3.1. As $\mathbf{G}$ is equipped with a Frobenius endomorphism so is $\mathfrak{g}$ and we will denote this again by $F:\mathfrak{g}\rightarrow \mathfrak{g}$ . Our main interest of study will be the variety of unipotent elements ${\mathcal{U}}(\mathbf{G})\subset \mathbf{G}$ and the nilpotent cone ${\mathcal{N}}(\mathfrak{g})\subset \mathfrak{g}$ . If there is no ambiguity over the ambient group, respectively, Lie algebra, then we will simply write ${\mathcal{U}}$ , respectively, ${\mathcal{N}}$ , for ${\mathcal{U}}(\mathbf{G})$ , respectively, ${\mathcal{N}}(\mathfrak{g})$ . To relate these two objects we will need the existence of a Springer isomorphism, which is a $\mathbf{G}$ -equivariant isomorphism of varieties $\unicode[STIX]{x1D719}_{\text{spr}}:{\mathcal{U}}\rightarrow {\mathcal{N}}$ compatible with the Frobenius endomorphisms on $\mathbf{G}$ and $\mathfrak{g}$ . Note that this is $\mathbf{G}$ -equivariant in the sense that

$$\begin{eqnarray}\unicode[STIX]{x1D719}_{\text{spr}}\circ \text{Inn}\,g=\text{Ad}\,g\circ \unicode[STIX]{x1D719}_{\text{spr}}\end{eqnarray}$$

for all $g\in \mathbf{G}$ where $\text{Ad}\,g=d(\text{Inn}\,g):\mathfrak{g}\rightarrow \mathfrak{g}$ is the differential of $\text{Inn}\,g:\mathbf{G}\,\rightarrow \,\mathbf{G}$ , see, 2.2.

3.2. We will need the following well known result which states that a separable isogeny between connected algebraic groups restricts to an isomorphism between unipotent varieties. This result was stated by Springer in [Reference SpringerSpr1] but he omits the proof. An argument is given in [Reference HumphreysHum, Section 6.2] for this statement but this seems only to show that the restriction is a birational morphism. If one knew the target unipotent variety was normal then one could conclude that this was indeed an isomorphism but usually one uses such an isomorphism to deduce the unipotent variety is normal so this is circular. The following argument was communicated to us by George McNinch; we thank him for kindly allowing us to include it here.

Proof. Let us denote by $Z$ the kernel of $\unicode[STIX]{x1D719}$ , which is by assumption a finite subgroup of the center $Z(\mathbf{H})$ . We start by noting that $(\mathbf{H}^{\prime },\unicode[STIX]{x1D719})$ is the affine quotient of $\mathbf{H}$ by $Z$ in the sense of [Reference GeckGec2, 2.5.8]. Indeed, if $\unicode[STIX]{x1D70B}:\mathbf{H}\rightarrow \mathbf{H}/Z$ is the natural projection map then there exists a unique morphism $\unicode[STIX]{x1D6FD}:\mathbf{H}/Z\rightarrow \mathbf{H}^{\prime }$ such that $\unicode[STIX]{x1D719}=\unicode[STIX]{x1D70B}\circ \unicode[STIX]{x1D6FD}$ by the universal property of the quotient. Clearly $\unicode[STIX]{x1D6FD}$ is bijective. Moreover, by assumption, the field extension $\mathbb{K}(\mathbf{H})/\unicode[STIX]{x1D719}^{\ast }(\mathbb{K}(\mathbf{H}^{\prime }))$ is separable hence so is $\unicode[STIX]{x1D70B}^{\ast }(\mathbb{K}(\mathbf{H}/Z))/\unicode[STIX]{x1D719}^{\ast }(\mathbb{K}(\mathbf{H}^{\prime }))$ but as $\unicode[STIX]{x1D70B}^{\ast }$ is injective this implies $\mathbb{K}(\mathbf{H}/Z)/\unicode[STIX]{x1D6FD}^{\ast }(\mathbb{K}(\mathbf{H}^{\prime }))$ is separable. Hence $\unicode[STIX]{x1D6FD}$ is bijective and separable and so is an isomorphism, see, [Reference SpringerSpr2, 4.3.7(iii), 5.3.3(ii)].

Now let us denote by $\mathbf{V}$ the preimage $\unicode[STIX]{x1D719}^{-1}({\mathcal{U}}(\mathbf{H}^{\prime }))$ of the unipotent variety under $\unicode[STIX]{x1D719}$ . The Jordan decomposition shows that $\mathbf{V}$ is the disjoint union $\bigsqcup _{z\in Z}z{\mathcal{U}}(\mathbf{H})$ and clearly each $z{\mathcal{U}}(\mathbf{H})$ is a closed subvariety isomorphic to ${\mathcal{U}}(\mathbf{H})$ . The kernel $Z$ acts simply transitively on $\mathbf{V}$ by left multiplication and we claim that $({\mathcal{U}}(\mathbf{H}^{\prime }),\unicode[STIX]{x1D719})$ is the affine quotient of $\mathbf{V}$ by $Z$ . Certainly the fibers of $\unicode[STIX]{x1D719}$ are $Z$ -orbits and $\unicode[STIX]{x1D719}$ is an open map by [Reference GeckGec2, 2.5.6(b)]. We just need to show that the map $\unicode[STIX]{x1D719}^{\ast }:\mathbb{K}[{\mathcal{U}}(\mathbf{H}^{\prime })]\rightarrow \mathbb{K}[\mathbf{V}]^{Z}$ is surjective. If $\unicode[STIX]{x1D704}_{\mathbf{V}}:\mathbf{V}\rightarrow \mathbf{H}$ and $\unicode[STIX]{x1D704}_{{\mathcal{U}}(\mathbf{H}^{\prime })}:{\mathcal{U}}(\mathbf{H}^{\prime })\rightarrow \mathbf{H}^{\prime }$ are the natural closed embeddings then we have a commutative square

where the vertical arrows are surjective. By the first part the top arrow is surjective, which proves the claim.

Now let $\unicode[STIX]{x1D6FC}:{\mathcal{U}}(\mathbf{H})\rightarrow \mathbf{V}/Z$ be the morphism obtained as the composition of the natural closed embedding ${\mathcal{U}}(\mathbf{H})\rightarrow \mathbf{V}$ and the natural projection $\mathbf{V}\rightarrow \mathbf{V}/Z$ . As $\unicode[STIX]{x1D719}:{\mathcal{U}}(\mathbf{H})\rightarrow {\mathcal{U}}(\mathbf{H}^{\prime })$ factors as a composition of $\unicode[STIX]{x1D6FC}$ and an isomorphism $\mathbf{V}/Z\rightarrow {\mathcal{U}}(\mathbf{H}^{\prime })$ it suffices to show that $\unicode[STIX]{x1D6FC}$ is an isomorphism. As $\mathbf{V}=\bigsqcup _{z\in Z}z{\mathcal{U}}(\mathbf{H})$ we have a $\mathbb{K}$ -algebra isomorphism  $\mathbb{K}[\mathbf{V}]\rightarrow \prod _{z\in Z}\mathbb{K}[z{\mathcal{U}}(\mathbf{H})]$ defined by $f\mapsto (f|_{z{\mathcal{U}}(\mathbf{H})})_{z\in Z}$ . This clearly maps $\mathbb{K}[\mathbf{V}]^{Z}$ onto the subalgebra $\{(f_{z})\in \prod _{z\in Z}\mathbb{K}[z{\mathcal{U}}(\mathbf{H})]\mid f_{z}(zu)=f_{z}(u)\text{ for all }z\in Z$ and $u\in {\mathcal{U}}(\mathbf{H})\!\}$ . Moreover, under this identification the comorphism of the closed embedding ${\mathcal{U}}(\mathbf{H})\rightarrow \mathbf{V}$ is simply the projection onto $\mathbb{K}[{\mathcal{U}}(\mathbf{H})]$ , that is, the factor indexed by the identity. With this it is clear that the comorphism $\unicode[STIX]{x1D6FC}^{\ast }:\mathbb{K}[\mathbf{V}/Z]\rightarrow \mathbb{K}[{\mathcal{U}}(\mathbf{H})]$ is an isomorphism, hence $\unicode[STIX]{x1D6FC}$ is an isomorphism as desired.◻

Proof. First let us note that ${\mathcal{U}}(\mathbf{G})={\mathcal{U}}(\mathbf{G}_{\text{der}})$ and ${\mathcal{N}}(\mathfrak{g})={\mathcal{N}}(\mathfrak{g}_{\text{der}})$ , for example, see [Reference PremetPre1, p. 2966]. As $\unicode[STIX]{x1D719}_{\text{sc}}:\mathbf{G}_{\text{sc}}\rightarrow \mathbf{G}_{\text{der}}$ is assumed to be a separable isogeny we have by Lemma 3.3 that it restricts to an equivariant isomorphism $\unicode[STIX]{x1D6FC}:{\mathcal{U}}(\mathbf{G}_{\text{sc}})\rightarrow {\mathcal{U}}(\mathbf{G}_{\text{der}})$ . Now we may view the vector spaces $\mathfrak{g}_{\text{sc}}$ and $\mathfrak{g}_{\text{der}}$ as affine varieties, in which case the nilpotent cones are closed subvarieties. By Proposition 2.13 the differential $d\unicode[STIX]{x1D719}_{\text{sc}}$ is an isomorphism of vector spaces which must therefore also be an isomorphism of varieties. This is because the inverse of $d\unicode[STIX]{x1D719}_{\text{sc}}$ is a linear map and any linear map between vector spaces is clearly a morphism of varieties. In particular, we have $d\unicode[STIX]{x1D719}_{\text{sc}}$ restricts to an equivariant isomorphism $\unicode[STIX]{x1D6FD}:{\mathcal{N}}(\mathfrak{g}_{\text{sc}})\rightarrow {\mathcal{N}}(\mathfrak{g}_{\text{der}})$ .

