Lemma 39.9.6. Let $k$ be a field. Let $A$ be an abelian variety over $k$. Let $\mathcal{L}$ be an invertible $\mathcal{O}_ A$-module. Then there is an isomorphism

$m_{1, 2, 3}^*\mathcal{L} \otimes m_1^*\mathcal{L} \otimes m_2^*\mathcal{L} \otimes m_3^*\mathcal{L} \cong m_{1, 2}^*\mathcal{L} \otimes m_{1, 3}^*\mathcal{L} \otimes m_{2, 3}^*\mathcal{L}$

of invertible modules on $A \times _ k A \times _ k A$ where $m_{i_1, \ldots , i_ t} : A \times _ k A \times _ k A \to A$ is the morphism $(x_1, x_2, x_3) \mapsto \sum x_{i_ j}$.

Proof. Apply the theorem of the cube (More on Morphisms, Theorem 37.32.8) to the difference

$\mathcal{M} = m_{1, 2, 3}^*\mathcal{L} \otimes m_1^*\mathcal{L} \otimes m_2^*\mathcal{L} \otimes m_3^*\mathcal{L} \otimes m_{1, 2}^*\mathcal{L}^{\otimes -1} \otimes m_{1, 3}^*\mathcal{L}^{\otimes -1} \otimes m_{2, 3}^*\mathcal{L}^{\otimes -1}$

This works because the restriction of $\mathcal{M}$ to $A \times A \times e = A \times A$ is equal to

$n_{1, 2}^*\mathcal{L} \otimes n_1^*\mathcal{L} \otimes n_2^*\mathcal{L} \otimes n_{1, 2}^*\mathcal{L}^{\otimes -1} \otimes n_1^*\mathcal{L}^{\otimes -1} \otimes n_2^*\mathcal{L}^{\otimes -1} \cong \mathcal{O}_{A \times _ k A}$

where $n_{i_1, \ldots , i_ t} : A \times _ k A \to A$ is the morphism $(x_1, x_2) \mapsto \sum x_{i_ j}$. Similarly for $A \times e \times A$ and $e \times A \times A$. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).