I come quite late with my comment, how to prove the very last claim?

I did. In that case we have the action on $\text{SL}(2,\Bbb Z)$ on $\Bbb R\times\Bbb R\cong\Bbb R^2$. So there are lattices of course. If you take a vector $v\in\Bbb R^2$ such that $v$ is not a multiple of an integral vector (all entries are integers) then the orbit is dense in $\Bbb R^2$. If the entries are rational multiples of integers, instead, the orbit is discrete.

Yes, I am interested in closed subspaces in particular. Right, I forgot to write it.

Yes, maybe the term "subspace" has been misleading. I actually mean topological subspaces, where $\Bbb R^n\times \Bbb R^n$ is endowed with the product of the standard euclidean topology. The point Is that I have examples with linear subspaces and lattices.

Hi @YCor, thanks for the answer first. I have a couple of doubts. Why do you say that the invariant subspaces for the $\text{SL}(2\Bbb Z)$ action are the same as for the $G$-action? The $n$-times product of lattice is not $G$-invariant. Am I missing something? The second doubt is about the splitting, the action is diagonal on $\Bbb R^2\times\cdots\times \Bbb R^2$. So the splitting may contain factors greater than $W_1$.

@EFinat-S yes, thank a lot for point me out the reference!

I see, I check that book! Thanks a lot!

@EFinat-S, Thanks for your answer. I think you are right but I am not able to find a proof. I have checked two books: Integral Matrices by Newmann and Lectures on the Geometry of numbers by Siegel. Is the book you have mentioned one of them? I didn't find a similar result but I may miss it.