I am asking this question about Chern-Simons theory from the paper "Quantum Field Theory and Jones Polynomial" by Edward Witten.

Let $M$ be a closed three dimensional manifold, and $P\rightarrow M$ is a principal $G$-bundle over $M$, with connection $1$-form $A\in\Omega^{1}(M;\mathfrak{g})$. The Chern-Simons action is then

$$S[A]=\frac{k}{4\pi}\int_{M}\mathrm{Tr}\left(A\wedge dA+\frac{2}{3}A\wedge A\wedge A\right)$$

where $k\in\mathbb{Z}$.

Under a gauge transformation

$$A[U]=U^{-1}dU+U^{-1}AU$$ the action transforms in the following way

$$S[A[U]]=\frac{k}{4\pi}\int_{M}\mathrm{Tr}\left(A\wedge dA+\frac{2}{3}A\wedge A\wedge A\right)+$$ $$-\frac{k}{12\pi}\int_{M}\mathrm{Tr}\left(U^{-1}dU\wedge U^{-1}dU\wedge U^{-1}dU\right)$$

The last term is called Wess-Zumino term. It can be shown that the last term takes value in $2\pi\mathbb{Z}$, so that the action is gauge-invariant in the Feynman path-integral


In other words, one has


By doing variation, one finds that the equation of motion of the classical action is

$$F=dA+A\wedge A=0$$

Thus, classically the gauge fields $A$ are flat connections.

In Witten's paper, on page 357, he computed the path-integral by splitting the gauge field $A$ in two parts: $A=a+B$, where $a$ is the flat connection satisfying the classical equation of motion, and $B$ is the fluctuation around the classical motion. i.e. $da+a\wedge a=0$. Then, the original action takes the following form

$$S[A]=\frac{k}{4\pi}\int_{M}\mathrm{Tr}\left(a\wedge da+\frac{2}{3}a\wedge a\wedge a\right)+$$ $$+\frac{k}{4\pi}\int_{M}\mathrm{Tr}(B\wedge D_{a}B)+\frac{k}{6\pi}\int_{M}\mathrm{Tr}(B\wedge B\wedge B)$$

where $D_{a}$ is the covariant differential with respect to the flat background $a$, i.e.


For a fixed flat connection $a$, the perturbation part

$$S[a;B]=\frac{k}{4\pi}\int_{M}\mathrm{Tr}(B\wedge D_{a}B)+\frac{k}{6\pi}\int_{M}\mathrm{Tr}(B\wedge B\wedge B)$$

is simply a "Chern-Simons action" with de-Rham differential replaced by covariant differential $D_{a}$. Since $D_{a}$ is also nilpotent, one can define the twisted de-Rham complex


and for the same reason, by replacing de-Rham differential $d$ by covariant differential $D_{a}$, under the following gauge transformations

$$a[U]=a,\quad B[U]=U^{-1}D_{a}U+U^{-1}BU$$

where $U^{-1}D_{a}U=U^{-1}dU+U^{-1}aU$, the perturbative part $S[a;B]$ transforms to

$$\frac{k}{4\pi}\int_{M}\mathrm{Tr}(B\wedge D_{a}B)+\frac{k}{6\pi}\int_{M}\mathrm{Tr}(B\wedge B\wedge B)+$$ $$-\frac{k}{12\pi}\int_{M}\mathrm{Tr}\left(U^{-1}D_{a}U\wedge U^{-1}D_{a}U\wedge U^{-1}D_{a}U\right)$$

The last term is then a generalized Wess-Zumino term.

My question is: does that last term also take value in $2\pi\mathbb{Z}$?

Does it make sense to talk about "twisted curvature" like $F_{a}=D_{a}A+A\wedge A$?

  • 2
    $\begingroup$ Crossposted to physics.stackexchange.com/q/420415/2451 $\endgroup$
    – Qmechanic
    Jul 31, 2018 at 16:56
  • 1
    $\begingroup$ What's the meaning of $U^{-1}D_a U$ exactly? Assume that $G$ is simply connected and fix a trivialization of the bundle $P$. $U$ is a map from $M$ to $G$ ($G$-valued function on $M$). $U^{-1}dU$ is originally defined to be the pullback of the Maurer-Cartan 1-form on $G$ via $U$. But now you want to apply covariant derivative $D_a$ on $U$. $U$ is not $\mathfrak{g}$-valued but $G$-valued. So you might want to apply infinitesimal gauge transformation, rather than an actual gauge transformation. Then you're gauge transformation will be $\mathfrak{g}$-valued. $\endgroup$
    – Henry
    Aug 2, 2018 at 18:06
  • $\begingroup$ So it seems that your question should be whether the infinitesimal generalized Wess-Zumino term vanishes? $\endgroup$
    – Henry
    Aug 2, 2018 at 18:08
  • $\begingroup$ $U^{-1}D_{a}U=U^{-1}dU+U^{-1}aU$. This is Lie algebra-valued. $\endgroup$ Aug 2, 2018 at 19:31
  • $\begingroup$ Do you think it makes sense to talk about the "generalized" second Chern-class $\mathrm{Tr}((D_{a}B+B\wedge B)\wedge(D_{a}B+B\wedge B))$? $\endgroup$ Aug 2, 2018 at 19:35


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