Lie Algebras and Infinitesimal Group Schemes

Classification over an arbitrary $R$. Put $B={\mathcal{O}}_{G,e}$ and $\mathfrak{m}=\ker (B\to k)$. Every morphism $\mathrm{Spec}R[\epsilon ]\to G$ reducing to the constant identity has underlying image $e$, since the nilpotent thickening has the same points as $\mathrm{Spec}R$; it therefore factors through every affine neighbourhood of $e$. Every element of a chart algebra outside the prime of $e$ maps to a unit: its reduction is a nonzero scalar in $k$, and $c+\epsilon r$ has inverse $c^{-1}-\epsilon c^{-2}r$. Consequently such morphisms correspond exactly to $k$-algebra maps $B\to R[\epsilon ]$ of the form $b\mapsto ϵ(b)+\epsilon D(b)$, where $ϵ:B\to k\to R$ is augmentation and $D:B\to R$ is $k$-linear with $D(b{b}^{\mathrm{\prime }})=ϵ(b)D(b^{\mathrm{\prime }})+ϵ(b^{\mathrm{\prime }})D(b)$. The splitting $B=k\oplus \mathfrak{m}$ shows that these $D$ correspond exactly to $k$-linear maps $\mathfrak{m}\mathrm{/}{\mathfrak{m}}^2\to R$: the product rule kills ${\mathfrak{m}}^2$, and conversely that rule follows by multiplying the two decompositions into scalar and augmentation parts. This gives a natural bijection ${\beta }_R:\mathrm{Lie}(G)(R)\to {\mathrm{Hom}}_k(\mathfrak{m}\mathrm{/}{\mathfrak{m}}^2,R)$, valid also for $R=0$. For $R=k$ it is the coefficient bijection of [F1] and [F5].

Finite-dimensional tensor identification. Write $V=\mathfrak{m}\mathrm{/}{\mathfrak{m}}^2$. It is finite-dimensional by [F6], and a basis is obtained by finite elimination as in [F7]. In that basis the natural map $V^{\vee }{\otimes }_{k}R\to {\mathrm{Hom}}_k(V,R)$, $\theta \otimes r\mapsto (v\mapsto r\theta (v))$, identifies both sides with $R^{{\dim }_{k}V}$ and is an $R$-linear isomorphism. Since $V^{\vee }=\mathfrak{g}$ by [F2], transporting this module structure through step 1.1 gives a natural $R$-module structure and a natural identification $\mathrm{Lie}(G)(R)\cong \mathfrak{g}{\otimes }_{k}R$. The pullback of the cotangent sheaf along the identity section is $V{\otimes }_{k}R$ by [F4] and [F5]; this uses a section pullback, not a local ring at a purported general $R$-point.

Multiplication is addition. Apply step 1.1 to $G{\times }_{k}G$ at $(e,e)$. A dual-number morphism to this product is a pair of such morphisms to $G$, so the cotangent fibre of the product is $V\oplus V$: this also follows from the universal product derivation formula $d(a\otimes b)=bda+adb$ on affine charts, followed by augmentation. The cotangent map induced by multiplication is $V\to V\oplus V$ and has both components the identity, since multiplication restricted to $(\mathrm{id},e)$ and $(e,\mathrm{id})$ is the identity by [F3]. Under the coefficient classification of step 1.1, composing two lifts with multiplication therefore takes the pair $(D,D^{\mathrm{\prime }})$ to $D+D^{\mathrm{\prime }}$. Thus the group product on the kernel is precisely addition in the module of step 2.1, for every $R$; it is abelian, its identity is the zero coefficient, and inversion negates the coefficient.

Scalars and the field case. For $c\in R$, the endomorphism $R[\epsilon ]\to R[\epsilon ]$ sending $\epsilon \mapsto c\epsilon$ takes the coefficient $D$ of step 1.1 to $cD$. These operations are the scalar multiplication of the $R$-module of step 2.1 and are natural under $k$-algebra maps $R\to R^{\mathrm{\prime }}$. Taking $R=k$, the dual-number bijection and the cotangent description are isomorphisms of finite-dimensional $k$-vector spaces, as asserted in (a).

Functoriality and closed immersions. A group-scheme morphism $f:G\to H$ preserves identities, so it induces a local homomorphism ${\mathcal{O}}_{H,e_H}\to {\mathcal{O}}_{G,e_G}$ and the corresponding $k$-linear cotangent map $V_H\to V_G$ by [F9]. Precomposition with this map carries a coefficient $D:V_G\to R$ to the coefficient of $f\circ e^{\epsilon X}$ by step 1.1; it is $R$-linear and, under step 2.1, is the scalar extension of its $k$-linear dual $\mathrm{Lie}(f)$. Identity and composite maps give the stated functorial equalities, and $f_R(e^{\epsilon X})=e^{\epsilon \mathrm{Lie}(f{)}_R(X)}$. If $f$ is a closed immersion, [F10] makes its map on $R[\epsilon ]$-points injective for every $R$, hence also on these kernels; taking $R=k$ proves injectivity of $\mathrm{Lie}(f)$. All basis selections are finite, so no choice principle is used.

