Varying the Chern-Simons action

Question 1

Summary/TL;DR

I want a detailed calculation of the derivation of classical equations of motion from the Chern-Simons action using differential forms, using variational derivatives. I mentioned "using variational derivatives" because the Lagrangian density might be case as an exact form (which I know how to derive) from which EOM can be found out easily, maybe: although I don't know how to do that. I would love to know finding EOM in that method, but that should be a separate question. Here I just want to know how$(1)$ and$(2)$ below are written, if correct at all.

Introduction

The Chern-Simons action is given by$$S=\int_N \operatorname{tr}(A \wedge dA+\frac{2}{3}A \wedge A \wedge A)$$ The equation of motion is well known$F:=dA+A^2=0$. I am having great trouble deriving this. From the proof of Proposition$(3.12)$ on pg.5 ofthis article extracted from"The geometry of infinite-dimensional groups Springer" by Khesin and Wendt (2008) in which it appears on pg. 185.

$$\delta S =\int_N \operatorname{tr}(\delta A \wedge d A)+\int_N \operatorname{tr}(A \wedge d \delta A)+2 \int_N \operatorname{tr}(\delta A \wedge A \wedge A) \tag{1} $$$$=\int_N d \operatorname{tr}(A \wedge \delta A)+2 \int_N \operatorname{tr}(\delta A \wedge(d A+A \wedge A)) \tag{2}$$$$=\int_{\partial N} \operatorname{tr}(A \wedge \delta A)+2 \int_N \operatorname{tr}(\delta A \wedge(d A+A \wedge A)) \tag{3} $$ Therefore, killing the boundary term we get,$F=0$ as expected.

What I have tried

In writing 1st two terms of (1) Leibnitz rule for$\delta$ has been used, and for writing the third term cyclicity of trace and anticommutativity of the basis 1-form$dx^\mu$ are used in addition. I have assumed the Leibnitz rule of$\delta$ to be ungraded unlike the exterior derivative$d$, although I am not sure whether that is true or not.

In (2)$d \operatorname{tr}(A \wedge \delta A)=\operatorname{tr}(dA \wedge \delta A)-\operatorname{tr}(A \wedge \delta d A)$

I don't understand$(2)$. Naively I thought$$d \operatorname{tr}(A \wedge \delta A)=\operatorname{tr}(dA \wedge \delta A)+\operatorname{tr}(A \wedge \delta d A)\tag{4}$$$$=-\operatorname{tr}(\delta A \wedge d A)+\operatorname{tr}(A \wedge \delta dA) \tag{5}$$ which when put in$(2)$ gives back$(1)$. But this is wrong because the Leibnitz rule for$d$ operator is graded$d (\omega \wedge \eta)=d \omega \wedge \eta+ (-1)^{\text{dim}(\omega)}\omega \wedge d \eta$, so there should be an extra minus sign (since$A$ is a 1-form) giving$$d \operatorname{tr}(A \wedge \delta A)=\operatorname{tr}(dA \wedge \delta A)-\operatorname{tr}(A \wedge \delta d A) \tag{6}$$ In writing$(5)$ I used antisymmetry of wedge product which is again wrong because of the grading$\omega \wedge \eta= (-1)^{d_{\omega \eta}} \eta \wedge \omega$ where$d_{\omega \eta}$ is the product of dimensions of the forms$\omega$ and$\eta$. There should be no extra minus sign as in$(5)$ because the dimension of$\delta A$ is lesser than$dA$ by$1$, so their product is an even number: the wedge product is actually symmetric, not antisymmetric. This finally gives from$(6)$$$d \operatorname{tr}(A \wedge \delta A)=\operatorname{tr}(\delta A \wedge d A)-\operatorname{tr}(A \wedge \delta d A) \tag{7}$$ But putting this carefully calculated expression in$(2)$ does not give back$(1)$. One way I could probably "make this work" if$\delta$ itself also had a grading, or maybe the$\delta$ and$d$ anticommutes? But if that was the case I don't know how I would obtain the 2nd and 3rd term in$(1)$. Moreover, the boundary term has opposite sign in equation (4.15) pg. 9 ofthis article"1602.09021 Asymptotic dynamics of three-dimensional gravity" by Laura Donnay , so I am not sure which computation is correct and why.

Question 2

Related:physics.stackexchange.com/q/811913/2451

Question 3

Let$(P, M, \pi, G)$ be a principal bundle such that$M$ is a closed oriented three dimensional smooth manifold and$G$ is a compact, connected, simply-connected, and simple Lie group. The orientation of$M$ necessary to integrate over$M$. That$G$ is compact and simply-connected implies that$G$ is non-abelian. That$G$ is simply connected and$\dim M \leq 3$ imply that$P \xrightarrow{\pi} M$ admits global sections, i.e.$P$ is a trivial bundle. That$G$ is simple means$\mathfrak{g}$ is simple, which implies that all$\text{ad}$-invariant symmetric bilinear forms are proportional to the Killing form.

