Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
add comment

2 Answers

up vote 7 down vote accepted

Inkspot is indeed correct that the component graphs are indeed not generally trees.

As you seem to have deduced for yourself, Cerednik-Drinfeld uniformization is a highly nontrivial concept, and it really helps to have some examples to set it in your mind. Most helpful in this direction is Ogg's "Mauvaise réduction des courbes de Shimura" where he draws out a few of these dual graphs.

In general the dual graphs have $2h$ vertices $x$ where $h$ is the class number of $\mathcal{O}_x$, a level $H$ Eichler order in $D$ (your totally definite quaternion algebra, so note there's a choice of which $x$ to make here, but as long as the level is squarefree all orders are hereditary and it doesn't make a difference).

An orbit-stabilizer theorem computation then shows that for instance when $B$ is a quaternion algebra over $\mathbf{Q}$, $p+1$ (the size of the set of edges $y$ stemming from a particular vertex $x$ in the Bruhat-Tits tree) is equal to $\sum_{e(y) = x} \frac{\mathcal{O}_x^\times}{\mathcal{O}_y^\times}$. So it's not just that there are edges, but we know exactly how many there are! (For a readable account of details of this, see Kurihara's paper on Equations defining Shimura Curves)

Also, if I may take issue with 1.(ii) and 1.(iii), you've given a good description of the special fiber over $\overline \kappa_v$ (which is what I'm taking the dual graph to represent the data for), not necessarily $\kappa_v$. What you've claimed is that the special fiber is a Mumford Curve, that is, the transverse union of a number of copies of $\mathbb{P}^1$'s. The truth is that the special fiber is a quadratic twist of a Mumford curve, where the Galois action is not simply given by the $| \kappa_v|$-Frobenius, but where the action of Frobenius is identified with the action of the Atkin-Lehner operator $w_p$, which interchanges some of the components (if you want to think about the graph, its vertex set can be partitioned into $ \{x_1, \dots , x_h, x_1', \dots, x_h'\}$ where $w_p(x_i) = x_i'$).

All that said, some of the best advice I've heard for trying to understand this stuff is to first completely understand what happens when $v$ divides the LEVEL because in that case the moduli problem is much easier (if an abelian variety here is isogenous to a product of supersingular elliptic curves, it's isomorphic to a product of supersingular elliptic curves)

Here are a few additional references:

http://www.math.mcgill.ca/cfranc/documents/bctranslation.pdf (a translation of Boutot-Carayol)

http://math.berkeley.edu/~ribet/Articles/bimodules.pdf (this includes a somewhat more intuitive description of the components of the Mumford curve)

http://math.uga.edu/~pete/thesis.pdf (a comprehensive introduction to Shimura Curves and the action of the Atkin-Lehner group)

http://www.springerlink.com/content/gj8365486214l141/ (this is Oort's "which abelian surfaces are products of supersingular elliptic curves", see also his book on moduli of supersingular abelian varieties, as when $v$ ramifies in $B$, you're asking a question about moduli of supersingular abelian varieties, see the appendix on Honda-Tate theory to the thesis above)

share|improve this answer
    
Thanks for this helpful answer! I assume that by dividing the level, you mean not dividing the discriminant of the quaternion algebra? In any case, I am interested specifically in the setting of general totally real fields, where there does not appear to be a moduli theoretic description of the Shimura curve ... which is why I also asked about the applicability of Drinfeld's work in this setting. –  jvo Jan 13 '11 at 20:04
    
Dear stankewicz @ Is there a English translation of Carayol's 1986 paper: Bad reducton of Shimura curves ? Thank you. –  user4245 Jan 14 '11 at 2:25
    
First an apology: I deleted my above comment because my remark on level was nonsense, I reproduce it with correction here: By level I mean the smallest integer $N$ such that $U (N) \subset H$ where $U (N)$ is as in Boutot-Carayol. As for a moduli-theoretic description there is something which is related to the Shimura curves you describe (with which I unfortunately hold only passing familiarity) see the following paper of Carayol, especially the section on "Un Mod`ele \'Etrange" in archive.numdam.org/ARCHIVE/CM/CM_1986__59_2/CM_1986__59_2_151_0/… –  stankewicz Jan 14 '11 at 10:18
    
As for whether there is a translation: I don't know of one. These translations generally pop up when someone is studying the original and decides if they are going to take notes on it in their native tongue they may as well translate it along the way. It's also generally less likely for this to happen with French papers because mathematical French is so close to mathematical English (case in point: EGA). –  stankewicz Jan 14 '11 at 10:28
add comment

"Now, since $\Delta$ is a tree, each component graph ... is a tree." This isn't convincing, since the universal cover of any graph is a tree. In fact, your argument would prove that every curve possessing a Mumford-Kurihara uniformization has genus zero.

share|improve this answer
    
Thanks for pointing this out! You are absolutely right. –  jvo Jan 13 '11 at 20:00
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.