Imaginary quadratic field contained in Hecke orbit field? Let $\tau$ in the upper half plane lie in an imaginary quadratic field $K$. 
Then is $K \subset \mathbb{Q}(\{j(g \tau) \ | \ g \in GL_2^+(\mathbb{Q}) \})$?
(here $j$ is the modular $j$-function, and $GL_2^+(\mathbb{Q})$ means positive determinant - i.e. we adjoined all $j$ values of elliptic curves with CM by an order in $K$).
If the above isn't true, then is $K$ contained in $\mathbb{Q}(S)$ where $S$ is the set of $j$ values of all CM elliptic curves?
I'm also interested in whether the appropriate statement is true for a general Shimura variety.
 A: "No" for both questions about CM elliptic curves and "I'm not even sure I know what the question would be" about general Shimura varieties.
Basic idea: If $j$ is the $j$-invariant of a CM elliptic curve, then there is some imaginary quadratic discriminant $\Delta \equiv 0$ or $1 \bmod 4$ such that $\mathbf{Q}(j) \cong \mathbf{Q}[X]/H_\Delta(X)$ where $H_\Delta(X) \in \mathbf{Z}[X]$ is the Hilbert Class Polynomial of discriminant $\Delta$, whose roots are the $j$-invariants of elliptic curves over the complex numbers (actually $\overline{\mathbf{Q}}$ is enough) with CM by $\mathbf{Z}\left[\dfrac{ \Delta + \sqrt \Delta}{2}\right]$.
The point is that now we can see that there is an embedding $\mathbf{Q}(j) \hookrightarrow \mathbf{R}$, for all possible $j$. Therefore there is an embedding $\mathbf{Q}(S) \hookrightarrow \mathbf{R}$. To see this, it's enough to note that for any two CM $j$-invariants $j_1$ and $j_2$ that there exists an embedding $\mathbf{Q}(j_1,j_2)\hookrightarrow \mathbf{R}$. Let $J_1$ and $J_2$ be the canonical image of $j_1$ and $j_2$ in the real numbers. Then $\mathbf{Q}(j_1)$ embeds into the real numbers as $\mathbf{Q} + \mathbf{Q}J_1 + \dots + \mathbf{Q}J_1^{h_1 -1}$ and $\mathbf{Q}(j_2)$ embeds into the real numbers as $\mathbf{Q} + \mathbf{Q}J_2 + \dots + \mathbf{Q}J_2^{h_2 -1}$. Therefore $\mathbf{Q} + \mathbf{Q}J_1 + \mathbf{Q}J_2 + \dots + \mathbf{Q}J_1^{h_1 -1}J_2^{h_2 -1}$ is a copy of $\mathbf{Q}(j_1,j_2)$ inside of $\mathbf{R}$. Notice that I didn't use direct sums because $\mathbf{Q}(j_1)$ and $\mathbf{Q}(j_2)$ might not be linearly disjoint over $\mathbf{Q}$! This is also the reason I didn't use a tensor product argument. In any case, this inductive step allows us to work with direct limits and embed $\mathbf{Q}(S)$ into $\mathbf{R}$.
Therefore if we assume that there is an embedding $K\hookrightarrow \mathbf{Q}(S)$ then there must be an embedding $K\hookrightarrow\mathbf{Q}(S) \hookrightarrow \mathbf{R}$, which is absurd. Therefore, there is no embedding $K\hookrightarrow \mathbf{Q}(S)$.
To show that we have an embedding $\mathbf{Q}(j)\hookrightarrow\mathbf{R}$, consider that there is some $\tau \in \mathcal{H}$ of the form $\dfrac{1}{2}\sqrt\Delta$ or $\dfrac{ 1 + \sqrt \Delta}{2}$ such that $j(\tau)$ is a root of $H_\Delta(X)$. But then $j(\tau)$ is real, because the inverse image of the reals under the $j$-function contains the lines $\lbrace iy : y \ge 1\rbrace$ and $\lbrace 1/2 + iy : y \ge (1/2)\sqrt 3\rbrace$. Therefore we have our embedding $\mathbf{Q}(j) \hookrightarrow \mathbf{R}$.