Let us denote by $F_{\text{sc}}:\mathbf{G}_{\text{sc}}\rightarrow \mathbf{G}_{\text{sc}}$ a Frobenius endomorphism such that $F\circ \unicode[STIX]{x1D719}_{\text{sc}}=\unicode[STIX]{x1D719}_{\text{sc}}\circ F_{\text{sc}}$ , which exists by [Reference SteinbergSte1, 9.16]. By [Reference JantzenJan, Corollary 8.5] we know that the nilpotent cone ${\mathcal{N}}(\mathfrak{g}_{\text{sc}})$ is a normal variety. Hence, by [Reference Springer and SteinbergSS, III, 3.12] and the remark following the statement of the theorem there exists a Springer isomorphism $\tilde{\unicode[STIX]{x1D719}}_{\text{spr}}:{\mathcal{U}}(\mathbf{G}_{\text{sc}})\rightarrow {\mathcal{N}}(\mathfrak{g}_{\text{sc}})$ with respect to $F_{\text{sc}}$ , see also [Reference HumphreysHum, Section 6.20]. Clearly one then also has that $\unicode[STIX]{x1D719}_{\text{spr}}=\unicode[STIX]{x1D6FD}\circ \tilde{\unicode[STIX]{x1D719}}_{\text{spr}}\circ \unicode[STIX]{x1D6FC}^{-1}:{\mathcal{U}}(\mathbf{G}_{\text{der}})\rightarrow {\mathcal{N}}(\mathfrak{g}_{\text{der}})$ is an equivariant isomorphism as it is a composition of equivariant isomorphisms. Finally, as $\mathbf{G}=\mathbf{G}_{\text{der}}Z(\mathbf{G})$ and $Z(\mathbf{G})$ acts trivially on both ${\mathcal{U}}$ and ${\mathcal{N}}$ we may consider $\unicode[STIX]{x1D719}_{\text{spr}}$ to be $\mathbf{G}$ -equivariant. Hence, we have $\unicode[STIX]{x1D719}_{\text{spr}}$ is a Springer isomorphism.◻

3.6. An issue for us in this article will also be the so-called separability of centralizers. For any subset $\mathfrak{h}\subseteq \mathfrak{g}$ we define

$$\begin{eqnarray}\displaystyle C_{\mathbf{G}}(\mathfrak{h}) & = & \displaystyle \{g\in \mathbf{G}\mid \text{Ad}\,g(x)=x\text{ for all }x\in \mathfrak{h}\},\nonumber\\ \displaystyle \mathfrak{c}_{\mathfrak{g}}(\mathfrak{h}) & = & \displaystyle \{y\in \mathfrak{g}\mid [y,x]=0\text{ for all }x\in \mathfrak{h}\}.\nonumber\end{eqnarray}$$

If $\mathfrak{h}=\{x\}$ then we simply write $C_{\mathbf{G}}(x)$ , respectively, $\mathfrak{c}_{\mathfrak{g}}(x)$ , for $C_{\mathbf{G}}(\mathfrak{h})$ , respectively, $\mathfrak{c}_{\mathfrak{g}}(\mathfrak{h})$ . Furthermore we will denote by $\mathfrak{z}(\mathfrak{g})=\bigcap _{x\in \mathfrak{g}}\mathfrak{c}_{\mathfrak{g}}(x)$ the center of the Lie algebra. Note that, in general, we do not have $\mathfrak{z}(\mathfrak{g})=\text{Lie}(Z(\mathbf{G}))$ . We will be interested in knowing when $C_{\mathbf{G}}(x)$ is separable in the sense that $\text{Lie}(C_{\mathbf{G}}(x))=\mathfrak{c}_{\mathfrak{g}}(x)$ . The following gives some equivalent characterizations of separability.

3.8. In [Reference HerpelHer] Herpel has considered when the scheme-theoretic centralizer of a closed subgroup scheme of $\mathbf{G}$ is smooth and shown that this is related to the notion of separability. To apply his results to centralizers we will need to recall his elegant notion of a pretty good prime.

Proof. Let $\mathfrak{h}\subseteq \mathfrak{g}$ be the 1-dimensional subalgebra generated by $x$ then it is clear that we have $C_{\mathbf{G}}(x)=C_{\mathbf{G}}(\mathfrak{h})$ and $\mathfrak{c}_{\mathfrak{g}}(x)=\mathfrak{c}_{\mathfrak{g}}(\mathfrak{h})$ . In particular, to show that $C_{\mathbf{G}}(x)$ is separable it suffices to show that $C_{\mathbf{G}}(\mathfrak{h})$ is separable. In [Reference HerpelHer, Lemma 3.1(ii)] Herpel constructs a closed subgroup scheme $\mathbf{H}^{\prime }\leqslant \mathbf{G}$ and shows that $C_{\mathbf{G}}(\mathfrak{h})$ is separable if and only if the scheme-theoretic centralizer of $\mathbf{H}^{\prime }$ in $\mathbf{G}$ is smooth. By [Reference HerpelHer, Theorem 1.1] the scheme-theoretic centralizer of any closed subgroup scheme of $\mathbf{G}$ is smooth if $p$ is a pretty good prime. Hence we can conclude that $C_{\mathbf{G}}(x)$ is smooth.◻

3.12. We will denote by $\mathbb{G}_{m}$ the set $\mathbb{K}\smallsetminus \{0\}$ viewed as an algebraic group under multiplication. It will be useful to recall here some properties of $\mathbb{G}_{m}$ -actions that we will use several times. Assume $\mathbf{X}$ is an affine $\mathbb{G}_{m}$ -variety, in the sense of [Reference SpringerSpr2, Section 2.3.1], with action map $\unicode[STIX]{x1D6FC}:\mathbb{G}_{m}\times \mathbf{X}\rightarrow \mathbf{X}$ . The affine algebra $\mathbb{K}[\mathbf{X}]$ of $\mathbf{X}$ then becomes an abstract $\mathbb{G}_{m}$ -module by setting

$$\begin{eqnarray}(k\cdot f)(x)=f(\unicode[STIX]{x1D6FC}(k^{-1},x))\end{eqnarray}$$

for all $k\in \mathbb{G}_{m}$ , $f\in \mathbb{K}[\mathbf{X}]$ and $x\in \mathbf{X}$ . By [Reference SpringerSpr2, 2.3.9(i), 3.2.3(c)] we have $\mathbb{K}[\mathbf{X}]=\bigoplus _{n\in \mathbb{Z}}\mathbb{K}[\mathbf{X}]_{n}$ where we have

$$\begin{eqnarray}\mathbb{K}[\mathbf{X}]_{n}=\{f\in \mathbb{K}[\mathbf{X}]\mid k\cdot f=k^{n}f\text{ for all }k\in \mathbb{G}_{m}\}\end{eqnarray}$$

is the corresponding weight space. For any $x\in \mathbf{X}$ we have a morphism $\unicode[STIX]{x1D6FC}_{x}:\mathbb{G}_{m}\rightarrow \mathbf{X}$ defined by setting $\unicode[STIX]{x1D6FC}_{x}(k)=\unicode[STIX]{x1D6FC}(k,x)$ for all $k\in \mathbb{G}_{m}$ . We say the limit $\lim _{k\rightarrow 0}\unicode[STIX]{x1D6FC}_{x}(k)$ exists if the morphism $\unicode[STIX]{x1D6FC}_{x}:\mathbb{G}_{m}\rightarrow \mathbf{X}$ extends to a morphism $\tilde{\unicode[STIX]{x1D6FC}}_{x}:\mathbb{A}^{1}\rightarrow \mathbf{X}$ ; note such an extension is unique by [Reference SpringerSpr2, 1.6.11(ii)]. Moreover we write $\lim _{k\rightarrow 0}\unicode[STIX]{x1D6FC}_{x}(k)=z$ if $\tilde{\unicode[STIX]{x1D6FC}}_{x}(0)=z$ . Now we say the $\mathbb{G}_{m}$ -action is contracting if there exists a fixed point $x_{0}\in \mathbf{X}^{\mathbb{G}_{m}}$ such that $\lim _{k\rightarrow 0}\unicode[STIX]{x1D6FC}_{x}(k)=x_{0}$ for all $x\in \mathbf{X}$ . We will also say that the action is a contraction to $x_{0}$ . With this we have the following.

