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Let $A$ be an abelian surface and $\text{Km}(A)$ be the Kummer surface of $A$. If I remember correctly, the Picard number $\rho(\text{Km}(A))$ is equal to $16+\rho(A)$.

Does anyone know any reference or proof for this fact?

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In the fourth section of this paper http://www.staff.uni-bayreuth.de/~bt270951/quart9h.pdf they sketch the proof.

The idea is that the orthogonal complement of the subspace generated by the sixteen $-2$ curves in $\mathrm{H}^2(Km(A),\mathbb{Z})$ is isomorphic to $\mathrm{H}^2(A,\mathbb{Z})$.

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Another way of phrasing this is to look at the transcendental lattices, which are the orthogonal complements of the Picard lattices.

One obtains in particular a morphism $T(A) \to T(Km(A))$ (by looking at the usual Kummer construction) which multiplies the intersection form by 2, but is an isomorphism as free abelian groups (i.e. they have the same rank). Since the rank of $T(A)$ is given by $6 - pic(A)$ and the rank of $T(Km(A))$ is given by $22 - pic(Km(A))$, the conclusion follows.

Edit: The rank of $T(A)$ is $6 - pic(A)$, since $T(A)$ is the orthogonal complement of $Pic(A)$ in the lattice $H^2(A)$, which is of rank 6. Similarly, since $H^2(Km(A))$ is of rank 22, it follows that the rank of $T(Km(A))$ is $22 - pic(Km(A))$.

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  • $\begingroup$ Simon, could you please elaborate you answer a little bit? Namely, why the rank of $T(A)$ is equal to $6-\rho(A)$? And the same question about the rank of $T(Km(A))$. $\endgroup$ – guest31 Dec 17 '15 at 12:13

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