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By Borel's theorem, lattices in simple Lie groups are Zariski dense. I expect that a small (in metric sense) deformation of a lattice in a Lie group is also Zariski dense.

Suppose we have a Zariski dense subgroup $\Gamma$ in a non-compact simple algebraic Lie group $G$. Suppose that we took a subset $\Lambda\subset G$ such that its $r$-neighbourhood contains $\Gamma$. Is the subgroup generated by $\Lambda$ Zariski dense in $G$?

More formally,

CONJECTURE: Let $\Gamma \subset G$ be a Zariski dense subgroup of a non-compact simple algebraic Lie group, and $B_r\subset G$ an open ball in a left-invariant metric. Let $\Lambda\subset G$ be a subset such that $B_r\!\cdot\!\Lambda$ contains $\Gamma$. I conjecture that the group generated by $\Lambda$ is Zariski dense in $G$.

Is it known? I have several ideas how it can be proven, but the "proof" that I have looks ugly, and I have a feeling that the result is known.

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This is an update, sorry for the trouble.

Let $M$ be a metric space. We say that $X \subset M$ is coarse equivalent to $Y \subset M$ if the the Hausdorff distance $d_H(X, Y)$ is finite, that is, there exists $C>0$ such that the $C$-neighbourhood of $X$ contains $Y$, and the $C$-neighbourhood of $Y$ contains $X$.

Let $G$ be a non-compact, simple real algebraic Lie group, and $d$ a right-invariant Riemannian metric, which is a posteriori complete on $G$. A subgroup $\Lambda\subset G$ is called coarse Zariski dense if any coarse equivalent subset $\Lambda' \subset G$ generates a Zariski dense subgroup in $G$.

In this language, my conjecture is equivalent to the following.

CONJECTURE: Any Zariski dense subgroup of a simple, non-compact algebraic Lie group is coarse Zariski dense.

This conjecture is false!

I have re-done the calculations and found that this conjecture is false, for example, for $SL(2,R)$, because its Borel subgroup is dense and coarse equivalent to $SL(2,R)$ (unless I made an error in this computation). I guess it is false for most or all simple groups as well. However, it seems to be true for all nilpotent and some or all solvable algebraic groups.

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    $\begingroup$ No Borel subgroup is not coarse equivalent to SL(2,R). Your conjecture is easily seen to be true for all real rank 1 Lie groups. It is most likely true in general as well. $\endgroup$ Commented Aug 4, 2022 at 0:18
  • $\begingroup$ thanks! but how can this happen if the quotient $SL(2,R)/B_+$ is compact? Also, I would be very grateful for any reference to anything related to the stuff $\endgroup$ Commented Aug 4, 2022 at 14:00

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