Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 15 / 2

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 Weierstrass elliptic function and its derivative give the maponto the Weierstrass equation of an elliptic curve $Y^{2}Z=4X^3-g_{2}X{Z}^2-g_3{Z}^3$. Near zero, multiplying these coordinates by $z^3$ gives $[z+O(z^5):-2+O(z^4):z^3]$, so the map extends holomorphically and is an immersion at the origin.

The meromorphic map $\wp :E\to {\mathbb{P}}^1$ has degree two, since its only pole on the torus has order two. Its evenness implies $\wp (u)=\wp (v)$ exactly when $u\equiv v$ or $u\equiv -v$. Oddness of ${\wp }^{\prime }$ distinguishes these two alternatives unless they are the same point class. At each nonzero half-period, ${\wp }^{\prime }=0$. These three points exhaust its zeros, since its only pole has order three. The three half-period values are distinct: equality of two of them would give two distinct zeros each of order at least two for $\wp -e$, contradicting its total pole order two. They are roots of $4X^3-g_{2}X-g_3$, so the polynomial has distinct roots and the cubic is smooth. At a half-period $h$ with value $e$, ${\wp }^{\prime \prime }(h)=2{\prod }_{e^{\prime }\ne e}(e-e^{\prime })\ne 0$. Elsewhere ${\wp }^{\prime }$ is nonzero. The coordinate pair therefore has nonzero derivative at every finite point, and the map is injective. Compactness and the smooth cubic equation then identify it as an embedding onto the cubic.

The hyperplane section at infinity has divisor $3[O]$. Thus the intersection divisor $D_{\ell }$ of any line with this cubic is linearly equivalent to $3[O]$. Applying the principality criterion on a complex elliptic curve to $D_{\ell }-3[O]$ shows that its three points, counted with intersection multiplicity, have sum zero in $E$.

Conversely, if the degree-three divisor $D=[u]+[v]+[w]$ has point sum zero, Question 1 gives a meromorphic function with divisor $D-3[O]$. A meromorphic function with no poles away from $O$ and pole order at most three lies inTo prove this description, subtract multiples of ${\wp }^{\prime }$ and $\wp$ to cancel the cubic and quadratic Laurent principal parts. The remaining possible simple-pole coefficient is a residue, which must be zero because it is the only pole on a compact torus. The remaining holomorphic elliptic function is constant. Thus the constructed function is a linear form in the projective coordinates, and its line has intersection divisor exactly $D$. This proves the chord-and-tangent collinearity criterion is $u+v+w=0$ in $E$, with multiplicities understood.

For three pairwise distinct point classes away from $O$, ordinary collinearity is equivalent to linear dependence of the coordinate columns, soThe original PDF has this three-column determinant; the TeX's extra copied column is a transcription defect.

There is also a genuine implicit distinctness condition in the printed determinant assertion. If $u=v={\omega }_1/10$ and $w={\omega }_1/5$, all entries are finite and the determinant is zero because two columns coincide, but $u+v+w=2{\omega }_1/5\notin \Lambda$. Thus the unrestricted literal equivalence is false. To specify repeated intersections, use the confluent elliptic collinearity determinant: for a double point at $u$ replace the second column by $c^{\prime }(u)$, where $c(z)=(1,\wp (z),{\wp }^{\prime }(z){)}^t$; for a triple point use $c(u),c^{\prime }(u),c^{\prime \prime }(u)$. These impose tangency or inflection, and recover the correct conditions $2u+w\in \Lambda$ or $3u\in \Lambda$. If a point is $O$, use the homogeneous coordinates or a local homogeneous lift, since the displayed affine entries have poles there.