The Lie bracket from infinitesimals and the adjoint action

The endomorphism-valued differential. If $\mathfrak{g}=0$, its automorphism group is the trivial group, its endomorphism space is zero, and all bracket and commutator assertions are immediate; hence suppose $\mathfrak{g}\mathrm{\ne }0$. The adjoint representation is a morphism $\mathrm{Ad}:G\to {\mathrm{GL}}_{\mathfrak{g}}$, so it has a $k$-linear differential $\mathrm{ad}=\mathrm{Lie}(\mathrm{Ad}):\mathfrak{g}\to \mathrm{Lie}({\mathrm{GL}}_{\mathfrak{g}})$ by [F2], and $\mathrm{Lie}({\mathrm{GL}}_{\mathfrak{g}})\cong \mathrm{End}(\mathfrak{g})$ via $A\mapsto \mathrm{id}+\epsilon A$ by [F3]. In particular, for $X\in \mathfrak{g}$ and ${\epsilon }^2=0$, the identity of [F1] and the exponential identity of [F2] give $\mathrm{Ad}(e^{\epsilon X})=e^{\epsilon \mathrm{ad}(X)}=\mathrm{id}+\epsilon \mathrm{ad}(X)$ inside ${\mathrm{GL}}_{\mathfrak{g}}(k[\epsilon ])$.

The commutator formula (c). Let $R=k[t]\mathrm{/}(t^2)$ and regard the dual-number point $e^{tX}\in G(R)$ reducing to the identity in $G(k)$; applying [F1] with this $x$ and with $Y$, in the ring $R[\epsilon ]=k[t,\epsilon ]\mathrm{/}(t^2,{\epsilon }^2)$ one has $e^{tX}{e}^{\epsilon Y}{e}^{-tX}=e^{\epsilon \mathrm{Ad}(e^{tX})Y}$. By step 1.1 applied after the base change $\epsilon \mapsto t$, $\mathrm{Ad}(e^{tX})=\mathrm{id}+t\mathrm{ad}(X)$ in ${\mathrm{GL}}_{\mathfrak{g}}(R)$, so $\mathrm{Ad}(e^{tX})Y=Y+t[X,Y]$; since the exponential correspondence is additive by [F2], $e^{\epsilon (Y+t[X,Y])}=e^{\epsilon Y}{e}^{\epsilon t[X,Y]}$. Multiplying by $e^{-\epsilon Y}$ and renaming $\epsilon$ as $t^{\mathrm{\prime }}$ gives $e^{tX}{e}^{t^{\mathrm{\prime }}Y}{e}^{-tX}{e}^{-t^{\mathrm{\prime }}Y}=e^{t{t}^{\mathrm{\prime }}[X,Y]}$ in $G(k[t,t^{\mathrm{\prime }}]\mathrm{/}(t^2,t^{\mathrm{\prime }2}))$, the unique such element because its coefficient on every local function is $t{t}^{\mathrm{\prime }}{D}_Z(a)$, and $t{t}^{\mathrm{\prime }}$ is a nonzero $k$-basis monomial, so $e^{t{t}^{\mathrm{\prime }}Z}=e$ forces $D_Z=0$ by [F2]; the "equivalently" statement is exactly this identity read as the image under $\epsilon \mapsto t{t}^{\mathrm{\prime }}$.

The case $G={\mathrm{GL}}_n$ (d). Under the identification $\mathrm{Lie}({\mathrm{GL}}_n)={\mathfrak{g}\mathfrak{l}}_n$ of [F3], the element $X$ corresponds to the point $\mathrm{id}+\epsilon X$, and by [F3] its adjoint action is conjugation: $\mathrm{Ad}(\mathrm{id}+\epsilon X)Y=(\mathrm{id}+\epsilon X)Y(\mathrm{id}-\epsilon X)=Y+\epsilon (XY-YX)$, using $(\mathrm{id}+\epsilon X{)}^{-1}=\mathrm{id}-\epsilon X$ from [F5]. Comparing with step 1.1, which writes the same operator as $Y+\epsilon \mathrm{ad}(X)Y$, gives $[X,Y]=\mathrm{ad}(X)Y=XY-YX$.

