Unitary representations of free groups I was wondering if there existed finite representations of a free group of rank n? Moreover, I was wondering if there existed unitary representations of free groups of rank n? I am looking to explicitly construct the homomorphism. One idea I had was to obtain a faithful representation by explicitly mapping generators of the free group of rank n, to generators of free subgroups of rank n in $\mathrm{GL}_n(\mathbb{C})$. Its not so clear if the same type of argument would hold for $\mathrm{U}_n(\mathbb{C})$, as I am unsure if $\mathrm{U}_n(\mathbb{C})$ is a free group of rank n.
Any help would be greatly appreciated.
 A: Here is  another concrete example of a free group in $U_4$.
Let $\cos\theta$ be a transcendental number and let
$$
A=
\begin{pmatrix}
  \cos\theta & \sin\theta & 0 & 0\\
  -\sin\theta & \cos\theta & 0 & 0\\
  0 & 0 & \cos\theta & -\sin\theta \\
  0 & 0 & \sin\theta &  \cos\theta     
\end{pmatrix},\quad
B=
\begin{pmatrix}
  \cos\theta & 0 & \sin\theta & 0\\
  0 & \cos\theta & 0 & \sin\theta\\
  -\sin\theta & 0 & \cos\theta & 0 \\
  0 & -\sin\theta &0 &  \cos\theta
\end{pmatrix}.
$$
The matrices $A$ and $B$ are matrices in the canonical basis $1,i,j,k$ of the following quaternions:
$$
a=\cos\theta+i\sin\theta,\quad b=\cos\theta+j\sin\theta.
$$
The group $\operatorname{gr}(A,B)$ is free.
Of course, it is necessary to find $\theta$ such that the number $\cos\theta$ is transcendent.
Addendum.
This free group was constructed in 1956 by Th. J. Dekker in Decompositions of sets and spaces II, Indagationes Math. 18, 581-595. In the same paper it is proved that the group $\operatorname{gr}(C,D)$ is also a free group:
$$
C=
\begin{pmatrix}
  \cos\theta & \sin\theta & 0 \\
  -\sin\theta & \cos\theta & 0 \\
  0 & 0 & 1 \\
\end{pmatrix},\quad
D=
\begin{pmatrix}
  1 & 0 &\\
  0 & \cos\theta & \sin\theta \\
  0 & -\sin\theta & \cos\theta 
\end{pmatrix}.
$$
