Suppose we have a PEL type $(H,\phi ,*;T,O,V)$ where H is a rational nonsplit quaternion algebra, $\phi$is an embedding of Qalgebra $\phi : H>M(2,R)$, 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 $\mathrm{GSP}(4,\mathbb{Q})$?

1$\begingroup$ You should look at the paper of Goresky and Tai, "Antiholomorphic multiplication and a real algebraic modular variety", J. Differential Geom. vol. 65 no. 3. $\endgroup$ – David Hansen Nov 12 '10 at 16:30
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.
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 $\mathcal{A}_2$.

$\begingroup$ 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 '10 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. $\endgroup$ – Pete L. Clark Nov 12 '10 at 22:55