Haar measure of a compact Lie group

Let $𝐺$ be an $𝑛$ -dimensional Compact Lie group, $𝔤$ be its corresponding Lie algebra¹ with basis ${𝑋 𝑗} 𝑛 𝑗 = 1$ , and $(𝑈, 𝜑)$ be a coördinate chart with $𝜑 (𝑒) = ⃗ 𝟎$ . For each $𝜑 (𝑥) \in 𝑈 \subseteq 𝐺$ define $𝐴 𝜑 𝑗 𝑘 (𝑥)$ so that

- 𝑖 𝜑 - 1 (𝑥) - 1 𝜕 𝑗 𝜑 - 1 (𝑥) = 𝑛 \sum 𝑘 = 1 𝑋 𝑘 𝐴 𝜑 𝑘 𝑗 (𝑥)

where both sides are clearly elements of $𝔤$ . Then for any Borel set $𝐵 \subseteq 𝑈$ , the unique left and right Haar measure $𝜇 (𝐵)$ is given by group

𝜇 (𝐵) = \int 𝐵 𝑑 𝜇 (𝑔) = \int 𝜑 (𝐵) 𝛼 | d e t 𝐀 𝜑 (𝑥) | 𝑑 𝑛 𝑥

where $𝛼$ is a normalisation constant. The Haar measure is defined for the whole of $𝐺$ by translating the chart (enabled by invariance).

Proof this is a Haar measure

Let $(𝑉, 𝜓)$ define a chart with $𝑈 \cap 𝑉 \neq \emptyset$ , and let $𝑓$ be a transition map such that $𝜓 (𝑥) = 𝜑 (𝑓 (𝑥))$ for all $𝑥 \in 𝜓 (𝑈 \cap 𝑉)$ . Then
$𝑛 \sum 𝑘 = 1 𝑋 𝑘 𝐴 𝜓 𝑘 𝑗 (𝑥) = - 𝑖 𝜓 - 1 (𝑥) - 1 𝜕 𝑗 𝜓 - 1 (𝑥) = - 𝑖 [𝜑 - 1 \circ 𝑓] (𝑥) - 1 𝜕 𝑗 [𝜑 - 1 \circ 𝑓] (𝑥) = - 𝑖 [𝜑 - 1 \circ 𝑓] (𝑥) - 1 𝑛 \sum ℓ = 1 𝜕 𝑗 𝑓 ℓ (𝑥) [𝜕 ℓ 𝜑 - 1 \circ 𝑓] (𝑥) = 𝑛 \sum ℓ, 𝑘 = 1 𝑋 𝑘 𝐴 𝜑 𝑘 ℓ (𝑓 (𝑥)) 𝜕 𝑗 𝑓 ℓ (𝑥)$
i.e. $𝐀 𝜓 (𝑥) = 𝐀 𝜑 (𝑓 (𝑥)) 𝐃 𝑓 (𝑥)$ and thus
$∣ d e t 𝐴 𝜓 (𝑥) ∣ = | d e t 𝐴 𝜑 (𝑓 (𝑥)) | | d e t 𝐃 𝑓 |$
as required.

For $𝑔 \in 𝐺$ we define the chart
$𝜑 𝑔 : 𝑔 𝑈 \to 𝜑 𝑔 (𝑈) ℎ \mapsto 𝜑 (𝑔 - 1 ℎ)$
then
$- 𝑖 𝜑 - 1 𝑔 (𝑥) - 1 𝜕 𝑗 𝜑 - 1 𝑔 (𝑥) = - 𝑖 (𝑔 𝜑 - 1 (𝑥)) - 1 𝑔 𝜕 𝑗 𝜑 - 1 (𝑥) = - 𝑖 𝜑 - 1 (𝑥) - 1 𝜕 𝑗 𝜑 - 1 (𝑥)$
which gives left-invariance.

