# What is the Theorem of the Cube?

What is the "theorem of the cube" for abelian varieties? What is the statement and how should I think about it?

If you have a line bundle trivial on 3 "surfaces" of a "cube" $A\times B\times C$ where $A$, $B$, $C$ are abelian varieties, then this line bundle in trivial on the whole "cube".

See wikipedia.

• NB: it's valid not only for abelian varieties, but for any complete varieties... – Michael Thaddeus Jun 7 '10 at 11:52

One application of the theorem of the cube is to study the map from an abelian variety A to its dual abelian variety; the map is defined in terms of line bundles and the key technical theorem one uses to prove anything (e.g. that the map to the dual is a homomorphism) is the theorem of the cube. See Mumford's Abelian Varieties book or Martin Olsson's notes from this summer's Hangzhou workshop.

I'm a bit late to the party, but since these question are clearly still getting views, I'll answer the second question a bit. Given a nice category, one can form the pointed category $C$, and consider functors $F:C\to Ab$. There are cannonical maps $\beta:F(X_0\times\dots \times X_n)\to \prod_i F(X_0\times \dots \times X_{i-1}\times X_{i+1}\times \dots \times X_n)$ and $\alpha:\prod_i F(X_0\times \dots \times X_{i-1}\times X_{i+1}\times \dots \times X_n)\to F(X_0\times\dots \times X_n)$. We have $F(X_0\times\dots \times X_n)=Ker(\beta)\oplus Im(\alpha)$, and we say that $F$ is of order $n$ if either $Ker(\beta)=0$ or alternatively, if $\alpha$ is surjective. Now if you have an exact sequeunce of functors, $T_1\to T_2\to T_3$, and $T_1, T_3$ are of order $n$, then $T_2$ is by the Snake Lemma. Additionally, if $T$ is of order $n$, then it is of order $m$ for $m>n$. Lastly, we note that $H^n(X; \mathcal{F})$ is of order $n$ by the Kunneth theorem.

Now note that the Theorem of a Cube is just the statement that $Pic$ is of order $2$. Now in the complex case, the exponential exact sequence $0\to \mathbb{Z}\to \mathcal{O}_X\to \mathcal{O}_{X}^*\to 0$ gives an exact sequence $H^1(X; \mathbb{Z})\to H^1(X; \mathcal{O}_{X}^*)\to H^2(X;\mathcal{O}_X)$. Now the left-hand and right-hand term are both of order $2$, so we have derived the theorem of the cube in the complex case.

So when thinking of the theorem of the cube, I usually think that it expresses the fact that $\mathcal{O}_X^*$ is resolved with two sheaves, in some sense, and this generalizes to the non-complex case in spirit.