Adjoint-irreducible subgroup is either finite or Zariski dense $ G $ is a linear algebraic group whose real points are connected.
$ H $ is a subgroup.
The adjoint action of $ H $ on the Lie algebra of $ G $ is irreducible.
Is it true that $ H $ is either finite or it must be Zariski dense in $ G $?
Some background:
This seems to be true for $ SU_2, SO_3 $ with the only ad irreducible subgroups being the three symmetry groups of the platonic solids (which are finite) and then the whole group.
It also seems true for non compact groups like $ SL_2 $ since I think every Ad-irreducible subgroup of $ SL_2 $ is Zariski dense
 A: Here is my attempt at an answer using the hint from Moishe Kohan.
Let $ H $ be an Ad-irreducible subgroup of a connected group $ G $. If $ H $ is not already an algebraic subgroup then take its Zariski closure $ \overline{H} $. Since $ H $ is Ad-irreducible so is $ \overline{H} $.
An algebraic subgroup of positive dimension and infinite index cannot be Ad-irreducible. Since $ \overline{H} $ is an Ad-irreducible algebraic subgroup then we must have that either $ {\rm dim}(\overline{H})=0 $ or the index $ [G:\overline{H}] $ is finite.
Since $ \overline{H} $ is an algebraic group it can only have finitely many connected components. But since $ \overline{H} $ is discrete every element is a connected component. Thus ${\rm dim}(\overline{H})=0 $ implies $ \overline{H} $ finite which implies $ H $ finite which implies ${\rm dim}(\overline{H})=0 $ (since finite sets are Zariski closed) thus all three statements are equivalent.
Now consider the condition $ [G:\overline{H}] $ is finite. Since we assumed $ G $ is connected that implies $ G/\overline{H} $ is a finite connected manifold thus trivial. So in fact $ G=\overline{H} $. Which is equivalent to saying that $ H $ is Zariski dense.
Thus we have that an ad irreducible subgroup $ H $ of a connected group $ G $ is either finite or Zariski dense.
Stated in a slightly more general way, this shows that an ad irreducible subgroup $ H $ of $ G $ is either finite or $ H $ is Zariski dense in every component of $ G $ that intersects $ H $.
EDIT:
It is interesting to note that for the special case that $ G $ is compact then corollary 3.5 of this paper
https://arxiv.org/pdf/1609.05780.pdf
states that a subgroup $ H $ of a connected compact simple group $ G $ is dense if and only if it is infinite and Ad-irreducible.
So if $ H $ is an ad irreducible subgroup of a connected compact group $ G $ then $ H $ ad irreducible implies $ G $ ad irreducible implies $ G $ simple. So now we have that $ G $ is simple compact and connected. So $ H $ is infinite if and only if it is dense.
In other words, either $ H $ is finite or it is dense and dense implies Zariski dense. Thus corollary 3.5 provides a proof of a stronger but more restricted result (if we strength the assumption to $ G $ compact then in the conclusion of the theorem we can replace Zariski dense with dense(in the manifold topology).