Proof. (i) Assume $x_{0}^{\prime }\in X^{\mathbb{G}_{m}}$ was another fixed point and let $\tilde{\unicode[STIX]{x1D6FC}}_{x_{0}^{\prime }}:\mathbb{A}^{1}\rightarrow \mathbf{X}$ be the morphism extending $\unicode[STIX]{x1D6FC}_{x_{0}^{\prime }}:\mathbb{G}_{m}\rightarrow \mathbf{X}$ . Let $f_{x_{0}^{\prime }}:\mathbb{A}^{1}\rightarrow \mathbf{X}$ be the morphism defined by $f_{x_{0}^{\prime }}(k)=x_{0}^{\prime }$ then as $x_{0}^{\prime }$ is fixed by the $\mathbb{G}_{m}$ action we have $\tilde{\unicode[STIX]{x1D6FC}}_{x_{0}^{\prime }}(k)=f_{x_{0}^{\prime }}(k)$ for all $k\in \mathbb{G}_{m}$ . As noted above, the extending morphism is unique so we must have $\tilde{\unicode[STIX]{x1D6FC}}_{x_{0}^{\prime }}=f_{x_{0}^{\prime }}$ . However, by assumption $\tilde{\unicode[STIX]{x1D6FC}}_{x_{0}^{\prime }}(0)=x_{0}$ which implies $x_{0}^{\prime }=x_{0}$ .

(ii) For any $x\in \mathbf{X}$ we have the morphism $\unicode[STIX]{x1D6FC}_{x}:\mathbb{G}_{m}\rightarrow \mathbf{X}$ extends to $\mathbb{A}^{1}$ if and only if the image of the comorphism $\unicode[STIX]{x1D6FC}_{x}^{\ast }:\mathbb{K}[\mathbf{X}]\rightarrow \mathbb{K}[\mathbb{G}_{m}]=\mathbb{K}[T,T^{-1}]$ is contained in $\mathbb{K}[T]$ . For any $f\in \mathbb{K}[\mathbf{X}]$ we may write $f=\sum _{n\in \mathbb{Z}}f_{n}$ with $f_{n}\in \mathbb{K}[\mathbf{X}]_{n}$ . It is easy to see that we have

$$\begin{eqnarray}\unicode[STIX]{x1D6FC}_{x}^{\ast }(f)=\mathop{\sum }_{n\in \mathbb{Z}}f_{n}(x)T^{n}.\end{eqnarray}$$

Hence for any $n<0$ we must have $f_{n}(x)=0$ for all $x\in \mathbf{X}$ and so $\mathbb{K}[\mathbf{X}]_{n}=\{0\}$ for any $n<0$ .

(iii) By part (ii) we need only show that $\dim _{\mathbb{K}}\mathbb{K}[\mathbf{X}]_{n}<\infty$ when $n\geqslant 0$ . Moreover, as $\mathbb{K}[\mathbf{X}]_{0}$ is nothing other than the affine algebra $\mathbb{K}[\mathbf{X}^{\mathbb{G}_{m}}]$ of the fixed points we have $\mathbb{K}[\mathbf{X}]_{0}=\mathbb{K}$ by (i) so we may assume $n>0$ . Let $\{g_{1},\ldots ,g_{k}\}\subseteq \mathbb{K}[\mathbf{X}]$ be a generating set for the algebra then the set ${\mathcal{B}}=\{g_{1}^{i_{1}}\cdots g_{k}^{i_{k}}\mid i_{1},\ldots ,i_{k}\geqslant 0\}$ is a $\mathbb{K}$ -basis of the algebra, where we assume that not all powers are 0. Let us assume that $g_{j}$ is contained in the weight space $\mathbb{K}[\mathbf{X}]_{n_{j}}$ then by (i) and (ii) we have $n_{j}>0$ . The subset ${\mathcal{B}}_{n}=\{g_{1}^{i_{1}}\cdots g_{k}^{i_{k}}\mid n=n_{1}i_{1}+\cdots +n_{k}i_{k}\}\subseteq {\mathcal{B}}$ is obviously a basis for the weight space $\mathbb{K}[\mathbf{X}]_{n}$ . However as $n,n_{1},\ldots ,n_{k}>0$ are all strictly positive we see that the set ${\mathcal{B}}_{n}$ must be finite.◻

3.15. We now recall some results from [Reference PremetPre2]. Assume $\unicode[STIX]{x1D706}\in Y(\mathbf{G})=\text{Hom}(\mathbb{G}_{m},\mathbf{G})$ is a cocharacter of $\mathbf{G}$ then $\unicode[STIX]{x1D706}$ defines a $\mathbb{Z}$ -grading $\mathfrak{g}=\bigoplus _{i\in \mathbb{Z}}\mathfrak{g}(\unicode[STIX]{x1D706},i)$ on the Lie algebra by setting

$$\begin{eqnarray}\mathfrak{g}(\unicode[STIX]{x1D706},i)=\{x\in \mathfrak{g}\mid (\text{Ad}\unicode[STIX]{x1D706}(k))(x)=k^{i}x\text{ for all }k\in \mathbb{G}_{m}\}.\end{eqnarray}$$

From the definition we see immediately that

(3.16) $$\begin{eqnarray}[\mathfrak{g}(\unicode[STIX]{x1D706},i),\mathfrak{g}(\unicode[STIX]{x1D706},j)]\subseteq \mathfrak{g}(\unicode[STIX]{x1D706},i+j)\end{eqnarray}$$

for all $i,j\in \mathbb{Z}$ . We note the following useful observation concerning the weight spaces. It is clear that we have $\mathfrak{g}=\mathfrak{g}_{\text{der}}+\text{Lie}(Z(\mathbf{G}))$ but this sum need not be direct. Furthermore, it is also obvious that $\text{Lie}(Z(\mathbf{G}))\subseteq \mathfrak{g}(\unicode[STIX]{x1D706},0)$ so we must have $\mathfrak{g}(\unicode[STIX]{x1D706},i)\subseteq \mathfrak{g}_{\text{der}}$ for any $i\neq 0$ . In particular, if $\unicode[STIX]{x1D706}\in Y(\mathbf{G}_{\text{der}})\subseteq Y(\mathbf{G})$ then we have $\mathfrak{g}(\unicode[STIX]{x1D706},i)=\mathfrak{g}_{\text{der}}(\unicode[STIX]{x1D706},i)$ for all $i\neq 0$ .

3.17. To any cocharacter $\unicode[STIX]{x1D706}\in Y(\mathbf{G})$ we assign a parabolic subgroup $\mathbf{P}(\unicode[STIX]{x1D706})\leqslant \mathbf{G}$ with unipotent radical $\mathbf{U}(\unicode[STIX]{x1D706})$ given by

$$\begin{eqnarray}\displaystyle \mathbf{P}(\unicode[STIX]{x1D706}) & = & \displaystyle \left\{x\in \mathbf{G}\mid \lim _{k\rightarrow 0}\unicode[STIX]{x1D706}(k)x\unicode[STIX]{x1D706}(k)^{-1}\text{ exists}\right\}\nonumber\\ \displaystyle \mathbf{U}(\unicode[STIX]{x1D706}) & = & \displaystyle \left\{x\in \mathbf{G}\mid \lim _{k\rightarrow 0}\unicode[STIX]{x1D706}(k)x\unicode[STIX]{x1D706}(k)^{-1}=1\right\},\nonumber\end{eqnarray}$$

see [Reference SpringerSpr2, 3.2.15, 8.4.5]. We then have a Levi decomposition $\mathbf{P}(\unicode[STIX]{x1D706})=\mathbf{L}(\unicode[STIX]{x1D706})\mathbf{U}(\unicode[STIX]{x1D706})$ by setting $\mathbf{L}(\unicode[STIX]{x1D706})=C_{\mathbf{G}}(\unicode[STIX]{x1D706}(\mathbb{G}_{m}))$ . The Lie algebras of these subgroups are given by

$$\begin{eqnarray}\mathfrak{p}(\unicode[STIX]{x1D706})=\mathop{\bigoplus }_{i\geqslant 0}\mathfrak{g}(\unicode[STIX]{x1D706},i)\quad \mathfrak{l}(\unicode[STIX]{x1D706})=\mathop{\bigoplus }_{i=0}\mathfrak{g}(\unicode[STIX]{x1D706},i)\quad \mathfrak{u}(\unicode[STIX]{x1D706})=\mathop{\bigoplus }_{i>0}\mathfrak{g}(\unicode[STIX]{x1D706},i).\end{eqnarray}$$

For any $i\in \mathbb{Z}_{{>}0}$ we will also need the Lie subalgebra $\mathfrak{u}(\unicode[STIX]{x1D706},i)=\bigoplus _{j\geqslant i}\mathfrak{g}(\unicode[STIX]{x1D706},j)$ and its corresponding closed connected unipotent subgroup $\mathbf{U}(\unicode[STIX]{x1D706},i)\leqslant \mathbf{U}(\unicode[STIX]{x1D706})$ .

