Batalin-Vilkovisky (BV) form of the Chern-Simons Action

Question

As seen in Section 4 of Chapter 5 of Costello, K. "Renormalization and Effective Field Theory", or in section 5.2 $L_\infty$-Algebras of Classical Field Theories and the Batalin-Vilkovisky Formalism, the BV form of the Chern-Simons action is $$S=\frac{1}{2}\langle A,dA\rangle+\frac{1}{6}\langle A,[A\wedge A]\rangle+\langle A^*,D_Ac\rangle+\frac{1}{2}\langle c^*,[c,c]\rangle,\tag{1}$$ with $c\in\Omega^0(M)\otimes\mathfrak{g}[1]$, $A\in\Omega^1(M)\otimes\mathfrak{g}$, $A^*\in\Omega^2(M)\otimes\mathfrak{g}[-1]$, and $c^*\in\Omega^3(M)\otimes\mathfrak{g}[-2]$. In here $\mathfrak{g}$ is a Lie algebra equipped with an invariant non-degenerate pairing $\langle\cdot,\cdot\rangle$. However, in the first reference it is also claimed that this action can be put into the form $$S=\frac{1}{2}\langle e,de\rangle+\frac{1}{6}\langle e,[e\wedge e]\rangle\tag{2}$$ for some field $e$. I don't see how this is possible.

Let me explain my reasoning. Let us first assume $e=c+A+A^*+c^*$. Note that $\langle\alpha,\beta\rangle=0$ if $\alpha\in\Omega^p(M)\otimes\mathfrak g$ and $\beta\in\Omega^q(M)\otimes\mathfrak g$ with $p+q\neq 3$. We can use this to expand $\langle e,d{e}\rangle$. For example, the only term that can be coupled with the $A$ coming from the left $e$ is the $d{A}$ coming from $d{e}$. We conclude that \begin{equation} \frac{1}{2}\langle e,d{e}\rangle=\frac{1}{2}\langle c,d{A^*}\rangle+\frac{1}{2}\langle A,d{A}\rangle+\frac{1}{2}\langle A^*,d{c}\rangle.\tag{3} \end{equation} Now, remembering that $A^*$ and $c$ are fermionic, we have \begin{equation} \begin{aligned} \langle c,d{A^*}\rangle&=\int c^ad{A^{*b}}\langle T_a,T_b\rangle_{\mathfrak g}=-\int d{A^{*b}}c^a\langle T_a,T_b\rangle_{\mathfrak g}\\ &=-\int d{(A^{*b}c^a)}\langle T_a,T_b\rangle_{\mathfrak g}+\int A^{*b}d{c^a}\langle T_a,T_b\rangle_{\mathfrak g}. \end{aligned}\tag{4} \end{equation} Thus, up to total derivatives we have \begin{equation} \frac{1}{2}\langle e,d{e}\rangle=\frac{1}{2}\langle A,d{A}\rangle+\langle A^*,d{c}\rangle.\tag{5} \end{equation}

To expand the term $\langle e,[e\wedge e]\rangle$, note that $[e\wedge e]$ can only have even forms. Indeed, an odd form in the expansion of $[e\wedge e]$ must come from the coupling $[\alpha\wedge\beta]$ of an odd form $\alpha$ and an even form $\beta$ in $e$. Since they are different, the term $[\beta\wedge\alpha]$ also appears in the expansion of $e$. Now, all even forms in $e$ are fermionic while all odd forms in $e$ are bosonic. We conclude that $\alpha$ is bosonic while $\beta$ is fermionic. Therefore \begin{equation} [\alpha\wedge\beta]=\alpha^a\wedge\beta^b[T_a,T_b]=\beta^b\wedge\alpha^a[T_a,T_b]=-\beta^b\wedge\alpha^a[T_b,T_a]=-[\beta\wedge\alpha].\tag{6} \end{equation} Therefore the terms $[\alpha\wedge\beta]$ and $[\beta\wedge\alpha]$ cancel. By the same token, the rest of the surviving terms in the expansion of $[e\wedge e]$ are symmetric $[\alpha\wedge\beta]=[\beta\wedge\alpha]$. Given that we are in three dimensions, they have to either be 0-forms or 2-forms. We conclude that \begin{equation} [e\wedge e]=[c\wedge c]+2[c\wedge A^*]+[A\wedge A].\tag{7} \end{equation} Of course, for $0$-forms we have $[c\wedge c]=[c,c]$. The second term is then
\begin{equation} \frac{1}{6}\langle e,[e\wedge e]\rangle=\frac{1}{6}\langle A,[A\wedge A]\rangle+\frac{1}{3}\langle A,[c\wedge A^*]\rangle+\frac{1}{6}\langle c^*,[c\wedge c]\rangle.\tag{8} \end{equation}

