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I am an undergraduate learning about gauge theory and I have been tasked with working through the two examples given on pages 65 and 66 of "Characteristic forms and geometric invariants" by Chern and Simon. I will recount the examples and my progress at a solution. For ease here is the relevant text:

Example 1. Let $M = \mathbb{R}P^3 = SO(3)$ together with the standard metric of constant curvature 1. Let $E_1, E_2, E_3$ be an orthonormal basis of left invariant fields on $M$, oriented positively. Then it is easily seen that $\nabla_{E_1}E_2 = E_3, \nabla_{E_1}E_3 = - E_2, \text{ and } \nabla_{E_2}E_3 = E_1$. Let $\chi : M \rightarrow F(M)$ be the cross-section determined by this frame. $$\Phi(SO(3)) = \frac{1}{2}.$$

Example 2. Again let $M = SO(3)$, but this time with left invariant metric $g_{\lambda}$, with respect to which $\lambda E_1, E_2, E_3$ is an orthonormal frame. Direct calculation shows $$\Phi(SO(3),g_{\lambda}) = \frac{2\lambda^2 - 1}{2\lambda^4}.$$

For each of these examples I am expected to calculate
$$\Phi(M) = \int_{\chi(M)} \frac{1}{2} TP_1(\theta)$$ which lies in $\mathbb{R}/\mathbb{Z}$. Previously in the paper they give an explicit formulation of $TP_1(\theta)$ in terms of the "component" forms of the connection $\theta$ and its curvature $\Omega$, $$TP_1(\theta) = \frac{1}{4\pi^2}\left( \theta_{12}\wedge\theta_{13}\wedge\theta_{23} + \theta_{12}\wedge\Omega_{12} + \theta_{13}\wedge\Omega_{13} + \theta_{23}\wedge\Omega_{23}\right).$$

I have verified this formula for myself given the information in the paper. Using the structural equation $\Omega = d\theta + \theta\wedge\theta$ I am able to reduce the expression for $TP_1(\theta)$ to $$TP_1(\theta) = \frac{-1}{2\pi^2}\left( \theta_{12}\wedge\theta_{13}\wedge\theta_{23} \right).$$

I don't believe I have assumed anything about the structure of $M$ during that reduction so I believe it should hold for both examples. I continue by claiming that since $E_1, E_2, E_3 \in so(3)$, the Lie algebra of $SO(3)$ I should be able to compute $\theta$ by considering $$\nabla_{E_i}E_j := (\nabla E_j)(E_i) = \sum_k E_k \otimes \theta^{k}{}_{ij}(E_i)$$ and comparing it with the given derivatives.

For example one this yielded for me $\theta_{12} = E^3, \theta_{13} = -E^2, \theta_{23} = E^1$ where $E^i$ are the 1-forms dual to the basis $E_i$. Then I think that $\chi^*$ should act trivially on $TP_1(\theta)$ as it is a horizontal form in $\Lambda^*(T^*F(M))$. Therefore I find that $\chi^*(TP_1(\theta)) = \frac{1}{2\pi^2}\omega$, where $\omega$ is the volume form of $M$, and when integrated this yields the correct answer of $\frac{1}{2}$ for the first example.

However, my approach fails completely for the second example. I assume that the set $\lambda E_1, E_2, E_3$ obeys the same derivate relationships as given in the first example, but this does not seem to give me enough factors of $\lambda$. I suspect that I am not handling the computation of the $\theta_{ij}$ forms or the application of $\chi^*$ correctly, however I am uncertain what my exact issue is. Is there a fundamental flaw in my understanding? I am hoping someone with more experience can point me in the right direction.

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In simplifying the expression for $T_1(\theta)$ using the structure equation, what happens to the terms that are like $\theta_{12} \wedge d\theta_{12}$? – Eric O. Korman Dec 10 '10 at 18:49
This has also been asked on MO where it has received a partial answer. – Michael Albanese Jan 4 '15 at 21:20
The problem lies in the deformation you have applied to the sphere, which messes up the covariant derivative and the volume form. If everything is calculated with this in mind, you get the right answer. – Pax Kivimae Sep 30 '15 at 21:20
I know this post is old, but, I tried the calculation, and I got $J=-1/2$ for the "standard" round metric on $SO(3)$, which is fine, since it is modulo $Z$. However, for the Berger metrics, I obtained: $J=-\lambda^{-2}(\lambda^4-(5/4)\lambda^2 + 3/4)$, which is not the same as what is written in the Chern-Simons paper... I probably made some mistake somewhere. – Malkoun Jun 14 at 6:46

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