Assume now that $\unicode[STIX]{x1D706}\in Y(\mathbf{T}_{0})$ and let $\mathbf{X}_{\unicode[STIX]{x1D6FC}}\leqslant \mathbf{G}$ be the 1-dimensional unipotent root subgroup whose Lie algebra is the root space $\mathfrak{g}_{\unicode[STIX]{x1D6FC}}$ for any $\unicode[STIX]{x1D6FC}\in \unicode[STIX]{x1D6F7}$ . If $\mathbf{P}(\unicode[STIX]{x1D706})$ is standard, that is, it contains $\mathbf{T}_{0}\leqslant \mathbf{B}_{0}$ , then we may also write the above as

$$\begin{eqnarray}\begin{array}{@{}ll@{}}\displaystyle \mathbf{P}(\unicode[STIX]{x1D706})=\langle \mathbf{T}_{0},\mathbf{X}_{\unicode[STIX]{x1D6FC}}\mid \langle \unicode[STIX]{x1D6FC},\unicode[STIX]{x1D706}\rangle \geqslant 0\rangle & \displaystyle \mathfrak{p}(\unicode[STIX]{x1D706})=\mathfrak{t}_{0}\oplus \mathop{\bigoplus }_{\langle \unicode[STIX]{x1D6FC},\unicode[STIX]{x1D706}\rangle \geqslant 0}\mathfrak{g}_{\unicode[STIX]{x1D6FC}}\\ \displaystyle \mathbf{L}(\unicode[STIX]{x1D706})=\langle \mathbf{T}_{0},\mathbf{X}_{\unicode[STIX]{x1D6FC}}\mid \langle \unicode[STIX]{x1D6FC},\unicode[STIX]{x1D706}\rangle =0\rangle & \displaystyle \mathfrak{l}(\unicode[STIX]{x1D706})=\mathfrak{t}_{0}\oplus \mathop{\bigoplus }_{\langle \unicode[STIX]{x1D6FC},\unicode[STIX]{x1D706}\rangle =0}\mathfrak{g}_{\unicode[STIX]{x1D6FC}}\\ \displaystyle \mathbf{U}(\unicode[STIX]{x1D706})=\langle \mathbf{X}_{\unicode[STIX]{x1D6FC}}\mid \langle \unicode[STIX]{x1D6FC},\unicode[STIX]{x1D706}\rangle >0\rangle & \displaystyle \mathfrak{u}(\unicode[STIX]{x1D706})=\mathop{\bigoplus }_{\langle \unicode[STIX]{x1D6FC},\unicode[STIX]{x1D706}\rangle >0}\mathfrak{g}_{\unicode[STIX]{x1D6FC}},\end{array}\end{eqnarray}$$

as $\mathfrak{t}_{0}\subseteq \mathfrak{g}(\unicode[STIX]{x1D706},0)$ . With this we see that we have an analogue of (3.16) which follows immediately from Chevalley’s commutator relation [Reference SpringerSpr2, 8.2.3]. Namely,

(3.18) $$\begin{eqnarray}[\mathbf{U}(\unicode[STIX]{x1D706},i),\mathbf{U}(\unicode[STIX]{x1D706},j)]\subseteq \mathbf{U}(\unicode[STIX]{x1D706},i+j)\end{eqnarray}$$

for any $i,j\geqslant 1$ . Note this clearly holds for all cocharacters.

3.19. Let us assume temporarily that $\mathbf{G}$ is semisimple and simply connected. We will denote by $\mathbf{G}_{\mathbb{C}}$ an algebraic group over $\mathbb{C}$ and by $\mathbf{T}_{\mathbb{C}}\leqslant \mathbf{B}_{\mathbb{C}}\leqslant \mathbf{G}_{\mathbb{C}}$ a maximal torus and Borel subgroup of $\mathbf{G}_{\mathbb{C}}$ such that the root datum, and simple roots, of $\mathbf{G}_{\mathbb{C}}$ with respect to $\mathbf{T}_{\mathbb{C}}\leqslant \mathbf{B}_{\mathbb{C}}$ is the same as that of $\mathbf{G}$ . Fix a Chevalley basis ${\mathcal{B}}_{\mathbb{C}}=\{X_{\unicode[STIX]{x1D6FC}}\mid \unicode[STIX]{x1D6FC}\in \unicode[STIX]{x1D6F7}\}\cup \{H_{\unicode[STIX]{x1D6FC}}\mid \unicode[STIX]{x1D6FC}\in \unicode[STIX]{x1D6E5}\}$ of $\mathfrak{g}_{\mathbb{C}}$ and denote by $\mathfrak{g}_{\mathbb{Z}}\subset \mathfrak{g}_{\mathbb{C}}$ the $\mathbb{Z}$ -span of ${\mathcal{B}}_{\mathbb{C}}$ . As $\mathbf{G}$ is simply connected we may identify $\mathfrak{g}$ with $\mathfrak{g}_{\mathbb{Z}}\otimes _{\mathbb{Z}}\mathbb{K}$ as Lie algebras. Setting $e_{\unicode[STIX]{x1D6FC}}=X_{\unicode[STIX]{x1D6FC}}\otimes 1$ and $h_{\unicode[STIX]{x1D6FC}}=H_{\unicode[STIX]{x1D6FC}}\otimes 1$ for each $\unicode[STIX]{x1D6FC}\in \unicode[STIX]{x1D6F7}$ we have ${\mathcal{B}}=\{e_{\unicode[STIX]{x1D6FC}}\mid \unicode[STIX]{x1D6FC}\in \unicode[STIX]{x1D6F7}\}\cup \{h_{\unicode[STIX]{x1D6FC}}\mid \unicode[STIX]{x1D6FC}\in \unicode[STIX]{x1D6E5}\}$ is a basis for $\mathfrak{g}$ and the corresponding root space $\mathfrak{g}_{\unicode[STIX]{x1D6FC}}$ of $\mathfrak{g}$ is simply $\mathbb{K}e_{\unicode[STIX]{x1D6FC}}$ .

By the Jacobson–Morozov theorem, see [Reference CarterCar2, 5.3.2], any nilpotent element $e\in \mathfrak{g}_{\mathbb{C}}=\text{Lie}(\mathbf{G}_{\mathbb{C}})$ is contained in an $\mathfrak{s}\mathfrak{l}_{2}$ -triple $\{e,h,f\}\subset \mathfrak{g}_{\mathbb{C}}$ . Arguing as in [Reference PremetPre2, p. 344] we may assume, after possibly replacing $\{e,h,f\}$ by $\{\text{Ad}\,g(e),\text{Ad}\,g(h),\text{Ad}\,g(f)\}$ for some $g\in \mathbf{G}$ , that $h=\sum _{\unicode[STIX]{x1D6FC}\in \unicode[STIX]{x1D6E5}}q_{\unicode[STIX]{x1D6FC}}H_{\unicode[STIX]{x1D6FC}}$ with $q_{\unicode[STIX]{x1D6FC}}\in \mathbb{Z}$ and $\unicode[STIX]{x1D6FC}(h)\geqslant 0$ for all $\unicode[STIX]{x1D6FC}\in \unicode[STIX]{x1D6E5}$ . The function $d:\unicode[STIX]{x1D6F7}\rightarrow \mathbb{Z}$ given by $\unicode[STIX]{x1D6FC}\mapsto \unicode[STIX]{x1D6FC}(h)$ is called the weighted Dynkin diagram of the nilpotent orbit ${\mathcal{O}}\subset \mathfrak{g}_{\mathbb{C}}$ containing  $e$ . We denote by ${\mathcal{D}}_{\unicode[STIX]{x1D6F7}}$ the set of weighted Dynkin diagrams.

