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An uncountable cardinal $\kappa$ is a measurable cardinal if there is a nonprincipal $\kappa$-complete ultrafilter $U$ on $\kappa$: intersections of fewer than $\kappa$ members of $U$ remain in $U$. Such an ultrafilter is uniform. No singleton belongs to it, and intersecting the complements of fewer than $\kappa$ singletons shows that no set of cardinality below $\kappa$ belongs to it.

Form the ultrapower of the universe by $U$, with classes $[f]$ of functions $f:\kappa \to V$, and $[f]{\in }_U[g]$ exactly when $\{\xi :f(\xi )\in g(\xi )\}\in U$. The Łoś theorem proves that the map taking $x$ to the constant function with value $x$ is elementary. The ultrapower relation is well-founded: from an infinite descending sequence $[f_0]{\ni }_U[f_1]{\ni }_U\cdots$, countable completeness gives a coordinate satisfying all the membership relations, producing an actual infinite descending membership sequence and contradicting foundation. It is also set-like. Predecessors of $[g]$ can be represented by functions whose value at $\xi$ lies in $g(\xi )\cup \{\varnothing \}$; these functions form a set. The Mostowski collapse theorem therefore gives a transitive class $N$ and an elementary embeddingThe usual class-ultrapower notation is understood via set-sized representatives and this set-like collapse.

For $\alpha <\kappa$, any function into $\alpha$ is constant on a $U$-large set: if none of its fewer-than-$\kappa$ fibers belonged to $U$, intersect their complements to get the empty set in $U$. Consequently every member of the ultrapower ordinal represented by the constant $\alpha$ is represented by a constant ordinal below $\alpha$. Induction on $\alpha$ now gives $j(\alpha )=\alpha$.

On the other hand the identity function $d(\xi )=\xi$ represents an ordinal below $j(\kappa )$. For every $\alpha <\kappa$, the tail $\{\xi :\xi >\alpha \}$ belongs to $U$, so the collapsed ordinal $[d]$ is above $j(\alpha )=\alpha$. HenceThis proves nonidentity explicitly. The collapsed identity class is at least $\kappa$; it need not equal $\kappa$ unless an additional normality convention is imposed on $U$.

Conversely, an elementary embedding $j:V\to N$ into a transitive class with critical point $\kappa$ yieldsFor the converse use the usual amenability assumption on the class embedding, so this collection is a set. Since $j(\kappa )>\kappa$, exactly one of $j(X)$ and $j(\kappa \setminus X)$ contains $\kappa$, proving the ultrafilter condition. Singletons are excluded because $j(\alpha )=\alpha <\kappa$. If $\lambda <\kappa$ and $X_i\in U$ for $i<\lambda$, then $j$ fixes $\lambda$ and every index below it, giving $j({\bigcap }_{i<\lambda }{X}_i)={\bigcap }_{i<\lambda }j(X_i)$; the intersection still contains $\kappa$. Thus $U$ is $\kappa$-complete and $\kappa$ is measurable.

In particular the first moved ordinal is a strongly inaccessible cardinal. If $\mathrm{cf}(\kappa )<\kappa$, partition $\kappa$ into fewer than $\kappa$ bounded intervals. Each is too small to lie in $U$, contradicting completeness. Thus $\kappa$ is regular. If $2^{\lambda }\ge \kappa$ for some $\lambda <\kappa$, choose distinct subsets $A_{\xi }\subseteq \lambda$ for $\xi <\kappa$. For each $\eta <\lambda$, choose the $U$-large side of the question $\eta \in A_{\xi }$. Intersect these fewer-than-$\kappa$ large sides. The intersection belongs to $U$, but all its indices have identical $A_{\xi }$, so it contains at most one index, contradicting nonprincipality. Therefore $2^{\lambda }<\kappa$ for every $\lambda <\kappa$, as required.

On complexified Minkowski spacetime, use the twistor incidence relationThe harmless factor $i$ is a conventional choice; omitting it corresponds to rescaling the complexified coordinates. For fixed $x$, projectivizing the nonzero two-component spinors $\pi$ gives a twistor line $L_x\simeq {\mathbb{CP}}^1$. The two stated patches cover all these incidence lines; they describe the affine-space-time twistor domain with the line $\pi =0$ removed from full projective twistor space.

