Timeline for Batalin-Vilkovisky (BV) form of the Chern-Simons Action
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
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| Oct 27, 2020 at 2:38 | comment | added | Ivan Burbano | 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 2:33 | vote | accept | Ivan Burbano | ||
| Sep 24, 2020 at 0:41 | comment | added | Ivan Burbano | I wanted to comment that in fully component notation this problem is even more preoccupying. Namely, if we write down things fully in coordinates $\langle A^*,[A,c]\rangle=\int d^3x\epsilon^{\mu\nu\rho}A^{*a}_{\mu\nu}A^b_\rho c^c\langle T_a,[T_b,T_c]\rangle=\int d^3x\epsilon^{\mu\nu\rho}A^{*a}_{\mu\nu}c^cA^b_\rho \langle T_a,-[T_c,T_b]\rangle=-\langle A^*,[c,A]\rangle$. Then, commuting $A^b_\rho$ past $c^c$ does not yield a sign change and commuting $[T_b,T_c]$ clearly does. | |
| Jul 24, 2020 at 9:06 | history | edited | Qmechanic♦ | CC BY-SA 4.0 |
Corrected eq. (E)
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| Jul 23, 2020 at 16:45 | comment | added | Ivan Burbano | So, do you agree that the expression $S=\frac{1}{2}\langle e,de\rangle+\frac{1}{6}\langle e,[e\wedge e]\rangle$ is not correct? Or is my handling of the degrees wrong? | |
| Jul 23, 2020 at 15:48 | history | edited | Qmechanic♦ | CC BY-SA 4.0 |
Added explanation
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| Jul 23, 2020 at 15:48 | comment | added | Qmechanic♦ | Yes, that seems to be the culprit. | |
| Jul 23, 2020 at 14:20 | comment | added | Ivan Burbano | This is clear if one writes the field down in components. Then the components carry the cohomological degree while the basis elements $dx^\mu$ carry the form degree. The root of the problem is the when one commutes a fermionic scalar $c$ past $dx^\mu$, there is no sign $c dx^\mu=dx^\mu c$ | |
| Jul 23, 2020 at 14:18 | comment | added | Ivan Burbano | I just studied your answer better and I found the root of the problem in your answer. Let me denote by $\deg$ the homological degree of the field mod 2, so that $\deg A=\deg c^*=0$ and $\deg c=\deg A^*=1$. On the other hand, let me denote by $|\cdot|$ the form degree of the field mod 2, so that $|A|=|c^*|=1$ and $|c|=|A^*|=0$. Your argument seems to use something like $FG=(-1)^{(|F|+\deg F)(|G|+\deg G)}GF$. That is, a field $F$ is bosonic or fermionic according to its total degree. That is not the case here however. Instead $FG=(-1)^{|F||G|+\deg F\deg G}GF$ is the correct formula. | |
| Jul 23, 2020 at 13:25 | comment | added | Ivan Burbano | 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 13:19 | comment | added | Ivan Burbano | 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 12:35 | history | edited | Qmechanic♦ | CC BY-SA 4.0 |
Added explanation
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| Jul 23, 2020 at 6:57 | comment | added | Qmechanic♦ | Terms add up rather than cancel. I updated the answer. | |
| Jul 22, 2020 at 23:03 | comment | added | Ivan Burbano | 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 22:30 | history | answered | Qmechanic♦ | CC BY-SA 4.0 |