Use integral singular homology. The singular chain group $C_n(B)$ is the free abelian group on continuous maps ${\Delta }^n\to B$, and its boundary operator isFor $i<j$, deleting vertex $j$ and then vertex $i$ has sign $(-1{)}^{i+j}$, while deleting $i$ and then the shifted vertex $j-1$ has sign $(-1{)}^{i+j-1}$. Both yield the same codimension-two face. Pairing these terms proves ${\partial }^2=0$. The inclusion $A\subset B$ identifies $C_n(A)$ with the subgroup spanned by those simplices having image in $A$, so it is injective and preserved by $\partial$.

The relative chain complex is $C_n(B,A)=C_n(B)/C_n(A)$ with induced boundary $\partial \bar{b}=\overline{\partial b}$. Thus there is a degreewise short exact sequence of chain complexesInclusion and quotient take homological cycles to homological cycles and boundaries to boundaries, hence define the group homomorphisms $i_{\ast }$ and $q_{\ast }$ on homology. We construct and prove exactness of the resulting sequence directly, rather than invoke an unproved algebraic lemma:ending in $H_0(A)\to H_0(B)\to H_0(B,A)\to 0$.

For the relative homology connecting homomorphism, lift a relative cycle $\bar{b}\in C_n(B,A)$ to $b\in C_n(B)$. Since $\partial \bar{b}=0$, the chain $\partial b$ belongs to $C_{n-1}(A)$; it is a homological cycle because ${\partial }^2=0$. DefineChanging the lift by $a\in C_n(A)$ changes the boundary by $\partial a$, hence does not change its homology class. More generally, if a different relative cycle represents the same relative homology class, lifts satisfy $b^{\prime }-b=\partial c+a$ for some $c\in C_{n+1}(B)$ and $a\in C_n(A)$. Then $\partial b^{\prime }-\partial b=\partial a$. This proves independence both of the lift and of the homology representative. Lifting a sum by the sum of lifts proves additivity, so $\delta$ is a well-defined group homomorphism.

At $H_n(A)$, an element in the image of $\delta$ is represented by $\partial b$ for some $b\in C_{n+1}(B)$, so its image in $H_n(B)$ is zero. Conversely, if a homological cycle in $A$ $a$ has $i_{\ast }[a]=0$, then $a=\partial b$ for a $B$-chain $b\in C_{n+1}(B)$. Its quotient $\bar{b}$ is a relative cycle and $\delta [\bar{b}]=[a]$. ThusFor completeness, the other terms are exact by the same elementary chain argument. At $H_n(B)$, clearly $q_{\ast }{i}_{\ast }=0$. If a homological cycle in $B$ $b$ has zero relative class, then $\bar{b}=\partial \bar{c}$ for a quotient chain $\bar{c}$. Lifting $c$ gives $a=b-\partial c\in C_n(A)$, and $\partial a=0$. Hence $[b]=i_{\ast }[a]$, proving $\ker q_{\ast }=\mathrm{im}{i}_{\ast }$. At $H_n(B,A)$, absolute homological cycles have zero connecting image. Conversely, if $\delta [\bar{b}]=0$, then $\partial b=\partial a$ for some $a\in C_n(A)$; the chain $b-a$ is an absolute homological cycle and maps to $[\bar{b}]$. Therefore $\ker \delta =\mathrm{im}{q}_{\ast }$. In degree zero every quotient chain lifts to an absolute homological cycle, because the differential to degree $-1$ is zero, so $q_{\ast }:H_0(B)\to H_0(B,A)$ is surjective. These proofs establish the whole long exact sequence in relative homology and all the algebraic facts used in its construction.

A simply connected manifold is connected and orientable: its orientation character is a group homomorphism ${\pi }_1(X)\to \{\pm 1\}$, and this group is trivial. Choose an orientation. The Hurewicz theorem in degree one gives $H_1(X;\mathbb{Z})=0$. The universal coefficient theorem for cohomology then gives $H^1(X;\mathbb{Z})=0$ andbecause the possible $\mathrm{Ext}(H_1(X),\mathbb{Z})$ term is zero.

