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For group homomorphisms $f,g:A\to B$, this Picard group class is symmetric and additive in both arguments. The Theorem of the Cube says precisely that the third additive difference of $h\mapsto [h^{\ast }L]$ vanishes, proving additivity of the cross-effect. The Poincaré line bundle expresses it as $(f,{\varphi }_{L}g{)}^{\ast }{P}_B$, so it depends only on the Néron-Severi group class of $L$. Its associated homomorphism is $\hat{f}{\varphi }_{L}g+\hat{g}{\varphi }_{L}f$.

If a line bundle on $E^3$ is trivial on each zero-coordinate face, its First Chern class vanishes: the integral second cohomology consists of single-factor second cohomology and cross-products of first cohomology from pairs, and each summand is detected on a coordinate face. The holomorphic exponential sequence then puts the bundle in ${\mathrm{Pic}}^0(E^3)$. The decomposition of first cohomology, including Dolbeault cohomology, identifies this group with ${\mathrm{Pic}}^0(E{)}^3$. Triviality on the three coordinate axes makes all three factors trivial. Applying this cube principle to the alternating partial-sum pullbacks proves the usual addition-map Theorem of the Cube.

One form of the Theorem of the Cube is the cube principle: a line bundle $M$ on $E^3$ which is trivial on each of $E\times E\times \{O\}$, $E\times \{O\}\times E$, and $\{O\}\times E\times E$ is trivial. Here is a proof for a complex elliptic curve.

Since $E$ is a real two-dimensional torus, its integral cohomology has no torsion. The Künneth theorem decomposesEvery single-factor summand is detected on a face containing that factor, and every cross summand is detected on the corresponding two-factor face. Restriction of $M$ to those faces is trivial, so its First Chern class has zero component in every summand. Thus $c_1(M)=0$.

The holomorphic exponential sequence identifies the kernel of $c_1$ in the holomorphic Picard group with $H^1(E^3,\mathcal{O})/\mathrm{im}{H}^1(E^3,\mathbb{Z})$. First cohomology decomposes factorwise here: $H^1(E^3,\mathcal{O})={⨁}_j{p}_j^{\ast }{H}^1(E,\mathcal{O})$ and likewise for integral first cohomology. Analytically the former decomposition says that the harmonic representatives of $(0,1)$-cohomology on the flat torus are the constant forms $d{\overline{z}}_j$, one from each factor. Consequently this kernel is ${\mathrm{Pic}}^0(E{)}^3$, andfor degree-zero line bundles $M_j$ on $E$. Restrict to each coordinate axis. The hypothesis makes every $M_j$ trivial, so $M$ is trivial. This proves the cohomological proof of the cube theorem for elliptic curves; the holomorphic isomorphism is algebraic on these projective varieties by GAGA.

To state its usual addition-map consequence, write $m_I:E^3\to E$ for summation of the coordinates indexed by $I$ and $p_j=m_{\{j\}}$. For any line bundle $L$ on $E$,More canonically, include the inverse of the constant bundle with fibre $L_O$ in the alternating product. Its restriction to a zero-coordinate face has pairwise cancellations, leaving the trivial bundle. The cube principle proves its triviality, which gives the displayed isomorphism after choosing a trivialization of the constant fibre. Pullback along any triple of maps to $E$ gives the corresponding alternating identity for their seven nonempty sums.

A pointwise divisor version can also be seen directly from Question 1. For a divisor $D$ and fixed points $a,b,c$, the alternating translation divisorhas degree zero and point sum zero: translation by $t$ subtracts $\mathrm{deg}(D)t$ from its point sum, and all the alternating terms cancel. The principality criterion proves that this divisor is principal.

Over a number field $K$, use algebraic line bundles and maps defined over $K$, and use Galois-invariant divisors rather than arbitrary individual complex points. The geometric line-bundle cube identity itself is still exact over $K$. To see descent in this setting, first pass to an algebraic closure, or an embedding in $\mathbb{C}$. The alternating bundle becomes trivial there by the preceding proof. Choose its nowhere-zero section with value one in a chosen $K$-basis at $(O,O,O)$. Any Galois conjugate of that section differs by an invertible global function, hence a constant on the connected projective variety; its normalized value at the rational origin forces that constant to be one. The section therefore descends, giving a $K$-defined trivialization.

For the arithmetic formulation used with heights, the necessary modification is a uniformly bounded error. A chosen Weil height machine representative $h_L$ satisfiesuniformly for $P,Q,R\in E(K)$. Indeed, pullback and tensor product of line bundles correspond to composition and addition of heights only modulo bounded functions. This is the arithmetic cube identity for Weil heights. Arbitrary naive heights do not satisfy an exact numerical cube identity; the canonical height of an elliptic curve removes the bounded error, as explained in Question 4. This arithmetic qualification does not weaken the exact geometric theorem.