Conjugation preserves the kernel. For $x\in G(R)$ define $c_x:G(R[\epsilon ])\to G(R[\epsilon ])$ by $c_x(y)=xy{x}^{-1}$, using the group structure of [F3]; it is an automorphism with inverse $c_{x^{-1}}$, and it is induced by the automorphism of the functor $G$ given by conjugation with the image of $x$ under $G(R)\to G(R[\epsilon ])$. Since the reduction $\rho :G(R[\epsilon ])\to G(R)$ is a homomorphism of groups and the image of $x$ in $G(R[\epsilon ])$ reduces to $x$, one has $\rho (c_x(y))=x\rho (y)x^{-1}$; hence $c_x$ maps $\mathrm{Lie}(G)(R)=\ker \rho$ into itself, and so does $c_{x^{-1}}$. Define $\mathrm{Ad}(x):=c_x{\mathrm{\mid }}_{\mathrm{Lie}(G)(R)}$.

The maps $\mathrm{Ad}(x)$ and the map $x\mapsto \mathrm{Ad}(x)$. Each $\mathrm{Ad}(x)$ is a group automorphism of $\mathrm{Lie}(G)(R)$ by step 1.1, hence is additive because the group law there is addition by [F1]; it is $R$-linear because the scalar action of $c\in R$ on $\mathrm{Lie}(G)(R)$ is induced by the algebra endomorphism $\epsilon \mapsto c\epsilon$ of $R[\epsilon ]$, which commutes with the conjugation $c_x$ since the image of $x$ in $G(R[\epsilon ])$ is fixed by that endomorphism. Moreover $\mathrm{Ad}(xy)=\mathrm{Ad}(x)\circ \mathrm{Ad}(y)$ and $\mathrm{Ad}(e)=\mathrm{id}$ because $c_{xy}=c_x\circ c_y$, and the construction is natural in $R$ because both the group structures and the reduction maps are. Thus $x\mapsto \mathrm{Ad}(x)$ is a natural homomorphism from the group-valued functor $h_G$ to the functor $R\mapsto {\mathrm{Aut}}_R(\mathfrak{g}{\otimes }_{k}R)$, which under the identification of [F1] is exactly the group of $R$-linear automorphisms of $\mathrm{Lie}(G)(R)$.

Clause (b). For $X\in \mathfrak{g}{\otimes }_{k}R$ the element $e^{\epsilon X}$ of $\mathrm{Lie}(G)(R)$ is fixed by the identification of [F2], and by definition $\mathrm{Ad}(x)$ is the restriction of conjugation by $x$; hence $x{e}^{\epsilon X}{x}^{-1}=e^{\epsilon \mathrm{Ad}(x)X}$ in $G(R[\epsilon ])$.

Clause (c), naturality. Let $f:G\to H$ be a morphism of affine group schemes of finite type over $k$ and let $x\in G(R)$, $X\in \mathfrak{g}{\otimes }_{k}R$. Applying the group homomorphism $f_{R[\epsilon ]}$ to the identity of clause (b) for $G$ gives $f(x)f(e^{\epsilon X})f(x{)}^{-1}=f(e^{\epsilon {\mathrm{Ad}}_G(x)X})$; by the functoriality of $f$ on points and the exponential identity of [F1] this reads $e^{\epsilon {\mathrm{Ad}}_H(f(x))\mathrm{Lie}(f)X}=e^{\epsilon \mathrm{Lie}(f){\mathrm{Ad}}_G(x)X}$, and the exponential correspondence $X\mapsto e^{\epsilon X}$ is injective, so ${\mathrm{Ad}}_H(f(x))\circ \mathrm{Lie}(f)=\mathrm{Lie}(f)\circ {\mathrm{Ad}}_G(x)$.

The adjoint representation is a morphism. If $\mathfrak{g}=0$, its automorphism functor is the one-element functor, represented by the trivial group $\mathrm{Spec}k$, and $\mathrm{Ad}$ is its unique morphism. Otherwise, by [F1] the Lie algebra $\mathfrak{g}$ is finite-dimensional, so by [F5] it has a finite $k$-basis and the functor $R\mapsto {\mathrm{Aut}}_R(\mathfrak{g}{\otimes }_{k}R)$ is naturally identified with ${\mathrm{GL}}_n$ for $n={\dim }_k\mathfrak{g}$; by The general linear group scheme and its coordinate ring this functor is the functor of points of the affine group scheme ${\mathrm{GL}}_{\mathfrak{g}}={\mathrm{GL}}_n$. The natural transformation of step 2.1 is therefore a natural transformation $h_G\to h_{{\mathrm{GL}}_{\mathfrak{g}}}$, and by [F4] it is induced by a morphism of $k$-schemes $\mathrm{Ad}:G\to {\mathrm{GL}}_{\mathfrak{g}}$, which is a morphism of group schemes because the transformation is a natural homomorphism of group-valued functors.

