I’d like to use this post to try to understand the “theorem of the cube,” following Mumford’s “Abelian varieties.”

Theorem 1 Let ${X, Y}$ be proper varieties over an algebraically closed field ${k}$, and let ${Z}$ be a connected variety. Let ${\mathcal{L} \in \mathrm{Pic}(X \times Y \times Z)}$ be a line bundle. Suppose there exist ${k}$-valued points ${x_0, y_0, z_0}$ in ${X, Y, Z}$ such that ${\mathcal{L}}$ is trivial when restricted to ${\left\{x_0\right\} \times Y \times Z, X \times \left\{y_0\right\} \times Z, X \times Y \times \left\{z_0\right\}}$. Then ${\mathcal{L}}$ is trivial.

This theorem is extremely useful in analyzing the behavior of line bundles on abelian varieties; see these posts for instance.

I’ve found the proof due to Weil and Murre in the second chapter of Mumford to be quite impenetrable; the argument makes sense line by line but I have never been able to see a larger picture. Fortunately, it turns out that the third chapter has a more scheme-theoretic (nilpotents will be used!) argument which is much more transparent.

1. Infinitesimal trivializations

Consider the set of ${z \in Z}$ such that ${\mathcal{L}_z \equiv \mathcal{L}|_{X \times Y \times \left\{z\right\}}}$ is trivial. Then this is a closed set: in fact, a point ${z}$ belongs to this set if and only if

$\displaystyle h^0( \mathcal{L}_z) , h^0(\mathcal{L}_z^{-1}) > 0.$

Now, the semicontinuity theorem implies that this set is closed. So far, we know that it is nonempty: it contains ${\left\{z_0\right\}}$. Using an infinitesimal thickening argument, we will show that it consists of all of ${Z}$, from which we will be able to conclude easily.

Let ${A}$ be an local finite-dimensional ${k}$-algebra, and let ${\mathrm{Spec} A \rightarrow Z}$ be a morphism which is set-theoretically at ${z_0}$ (in other words, an “infinitesimal thickening” of the point at ${z_0}$). We will show that

$\displaystyle \mathcal{L}|_{X \times Y \times \mathrm{Spec} A}$

is trivial. To see this claim, we use induction on the dimension of ${A}$. When ${A}$ is one-dimensional, this is the assumption. In general, assume the claim for local artinian ${k}$-algebras of smaller dimension. Given ${A}$, there is an element ${a \in A \setminus \left\{0\right\}}$ annihilated by the maximal ideal of ${A}$. Then we have a surjection of ${k}$-algebras

$\displaystyle A \twoheadrightarrow A/ ka,$

and an exact sequence of sheaves (on a one-point space)

$\displaystyle 0 \rightarrow \mathcal{O}_{\mathrm{Spec} k} \rightarrow \mathcal{O}_{\mathrm{Spec} A} \rightarrow \mathcal{O}_{\mathrm{Spec} A/a} \rightarrow 0.$

This gives an exact sequence of sheaves

$\displaystyle 0 \rightarrow \mathcal{O}_{X \times Y \times \mathrm{Spec} k} \rightarrow \mathcal{O}_{X \times Y \times \mathrm{Spec} A} \rightarrow \mathcal{O}_{X \times Y \times \mathrm{Spec} A/a} \rightarrow 0.$

Tensoring with ${\mathcal{L}}$ gives an exact sequence of sheaves on ${ X \times Y \times \mathrm{Spec} A}$,

$\displaystyle 0 \rightarrow \mathcal{L}|_{X \times Y \times \mathrm{Spec} k} \rightarrow \mathcal{L}|_{X \times Y \times \mathrm{Spec} A} \rightarrow \mathcal{L}|_{X \times Y \times \mathrm{Spec} A/a} \rightarrow 0.$

The goal is now to show that there exists an element ${s \in H^0(\mathcal{L}|_{X \times Y \times \mathrm{Spec} A})}$ which restricts to a generator at each closed point, for this will give a trivialization of ${\mathcal{L}|_{X \times Y \times \mathrm{Spec} A}}$. By the inductive hypothesis on the dimension, we know that there exists a section ${\overline{s} \in H^0(\mathcal{L}|_{X \times Y \times \mathrm{Spec} A/a}}$ with this property. It suffices to show that ${\overline{s}}$ lifts to a section of ${H^0(\mathcal{L}|_{X \times Y \times \mathrm{Spec} A})}$, as that will automatically by a generator at each closed point.

In order to do this, we have to consider the connecting homomorphism

$\displaystyle \delta: H^0(\mathcal{L}|_{X \times Y \times \mathrm{Spec} A/a}) \rightarrow H^1(\mathcal{L}|_{X \times Y \times \mathrm{Spec} k}) = H^1(\mathcal{L}|_{X \times Y \times \left\{z_0\right\}}). \ \ \ \ \ (1)$

If ${\delta s = 0}$, then ${\overline{s}}$ can be lifted to a section of ${\mathcal{L}|_{X \times Y \times \mathrm{Spec} A}}$, and conversely.

Now, to show that ${\delta s = 0}$, we use the fact that

$\displaystyle H^1(\mathcal{L}|_{X \times Y \times \left\{z_0\right\}}) \simeq H^1(\mathcal{L}|_{X \times \left\{y_0\right\} \times \left\{z_0\right\}}) \oplus H^1(\mathcal{L}|_{\left\{x_0\right\} \times Y \times \left\{z_0\right\}}),$

by the Künneth formula, since ${\mathcal{L}}$ is trivial there. Consequently, to show that ${\delta s = 0}$, it suffices to show that after replacing ${Y}$ by ${\left\{y_0\right\}}$ and after replacing ${X}$ by ${\left\{x_0\right\}}$.

