When minimizing a scalar potential how come we are allowed to set the VEV of charged components to zero?

I stumbled upon this when studying the type-II seesaw model in this paper (though the question should be self-contained). The potential couples a Higgs scalar, $ \Phi = \left( \phi ^+ , \phi ^0 \right) $ with the new isospin triplet, $ \xi = \left( \xi ^{ + + } , \xi ^+ , \xi ^0 \right) $. The charges on the scalars indicate their $ T _3 + Y $ values and hence their charge after EW symmetry breaking. The most general potential is,

\begin{align}

V &= m ^2 \Phi ^\dagger \Phi + M ^2 \xi ^\dagger \xi + \frac{1}{2} \lambda _1 ( \Phi ^\dagger \Phi ) ^2 + \frac{1}{2} \lambda _2 \left( \xi ^\dagger \xi \right) ^2 + \lambda _3 \left( \Phi ^\dagger \Phi \right) \left( \xi ^\dagger \xi \right) \\

& \qquad + \mu \left( \bar{\xi} ^0 \phi ^0 \phi ^0 + \sqrt{2} \xi ^- \phi ^+ \phi ^0 + \xi ^{ - - } \phi ^+ \phi ^+ h.c. \right)

\end{align}

Then instead of doing a 10 (4 degrees of freedom in $ \Phi $ and 6 in $ \xi $) field minimization, the authors write $ \left\langle \phi ^0 \right\rangle = v $ and $ \left\langle \xi ^0 \right\rangle = u $. They completely ignore the possibility of the charged components getting a VEV. Now of course, these should not get a VEV to break $ SU(2) \times U(1) _Y \rightarrow U(1) _{ EM } $, but shouldn't this be a output of the model instead of something that is forced upon it? Otherwise I think the resulting potential could be unstable since it may not be symmetric around the charged components.

Note that this situation is distinct from the Higgs field where we are able to choose where the VEV's go since the Higgs is the only scalar and can be transformed using the SU(2)xU(1) gauge to get a VEV in whichever component we please.

**Edit: **I think they are also setting the VEV of the imaginary component of each uncharged field to zero. Is there a reason why this would be justified as well?