A symmetric operator on a Hilbert space, with deficiency indices (m; m) has self-adjoint extensions. These are ‘highly reducible’. The original operator may be irreducible, (see example (i), below). Can the mechanism whereby reducibility is achieved be understood? The concrete examples most readily studied are those associated with differential operators. It is easy to obtain operators, associated with a formal linear differential operator, having deficiency indices (m; m). What of reducibility? Nothing seems to be known. In the case of the first-order operator we were able, using the Volterra operator, to establish irreducibility of the associated minimal operator. To investigate symmetric operators associated with a second-order differential operator, different methods had to be developed. They apply also to the first-order operator, and we employ them to demonstrate the irreducibility of the associated minimal operator. In the second-order case the minimal operator proves reducible, and we also exhibit examples of reducibility of associated symmetric operators. It would clearly be of interest to elucidate the influence of the boundary conditions on reducibility.