solution is equal to the topological charge associated with $H^2$, as follows from the second equation of . The ‘selfdual’ instanton (upper sign in ) is taken for negative $Q^{(2)}$, the ‘anti-selfdual’ instanton for positive $Q^{(2)}$.

Since, up to proportionality factors, there is only a unique spherically symmetric harmonic function, $\chi$ must be of the form $\chi=\chi_1
{\mathrm{e}}^{-\phi}+\chi_0$ with $\chi_0$, $\chi_1$ constant. It then follows from that $\chi_0$ is determined by $$\chi_0 = {\frac{\raisebox{-2pt}{$Q^{(1)}$}}{Q^{(2)}}}\ ,$$ and this relation is consistent with the shift symmetries and , since the charges transform non-trivially. The instanton action for $O_1$ is given by $$S_1^E = \mp \int_{{\partial}{\mathcal{M}}} \big[ \chi H^1 + ({\mathrm{e}}^{\phi} - {\tfrac{1}{2}}\chi^2)
  H^2 \big]\ ,$$ where the boundary consists of the disjoint union of two spheres, ${\partial}{\mathcal{M}}=S^{D-1}_\infty\,\cup\,S^{D-1}_0$, with radii as indicated. The terms involving $\chi$ will diverge on $S^{D-1}_0$ since $\chi$ is harmonic, so in order to obtain a finite action we have to take $\chi=
\chi_0$ constant. This was already anticipated from the asymptotic behaviour of the fields, discussed in the beginning of this section. The action then reads $$S_1^E = \frac{\big| Q^{(2)} \big|}{g_s^2}\ .$$ This solution was also found in [@BB; @GS1], and should correspond to the fivebrane wrapping the 