$ and beyond $b=2$, where ignoring free hydronium and hydroxide would fail. At the first pKa, the full model gives $b\approx 0.436$, rather than $0.5$: appreciable free hydronium makes that half-equivalent shortcut inaccurate. At the second pKa, $b\approx 1.502$. Known acid inventory, dilution, and thermodynamic activity assumptions are therefore part of any quantitative interpretation, not optional corrections inferred from the curve's shape.

In an ideal aqueous chemical solution, set $h=[{\mathrm{H}}_3{\mathrm{O}}^{+}]/c^{\circ }=10^{-\mathrm{pH}}$, with $c^{\circ }=1\mathrm{mol}{\mathrm{L}}^{-1}$, and approximate activity coefficients by one. Put $K_1=10^{-2.35}$ and $K_2=10^{-9.78}$ for the stipulated dimensionless acid dissociation constants. The Henderson-Hasselbalch equation gives the adjacent-state ratiosWith analytical concentration $C_T=[G^{+}]+[G^0]+[G^{-}]$, normalization gives the three fractions and mean electric charge in units of the elementary charge:All three fractions are nonnegative and sum to one. At pH $7$, the model gives about $99.832\%$ zwitterion, $0.00224\%$ cation and $0.1657\%$ anion; $\bar{z}\approx -0.00163$. Predominantly zwitterionic therefore need not mean exactly at the isoelectric point.

Setting $\bar{z}=0$ requires $h^2=K_1{K}_2$. Taking the logarithm yieldsAt this model isoelectric point, ${\alpha }_{+}={\alpha }_{-}\approx 0.0001927$ and ${\alpha }_0\approx 0.9996146$: about $0.0193\%$ of each charged form remains. At either pKa, the relevant adjacent fractions are equal, but the third fraction is not mathematically zero. This three-state model neglects the minor uncharged form ${\mathrm{H}}_2\mathrm{N}-\mathrm{C}{\mathrm{H}}_2-\mathrm{COOH}$; macroscopic pKa values alone do not determine its fraction separately from the zwitterion.

If there are fewer left owners than right owners, deleting any right owner leaves the grand-coalition value unchanged. The complement's core constraint and efficiency force that owner's payoff to be zero. Mixed-pair inequalities then require every left owner to receive at least $\omega$, and efficiency makes all these payoffs exactly $\omega$. Reverse the types when right owners are scarce. If both counts equal $k$, sum all $k^2$ mixed-pair inequalities. Their total is exactly $k$ times the grand-coalition payoff, so each pair sum is $\omega$. Hence all left owners receive a common $\theta \in [0,\omega ]$ and all right owners $\omega -\theta$. Such allocations meet every coalition constraint because $\theta u+(\omega -\theta )v\ge \omega \min (u,v)$. This proves the complete core description.

Let $F$ be a globally generated vector bundle of rank $r\ge 2$ on a projective curve over an algebraically closed field $k$. Then some global section of $F$ vanishes nowhere and defines a line subbundle ${\mathcal{O}}_C\subset F$ with locally free quotient.

Put $W=H^0(C,F)$. The incidence pairs $(p,s)$ satisfying $s(p)=0$ form the total space of the kernel of the surjective evaluation vector bundle morphism $W\otimes {\mathcal{O}}_C\to F$. Their dimension is $\dim W+1-r<\dim W$. Projection to $W$ is closed because projection from the product with $C$ is a proper morphism, so the sections vanishing somewhere lie in a proper closed subset of $W$. Choose a point outside it. Its fiberwise nonzero values extend locally to frames, giving the asserted line subbundle and locally free quotient.

The same rational-point choice works over an infinite field, after forming the bad locus geometrically. Over a finite field, a proper closed subset of an affine affine space can contain every rational point; the dimension argument alone does not provide a $k$-defined nowhere-vanishing global section.

