# A group in a neighbourhood of a Zariski dense subgroup

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.

## 1 Answer

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.

• 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. Commented Aug 4, 2022 at 0:18
• 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 Commented Aug 4, 2022 at 14:00