  We analyze here the consequence of local rotational-symmetry breaking in the
quantum spin (or phase) glass state of the quantum random rotor model. By
coupling the spin glass order parameter directly to a vector potential, we are
able to compute whether the system is resilient (that is, possesses a phase
stiffness) to a uniform rotation in the presence of random anisotropy. We show
explicitly that the O(2) vector spin glass has no electromagnetic response
indicative of a superconductor at mean-field and beyond, suggesting the absence
of phase stiffness. This result confirms our earlier finding (PRL, {\bf 89},
27001 (2002)) that the phase glass is metallic, due to the main contribution to
the conductivity arising from fluctuations of the superconducting order
parameter. In addition, our finding that the spin stiffness vanishes in the
quantum rotor glass is consistent with the absence of a transverse stiffness in
the Heisenberg spin glass found by Feigelman and Tsvelik (Sov. Phys. JETP, {\bf
50}, 1222 (1979).
