Spontaneous symmetry breaking in inflationary cosmology: on the fate of
  Goldstone Bosons
  We argue that in an inflationary cosmology a consequence of the lack of time
translational invariance is that spontaneous breaking of a continuous symmetry
and Goldstone's theorem \emph{do not} imply the existence of \emph{massless}
Goldstone modes. We study spontaneous symmetry breaking in an O(2) model, and
implications for O(N) in de Sitter space time. The Goldstone mode acquires a
radiatively generated mass as a consequence of infrared divergences, and the
continuous symmetry is spontaneously broken for any finite $N$, however there
is a \emph{first order phase transition} as a function of the Hawking
temperature $T_H=H/2\pi$. For O(2) the symmetry is spontaneously broken for
$T_H < T_c= \lambda^{1/4} v/2.419$ where $\lambda$ is the quartic coupling and
$v$ is the tree level vacuum expectation value and the Goldstone mode acquires
a radiatively generated mass $\mathcal{M}^2_\pi \propto \lambda^{1/4} H$. The
first order nature of the transition is a consequence of the strong infrared
behavior of minimally coupled scalar fields in de Sitter space time, the jump
in the order parameter at $T_H=T_c$ is $\sigma_{0c} \simeq 0.61\,
{H}/{\lambda^{1/4}}$. In the strict $N\rightarrow \infty$ the symmetry cannot
be spontaneously broken. Furthermore, the lack of kinematic thresholds imply
that the Goldstone modes \emph{decay} into Goldstone and Higgs modes by
emission and absorption of superhorizon quanta.