Since $E{|}_{L_x}$ is holomorphically trivial, choose a global bundle frame on that line. Comparing its columns with the bundle frames in the two patch trivializations gives invertible matrices $H(x,\pi )$ and $\tilde{H}(x,\pi )$, holomorphic on their respective patches, withThis is the holomorphic splitting of a twistor patching matrix. It follows directly from existence of the global bundle frame, rather than from an arbitrary factorization of an arbitrary matrix: nontrivial line bundles would obstruct such a nonsingular splitting.

The factors can be chosen holomorphically in $x$ near any given $x_0$. An explicit local argument is useful. Normalize the known splitting at $x_0$ and set $Q(x,\lambda )=\tilde{H}(x_0,\lambda {)}^{-1}F(x,\lambda )H(x_0,\lambda )$ on a fixed circular contour in the overlap. It is close to the identity when $x$ is close to $x_0$. Let $P_{+}$ select strictly positive Laurent powers, and seek $q=1+$ positive powers satisfyingIn a holomorphic contour-function norm the operator has norm below one for sufficiently small variation, so its Neumann series converges and is holomorphic in $x$. Then $Qq=\tilde{q}$ has only nonpositive powers and is holomorphic on the exterior patch including infinity. Both factors are close to the identity and invertible. They extend over their full patches through $Q$ and its inverse. Taking $H=H(x_0)q$ and $\tilde{H}=\tilde{H}(x_0)\tilde{q}$ gives the required local family. This is also the elementary Laurent series reason that $H^1({\mathbb{CP}}^1,\mathcal{O})=0$ leaves no infinitesimal obstruction.

Raise primed two-component spinor indices with the antisymmetric two-component spinor form and define $V_A={\pi }^{A^{\prime }}{\partial }_{A{A}^{\prime }}$, holding $\pi$ fixed. Incidence givesThus $V_{A}F=0$ for a pulled-back function depending only on twistor coordinates. Differentiating the splitting and multiplying by the inverse factors givesThe left expression is holomorphic on one two-component spinor patch, the right on the other, and each is homogeneous of degree one in $\pi$. They therefore glue to a global matrix-valued section of $\mathcal{O}(1)$ on the line.

Such a section is linear in its homogeneous coordinates. To see this explicitly, set $\lambda ={\pi }_{0^{\prime }}/{\pi }_{1^{\prime }}$ on $U$. The normalized expression is an entire matrix function of $\lambda$; on the other patch its quotient by $\lambda$ is holomorphic at infinity. Hence it has at most linear growth. Cauchy estimates force all its derivatives of order at least two to vanish. Entry by entry it is $a_A(x)+\lambda b_A(x)$, or invariantlyThe coefficients are holomorphic in $x$. This proves existence of the desired space-time one-form, with no remaining dependence on $\pi$ in its coefficients.

A second splitting must differ by right multiplication $H\mapsto Hg(x)$ and $\tilde{H}\mapsto \tilde{H}g(x)$. Indeed the comparison matrix is holomorphic on both patches of a compact complex projective line and therefore constant along the line. The resulting transformation isthe gauge connection transformation law. Local coefficient matrices consequently glue to a holomorphic gauge connection on the bundle whose fiber at $x$ is the space of holomorphic sections of $E{|}_{L_x}$. If a single global matrix-valued one-form is wanted on all $M_{\mathbb{C}}\simeq {\mathbb{C}}^4$, choose a basis at the origin and parallel-transport it along the straight radial paths to each $x$. Linear transport exists on every finite path and varies holomorphically with $x$; this gives a global holomorphic bundle frame. The assertion does not require flat connection in all four directions, only a fixed choice of paths for a trivialization. On more general space-time domains the intrinsic object is a gauge connection with these gauge-related local one-forms.

