The Lie functor: exactness, fixed points and generation

On every $k$-algebra $R$, the functor of points of a group-scheme limit is the limit of the point functors. The Lie functor is the kernel of the reduction map from $R[\epsilon ]$ to $R$, and kernels commute with limits: a compatible tuple reduces to the identity exactly when each component does. Hence . Applying this to a kernel or a fiber product gives left exactness and the displayed fiber-product equality. For subgroup inclusions their fiber product is their scheme intersection, and the vector-space fiber product is the intersection of the Lie subspaces.

Put $\mathfrak{h}=\mathrm{Lie}(H)$. By [F1], $e^{\epsilon X}$ lies in the centralizer precisely when it commutes with $H(S)$ for every $k[\epsilon ]$-algebra $S$. This is equivalent to $X$ being invariant in the rational adjoint representation: invariance tested on any $k$-algebra $R$ gives $h{e}^{{\epsilon }^{\mathrm{\prime }}X}{h}^{-1}=e^{{\epsilon }^{\mathrm{\prime }}X}$ in $G(R[{\epsilon }^{\mathrm{\prime }}])$ for all $h\in H(R)$; taking $R=S$ and specializing ${\epsilon }^{\mathrm{\prime }}$ to the given nilpotent $\epsilon \in S$ proves the centralizer condition. Conversely take $S=R[\epsilon ]$ and $h\in H(R)$ to recover invariance. Thus $\mathrm{Lie}(C_G(H))={\mathfrak{g}}^H$. The tangent dimension criterion [F2] gives the stated dimension inequality and equality case.

Assume AC for part (b). Suppose $\mathrm{Lie}(H)=\mathrm{Lie}(G)$ with $H$ smooth. Since $H\subseteq G$ we have $\dim H\le \dim G$, and by [F2] $\dim G\le \dim \mathrm{Lie}(G)=\dim \mathrm{Lie}(H)=\dim H$, so equality holds throughout; thus $G$ is smooth and $\dim G=\dim H$. A closed subgroup of the connected smooth group $G$ of dimension $\dim G$ equals $G$ by [F2], so $H=G$. Now let $H_1,\dots ,H_n$ be smooth subgroups whose Lie algebras generate $\mathfrak{g}$, and let $H$ be the algebraic subgroup they generate, which is the scheme-theoretic closure of the union of finite product maps from the $H_i$. The product maps and their inverses are stable under multiplication and inversion, so their closure is a subgroup. These maps have geometrically reduced sources and are schematically dominant as a family onto that closure, hence the closure is geometrically reduced and therefore smooth as a finite-type group scheme; by functoriality of $\mathrm{Lie}$, $\mathrm{Lie}(H)$ is a Lie subalgebra of $\mathfrak{g}$ containing each $\mathrm{Lie}(H_i)$, hence containing the subalgebra they generate, which is all of $\mathfrak{g}$; so $\mathrm{Lie}(H)=\mathfrak{g}=\mathrm{Lie}(G)$ and the first part gives $H=G$.

For the normalizer, the same calculation gives $e^{\epsilon X}h{e}^{-\epsilon X}{h}^{-1}=e^{\epsilon (X-\mathrm{Ad}(h)X)}\mathrm{.}$ If $e^{\epsilon X}$ normalizes $H$, then for every $R$ and $h\in H(R)$ this commutator lies in $H(R[\epsilon ])$ and reduces to the identity; by [F1] this means $X-\mathrm{Ad}(h)X\in {\mathfrak{h}}_R$. Hence the class of $X$ in $\mathfrak{g}\mathrm{/}\mathfrak{h}$ is $H$-invariant. Conversely, if that class is invariant, take any $k[\epsilon ]$-algebra $S$ and $h\in H(S)$. The vector $X-\mathrm{Ad}(h)X$ lies in ${\mathfrak{h}}_S$, so the corresponding dual-number point of $H(S[{\epsilon }^{\mathrm{\prime }}])$ specializes to the displayed commutator in $H(S)$ under ${\epsilon }^{\mathrm{\prime }}\mapsto \epsilon$. This proves conjugation by $e^{\epsilon X}$ maps $H(S)$ into itself; applying the argument to $-X$ gives equality, so $e^{\epsilon X}$ normalizes $H$. Thus $\mathrm{Lie}(N_G(H))$ is the inverse image of $(\mathfrak{g}\mathrm{/}\mathfrak{h}{)}^H$, and its quotient by $\mathfrak{h}$ is that invariant space. This is a quotient of Lie algebras; it is not a claim about $\mathrm{Lie}(N_G(H)\mathrm{/}H)$ for nonsmooth $H$. These calculations prove all assertions.