In 3d, one can write down the $SO(N)$ Chern-Simons action to be $$S(A)=\frac{k}{192\pi}\int_{M}\text{Tr}(A d A +\frac{2}{3}A^3),$$ where $A$ is an $SO(N)$ connection. The level quantization can be derived as follows:

Let $M^{\prime}$ be a bounding 4-manifold of $M$. We can always find such $M^{\prime}$ since $\Omega^{SO}_3=0$. Extend $A$ to $M^{\prime}$ and define $$S(A)=\frac{k}{192\pi}\int_{M^{\prime}}\text{Tr}(F \wedge F),$$ where $F$ is the curvature 2-form of $A$. We need $\exp(iS_M(A))$ to the independent of the choice of $M^{\prime}$, and the extension of $A$ from $M$ to $M^{\prime}$. Let $M^{\prime\prime}$ be another bounding manifold of $M$, then the difference of $S$ is $$\delta S = \frac{k}{192\pi}\int_{M^{\prime}\cup \bar{M}^{\prime\prime}}\text{Tr}(F \wedge F),$$ where $\bar{M}^{\prime\prime}$ denotes the orientation reversal of $M^{\prime\prime}$. $\delta S$ can be rewritten as $$\delta S = \frac{k\pi}{24}p_1(M^{\prime}\cup \bar{M}^{\prime\prime}) = \frac{k\pi}{8}\sigma(M^{\prime}\cup \bar{M}^{\prime\prime}),$$ where $p_1$ is the first Pontryagin number, and $\sigma$ is the signature of a 4-manifold. We also used the Hirzbruch signature theorem $\sigma(X)=p_1(X)/3$ for 4-manifolds $X$. Since $\sigma(X)$ is an integer, $exp(iS_M(A))$ is well-defined for $k$ equals multiples of 16.

One can use the above argument, together with the fact that $\Omega^{spin}_3=0$ and the Rohlin theorem which implies that the signature of a closed spin 4-manifold is divisible by 16, to argue that for a spin 4-manifold, $\exp(iS)$ is well-defined for $k\in \mathbb{Z}$.

I'm trying to derive the quantization condition of $k$ using similar arguments as above, for 7d $SO(N)$ Chern-Simons action (simply replace $M$ by a 7-manifold, and $A$ by 3-form ). The following facts may be helpful: $\Omega^{SO}_7=0$, $\Omega^{spin}_7=0$, $$\sigma(X) = (7p_2(X)-p_1^2(X))/45$$ for 8-manifold $X$.

This post imported from StackExchange Physics at 2014-09-15 21:05 (UCT), posted by SE-user Zitao Wang