The homology groups are finitely generated for a closed manifold. The displayed dual group is therefore free abelian. Poincare duality identifies $H_2(X;\mathbb{Z})$ with $H^2(X;\mathbb{Z})$, so $H_2$ itself has no torsion. Since the second Betti number is one, both groups are isomorphic to $\mathbb{Z}$. This torsion argument is essential: rank one by itself would not rule out finite summands.

Again by Poincare duality, $H^3(X;\mathbb{Z})\cong H_1(X;\mathbb{Z})=0$, and $H^4(X;\mathbb{Z})\cong H_0(X;\mathbb{Z})=\mathbb{Z}$. Thus the only nonzero cohomology groups are $\mathbb{Z}$ in degrees $0,2,4$.

Let $u$ generate $H^2$. The integral intersection formis unimodular. Indeed, cap product with $[X]$ is the Poincare duality isomorphism to $H_2$, and the preceding universal-coefficient identification makes evaluation the full integral dual. Composing these isomorphisms identifies the pairing with an isomorphism $H^2\to \mathrm{Hom}(H^2,\mathbb{Z})$. In rank one its matrix must therefore be $[1]$ or $[-1]$. Consequently $⟨u^2,[X]⟩=\pm 1$, so $u^2$ generates $H^4$.

All products in degrees above four vanish. Hence the assignment of the degree-two generator giveswhich is the cohomology ring of complex projective space for ${\mathbb{CP}}^2$. If the intersection form is negative, this ring isomorphism reverses the chosen top orientation generator; preserving that additional orientation is not part of the requested ring isomorphism.

We state the integer form of Freiman theorem for integer sets: for every $K\ge 1$ there are finite constants $r(K),C(K)$ such that every finite nonempty $A\subset \mathbb{Z}$ with doubling at most $K$ lies in a proper generalized arithmetic progression of rank at most $r(K)$ and cardinality at most $C(K)|A|$. An arbitrary abelian group requires the coset-progression version, allowing a subgroup factor. We prove the integer statement, with no claim of optimized constants.

We use the allowed Plünnecke inequality and basic Freiman isomorphism facts, and explicitly prove the modelling, Fourier and covering steps. Two geometry-of-numbers consequences are used as permitted: a cyclic rank-$d$ phase-radius-$\rho$ Bohr set contains a proper symmetric progression of rank at most $d+1$ and size at least $c(d,\rho )$ times the cyclic group order; and a rank-$r$ integer progression with parameter-box volume $V$ can be contained in a proper progression of rank at most $r$ and size at most $c_{r}V$. These follow from successive minima and lattice basis reduction. All these constants depend only on the displayed rank and radius, not on the ambient order or step sizes.

First construct a dense cyclic model of a large subset. Put $D=8A-8A$ and $q=8|D|$. Plünnecke gives $|D|\le K^{16}|A|$. Choose an auxiliary prime $p$ so large that every nonzero integer in $D$ stays nonzero modulo $p$, and $p>2q$. Arbitrarily large primes exist: a purported finite list would miss a prime divisor of one plus their product. For a uniform nonzero multiplier $\lambda (\mathrm{mod}p)$, each nonzero $d\in D$ has $\lambda d$ uniformly distributed among the nonzero residues. The forbidden residues are those represented by nonzero multiples of $q$ with absolute value less than $p$; there are at most $2(p-1)/q$ of them. The union bound gives failure probability at mostThus choose $\lambda$ avoiding them for every nonzero $d\in D$.

Represent the dilated elements of $A$ in $[0,p-1]$, divide this interval into nine consecutive subintervals, and retain a most-populated one. It defines $A^{\prime }\subseteq A$ with $|A^{\prime }|\ge |A|/9$. If $u(a)$ is the chosen representative of $\lambda a$, eight-term sums of representatives from this subinterval have a range shorter than $p$: eight times its width is less than $8p/9$. Consequently every original eight-term relation lifts from equality modulo $p$ to equality of the representative sums as integers.

