1
$\begingroup$

Let $A$ be an abelian variety of dimension $2$ over a $p$-adic field $K$ with (additive) valuation $v$. Assuming $A$ has multiplicative reduction, the theory of $p$-adic theta functions gives us an isomorphism $A \cong (\bar{K}^\times \times \bar{K}^\times)/(q_1,q_2)$ where $q_1,q_2=(q_{11},q_{12}),(q_{21},q_{22}) \in \bar{K}^\times \times \bar{K}^\times$ is such that the "valuation matrix":

$$\begin{pmatrix} v(q_{11}) & v(q_{12})\\ v(q_{21}) & v(q_{22}) \end{pmatrix} $$

has nonzero determinant (for this description, see e.g. Ribet's Ph.D thesis http://www.jstor.org/stable/10.2307/2373815).

In the case of dimension $1$, this is Tate's $p$-adic uniformization of elliptic curves. In that case $v(q)=v(q_{11})$ is negative the valuation of the $j$-invariant of the curve $E$, which is the number of components of the special fiber.

My question is: can we find a similar description of the above matrix in terms of the component group of the special fiber of the Neron model? I'm hoping it will end up being the determinant.

My best guess is to relate the (algebraic) Neron model to the (rigid analytic) parametrized abelian variety using the connection between formal schemes and rigid analysis, but I don't know enough about these topics to do this.

Improve this question
$\endgroup$
4
  • $\begingroup$ Since $GL_2(\mathbb Z)$ acts on your matrix on the right without changing the geometry of the abelian variety, only the image of the matrix should carry geometric meaning. In particular the quotient of $\mathbb Z^2$ by the image of the valuation matrix should be the group of components. This would follow from the fact that $\mathcal O_K^{\times} \times \mathcal O_K^{\times}$ is the group of sections which have good reduction at $0$, that is, lie in the identity component. $\endgroup$
    – Will Sawin
    Commented Jul 16, 2012 at 21:39
  • 2
    $\begingroup$ I don't understand exactly what is the question, but observe that the component group of the Neron model is isomorphic (as abstract group) to the cokernel of the map given by the valuation matrix (from $\mathbb{Z}^2$ to itself). This is shown, if I remember well, in the SGA7, Expose IX, and it is due to Raynaud. $\endgroup$
    – Xarles
    Commented Jul 26, 2012 at 21:32
  • 1
    $\begingroup$ And observe that your last coment is directly related to what I did together with Siegfried Bosch in the paper Component groups of Néron models via rigid uniformization, Math. Ann. Volume 306, Number 1 (1996), 459-486. $\endgroup$
    – Xarles
    Commented Jul 26, 2012 at 21:35
  • $\begingroup$ It seems that I wasn't reading carefully enough. My question is actually answered in the paragraph after Theorem 3.2.2 in Ribet's thesis. $\endgroup$
    – David Corwin
    Commented Dec 18, 2012 at 19:50

0

Reset to default

You must log in to answer this question.