Is every lattice in a Lie group Ad-irreducible?
No. This is false for $ G $ compact because any finite subgroup is a lattice. And it is certainly false if $ G $ is not simple since a group can only contain an Ad-irreducible subgroup if it is itself simple. So I'm asking...
Is every lattice in a non compact simple Lie group Ad-irreducible?
In other words, let $ G $ be a non compact simple Lie group. Let $ \Gamma $ be a lattice in $ G $. Must the the conjugation action of $ \Gamma $ on the lie algebra of $ G $ be an irreducible representation?
For example, I think it is true that all lattices in $ SL_2(\mathbb{R}) $ and $ SL_2(\mathbb{C}) $ are Ad-irreducible.
The desired fact follows from a Borel density theorem type result. Thm 3 of
https://www3.nd.edu/~andyp/notes/BorelDensity.pdf
states "Let G be a connected semisimple $\mathbb{R}$-algebraic group and let Γ < G be a lattice. Let ϕ : G → $GL_m(\mathbb{R})$ be an irreducible polynomial representation of G. Then the restriction of ϕ to Γ is irreducible." (this theorem should also assume that $ G $ has no compact factors, but that is sort of implied by context, for example theorem 1 earlier in the reference specifies that $ G $ has no compact factors)
Every noncompact simple Lie group, other than the universal cover of $ SL_2(\mathbb{R}) $, is a connected semisimple $\mathbb{R}$-algebraic group. For a simple group the adjoint representation is irreducible by definition. Also the adjoint representation is always a polynomial representation. Thus we can apply the theorem to conclude that the adjoint representation restricted to $ \Gamma $ must be irreducible.
The last case to consider is when $ G $ is not algebraic because it is the universal cover of $ SL_2(\mathbb{R}) $. In this case the adjoint representation still factors through actual $ SL_2(\mathbb{R}) $ and any lattice $ \tilde{\Gamma} $ in the universal cover factors through a lattice $ \Gamma $ in actual $ SL_2(\mathbb{R}) $. So the result is also true for all lattices in the universal cover of $ SL_2(\mathbb{R}) $.