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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?

Arctic Char
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user45765
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  • What do you mean by "algebraic"? Is it: the complex points of an locally closed algebraic subvariety? – Bart Michels Mar 01 '20 at 19:57
  • @punctureddusk I mean algebraic as a variety over field $k$ equipped with hopf algebra over structure on ring of functions over base field $k$. – user45765 Mar 01 '20 at 19:58
  • Ok, but you certainly want to impose some compatibility with the algebraic variety structure of $GL_n(\mathbb C)$. What exactly do you mean by "as a variety"? – Bart Michels Mar 01 '20 at 20:00
  • @punctureddusk For an affine variety $V$, suppose $k[V]$ ring of functions is equipped with hopf algebra. Then $V$ is called affine algebraic group. Now any closed subgroup of $V$ is also affine algebraic group. $GL_n(C)$ has affine structure by embedding as hypersurface of $k^l$ for some $l$. – user45765 Mar 01 '20 at 20:03
  • Aha, ok, so you want a closed algebraic subgroup. Thanks. – Bart Michels Mar 01 '20 at 20:05
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    A countable infinite subgroup is never algebraic (the set of complex points of an arbitrary variety is never infinite countable). – YCor Mar 01 '20 at 21:37

3 Answers3

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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.

Bart Michels
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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.

KReiser
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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.

KReiser
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Tsemo Aristide
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  • Your second link does not have any (obvious) relation to the claim immediately before it, which also requires adjustment: you should have $\Bbb C^n$ there, and you should also specify the topology you're working with - this statement is not true in the context of schemes. – KReiser Mar 01 '20 at 19:57
  • I would say that the ring of regular functions $k[x_1,\ldots,x_m]/I$ is an algebraic extension of $S=k[x_{e_1},\ldots,x_{e_n}],n\ge 1$. Let $U$ be $\Bbb{C}^n$ minus the zeros of the discriminants and of the denominators of the coefficients of the $Frac(S)$ monic minimal polynomials of the $x_j$. Let $f$ be the map sending $a\in G$ to $(a_{e_1},\ldots,a_{e_n})$. Then $f$ is a finite analytic covering $f^{-1}(U)\to U$. – reuns Mar 01 '20 at 20:07
  • @reuns I am very familiar with how the statement could be made to work, but this may not necessarily be the case for every reader. The point is that as currently written in this answer, the explanation could use some clarification. – KReiser Mar 01 '20 at 20:10