entire Calabi-Yau [@BBS]. The instanton action is positive and hence does not distinguish instantons from anti-instantons. Imaginary theta-angle-like terms will have to be added to make this distinction.

Turning to $O_2$, we have the BPS condition $$\label{H-O2}
  * H = \pm\, {\mathrm{d}}\begin{pmatrix} {\mathrm{e}}^{-\phi} \chi \\[2pt] {\mathrm{e}}^{-\phi}
  \end{pmatrix}\ .$$ Again, ${\mathrm{e}}^{-\phi}$ is harmonic, and the same now applies to ${\mathrm{e}}^{-\phi}
\chi$. If one imposes rotational symmetry then $$\label{O2-sol}
  O_2:\quad {\mathrm{d}}*\! {\mathrm{d}}{\mathrm{e}}^{-\phi} = 0\ ,\quad \chi = \chi_1 {\mathrm{e}}^{\phi} +
  \chi_0\ ,$$ and from , it follows again that $Q^{(1)}=\chi_0 Q^{(2)}$. Notice that the field $\chi$ is now completely regular everywhere, and interpolates between the boundaries according to $$\Delta \chi \equiv \chi_\infty - \chi_0 = {\frac{\raisebox{-2pt}{$\chi_1$}}{g_s^2}}\ .$$ The complete solution agrees with the asymptotics derived in and .

For this solution, with the dilaton again given by , the instanton action then becomes $$\label{S2-inst}
  S_2^E = \big| Q^{(2)} \big|\, \Big( {\frac{\raisebox{-2pt}{$1$}}{g_s^2}} + {\tfrac{1}{2}}\,
  (\Delta \chi)^2 \Big)\ .$$ For the particular case of $\Delta \chi=0$, the solution and instanton action are the same as for the $O_1$ solution. Notice also that both terms are positive and invariant under the shift symmetries and , as guaranteed by the properties of the original action. For 