SL(n) as a closed Lie subgroup of GL(n)

The locus $\det \mathrm{\ne }0$ is open in the finite-dimensional real vector space underlying $M_n(\mathbb{F})$. Matrix multiplication is polynomial there, and the adjugate formula $A^{-1}=\mathrm{adj}(A)\mathrm{/}\det A$ makes inversion smooth, so this locus is the Lie group ${\mathrm{GL}}_n(\mathbb{F})$ in the sense of [F1]. The determinant is polynomial, hence smooth, and multiplicativity makes $\det :{\mathrm{GL}}_n(\mathbb{F})\to {\mathbb{F}}^{\times }$ a Lie-group homomorphism. Its identity fibre is exactly ${\mathrm{SL}}_n(\mathbb{F})$, so [A1] makes this fibre a closed embedded normal Lie subgroup.

In the Leibniz expansion of $\det (I+tX)$, the identity permutation contributes $1+t{\sum }_i{X}_{ii}+O(t^2)$, while every nonidentity permutation needs at least two off-diagonal factors and contributes $O(t^2)$. Thus $d(\det {)}_I(X)=\mathrm{tr}X$. This differential is onto $\mathbb{F}$: the matrix $\mathrm{diag}(z,0,\dots ,0)$ has trace $z$. Left multiplication transports surjectivity to every point of the identity fibre, so the fibre is regular and [F2] gives the same tangent kernel.

Combining steps 1.1 and 1.2 with [A1] yields $\mathrm{Lie}({\mathrm{SL}}_n(\mathbb{F}))=\ker d(\det {)}_I=\{X:\mathrm{tr}X=0\}$. For $n=1$ the subgroup and Lie algebra are both trivial; singular matrices $X$ are allowed as tangent vectors. The complex case is read as a real Lie group, and the complex-linear trace map is also onto as a real map. No endpoint or metric choice occurs. ${\mathrm{A}\mathrm{C}}_{\omega }$ is inherited exactly through [A1].