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The group $SU(n, \mathbb{C})$ is the set of $n \times n$ complex matrices $Q$ such that $\det Q = 1$ and $Q\overline{Q}^{T} = 1$. The Lie algebra is the set of traceless anti-Hermitian matrices.

To compute the Lie algebra, we have: Write $Q = I + \epsilon K$ be a matrix in $SU(n, \mathbb{C})$. Then as $Q\overline{Q}^{T} = I$, $I + \epsilon(\overline{K}^{T} + K) + \epsilon^{2}K\overline{K}^{T} = I$. Letting $\epsilon \rightarrow 0$, we have $\overline{K}^{T} + K = 0$ and $K$ must be anti-Hermitian. How do I find that the matrices in the Lie algebra must also be traceless?

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A matrix tangent to $SL(n)$ can be given by a smooth curve $$ c:]-\epsilon;+\epsilon[ \rightarrow M_n \quad \operatorname{det}(c(t))=1 \text{ for all t and } c(0)=E_n $$ det is multilinear and therefore has a derivative that is easy to compute $$ 0=\operatorname{trace}(c'(0)) $$

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