No. It's irreducible. The element $U$ generates the maximal abelian subalgebra $L^\infty({\mathbb T})$ and hence one computes the commutant:
$$\{U,V\}'=\{U\}'\cap\{V\}'=L^\infty({\mathbb T})\cap\{V\}'={\mathbb C}1.$$
By the way, the invariant subspace problem for $UV$ is still open in full generality. 
https://mathscinet.ams.org/mathscinet-getitem?mr=353015