Fix an elliptic curve $E/K$ with origin $O$ and a Weierstrass equation of an elliptic curve. Define the naive logarithmic $x$-height by $h_x(P)=h([x(P):1])$ for $P\ne O$ and $h_x(O)=0$. The $x$-map has degree two and pulls the projective hyperplane bundle back to ${\mathcal{O}}_E(2[O])$. Thusis a Weil height machine representative for the symmetric ample line bundle $L={\mathcal{O}}_E([O])$. This factor of one half fixes the convention for the normalized height on an elliptic curve used below.

The Theorem of the Cube gives the quadratic geometric identities. Pull its alternating line-bundle identity back along $(P,Q)\mapsto (P,Q,-Q)$ and use $[-1{]}^{\ast }L\cong L$. For the addition and subtraction maps $m,d:E^2\to E$, this givesOn heights, therefore,uniformly on $E(K{)}^2$. In particular, restriction to the diagonal gives $[2{]}^{\ast }L\cong L^{\otimes 4}$ and a uniform constant $C$ with $|h_E(2P)-4h_E(P)|\le C$. The general multiplication formula is $[n{]}^{\ast }L\cong L^{\otimes n^2}$: the cube identity applied to $[n],[1],[-1]$ gives a second-difference recurrence, whose initial classes are $[0{]}^{\ast }L=0$ and $[1{]}^{\ast }L=L$ in the Picard group.

Define the canonical height of an elliptic curve byThe existence and bounded comparison are elementary: consecutive terms differ by at most $C{4}^{-n-1}$, so their telescoping series converges andThe bound is uniform in $P$, and the limit is nonnegative. Shifting the index gives $\hat{h}(2P)=4\hat{h}(P)$. This is the unique function within bounded distance of $h_E$ with that doubling rule: the difference of two such functions at $P$ is $4^{-n}$ times their bounded difference at $2^{n}P$, and hence zero. A change of Weierstrass model changes the naive height only by a bounded function, so the normalized limit is independent of that model.

Apply the approximate parallelogram identity to $2^{n}P,2^{n}Q$, divide by $4^n$, and pass to the limit. The bounded error disappears, yielding the exact height parallelogram identityEvenness follows already from $x(-P)=x(P)$. The second-difference recurrence along $nP$, with initial values at $0,P$, consequently gives $\hat{h}(nP)=n^2\hat{h}(P)$ for every integer $n$. The arithmetic cube identity from 2(i), scaled in the same way, also becomes an exact third-difference identity for $\hat{h}$.

The canonical height pairingis symmetric and biadditive: the difference $⟨P+Q,R⟩-⟨P,R⟩-⟨Q,R⟩$ is half the vanishing third difference. Its diagonal is $\hat{h}(P)$. Positivity gives the Cauchy-Schwarz inequality for this pairing without first assuming finite generation. Indeed $\hat{h}(mP+nQ)\ge 0$ for integers $m,n$; dividing by $n^2$ and using density of rational $m/n$ shows that the associated quadratic polynomial is nonnegative on the real line. Its discriminant therefore gives $⟨P,Q{⟩}^2\le \hat{h}(P)\hat{h}(Q)$.

The Northcott theorem and the bounded comparison make $\{P\in E(K):\hat{h}(P)\le B\}$ finite. The map $x:E\to {\mathbb{P}}^1$ has finite fibres, so bounded $x$-height indeed controls the number of points, including $O$. This proves the important criterionFor a torsion point the integer quadratic rule gives zero. Conversely, height zero makes all $2^{n}P$ belong to one finite bounded-height set; two coincide, forcing $(2^j-2^i)P=O$ for some $j>i$. This argument also proves torsion finiteness over the fixed number field.

The normalized height is functorial for an isogeny of elliptic curves $\varphi :E\to E^{\prime }$:The pullback of ${\mathcal{O}}_{E^{\prime }}([O^{\prime }])$ has degree $\mathrm{deg}\varphi$ and is symmetric. Its difference from that many copies of $[O]$ is a symmetric degree-zero line bundle, hence a two-torsion Picard class. Its height is bounded, so the ordinary height relation has a bounded error. Commuting the isogeny with $[2^n]$ and taking the normalized limit removes that error. This explains the natural quadratic scaling under multiplication and the usefulness of normalized heights in comparing isogenous curves.

