Deriving the Topological Descent Equations I am trying to show that in a cohomological TQFT, given a physical operator $\phi^{(0)}$, one can construct a chain of non-local physical operators. In doing so, I need to show that a certain set of operators $\phi^{(n)}$ obey the so called topological descent equations. I am following these notes, as well as these ones. This unanswered StackExchange question is similar, but somewhat more general and unanswered. I will be following the conventions of my first reference.
The setup is as follows. We consider a TQFT with a nilpotent fermionic (supercharge), operator $Q$. Physical operators are those for which $[Q,\mathcal{O}\}=0$, where $[\cdot,\cdot\}$ is the graded commutator (the ordinary commutator unless both arguments are fermionic). Being a TQFT, we also have that the momentum operator is $P_{\mu}=\{Q,G_{\mu}\}$ for some fermionic operator $G_{\mu}$.
We take a (for simplicity), bosonic physical operator $\phi^{(0)}$ , so that $[Q,\phi^{(0)}]=0$. From here, we define descendent operators:
\begin{align*}
&
\phi^{(1)}=\phi_{\mu}^{(1)}dx^{\mu}=[G_{\mu},\phi^{(0)}]dx^{\mu}
\\
&
\phi^{(2)}=\phi_{\mu\nu}^{(2)}dx^{\mu}\wedge dx^{\nu}=\{G_{\mu},[G_{\nu},\phi^{(0)}]\}dx^{\mu}\wedge dx^{\nu}=\{G_{\mu},\phi^{(1)}_{\nu}\}dx^{\mu}\wedge dx^{\nu}
\end{align*}
and so on. The topological descent equations are:
$$
d\phi^{(n)}=i[Q,\phi^{(n+1)}\}\text{ , for all }n\geq 0\text{.}
$$
For $n=0$, things are relatively easy because $\phi^{(0)}$ is physical. My second reference shows this case in detail. However, I am having trouble showing the the next-simplest case, $n=1$. I am able to obtain $d\phi^{(1)}=i[Q,\phi^{(2)}]$ up to a second term, but cannot see why the second term vanishes. I will repeat my calculation here.
\begin{align*}
d\phi^{(1)}
&=
\frac{\partial\phi_{\mu}^{(1)}}{\partial x^{\nu}}dx^{\nu}\wedge dx^{\mu}
\\
&=
i[P_{\nu},\phi^{(1)}_{\mu}]dx^{\nu}\wedge dx^{\mu}
\\
&=
i[\{Q,G_{\nu}\},\phi^{(1)}_{\mu}]dx^{\nu}\wedge dx^{\mu}
\\
&=
i
\left(
-
[\{G_{\nu},\phi^{(1)}_{\mu}\},Q]
-
[\{\phi^{(1)}_{\mu},Q\},G_{\nu}]
\right)
\\
&=
i[Q,\phi^{(2)}]-i[\{\phi^{(1)}_{\mu},Q\},G_{\nu}]dx^{\nu}\wedge dx^{\mu}
\end{align*}
Where in the second equality I have used that momentum is the generator of translations, and in the fourth equality I have used the Jacobi-like identity $[\{A,B\},C]=\{[B,C],A\}-\{[C,A],B\}$.
The first term is as desired, but I can see no reason why the term $-i[\{\phi^{(1)}_{\mu},Q\},G_{\nu}]dx^{\nu}\wedge dx^{\mu}$ should vanish. I have fiddled around with Jacobi-like identities and tried to relate it to $0=d^{2}\phi^{(0)}$, but with no luck. Any pointers would be much appreciated.
 A: To fix notation, the "b-ghost" $b_{\alpha}$ satisfy $[Q,b_{\alpha}]=\partial_{\alpha}$ and we have
$$
[Q,\phi^{(0)}]=0,\qquad \phi_{\alpha}^{(1)}=[b_{\alpha},\phi^{(0)}],\qquad [Q,\phi^{(1)}_{\alpha}]=\partial_{\alpha}\phi^{(0)}
$$
Now consider $\phi_{\alpha\beta}^{(2)}=\frac{1}{2}[b_{[\alpha},[b_{\beta]},\phi^{(0)}]]=\frac{1}{2}[b_{[\alpha},\phi^{(1)}_{\beta]}]$, then
$$
[Q,\phi^{(2)}_{\alpha\beta}]=\frac{1}{2}[Q,[b_{[\alpha},b_{\beta]},\phi^{(0)}]]=\frac{1}{2}\partial_{[\alpha}\phi^{(1)}_{\beta]}-\frac{1}{2}[b_{[\alpha},\partial_{\beta]}\phi^{(0)}]=\partial_{[\alpha}\phi_{\beta]}^{(1)}
$$
where I used that
$$
-\frac{1}{2}[b_{[\alpha},\partial_{\beta]}\phi^{(0)}]=-\frac{1}{2}\partial_{[\beta}[b_{\alpha]},\phi^{(0)}]=-\frac{1}{2}\partial_{[\beta}\phi_{\alpha]}^{(1)}=+\frac{1}{2}\partial_{[\alpha}\phi_{\beta]}
$$
Note that the anti-symetrization of the indices was important to get the right sign.
Now, you are trying to do the reverse, i.e. find $\phi^{(2)}_{\alpha\beta}$ by playing with $\phi^{(1)}_{\alpha}$, here we go:
$$
\partial_{[\alpha}\phi_{\beta]}^{(1)}=[[Q,b_{[\alpha}],\phi_{\beta]}^{(1)}]=[Q,[b_{[\alpha},\phi^{(1)}_{\beta]}]]+[b_{[\alpha},[Q,\phi_{\beta}^{(1)}]]
$$
the last term can be worked out by noting that $[Q,\phi_{\beta}^{(1)}]=\partial_{\beta}\phi^{(0)}$:
$$
[b_{[\alpha},[Q,\phi_{\beta]}^{(1)}]]=[b_{[\alpha},\partial_{\beta]}\phi^{(0)}]=\partial_{[\beta}[b_{\alpha]},\phi^{(0)}]=\partial_{[\beta}\phi^{(1)}_{\alpha]}=-\partial_{[\alpha}\phi_{\beta]}^{(1)}
$$
so we get
$$
\partial_{[\alpha}\phi_{\beta]}^{(1)}=[Q,[b_{[\alpha},\phi^{(1)}_{\beta]}]]-\partial_{[\alpha}\phi_{\beta]}^{(1)}\implies [Q,\frac{1}{2}[b_{[\alpha},\phi^{(1)}_{\beta]}]]=\partial_{[\alpha}\phi^{(1)}_{\beta]}
$$
and so $\phi^{(2)}_{\alpha\beta}=\frac{1}{2}[b_{[\alpha},\phi^{(1)}_{\beta]}]=\frac{1}{2}[b_{[\alpha},[b_{\beta]},\phi^{(0)}]]$.
