I wonder if there is a simple instance of the following phenomena : an abstract subgroup S $\subset GL\_n(\mathbb{C})$ whose $\mathbb{Q}$-Zariski closure isn't a group ?


Is there some criteria to ensure this is not happening ? (Maybe under some reductivity hypothesis, e.g. Mumford-Tate groups ?)


(there is a non intuitive example in exercice 11 of [https://www2.math.upenn.edu/~chai/620\_00/hw1.pdf](https://www2.math.upenn.edu/%7Echai/620_00/hw1.pdf)).


Thanks !



$\mathbb{Q}$-Zariski Closure not equal to smallest Q-subgroup
Sure. Take some elements $A\_1,\dots, A\_n$ that generate a dense subset $\Gamma$ of a non-abelian connected reductive group $G \subseteq GL\_n$.


For instance, I can take $$A\_1=\begin{pmatrix} 1 &1 \\ 0 & 1 \end{pmatrix}, A\_2 = 
 \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$$ which generate $\Gamma = SL\_2(\mathbb Z)$ which is Zariski dense in $G = SL\_2$.


Now take a parametric curve $A(t)$ in $G$ which passes through $A\_1, \dots ,A\_n$. For example, I can take $$ A(t) =\begin{pmatrix} t & 1 \\ t^2-1 & t \end{pmatrix}$$ which gives $A\_1$ for $t=1$ and $A\_2$ for $t=0$.


Now just take the group generated by $A(\pi)$, or use your favorite transcendental number.


Let $T$ be the torus generated by $A(\pi)$, which is defined over $\pi$, and defines a point in the space $G/ N(T)$ of toruses in $G$ conjugate to $T$. The $\mathbb Q$-Zariski closure of this point in $G/N(T)$ is a one-dimensional curve, since the point is defined over $\mathbb Q(\pi)$, which has transcendence degree $1$. If we take the inverse image of this curve in the universal family of tori and project to $G$, we get a $\dim T +1$-dimensional subset of $G$, defined over $\mathbb Q$, containing all powers of $A(\pi)$, and therefore the $\mathbb Q$-Zariski closure of the subgroup generated by $A(\pi)$ has dimension $\leq \dim T+1$ and can't contain all of $G$.


But by construction, the minimal $\mathbb Q$-subgroup containing $A(\pi)$ contains $A\_1,\dots, A\_n$ and thus is $G$.


For example, in the $SL\_2$ case, the Zariski closure consists of matrices whose two diagonal entries are equal, a set which is closed under taking powers, but not products.


Because this works for any non-abelian connected reductive group, there is no reasonable condition on the Mumford-Tate group that rules it out, although $G = \mathbb G\_m^n$ is OK because all its Zariski closed subgroups are defined over $\mathbb Q$.