Now reduce these representatives modulo $q$. If their two eight-term sums agree modulo $q$, their integer difference is a multiple of $q$ of absolute value below $p$. A nonzero such difference would give a forbidden residue $\lambda d$ for some nonzero $d\in D$. If $d=0$, its representative difference is already zero. Hence there are no new relations either. The map is therefore a Freiman s-isomorphism of order eight onto $B\subseteq \mathbb{Z}/q\mathbb{Z}$. It is injective, by padding shorter relations to eight terms with a fixed retained point. Its density satisfiesThis proves the cyclic Freiman model of a small-doubling integer set needed below.

Next we prove the cyclic Bogolyubov lemma. Use normalized finite-group Fourier analysis on $\mathbb{Z}/q\mathbb{Z}$. The convolutionhas Fourier expansion $r(x)={\sum }_{\chi }|\widehat{1_B}(\chi ){|}^4\chi (x)$. Let $\Gamma$ consist of those characters whose Fourier coefficient has modulus at least ${\beta }^{3/2}/\sqrt{2}$. Parseval gives $|\Gamma |\le 2{\beta }^{-2}$, and the fourth-power tail outside $\Gamma$ is at most ${\beta }^4/2$. The zero character belongs to $\Gamma$ and contributes ${\beta }^4$.

For the phase-distance Bohr set $\mathcal{B}=\{x:\Vert \arg \chi (x)/(2\pi )\Vert \le 1/10\text{ for all }\chi \in \Gamma \}$, every retained character has real part at least $\cos (\pi /5)>3/4$. ThereforeThe support of this convolution is $2B-2B$, so $\mathcal{B}\subseteq 2B-2B$. Apply the stated progression in a Bohr set from successive minima consequence. It supplies a symmetric proper progression $P_B\subseteq 2B-2B$ of rank bounded by a function of $K$ and size at least $c(K)q$, hence at least $c(K)|A|$ after adjusting the constant.

Lift this progression back to the integers. The inverse Freiman map inducesEquality of two representations uses a four-term relation on each side, so $\psi$ is well defined and injective. If $y,z,y+z$ all lie in this difference set, the relation between their representations involves at most six original terms on either side. The order-eight model therefore gives $\psi (y+z)=\psi (y)+\psi (z)$ and $\psi (0)=0$. Along every coordinate path within the symmetric progression box, all partial sums remain in $P_B$. It follows thatThus $P=\psi (P_B)$ is a symmetric proper integer progression of the same rank and cardinality, and $P\subseteq 2A-2A$. This proves the Freiman lifting of a progression step, including why a sufficiently high-order model was selected.

It remains to cover all of $A$, not merely its large modeled subset. Since $P\subseteq 2A-2A$, Plünnecke givesChoose $X\subseteq A$ maximal such that the translates $x+P$ are pairwise disjoint. Their union lies in $A+P$, so $|X|\le K^5|A|/|P|\le C_1(K)$. For every $a\in A$, maximality makes $a+P$ meet some $x+P$, giving $a\in x+P-P$. ConsequentlyThis proves the needed Ruzsa covering lemma directly.

Write $P=\{{\sum }_{i=1}^r{n}_i{w}_i:|n_i|\le L_i\}$ and $X=\{x_1,\dots ,x_s\}$. The setcontains $X+(P-P)$ and hence $A$. Its rank is at most $r+s$, bounded only by $K$, and its parameter-box volume is at most $2^{r+s}|P|$. Also $|P|\le |2A-2A|\le K^4|A|$, so this volume is at most $C_2(K)|A|$. The allowed integer properification consequence now contains $Q$ in a proper progression of bounded rank and size at most $C(K)|A|$. ThereforeThis proves Freiman's theorem, using geometry of numbers only in the two explicitly stated places.

