A constructive proof of the theorem of the cube

Question

Do you know a constructive proof of the theorem of the cube ? More precisely, let $X$, $Y$, $Z$ be projective varieties (e.g., over an algebraically closed field $k$) with points $x$, $y$, $z$ respectively and let $D$ be a divisor on the direct product $V := X \times Y \times Z$. Assume that the restrictions of $D$ to $$ V_0 := \{x\} \times Y \times Z, \qquad V_1 := X \times \{y\} \times Z, \qquad V_2 := X \times Y \times \{z\} $$ are the principal divisors of some functions $f_i \in k(V_i)$. In other words, $D \cdot V_i = \mathrm{div}(f_i)$. The theorem of the cube claims that $D$ is principal. How to construct explicitly a function $f \in k(V)$ such that $D = \mathrm{div}(f)$, given the functions $f_i$? I am mainly interested in the case $X = Y = Z$ is an elliptic curve and $x = y = z$ is the zero point on it.

Answer 1

This is not really an answer, but a rephrasing together with some comments on why this is difficult. I end with one example where you can actually compute something (purely algebraically) on $E \times E \times E$.

In summary, the question reduces to the case where the three restrictions of $D$ are trivial as divisors (as opposed to merely principal), i.e. the $f_i$ are all $1$. It's actually convenient for me to change the notation a little:

Notation. We assume throughout that $(X,x)$, $(Y,y)$, and $(Z,z)$ are pointed projective varieties over a field $k$ (for me, variety means geometrically integral). It will occasionally be convenient to denote these $(V_1,v_1)$, $(V_2,v_2)$, and $(V_3,v_3)$ respectively. Denote by $V$ the product $X \times Y \times Z$, and for $I \subseteq \{1,2,3\}$ write $$V_I = \prod_{i \in I} V_i.$$ Let $\pi_I \colon V \to V_I$ be the projection, for which we use shorthands like $\pi_{1,2} = \pi_{\{1,2\}} \colon V \to X \times Y$. It has a section $s_I \colon V_I \to V$ given by the points $v_j$ for $j \not\in I$, whose image we also denote $V_I \subseteq V$ by abuse of notation (the OP calls this $V_{I^{\text{c}}} \subseteq V$). For a subscheme $W \subseteq P$, denote by $W_I \subseteq V_I$ the intersection $W \cdot V_I$ (there are problems with intersection products on the level of divisors; see the difficulties below).

---

Firstly, note that the theorem of the cube is equivalent to the following:

Lemma. Let $D$ be a Cartier divisor on $V$. Then the divisor \begin{align*} \bar D :=& \ D - \pi_{1,2}^*D_{1,2} - \pi_{1,3}^*D_{1,3} - \pi_{2,3}^*D_{2,3} + \pi_1^*D_1 + \pi_2^*D_2 + \pi_3^*D_3 \\ =& \sum_{I \subseteq \{1,2,3\}} (-1)^{\lvert I \rvert+1} \pi_I^*D_I \end{align*} on $V$ is principal.

Note that we omitted $\pi_{\varnothing}^*D_{\varnothing}$ since it is trivial (it's pulled back from $\operatorname{Spec} k$).

Proof. It suffices to show this when $D$ is effective. For any $I \subseteq \{1,2\}$, the restrictions of $\pi_I^*D_I$ and $\pi_{I \cup \{3\}}D_{I \cup \{3\}}$ to $V_{1,2}$ agree, so we see that $\bar D \cdot V_{1,2}$ is trivial (as divisor!), and likewise for $\bar D \cdot V_{1,3}$ and $\bar D \cdot V_{2,3}$. The theorem of the cube says that $\bar D$ is principal. Conversely, if this lemma holds, applying it to the case where $D_I \sim 0$ on $V_I$ for any $I \subsetneq \{1,2,3\}$ gives $D \sim \bar D \sim 0$, proving the theorem of the cube. $\square$

The situation in the lemma shows what can happen: if we're presented with $D = \bar E$ for some divisor $E$, then the restrictions to $V_I$ for $\lvert I \rvert = 2$ are trivial as divisors. Thus, the $f_I \in k(V_I)$ witnessing the rational triviality of $D_I$ can be taken $1$, yet on $V$ there is something nontrivial happening.

In fact, since $\operatorname{Div}(P)$ is a free abelian group, the lemma shows that there exists some homomorphism $$\phi \colon \operatorname{Div}(X \times Y \times Z) \to k(X\times Y \times Z)^\times$$ such that $(\phi(D)) = \bar D$ for all $D \in \operatorname{Div}(X \times Y \times Z)$. The task at hand is to "compute" this map: if we can find an explicit $f$ such that $(f) = \bar D$, then the assumption $D_I = (f_I)$ for all $I \subsetneq \{1,2,3\}$ gives $$D = \bar D + \sum_{U \subsetneq \{1,2,3\}} (-1)^I \pi_I^*D_I = (f) + \sum_{U \subsetneq \{1,2,3\}} (-1)^{\lvert I \rvert} (\pi_I^*f_I),$$ explicitly exhibiting $D$ as a principal divisor.

---

Now we come to the difficulties.

Difficulty 1. The first issue is that $\operatorname{Div}$ has fewer functoriality properties than $\operatorname{Pic}$. Notably, the intersection product $D \mapsto D \cdot V_I$ is not defined on $\operatorname{Div}$ in general: what to do if $D$ contains $V_I$?

