I recently tried asking this question on math.stackexchange.com but I have not yet received any feedback so I thought I should try asking here. I apologize if this question is too basic. The question is reproduced below:

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.


Not going into details, just would like to note that connection forms do depend on $\lambda$. If I take $t=\lambda^{-1}$, then from Coshul formula $$2(\nabla_{E_1}E_2,E_3) = E_1(E_2,E_3) + E_2(E_1,E_3) - E_3(E_1,E_3) - (E_1,[E_2,E_3]) - (E_2,[E_1,E_3]) + (E_3,[E_1,E_2])$$ it follows $2(\nabla_{E_1}E_2,E_3) = -t^2 + 2$ (from $(E_1,E_1)=t^2$). The double cover of $SO(3)$ with this metric $g_t$ is the well known "Berger sphere" which collapses to the base of the Hopf fibration $S^1 \to \pi: S^3 \to S^2$ when $t$ goes to $0$, and the fiber $S^1$ as t goes to $\infty$. You may find complete formulas for the connection $\nabla^t$ of $g_t$ by searching Berger sphere, I think.

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