# Computing variation of triple wedge term in Chern-Simons action

For simplicity, assume $$G$$ is simple, compact, connected, and simply-connected. The Chern-Simons action for a non-abelian structure group $$G$$ on a trivial bundle with closed base manifold is given by $$S \propto \int_M\kappa(A \wedge dA + \frac{2}{3}A\wedge A \wedge A)$$ where $$\kappa$$ is the Killing form on $$\mathfrak{g}$$. I am trying to compute the variation of the triple wedge term $$T = \frac{2}{3}(A \wedge A \wedge A)$$: $$\delta T = \frac{2}{3}\kappa([\delta A \wedge A] \wedge A + [A \wedge \delta A] \wedge A + [A \wedge A] \wedge \delta A).$$ I see that two of these terms are the same, but I keep getting that the third term is the opposite sign of the other two. In particular, $$\kappa([A \wedge A ] \wedge \delta A) = - \kappa(\delta A \wedge [A \wedge A]) = -\kappa([\delta A \wedge A] \wedge A)$$ by symmetry of $$\kappa$$, cyclicity of $$\kappa$$, and antisymmetry of $$\wedge$$ counteracting antisymmetry of $$[-,-]$$, and $$\kappa([A \wedge \delta A] \wedge A) = \kappa([\delta A \wedge A] \wedge A)$$ by antisymmetry of $$\wedge$$ counteracting antisymmetry of $$[-,-]$$.

However, this leads to the erroneous result that $$\delta T =\frac{2}{3}\kappa(\delta A \wedge A \wedge A).$$ I believe that I have an incorrect understanding of manipulating these $$\mathfrak{g}$$-valued differential forms that is leading to this error. Hence, my question is how to correctly compute the variation of this triple wedge term.

1. The Lie-bracket $$[\cdot,\cdot]:\mathfrak{g}\times \mathfrak{g}\to \mathfrak{g}$$ and the Killing form $$\kappa:\mathfrak{g}\times \mathfrak{g}\to \mathbb{C}$$ are extended to Lie algebra-valued differential forms so that \begin{align} [A\stackrel{\wedge}{,}B]~=~&-(-1)^{|A||B|}[B\stackrel{\wedge}{,}A],\cr 0~=~&\sum_{A,B,C\text{ cycl.}}(-1)^{|A||C|}[A\stackrel{\wedge}{,}[B\stackrel{\wedge}{,}C]] \cr \kappa(A\stackrel{\wedge}{,}B)~=~&(-1)^{|A||B|}\kappa(B\stackrel{\wedge}{,}A),\cr \kappa(A\stackrel{\wedge}{,}[B\stackrel{\wedge}{,}C])~=~&\kappa([A\stackrel{\wedge}{,}B]\stackrel{\wedge}{,}C), \cr A,B,C~\in~&\Omega(M)\otimes\mathfrak{g}, \end{align}\tag{1} where $$|\cdot|$$ denotes the form degree.
2. In this answer we will work manifestly in the context of an abstract Lie algebra $$\mathfrak{g}$$, i.e. the only product of Lie algebra elements that we allow is the Lie product $$[\cdot,\cdot]$$. E.g., we will write the Lie algebra-valued 2-form field strength as $$F~=~\frac{1}{2}[\mathrm{d}+A\stackrel{\wedge}{,}\mathrm{d}+A]~=~\mathrm{d}A +\frac{1}{2}[A\stackrel{\wedge}{,}A] \tag{2}$$ rather than $$F=\mathrm{d}A +A\wedge A$$.
3. The non-Abelian Chern-Simons Lagrangian 3-form is in this context $$\mathbb{L}~=~\kappa\left(A\stackrel{\wedge}{,}~\mathrm{d}A +\frac{1}{3}[A\stackrel{\wedge}{,}A]\right).\tag{3}$$
4. An infinitesimal variation is \begin{align} \delta\mathbb{L}~\stackrel{(5)+(6)}{=}~&I+II\cr ~\stackrel{(5)+(6)}{=}~&\kappa\left(\underbrace{2\mathrm{d}A +[A\stackrel{\wedge}{,}A]}_{=2F}~\stackrel{\wedge}{,}\delta A\right)\cr &-\mathrm{d}\kappa(A\stackrel{\wedge}{,}\delta A),\end{align} \tag{4} where $$I~:=~\kappa\left(\delta A\stackrel{\wedge}{,}~\mathrm{d}A +\frac{1}{3}[A\stackrel{\wedge}{,}A]\right),\tag{5}$$ and \begin{align} II~:=~&\kappa\left( A\stackrel{\wedge}{,}~\mathrm{d}\delta A +\frac{2}{3}[A\stackrel{\wedge}{,}\delta A]\right)\cr ~=~&-\mathrm{d}\kappa(A\stackrel{\wedge}{,}\delta A)+\kappa(\mathrm{d}A\stackrel{\wedge}{,}\delta A)\cr &+\frac{2}{3}\kappa([A\stackrel{\wedge}{,}A]\stackrel{\wedge}{,}\delta A]).\end{align} \tag{6}
• Would you say that $\frac{1}{2}[A \stackrel{\wedge}{,} A]$ is the right way to write down this term? As opposed to $A \wedge A$? Am I understanding correctly that $\frac{1}{2}[A \stackrel{\wedge}{,} A] \to A \wedge A$ only when products between Lie algebra elements are defined? Commented Apr 27 at 21:40
• $\uparrow$ Yes. Commented Apr 27 at 22:02