Let $F_2$ be the free group on two generators. If $\pi$ is the universal unitary representation of $F_2$ on a Hilbert space ${\cal H}$, then the group $C^*$-algebra $C^*(F_2)$ is the $C^*$-subalgebra of $B({\cal H})$ generated by $\{\pi(g):g\in F_2\}$.
Alternatively:
Let $U,V$ be two unitary operators on a Hilbert space ${\cal H}$. Then $(U,V)$ is a universal pair of unitaries if for each pair of unitary operators $(U_1,V_1)$ on a Hilbert space ${\cal H}_1$, the assignment $U\mapsto U_1$ and $V\mapsto V_1$ extends to a $*$-homomorphism from $C^*(U,V)$ onto $C^*(U_1,V_1)$. In this case, $C^*(F_2)$ is the $C^*$-algebra generated by a universal pair of unitaries.
For the existence of such universal pair, M.-D. Choi said that we may simply let $U=\oplus U_\alpha$ and $V=\oplus V_\alpha$ where $(U_\alpha,V_\alpha)$ runs through all possible pairs of unitary operators on a fixed separable Hilbert space.
Question: In the first definition, in my mind we can safely assume that the two generators of $F_2$ do not have dependence. How do we make sure that this is also case for the direct sum construction above? In particular, say if I already know $U=\oplus U_\alpha$, then how do I form $V$? I think we cannot have $V$ to be a multiple of $U$ for example.