Write $e(t)=e^{2\pi it}$ and $N=n+1$. A convenient form of the quadratic Weyl inequality is the following: if $q\ge 1$, $(a,q)=1$ and $|\alpha -a/q|\le q^{-2}$, thenThe implied constant is absolute. In particular the familiar square-root saving is $|S|\ll (N^2/q+N+q{)}^{1/2}\log (2q{)}^{1/2}$. We prove all estimates used in this statement.

First, the finite geometric series identity giveswhere $\Vert \theta \Vert$ denotes distance to the nearest integer and the second bound is interpreted as infinite when that distance is zero. Indeed the sum is $(1-e(L\theta ))/(1-e(\theta ))$, and $|1-e(\theta )|=2|\sin \pi \theta |\ge 4\Vert \theta \Vert$. This proves the exponential geometric sum bound.

Expand the squared quadratic exponential sum by the difference $h$ between its indices. Since $(x+h{)}^2-x^2=2hx+h^2$, the diagonal and the paired off-diagonals giveFor $q\ge 16$, partition the $h$ into consecutive blocks of length at most $L=\lfloor q/8\rfloor$. Within one block, two distinct indices have difference $0<|d|<q/2$. Since the denominator of $2a/q$ after reduction is at least $q/2$, $\Vert 2ad/q\Vert \ge 1/q$. ThereforeThus the points $2\alpha h$ in a block are separated on the circle by at least $3/(4q)$.

For any such separated set, at most an absolute number of points lie in each pair of arcs whose distances from zero are between $j/(4q)$ and $(j+1)/(4q)$. The closest few contribute $O(N)$ to the capped reciprocal sum; the remaining arcs contribute $O(q/j)$ for $1\le j\le 4q$. Summing the harmonic series, using ${\sum }_{j\le m}{j}^{-1}\le 1+\log m$, proves a block bound $O(N+q\log (2q))$. This is the required separated reciprocal-distance sum estimate, with its proof supplied here. There are $O(N/q+1)$ blocks, soFor $q<16$, the trivial bound $|S|\le N$ already implies this estimate after increasing the absolute constant. This completes the proof, including the small denominators and the possible common factor of $2$ and $q$.

We prove the three-term Roth theorem by a density increment, keeping the interval boundary rather than replacing the interval by a cyclic group with a different density. The case $\delta >1$ is vacuous; assume $0<\delta \le 1$.

We first prove the following alternative with absolute constants $c,C>0$: if $A\subseteq [1,N]$ has density $\eta$ and no nonconstant three-term arithmetic progression, and $N\ge C{\eta }^{-4}$, then some integer progression $P\subseteq [1,N]$ hasFor this purpose take $M=6\lceil 4N/6\rceil +1$, so $4N<M\le 5N$ for large $N$ and $2,3$ are invertible modulo $M$. On $G=\mathbb{Z}/M\mathbb{Z}$, put $I=1_{[1,N]}$, $a=1_A$, $g=\eta I$ and $f=a-g$. In particular $\sum f=0$, and $f$ vanishes outside the interval. Use normalized Fourier coefficients on a finite abelian group, $\hat{u}(r)=M^{-1}{\sum }_{x}u(x)e(-rx/M)$. Character orthogonality, obtained by summing a finite geometric series, gives Fourier inversion on a finite group and Parseval identity on a finite group.

For $\Lambda (u,v,w)={\mathbb{E}}_{x,d}u(x)v(x+d)w(x+2d)$ these identities giveThere is no wrapping ambiguity when all three entries lie in $[1,N]$: the integer $x-2y+z$ has absolute value below $M$, so a modular relation is an integer relation. If no nonconstant progression exists, $\Lambda (a,a,a)=|A|/M^2$. Meanwhile the choices $1\le x\le \lfloor N/4\rfloor$ and $1\le d\le \lfloor N/8\rfloor$ give $\Lambda (I,I,I)\ge 1/10000$ for sufficiently large $N$.

