Complex matrix Lie algebra

Let $𝐺 \subseteq G L 𝑛 (ℂ)$ be a Matrix Lie group, and $𝒞$ denote the set of curves in $𝐺$ intersecting identity at zero, i.e.

𝒞 = {𝑐 \in 𝖬 𝖺 𝗇 𝜔 (ℝ, 𝐺) : 𝑐 (0) = 𝟏 𝑛}

The matrix Lie algebra $𝔤$ is given by

𝔤 = {𝑐' (0) : 𝑐 \in 𝒞} \subseteq ℂ 𝑛 \times 𝑛

i.e. the tangent space at identity, with the Lie bracket given by the Linear commutator:

[𝑋, 𝑌] = 𝑋 𝑌 - 𝑌 𝑋

Proof of Lie algebra

By properties of the Linear commutator, $- 𝑖 [-, -]$ is alternating bilinear and satisfies the Jacobi identity. All that remains to show is that $𝔤$ is closed under this operation. To this end let $𝑥, 𝑦 \in 𝒞$ so that $𝑋 = - 𝑖 𝑥' (0) \in 𝔤$ and $𝑌 = - 𝑖 𝑦' (0) \in 𝔤$ . Now define a curve $˜ 𝑋 : ℝ \to 𝔤$ by $˜ 𝑋 (𝑡) = 𝑦 (𝑡) 𝑋 𝑦 (𝑡) - 1$ . Since $𝔤$ is a vector space and the Tangent space at any point in a vector space is the vector space, $˜ 𝑋' (0) \in 𝔤$ . From the product rule

𝑑 𝑑 𝑡 ˜ 𝑋 (𝑡) = 𝑦 (𝑡) 𝑑 𝑑 𝑡 𝑦 (𝑡) - 1 + 𝑦 (𝑡) - 1 𝑑 𝑑 𝑡 𝑦 (𝑡)

and

𝑑 𝑑 𝑡 𝑦 (𝑡) 𝑦 (𝑡) - 1 ∣ 𝑡 = 0 = 𝑦 (0) 𝑑 𝑑 𝑡 𝑦 (𝑡) - 1 ∣ 𝑡 = 0 + 𝑦 (0) - 1 𝑑 𝑑 𝑡 𝑦 (𝑡) ∣ 𝑡 = 0 = 𝑑 𝑑 𝑡 𝑦 (𝑡) - 1 ∣ 𝑡 = 0 + 𝑖 𝑌 = 0

it follows that $𝑑 𝑑 𝑡 𝑦 (𝑡) - 1 ∣ 𝑡 = 0 = - 𝑖 𝑌$ and thus

𝑑 𝑑 𝑡 ˜ 𝑋 (𝑡) ∣ 𝑡 = 0 = 𝑦 (0) 𝑋 𝑑 𝑑 𝑡 𝑦 (𝑡) - 1 ∣ 𝑡 = 0 + 𝑑 𝑑 𝑡 𝑦 (𝑡) ∣ 𝑡 = 0 𝑋 𝑦 (0) - 1 = - 𝑖 𝑋 𝑌 + 𝑖 𝑌 𝑋 = - 𝑖 [𝑋, 𝑌]

hence $𝔤$ is closed under the Lie bracket.

Basis

A good choice of basis uses coördinate lines in a chart containing the identity matrix. Let $(𝑈, 𝜑)$ be a $𝑛$ -dimensional coördinate chart with $𝜑 (⃗ 𝟎) = 𝑒$ . Then

𝑋 𝑖 = 𝜕 𝜕 𝑥 𝑖 𝜑 - 1 (𝑥 1, \dots, 𝑥 𝑛) ∣ ⃗ 𝑥 = ⃗ 𝟎