I would like to ask your opinion on the point that looks simple.
Consider the group of orthogonal matrices of order n over the field R of reals, equipped with the topology induced by the Euclidean norm of matrices.
Let g be one of such matrix and denote by X the topological closure of the cyclic group generated by g.
My question: May I find a matrix g with the property that:
if the identity matrix I is the accumulation point in X of some sequence of powers of g, then the sequence is definitely trivial, that is, all the elements (except finitely many) equal I?
In other words, can I exclude that, for a suitable g, the identity I is obtained as the accumulation point of a non trivial sequence in X?
I don't think such g exists but I'm not be able to exclude it with a direct argument.
Thank you very much for help.
Let $G$ be a compact Hausdorff group (in your case the group of orthogonal matrices) and let $g\in G$. Of course $\overline{\langle g\rangle}$ is a compact subgroup of $G$.
If $|g|<\infty$ then $\overline{\langle g\rangle}=\langle g\rangle$ is finite and therefore the identity (and every other element of $\overline{\langle g\rangle}$) is isolated.
On the other hand, if $|g|=\infty$ then no element of $\overline{\langle g\rangle}$ can be isolated. That's because in general if a topological group has an isolated point then it is discrete. And discrete compact group is finite, but $\langle g\rangle$ is not finite. Therefore if $|g|=\infty$ then the identity (and every other element of $\overline{\langle g\rangle}$) can be approximated by non-constant sequence from $\langle g\rangle$.
An important example of those situations is with the group of rotations of plane, which is a subgroup of $O(2)$ topologically isomorphic to the circle group $S^1$. For a given real $r\in\mathbb{R}$ define $g_r$ to be the rotation by $2\pi r$ radians. Then $\langle g_r\rangle$ is finite if and only if $r\in\mathbb{Q}$. And if $r\in\mathbb{R}\backslash\mathbb{Q}$ then $\overline{\langle g_r\rangle}=S^1$.