 Solution

The axion variation gives $\partial \mathcal{L}/\partial a=-m_a^{2}a+g_a{F}_{\mu \nu }{\tilde{F}}^{\mu \nu }/4$, soFor the photon variation, the dual electromagnetic field tensor obeys $\delta (F_{\rho \sigma }{\tilde{F}}^{\rho \sigma })=2{\tilde{F}}^{\rho \sigma }\delta F_{\rho \sigma }$. Antisymmetry then givesThere is no undifferentiated $A_{\nu }$, henceThe last equality uses the electromagnetic Bianchi identity ${\partial }_{\mu }{\tilde{F}}^{\mu \nu }=0$. These signs follow from the positive $aF\tilde{F}$ interaction in the source; reversing that interaction reverses both source terms. The induced electromagnetic current is identically conserved: its divergence contracts a symmetric second derivative of $a$ with an antisymmetric tensor, while the remaining term vanishes by the same Bianchi identity.