Let $G$ be a discrete infinite subgroup of $GL_n(\mathbb C)$. Is $G$ not algebraic in general? Let $G$ be a discrete subgroup of $GL_n(\mathbb C)$.
Q: If $G$ has infinite order elements, is $G$ algebraic?
I do not think this is the case and it seems that $G$ fails to be algebraic for most of the classical groups I have seen. Consider $SL_2(\mathbb Z)\leq GL_n(\mathbb C)$. Clearly the same defining equations for $SL_2(\mathbb Z)$ will pin down all elements of $SL_2(\mathbb C)$ as the equations have real coefficients. In other words, having integer entries is not an algebraic condition. So any level $N$, the level $N$ congruence subgroup $\Gamma_N\leq SL_2(\mathbb Z)$ is not algebraic as well.
Q': Suppose $G$ is a non-discrete subgroup of $GL_n(\mathbb C)$. Say $G = SL_2(S)$ where $S$ is the integral closure of $\mathbb Q(\sqrt{5})$. Then I do not think this $G$ will be algebraic either. How do I see it is not algebraic?
 A: It is true that $\operatorname{SL}_n(\mathbb Z)$ cannot be the group of complex points of a closed algebraic subgroup of $\operatorname{SL}_n(\mathbb C)$. This is because $\operatorname{SL}_n(\mathbb Z)$ is Zariski-dense in $\operatorname{SL}_n(\mathbb C)$. This boils down to the following fact:
If $P \in \mathbb C[X_1, \ldots, X_n]$ is a polynomial that vanishes on $\mathbb Z^n$, then $P = 0$.
You can prove this by induction on $n$.
For the other example, given a (sufficiently natural) embedding $\operatorname{SL}_n(\mathcal O_{\mathbb Q(\sqrt 5)}) \subset \operatorname{SL}_n(\mathbb C)$ you can use a similar density argument to show that the image cannot be the group of complex points of a closed algebraic subgroup of $\operatorname{SL}_n(\mathbb C)$.
A completely different question is whether those groups are arithmetic.
A: Let $R$ be a noetherian ring. Then $GL_n(R)=\operatorname{Spec} R[x_{ij}]_{\det}$ is a noetherian scheme, and therefore every subset of $GL_n(R)$ with the subspace topology is again noetherian. In particular, this means any immersed discrete set must be finite. So we see immediately that in your case no discrete subgroup with an element of infinite order is algebraic.
For question 2, we can use the same idea if we're clever about it. Consider the intersection of $SL_2(S)$ with the closed set $L\subset GL_2(\Bbb C)$ given by $x_{11}=x_{22}=1$, $x_{21}=0$. If $SL_2(S)$ was a closed subspace, then it's intersection with $L$ would also be closed inside $L$. But this would imply that this intersection either is $L$ or is only has finitely many points inside $L$, both of which are clearly not the case. So $SL_2(S)\subset GL_2(\Bbb C)$ cannot be an algebraic subgroup.
A: Let $G$ be an algebraic group, if $\dim G=0$ then $G$ is finite.
algebraic variety of dimension 0
If $\dim G>0$, then there exists a subset of $G$ homeorphic to an open subset of $\mathbb{C}$ https://en.wikipedia.org/wiki/Complex_algebraic_variety and $G$ is not discrete, therefore, a discrete infinite complex group is not algebraic.
