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.


1 Answer 1


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 $\phi^{(2)}_{\alpha\beta}=\frac{1}{2}[b_{[\alpha},\phi^{(1)}_{\beta]}]=\frac{1}{2}[b_{[\alpha},[b_{\beta]},\phi^{(0)}]]$.

  • $\begingroup$ Thanks for your comment! To check my understanding, are you saying that the topological descent equations hold only up to $Q$-cohomology? That would make a lot of sense actually. Furthermore, I don't quite understand how you obtained the second equality on your third line. Applying a Jacobi-type identity to this expression again leaves me with an extra term of the form $[[b,Q],[b,\phi^{(0)}]]$. $\endgroup$
    – CoffeeCrow
    May 10, 2021 at 5:15
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    $\begingroup$ @CoffeeCrow I appologize for the notation, I actually did a mistake in my previous computation, but now I fixed and made all the indices explicit. You was right about the extra term of the Jacobi identity, There was also a sing mistake in my previous answer. I hope this update of my answer will settle the question. $\endgroup$
    – Nogueira
    May 10, 2021 at 22:51
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    $\begingroup$ @CoffeeCrow, No problem. Yes, I can explain better the notation: $[b_{[\alpha},\phi^{(1)}_{\beta]}]\equiv [b_{\alpha},\phi^{(1)}_{\beta}]-[b_{\beta},\phi^{(1)}_{\alpha}]$. Note that you can put a factor of one over two or not, it is a matter of taste $\endgroup$
    – Nogueira
    May 11, 2021 at 12:14
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    $\begingroup$ Ok, $b_{\alpha}$ is suppose to be a mode of the $b_{\alpha}(z)$, not the whole tower of mode, so it is constant, it commutes with derivatives. $\endgroup$
    – Nogueira
    May 12, 2021 at 21:52
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    $\begingroup$ Thank you very much, that makes a lot of sense. In the way I'm thinking about this problem, that fact follows from $G_{\mu}$ being a Noether charge, which at the algebra level corresponds to $b_{\alpha}$ being a single mode! Thanks again for your patience and all of your help! $\endgroup$
    – CoffeeCrow
    May 12, 2021 at 23:51

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