Circle reduction of eleven-dimensional supergravity ($d{s}_{11}^2=e^{-2\Phi /3}d{s}_s^2+e^{4\Phi /3}(dy-C_1{)}^2$)

For circle-independent fields, the metric gives the type-IIA string-frame metric, dilaton and RR one-form. Decomposing $A_3=C_3+B_2\wedge dy$ gives the RR three-form and Kalb–Ramond field. With $H_3=d{B}_2$, $F_2=d{C}_1$, the horizontal four-form is ${\tilde{F}}_4=d{C}_3-C_1\wedge H_3$. The Einstein term reduces, modulo a total derivative, to $\sqrt{-g}[e^{-2\Phi }(R+4(\nabla \Phi {)}^2)-\frac{1}{4}{F}_{2,\mu \nu }{F}_2^{\mu \nu }]$. The eleven-dimensional form norm gives $\sqrt{-g}[e^{-2\Phi }|H_3{|}^2+|{\tilde{F}}_4{|}^2]$, where $|F_k{|}^2=F^2/k!$. Ordinary reduction generates massless type IIA supergravity, not Romans mass.

Put $A_3=C_3+B_2\wedge dy$ and $F_4=G_4+H_3\wedge dy$, where $G_4=d{C}_3$. The circle component of $A_3\wedge F_4\wedge F_4$ is $[B_2\wedge G_4\wedge G_4+2C_3\wedge G_4\wedge H_3]\wedge dy$. Since $d(C_3\wedge B_2\wedge G_4)=B_2\wedge G_4\wedge G_4-C_3\wedge H_3\wedge G_4$, its integral equals three copies of $B_2\wedge G_4\wedge G_4$ up to a boundary term. The coefficient therefore becomes $-1/2$ after circle integration. Boundary terms and global potential patching must be retained when the spacetime has boundary or nontrivial flux topology.