Alternation, skew-symmetry and Jacobi. Put $A=\mathcal{O}(G)$ with augmentation $ϵ$ and comultiplication $\mathrm{\Delta }$ from The coordinate Hopf algebra of an affine group scheme. The point $e^{tX}$ is the algebra map $a\mapsto ϵ(a)+t{D}_X(a)$, where $D_X:A\to k$ is the tangent coefficient. The product of the two same-vector lifts $e^{tX},e^{t^{\mathrm{\prime }}X}$ evaluates $a$ as $ϵ(a)+(t+t^{\mathrm{\prime }})D_X(a)+t{t}^{\mathrm{\prime }}(D_X\otimes D_X)\mathrm{\Delta }(a)$, by the counit identities. Reversing the two lifts gives the identical formula, so they commute. Step 2.1 with $Y=X$ now gives $e^{t{t}^{\mathrm{\prime }}[X,X]}=e$, hence $[X,X]=0$: the map $Z\mapsto e^{t{t}^{\mathrm{\prime }}Z}$ is injective because its coefficient is $t{t}^{\mathrm{\prime }}{D}_Z(a)$ and $1,t,t^{\mathrm{\prime }},t{t}^{\mathrm{\prime }}$ are linearly independent over $k$. This proves alternation also in characteristic two. The bracket is bilinear since $\mathrm{ad}$ and its values are linear; expanding $[X+Y,X+Y]=0$ gives $[X,Y]+[Y,X]=0$. Applying the group homomorphism $\mathrm{Ad}$ to step 2.1 and using [F5] gives $\mathrm{ad}([X,Y])=\mathrm{ad}X\circ \mathrm{ad}Y-\mathrm{ad}Y\circ \mathrm{ad}X$, by comparison of $t{t}^{\mathrm{\prime }}$-coefficients. Evaluating this identity at $Z$ and using skew-symmetry gives the Jacobi identity. It also gives ${\mathrm{ad}}_X([Y,Z])=[[X,Y],Z]+[Y,[X,Z]]$, so every ${\mathrm{ad}}_X$ is a derivation. This proves (a) without a faithful embedding; the coefficient calculation adds no choice use beyond the finite-type suppliers.

Functoriality (b). Let $f:G\to H$ be a morphism of affine group schemes of finite type over $k$ and let $X,Y\in \mathfrak{g}$. Applying the homomorphism $f$ on points to the identity of step 2.1 and using $f_R(e^{\epsilon Z})=e^{\epsilon \mathrm{Lie}(f{)}_R(Z)}$ from [F2] gives $e^{t\mathrm{Lie}(f)X}{e}^{t^{\mathrm{\prime }}\mathrm{Lie}(f)Y}{e}^{-t\mathrm{Lie}(f)X}{e}^{-t^{\mathrm{\prime }}\mathrm{Lie}(f)Y}=e^{t{t}^{\mathrm{\prime }}\mathrm{Lie}(f)[X,Y]}$. The same commutator formula applied in $H$ identifies the left-hand side with $e^{t{t}^{\mathrm{\prime }}[\mathrm{Lie}(f)X,\mathrm{Lie}(f)Y]}$, so injectivity of the exponential correspondence [F2] gives $\mathrm{Lie}(f)[X,Y]=[\mathrm{Lie}(f)X,\mathrm{Lie}(f)Y]$.

Conclusion and uniqueness (e). By step 3.1 the bracket is a Lie bracket with every $\mathrm{ad}(X)$ a derivation; step 3.2 says $\mathrm{Lie}(f)$ is a homomorphism for every morphism; step 2.2 identifies the bracket on ${\mathrm{GL}}_n$; and step 2.1 is (c). For (e), let $i:G\hookrightarrow {\mathrm{GL}}_V$ be a closed immersion; by [F2], $\mathrm{Lie}(i)$ is injective, and by step 3.2 it is a homomorphism of Lie algebras for the present bracket. If $[-,-{]}^{\mathrm{\prime }}$ is another functorial bracket assignment agreeing with the matrix commutator on ${\mathrm{GL}}_n$, then for $X,Y\in \mathfrak{g}$ both $\mathrm{Lie}(i)([X,Y])$ and $\mathrm{Lie}(i)([X,Y{]}^{\mathrm{\prime }})$ equal $[\mathrm{Lie}(i)X,\mathrm{Lie}(i)Y{]}_{{\mathrm{GL}}_V}$, and injectivity gives $[X,Y]=[X,Y{]}^{\mathrm{\prime }}$; the closed immersion $i$ exists for every affine $G$ of finite type over $k$ by A finitely generated affine group scheme has a faithful finite-dimensional representation, which is the additional use of Choice in (e).