# 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 $$S=\frac{1}{2}\langle A,dA\rangle+\frac{1}{6}\langle A,[A\wedge A]\rangle+\langle A^*,D_Ac\rangle+\frac{1}{2}\langle c^*,[c,c]\rangle,\tag{1}$$ with $$c\in\Omega^0(M)\otimes\mathfrak{g}$$, $$A\in\Omega^1(M)\otimes\mathfrak{g}$$, $$A^*\in\Omega^2(M)\otimes\mathfrak{g}[-1]$$, and $$c^*\in\Omega^3(M)\otimes\mathfrak{g}[-2]$$. In here $$\mathfrak{g}$$ is a Lie algebra equipped with an invariant non-degenerate pairing $$\langle\cdot,\cdot\rangle$$. However, in the first reference it is also claimed that this action can be put into the form $$S=\frac{1}{2}\langle e,de\rangle+\frac{1}{6}\langle e,[e\wedge e]\rangle\tag{2}$$ 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 $$\langle\alpha,\beta\rangle=0$$ if $$\alpha\in\Omega^p(M)\otimes\mathfrak g$$ and $$\beta\in\Omega^q(M)\otimes\mathfrak g$$ with $$p+q\neq 3$$. 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_b\rangle_{\mathfrak g}=-\int d{A^{*b}}c^a\langle T_a,T_b\rangle_{\mathfrak g}\\ &=-\int d{(A^{*b}c^a)}\langle T_a,T_b\rangle_{\mathfrak g}+\int A^{*b}d{c^a}\langle T_a,T_b\rangle_{\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 $$\langle e,[e\wedge e]\rangle$$, note that $$[e\wedge e]$$ can only have even forms. Indeed, an odd form in the expansion of $$[e\wedge e]$$ must come from the coupling $$[\alpha\wedge\beta]$$ of an odd form $$\alpha$$ and an even form $$\beta$$ in $$e$$. Since they are different, the term $$[\beta\wedge\alpha]$$ 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 $$\begin{equation} [\alpha\wedge\beta]=\alpha^a\wedge\beta^b[T_a,T_b]=\beta^b\wedge\alpha^a[T_a,T_b]=-\beta^b\wedge\alpha^a[T_b,T_a]=-[\beta\wedge\alpha].\tag{6} \end{equation}$$ Therefore the terms $$[\alpha\wedge\beta]$$ and $$[\beta\wedge\alpha]$$ cancel. By the same token, the rest of the surviving terms in the expansion of $$[e\wedge e]$$ are symmetric $$[\alpha\wedge\beta]=[\beta\wedge\alpha]$$. Given that we are in three dimensions, they have to either be 0-forms or 2-forms. We conclude that $$\begin{equation} [e\wedge e]=[c\wedge c]+2[c\wedge A^*]+[A\wedge A].\tag{7} \end{equation}$$ Of course, for $$0$$-forms we have $$[c\wedge c]=[c,c]$$. The second term is then
$$\begin{equation} \frac{1}{6}\langle e,[e\wedge e]\rangle=\frac{1}{6}\langle A,[A\wedge A]\rangle+\frac{1}{3}\langle A,[c\wedge A^*]\rangle+\frac{1}{6}\langle c^*,[c\wedge c]\rangle.\tag{8} \end{equation}$$

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 $$\langle A^*,dc\rangle$$ 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 $$A^*$$.

• Which page in the first reference for claim? Which eqs? Jul 22, 2020 at 20:53
• I realized that it is not only Section 4.1 but section 4. All that I am saying is in page 161 of bookstore.ams.org/surv-170. Jul 22, 2020 at 21:12
• As another problem that I just realized, the term $\langle A,[c\wedge A^*]\rangle$ differs from the correct $\langle A^*,[A,c]\rangle$ by a sign. If one looks at the second reference, they do take that sign into account. But then the sign of their $\langle A^*,dc\rangle$ term is wrong. Jul 22, 2020 at 21:17
• For starters, which coefficients are off by more than a sign? Jul 22, 2020 at 21:33
• The coefficient $\langle A,[c\wedge A^*]\rangle$ appears with a $1/3$ instead of a $1$ in the action of the field $e$. Similarly, the coefficient of $\langle c^*,[c\wedge c]\rangle$ appears with a $1/6$ instead of a $1/2$. Jul 22, 2020 at 21:35

In this answer we will focus on the cubic term, which seems to be OP's main question.

1. The trilinear form $$t\equiv\langle\cdot,[\cdot,\cdot]\rangle: \mathfrak{g}\times \mathfrak{g}\times\mathfrak{g}\to \mathbb{C}\tag{A}$$ is totally antisymmetric, because the bilinear form $$\langle\cdot,\cdot\rangle$$ is invariant/associative.

