6
$\begingroup$

I'm a bit puzzled about the following considerations, and am looking for some explanations or maybe some references about it.

Setting: Let $E/F$ be a CM extension of number fields ($F$ being totally real) and let $(V, \langle\cdot{},\cdot{}\rangle)$ be a $n$-dimensional nondegenerate $E/F$-hermitian space (with respect to the complex conjugation $c\in\mathrm{Gal}(E/F)$), for some positive integer $n$. Lattices will all be of full rank.

One defines a (connected) reductive group scheme $\mathrm{U}(V)$ over $\mathrm{Spec}\,F$ by setting, for all $F$-algebra $R$, $$\mathrm{U}(V)(R) =\{g\in \mathrm{GL}(V\otimes_F R);\, \langle g\cdot{}v,g\cdot{}w\rangle =\langle v,w\rangle,\, \forall v,\,w \in V\otimes_F R\}.$$

By definition, one gets a closed immersion between (affine) $F$-group schemes $\mathrm{U}(V)\hookrightarrow \mathrm{GL}(V_F)=\mathrm{Res}_{E/F}\mathrm{GL}(V)$

Let $\tau$ denote any finite place of $F$. If $L$ is an $\mathcal{O}_E$-lattice in $V$ (resp. an $\mathcal{O}_{E_{\tau}}$–lattice in $V_{\tau}$, with $\mathcal{O}_{E_{\tau}}:=\mathcal{O}_E\otimes_{\mathcal{O}_F}\mathcal{O}_{F_{\tau}}$ and $V_{\tau}=V\otimes_F F_{\tau}$. e.g., if $\tau$ is inert in $E/F$ then $\mathcal{O}_{E_{\tau}}$ is just the ring of integers of the quadratic extension $E_{\tau}=E\otimes_F F_{\tau}$ of $F_{\tau}$, and $V_{\tau}$ is the $n$-dimensional $E_{\tau}$-vector space $V\otimes_E E_{\tau}$) one defines its dual by setting $$L^{\vee}=\{z\in V;\,\langle z, L\rangle \in \mathcal{O}_E\}$$ (resp. $L^{\vee}=\{z\in V_{\tau};\,\langle z, L\rangle \in \mathcal{O}_{E_{\tau}}\}$)

Now, take any $\mathcal{O}_E$-lattice $L$ and assume that $L\subset L^{\vee}$ and set $N=[L^{\vee}:L]$. Take $S$ to be the finite set of finite places of $F$ containing the prime divisors of $N\mathcal{O}_F$, and set $\mathcal{O}_F^S$ to be the ring of $S$-integers in $F$, $$\mathcal{O}_F^S=\{x\in F;\, \mathrm{ord}_v(x)\ge 0, \, \forall v\notin S\}.$$ Away from $S$, the local lattice $L_{\tau}:=L\otimes_{\mathcal{O}_E}\mathcal{O}_{E_{\tau}}$ is self-dual. Accordingly, the local pairing

$$\langle \cdot{},\cdot{}\rangle_{\mathcal{O}_{E_{\tau}}}:\, L_{\tau}\times L_{\tau}\rightarrow \mathcal{O}_{E_{\tau}}$$ is perfect, for all $\tau\notin S$.

  1. My first question. Is the functor which goes from the category of $\mathcal{O}_F^S$-algebras to the category of groups, and sends any $R/\mathcal{O}_F^S$ to the group $$\underline{\mathrm{U}}_V(R):=\{g\in \mathrm{GL}(L\otimes_{\mathcal{O}_F}R);\, \langle g\cdot{}v,g\cdot{}w \rangle=\langle v,w \rangle\in\mathcal{O}_E\otimes_{\mathcal{O}_F}R, \,\forall v,\,w\in L\otimes_{\mathcal{O}_F}R\}$$ (whose restriction to $F$-algebras coincides with $\mathrm{U}(V)$), representable by a smooth $\mathrm{Spec}\,\mathcal{O}_F^S$-group scheme $\underline{\mathrm{U}}_V$, such that $\underline{\mathrm{U}}_V\times\mathrm{Spec}\,F\simeq \mathrm{U}(V)$ ? Moreover, are the fibers $(\underline{\mathrm{U}}_V)_v$ of $\underline{\mathrm{U}}_V$ at $v$ already reductive for all $v\notin S$, or do we possibly need to enlarge $S$ ?