For each $𝑔 \in 𝐺$ let $𝑀 𝑘 ℓ (𝑔)$ so that $𝑔 - 1 𝑋 𝑘 𝑔 = 𝑋 ℓ Δ ℓ 𝑘 (𝑔)$ . Letting $𝜓 (ℎ) = 𝜓 (ℎ 𝑔 - 1)$ , i.e. $𝜓 - 1 (𝑥) = 𝜑 - 1 (𝑥) 𝑔$ , then
$- 𝑖 𝜓 - 1 (𝑥) - 1 𝜕 𝑗 𝜓 - 1 (𝑥) = - 𝑖 𝑔 - 1 𝜑 - 1 (𝑥) - 1 𝜕 𝑗 𝜑 - 1 (𝑥) 𝑔 = 𝑛 \sum 𝑘 = 1 𝑔 - 1 𝑋 𝑘 𝑔 𝐴 𝜑 𝑘 𝑗 (𝑥) = 𝑛 \sum 𝑘, ℓ = 1 𝑋 ℓ 𝑀 ℓ 𝑘 (𝑔) 𝐴 𝜑 𝑘 𝑗 (𝑥)$
i.e. $𝐴 𝜓 = 𝑀 (𝑔) 𝐴 𝜑$ . But
$\int 𝐺 𝑑 𝜇 (𝑔) = \int 𝐺 𝑑 𝜇 (𝑔' ℎ - 1) = ∣ d e t 𝑀 (ℎ - 1) ∣ \int 𝐺 𝑑 𝜇 (𝑔)$
and since $𝐺$ is compact all integrals are finite, thus the $∣ d e t 𝑀 (ℎ - 1) ∣ = 1$ , i.e. $𝐺$ is unimodular. Therefore $𝜇$ is right-invariant.

Proof of uniqueness

Let $𝜇, 𝜈$ both be two sided Haar measures normalised such that $𝜇 (𝐺) = 𝜈 (𝐺) = 1$ . Then for any $𝑓 \in ℂ [𝐺]$
$\int 𝐺 𝑓 (𝑔) 𝑑 𝜇 (𝑔) = \int 𝐺 \int 𝐺 𝑓 (𝑔) 𝑑 𝜇 (𝑔) 𝑑 𝜈 (ℎ) = \int 𝐺 \int 𝐺 𝑓 (ℎ 𝑔) 𝑑 𝜇 (𝑔) 𝑑 𝜈 (ℎ) = \int 𝐺 \int 𝐺 𝑓 (ℎ) 𝑑 𝜇 (𝑔) 𝑑 𝜈 (ℎ) = \int 𝐺 𝑓 (ℎ) 𝑑 𝜈 (ℎ)$
thus $𝜇 = 𝜈$ .

Properties

Properties under integration

Let $𝑓 \in ℂ [𝐺]$ and $𝜙 : 𝐺 \to 𝐺$ be a bijection (e.g. taking the inverse of each group element)

\int 𝐺 𝑓 (ℎ 𝑔) 𝑑 𝜇 (𝑔) = \int 𝐺 𝑓 (𝑔) 𝑑 𝜇 (ℎ - 1 𝑔) = \int 𝐺 𝑓 (𝑔) 𝑑 𝜇 (𝑔)

\int 𝐺 𝑓 (𝑔 ℎ) 𝑑 𝜇 (𝑔) = \int 𝐺 𝑓 (𝑔) 𝑑 𝜇 (𝑔 ℎ - 1) = \int 𝐺 𝑓 (𝑔) 𝑑 𝜇 (𝑔)

\int 𝐺 𝑓 (𝜙 (𝑔)) 𝑑 𝜇 (𝑔) = \int 𝐺 𝑓 (𝑔) 𝑑 𝜇 (𝑔)

Proof of property 3

From properties 1 and 2
$\int 𝐺 𝑓 (𝜙 (𝑔)) 𝑑 𝜇 (𝑔) = \int 𝐺 𝑓 (ℎ 𝜙 (𝑔)) 𝑑 𝜇 (𝑔) = \int 𝐺 \int 𝐺 𝑓 (ℎ 𝜙 (𝑔)) 𝑑 𝜇 (ℎ) 𝑑 𝜇 (𝑔) = \int 𝐺 \int 𝐺 𝑓 (ℎ) 𝑑 𝜇 (ℎ) 𝑑 𝜇 (𝑔) = \int 𝐺 𝑑 𝜇 (𝑔)$
where we used the normalisation of the group to 1.

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Here we use Keppeler’s Lie algebra convention ↩

Table of Contents

Haar measure of a compact Lie group

Properties

Properties under integration

Graph View

Backlinks

Table of Contents

Haar measure of a compact Lie group

Properties

Properties under integration

Footnotes

Graph View

Backlinks