Let $C$ be a smooth projective curve over an algebraically closed field, let $p_0\in C$, and let $E$ be a vector bundle of finite positive rank. Then $E\otimes {\mathcal{O}}_C(m{p}_0)$ is globally generated for all sufficiently large integers $m$. This is the curve version of generation by sufficiently positive twists of an ample line bundle.^[ref]

For a uniform bound, take an upper degree bound for line subbundles on a smooth curve $B$ for $E^{\ast }\otimes K_C$. If $m-1>B$, a nonzero map ${\mathcal{O}}_C(m{p}_0-p)\to E^{\ast }\otimes K_C$ would have saturated image of degree at least $m-1$, a contradiction. Serre duality gives $H^1(E(m{p}_0-p))=0$ for every point $p$. The evaluation exact sequence then makes $H^0(E(m{p}_0))\to E(m{p}_0){|}_p$ surjective at every point.

On the projective line, $E=\mathcal{O}(a)$ gives the concrete bound $m+a\ge 0$: the degree-$m+a$ homogeneous polynomials generate the twisted line bundle. The chosen field hypothesis makes every closed point have degree one, as used in the bound $m-1$.

Orient the upper-quadrant contour from $i\infty$ down to zero and then to $+\infty$. The reflected half-line transform has zero inverse contribution for positive $x$. Adding $c$ times this contribution, and deforming boundary integrands through the second quadrant where $|e^{ikx-i{k}^2(t-\tau )}|\le e^{-x\mathrm{Im}k}$, gives boundary coefficients $(1-c)k{\tilde{g}}_0-i(1+c){\tilde{g}}_1$. Thus $c=-1$ removes the unknown Neumann trace for a prescribed Dirichlet value, while $c=1$ removes the unknown Dirichlet trace for a prescribed Neumann value. The initial reflected term has coefficient $c$ and cannot be discarded after the time-dependent spectral decomposition.

For the linear dispersive Stokes equation $q_t+q_x+q_{xxx}=0$, let $\omega (k)=i(k-k^3)$. Its local relation is obtained with $X=q_{xx}+ik{q}_x+(1-k^2)q$. Integrating ${\partial }_t(e^{-ikx+\omega t}q)+{\partial }_x(e^{-ikx+\omega t}X)=0$ over the half-line gives the displayed global relation, with negative-exponential Half-range Fourier transforms and $G_j(k,t)={\int }_0^t{e}^{\omega (k)s}{\partial }_x^{j}q(0,s)ds$. The spatial transforms are holomorphic for $\mathrm{Im}k<0$, whereas the finite-time boundary transforms are entire functions.

For $q_t=q_{xx}+\alpha q_x$ on $x>0$, write $Q(k)={\int }_0^{\infty }{e}^{-ikx}q(x,0)dx$ and $G_j(k,t)={\int }_0^t{e}^{\omega (k)s}{\partial }_x^{j}q(0,s)ds$. Twice integrating by parts gives $e^{\omega (k)t}\hat{q}(k,t)=Q(k)-G_1(k,t)-(ik+\alpha )G_0(k,t)$ in the lower transform half-plane. The dispersion relation is invariant under $k\mapsto i\alpha -k$. This global relation eliminates one unknown boundary trace after the symmetry maps the inversion contour into the transform's analytic domain.

Let $q$ be a real harmonic function in a semistrip, $g_0(x)=q(x,0)$, $g_l(x)=q(x,l)$ and $h(y)=q(0,y)$. Let $G_j,N_j$ be the negative-exponential Half-range Fourier transforms of $g_j$ and $q_y(x,j)$, and put $H(k)={\int }_0^l{e}^{ky}h(y)dy$, $W(k)={\int }_0^l{e}^{ky}{q}_x(0,y)dy$. The closed differential form $e^{-ikz}{q}_{z}dz$ has zero boundary integral. Integration by parts of tangential derivatives cancels all corner values and gives the displayed global relation, for $\mathrm{Im}k\le 0$.

// Target: analysis.bigb

For the Airy equation $q_t+q_{xxx}=0$ on $0<x<L$, define $F_j={\int }_0^t{e}^{-i{k}^{3}s}{\partial }_x^{j}q(0,s)ds$ and $G_j$ analogously at $L$. Set $\mathcal{F}=k^2{F}_0-ik{F}_1-F_2$ and $\mathcal{G}=k^2{G}_0-ik{G}_1-G_2$. Integrating the local relation with flux $-q_{xx}-ik{q}_x+k^{2}q$ proves the displayed global relation. The dispersion relation is invariant under multiplication of $k$ by a cube root of unity; the resulting three equations determine the three missing boundary transforms for data $q(0,t),q(L,t),q_x(L,t)$.