It's known that if $A$ is an abelian variety of totally multiplicative reduction over a p-adic field K, then, after taking a finite field extension, it becomes isomorphic, as a rigid analytic group, to $G_m^g/\prod_{i=1}^g q_i^{\mathbb{Z}}$ where $q_i$ are points (after a finite field extension) of $G_m^g$ which generate a discrete subgroup.

My question is, what can be said when $A$ is not totally multiplicative or good reduction, but is some semistable abelian variety of mixed multiplicative-good reduction. I would guess that instead of being a quotient of $G_m$ by a discrete subgroup, $A$ will be (as a rigid analytic variety) a quotient of a good-reduction semi-abelian variety of good reduction (i.e. a variety which is an extension by $G_m^m$ of an abelian variety of good reduction) by a free discrete subgroup. Does anyone know whether this is true, or whether there's something else ok90o,lreplacing the Tate uniformization in this case?

It's known that if $A$ is an abelian variety of totally multiplicative reduction over a p-adic field K, then, after taking a finite field extension, it becomes isomorphic, as a rigid analytic group, to $G_m^g/\prod_{i=1}^g q_i^{\mathbb{Z}}$ where $q_i$ are points (after a finite field extension) of $G_m^g$ which generate a discrete subgroup.

My question is, what can be said when $A$ is not totally multiplicative or good reduction, but is some semistable abelian variety of mixed multiplicative-good reduction. I would guess that instead of being a quotient of $G_m$ by a discrete subgroup, $A$ will be (as a rigid analytic variety) a quotient of a good-reduction semi-abelian variety (i.e. a variety which is an extension by $G_m^m$ of an abelian variety of good reduction) by a free discrete subgroup. Does anyone know whether this is true, or whether there's something else replacing the Tate uniformization in this case?