# Chern-Simons theory

The Chern-Simons 3-form is given by

$\omega_3={\rm Tr} \left[ A\wedge dA+\frac{2}{3}A\wedge A\wedge A\right]$

where $A$ is a connection one-form in the adjoint representation of a non-Abelian gauge group.

My differential geometry is rather rusty (and this is new to me too) hence my questions;

$A$ is a 1-form. By definition of the wedge product between a $p$ form $\alpha$ and $q$ form $\beta$ we have $\alpha\wedge\beta=(-1)^{pq}\beta\wedge\alpha$. So we should have $A\wedge A=-A\wedge A=0$.

Why is this not the case?

Next question; I want to calculate $d\omega_3$ Does the fact that everything is inside the trace effect my calculation? In other words does the differential operator pass through the trace and only act on the forms?

## 2 Answers

That's because you are forgetting that $A$ has a Yang-Mills index. You better write this in components, which reads

$\epsilon^{\mu\nu\rho} g_{IJ} \Big( A^I_\mu \partial_\nu A_\rho^J + \frac{1}{3} f^J{}_{KL} A^I_\mu A^K_\nu A^L_\rho \Big)$

This component notation also answers your second question.

• Right so each $A^I$ is a one-form in the sense of my original post which satisfies the wedge product property I listed? i.e $A^I\wedge A^J=-A^J\wedge A^I$ – Okazaki Jan 23 '16 at 20:13
• Yes. That's why "$A^I$ is a connection one-form in the adjoint representation of a non-Abelian gauge group". In the case of Abelian, that term just vanishes. – John Doe Jan 23 '16 at 20:18
• Is something like $A\wedge A$ written in component form as $g_{IJ}A^I\wedge A^J$ or do I need to include the Lorentz indicies too? – Okazaki Jan 23 '16 at 21:20
• $A \wedge A = A_\mu^I A_\nu^J d x^\mu \wedge dx^\nu T_I T_J = A_\mu^I A_\nu^J d x^\mu \wedge dx^\nu [T_I, T_J] = A_\mu^I A_\nu^J ( d x^\mu \wedge dx^\nu ) f^K{}_{IJ} T_K$. Note that I was not careful with the numerical factors. – John Doe Jan 23 '16 at 22:37
• You can read the (unofficial) notes by Malcolm Perry on Applications of Differential Geometry to Physics, link: aei.mpg.de/~gielen/diffgeo.pdf – John Doe Jan 23 '16 at 22:39

This is an example of lie-algebra valued 1-forms. Actually you may write explicitly, $A = A_{\mu} ^a T^a dx^{\mu}$. Since the generators also anti-commute so we get the result. And for the same reason sometimes you will find expressions like $[A,A]$ in literature for your term.