Finally put $D_{A{A}^{\prime }}={\partial }_{A{A}^{\prime }}+{\Phi }_{A{A}^{\prime }}$ and ${\mathcal{D}}_A(\pi )={\pi }^{A^{\prime }}{D}_{A{A}^{\prime }}$. From the defining identity,Thus the columns of $H^{-1}$ are an invertible parallel bundle frame in the two alpha-plane directions. Their matrix commutator must annihilate that bundle frame, soUse the spinor decomposition of gauge curvaturewhere both gauge curvature two-component spinors are symmetric. The second term vanishes under contraction with ${\pi }^{A^{\prime }}{\pi }^{B^{\prime }}$. The first is a quadratic polynomial in the two two-component spinor components; vanishing for every two-component spinor forces each of its coefficients to vanish. Thereforewhich is precisely the ASDYM equation in the convention where the primed symmetric two-component spinor is the self-dual part. Equivalently, the gauge connection is flat when restricted to every alpha-plane. This completes the forward Penrose-Ward correspondence, including both the existence of the gauge connection and its anti-self-duality of gauge curvature.

For a continuous $2\pi$-periodic function, define the Fourier coefficients and Fourier partial sum byThe Fejér sum is the arithmetic average of the first $n$ Fourier partial sums:Thus $s_n$ has degree at most $n$, whereas ${\sigma }_n$ has degree at most $n-1$. These indexing conventions matter in the identity in (b). Throughout, $m$ is a positive integer; when $n=0$, any term $n{\sigma }_n$ is interpreted as zero without introducing ${\sigma }_0$.

For later estimates, the Fejér kernel in the PDF's normalization isThe apparent singularities are removable. The squared modulus makes $F_n$ nonnegative, and integrating its finite exponential expansion leaves only the $n$ diagonal terms, so ${\pi }^{-1}\int F_n=1$. In particular Fejér summation is a uniform-norm contraction:

Use the periodic modulus of continuitySubdividing a displacement $t$ into at most $\lceil |t|/\delta \rceil$ steps of size at most $\delta$ proves the chaining inequalityFor $|t|\le \delta$, the sharper bound $\omega (f,\delta )$ holds directly.

Set $\delta =n^{-\alpha }$, where $0<\alpha <1$ and $n\ge 1$. Positivity and unit normalized mass of the Fejér kernel giveThe portion $|t|\le \delta$ is at most $\omega (f,\delta )$. For $0<|t|\le \pi$, $\sin (|t|/2)\ge |t|/\pi$ and $|\sin (nt/2)|\le 1$, soApply this bound and the chaining inequality to the two tails:Let $\beta =1-\alpha >0$. Calculus gives ${\sup }_{u\ge 1}{u}^{-\beta }\log u=1/(e\beta )$, attained at $u=e^{1/\beta }$, and $u^{-\beta }\le 1$. Thus a possible explicit constant for the fractional-scale Fejér approximation bound isThe constant depends only on $\alpha$, not on $f$ or $n$. Since a continuous periodic function is uniformly continuous, the right-hand side tends to zero. The logarithmic tail estimate is what permits every $\alpha <1$; a crude global-oscillation bound would not prove the whole requested range.

Use the Wirtinger derivative $f=q_z=(q_x-i{q}_y)/2$. Since $q$ satisfies the Laplace equation, ${\partial }_{\bar{z}}f=\Delta q/4=0$, so $f$ is holomorphic. The differential relation is understood as a one-form identity:The differential $dz$ is needed on its right-hand side. Because this form is closed, its integral is independent of path within the semistrip.

Three spectral primitives are obtained by integrating from the two finite corners and from the infinite end:The last integral is taken horizontally and defines the infinite-end primitive for $\mathrm{Im}k<0$. Their differences are $e^{ikz}$ times the three boundary spectral functions. Using counterclockwise boundary orientation, defineThen ${\mu }_0-{\mu }_{\infty }=e^{ikz}{\rho }_B$, ${\mu }_{\infty }-{\mu }_l=e^{ikz}{\rho }_T$, and ${\mu }_l-{\mu }_0=e^{ikz}{\rho }_L$. Their sum gives the global relation ${\rho }_L+{\rho }_B+{\rho }_T=0$ for $\mathrm{Im}k\le 0$.

