Computing the Chern-Simons invariant of $SO(3)$

Question

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.

Answer 1

I will use co-frames. If $\bar{\theta}_1$, $\bar{\theta}_2$ and $\bar{\theta}_3$ is the co-frame dual to $E_1$, $E_2$ and $E_3$, then the co-frame dual to $\lambda E_1$, $E_2$ and $E_3$ is:

$\theta_1 = \lambda^{-1} \bar{\theta}_1$, $\theta_2 = \bar{\theta}_2$ and $\theta_3 = \bar{\theta}_3$.

We then have: $d \theta_1 + 2\lambda^{-1} \theta_2 \wedge \theta_3 = 0$, $d \theta_2 +2 \lambda \theta_3 \wedge \theta_1 = 0$ and $d \theta_3 + 2 \lambda \theta_1 \wedge \theta_2 = 0$.

From Cartan's first structure equation $d \theta_i + \sum_j \theta_{ij} \wedge \theta_j = 0$, we then deduce that $\theta_{12} = -\lambda^{-1} \theta_3$, $\theta_{13} = \lambda^{-1} \theta_2$ and $\theta_{23} = (\lambda^{-1}-2\lambda) \theta_1$.

We then compute the curvature $2$-forms $\Omega_{ij} = d\theta_{ij} + \sum_k \theta_{ik} \wedge \theta_{kj}$. We obtain

$\Omega_{12} = \lambda^{-2} \theta_1 \wedge \theta_2$, $\Omega_{13} = \lambda^{-2} \theta_1 \wedge \theta_3$ and $\Omega_{23} = (4-3\lambda^{-2}) \theta_2 \wedge \theta_3$.

We then get that $\frac{1}{2} TP_1(\theta) = \frac{1}{2\pi^2}(-\lambda^{-4}+2\lambda^{-2}-2)Vol_{SO(3)}$, where $Vol_{SO(3)}$ denotes the volume form of the round $SO(3)$ (i.e. corresponding to $\lambda = 1$).

Since the volume of the round $SO(3)$ is $\pi^2$, the formula for the Chern-Simons invariant of $g_\lambda$ now follows.

Edit 1: I will provide more details. Using the formula (I think $6.1$ in that Chern-Simons paper):

$$TP_1(\theta) = \frac{1}{4\pi^2}\left( \theta_{12}\wedge\theta_{13}\wedge\theta_{23} + \theta_{12}\wedge\Omega_{12} + \theta_{13}\wedge\Omega_{13} + \theta_{23}\wedge\Omega_{23}\right),$$

we get that

\begin{align*} \frac{1}{2} TP_1(\theta) = &\frac{1}{8 \pi^2} \left(-\lambda^{-1}\lambda^{-1}(\lambda^{-1}-2\lambda) \, \theta_3 \wedge \theta_2 \wedge \theta_1 \right. \\ &-\lambda^{-1} \lambda^{-2} \, \theta_3 \theta_1 \theta_2 + \lambda^{-1} \lambda^{-2} \, \theta_2 \wedge \theta_1 \wedge \theta_3 \\ &\left.\,+(\lambda^{-1} - 2\lambda)(4 - 3\lambda^{-2}) \, \theta_1 \wedge \theta_2 \wedge \theta_3\right) \\ & = \frac{1}{2 \pi^2}(-\lambda^{-3} + 2\lambda^{-1} - 2\lambda) \, \theta_1 \wedge \theta_2 \wedge \theta_3 \end{align*} But remember the formulas $\theta_1 = \lambda^{-1} \bar{\theta}_1$, $\theta_2 = \bar{\theta}_2$ and $\theta_3 = \bar{\theta}_3$. We thus pick up an extra factor of $\lambda^{-1}$ when written in terms of the standard volume form on $SO(3)$. This is how I obtained:

$$\frac{1}{2} TP_1(\theta) = \frac{1}{2\pi^2}(-\lambda^{-4}+2\lambda^{-2}-2)\,Vol_{SO(3)}.$$

Finally, since the volume of $SO(3)$ is $\pi^2$, we get that

$$\Phi(g_{\lambda}) \equiv \frac{1}{2}(-\lambda^{-4} + 2 \lambda^{-2} -2) \equiv -\frac{1}{2} \lambda^{-4} + \lambda^{-2} \quad \text{(mod $\mathbb{Z}$)},$$

which is what Chern and Simons wrote, up to a minor algebraic manipulation. Yes, it is a tricky calculation!

Edit 2: I will explain how to obtain the connection $1$-forms $\theta_{ij}$. We should have

$$d\theta_1 + \theta_{12} \wedge \theta_2 + \theta_{13} \wedge \theta_3 = 0.$$

We have that $d\theta_1 + 2 \lambda^{-1} \theta_2 \wedge \theta_3 = 0$.

It can be shown that $\theta_{12} = h \theta_3$. Basically, if $\theta_{12}$ has a non-zero $\theta_1$ component, then this would lead to a non-zero $\theta_1 \wedge \theta_2$ component of $d\theta_1$, which is a contradiction. It can similarly be shown that $\theta_{12}$ has no $\theta_2$ component either.

A similar reasoning gives that: $$\theta_{12} = h\, \theta_3 \text{ , } \theta_{31} = g \,\theta_2 \text{ and } \theta_{23} = f \,\theta_1.$$

We therefore have, using the first two equations in edit 2, that:

$$ -g - h = 2 \lambda^{-1} .$$

Using similar equations for $d\theta_2$ and $d\theta_3$, we get that:

$$ -f - h = 2 \lambda, $$ $$ -f - g = 2 \lambda. $$

Solving these $3$ linear equations in $f, g$ and $h$ yields

$$f = \lambda^{-1} - 2\lambda \text{ , } g = -\lambda^{-1} \text{ , } h = -\lambda^{-1}.$$

Therefore, we have that

$\theta_{12} = -\lambda^{-1} \theta_3$, $\theta_{13} = \lambda^{-1} \theta_2$ and $\theta_{23} = (\lambda^{-1}-2\lambda) \theta_1.$