But the coboundary map

$\displaystyle H^0(\mathcal{L}|_{X \times \left\{y_0\right\} \times \mathrm{Spec} A}) \rightarrow H^1(\mathcal{L}|_{X \times \left\{y_0\right\} \times \mathrm{Spec} k})$

is trivial because the line bundle ${\mathcal{L}|_{X \times \left\{y_0\right\} \times Z}}$ is trivial and thus (by the Künneth formula) any section over ${X \times \left\{y_0\right\} \times \mathrm{Spec} k}$ can be lifted to ${\mathcal{L}|_{X \times \left\{y_0\right\} \times \mathrm{Spec} A}}$. Similarly for the analogous one when ${Y}$ is replaced by a point.

It follows now that in our original situation, the coboundary ${\delta s}$ vanishes when restricted to ${H^1(\mathcal{L}|_{X \times \left\{y_0\right\} \times \mathrm{Spec} k})}$ and dually when restricted to ${H^1(\mathcal{L}|_{\left\{x_0\right\} \times Y \times \mathrm{Spec} k})}$.

2. Extending to a full neighborhood

To complete the proof of the theorem of the cube, we now need to pass from arbitrary nilpotent thickenings to the whole scheme. In other words, we need to show the following:

Proposition 2 Let ${X}$ be a proper ${k}$-variety, and let ${T}$ be a connected variety. Let ${\mathcal{L} \in \mathrm{Pic}(X \times T)}$. Suppose there exists a closed point ${t \in T}$ such that for every local finite-dimensional ${k}$-algebra ${A}$ and map ${\mathrm{Spec} A \rightarrow T}$ thickening ${t}$, the restriction

$\displaystyle \mathcal{L}|_{X \times \mathrm{Spec} A}$

is trivial. Then ${\mathcal{L} \simeq \pi^* \mathcal{M}}$ for a suitable line bundle ${\mathcal{M} \in \mathrm{Pic}(T)}$ for ${\pi: X\times T \rightarrow T}$ the projection.

Assume ${T}$ irreducible, without loss of generality.

We will in fact take ${\mathcal{M} = \pi_* \mathcal{L}}$. We observe that ${\mathcal{M}}$ is a coherent sheaf on ${X}$, so that ${\mathcal{M}_t}$ is a finitely generated ${\mathcal{O}_{T, t}}$-module. Moreover, by the formal function theorem, we have

$\displaystyle \widehat{\mathcal{M}_t} \simeq \varprojlim H^0( \mathcal{L}|_{\mathrm{Spec} \mathcal{O}_{T, t}/\mathfrak{m}_{T, t}^n}) \simeq \widehat{\mathcal{O}_{T, t}},$

where the last step used the fact that ${\mathcal{L}|_{X \times \mathrm{Spec} \mathcal{O}_{T, t}/\mathfrak{m}_{T, t}^n}}$ is a trivial line bundle. It follows that the completion of ${\mathcal{M}_t}$ is free of rank one, so ${\mathcal{M}_t}$ is free of rank one.

In a neighborhood of ${t}$, we find that ${\mathcal{M}}$ is a line bundle. We also find from this that ${\mathcal{M}_t \rightarrow H^0(\mathcal{L}|_{X \times \left\{t\right\}}}$ is surjective. In particular, the unit element of ${H^0(\mathcal{L}_{X \times \left\{t\right\}}) \simeq k}$ (recall triviality of ${\mathcal{L}}$ here) prolongs to a section ${s}$ over ${X \times U}$ for some open neighborhood ${U}$ of ${t}$. Shrinking ${U}$ (and using properness of ${X}$), we may assume ${s}$ is invertible at all points of ${X \times U}$.

In any event, we find that ${\mathcal{L}|_{X \times U}}$ is trivial and that the map

$\displaystyle \pi^*\pi_* \mathcal{L} \rightarrow \mathcal{L}$

is an isomorphism on ${ X \times U}$. In particular, ${\mathcal{L}|_{X\times \left\{u\right\}}}$ is trivial for ${u \in U}$.

But the set of points ${t' \in T}$ such that ${\mathcal{L}|_{X \times \left\{t;\right\}}}$ is trivial is a closed subset of ${T}$ (by the reasoning earlier), and consequently ${\mathcal{L}|_{X \times \left\{t'\right\}}}$ is trivial for all ${t'}$. From this, one can appeal to the general machine “cohomology and base change” to argue that ${\pi^* \pi_* \mathcal{L} \rightarrow \mathcal{L}}$ is an isomorphism in general.

3. Conclusion of the argument

Let’s now finish the proof of the theorem of the cube. Given ${X, Y, Z}$ as in the statement, and ${\mathcal{L} \in \mathrm{Pic}(X \times Y \times Z)}$ satisfying the desired conditions, we find that ${\mathcal{L}|_{X \times Y \times \left\{z\right\}}}$ is trivial for all ${z \in Z}$. Moreover, ${\mathcal{L}_{X \times Y \times Z}}$ is the pull-back of a line bundle on ${\mathcal{L}|_{X \times Y}}$. Restricting to ${X \times Y \times \left\{z_0\right\}}$, we find that this line bundle must be trivial, so ${\mathcal{L}}$ is trivial.