TITLE: Zariski density of conjugates of $SL_2(\mathbb{Z})$ in $Sp_{2g}$
QUESTION [6 upvotes]: Let $Sp_2g$ be the symplectic group defined over $\mathbb{Q}$. Consider $SL_2(\mathbb{Z})$ as a subgroup of $Sp_{2g}(\mathbb{Z})$ (the embedding that I have in mind is $A\to \begin{pmatrix} 
A& 0\\
0& I_{2g-2}
\end{pmatrix}$, but you can take any algebraic group embedding which yields a positive answer to the following question).
Is it true that there eixst finitely many $g_i\in Sp_{2g}(\mathbb{Q})$ such that the $\mathbb{Q}$-Zariski closure of $\cup_i g_i SL_2(\mathbb{Z})g_i^{-1}$ is equal to $Sp_{2g}$? i.e. that this union is Zariski-dense in $Sp_{2g}$? If yes, what conditions and restrictions should these elements satisfy? How should the $g_i$ be chosen? I would appreciate any result, comments or reference in this direction.

REPLY [5 votes]: Even for any algebraic group embedding of $SL_2$ into $Sp_{2g}$, with an arbitrary amount of conjugation by arbitrary elements, the answer is no. Consider the matrix $\begin{pmatrix} \lambda & 0 \\ 0 & \lambda^{-1} \end{pmatrix}$ in $SL_2$. Its image under the algebraic group embedding is some $2g \times 2g$ matrix whose eigenvalues are powers of $\lambda$. The characteristic polynomial of this matrix is some $2g$-tuple of polynomials in $\lambda$, all defined over $\mathbb Z$, whose image is a $1$-dimensional subvariety, defined over $\mathbb Q$, of the space of polynomials.
Because elements conjugate to this one are dense in $SL_2(\mathbb C)$, the image of every element of $SL_2(\mathbb C)$ under the characteristic polynomial map is contained in this $1$-dimensional $\mathbb Q$-subvariety. The same is true for any number (even an infinite number) of conjugate copies of $SL_2$.  Because the characteristic polynomials of elements of $Sp_{2g}$ form a $g$-dimensional variety, there is no way this union can be $\mathbb Q$-Zariski dense.