FromQmechanic's answer linked in the comments, we have the defining properties of the Lie bracket, Killing form, and wedge product extended from their usual domains to the domain of$\mathfrak{g}$-valued differential forms:$$[X \stackrel{\wedge}{,} Y] = -(-1)^{\lvert X \lvert \lvert Y \lvert}[Y \stackrel{\wedge}{,} X] \\ \kappa(X \stackrel{\wedge}{,} Y) = (-1)^{\lvert X \lvert \lvert Y \lvert}\kappa(Y \stackrel{\wedge}{,} X) \\ \kappa(X \stackrel{\wedge}{,} [Y \stackrel{\wedge}{,} Z]) = \kappa([X \stackrel{\wedge}{,} Y] \stackrel{\wedge}{,} Z) \tag{1}$$where$X, Y, Z \in \mathfrak{g} \otimes \Omega(M)$ and$\lvert X \lvert$ is the degree of the form.

The classical non-abelian Chern-Simons action on a trivial bundle is given by$$S = \frac{k}{8\pi^2}\int_M \kappa\left( A \stackrel{\wedge}{,} dA + \frac{1}{3} [A \stackrel{\wedge}{,} A] \right), \quad k \in \mathbb{R}\backslash\{0\}. \tag{2}$$The equations of motion are found by stationizing the action to first order (that is, by ignoring second order and higher variations). An arbitrary variation of the action is given by$$\delta S = S[A + \delta A] - S[A] \\ = \frac{k}{8\pi^2}\int_M \kappa\left( A+\delta A \stackrel{\wedge}{,} d(A+\delta A) + \frac{1}{3} [A + \delta A \stackrel{\wedge}{,} A + \delta A] \right) - \frac{k}{8\pi^2}\int_M \kappa\left( A \stackrel{\wedge}{,} dA + \frac{1}{3} [A \stackrel{\wedge}{,} A] \right) \\ = \frac{k}{8\pi^2} \int_M \biggl(\kappa(A \stackrel{\wedge}{,} dA) + \kappa(A \stackrel{\wedge}{,} \delta dA) + \kappa(\delta A \stackrel{\wedge}{,} dA) + \kappa(\delta A \stackrel{\wedge}{,} \delta dA) + \kappa(A \stackrel{\wedge}{,} \frac{1}{3}[A + \delta A \stackrel{\wedge}{,} A + \delta A]) \\ \quad + \kappa(\delta A \stackrel{\wedge}{,} \frac{1}{3}[A + \delta A \stackrel{\wedge}{,} A + \delta A]) - \kappa(A \stackrel{\wedge}{,} dA) - \kappa(A \stackrel{\wedge}{,} \frac{1}{3}[A \stackrel{\wedge}{,} A]) \biggr) \\ = \frac{k}{8\pi^2} \int_M \biggl( \kappa(A \stackrel{\wedge}{,} \delta d A) + \kappa(\delta A \stackrel{\wedge}{,} dA) + \kappa(A \stackrel{\wedge}{,} \frac{2}{3} [A \stackrel{\wedge}{,} \delta A]) + \kappa(\delta A \stackrel{\wedge}{,} \frac{1}{3}[A \stackrel{\wedge}{,} A])\biggr) \\ = \frac{k}{8\pi^2} \int_M \biggl( 2\kappa(dA \stackrel{\wedge}{,} \delta A) - d\kappa(A \stackrel{\wedge}{,} \delta A) + \kappa([A \stackrel{\wedge}{,} A] \stackrel{\wedge}{,} \delta A) \biggr) \\ = \frac{k}{8\pi^2} \int_M \kappa(2dA + [A \stackrel{\wedge}{,} A] \stackrel{\wedge}{,} \delta A)$$where the third equality is obtained using the defining rules$(1)$ and the fifth equality is obtained using generalized product rule, generalized Stokes' theorem, and the fact that$M$ is a closed manifold (in particular, that$M$ has no boundary).

Requiring the action to be stationized by$A$,$$ \delta S = \frac{k}{8\pi^2} \int_M \kappa(2dA + [A \stackrel{\wedge}{,} A] \stackrel{\wedge}{,} \delta A) = 0 \quad \forall \delta A\\ \implies \frac{k}{8\pi^2}(2dA + [A \stackrel{\wedge}{,} A]) = 0. \tag{3}$$We define the field strength of$A$ by$$ F = dA + \frac{1}{2}[A \stackrel{\wedge}{,} A],$$so the equation of motion is$$ dA + \frac{1}{2}[A \stackrel{\wedge}{,} A] = F = 0 \tag{4}$$where we have multiplied both sides of$(3)$ by$\frac{4\pi^2}{k}$.

Silly Goose 4,1131 gold badge7 silver badges29 bronze badges · Accepted Answer · 2024-06-17 06:35:24Z

Let$(P, M, \pi, G)$ be a principal bundle such that$M$ is a closed oriented three dimensional smooth manifold and$G$ is a compact, connected, simply-connected, and simple Lie group. The orientation of$M$ necessary to integrate over$M$. That$G$ is compact and simply-connected implies that$G$ is non-abelian. That$G$ is simply connected and$\dim M \leq 3$ imply that$P \xrightarrow{\pi} M$ admits global sections, i.e.$P$ is a trivial bundle. That$G$ is simple means$\mathfrak{g}$ is simple, which implies that all$\text{ad}$-invariant symmetric bilinear forms are proportional to the Killing form.