Expand the difference by telescoping,If $\rho ={\max }_r|\hat{f}(r)|$, each term on the right has modulus at most $\rho \eta N/M$. Indeed multiplication of frequencies by $2$ permutes them, and Cauchy-Schwarz with Parseval identity bounds the two other Fourier factors by their $L^2$ norms; each squared norm is at most $\eta N/M$. For $N\ge 10000{\eta }^{-2}$, the diagonal count is less than half the main density term. ThereforeIn particular there is a nonzero frequency $r$ withThe frequency is nonzero because $\hat{f}(0)=0$.

Here is the conversion of this coefficient into a density increment. Put $Q=\lfloor \sqrt{N}\rfloor$. Of the $Q+1$ numbers $0,\theta ,\dots ,Q\theta$ modulo one, two lie in the same one of $Q$ equal arcs. Their difference gives $1\le q\le Q$ with $\Vert q\theta \Vert \le 1/Q$, where $\theta =-r/M$. This is the needed Dirichlet approximation theorem, proved by the pigeonhole principle.

Partition each residue class modulo $q$ in $[1,N]$ into consecutive progressions of length $L=\lfloor \gamma Q/(100\pi )\rfloor$, leaving its final short tail. Their total tail size is at most $qL\le \gamma N/(100\pi )$. On each full progression the character differs from its value at the first point by at most $2\pi L/Q\le \gamma /50$. Hence removing tails and replacing the character by its initial value showsThe sum of $f$ over all full progressions is the negative of its sum over the tails, so its absolute value is at most $\gamma N/(100\pi )$. Consequently the sum of the positive block sums is at least $\gamma N/5$. Since these blocks contain at most $N$ points in total, at least one has average $f$ at least $\gamma /5$. It gives a density increment at least ${\eta }^2/100000$. For $N\ge C{\eta }^{-4}$ with a sufficiently large absolute $C$, $L\ge c{\eta }^2\sqrt{N}$. This proves the alternative, with one small common absolute $c$.

Reidentify $P$ with $[1,|P|]$ by its affine parametrization. Absence of a nonconstant progression is preserved, while density increases. At every step the density is at least the original $\delta$, so the increment is at least $c{\delta }^2$ and the new interval length is at least $c{\delta }^2\sqrt{N_{\mathrm{old}}}$. After at most $T=\lceil 1/(c{\delta }^2)\rceil +1$ steps, the density would exceed one. Choose the original $N$ so large that the length stays above $C{\delta }^{-4}$ throughout these finitely many steps. Such a choice exists explicitly by working backwards: a required next length $L_0$ is ensured by previous length at least $(L_0/(c{\delta }^2){)}^2$. The alternative could then be applied at every step, giving a contradiction. ThusThis is the Roth density increment on an integer progression proof.

For the pattern with offsets $0,1,3$, replace the counting functional by ${\mathbb{E}}_{x,d}u(x)v(x+d)w(x+3d)$. Its Fourier expression iscoming from the relation $2x-3y+z=0$. Both multipliers are invertible modulo the chosen $M$, so exactly the same Parseval estimates apply. This relation has no wrap in $[1,N]$, since its magnitude is at most $3N<M$, and the earlier restricted choices of $x,d$ still give a fixed positive density main term. The same increment and iteration find a nonconstant triple $\{a,a+d,a+3d\}$. Negative $d$ is allowed, as requested, and the affine reparametrization used in the iteration preserves this pattern as well.

We give a spectral proof of square-difference recurrence for finite colourings, proving the auxiliary results it uses. Extend the colouring arbitrarily to nonpositive integers, and let $f_j$ be the indicator of colour $j$. By successively taking subsequences of the interval lengths, followed by a diagonal subsequence, we may arrange that all the limitsexist, for every integer $h$. This uses only sequential compactness of bounded real sequences: list the countably many quantities, choose a convergent subsequence for each in turn, and take the diagonal. All limits below are along this common subsequence. We have $c(0)=1$, $c(-h)=c(h)$, $c(h)\ge 0$ and ${\sum }_j{\delta }_j=1$. Translating a fixed interval endpoint changes a normalized average by $O(|h|/L)$.

