  The ground state energy of infinite symmetric nuclear matter is usually
described by strongly interacting nucleons obeying the Pauli exclusion
principle. We can imagine a unitary transformation which groups four non
identical nucleons (i.e. with different spin and isospin) close in coordinate
space. Those nucleons, being non identical, do not obey the Pauli principle,
thus their relative momenta are negligibly small (just to fulfill the
Heisenberg principle). Such a cluster can be identified with an $\alpha$ boson.
But in dense nuclear matter, those $\alpha$ particles still obey the Pauli
principle since are constituted of Fermions. The ground state energy of nuclear
matter $\alpha$ clusters is the same as for nucleons, thus it is degenerate. We
could think of $\alpha$ particles as vortices which can now braid, for instance
making $^8Be$ which leave the ground state energy unchanged. Further braiding
to heavier clusters ($^{12}C$, $^{16}O$..) could give a different
representation of the ground state at no energy cost. In contrast d-like
clusters (i.e. N=Z odd-odd nuclei, where N and Z are the neutron and proton
number respectively) cannot describe the ground state of nuclear matter and can
be formed at high excitation energies (or temperatures) only. We show that
even-even, N=Z, clusters could be classified as non-Abelian states of matter.
As a consequence an $\alpha$ condensate in nuclear matter might be hindered by
the Fermi motion, while it could be possible a condensate of $^8Be$ or heavier
clusters.