2. Consider fields $${\bf e}$$ that are both Lie-algebra-valued, form-valued & supernumber-valued. Note that in OP's references the $$n$$-forms are (implicitly) interpreted as carrying Grassmann-degree $$n$$ (modulo 2). The total Grassmann-parity of the fields $${\bf e}$$ is assumed to be odd, so that such fields anti-commute (in the appropriate graded symmetric tensor algebra). The trilinear form $$t$$ therefore becomes totally symmetric wrt. such fields.

3. In BV-CS theory (before gauge-fixing), we consider a minimal field $${\bf e} ~=~ c ~+~\underbrace{A_{\mu}\mathrm{d}x^{\mu}}_{=~{\bf A}}~+~\underbrace{A^{\ast\mu}(\star \mathrm{d}x)_{\mu}}_{=~{\bf A}^{\ast}} ~+~\underbrace{c^{\ast}\Omega}_{=~{\bf c}^{\ast}} \tag{B}$$ of above type, where $$(\star \mathrm{d}x)_{\mu}~:=~\frac{1}{2}\epsilon_{\mu\nu\lambda}\mathrm{d}x^{\nu}\wedge \mathrm{d}x^{\lambda}\tag{C}$$ and where $$\Omega~:=~\frac{1}{6}\epsilon_{\mu\nu\lambda}\mathrm{d}x^{\mu}\wedge\mathrm{d}x^{\nu}\wedge \mathrm{d}x^{\lambda} ~=~\frac{1}{3}\mathrm{d}x^{\mu}\wedge(\star \mathrm{d}x)_{\mu}.\tag{D}$$ (The wedges will be not be written explicitly from now on.)

4. The cubic action term is a multinomial expression \begin{align} \left. \frac{1}{6} t({\bf e},{\bf e},{\bf e})\right|_{\text{top-form}}~=~& \frac{1}{6}t({\bf A},{\bf A},{\bf A})+ t({\bf A}^{\ast},{\bf A},c) +\frac{1}{2}t({\bf c}^{\ast},c,c)\cr ~=~&\left( t(A_1,A_2,A_3)+ t(A^{\ast\mu},A_{\mu},c) +\frac{1}{2}t(c^{\ast},c,c)\right) \Omega.\end{align}\tag{E} Note that the (reciprocal) coefficient of each term of eq. (E) is precisely its symmetry factor. Eq. (E) agrees with OP's eq. (1).

• The problem with that hint is that some of these terms cancel. For example, in the expansion of $\langle e,[e\wedge e]\rangle$ we get both the term $\langle A^*, [A\wedge c]\rangle$ and the term $\langle A^*,[c\wedge A]\rangle$. However these two terms cancel. Indeed, in a basis of $\mathfrak{g}$ we have $A^ac^b[T_a,T_b]=c^bA^a[T_a,T_b]=-c^bA^a[T_b,T_a]$, so that $[A\wedge c]=-[c\wedge A]$. Jul 22, 2020 at 23:03
• Terms add up rather than cancel. I updated the answer. Jul 23, 2020 at 6:57
• I still don't understand the answer. Let us focus on the cubic $A^*,A,c$ terms. We have $t(e,e,e)=\cdots t(A^*,A,c)+t(A^*,c,A)+t(c,A^*,A)+t(c,A,A^*)+t(A,c,A^*)+t(A,A^*,c)\cdots$. It seems to me that you are claiming that these 6 terms give the same result. This is not the case. The first two terms cancel because of my first comment. The second two terms cancel because of a similar reason. Finally, the last two terms are equal (the antisymmetry of $t$ on the Lie algebra does not remain intact at the level of the forms since we have the super graduation and the form graduation). Jul 23, 2020 at 13:19
• Just to exaplain myself better, the last two terms survive because, while the Lie algebra part gives a minus sign, the fact that $A^*$ and $c$ are fermions also gives a minus sign. The overall sign is then positive when exchanging these two terms. Jul 23, 2020 at 13:25
• After reading the footnote in page 7 of Axelrod and Singer arxiv.org/abs/hep-th/9110056 I realized what is going on. Indeed, as you said, there is the implicit assumption that the parity of a field is given by the sum of its de Rham and cohomological degree. If one wants to stick to the formulation I used above, one has to be careful of defining $\langle\alpha,\beta\rangle=(-1)^{|\beta|\deg \alpha}\int \alpha^a\wedge\beta^b\langle T_a,T_b\rangle$. This solves all of the sign issues and the coefficients. Thanks! Oct 27, 2020 at 2:38