The sequence is a positive-definite function on $\mathbb{Z}$. Indeed, for finitely many complex numbers $z_u$,We next prove the required Bochner–Herglotz theorem for this sequence. The trigonometric polynomialis nonnegative by positive definiteness, and its integral on the circle is $c(0)=1$. Thus $F_H(t)dt$ is a probability measure. Such measures on a compact interval have a weakly convergent subsequence. For completeness, take a diagonal subsequence on all rational points of their nondecreasing distribution functions. The right-continuous extension of the limiting rational values is a distribution function and defines a probability measure by its interval increments. At its continuity points the original distribution functions converge, by sandwiching between rational endpoints. Approximating continuous functions by step functions with such endpoints proves convergence of their integrals. Identify the interval endpoints to obtain a measure on the circle. This proves the compactness of probability measures on a compact metric space in the case needed here.

For a fixed $h$, the Fourier coefficient of $F_{H}dt$ is $(1-|h|/H)c(h)$ once $H>|h|$. Taking the weak limit therefore supplies a positive probability measure $\mu$ such thatNo spectral representation has been assumed without proof.

Its atom at zero has positive mass. For each $H$,The last inequality is Cauchy-Schwarz on each colour's interval average. The integrand is at most one and tends to zero for every nonzero point of the circle, by the geometric-series formula, while it is one at zero. Bounded convergence gives

We also prove that quadratic phases with an irrational coefficient average to zero. For $z_0,\dots ,z_{L-1}$ bounded by one, extended by zero, Cauchy-Schwarz applied to ${\sum }_x{\sum }_{u=0}^{H-1}{z}_{x+u}=H{\sum }_x{z}_x$ yieldsThis identity follows by expanding the square of the inner shifted sum and grouping pairs by their difference; it is the needed finite Van der Corput inequality for finite scalar sequences. If $z_x=e(\theta x^2)$ and $\theta$ is irrational, each fixed nonzero lag correlation is a geometric sum with frequency $2\theta h$, hence is bounded independently of $L$. The geometric bound follows from $(1-z^L)/(1-z)$ with $z=e(2\theta h)\ne 1$. Dividing by $H^2{L}^2$ and letting $L$ tend to infinity gives $\mathrm{limsup}|L^{-1}\sum z_x{|}^2\le 1/H$. Letting $H$ tend to infinity proves the asserted cancellation. Rational coefficients instead give periodic sequences and therefore have convergent averages.

Choose a finite set $T$ of rational points of the circle, including zero, such that the total mass of the remaining rational atoms is below $1/(2r)$. This is possible because the rational points are countable and $\mu$ is finite. Let $q$ be a common multiple of their denominators. For $t\in T$, $e(t{q}^2{m}^2)=1$ for every integer $m$. For irrational $t$ the preceding cancellation applies to $q^{2}t$; at every other rational point the average exists by periodicity and has modulus at most one. Bounded convergence consequently givesHere the inequality takes real parts; $c$ itself is real. Thus some $b=qm>0$ has $c(b^2)>0$. By its definition as a limit of nonnegative averages, at least one actual positive $a$ has ${\sum }_j{f}_j(a)f_j(a+b^2)>0$. The same colour occurs at both points, provingThe arbitrary extension to nonpositive integers played no role in this final pair.

The analytic input for these Mellin transforms is elementary cusp decay. For a level-one weight-$k$ cusp form, the invariant norm of a modular form $y^{k/2}|f(x+iy)|$ is invariant under the modular group. It is bounded on the standard fundamental region, because the cusp expansion decays exponentially, and hence bounded on all of $\mathbb{H}$. Fourier inversion givesTaking $y=1/n$ proves the Fourier coefficient bound for a cusp form $c_n=O(n^{k/2})$. In particular rational translates have polynomially bounded coefficients and exponential decay as $y\to \infty$, uniformly in their real parts. These facts justify the initial Mellin integrations and the later entire continuation.