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# "Bad" reduction of Shimura curves via dual graphs

I have the following naive (and inexpert) question about the reduction of Shimura curves at primes dividing the discriminant of the underlying quaternion algebra. It requires some background to state. That is, let $F$ be a totally real field of degree $d$. Fix a real place $\tau_1$ in the set of real places $\lbrace \tau_1, \ldots, \tau_d \rbrace$ of $F$. Let $B$ be a quaternion algebra over $F$ that is split at $\tau_1$ and ramified at $\tau_2, \ldots, \tau_d$. Let $H \subset \widehat{B}~(= B \otimes \widehat{Z})$ be a compact open subgroup. Let $M_H$ denote the Shimura curve over $F$ of level $H$, with complex points given by \begin{align*} M_H({\bf{C}}) &= B^{\times}\backslash \widehat{B}^{\times} \times \left({\bf{C}}-{\bf{R}} \right)/H.\end{align*} Fix a prime $v \subset \mathcal{O}_{F}$. Assume that $H$ can be factored as $H^v \times H_v$, with $H_v \subset B_v^{\times}~(= B^{\times} \otimes F_v)$ maximal. If $v$ does not divide the discriminant of $B$, then it is known by work of Morita and Carayol that $M_H$ has good reduction over $v$, hence that there exists a smooth model ${\bf{M}}_H$ of $M_H$ over $\mathcal{O}_{(v)}$. If $v$ divides the discriminant of the quaternion algebra $B$, then it is known by work of Varshavsky for instance that there exists an integral model ${\bf{M}}_{H}^V$ of $M_H$ over $\mathcal{O}_{(v)}$. (N.B. there is apparently also a model due to Drinfeld, described extensively in the literature for the case of $F={\bf{Q}}$, though it is not clear to me why Drinfeld's work, which seems to require a moduli theoretic description of $M_H$, extends to the general totally real fields setting). Anyhow, let $F_v$ denote the completion of $F$ at $v$, with $\kappa_v$ the residue field and $\pi_v$ a uniformizer. By Cerednik's theorem, the completion of ${\bf{M}}_H^V$ along its closed fibre is canonically isomorphic to the product \begin{align*} GL(F_v)\backslash \widehat{\Omega}^{unr} \times D^{\times}\backslash \widehat{D}^{\times}/\overline{H}^v. \end{align*} Here, $\widehat{\Omega}^{unr}$ denotes the product $\widehat{\Omega} \times_{\operatorname{Spf}\mathcal{O}_{F_v}} \operatorname{Spf} \mathcal{O}_{F_v}^{unr}$, where $\widehat{\Omega}$ denotes the $v$-adic upper half plane (viewed as a formal scheme), and $\mathcal{O}_{F_v}^{unr}$ the ring of Witt vectors with coefficients in $\overline{\kappa}_v$. The action of $\gamma \in GL(F_v)$ on $\widehat{\Omega}^{unr}$ is via the image of $\gamma$ in $PGL(F_v)$ on the component $\widehat{\Omega}$, and via multiplication by $\operatorname{Frob}_v^{n(\gamma)}$ on $\widehat{\mathcal{O}}_{F_v}^{unr}$, where $n(\gamma) = - ord_v \left( \det(\gamma) \right)$. As well, $D$ denotes the totally definite quaternion algebra over $F$ obtained from $B$ by switching invariants at $v$ and $\tau_1$, with $\overline{H}^v$ the compact open subgroup of $\widehat{D}^{\times v}$ corresponding to $H^v$ under a fixed isomorphism $B^{\times v} \cong D^{\times v}$. The theory of Mumford-Kurihara unifomization then gives the following information about this curve ${\bf{M}}_{H}^V$:

1. The curve ${\bf{M}}_{H}^V$ is an admissible curve over $\mathcal{O}_{F_v}$ in the sense of Jordan-Livne, i.e.

(i) ${\bf{M}}_H^V$ is a flat, proper curve over $\mathcal{O}_{F_v}$ with a smooth generic fibre.

(ii) The special fibre of ${\bf{M}}_H^V$ is reduced; the normalization of each of its irreducible components is isomorphic to ${\bf{P}}^1_{\kappa_v}$, and its only singular points are $\kappa_v$-rational, ordinary double points.

(iii) The local ring ${\bf{M}}_{H, x}$ at any singular point $x$ of the special fibre is isomorphic as an $\mathcal{O}_{F_v}$-algebra to $\mathcal{O}_{F_v}[[X,Y]]/(XY - \pi_v^{m(x)})$, for $m(x) \geq 1$ a uniquely determined integer.

1. The dual graph $\mathcal{G}({\bf{M}}_H^V) = (\mathcal{V}({\bf{M}}_H^V),\mathcal{E}({\bf{M}}_H^V))$ of the special fibre of ${\bf{M}}_H^V$ is
isomorphic to $GL(F_v)^{+} \backslash \left( \Delta \times D^{\times}\backslash \widehat{D}^{\times}/\overline{H}^v \right)$, minus any loops. Here, $GL(F_v)^+ \subset GL(F_v)$ denotes the collection of matrices with determinant having even $v$-adic valuation, and $\Delta = (\mathcal{V}(\Delta), \mathcal{E}(\Delta))$ the Bruhat-Tits tree of $SL(F_v)$.

My question is the following: why is the edgeset $\mathcal{E}({\bf{M}}_H^V)$ nonempty? The dual graph $\mathcal{G}({\bf{M}}_H^V)$ is clearly disconnected, and seen easily to be given by the disjoint union of connected graphs \begin{align*} \coprod_i \mathcal{G}_i &= \coprod_i \overline{\Gamma}_i \backslash \Delta. \end{align*} Here, each $\overline{\Gamma}_i$ denotes the image in $PGL(F_v)$ of a suitable arithmetic subgroup $\Gamma_i \subset D^{\times} \cong D_v^{\times} \cong GL(F_v)$. Each component graph $\mathcal{G}_i = (\mathcal{V}_i, \mathcal{E}_i)$ is connected. Now, since $\Delta$ is a tree, each component graph $\mathcal{G}_i = \overline{\Gamma}_i \backslash \Delta$ is a tree. It is then well known that each (first) Betti number $\beta(\mathcal{G}_i) := \vert \mathcal{E}_i \vert - \vert \mathcal{V}_i \vert + 1$ must vanish, i.e. $\vert \mathcal{E}_i \vert = \vert \mathcal{V}_i \vert -1 = 0$. If so, then the cardinality of the edgeset $\mathcal{E}({\bf{M}}_H^V)$ must also equal zero. i.e. the special fibre of ${\bf{M}}_H^V$ would have no singular points ... what have I missed here?