Note that, for each weighted Dynkin diagram $d\in {\mathcal{D}}_{\unicode[STIX]{x1D6F7}}$ , there exists a cocharacter $\unicode[STIX]{x1D706}_{d}^{\mathbf{G}}\in Y(\mathbf{T}_{0})\subseteq Y(\mathbf{G})$ such that $\text{Ad}\,\unicode[STIX]{x1D706}_{d}^{\mathbf{G}}(k)(e_{\unicode[STIX]{x1D6FC}})=k^{d(\unicode[STIX]{x1D6FC})}e_{\unicode[STIX]{x1D6FC}}$ and $\text{Ad}\,\unicode[STIX]{x1D706}_{d}^{\mathbf{G}}(k)(x)=x$ for all $\unicode[STIX]{x1D6FC}\in \pm \unicode[STIX]{x1D6E5}$ , $x\in \mathfrak{t}_{0}$ and $k\in \mathbb{G}_{m}$ . In fact this can be constructed by setting $\unicode[STIX]{x1D706}_{d}^{\mathbf{G}}=\sum _{\unicode[STIX]{x1D6FC}\in \unicode[STIX]{x1D6E5}}q_{\unicode[STIX]{x1D6FC}}\check{\unicode[STIX]{x1D6FC}}\in \mathbb{Z}\check{\unicode[STIX]{x1D6F7}}$ where $q_{\unicode[STIX]{x1D6FC}}\in \mathbb{Z}$ is as above and $\check{\unicode[STIX]{x1D6FC}}\in \check{\unicode[STIX]{x1D6F7}}$ is the coroot corresponding to $\unicode[STIX]{x1D6FC}\in \unicode[STIX]{x1D6E5}$ .

3.20. We now drop our assumption that $\mathbf{G}$ is semisimple and simply connected. Let us denote by $\mathbf{T}_{\text{sc}}\leqslant \mathbf{G}_{\text{sc}}$ the maximal torus such that $\unicode[STIX]{x1D719}_{\text{sc}}(\mathbf{T}_{\text{sc}})\leqslant \mathbf{T}_{0}$ . The isogeny $\unicode[STIX]{x1D719}_{\text{sc}}$ then induces a natural injection $Y(\mathbf{G}_{\text{sc}})\rightarrow Y(\mathbf{G})$ which maps $Y(\mathbf{T}_{\text{sc}})$ into $Y(\mathbf{T}_{0})$ . In 3.19 we have defined a cocharacter $\unicode[STIX]{x1D706}_{d}^{\mathbf{G}_{\text{sc}}}\in Y(\mathbf{T}_{\text{sc}})$ for each weighted Dynkin diagram $d\in {\mathcal{D}}_{\unicode[STIX]{x1D6F7}}$ . We now set $\unicode[STIX]{x1D706}_{d}^{\mathbf{G}}:=\unicode[STIX]{x1D719}_{\text{sc}}\circ \unicode[STIX]{x1D706}_{d}^{\mathbf{G}_{\text{sc}}}$ and denote by ${\mathcal{D}}_{\unicode[STIX]{x1D6F7}}(\mathbf{G})$ the set $\{\unicode[STIX]{x1D706}_{d}^{\mathbf{G}}\mid d\in {\mathcal{D}}_{\unicode[STIX]{x1D6F7}}\}$ of resulting cocharacters. Note that the definition of $\unicode[STIX]{x1D706}_{d}^{\mathbf{G}}$ does not depend upon the choice of simply connected cover $\unicode[STIX]{x1D719}_{\text{sc}}$ . Indeed, if $\unicode[STIX]{x1D719}_{\text{sc}}^{\prime }$ is another simply connected cover then $\unicode[STIX]{x1D719}_{\text{sc}}^{\prime }=\unicode[STIX]{x1D719}_{\text{sc}}\circ \text{Inn}\,t$ for some element $t\in \mathbf{T}_{\text{sc}}$ by Theorem 2.3. However, clearly $\text{Inn}\,t\circ \unicode[STIX]{x1D706}_{d}^{\mathbf{G}_{\text{sc}}}=\unicode[STIX]{x1D706}_{d}^{\mathbf{G}_{\text{sc}}}$ because the image is contained in $\mathbf{T}_{\text{sc}}$ , which is abelian. Moreover we will denote by ${\mathcal{D}}(\mathbf{G})$ the set

$$\begin{eqnarray}\{\text{Inn}\,g\circ \unicode[STIX]{x1D706}_{d}^{\mathbf{G}}\mid d\in {\mathcal{D}}_{\unicode[STIX]{x1D6F7}}~\text{and}~g\in \mathbf{G}\}.\end{eqnarray}$$

Finally we remark that if $\mathbf{G}$ is a proximate algebraic group then $\text{d}\unicode[STIX]{x1D719}_{\text{sc}}$ is an isomorphism so we may, and will, identify the Chevalley basis considered in 3.19 with a basis of $\mathfrak{g}_{\text{der}}$ .

3.21. We now have the following cheap generalization of [Reference PremetPre2], which gives a case-free proof of the classification of nilpotent orbits in good characteristic by weighted Dynkin diagrams. For historical remarks concerning this theorem see [Reference PremetPre2, Remark 2]. In fact, we will also give certain statements concerning centralizers of nilpotent elements but for this we will need to assume that $\mathbf{G}$ is proximate. This assumption cannot be dropped, in general, as is evident by the example given at the end of [Reference PremetPre1, Introduction].

Proof. (i) Assume $\unicode[STIX]{x1D711}:\mathbf{G}\rightarrow \mathbf{H}$ is a surjective homomorphism of algebraic groups such that $\text{Ker}(\unicode[STIX]{x1D711})\subseteq Z(\mathbf{G})$ and $\text{Ker}(\text{d}\unicode[STIX]{x1D711})\subseteq \mathfrak{z}(\mathfrak{g})$ . Let $\mathfrak{h}=\text{Lie}(\mathbf{H})$ then according to [Reference JantzenJan, 2.7, Proposition] the differential $\text{d}\unicode[STIX]{x1D711}:\mathfrak{g}\rightarrow \mathfrak{h}$ induces a bijection ${\mathcal{N}}(\mathfrak{g})/\mathbf{G}\rightarrow {\mathcal{N}}(\mathfrak{h})/\mathbf{H}$ . Now the restriction of $\unicode[STIX]{x1D711}$ to the derived subgroup $\mathbf{G}_{\text{der}}$ defines an isogeny between the derived subgroups $\mathbf{G}_{\text{der}}\rightarrow \mathbf{H}_{\text{der}}$ . Moreover $\unicode[STIX]{x1D711}\circ \unicode[STIX]{x1D719}_{\text{sc}}:\mathbf{G}_{\text{sc}}\rightarrow \mathbf{H}_{\text{der}}$ is a simply connected covering. As remarked in 3.20 the cocharacter $\unicode[STIX]{x1D706}_{d}^{\mathbf{H}}$ does not depend upon the choice of simply connected cover used to define it. Hence, we must have $\unicode[STIX]{x1D706}_{d}^{\mathbf{H}}=\unicode[STIX]{x1D711}\circ \unicode[STIX]{x1D706}_{d}^{\mathbf{G}}$ for any weighted Dynkin diagram $d\in {\mathcal{D}}_{\unicode[STIX]{x1D6F7}}$ . In particular, this shows that $\unicode[STIX]{x1D711}$ induces a bijection ${\mathcal{D}}_{\unicode[STIX]{x1D6F7}}(\mathbf{G})\rightarrow {\mathcal{D}}_{\unicode[STIX]{x1D6F7}}(\mathbf{H})$ . It is easily checked that $\unicode[STIX]{x1D711}(\mathbf{L}(\unicode[STIX]{x1D706}_{d}^{\mathbf{G}}))=\mathbf{L}(\unicode[STIX]{x1D706}_{d}^{\mathbf{H}})$ and $\text{d}\unicode[STIX]{x1D711}(\mathfrak{g}(\unicode[STIX]{x1D706}_{d}^{\mathbf{G}},2))=\mathfrak{h}(\unicode[STIX]{x1D706}_{d}^{\mathbf{H}},2)$ , see, 3.17 and [Reference SpringerSpr2, 4.4.11(7)]. Moreover $\text{d}\unicode[STIX]{x1D711}$ induces a bijection between the $\mathbf{L}(\unicode[STIX]{x1D706}_{d}^{\mathbf{G}})$ -orbits on $\mathfrak{g}(\unicode[STIX]{x1D706}_{d}^{\mathbf{G}},2)$ and the $\mathbf{L}(\unicode[STIX]{x1D706}_{d}^{\mathbf{H}})$ -orbits on $\mathfrak{h}(\unicode[STIX]{x1D706}_{d}^{\mathbf{H}},2)$ . Now $\text{d}\unicode[STIX]{x1D711}:{\mathcal{N}}(\mathfrak{g})\rightarrow {\mathcal{N}}(\mathfrak{h})$ is a homeomorphism as it is a bijective morphism between irreducible varieties, see, [Reference SpringerSpr2, 5.2.3, 5.2.6, 5.2.9(4)]. In particular, the restriction $\text{d}\unicode[STIX]{x1D711}:\mathfrak{g}(\unicode[STIX]{x1D706}_{d}^{\mathbf{G}},2)\rightarrow \mathfrak{h}(\unicode[STIX]{x1D706}_{d}^{\mathbf{H}},2)$ is also a homeomorphism so clearly $\text{d}\unicode[STIX]{x1D711}({\mathcal{O}}_{\mathbf{G}}(d))={\mathcal{O}}_{\mathbf{H}}(d)$ . With this we see that (i) holds in $\mathbf{G}$ if and only if (i) holds in  $\mathbf{H}$ .