To express these spectra in boundary data, write $g_0(x)=q(x,0)$, $g_l(x)=q(x,l)$, $h(y)=q(0,y)$, $n_0(x)=q_y(x,0)$, $n_l(x)=q_y(x,l)$, and $v(y)=q_x(0,y)$. Direct substitution into $f$ givesThese are transforms of tangential derivatives of Dirichlet boundary data and the corresponding Neumann boundary data. The outward derivatives are $-n_0$ at the bottom, $n_l$ at the top, and $-v$ at the left.

The boundary spectral representation of a holomorphic function on a semistrip isAll three $k$-rays are oriented outward from zero. To verify the spectral inversion, usewhenever the exponential decays. The positive real ray works for the bottom, the negative real ray for the top, and the positive imaginary ray for the left. Substitution of the boundary spectra therefore turns the boxed formula into $(2\pi i{)}^{-1}{\int }_{\partial \Omega }f(w)/(w-z)dw=f(z)$ by the Cauchy integral formula. This also fixes all orientation signs.

Let $G_0(k),G_l(k)$ denote the Half-range Fourier transforms of $g_0,g_l$, and let $N_0(k),N_l(k)$ denote those of the upward derivatives $n_0,n_l$. Also setIntegrate the tangential derivatives in the three spectral functions by parts. For example, ${\int }_0^{\infty }{e}^{-ikx}{g}_0^{\prime }=-h(0)+ik{G}_0$, and ${\int }_0^l{e}^{ky}{h}^{\prime }=e^{kl}h(l)-h(0)-kH$. All corner values cancel in ${\rho }_L+{\rho }_B+{\rho }_T=0$. The result is the semistrip Laplace spectral global relation

For $\kappa >0$, use the sine transformsReality of $q$ makes $W(\kappa )$ and $W(-\kappa )$ real, so taking imaginary parts removes the unknown left-side normal derivative. At $k=\kappa$ and $k=-\kappa$ the two consequences of the global relation areSubtracting these equations and then substituting back gives the semistrip Dirichlet-to-Neumann sine transforms entirely in terms of the prescribed Dirichlet boundary data:These are the requested transforms of $q_y$ itself. The bottom outward Neumann boundary data have the opposite sign. The apparent division at $\kappa =0$ is handled by a continuous limit whenever the corresponding moments exist; there is no singularity for any $\kappa >0$.

Since $q$ is harmonic, $q_z$ is holomorphic and the one-form $e^{-ikz}{q}_{z}dz$ is closed. Closing the physical quadrant at infinity gives the global relationIn this third spectral quadrant the exponential decays on both boundary axes. Set $\gamma ={\beta }_1+{\beta }_2$. The relation becomes

All four boundary functions are real. Consequently their transforms obeyand the same identities for $J_1,J_2$. Complex conjugation of the global relation therefore gives its upper-left consequenceIntroduce just one unknown function,It is holomorphic in the upper half-plane. Under the transform-admissible decay assumptions, $|\Phi (k)|\le {\int }_0^{\infty }(|h_2|+|j_2|)dx$, so it is bounded there.

For $k$ on the positive real ray, use both consequences of the global relation at $-k$. Adding them removes $J_1(-k)$ and givesFor $k$ on the positive imaginary ray, use the upper-left relation and $H_2(-k)+i{J}_2(-k)=2H_2(-k)-\Phi (k)$. The resulting single-function spectral elimination for a rational-angle quadrant isBoundary limits from the appropriate transform domains are understood. Every remaining unknown boundary derivative is contained in the one bounded upper-half-plane function $\Phi$.

First resolve the orientation sign in quadrant harmonic spectral inversion. The printed spectral definitions traverse the physical boundary clockwise, whereas both reconstruction rays run outward from zero. Integrating first in $k$ gives the kernel $i/(z-w)$. The resulting clockwise Cauchy integral is the negative of the desired holomorphic function. Thus, with exactly the spectral definitions used above, the consistent reconstruction isThe printed plus sign would reconstruct $-q_z$. For a concrete nonzero example, $q(z)=-2\mathrm{Re}[(1+i)/(z+1+i)]$ is real, harmonic, smooth on the closed quadrant and decays along both axes. With ${\beta }_1={\beta }_2=0$, it has $h_1(0)=h_2(0)=1$ and $q_z=(1+i)/(z+1+i{)}^2$. The printed orientation returns the negative of this derivative, so the issue persists even under the stated corner condition.