Conclusion. Steps 1.1, 2.1, 2.2, 3.1 and 3.2 prove (a), (b) and (c): conjugation restricts to an $R$-linear automorphism of the Lie algebra, the assignment is a natural group homomorphism and hence defines the morphism $\mathrm{Ad}:G\to {\mathrm{GL}}_{\mathfrak{g}}$ of $k$-group schemes, the identity of (b) is the definition of the restriction, and (c) is the differentiated naturality. The conjugation construction and finite basis selection are choice-free; the finite-type assertion for ${\mathrm{GL}}_{\mathfrak{g}}$ inherits Choice from The general linear group scheme and its coordinate ring.

The cotangent space and dual-number points. Set $y_{ij}=x_{ij}-{\delta }_{ij}$ and $J=(y_{ij})\subseteq A$. Then $A=k[y_{ij}][D^{-1}]$ with $D=\det ({\delta }_{ij}+y_{ij})\equiv 1<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>J$, so $A\mathrm{/}{J}^2\cong k[y_{ij}]\mathrm{/}(y_{ij}{)}^2$ has basis $1,[y_{ij}]$. Here $J$ is the augmentation ideal in $A$, and ${\mathfrak{m}}_e=J{A}_J$ is the maximal ideal in ${\mathcal{O}}_{G,e}=A_J$. Every element of $A\setminus J$ has nonzero constant term and is a unit modulo $J^2$, with inverse given by the square-zero formula; thus localization does not change this quotient and ${\mathfrak{m}}_e\mathrm{/}{\mathfrak{m}}_e^2$ has basis $[y_{ij}]$. Its dual is $M_n(k)$ via ${\theta }_X([y_{ij}])=X_{ij}$ by [F2] and [F5]. The coefficient correspondence sends ${\theta }_X$ to the algebra map $A\to k[\epsilon ]$, $y_{ij}\mapsto \epsilon X_{ij}$, since $\det (I+\epsilon X)=1+\epsilon \mathrm{tr}X$ is a unit by [F4]; hence $e^{\epsilon X}=I+\epsilon X$. By [F3] this identification extends naturally to $M_n(R)$ for every $R$. For $n=1$ the Lie-algebra generator is the functional taking $[t-1]$ to $1$, dual to the cotangent generator.

The commutator of the two lifts. In $B=k[t,t^{\mathrm{\prime }}]\mathrm{/}(t^2,t^{\mathrm{\prime }2})$ one has $t^2=t^{\mathrm{\prime }2}=0$, so $(I+tX{)}^{-1}=I-tX$ and $(I+t^{\mathrm{\prime }}Y{)}^{-1}=I-t^{\mathrm{\prime }}Y$ by [F4], and using associativity and distributivity one computes $(I+tX)(I+t^{\mathrm{\prime }}Y)(I-tX)(I-t^{\mathrm{\prime }}Y)=I+t{t}^{\mathrm{\prime }}(XY-YX)$: the terms linear in $t$ or $t^{\mathrm{\prime }}$ cancel in pairs and the only surviving second-order term is $t{t}^{\mathrm{\prime }}XY-t{t}^{\mathrm{\prime }}YX$, the other products $t^2,t^{\mathrm{\prime }2}$ vanishing.

The adjoint action. Let $R$ be a commutative $k$-algebra, $A\in {\mathrm{GL}}_n(R)$ and $X\in M_n(R)$. By step 1.1 applied over $R$, the element $e^{\epsilon X}\in \mathrm{Lie}({\mathrm{GL}}_n)(R)$ is the point $I+\epsilon X$ of ${\mathrm{GL}}_n(R[\epsilon ])$, and $A$ is invertible in $M_n(R)\subseteq M_n(R[\epsilon ])$; conjugating with $A$ and using the matrix arithmetic of [F4] gives $A(I+\epsilon X)A^{-1}=I+\epsilon (AX{A}^{-1})$. Comparing with the defining identity $A{e}^{\epsilon X}{A}^{-1}=e^{\epsilon \mathrm{Ad}(A)X}$ of [F6] identifies the $\epsilon$-coefficient, so $\mathrm{Ad}(A)X=AX{A}^{-1}$.

Conclusion. Step 1.1 identifies the tangent space at the identity with $M_n(k)$ via $X\mapsto I+\epsilon X$ and gives the $n=1$ case; step 1.2 computes the commutator of the two independent lifts as $I+t{t}^{\mathrm{\prime }}(XY-YX)$; and step 2.1 computes the adjoint representation as conjugation. This proves all the displayed claims.

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 $\epsilon \mapsto t{t}^{\mathrm{\prime }}$.