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Assume that we have a cohomological field theory, with an odd symmetry generated by an odd operator $Q$ and an exact energy momentum tensor $T_{\mu\nu}=[Q,G_{\mu\nu}]$. Then by integrating over an spatial slice we define $G_\mu=\int d^3\vec{x} G_{0\mu}$, so that $P_\mu=[Q,G_\mu]$. The first claim is that if $\mathcal{O}^{(0)}$ is a scalar operator then $\mathcal{O}^{(k)}_{\mu_1\cdots\mu_k}:=i^k[G_{\mu_1},[\cdots,[G_{\mu_k},\mathcal{O}^{(0)}]\cdots]$ is a $k$-form. I see that this would be a $k$-covariant tensor since $G_\mu$ is clearly a $1$-form. However, for antisymmetry I would need to have $[G_\mu,G_\nu]=0$ since for all operators one has $$[G_{\mu}[G_\nu,A]]=[[G_\mu,G_\nu],A]-[G_\nu,[G_\mu,A]].$$ I don't see however why $[G_\mu,G_\nu]=0$.

My second question is that I don't see why the claim $d\mathcal{O}^{(k)}=[Q,\mathcal{O}^{(k-1)}]$ is true if $[Q,\mathcal{O}^{0}]=0$. I see that this is clearly the case for $k=1$. Indeed, in that case we have $$d\mathcal{O}^{(0)}=\partial_\mu\mathcal{O}^{(0)}dx^\mu=i[P_\mu,\mathcal{O}^{(0)}]dx^\mu=i[[Q,G_\mu],\mathcal{O}^{(0)}]dx^\mu=i[Q,[G_\mu,\mathcal{O}^{(0)}]]dx^\mu+i[G_\mu,[Q,\mathcal{O}^{(0)}]]dx^\mu=[Q,\mathcal{O}^{(1)}],$$ where the second term vanishes since $[Q,\mathcal{O}^{(0)}]$. However, this is no longer true in for $k>1$. In that case one can repeat the computation identically. HOwever, in general one can only say that $[Q,\mathcal{O}^{(k-1)}]$ is a total derivative. I don't see why this would imply that $[G_\mu,[Q,\mathcal{O}^{(k-1)}]]=0$.

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