Let $a=\frac{1}{2}{e}^{-i{\beta }_1-i\gamma }$. In the expressions from the preceding part, the coefficient of $\Phi$ on the imaginary ray is $a$, and the coefficient on the real ray is $-a{e}^{4i\gamma }$. For each of the three allowed sums, $e^{4i\gamma }=1$. The unknown part is therefore proportional toIndeed, close the first spectral quadrant. For $z=x+iy$ with $x,y>0$, $|e^{ikz}|=e^{-(x\mathrm{Im}k+y\mathrm{Re}k)}$, so the large-arc contribution vanishes for bounded $\Phi$. The Cauchy integral theorem proves the equality and eliminates the single unknown function.

Substituting only the known pieces leavesThis already determines the answer from $h_1,h_2$ alone. Evaluating the elementary ray kernels gives the more direct rational-angle oblique derivative problem on a quadrant formulaNo denominator vanishes for an interior point of the first quadrant.

For explicit formulas in the three cases, defineCombining the paired fractions and using ${\beta }_2=\gamma -{\beta }_1$ yields

There is also a necessary corner compatibility for collinear oblique derivative data. When $\gamma =\pi /2$, the two prescribed derivative directions are opposites, so a continuous corner gradient requires $h_2(0)=-h_1(0)$. Together with the stipulated equality, this forces $h_1(0)=h_2(0)=0$ in that case. Existence was assumed, so admissible data must satisfy this additional consequence. Smoothness at the corner and decay at infinity also exclude the homogeneous corner ambiguity in a quadrant Laplace problem; no extra singular or growing term is permitted in the boxed reconstruction.

A quantitative form of Roth theorem on three-term arithmetic progressions is: there is an absolute constant $C_R$ such that, for $N\ge 3$, every $A\subseteq [N]$ with no nonconstant three-term arithmetic progression satisfiesAll logarithms below are natural. We prove the classical Roth bound for three-term progressions with an unoptimized $C_R=100$. The empty set is immediate; write $\delta =|A|/N>0$.

First establish a quantitative density increment. Suppose $N\ge 4\cdot 10^6{\delta }^{-4}$. Put $M=2N+1$, embed $I=[N]$ in $G={\mathbb{Z}}_M$, and extend all interval functions by zero. Let $a=1_A$, $u=\delta 1_I$ and $f=a-u$, the balanced subset indicator. DefineThe relation in the sum is modulo $M$. For interval elements its integer discrepancy has absolute value less than $M$, so it is precisely the integer relation. Thus $\Lambda (a,a,a)=|A|=\delta N$. The full interval count is the number of endpoint pairs of the same parity:Consequently, since $N\ge 4{\delta }^{-2}$,

Orthogonality of the additive characters givesBoth identities follow by expanding the sums and using ${\sum }_r{e}^{2\pi irt/M}=M$ for $t=0$ in $G$, and zero otherwise; the latter follows from a finite geometric series. Since $M$ is odd, $r\mapsto -2r$ permutes the frequencies. Telescope the difference asThe Cauchy-Schwarz inequality and the displayed Parseval identity on a finite group bound each term by $\delta N{\max }_r|\hat{f}(r)|$. Indeed, $\Vert a{\Vert }_2^2=\delta N$ and $\Vert u{\Vert }_2^2={\delta }^{2}N$, so every product of the two remaining L2 norms is at most $\delta N$. HenceThe zero coefficient vanishes because $\sum f=0$, so choose a nonzero frequency $\theta =r/M$ giving this bound.