We see that we have failed to recover our original action because of some factors. One could try to resolve this by combining the fields in $e$ with different numerical factors. However, since the action of $A$ already has the correct factors, we cannot rescale $A$. Indeed, any rescaling of $A$ would produce a mismatch in the scales of the quadratic and cubic terms in $A$. On the other hand, the term $\langle A^*,dc\rangle$ has also the correct factor, so that we must scale $c$ and $A^*$ inversely. This means that we will never get the correct factor for the cubic term in $c$, $A$, and $A^*$.

Answer 1

In this answer we will focus on the cubic term, which seems to be OP's main question.

1. The trilinear form $$t\equiv\langle\cdot,[\cdot,\cdot]\rangle: \mathfrak{g}\times \mathfrak{g}\times\mathfrak{g}\to \mathbb{C}\tag{A}$$ is totally antisymmetric, because the bilinear form $\langle\cdot,\cdot\rangle$ is invariant/associative.

2. Consider fields ${\bf e}$ that are both Lie-algebra-valued, form-valued & supernumber-valued. Note that in OP's references the $n$-forms are (implicitly) interpreted as carrying Grassmann-degree $n$ (modulo 2). The total Grassmann-parity of the fields ${\bf e}$ is assumed to be odd, so that such fields anti-commute (in the appropriate graded symmetric tensor algebra). The trilinear form $t$ therefore becomes totally symmetric wrt. such fields.

3. In BV-CS theory (before gauge-fixing), we consider a minimal field $$ {\bf e} ~=~ c ~+~\underbrace{A_{\mu}\mathrm{d}x^{\mu}}_{=~{\bf A}}~+~\underbrace{A^{\ast\mu}(\star \mathrm{d}x)_{\mu}}_{=~{\bf A}^{\ast}} ~+~\underbrace{c^{\ast}\Omega}_{=~{\bf c}^{\ast}} \tag{B}$$ of above type, where $$(\star \mathrm{d}x)_{\mu}~:=~\frac{1}{2}\epsilon_{\mu\nu\lambda}\mathrm{d}x^{\nu}\wedge \mathrm{d}x^{\lambda}\tag{C}$$ and where $$\Omega~:=~\frac{1}{6}\epsilon_{\mu\nu\lambda}\mathrm{d}x^{\mu}\wedge\mathrm{d}x^{\nu}\wedge \mathrm{d}x^{\lambda} ~=~\frac{1}{3}\mathrm{d}x^{\mu}\wedge(\star \mathrm{d}x)_{\mu}.\tag{D}$$ (The wedges will be not be written explicitly from now on.)

4. The cubic action term is a multinomial expression $$\begin{align} \left. \frac{1}{6} t({\bf e},{\bf e},{\bf e})\right|_{\text{top-form}}~=~& \frac{1}{6}t({\bf A},{\bf A},{\bf A})+ t({\bf A}^{\ast},{\bf A},c) +\frac{1}{2}t({\bf c}^{\ast},c,c)\cr ~=~&\left( t(A_1,A_2,A_3)+ t(A^{\ast\mu},A_{\mu},c) +\frac{1}{2}t(c^{\ast},c,c)\right) \Omega.\end{align}\tag{E}$$ Note that the (reciprocal) coefficient of each term of eq. (E) is precisely its symmetry factor. Eq. (E) agrees with OP's eq. (1).