Complexification of Néron models of Abelian varieties

Question

Let $A$ be an abelian variety of dimension $g$ over the quotient field $K$ of a DVR $R$ which is a subfield of the complex field $\mathbb{C}$. Then by a result of Grothendieck, we know that there is a finite extension $K\subseteq K^{\prime}$ such that the neutral component of the Néron model of $A^{\prime}=A\otimes_{K}K^{\prime}$, i.e., $N(A^{\prime})^{\circ}$ is semiabelian, so it fits in anexact sequence $0\to T^{\prime}\to N(A^{\prime})^{\circ}_s\to B^{\prime}\to 0$, where $T^{\prime}$ is a torus of dimension $r$ and $B^{\prime}$ is an abelian variety of dimension $g-r$ and $N(A^{\prime})^{\circ}_s$ is the special fiber of $N(A^{\prime})$. Let us construct the real torus $U=\mathbb{R}^g/\Lambda^{\prime}$, where $\Lambda^{\prime}=\mathbb{Z}^{g-r}\oplus\Lambda(T^{\prime})$ and $\Lambda(T^{\prime})=\operatorname{Hom}(T^{\prime}, \mathbb{G}_m)$ is the lattice of characters of $T^{\prime}$.

I would like to know if it is true that the complexification of $U_C$ is isomorphic (as a complex torus) to the complexification $A_C$ of $A$.

Answer 1

No. Take an abelian variety $A_0$ over $\mathbf{Q}$ and let $A = A_0\otimes K$ where $R = \mathbf{Q}[\pi]_{(\pi)}$ and $K = R[1/\pi]$. Then $K=K'$, $T'=0$, $N(A')_s^\circ = A\otimes R$. Then you are asking if $A_{\mathbf{C}}$ is isomorphic to the complexification of $\mathbf{R}^g/\mathbf{Z}^g$. Obviously this will not hold in general.