Let $Q=\lfloor \sqrt{N}\rfloor$. The pigeonhole principle applied to the $Q+1$ fractional parts $0,\theta ,\dots ,Q\theta$ in $Q$ intervals gives an integer $1\le q\le Q$ with $\Vert q\theta {\Vert }_{\mathbb{R}/\mathbb{Z}}\le 1/Q$. Partition each residue class modulo $q$ in $[N]$ into consecutive arithmetic progressions of exactlyelements, leaving fewer than $L$ remainder points per residue. The total remainder $R_0$ has size at most $qL\le {\delta }^{2}N/1000$. The size hypothesis gives $L\ge {\delta }^2\sqrt{N}/2000$. On a full arithmetic progression $P$, the phase $e^{-2\pi i\theta x}$ varies from its value at the first point by at most $2\pi L/Q\le 4\pi {\delta }^2/1000$.

Let $F_P={\sum }_{x\in P}f(x)$. Replacing the phase by its first value on each full arithmetic progression and estimating the remainder using $|f|\le 1$ givesAlso $|{\sum }_P{F}_P|=|{\sum }_{R_0}f|\le {\delta }^{2}N/1000$. Therefore the sum of the positive $F_P$ isSince the total length of these arithmetic progressions is at most $N$, at least one has $F_P/|P|\ge {\delta }^2/64$. Thus the step just proved isAn affine map taking this arithmetic progression to $[|P|]$ preserves the absence of nonconstant three-term arithmetic progressions.

Now iterate. Write $N_j,{\delta }_j$ for successive lengths and subset densities, with $N_0=N$ and ${\delta }_0=\delta$. Since every subset density is at most one,Consequently $T=\lceil 65/\delta \rceil$ successful steps are impossible. On the other hand, subset densities never decrease, so with $B=\log (2000/{\delta }^2)$,Suppose $\delta \log \log N>100$. Since $T\le 66/\delta$,Let $D=\log (4\cdot 10^6{\delta }^{-4})$. We have $2B+D<31+8\log (1/\delta )\le 39/\delta <e^{54/\delta }$. Therefore every length through step $T$ satisfies $\log N_j>D$, and hence $N_j\ge 4\cdot 10^6{\delta }^{-4}\ge 4\cdot 10^6{\delta }_j^{-4}$. The density increment lemma remains applicable for all $T$ steps, contradicting the reciprocal-density bound.

Thus $\delta \log \log N\le 100$, proving the boxed theorem. The proof includes the Fourier analysis on a finite abelian group detection, conversion to an integer arithmetic progression, and the length bookkeeping that produces the double logarithm.

Yes. The same telescoping argument proves the counting lemma for octahedrally quasirandom three-uniform hypergraphs for every fixed simple three-uniform hypergraph $F$.

Label its vertices of a hypergraph $1,\dots ,v$ and give vertex of a hypergraph $i$ a nonempty host part $V_i$. For each hypergraph edge $e=\{i,j,k\}$, let $h_e$ be the corresponding host hypergraph edge indicator function of subset density $p_e$, with $\Vert h_e-p_e{\Vert }_{{\square }^3}\le {\alpha }^{1/8}$. The normalized labelled hypergraph copy count isIf the host parts are disjoint, every such vertex of a hypergraph choice is automatically injective. Enumerate the $m$ hypergraph edges as $e_1,\dots ,e_m$ and telescopeIn the term for $e_j=\{a,b,c\}$, fix all variables outside this triple. Since $F$ is simple and three-uniform, every other hypergraph edge intersects $\{a,b,c\}$ in at most two vertices of a hypergraph. Its remaining factor therefore depends on at most two of $x_a,x_b,x_c$. Group all two-variable factors according to the pairs $(a,b),(a,c),(b,c)$. Absorb each one-variable factor into any pair containing its variable, and constants into any group. The three resulting pair functions remain bounded by one.

The balanced factor is now multiplied by exactly the sort of pair functions handled in part (ii). Its average has absolute value at most ${\alpha }^{1/8}$, independently of the fixed outside variables. Averaging those variables and summing the $m$ errors givesEquivalently,This counts labelled copies preserving the specified vertex of a hypergraph classes, for arbitrary intersection patterns among the hypergraph edges, and includes the three-uniform tetrahedron as a special case.