Regardless of the stability of the topological glass, one can always increase disorder enough to ensure that defects will definitely unbind. How does the system behave in this case?
In the CDW or simple XY case, the system is simply fully disordered. Because there are no symmetries to distinguish it, we expect that the high-temperature liquid phase simply extends down to T=0, with only a gradual decrease in fluctuations as T is lowered.
The three dimensional vortex lattice, however, is a different case. Consider the symmetry properties of the topological glass phase. We expect some sort of slow decay (power-law?) of the translational correlations,
This also implies long-range hexatic order,
What about the ``boson'' or ``vacancy'' order parameter b? Well, in the perfectly ordered lattice, vacancies and interstitials are excluded, since they cost some finite energy per unit length, which implies
Note that there are several symmetries which distinguish the
topological glass from the liquid phase, which has 
and 
. Thus, even if vortices unbind and render 
, the system can remain ordered in other senses. One can,
for instance, imagine a hexatic glass with 
and 
.
The most isotropic phase possible has 
, i.e. no spatial order whatsoever, but
still maintains . This possibility is known as
a vortex glass phase[14]. It may also be thought of in
the original Ginzburg-Landau model language. The vortex glass is then
a state with Edwards-Anderson order in the Ginzburg-Landau field,
We note in passing that some care must be taken to make the previous
statement gauge-invariant, but this can be taken care of
satisfactorily. M.P.A. Fisher and D.-H. Lee have shown that 
is
actually related via a duality transformation to b, which also
explains the existence of the complex b order parameter.
* Phase diagrams?