2
$\begingroup$

Suppose we have a PEL type $(H,\phi ,*;T,O,V)$ where H is a rational nonsplit quaternion algebra, $\phi$is an embedding of Q-algebra $\phi : H-->M(2,R)$, and * is a positive anti involution of H; O is the maximal order of H , and V level structure . Associate these datum a shimura curve parametric fake elliptic curves in a standar way. My question is :what is the group for the shimura datum ,is it the group H*, the invertible element of H ? And since this family parametirc abelian two folds , what is the map from this group to $\mathrm{GSP}(4,\mathbb{Q})$?

$\endgroup$
1
  • 1
    $\begingroup$ You should look at the paper of Goresky and Tai, "Anti-holomorphic multiplication and a real algebraic modular variety", J. Differential Geom. vol. 65 no. 3. $\endgroup$ Nov 12, 2010 at 16:30

2 Answers 2

11
$\begingroup$

Actually, fake elliptic curves are discussed in Chapter 9 of Lang's Introduction to algebraic and abelian functions.

In order to describe the map to $GSP(4,Q)$, recall that there is the standard antiinvolution $x\to x'$ on the quaternion $Q$-algebra $H$ such that both $tr(x)=x+x'$ and $Norm(x)=xx'=x'x$ are rational numbers for all $x\in H$. Every positive antiinvolution of $H$ is of the form $x \to \gamma^{-1}x'\gamma$ where $\gamma$ is a fixed nonzero (invertible) element of $H$ such that $\gamma^2$ is a negative rational number and therefore $\gamma'=-\gamma$. This gives rise to the alternating nondegenerate $Q$-bilinear form $$E: H \times H \to Q, x,y \mapsto tr(\gamma x y').$$ Now let us consider the following faithful action of the multiplicative group $G$ of $H$ on the $Q$-vector space $V=H$: $$u(x)=x u'$$ for $x \in V=H$ and $u \in G$. Then

$$E(u(x),u(y))=tr(\gamma x u'u y')=(u'u)tr(\gamma x y'),$$

which means that $$E(u(x),u(y))=Norm(u) E(x,y).$$

This gives us the embedding $G \to GSP(V,E)\cong GSP(4.Q)$.

The same construction over arbitrary commutative $Q$-algebras $R$ gives us the desired embedding of the corresponding $Q$-algebraic groups.

$\endgroup$
0
3
$\begingroup$

Question 1(what is the group for the Shimura datum):

Well, remember that $H^\times$ is just a bare group. A Shimura datum requires an algebraic group over $\mathbf{Q}$: that is, a functor from $\mathbf{Q}$-schemes to groups. Assuming you mean the group whose $\mathbf{Q}$ points, yes and you can see this in example 5.24 of http://www.jmilne.org/math/xnotes/svi.pdf although you can also use the algebraic group whose $\mathbf{Q}$ points are the norm one units of $H$ if you were interested in the connected Shimura datum (which is another example in milne's notes).

Question 2(what is the map from this group to the symplectic group):

I don't know. Is it even clear that a forgetful map of coarse moduli spaces which happen to be Shimura varieties induces a morphism of Shimura data? Either way your question sounds strongly related to the work of Victor Rotger's thesis which asks about the irreducibility of the quaternionic locus in $\mathcal{A}_2$.

$\endgroup$
2
  • $\begingroup$ I think that if you use the norm one units of $H$ than you get a "connected Shimura variety" rather than a Shimura variety proper, since the norm one units don't have enough room to accept a map from all of $Res^{\mathbb C}_{\mathbb R} \mathbb G_m$ (as is necessary for a Shimura datum); they can only accept a map from the norm one part, which gives a connected Shimura datum. $\endgroup$
    – Emerton
    Nov 12, 2010 at 14:21
  • $\begingroup$ @stankewicz: Regarding 2: as you can see from Professor Zarhin's answer, this is true in this case (and the use of additional endomorphisms to construct a Riemann form works in other situations as well, as he says). As a matter of general philosophy, I would say that yes, naturally defined morphisms between Shimura varieties do tend to be induced by homomorphisms of the Shimura data. $\endgroup$ Nov 12, 2010 at 22:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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