Now let $\mathbf{G}_{\text{ad}}=\text{Ad}(\mathbf{G})$ be the image of the adjoint representation $\text{Ad}:\mathbf{G}\rightarrow \text{GL}(V)$ , where $V=\mathfrak{g}$ . Certainly $\text{Ad}$ satisfies the above hypotheses, see, [Reference JantzenJan, 2.7], so applying the above argument we see that we need only prove (i) for $\mathbf{G}_{\text{ad}}$ . As $\mathbf{G}_{\text{ad}}$ is an adjoint semisimple group it is a direct product of simple algebraic groups. Hence, we clearly need only prove (i) in the case where $\mathbf{G}_{\text{ad}}$ is simple. Let us now assume this to be the case. If $\mathbf{G}_{\text{ad}}$ is of type $\mathsf{A}_{n}$ then we assume $\mathbf{G}=\text{GL}_{n+1}(\mathbb{K})$ and $\unicode[STIX]{x1D711}:\mathbf{G}\rightarrow \mathbf{G}_{\text{ad}}$ is the adjoint representation. Otherwise we assume that $\unicode[STIX]{x1D711}:\mathbf{G}\rightarrow \mathbf{G}_{\text{ad}}$ is a simply connected covering of $\mathbf{G}_{\text{ad}}$ . By [Reference PremetPre2, 2.6, 2.7] we then have (i) holds for $\mathbf{G}$ so by the above argument we have (i) holds for $\mathbf{G}_{\text{ad}}$ .

(ii) We assume $\unicode[STIX]{x1D706}^{\mathbf{G}}\in {\mathcal{D}}(\mathbf{G})$ is a cocharacter. Now, assume $\unicode[STIX]{x1D70B}:\mathbf{G}\rightarrow \widetilde{\mathbf{G}}$ is a closed embedding of algebraic groups such that $\widetilde{\mathbf{G}}$ is a connected reductive algebraic group and $\unicode[STIX]{x1D70B}$ restricts to an isomorphism between the derived subgroup of $\mathbf{G}$ and the derived subgroup of $\widetilde{\mathbf{G}}$ . In particular, $\widetilde{\mathbf{G}}=\unicode[STIX]{x1D70B}(\mathbf{G})Z(\widetilde{\mathbf{G}})$ and $\widetilde{\mathbf{G}}$ is proximate because we assume $\mathbf{G}$ is proximate. For convenience we will consider $\mathbf{G}$ as a subgroup of $\widetilde{\mathbf{G}}$ by identifying it with $\unicode[STIX]{x1D70B}(\mathbf{G})$ . Moreover we will identify $\mathfrak{g}$ with a subalgebra of $\widetilde{\mathfrak{g}}=\text{Lie}(\widetilde{\mathbf{G}})$ . By the remarks in 3.15 we see that ${\mathcal{N}}(\mathfrak{g})={\mathcal{N}}(\widetilde{\mathfrak{g}})$ and $\mathfrak{g}(\unicode[STIX]{x1D706}^{\mathbf{G}},2)=\widetilde{\mathfrak{g}}(\unicode[STIX]{x1D706}^{\widetilde{\mathbf{G}}},2)$ where $\unicode[STIX]{x1D706}^{\widetilde{\mathbf{G}}}:=\unicode[STIX]{x1D70B}\circ \unicode[STIX]{x1D706}^{\mathbf{G}}$ .

From the definition one can easily check that $\mathbf{P}(\unicode[STIX]{x1D706}^{\widetilde{\mathbf{G}}})=\mathbf{P}(\unicode[STIX]{x1D706}^{\mathbf{G}})Z(\widetilde{\mathbf{G}})$ and $\mathfrak{p}(\unicode[STIX]{x1D706}^{\widetilde{\mathbf{G}}})=\mathfrak{p}(\unicode[STIX]{x1D706}^{\mathbf{G}})+\text{Lie}(Z(\widetilde{\mathbf{G}}))$ . In particular, for any $e\in \mathfrak{g}(\unicode[STIX]{x1D706}^{\mathbf{G}},2)_{\text{reg}}=\widetilde{\mathfrak{g}}(\unicode[STIX]{x1D706}^{\widetilde{\mathbf{G}}},2)_{\text{reg}}$ we see that

$$\begin{eqnarray}\mathfrak{c}_{\mathfrak{g}}(e)\subseteq \mathfrak{p}(\unicode[STIX]{x1D706}^{\mathbf{G}})\Leftrightarrow \mathfrak{c}_{\mathfrak{ g}}(e)+\text{Lie}(Z(\widetilde{\mathbf{G}}))\subseteq \mathfrak{p}(\unicode[STIX]{x1D706}^{\mathbf{G}})+\text{Lie}(Z(\widetilde{\mathbf{G}}))\Leftrightarrow \mathfrak{c}_{\widetilde{\mathfrak{ g}}}(e)\subseteq \mathfrak{p}(\unicode[STIX]{x1D706}^{\widetilde{\mathbf{G}}}),\end{eqnarray}$$

which shows that (a) holds in $\mathbf{G}$ if and only if (a) holds in $\widetilde{\mathbf{G}}$ . As $\text{Lie}(Z(\widetilde{\mathbf{G}}))\subseteq \mathfrak{z}(\widetilde{\mathfrak{g}})$ we have $[e,\mathfrak{p}(\unicode[STIX]{x1D706}^{\widetilde{\mathbf{G}}})]=[e,\mathfrak{p}(\unicode[STIX]{x1D706}^{\mathbf{G}})]$ so the observations in 3.15 show that (b) holds in $\mathbf{G}$ if and only if (b) holds in $\widetilde{\mathbf{G}}$ . Finally we clearly have $(\text{Ad}\,\mathbf{P}(\unicode[STIX]{x1D706}^{\widetilde{\mathbf{G}}}))e=(\text{Ad}\,\mathbf{P}(\unicode[STIX]{x1D706}^{\mathbf{G}}))e$ hence (c) holds in $\mathbf{G}$ if and only if (c) holds in $\widetilde{\mathbf{G}}$ . Hence we have shown that (ii) holds in $\mathbf{G}$ if and only if (ii) holds in $\widetilde{\mathbf{G}}$ .

Applying this argument to the natural closed embedding $\mathbf{G}_{\text{der}}{\hookrightarrow}\mathbf{G}$ we see that we need only prove (ii) for $\mathbf{G}_{\text{der}}$ . Now assume (ii) holds for $\mathbf{G}_{\text{sc}}$ then we claim (ii) holds for $\mathbf{G}_{\text{der}}$ . Let $\unicode[STIX]{x1D706}^{\mathbf{G}_{\text{der}}}\in {\mathcal{D}}(\mathbf{G}_{\text{der}})$ be a cocharacter then by the definition of the set ${\mathcal{D}}(\mathbf{G}_{\text{der}})$ there exists a cocharacter $\unicode[STIX]{x1D706}^{\mathbf{G}_{\text{sc}}}\in {\mathcal{D}}(\mathbf{G}_{\text{sc}})$ such that $\unicode[STIX]{x1D706}^{\mathbf{G}}=\unicode[STIX]{x1D719}_{\text{sc}}\circ \unicode[STIX]{x1D706}^{\mathbf{G}_{\text{sc}}}$ . From the definition we see that $\unicode[STIX]{x1D719}_{\text{sc}}(\mathbf{P}(\unicode[STIX]{x1D706}^{\mathbf{G}_{\text{sc}}}))=\mathbf{P}(\unicode[STIX]{x1D706}^{\mathbf{G}_{\text{der}}})$ and as $\text{d}\unicode[STIX]{x1D719}_{\text{sc}}:\mathfrak{g}_{\text{sc}}\rightarrow \mathfrak{g}_{\text{der}}$ is an isomorphism it is clear that (c), (b) and the first part of (a) hold in $\mathbf{G}_{\text{der}}$ if they hold in $\mathbf{G}_{\text{sc}}$ . To see that the second part of (a) holds in $\mathbf{G}_{\text{der}}$ if it holds in $\mathbf{G}_{\text{sc}}$ one only needs to note that $\unicode[STIX]{x1D719}_{\text{sc}}(C_{\mathbf{G}_{\text{sc}}}(e))=C_{\mathbf{G}_{\text{der}}}(e)$ , where we identify $e$ with an element of $\mathfrak{g}_{\text{sc}}$ . This is because the kernel of $\unicode[STIX]{x1D719}_{\text{sc}}$ is contained in the center $Z(\mathbf{G}_{\text{sc}})$ which is the kernel of $\text{Ad}$ .