This problem goes away when $X$, $Y$, and $Z$ are curves: for $\lvert I \rvert = 2$, we get $D \subseteq V_I$ if and only if $D = V_I$. But since the normal bundle of a fibre is trivial, we can simply set $V_I \cdot V_I = 0 \in \operatorname{Div}(V_I)$. This also works for principal divisors: the specialisation $\operatorname{Prin}(V) \to \operatorname{Prin}(V_I)$ is then defined using the local ring $\mathcal O_{V,V_I}$ at the generic point of $V_I$. This is a DVR, so $k(V)^\times \cong \mathcal O_{V,V_I}^\times \times \mathbf Z$. The specialisation $k(V)^\times \to k(V_I)^\times$ is defined on $\mathcal O_{V,V_I}^\times$ by reduction modulo the maximal ideal, and identically zero on the factor $\mathbf Z$.

Note also that neither $\operatorname{Div}$ nor $\operatorname{Pic}$ have a pushforward along proper maps, although coherent sheaves in general do (these play a role in some proofs of the theorem of the cube ― one then shows that the pushforward is again locally free).

Difficulty 2. Even when $X=Y=Z=E$, it's not clear at all what the function $\phi$ above is. I don't think this is something that you can write down, for one because the function field of a product of elliptic curves is a kind of unwieldy thing to work with.

The theorem of the cube predates the six functor formalism, so the original proof is not the one you find in Mumford. I personally like the proof using Picard schemes: triviality along $X \times Y \times z$ gives a map $X \times Y \to \mathbf{Pic}^0_Z$ that is trivial on $X \times y$ and $x \times Y$, hence trivial by the rigidity lemma. But this does not lift to $X \times Y \to \mathbf{Div}_Z$ because the scheme of relative effective Cartier divisors has an additional flatness hypothesis. You might be able to work around this, but it gets messy.

Maybe the most suited language is the one in Weil's original proof, which you can read for instance in Lang's Abelian varieties, §III.2. But even that uses maps to Jacobians and is ultimately inexplicit. Notably, Weil uses the divisor language (as opposed to line bundles), so the fact that the old proofs are inexplicit suggests that it might be really hard.

---

Here's a first question:

Question. Can you do this for "simple" examples? For instance, if $E$ is an elliptic curve and $D = \{(a,b,c) \in E^3\ |\ a+b+c=0\}$, then do you know how to find $\phi(D)$? This will tell you why \begin{align*} \bar D =\ &\{a+b+c=0\} - \{a+b=0\} - \{a+c=0\} - \{b+c=0\} \\ &+ \{a=0\} + \{b=0\} + \{c=0\} \end{align*} is rationally trivial on $E \times E \times E$.

Edit: In this specific case you can actually write down the answer! If $$y^2z+a_1xyz+a_3yz^2=x^3+a_2x^2z+a_4xz^2+a_6z^3$$ is a homogeneous Weierstrass equation for $E$ in $\mathbf P^2$, then the locus $\{a+b+c=0\}$ is a component of the locus of collinear triples in $\mathbf P^2 \times \mathbf P^2 \times \mathbf P^2$. Denoting the coordinates on $\mathbf P^2 \times \mathbf P^2 \times \mathbf P^2$ by $[x_1:y_1:z_1]$, $[x_2:y_2:z_2]$, and $[x_3:y_3:z_3]$, the collinear points are given by the vanishing of $$\det\begin{pmatrix} x_1 & x_2 & x_3 \\ y_1 & y_2 & y_3 \\ z_1 & z_2 & z_3\end{pmatrix},$$ or even more explicitly, the locus $$x_1y_2z_3 - x_1z_2y_3 + y_1z_2x_3 - y_1x_2z_3 + z_1x_2y_3 - z_1y_2x_3 = 0.$$ When intersecting with $E \times E \times E$, this picks up three more components $\{a=b\}$, $\{b=c\}$, $\{c=a\}$ in addition to $\{a+b+c=0\}$.

On the other hand, the Weierstrass coordinates give an explicit rational function $f$ on $E \times E$ such that $$\operatorname{div}(f) = \{a=b\} + \{a+b=0\} - 2\{a=0\} - 2\{b=0\}.$$ Indeed, the section $y_1z_2-z_1y_2$ of $\mathcal O(1,1)$ vanishes at $\{a=b\}$, $\{a+b=0\}$, $\{a=0\}$, and $\{b=0\}$, whereas the section $z_1z_2$ vanishes at $3\{a=0\} + 3\{b=0\}$. So $f = \tfrac{y_1}{z_1}-\tfrac{y_2}{z_2}$ does the trick. (You could also have found this by pulling back the principal divisor $\Delta_{\mathbf P^1} - (\infty \times \mathbf P^1) - (\mathbf P^1 \times \infty) \in \operatorname{Div}(\mathbf P^1 \times \mathbf P^1)$ along the double cover $[y:z] \colon E \to \mathbf P^1$.)

Combining the above, we see that the rational function $$f = \frac{\big(x_1y_2z_3 - x_1z_2y_3 + y_1z_2x_3 - y_1x_2z_3 + z_1x_2y_3 - z_1y_2x_3\big)\big(z_1z_2z_3\big)}{\big(y_1z_2-z_1y_2\big)\big(y_2z_3-z_2y_3\big)\big(y_3z_1-z_3y_1\big)}$$ on $E \times E \times E$ has \begin{align*} \operatorname{div}(f) =\ &\{a+b+c=0\} + \{a=b\} + \{b=c\} + \{c=a\} \\ &+ 3\{a=0\} + 3\{b=0\} + 3\{c=0\} \\ &- \{a=b\} - \{a+b=0\} - \{a=0\} - \{b=0\} \\ &- \{b=c\} - \{b+c=0\} - \{b=0\} - \{c=0\} \\ &- \{c=a\} - \{c+a=0\} - \{c=0\} - \{a=0\}. \end{align*} Simplifying gives $\operatorname{div}(f) = \bar D$, as desired. $\square$