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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.

Identify ${\mathrm{Pic}}^{g-1}(X)$ with the Jacobian variety using the base point and the translation of Solution 4. A line bundle $L$ of degree $g-1$ corresponds to the point $z=u(D)-\kappa$ for any divisor representation $L\simeq {\mathcal{O}}_X(D)$. The Riemann singularity theorem isIn particular, the theta divisor is singular exactly at the degree-$g-1$ line bundles with at least two independent sections. We prove the equality, including the upper bound on multiplicity.

Choose a fixed effective divisor $A$ of degree $N\ge g$, and a local holomorphic family $L_{\eta }$ representing nearby points of ${\mathrm{Pic}}^{g-1}(X)$. Such a family can be constructed locally from the holomorphic exponential sequence, by varying line-bundle transition functions through exponentials of Čech cohomology representatives. By Serre duality, $H^1(L_{\eta }(A))$ is dual to $H^0(K_X\otimes L_{\eta }^{-1}(-A))$, whose degree is $g-1-N<0$, so it vanishes. The Riemann-Roch theorem gives $h^0(L_{\eta }(A))=N$. By cohomology and base change for line bundles on a curve, these spaces form a holomorphic vector bundle $E$ of rank $N$. The spaces $H^0(L_{\eta }(A){|}_A)$ form another holomorphic vector bundle $F$ of rank $N$. Evaluation gives a holomorphic square matrix $T(\eta )$ locally, with exact sequenceConsequently $\det T(\eta )=0$ exactly when $L_{\eta }$ has a nonzero section, namely on the support of the theta divisor.

At the point $L$, put $r=h^0(L)=h^1(L)$, the equality being Riemann-Roch theorem. An invertible minor of size $N-r$ persists near this point. Holomorphic row and column operations, or the Schur complement, givewhere $M$ is an $r\times r$ holomorphic matrix. This is the Schur complement presentation of a theta singularity. Its linear term in a tangent direction $\xi \in H^1({\mathcal{O}}_X)$ is the cup product mapTo see this concretely, vary transition functions $g_{ij}$ of $L$ to $g_{ij}(1+ϵ{\xi }_{ij})$. A section $s_i$ lifts to $s_i+ϵs_i^{\prime }$ precisely when its first-order compatibility equation can be solved; the obstruction is the cocycle ${\xi }_{ij}{s}_j$. This is $\xi ⌣s$. Passing to the kernel and cokernel of $T(0)$ gives exactly the derivative of the reduced matrix $M$, proving the assertion without assuming a multiplicity formula.

Before extracting the order of $\det M$, we verify that $\det T$ is a reduced equation of the theta divisor. Generic points of $W_{g-1}$ have $h^0(L)=1$ by Solution 3. At such a point, both $H^0(L)$ and $H^0(K_X\otimes L^{-1})$ are one-dimensional. Their nonzero sections $s,t$ have nonzero product $st\in H^0(K_X)$. Choose $\xi \in H^1({\mathcal{O}}_X)$ with $⟨\xi ,st⟩\ne 0$ under Serre duality. The scalar $d{M}_0(\xi )$ is nonzero, so $\det T$ has a simple zero generically on the irreducible theta divisor. Since its support is exactly that irreducible hypersurface, it defines the reduced divisor. Solution 4 independently proved the same reduced structure for the Riemann theta function. On the smooth ambient Jacobian variety, local equations of the same reduced divisor differ by a nonvanishing holomorphic unit. Hence the orders of the local theta equation and $\det M$ agree everywhere.

Every entry of $M$ vanishes at the point, so each term in its determinant has order at least $r$. This proves ${\mathrm{mult}}_z\Theta \ge r$. To obtain equality, we exhibit an invertible cup-product direction for a line bundle on a curve. Set $V=H^0(L)$ and $W=H^0(K_X\otimes L^{-1})$, both of dimension $r$. There are $r$ distinct points $p_1,\dots ,p_r$ at which evaluation is an isomorphism for $V$: choose them inductively, since a surviving nonzero section has only finitely many zeros. The same holds for $W$. The nonvanishing of each evaluation determinant defines a nonempty open subset of the irreducible variety $X^r$. Their intersection, also avoiding the diagonals, is therefore nonempty. Choose points in it and local coordinates and compatible trivializations of $L$ and $K_X\otimes L^{-1}$ at each point.

Let $\xi$ be the connecting homomorphism class of the meromorphic principal parts ${\lambda }_i/t_i$ at these points, with all ${\lambda }_i\ne 0$. The residue form of Serre duality givesIf $S$ and $U$ are the two invertible evaluation matrices, the matrix of this pairing is $S^t\mathrm{diag}({\lambda }_1,\dots ,{\lambda }_r)U$, and is invertible. Thus the cup product map $\xi ⌣-:H^0(L)\to H^1(L)$ is an isomorphism. Along the corresponding analytic parameter line,with nonzero leading coefficient. The order is at most $r$, completing the proof of the Riemann singularity theorem.

This argument also identifies the tangent cone to a theta divisor: its degree-$r$ equation is $\det (\xi ⌣-)$, which is not the zero polynomial. For $r=1$, the tangent hyperplane is $⟨\xi ,st⟩=0$, in agreement with the annihilator description of the derivative of the Abelian sum map. For $r\ge 2$ all first derivatives vanish, proving singularity. In genus $1$ the theta divisor is a single smooth point of multiplicity $1$. In genus $0$ the effective degree-$-1$ locus is empty, so there is no singularity statement at a point of it.

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.

The Dolbeault theorem identifies $H^2(M,{\mathcal{O}}_M)$ with $H_{\bar{\partial }}^{0,2}(M)$. It vanishes by hypothesis, and complex conjugation in the Hodge decomposition theorem for compact Kähler manifolds makes $H^{2,0}$ vanish too. Thus every real degree-two cohomology class has type $(1,1)$.

Let $\omega$ be a Kähler form. Choose a rational cohomology class $\eta$ sufficiently close to $[\omega ]$. More concretely, using harmonic differential forms as representatives for a fixed Kähler metric, the harmonic form representing $\eta$ is close to $\omega$ in every smooth norm: harmonic differential forms constitute a finite-dimensional vector space. It is a real closed $(1,1)$ form, and is positive if sufficiently close to $\omega$, by compactness. This is the openness of the Kähler cone. Multiplying by a positive integer gives an integral Kähler class $\kappa =N\eta$.

The holomorphic exponential sequence contains the cohomology segmentChoose an integral lift of $\kappa$. Exactness gives a holomorphic line bundle $\mathcal{L}$ having that First Chern class. We must ensure that it has a positive metric, rather than merely a positive representative of its class.

Choose any Hermitian metric $h_0$ and let ${\theta }_0$ be its normalized curvature form of a Hermitian holomorphic line bundle. If $\theta$ is the positive form representing $\kappa$, then $\theta -{\theta }_0$ is an exact real $(1,1)$ form. The ddbar lemma gives a real smooth function $u$ withReplace $h_0$ by $h=e^{-u}{h}_0$. Its normalized curvature is $\theta$, so $\mathcal{L}$ is a positive holomorphic line bundle. The Kodaira embedding theorem now applies: sufficiently many sections of a sufficiently high tensor power define the desired embedding. Therefore