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 1


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.

  • 2
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.