Here is their role in the Mordell-Weil theorem. The separate arithmetic input is the Weak Mordell-Weil theorem, asserting finiteness of $E(K)/2E(K)$. One route uses the Kummer exact sequence of an elliptic curve to inject this quotient into $H^1(K,E[2])$. The division torsors are unramified outside finitely many places containing those over $2$ and those of bad reduction. After a finite Galois extension splitting $E[2]$, their classes are pairs of square classes with even valuations outside this finite set. These square classes form a finite group: the obstruction to being represented by an S-unit lies in the finite ideal class group's two-torsion, and the S-unit group is finitely generated, hence finite modulo squares. The inflation-restriction kernel is a cohomology group of a finite group with finite coefficients, also finite. This supplies the finite quotient without using the full theorem in advance.

Choose representatives $R_1,\dots ,R_s$ for that finite quotient and put $H={\max }_i\hat{h}(R_i)$. Every point has the form $P=2Q+R_i$. Nonnegativity and the exact parallelogram identity giveIf $\hat{h}(P)>H+1$, this decreases the height by more than $1/2$. Repeating must reach the finite set $T=\{P:\hat{h}(P)\le H+1\}$. Unwinding the halving expressions writes the initial point as an integer linear combination of the $R_i$ and a point of $T$. Thus $E(K)$ is finitely generated, the canonical-height proof of Mordell-Weil finite generation. The normalized height supplies the descent and Northcott supplies its finite endpoint; neither by itself supplies the weak theorem's finite quotient.

After finite generation, $E(K)$ is a finite torsion subgroup plus a free abelian group of finite rank. The canonical height pairing is positive-definite on the associated real vector space. To justify the latter rather than assuming it, if its nonnegative quadratic form had a real null direction, rounding longer and longer multiples of that direction to lattice points would produce infinitely many points of bounded height: the null-direction cross terms vanish, leaving only uniformly bounded rounding errors. This contradicts Northcott finiteness. Hence the free part is a Euclidean height lattice. Heights can consequently bound point searches, certify independence through positive Gram determinants, and organize the arithmetic complexity of rational points.

We regard $N$ as positive and invertible in $k$; $N=1$ gives the trivial group and the same arguments apply. Put $A={\mathrm{Pic}}^0(C)$, an abelian variety of dimension $g$.

First compute the prime-to-characteristic torsion of an abelian variety. The differential of $[N]:A\to A$ at the origin is $N\mathrm{id}$, and translation gives the same assertion everywhere. Consequently $[N]$ is an étale morphism, its fibres have dimension zero, and properness makes it finite. Its image is closed of dimension $g$, hence all of $A$. Choose an ample line bundle $L_0$ on $A$ and put $L=L_0\otimes [-1{]}^{\ast }{L}_0$, which is symmetric and ample. The Theorem of the Cube gives the recurrenceAfter fixing a trivialization at the origin, induction from $n=0,1$ gives $[n{]}^{\ast }L\cong L^{\otimes n^2}$. Comparing top intersection products under the finite map $[N]$ givesThe intersection number is positive because $L$ is ample, so $\mathrm{deg}[N]=N^{2g}$. Since $[N]$ is étale and $k$ is algebraically closed field, its kernel has precisely $N^{2g}$ points.

For a prime power ${\ell }^r\mid N$, write the finite abelian group $A[{\ell }^r]$ as a product of cyclic groups of orders ${\ell }^{a_j}$, with $1\le a_j\le r$. Its subgroup killed by $\ell$ is $A[\ell ]$, so the number of factors is $2g$. Its total order is ${\ell }^{2gr}$, so $\sum a_j=2gr$ forces every $a_j=r$. Combining the prime-power factors by the Chinese remainder theorem provesThe same reasoning works for the dual abelian variety $\hat{A}$, also of dimension $g$.

To define the Weil pairing for an abelian variety on two copies of $A[N]$, we first justify the required theta identification with $\hat{A}$. Identify $X={\mathrm{Pic}}^{g-1}(C)$ with an $A$-torsor and use ordinary translation $T_q(x)=x+q$. The Theorem of the square, a consequence of the Theorem of the Cube, makesa group homomorphism $A\to \hat{A}$. Fix the negative Abel map of an algebraic curve $j:C\to A$, $P\mapsto [P_0-P]$, and let $\rho :\hat{A}\to {\mathrm{Pic}}^0(C)=A$ be pullback by $j$. Pullback by a translate of $j$ gives the same map: algebraically trivial line bundles on an abelian variety are translation-invariant as Picard classes, by the Theorem of the square.

