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

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.

• 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. – max Sep 30 '15 at 21:20

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 $$\theta_{23} = (\lambda^{-1}-2\lambda) \theta_1.$$

• How do you go from the expression involving $-\lambda^{-4} + 2\lambda^{-2} -2$ to the one in the original paper? You integrate the volume, but why is the final answer the addition of two instead of three terms? – Alonso Perez Lona Sep 9 '19 at 11:16
• @AlonsoPerezLona, it has been a while, I have to redo the calculation to remember. One moment. – Malkoun Sep 9 '19 at 11:43
• @AlonsoPerezLona, I edited my answer and provided more details. – Malkoun Sep 9 '19 at 14:46
• @AlonsoPerezLona. I edited again. Now my answer is more or less complete! I hope this helps. – Malkoun Sep 10 '19 at 10:14
• Well, you want to make sure that the integral converges. If you don't have compactness, you probably need to impose some decay conditions to make sure that the integral converges. – Malkoun Oct 25 '19 at 8:52