On the other hand, a more general construction related to reductive groups is described (for instance) in Chapter II of Getz-Hahn's book An Introduction to Automorphic Representations with a view toward Trace Formulae, whose online version can be found on this page https://services.math.duke.edu/~hahn/Chapter2.pdf

Namely, one may apply their Lemma 2.4.2 in the following way: the choice of $L$ defines an integral model $\mathrm{GL}(V)_{\mathcal{O}_F}$, namely, whose $R$-points are given by $\mathrm{GL}(L\otimes_{\mathcal{O}_F}R)$ for any $\mathcal{O}_F$-algebra $R$, hence one has a composition of closed immersions $$\mathrm{U}(V)\hookrightarrow \mathrm{GL}(V_F)\hookrightarrow \mathrm{GL}(V)_{\mathcal{O}_F}.$$

By setting $\underline{\mathrm{U}}_V\subset \mathrm{GL}(V)_{\mathcal{O}_F}$ to be the Zariski closure of $\mathrm{U}(V)$ in $\mathrm{GL}(V)_{\mathcal{O}_F}$, then there exists a finite set $S'$ of finite places of $F$, such that $(\underline{\mathrm{U}}_V)_{\mathcal{O}_F^{S'}}$ is a smooth $\mathcal{O}_F^{S'}$-group scheme which is a model for $\mathrm{U}(V)$.

Then, for its generic fiber $\mathrm{U}(V)$ is already reductive, one may apply Proposition 3.1.9 of Conrad's Reductive group schemes paper (using a so-called "spreading out" argument) to deduce that the fibers of $(\underline{\mathrm{U}}_V)_{\mathcal{O}_F^{S'}}$ are reductive outside a finite number of places, i.e., that there exists a finite set $S''$ of finite places of $F$ containing $S'$ and such that $(\underline{\mathrm{U}}_V)_{\mathcal{O}_F^{S''}}$ has reductive fibers.

  1. My second question. Is the above construction strictly equivalent to the (more explicit but less general) previous one ? Assuming the previous construction works with $S$, how can one compare $S$ and $S''$ in general ?

EDIT: One of the aims behind my question is to convince myself of the link between the notion of an hyperspecial maximal compact subgroup $K_{\tau}$ of $\mathrm{U}(V)(F_{\tau})$ on the one hand (which arises as the $\mathcal{O}_{F_{\tau}}$-points of my almost-everywhere reductive model obtained by the second construction), and the notion of stabilizer of a self-dual local $\mathcal{O}_{E_{\tau}}$-lattice (which arises as the $\mathcal{O}_{F_{\tau}}$-points of the expected model obtained by the first construction).

Many thanks for any kind of help, and please don't hesitate to relocate this question to Stackexchange if you feel it's not advanced enough.

Yoël.

$\endgroup$

2 Answers 2

1
$\begingroup$

Regarding the second part of question 1, which is what the fibres of the integral model given by a choice of L will look like: you might find the paper of Gan, Hanke and Yu, On an exact mass formula of Shimura, helpful. They assume that $L$ is maximal -- i.e. there is no larger lattice $L' \supsetneq L$ such that $\langle x, x \rangle \in \mathcal{O}_F$ for all $x \in L'$ -- and for such lattices they describe what the resulting group scheme (the thing you call $\underline{U}(V)$) looks like locally at the bad places. (It's not always reductive, but they describe its maximal reductive quotient explicitly).

$\endgroup$
2
  • $\begingroup$ Thank you David for your answer ! Do you mean that GTA's answer - stating that the fiber of $\underline{\mathrm{U}}_V$ at $v$ is reductive whenever $v\notin S$- is a bit too optimistic, but that we might rather enlarge $S$ by adding a finite set of bad primes ? $\endgroup$
    – Yoël
    Commented Aug 31, 2019 at 17:55
  • $\begingroup$ I meant that there is a recipe for describing the fibres at "bad" places $v$. As to whether or not the set $S$ you describe contains all the bad places, I'm not 100% sure, but the Gan--Hanke--Yu paper should give you enough tools to find out. $\endgroup$ Commented Aug 31, 2019 at 21:59
2
+100
$\begingroup$

For 1, the functor is representable just because it is a closed subscheme of GL(L) cut out by some set of equations. Seeing fibers as formal completions one knows that one can use the same bilinear form for defining fibers $(\underline{\text{U}}_{V})_{v}$ so it is reductive for $v\notin S$.

For 2, you see that Zariski closure is a closed subscheme of $\underline{\text{U}}_{V}$ over $\mathcal{O}_{F}^{S'}$ where both schemes are smooth and have the same dimension. So whether they are actually the same boils down to whether the larger scheme has an extra connected component missed by the smaller scheme, but this should also be observed at the generic fiber where we already know both schemes become the same.

$\endgroup$
2
  • $\begingroup$ Thank you GTA for your answer ! I'm trying to figure out the details, can you please elaborate a bit on what you mean by "Seeing fibers as formal completions" ? Do you mean that the functor obtained by base change from $\mathcal{O}^S_F$-algebras to $(\mathcal{O}_F^S)_v$-algebras is represented by the formal completion of $(\underline{\mathrm{U}}_V)_v$ ? $\endgroup$
    – Yoël
    Commented Aug 31, 2019 at 15:28
  • $\begingroup$ I meant, the formal completion of $\underline{\mathrm{U}}_V$ at $v$. But I think it's clear anyway :-) $\endgroup$
    – Yoël
    Commented Aug 31, 2019 at 15:51

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .