Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 126 / 3

Let $m_I:X^3\to X$ add the coordinates indexed by a nonempty subset $I$ of the three-element index set. The Theorem of the Cube says that for every line bundle $\mathcal{L}$ on an abelian variety,is trivial, up to the harmless constant line given by the fiber of $\mathcal{L}$ at the identity.

Pull this line bundle back along $(f,g,h):Y\to X^3$. Pullback commutes with tensor products and duals, and $m_I\circ (f,g,h)$ is the corresponding sum of morphisms. The resulting bundle is precisely ${\mathcal{M}}_{f,g,h}$, so it is trivial.

Take $Y=X$, $f={\mathrm{id}}_X$, and let $g,h$ be the constant maps with values $x,y$. All pullbacks along constant maps are trivial line bundles. The formula for ${\mathcal{M}}_{f,g,h}$ then becomesor equivalently

For a line bundle $\mathcal{L}$ on $X$, define the homomorphism associated to a line bundle on an abelian varietyThe translation identity from part (i) givesso ${\varphi }_{\mathcal{L}}$ is a homomorphism.

Suppose $\mathcal{M}\simeq T_y^{\ast }\mathcal{L}\otimes {\mathcal{L}}^{\vee }$ lies in its image. Thenwhich is trivial by the same translation identity. Hence ${\varphi }_{\mathcal{M}}=0$; the image of every ${\varphi }_{\mathcal{L}}$ lies in the Identity component of the Picard group ${\mathrm{Pic}}^{0}X$.

Directly from the definition,AlsoThe bundle in parentheses lies in ${\mathrm{Pic}}^{0}X$ by part (ii), so it is translation invariant. This proves ${\varphi }_{T_x^{\ast }\mathcal{L}}={\varphi }_{\mathcal{L}}$.

Finally, $[n]\circ T_x=T_{nx}\circ [n]$, and thereforeFor $\mathcal{N}\in {\mathrm{Pic}}^{0}X$, the multiplication pullback formula reduces to $[n{]}^{\ast }\mathcal{N}\simeq {\mathcal{N}}^{\otimes n}$. Consequentlyas required.