For any $q\in A$, choose a degree-$g$ class $D$ such that both $D$ and $D+q$ have one-dimensional section spaces. This is possible because the nonspecial degree-$g$ classes form a nonempty open subset of the irreducible Picard scheme of a curve, and its translate has a nonempty intersection with it. Since $T_q{\varphi }_D={\varphi }_{D+q}$, the restriction of the theta divisor to an Abel curve proved above givesEach ${\varphi }_D$ differs from $j$ by a translation. Thus $\rho \lambda ={\mathrm{id}}_A$. This equality on closed points is equality of morphisms, since $A$ is reduced and the target is separated. In particular $\lambda$ is a closed immersion, being a section of the separated morphism $\rho$. Its image has dimension $g=\dim \hat{A}$, so it is all of the reduced irreducible abelian variety $\hat{A}$. Therefore $\lambda$ is an isomorphism of schemes. This proves the needed theta autoduality of a Jacobian, including possible infinitesimal issues in positive characteristic.

Here is a definition of the Weil pairing for an abelian variety that also proves nondegeneracy. For $M\in \hat{A}[N]$, the Theorem of the Cube gives $[N{]}^{\ast }M\cong M^{\otimes N}\cong {\mathcal{O}}_A$. Choose a trivialization of $[N{]}^{\ast }M$. The finite étale cover $[N]:A\to A$ has deck group $G=A[N]$. Its natural descent action on $[N{]}^{\ast }M$ becomes multiplication by scalars under this trivialization: these scalars are constant because every invertible regular function on the proper connected abelian variety $A$ is constant. They form a characterwhose values lie in ${\mu }_N$, since $NP=0$ for every $P\in G$. Rescaling the trivialization changes none of these scalars. Translation gives multiplicativity in $P$, and tensor product gives ${\chi }_{M\otimes M^{\prime }}={\chi }_M{\chi }_{M^{\prime }}$. DefineThis fixes a convention for the Weil pairing for an abelian variety; reversing the convention for the descent action inverts all values and changes no nondegeneracy assertion.

If ${\chi }_M=1$, the trivialization is equivariant for the deck group. Descent then makes $M$ the trivial line bundle on the target. Thusis injective. The source and target both have $N^{2g}$ elements, since $k$ contains all $N$ distinct roots of unity and $A[N]\cong (\mathbb{Z}/N{)}^{2g}$. Hence it is an isomorphism. Characters of this group separate its points, so the Weil pairing for an abelian variety is nondegenerate in both variables. Composing with the isomorphism $\lambda$ proves the pairing on $A[N]$ is perfect.

The bundle is on $A$, so the printed, undefined target $\mathrm{Pic}(X)$ must be read as the Picard group $\mathrm{Pic}(A)$. Write this Picard group additively and set $q(h)=[h^{\ast }L]$. The zero morphism pulls back $L$ to a constant one-dimensional vector space tensored with ${\mathcal{O}}_A$, so $q(0)=0$. The Theorem of the Cube, pulled back along $(f,g,h):A\to B^3$, givesThis is the bilinear cross-effect of a line bundle on an abelian variety: set $d(f,g)=q(f+g)-q(f)-q(g)$. Substitution into this identity givesThe definition gives $d(f,g)=d(g,f)$, so additivity holds in both arguments. Additivity includes negative multiples and the zero morphism. Hence $D_L$ is a symmetric bilinear map with values in $\mathrm{Pic}(A)$. The equalities refer to isomorphism classes of line bundles; without chosen rigidifications they are not asserted to be specified canonical isomorphisms of bundles.

Let $m_I:X^3\to X$ add the coordinates indexed by a nonempty subset $I$ of the three-element index set. The Theorem of the Cube says that for every line bundle $\mathcal{L}$ on an abelian variety,is trivial, up to the harmless constant line given by the fiber of $\mathcal{L}$ at the identity.

Pull this line bundle back along $(f,g,h):Y\to X^3$. Pullback commutes with tensor products and duals, and $m_I\circ (f,g,h)$ is the corresponding sum of morphisms. The resulting bundle is precisely ${\mathcal{M}}_{f,g,h}$, so it is trivial.

Take $Y=X$, $f={\mathrm{id}}_X$, and let $g,h$ be the constant maps with values $x,y$. All pullbacks along constant maps are trivial line bundles. The formula for ${\mathcal{M}}_{f,g,h}$ then becomesor equivalently

If $N$ is invertible in the base field, multiplication by $N$ on an abelian variety is finite étale: its differential is $N$ times the identity, and properness turns its zero-dimensional fibres into finite fibres. The Theorem of the Cube gives $[N{]}^{\ast }L\cong L^{N^2}$ for a symmetric ample line bundle. Top intersection products give degree $N^{2\dim A}$. For each prime power ${\ell }^r\mid N$, the kernels killed by ${\ell }^s$ have orders ${\ell }^{2s\dim A}$; the classification of finite abelian groups forces $2\dim A$ cyclic factors of length $r$. Combining the prime powers gives the displayed group.