FromQmechanic's answer linked in the comments, we have the defining properties of the Lie bracket, Killing form, and wedge product extended from their usual domains to the domain of$\mathfrak{g}$-valued differential forms:$$[X \stackrel{\wedge}{,} Y] = -(-1)^{\lvert X \lvert \lvert Y \lvert}[Y \stackrel{\wedge}{,} X] \\ \kappa(X \stackrel{\wedge}{,} Y) = (-1)^{\lvert X \lvert \lvert Y \lvert}\kappa(Y \stackrel{\wedge}{,} X) \\ \kappa(X \stackrel{\wedge}{,} [Y \stackrel{\wedge}{,} Z]) = \kappa([X \stackrel{\wedge}{,} Y] \stackrel{\wedge}{,} Z) \tag{1}$$where$X, Y, Z \in \mathfrak{g} \otimes \Omega(M)$ and$\lvert X \lvert$ is the degree of the form.

The classical non-abelian Chern-Simons action on a trivial bundle is given by$$S = \frac{k}{8\pi^2}\int_M \kappa\left( A \stackrel{\wedge}{,} dA + \frac{1}{3} [A \stackrel{\wedge}{,} A] \right), \quad k \in \mathbb{R}\backslash\{0\}. \tag{2}$$The equations of motion are found by stationizing the action to first order (that is, by ignoring second order and higher variations). An arbitrary variation of the action is given by$$\delta S = S[A + \delta A] - S[A] \\ = \frac{k}{8\pi^2}\int_M \kappa\left( A+\delta A \stackrel{\wedge}{,} d(A+\delta A) + \frac{1}{3} [A + \delta A \stackrel{\wedge}{,} A + \delta A] \right) - \frac{k}{8\pi^2}\int_M \kappa\left( A \stackrel{\wedge}{,} dA + \frac{1}{3} [A \stackrel{\wedge}{,} A] \right) \\ = \frac{k}{8\pi^2} \int_M \biggl(\kappa(A \stackrel{\wedge}{,} dA) + \kappa(A \stackrel{\wedge}{,} \delta dA) + \kappa(\delta A \stackrel{\wedge}{,} dA) + \kappa(\delta A \stackrel{\wedge}{,} \delta dA) + \kappa(A \stackrel{\wedge}{,} \frac{1}{3}[A + \delta A \stackrel{\wedge}{,} A + \delta A]) \\ \quad + \kappa(\delta A \stackrel{\wedge}{,} \frac{1}{3}[A + \delta A \stackrel{\wedge}{,} A + \delta A]) - \kappa(A \stackrel{\wedge}{,} dA) - \kappa(A \stackrel{\wedge}{,} \frac{1}{3}[A \stackrel{\wedge}{,} A]) \biggr) \\ = \frac{k}{8\pi^2} \int_M \biggl( \kappa(A \stackrel{\wedge}{,} \delta d A) + \kappa(\delta A \stackrel{\wedge}{,} dA) + \kappa(A \stackrel{\wedge}{,} \frac{2}{3} [A \stackrel{\wedge}{,} \delta A]) + \kappa(\delta A \stackrel{\wedge}{,} \frac{1}{3}[A \stackrel{\wedge}{,} A])\biggr) \\ = \frac{k}{8\pi^2} \int_M \biggl( 2\kappa(dA \stackrel{\wedge}{,} \delta A) - d\kappa(A \stackrel{\wedge}{,} \delta A) + \kappa([A \stackrel{\wedge}{,} A] \stackrel{\wedge}{,} \delta A) \biggr) \\ = \frac{k}{8\pi^2} \int_M \kappa(2dA + [A \stackrel{\wedge}{,} A] \stackrel{\wedge}{,} \delta A)$$where the third equality is obtained using the defining rules$(1)$ and the fifth equality is obtained using generalized product rule, generalized Stokes' theorem, and the fact that$M$ is a closed manifold (in particular, that$M$ has no boundary).

Requiring the action to be stationized by$A$,$$ \delta S = \frac{k}{8\pi^2} \int_M \kappa(2dA + [A \stackrel{\wedge}{,} A] \stackrel{\wedge}{,} \delta A) = 0 \quad \forall \delta A\\ \implies \frac{k}{8\pi^2}(2dA + [A \stackrel{\wedge}{,} A]) = 0. \tag{3}$$We define the field strength of$A$ by$$ F = dA + \frac{1}{2}[A \stackrel{\wedge}{,} A],$$so the equation of motion is$$ dA + \frac{1}{2}[A \stackrel{\wedge}{,} A] = F = 0 \tag{4}$$where we have multiplied both sides of$(3)$ by$\frac{4\pi^2}{k}$.

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