# Integrality of the Chern-Simons form and normalization of the action

I'm somewhat confused about normalization of the Chern-Simons action (for arbitrary compact gauge group). If we have a trivial principal bundle we write $$S(A)=\frac{k}{8\pi^2}\int_M\text{Tr}\left(A\wedge dA+\frac23A\wedge[A\wedge A]\right)$$ Where $\text{Tr}$, as I understand it, is the appropriately normalized Killing form on the Lie-algebra $\mathfrak{g}$. If we can extend the connection to some bounding $B$, $\partial B=M$, then we can write $$S(A)= \frac{k}{8\pi^2}\int_B\text{Tr}\left(F\wedge F\right)\mod 1$$ with $F$ the curvature. In order for this to be well-defined, we require $\text{Tr}(F\wedge F)$ to be integral (so that it's integer for closed $B$ and doesn't depend on choice of $B$). This expression is more general than the former, and in fact there are also other reasons why we want integrality of the Chern-Simons form (cf. Dijkgraaf&Witten's paper).

I don't completely understand why this form is integral. It's a characteristic class in the sense of being in the image of the Weil-homomorphism, but this doesn't guarantee integrality; it depends on the choice of symmetric form $\text{Tr}$. It seems, following Chern & Simons original paper, that for the case $G=U(n),O(n)$ choosing ordinary trace of matrices we get respectively $c_2(B)$ and $p_1(B)$.

Now my idea is that we can embed $G\subset U(n)$ for some $n$. Then a principal $G$ bundle can be extended to some principal $U(n)$ bundle withe same transition functions. A connection on the principal $U(n)$ bundle gives us on the $G$ bundle, and we can restrict $\text{Tr}$ on $U(n)$ to $G$ (which gives some multiple of the Killing form on $\mathfrak g$, if I recall correctly). Through this way we obtain that $\text{Tr}(F\wedge F)$ is also a multiple of $c_2$ for a curvature on a principal $G$-bundle. Thus scaling appropriately we get an integral class.

This poses some problems for me however. Another point of view on the normalization is that the action is gauge-invariant, which boils down to the Wess-Zumino term $\frac{k}{24\pi^2}\int_M\text{Tr}(g^{-1}dg)^3$ being integral. How do we know that the normalization we chose here (which seems to depend on the choice of unitary representation) is compatible with the argument for gauge-invariance?

Another thing is that I'm not sure about generalization to higher dimensions -- can we show $\text{Tr}(F\wedge\cdots\wedge F)$ is integral using a similar argument? The general formula for the $k$-th Chern class is not as simple as the one for $c_2$, so I think $\text{Tr}(F\wedge\cdots\wedge F)\neq c_k$.

• I don't understand why would you need integrality isnt your formula for $S(B)$ simply an application of Stokes theorem. – Omar Mar 24 '17 at 22:01
• You mean to say that $\text{Tr}(F\wedge F)$ is closed, therefore integrating it over a closed manifold will give zero? I don't think this argument works in general; we want to write it as $d\text{Tr}\left(A\wedge dA+\frac23A\wedge[A\wedge A]\right)$, but this expression only makes sense if the principal bundle in which $A$ lives is trivial. The second definition of $S(A)$ is more general, and in fact we wish to generalize it even further (as is done in the paper by Dijkgraaf and Witten). For this generalization integrality of the Chern-Simons form is essential. I will edit my post for clarity. – Rik Voorhaar Mar 27 '17 at 9:42
• To understand the integrality of the $F \wedge F$ term, note that it's representing (via Chern-Weil) a class in $H^4(BG,\mathbb{Z})$, which is equal to $\mathbb{Z}$ as $\pi_3(G) \cong \mathbb{Z}$ (for G simple and simply connected). For the gauge invariance, it has to work out because the integral of the CS-form can be given a gauge-invariant definition as living in $\mathbb{R}/\mathbb{Z}$ (which you presumably know since you're citing Dijkgraaf-Witten). – Aaron Bergman Mar 27 '17 at 17:54