We now need only prove (ii) for $\mathbf{G}_{\text{sc}}$ . As $\mathbf{G}_{\text{sc}}$ is simply connected it is a direct product of simple groups, hence we may clearly assume that $\mathbf{G}_{\text{sc}}$ is simple. If $\mathbf{G}_{\text{sc}}$ is of type $\mathsf{A}_{n}$ then we choose a closed embedding $\unicode[STIX]{x1D70B}:\mathbf{G}_{\text{sc}}\rightarrow \widetilde{\mathbf{G}}$ such that $\widetilde{\mathbf{G}}=\text{GL}_{n+1}(\mathbb{K})$ and $\unicode[STIX]{x1D70B}(\mathbf{G}_{\text{sc}})$ is the derived subgroup, which clearly exists as we have an isomorphism $\mathbf{G}_{\text{sc}}\cong \text{SL}_{n+1}(\mathbb{K})$ . If $\mathbf{G}_{\text{sc}}$ is not of type $\mathsf{A}_{n}$ then we assume $\widetilde{\mathbf{G}}=\mathbf{G}_{\text{sc}}$ and $\unicode[STIX]{x1D70B}$ is the identity. By the previous argument we need only prove (ii) for $\widetilde{\mathbf{G}}$ . Let $\unicode[STIX]{x1D706}^{\widetilde{\mathbf{G}}}\in {\mathcal{D}}(\widetilde{\mathbf{G}})$ be a cocharacter. The first two parts (a) and (b) are given by [Reference PremetPre2, Theorem 2.3]. For part (c) we will use the fact that $C_{\mathbf{G}}(e)$ is separable because $p$ is a pretty good prime for $\widetilde{\mathbf{G}}$ , see Proposition 3.10. In particular, by (a) we have $C_{\widetilde{\mathbf{G}}}(e)=C_{\mathbf{P}(\unicode[STIX]{x1D706}^{\widetilde{\mathbf{G}}})}(e)$ so

$$\begin{eqnarray}\displaystyle \dim (\text{Ad}\,\mathbf{P}(\unicode[STIX]{x1D706}^{\widetilde{\mathbf{G}}})e) & = & \displaystyle \dim \mathbf{P}(\unicode[STIX]{x1D706}^{\widetilde{\mathbf{G}}})-\dim C_{\mathbf{P}(\unicode[STIX]{x1D706}^{\widetilde{\mathbf{G}}})}(e)\nonumber\\ \displaystyle & = & \displaystyle \dim \mathfrak{p}(\unicode[STIX]{x1D706}^{\widetilde{\mathbf{G}}})-\dim \mathfrak{c}_{\widetilde{\mathfrak{g}}}(e)\nonumber\\ \displaystyle & = & \displaystyle \dim [e,\mathfrak{p}_{\widetilde{\mathfrak{g}}}(\unicode[STIX]{x1D706})].\nonumber\end{eqnarray}$$

Combining this with (b) gives (c). ◻

3.23. To complete this picture we show that rational nilpotent elements are determined by rational cocharacters. Specifically let $F^{\ast }=q\unicode[STIX]{x1D70F}:X(\mathbf{T}_{0})\rightarrow X(\mathbf{T}_{0})$ with $\unicode[STIX]{x1D70F}$ a finite order automorphism, see, Theorem 2.3. If $F_{q}:\mathbb{G}_{m}\rightarrow \mathbb{G}_{m}$ denotes the Frobenius endomorphism defined by $F_{q}(k)=k^{q}$ then we define an action of $F$ on the set of cocharacters $Y(\mathbf{G})$ by setting

$$\begin{eqnarray}(F\cdot \unicode[STIX]{x1D706})(k)=F(\unicode[STIX]{x1D706}(F_{q}^{-1}(k)))\end{eqnarray}$$

for all $k\in \mathbb{G}_{m}$ and $\unicode[STIX]{x1D706}\in Y(\mathbf{G})$ . An easy calculation shows that for any integer $i\in \mathbb{Z}$ we have

$$\begin{eqnarray}F(\mathfrak{g}(\unicode[STIX]{x1D706},i))=\mathfrak{g}(F\circ \unicode[STIX]{x1D706},qi)=\mathfrak{g}(F\cdot \unicode[STIX]{x1D706},i).\end{eqnarray}$$

In particular, if $F\cdot \unicode[STIX]{x1D706}=\unicode[STIX]{x1D706}$ then we have $F(\mathfrak{u}(\unicode[STIX]{x1D706},i))=\mathfrak{u}(\unicode[STIX]{x1D706},i)$ and $F(\mathbf{U}(\unicode[STIX]{x1D706},i))=\mathbf{U}(\unicode[STIX]{x1D706},i)$ for any $i\in \mathbb{Z}_{{>}0}$ .

3.24. It is clear from the definition of the characters $\unicode[STIX]{x1D706}_{d}^{\mathbf{G}}\in {\mathcal{D}}_{\unicode[STIX]{x1D6F7}}(\mathbf{G})$ that we have $F\cdot \unicode[STIX]{x1D706}_{d}^{\mathbf{G}}=\unicode[STIX]{x1D706}_{d\circ \unicode[STIX]{x1D70F}}^{\mathbf{G}}$ . Moreover, the composition $d\circ \unicode[STIX]{x1D70F}$ is a weighted Dynkin diagram and so $F\cdot \unicode[STIX]{x1D706}_{d}^{\mathbf{G}}\in {\mathcal{D}}_{\unicode[STIX]{x1D6F7}}(\mathbf{G})$ . This implies that $F$ stabilizes the sets ${\mathcal{D}}_{\unicode[STIX]{x1D6F7}}(\mathbf{G})\subseteq {\mathcal{D}}(\mathbf{G})$ . It is in fact known that the bijection ${\mathcal{D}}_{\unicode[STIX]{x1D6F7}}(\mathbf{G})\rightarrow {\mathcal{N}}(\mathfrak{g})/\mathbf{G}$ restricts to a bijection ${\mathcal{D}}_{\unicode[STIX]{x1D6F7}}(\mathbf{G})^{F}\rightarrow ({\mathcal{N}}(\mathfrak{g})/\mathbf{G})^{F}$ between the $F$ -fixed cocharacters and the $F$ -stable nilpotent orbits, although we will not need this here. We will, however, need the following concerning rational cocharacters and rational nilpotent elements.

Proof. Let ${\mathcal{X}}_{e}$ denote the set of all cocharacters $\unicode[STIX]{x1D706}\in {\mathcal{D}}(\mathbf{G})$ such that $e\in \mathfrak{g}(\unicode[STIX]{x1D706},2)_{\text{reg}}$ . As any $\unicode[STIX]{x1D706}\in {\mathcal{X}}_{e}$ is conjugate to a cocharacter of the form $\unicode[STIX]{x1D706}_{d}^{\mathbf{G}}$ it follows from [Reference PremetPre2, 2.5] that $\unicode[STIX]{x1D706}$ is associated to $e$ in the sense of [Reference JantzenJan, 5.3, Definition]. We thus have by [Reference JantzenJan, 5.3, Lemma] that the connected centralizer $C_{\mathbf{G}}^{\circ }(e)$ of $e$ acts transitively on ${\mathcal{X}}_{e}$ via $(g,\unicode[STIX]{x1D706})\mapsto \text{Inn}\,g\circ \unicode[STIX]{x1D706}$ . As $F(e)=e$ we have $F$ restricts to a Frobenius endomorphism on $C_{\mathbf{G}}^{\circ }(e)$ . The existence of a fixed point $\unicode[STIX]{x1D706}\in {\mathcal{X}}_{e}^{F}$ is a standard application of the Lang–Steinberg theorem, see [Reference GeckGec2, 4.3.2(a)]. The stabilizer of $\unicode[STIX]{x1D706}$ is the closed subgroup $C_{\mathbf{G}}^{\circ }(e)\cap \mathbf{L}(\unicode[STIX]{x1D706})$ . One easily concludes from [Reference PremetPre2, 2.3], using similar arguments to those used in the proof of Theorem 3.22, that $C_{\mathbf{G}}^{\circ }(e)\cap \mathbf{L}(\unicode[STIX]{x1D706})=C_{\mathbf{L}(\unicode[STIX]{x1D706})}^{\circ }(e)$ is connected. Again applying [Reference GeckGec2, 4.3.2(a)] we see that $\unicode[STIX]{x1D706}$ is unique up to $G$ -conjugacy.◻

3.26. Assume now that $x\in \mathfrak{g}$ is any element of the Lie algebra. Following [Reference SlodowySlo, Section 5.1, p. 60] we say a locally closed subvariety $\unicode[STIX]{x1D6F4}\subseteq \mathfrak{g}$ is a transverse slice to $(\text{Ad}\,\mathbf{G})x$ at $x$ if the following hold:

1. (i) $x\in \unicode[STIX]{x1D6F4}$ ;

2. (ii) the action map $\mathbf{G}\times \unicode[STIX]{x1D6F4}\rightarrow \mathfrak{g}$ , defined by $(g,s)\mapsto \text{Ad}\,g(s)$ , is a smooth morphism;

3. (iii) $\dim \unicode[STIX]{x1D6F4}$ is minimal with respect to (i) and (ii).

