group. In the case without spin-orbit coupling, the relevant free energy for this representation can be constructed using the known free energy for the $E_u$ representation [@RevModPhys.63.239], we list below $f^{(2)}_{in}$, $f^{(4)}_{hom}$ and $f^{(4)}_{in}$. Before listing $f^{(2)}_{in}$, we note that this free energy term should respect the $C_4$ symmetry on the $xy$ plane only for the orbital degrees of freedom: $$(D_x, D_y, \Delta_{s,+}, \Delta_{s,-}) \to (D_y, -D_x,
i\Delta_{s,+}, -i\Delta_{s,-}).$$ However, for simplicity, we consider cylindrical symmetry in the orbital degrees of freedom, this does not significantly alter the arguments below. This symmetry gives us $f^{(2)}_{in} = \sum_s f^{(2,s)}_{in}$ where $$\begin{aligned}
f^{(2,s)}_{in} &=& K_1(|{\bf D}\Delta_{s,+}|^2 + |{\bf D}\Delta_{s,-}|^2)\nonumber\\
&+& K_2[\!\{\!(D_x \Delta_{s,+})\!(D_x \Delta_{s,-})^*\!-\!(D_y \Delta_{s,+})\!(D_y \Delta_{s,-})^*\!\}\!/2\nonumber\\
&+& \!\{\!(D_x \Delta_{s,-})\!(D_x \Delta_{s,+})^*\!-\!(D_y \Delta_{s,-})\!(D_y \Delta_{s,+})^*\!\}\!/2\nonumber\\
&+& i\!\{\!(D_x \Delta_{s,-})\!(D_y \Delta_{s,+})^*\!+\!(D_y \Delta_{s,-})\!(D_x \Delta_{s,+})^*\!\}\!/2\nonumber\\
&-& i\!\{\!(D_x \Delta_{s,+})\!(D_y \Delta_{s,-})^*\!+\!(D_y \Delta_{s,+})\!(D_x \Delta_{s,-})^*\!\}\!/2]\nonumber\\
&+& K_4 (|D_z\Delta_{s,+}|^2 + |D_z\Delta_{s,-}|^2).
\label{EQ:quadDeriv2}\end{aligned}$$ In addition, the following term is also allowed by symmetry $$\delta K \frac{2\pi}{\Phi_0}h \sum_s (-|\Delta_{s+}|^2 + |\Delta_{s-}|^2),
\label{EQ:chiralSplit}$$ and it stabilizes this in-plane chiral phase for strong enough magnetic field (note the similarity to the Zeeman term for the condensate spin degrees of freedom). As for the homogeneous quartic terms, $$\begin{aligned}
f^{(4)}_{hom} &=& \sum_s [\beta_1(|\Delta_{s+}|^4+|\Delta_{s-}|^4)/2 + \beta'_1|\Delta_{s+}|^2 |\Delta_{s-}|^2]\nonumber\\
&-&\sum_{\sigma=\pm}
(\beta_2|\Delta_{\uparrow\!\uparrow,\sigma}|^2
|\Delta_{\downarrow\!\downarrow,\sigma}|^2 + \beta'_2
