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A number is in BA if its orbit is bounded. Any such orbit closure must contain a full $A=\langle g_t\rangle$ orbit. By examining the possible subgroups, any such hypothetical $H$, as a stability group, must contain $A$ and can only be either $A$ or $B$ (the Borel subgroup) or all of $G$. As $B$ does not support a lattice, so cannot have a closed orbit, so if not $A$ it must be $G$. If the stability is whole of $G$, as $G$ acts transitively on $G/\Gamma$, the orbit is unbounded.

So we relaxed the problem of finding closed $A$ orbits (namely closed geodesics). This amounts to closed orbits of (real split) tori, which is well known to come from real quadratic fields.

In general, your hope of finding such non-trivial equidistribution results (pointwise) is slim as the action is Bernoulli (c.f. Katok's book), so in particular one can arrange an orbit closure of any arbitrary dimension...

A number is in BA if its orbit is bounded. Any such orbit closure must contain a full $A=\langle g_t\rangle$ orbit. By examining the possible subgroups, any such hypothetical $H$, as a stability group, must contain $A$ and can only be either $A$ or $B$ (the Borel subgroup) or all of $G$. As $B$ does not support a lattice, so cannot have a closed orbit, so if not $A$ it must be $G$. If the stability is whole of $G$, as $G$ acts transitively on $G/\Gamma$, the orbit is unbounded.

So we relaxed the problem of finding closed $A$ orbits (namely closed geodesics). This amounts to closed orbits of (real split) tori, which is well known to come from real quadratic fields.

In general, your hope of finding such non-trivial equidistribution results (pointwise) is slim as the action is Bernoulli (c.f. Katok's book), so in particular one can arrange an orbit closure of any arbitrary dimension...

P.S. You mentioned the pointwise ergodic theorem. It refers (in the standard formulation) to Haar-generic points, where in that case $H=G$, so in particular the orbit $H.x$ is unbounded (hence it does not tell us anything about BA points). Obviously you can apply the pointwise theorem for the case of a closed $A$ orbit (but then you are just having a translation over a torus and it is not really interesting).

A number is in BA if its orbit is bounded. Any such orbit closure must contain a full $A=\langle g_t\rangle$ orbit. By examining the possible subgroups, any such hypothetical $H$, as a stability group, must contain $A$ and can only be either $A$ or $B$ (the Borel subgroup) or all of $G$. As $B$ does not support a lattice, so cannot have a closed orbit, so if not $A$ it must be $G$. If the stability is whole of $G$, as $G$ acts transitively on $G/\Gamma$, the orbit is unbounded.

So we relaxed the problem of finding closed $A$ orbits (namely closed geodesics). This amounts to closed orbits of (real split) tori, which is well known to come from real quadratic fields.

In general, your hope of finding such non-trivial equidistribution results (pointwise) is slim as the action is Bernoulli (c.f. Katok's book).

A number is in BA if its orbit is bounded. Any such orbit closure must contain a full $A=\langle g_t\rangle$ orbit. By examining the possible subgroups, any such hypothetical $H$, as a stability group, must contain $A$ and can only be either $A$ or $B$ (the Borel subgroup) or all of $G$. As $B$ does not support a lattice, so cannot have a closed orbit, so if not $A$ it must be $G$. If the stability is whole of $G$, as $G$ acts transitively on $G/\Gamma$, the orbit is unbounded.

So we relaxed the problem of finding closed $A$ orbits (namely closed geodesics). This amounts to closed orbits of (real split) tori, which is well known to come from real quadratic fields.

In general, your hope of finding such non-trivial equidistribution results (pointwise) is slim as the action is Bernoulli (c.f. Katok's book), so in particular one can arrange an orbit closure of any arbitrary dimension...

A number is in BA if its orbit is bounded. Any such orbit closure must contain a full A orbit. By examining the possible subgroups, any such hypothetical $H$, as a stability group, must contain $A$ and can only be either $A$ or $B$ (the Borel subgroup) or all of $G$. As $B$ does not support a lattice, so cannot have a closed orbit, so if not $A$ it must be $G$. If the stability is whole of $G$, as $G$ acts transitively on $G/\Gamma$, the orbit is unbounded.

So we relaxed the problem of finding closed $A$ orbits (namely closed geodesics). This amounts to closed orbits of (real split) tori, which is well known to come from real quadratic fields.

In general, your hope of finding such non-trivial equidistribution results (pointwise) is slim as the action is Bernoulli (c.f. Katok's book).

A number is in BA if its orbit is bounded. Any such orbit closure must contain a full $A=\langle g_t\rangle$ orbit. By examining the possible subgroups, any such hypothetical $H$, as a stability group, must contain $A$ and can only be either $A$ or $B$ (the Borel subgroup) or all of $G$. As $B$ does not support a lattice, so cannot have a closed orbit, so if not $A$ it must be $G$. If the stability is whole of $G$, as $G$ acts transitively on $G/\Gamma$, the orbit is unbounded.

So we relaxed the problem of finding closed $A$ orbits (namely closed geodesics). This amounts to closed orbits of (real split) tori, which is well known to come from real quadratic fields.

In general, your hope of finding such non-trivial equidistribution results (pointwise) is slim as the action is Bernoulli (c.f. Katok's book).