Let $A_{K}$ be an abelian variety over the field $K$ of fractions
of
a discrete valuation ring $R$, and let $A^{\prime }_{K}$ be the dual
abelian variety of $A_{K}$. Then $A_{K}$ and $A_{K^{\prime}}$ have
a
natural integral structure over $R$, given by their N\'eron models
$A$ and $A^{\prime }$. An interesting question is to investigate the
duality between $A_{K}$ and $A^{\prime }_{K}$ in terms of integral
data, i.~e., on the level of associated N\'eron models. Grothendieck
has shown that the integral structure of this duality is encoded in
a
bilinear pairing
\[\Phi _{A} \times \Phi _{A^{\prime}} \longrightarrow \mathbbQ/\mathbb
Z\]
on the component groups $\Phi _{A},\Phi _{A^{\prime}}$ associated to
the N\'eron models $A,A^{\prime }$. In fact, the pairing represents
the obstruction of extending the Poincar\'e bundle $\mathcal P$ on
$A_{K} \times A^{\prime }_{K}$ to a corresponding object on $A
\times A^{\prime }$. Grothendieck conjectured that the pairing
defines always a perfect duality between $\Phi _{A}$ and $\Phi
_{A^{\prime}}$, and he was able to prove this in certain situations.
The conjecture has been established in a number of additional cases,
notably by work of Artin and Mazur (unpublished), B\'egueri, and
McCallum.
In the lecture we report about a new approach to Grothendieck's
pairing, which was developed in joint work with Dino Lorenzini. We
start with an interpretation of the pairing in terms of certain
vanishing orders of rational functions. Thereby we are able to relate
Grothendieck's pairing to N\'eron's local height pairing. The latter
has good functorial properties, in contrast to the behavior of
N\'eron models and their component groups. For Jacobians, we thereby
are able to transfer all ingredients of Grothedieck's pairing to the
underlying curve. Using Raynaud's characterization of component
groups of Jacobians, we arrive at an explicit formula for the
pairing, involving only combinatorial data of the curve. The formula
shows that, for Jacobians, Grothendieck's pairing is always perfect,
provided the residue field is perfect, and that there are
counterexamples otherwise.