Note that we have $\dim \unicode[STIX]{x1D6F4}\geqslant \dim C_{\mathbf{G}}(x)$ with equality if and only if $C_{\mathbf{G}}(x)$ is separable, see, [Reference SlodowySlo, Section 5.1, p. 61]. The following corollary of Theorem 3.22 shows that under some mild restrictions we can always find a transverse slice to a nilpotent orbit. This fact will be used later.

Proof. First, as $C_{\mathbf{G}}(e)$ is separable we have $T_{e}(\text{Ad}\,\mathbf{G}(e))=[\mathfrak{g},e]$ by Lemma 3.7 so $T_{e}(\text{Ad}\,\mathbf{G}(e))$ is stable under the $\mathbb{G}_{m}$ -action. Hence, one can find a graded, that is, $\mathbb{G}_{m}$ -invariant, complement $\mathfrak{s}$ of $T_{e}(\text{Ad}\,\mathbf{G}(e))$ in $\mathfrak{g}$ by simply picking a complement in each graded piece. By Theorem 3.22 we certainly have $\mathfrak{u}(\unicode[STIX]{x1D706},2)\subseteq [\mathfrak{g},e]$ . Now the map $\mathfrak{g}(\unicode[STIX]{x1D706},-1)\rightarrow \mathfrak{g}(\unicode[STIX]{x1D706},1)$ defined by $x\mapsto [x,e]$ is injective because $\mathfrak{g}(\unicode[STIX]{x1D706},-1)\cap \mathfrak{c}_{\mathfrak{g}}(e)=\{0\}$ by Theorem 3.22. However it must also be surjective because $\dim \mathfrak{g}(\unicode[STIX]{x1D706},-1)=\dim \mathfrak{g}(\unicode[STIX]{x1D706},1)$ , see, 5.8. This shows that $\mathfrak{u}(\unicode[STIX]{x1D706},1)\subseteq T_{e}(\text{Ad}\,\mathbf{G}(e))$ and so $\mathfrak{s}\subseteq \bigoplus _{i\leqslant 0}\mathfrak{g}(\unicode[STIX]{x1D706},i)$ .

Now the set $\unicode[STIX]{x1D6F4}$ is certainly locally closed and has dimension $\dim C_{\mathbf{G}}(e)$ , so we need only show that the action map $\unicode[STIX]{x1D70B}:\mathbf{G}\times \unicode[STIX]{x1D6F4}\rightarrow \mathfrak{g}$ is a smooth morphism. Recall that for a morphism of varieties $f:\mathbf{X}\rightarrow \mathbf{Y}$ we say $x\in \mathbf{X}$ is a smooth point of $f$ if $x$ and $f(x)$ are nonsingular and the differential $d_{x}f:T_{x}\mathbf{X}\rightarrow T_{f(x)}\mathbf{Y}$ is surjective. Now clearly $(1,e)$ is a smooth point of $\unicode[STIX]{x1D70B}$ because $\mathfrak{g}=\mathfrak{s}\oplus [\mathfrak{g},e]$ . In particular, by [Reference SpringerSpr2, Theorem 4.3.6], we have $\unicode[STIX]{x1D70B}$ is dominant and separable and the set $O$ of smooth points of $\unicode[STIX]{x1D70B}$ is nonempty and open because $\mathbf{G}\times \unicode[STIX]{x1D6F4}$ and $\mathfrak{g}$ are irreducible. By [Reference HartshorneHar, III, Section 10, 10.4] we then have $\unicode[STIX]{x1D70B}$ is a smooth morphism if and only if $O=\mathbf{G}\times \unicode[STIX]{x1D6F4}$ because $\mathbf{G}\times \unicode[STIX]{x1D6F4}$ and $\mathfrak{g}$ are nonsingular. To show this we argue as in [Reference SlodowySlo, p. 111].

We define a linear action $\unicode[STIX]{x1D6FC}:\mathbb{G}_{m}\times \mathfrak{g}\rightarrow \mathfrak{g}$ by setting $\unicode[STIX]{x1D6FC}(k,x)=\unicode[STIX]{x1D70C}(k)x$ , where $\unicode[STIX]{x1D70C}:\mathbb{G}_{m}\rightarrow \text{GL}(\mathfrak{g})$ is defined by $\unicode[STIX]{x1D70C}(k)(x)=k^{2}(\text{Ad}\,\unicode[STIX]{x1D706}(k^{-1}))(x)$ for all $x\in \mathfrak{g}$ and $k\in \mathbb{G}_{m}$ . Note that the $\mathbb{G}_{m}$ -action $\unicode[STIX]{x1D70C}$ preserves $\mathfrak{s}$ and as $e\in \mathfrak{g}(\unicode[STIX]{x1D706},2)$ we have $\unicode[STIX]{x1D70C}(k)(e+y)=e+\unicode[STIX]{x1D70C}(k)(y)$ for all $y\in \mathfrak{s}$ . In particular, the restriction of $\unicode[STIX]{x1D6FC}$ to $\mathbb{G}_{m}\times \unicode[STIX]{x1D6F4}$ is a contraction to $e$ , see, Remark 3.14. Now we can define an action of the direct product $\mathbf{G}\times \mathbb{G}_{m}$ on $\mathbf{G}\times \unicode[STIX]{x1D6F4}$ defined by $(h,k)\cdot (g,x)=(hg\unicode[STIX]{x1D706}(k),\unicode[STIX]{x1D70C}(k)(x))$ and on $\mathfrak{g}$ defined by $(h,k)\cdot y=k^{2}\text{Ad}\,h(y)$ . An easy calculation shows that $\unicode[STIX]{x1D70B}$ is equivariant with respect to these actions. Hence we need only show that $O$ meets every $\mathbf{G}\times \mathbb{G}_{m}$ -orbit of $\mathbf{G}\times \unicode[STIX]{x1D6F4}$ . Assume $n$ is the minimal integer such that $\mathfrak{g}(\unicode[STIX]{x1D706},n)\neq \{0\}$ then it is easy to see that every $\mathbf{G}\times \mathbb{G}_{m}$ -orbit of $\mathbf{G}\times \unicode[STIX]{x1D6F4}$ contains a set of the form

$$\begin{eqnarray}X_{y}=\left\{\left.\left(1,e+\mathop{\sum }_{i=n}^{0}a_{i}y_{i}\right)\right|a_{i}\in \mathbb{G}_{m}\right\}\end{eqnarray}$$

for some $y_{i}\in \mathfrak{g}(\unicode[STIX]{x1D706},i)$ because $\mathbb{K}$ is algebraically closed and $\mathfrak{s}\subseteq \bigoplus _{i\leqslant 0}\mathfrak{g}(\unicode[STIX]{x1D706},i)$ . Now $O\cap \overline{X_{y}}$ is a nonempty open set of the closure $\overline{X_{y}}$ because it contains $(1,e)$ , see, Remark 3.14. However as $\overline{X_{y}}$ is clearly irreducible, as $X_{y}$ is, we must have $O\cap X_{y}\neq \emptyset$ because $X_{y}$ is open in its closure. Thus we have shown that $\unicode[STIX]{x1D70B}$ is smooth.

We now consider the final point. Embedding $\mathbf{G}$ as a closed subgroup of some $\text{GL}_{n}(\mathbb{K})$ we may easily compute the differential $\text{d}_{(1,x)}\unicode[STIX]{x1D70B}:\mathfrak{g}\oplus T_{x}(\unicode[STIX]{x1D6F4})\rightarrow T_{x}(\mathfrak{g})$ using the framework of dual numbers, see, [Reference BorelBor, AG, 16.2]. Indeed, one readily checks that we have $\text{d}_{(1,x)}\unicode[STIX]{x1D70B}(g,y)=y+[g,x]$ . Now, by the above we know the differential is surjective so we must have $T_{x}(\mathfrak{g})=T_{x}(\unicode[STIX]{x1D6F4})+[\mathfrak{g},x]$ ; so the statement follows from Lemma 3.7.◻