Notation 2.1. We will assume from now on that L is a Lie algebra over k generated by its pure extremal elements. As in Definition 1.2, we will write E for the set of extremal elements of L and ${\mathcal {E}}$ for the set of extremal points of L. We let g be the symmetric bilinear form from Proposition 1.7. If V is any subspace of L, we write

$$\begin{align*}V^\perp := \{ u \in L \mid g(u,v) = 0 \text{ for all } v \in V \}. \end{align*}$$

Proof. This is shown in the proof of [Reference Cohen and IvanyosCI06, Theorem 28].

Proof. By Proposition 2.5, $E_{-1}(x) = E \cap (kx + L_{-1}) \setminus kx$ . Now observe that an element $\lambda x + e$ (with $e \in L_{-1}$ ) is extremal if and only if e is extremal, because if one of them is extremal, it is collinear with x, and hence, all points on the line through x are extremal points.

Proof. By Proposition 2.4(iii), we can write $a = [y,c]$ for some $c \in L_{-1}$ . By the Premet identity (2) and Proposition 2.4(v), we then get

$$ \begin{align*} [a, [y,b]] &= [[y,c], [y,b]] \\ &= g_y([c,b])y + g_y(b) [y,c] - g_y(c) [y,b] \\ &= g(y, [c,b])y = g([y,c], b)y = g(a,b)y. \end{align*} $$

Proof. Since $g_x(y)=g_y(x)=1$ , we get $\exp (x)(y) = y + [x,y] + x$ and $\exp (y)(x) = x - [x,y] + y$ . Since $g_x([x,y]) = 0$ and $g_y([x,y]) = 0$ , it now follows that $\varphi (x)=y$ and $\varphi (y)=x$ .

Next, let $l_1\in L_1$ . Then $\exp (y)(l_1) = l_1$ and $\exp (x)(l_1) = l_1 + [x, l_1]$ because $g_x(L_1) = 0$ (by Proposition 2.4(v)). By the same Proposition 2.4(v), we also have $g_y(L_{-1}) = 0$ , and hence,

$$\begin{align*}\varphi(l_1)=\exp(y)([x,l_1]+l_1)=[x,l_1]+ \big( l_1+[y,[x,l_1]] \big) = [x,l_1] , \end{align*}$$

by Proposition 2.4(iii). Similarly, $\varphi (l_{-1})=[y,l_{-1}]$ for all $l_{-1}\in L_{-1}$ .

Finally, let $l_0\in L_0$ . Let $\lambda ,\mu \in k$ be such that $[x,l_0]=\lambda x$ and $[y,l_0]=\mu y$ . Then by the fact that g associates with the Lie bracket, we have

$$\begin{align*}\mu = g_x(\mu y) = g_x([y,l_0]) = -g_y([x,l_0]) = -g_y(\lambda x) = -\lambda. \end{align*}$$

Hence,

$$ \begin{align*} \varphi(l_0) &= \exp(y)\exp(x)(l_0 - \lambda y) \\ &= \exp(y) (l_0 + \lambda x) - \lambda \varphi(y) \\ &= (l_0 - \lambda y) + \lambda (x - [x,y] + y) - \lambda x \\ &= l_0 - \lambda [x,y] = l_0 + [x,[y,l_0]].\\[-34pt] \end{align*} $$

Proof. Clearly, $\varphi _\lambda $ is bijective. Consider $l_i\in L_i$ and $l_j\in L_j$ arbitrary, with $i,j\in [-2, 2]$ . Then

$$\begin{align*}\varphi_\lambda ([l_i,l_j])=\lambda^{i+j}[l_i,l_j]=[\lambda^il_i,\lambda^jl_j]=[\varphi_\lambda(l_i),\varphi_\lambda(l_j)].\\[-37pt] \end{align*}$$

Proof. Let $z\in Z_{-1}$ and $a\in L_{-1}$ . Then $[a,z] = 0$ by definition of $Z_{-1}$ , and $g_y(z) = g_y(a) = 0$ by Proposition 2.4(v). By the Premet identities, we then have

$$\begin{align*}[[y,a],[y,z]]=g_y([a,z])+g_y(z)[y,a]-g_y(a)[y,z]=0. \end{align*}$$

Since $[y, L_{-1}] = L_1$ by Proposition 2.4(iii), this shows that $[y,Z_{-1}]\leq Z_1$ . Similarly, $[x, Z_1] \leq Z_{-1}$ , but then $Z_{-1} = [x, [y, Z_{-1}]] \leq [x, Z_1] \leq Z_{-1}$ , so in fact,

$$\begin{align*}[y, Z_{-1}] = Z_1 \quad \text{and} \quad [x, Z_1] = Z_{-1}. \end{align*}$$

