Let $ SU_n(O_d) $ denote an integral unitary group of $ n \times n $ matrices over a totally real number field $ K_d:=\mathbb{Q}(\cos(\frac{2\pi }{d})) $ where $ O_d $ is the ring of integers of $ K_d $.
Question:
For which $ d $ is $ SU_n(O_d) $ dense (with respect to the manifold topology) in the full group of $ n \times n $ unitary matrices $ SU_n $?
And is it true that for every $ d $ either $ SU_n(O_d) $ is finite or $ SU_n(O_d) $ is dense in $ SU_n $?
Context:
For $ d=1,2 $ then $ O_d = \mathbb{Z} $ and $ SU_n(O_d) $ is finite.
For an example where $ SU_n(O_d) $ is dense in $ SU_n $ consider $ n=2,d=16 $ then the matrices $$ \frac{i}{\sqrt{2}}\begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} $$ and $$ \begin{bmatrix} \overline{\zeta_{16}} & 0 \\ 0 & \zeta_{16} \end{bmatrix} $$ generate a dense subgroup of $ SU_2 $. Similar examples exist for $ n $ higher powers of $ 2 $ (and one can still choose $ d=16 $).
I could have asked the same question about $ U_n(O_d) $. But I think that $ U_n(O_d) $ would never be dense in $ U_n $ because if it was then the image of $ U_n(O_d) $ in $ U_1 $ under the determinant mapping would also be dense. But I don't think that's possible since the determinant of a matrix in $ U_n(O_d) $ is in $ U_1(O_d) $ which is finite and thus not dense in $ U_1 $.
