# Topological Descent Equation

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 $$[G_\mu,[Q,\mathcal{O}^{(k-1)}]]=0$$.