Kaluza-Klein reducing the Chern-Simons term in 11D supergravity

Question

In 11D supergravity we have the Chern-Simons term of the 3-form field $C$ \begin{equation} \int C \land d C \land d C \end{equation} I want to consider this on a spacetime $\mathbb{R}^{1,6}\times S$ where $S$ is a resolution of an $A_n$ singularity. Then $H_2(S)=\mathbb{R}^n$, and I decompose the 3-form as \begin{equation} C = \sum_i^n \omega_i \land A_i \end{equation} where $\omega_i$ form a basis of the harmonic 2-forms on $S$ and $A_i$ are 1-forms on $\mathbb{R}^{1,6}$. Now plugging this into the Chern-Simons term I get \begin{equation} \int_{\mathbb{R}^{1,6}} \int_S \sum_{ijk}^n \omega_i \land A_i \land\omega_j \land dA_j \land\omega_k \land dA_k \overset{?}{=} \sum_{ijk}^n \int_{\mathbb{R}^{1,6}} \int_S \omega_i \land\omega_j\land\omega_k \land A_i \land dA_j \land dA_k \end{equation} at which point I'm not sure how to integrate over $S$, especially given that $\omega_i \land\omega_j\land\omega_k$ is a 6-form in a 4-dimensional space.

Any help would be greatly appreciated.

Answer 1

$\newcommand{\d}{\mathrm{d}}\newcommand{\w}{\wedge}\newcommand{\R}{\mathbb{R}}$That integral vanishes because the differential form $\omega_i\w\omega_j\w\omega_k\in\Omega^6(S_4)$ vanishes (you cannot antisymmetrise more that $d$ indices in a $d$-dimensional manifold).

What you could do, instead, is use the fact that $$\Omega^n(X\times Y)=\bigoplus_{p+q=n}\Omega^p(X)\w\Omega^q(Y),$$ to write $$ \Omega^3\!\left(\R^{1,6}\times S_4\right) \ni C= \sum_{\begin{smallmatrix} 0\leq \!\!\!\!&p&\!\leq 3 \\ 0\leq \!\!\!\!&q&\!\!\leq 3\\ p+q&=&3\end{smallmatrix}} A_{[p]} \w \omega_{[q]}, $$ where $A_{[p]}\in\Omega^p\!\left(\R^{1,6}\right)$ and $\omega_{[q]}\in\Omega^q(S_4)$. Suppose that, for some reason, you only consider $\omega_{[q]}\in\mathrm{H}^q(S_4)$ (following e.g. Compactification of D=11 supergravity on spaces of exceptional holonomy that was linked in a comment by the OP). Then the integral $$\int_{\R^{1,6}\times S_4} C\w\d C\w \d C,$$ receives contributions from $$ \omega_{[q]}\w \omega_{[q']}\w \omega_{[q'']} $$ with $q+q'+q''=4$.

Let's expand $\omega_{[q]}$ in an orthonormal basis of the $q$-th cohomology group $\left\{\omega_{I_q}\right\}$, with $I_q\in \left\{1,2,\cdots, b_q=\mathrm{\dim}(\mathrm{H}^q(S_4))\right\}$: $$ \omega_{[q]} = \left<\omega_{[q]},\omega^{I_q}\right>\ \omega_{I_q},$$ so that $$C = \sum_{p+q=3} A^{I_q}\w \omega_{I_q},$$ with implicit summation over $I_q$ and with the coefficients $\left<\omega_{[q]},\omega^{I_q}\right>$ soaked up by $A^{I_q}$.

The desired integral decomposes, then, as \begin{align} \int_{\R^{1,6}\times S_4} C\w\d C\w \d C &= \sum_{\begin{smallmatrix}p+q&=&3 \\p'+q'&=&3 \\ p''+q'' &=&3 \end{smallmatrix}}\int_{\R^{1,6}\times S_4} A^{I_q}\w \omega_{I_q} \w \d A^{J_{q'}}\w \omega_{J_{q'}}\w \d A^{K_{q''}}\w \omega_{K_{q''}} = \\ &= \sum_{\begin{smallmatrix}p+q&=&3 \\p'+q'&=&3 \\ p''+q'' &=&3 \\ q+q'+q''&=&4 \end{smallmatrix}} C_{I_{q},J_{q'},K_{q''}} \int_{\R^{1,6}} A^{I_{q}}\w\d A^{J_{q'}}\w\d A^{K_{q''}}, \end{align} where $$ C_{I_{q},J_{q'},K_{q''}} :=\int_{S_4} \omega_{I_q}\w \omega_{J_{q'}}\w \omega_{K_{q''}}, $$ generalise $C_{IJK}$ (eq. (3)) and $C_{IJk}$ (eq. (9)) of the linked reference. If, furthermore, in your case at hand some of the $C_{I_{q},J_{q'},K_{q''}}$ vanish you can simplify your result further.