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

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
$$S=frac{1}{2}langle A,dArangle+frac{1}{6}langle A,[Awedge A]rangle+langle A^*,D_Acrangle+frac{1}{2}langle c^*,[c,c]rangle,tag{1}$$
with $cinOmega^0(M)otimesmathfrak{g}[1]$, $AinOmega^1(M)otimesmathfrak{g}$, $A^*inOmega^2(M)otimesmathfrak{g}[-1]$, and $c^*inOmega^3(M)otimesmathfrak{g}[-2]$. In here $mathfrak{g}$ is a Lie algebra equipped with an invariant non-degenerate pairing $langlecdot,cdotrangle$. However, in the first reference it is also claimed that this action can be put into the form
$$S=frac{1}{2}langle e,derangle+frac{1}{6}langle e,[ewedge e]rangletag{2}$$
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 $langlealpha,betarangle=0$ if $alphainOmega^p(M)otimesmathfrak g$ and $betainOmega^q(M)otimesmathfrak g$ with $p+qneq 3$. 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_brangle_{mathfrak g}=-int d{A^{*b}}c^alangle T_a,T_brangle_{mathfrak g}
&=-int d{(A^{*b}c^a)}langle T_a,T_brangle_{mathfrak g}+int A^{*b}d{c^a}langle T_a,T_brangle_{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 $langle e,[ewedge e]rangle$, note that $[ewedge e]$ can only have even forms. Indeed, an odd form in the expansion of $[ewedge e]$ must come from the coupling $[alphawedgebeta]$ of an odd form $alpha$ and an even form $beta$ in $e$. Since they are different, the term $[betawedgealpha]$ 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}
[alphawedgebeta]=alpha^awedgebeta^b[T_a,T_b]=beta^bwedgealpha^a[T_a,T_b]=-beta^bwedgealpha^a[T_b,T_a]=-[betawedgealpha].tag{6}
end{equation}
Therefore the terms $[alphawedgebeta]$ and $[betawedgealpha]$ cancel. By the same token, the rest of the surviving terms in the expansion of $[ewedge e]$ are symmetric $[alphawedgebeta]=[betawedgealpha]$. Given that we are in three dimensions, they have to either be 0-forms or 2-forms. We conclude that
begin{equation}
[ewedge e]=[cwedge c]+2[cwedge A^*]+[Awedge A].tag{7}
end{equation}
Of course, for $0$-forms we have $[cwedge c]=[c,c]$. The second term is then
begin{equation}
frac{1}{6}langle e,[ewedge e]rangle=frac{1}{6}langle A,[Awedge A]rangle+frac{1}{3}langle A,[cwedge A^*]rangle+frac{1}{6}langle c^*,[cwedge 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 $langle A^*,dcrangle$ 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 $A^*$.