Observe now that by the Jacobi identity, $[[L_0,Z_{-1}],L_{-1}] \leq [L_0,[Z_{-1},L_{-1}]]+[Z_{-1},[L_0,L_{-1}]]=0$ , and thus, $[L_0,Z_{-1}]\leq Z_{-1}$ ; similarly, we have $[L_0,Z_1]\leq Z_1$ . Moreover,

$$\begin{align*}[y,[Z_{-1},L_1]]=[[y,Z_{-1}],L_1]=[Z_1,L_1]=0 , \end{align*}$$

and hence, the map $\varphi $ from Lemma 2.8 fixes $[Z_{-1}, L_1]$ . However, $\varphi (Z_{-1}) = [y, Z_{-1}] = Z_1$ and $\varphi (L_1) = [x, L_1] = L_{-1}$ , and therefore, $[Z_{-1},L_1] = [Z_1,L_{-1}]$ . We can now see that

$$\begin{align*}Z_{-1}\oplus [Z_{-1},L_1]\oplus Z_1 \end{align*}$$

is an ideal of L. Because L is simple, we conclude that $Z_{-1}=0$ and $Z_1=0$ .

Proof. By assumption, there exist $\lambda ,\mu \in k^\times $ such that $\varphi (x)=\lambda a$ and $\varphi (y)=\mu b$ . We have $L_0=N_L(x)\cap N_L(y)$ and $L^{\prime }_0=N_L(a)\cap N_L(b)$ by Proposition 2.4, so $L_0$ is mapped to $L^{\prime }_0$ by $\varphi $ .

Next, let $U = \langle x, y, [x,y] \rangle ^\perp $ and $U' = \langle a, b, [a,b] \rangle ^\perp $ . By Remark 1.3 and Proposition 1.7(ii), the form g is uniquely determined by the Lie algebra L, so $\varphi (U) = U'$ . (Recall that ‘ $\perp $ ’ is with respect to g.) It follows that $\varphi $ maps $L_{-1} = [x,U]$ to $L^{\prime }_{-1} = [a,U']$ and $L_1 = [y,U]$ to $L^{\prime }_1 = [b, U']$ .

Proof. By Remark 2.2, L does not contain sandwich elements. By [Reference Cohen and IvanyosCI06, Theorem 28], $({\mathcal {E}}, \mathcal {F})$ is a so-called root filtration space, which is either nondegenerate or has an empty set of lines. By our assumption, the latter case does not occur.

In particular, the statements of [Reference Cohen and IvanyosCI06, Lemmas 1 and 4] and [Reference Cohen and IvanyosCI07, Lemma 8] hold. Now claim (i) is [Reference Cohen and IvanyosCI06, Lemma 4(ii)] and claim (ii) is [Reference Cohen and IvanyosCI07, Lemma 8(ii)]. To show claim (iii), let $z := [x,y]$ , so $(z,x) \in E_{-1}$ . By (i), we can find some $u \in E_{-1}(x) \cap E_1(u)$ . We can now invoke [Reference Cohen and IvanyosCI06, Lemma 1(v)] on the pairs $(y,x)$ and $(z,u)$ to conclude that $u \in E_{-1}(x) \cap E_2(y)$ .

Finally, to show (iv), we first observe that the nondegeneracy of the root filtration space implies that $E_{-1}(x) \neq \emptyset $ (this is condition (H) on [Reference Cohen and IvanyosCI06, p. 439]) and hence, there exists a line $\ell $ through $\langle x \rangle $ . We now use the defining properties (D) and (F) of a root filtration space (see [Reference Cohen and IvanyosCI06, p. 435]) to see that ${\mathcal {E}}_1(y)$ has nonempty intersection with $\ell $ , as required. The final statement $E_{-1}(x) \cap E_{\leq 0}(y) = \emptyset $ is just the defining property (D) itself.

Proof. By Proposition 2.5, we have $E_{-1}(x) = (E \cap L_{\leq -1}) \setminus L_{-2}$ . Similarly (e.g., by applying Lemma 2.8; see Remark 2.9), we have $E_1(y) = (E \cap L_{\geq {-1}}) \setminus L_{\geq 0}$ . We conclude that $E_{-1}(x) \cap E_1(y) = E \cap L_{-1}$ .

Proof. By Proposition 2.4(v), we have $g_e(x)=0$ and $g_e(y)=0$ . Moreover, for any $l\in L_1$ , there exists $l'\in L_{-1}$ such that $l=[y,l']$ by Proposition 2.4. Since g associates with the Lie bracket, we have

$$\begin{align*}g_e(l) = g_e([y,l']) = g([e,y],l') = 0. \end{align*}$$

Finally, let $l \in L_0$ . By Proposition 2.4 again, there exist $e'\in L_{-1}$ such that $e=[y,e']$ . Let $\lambda \in k$ such that $[y,l]=\lambda y$ . Then

$$\begin{align*}g_e(l) = g([y,e'],l)=-g(e',[y,l])=\lambda g_y(e')=0.\\[-37pt] \end{align*}$$

Proof. By Proposition 2.13(iv), there exists an $a \in E_{-1}(x) \cap E_{1}(y)$ . By Lemma 2.14, $a\in E \cap L_{-1}$ .

Now, by Proposition 2.13(i), we can find some $b \in E_{-1}(x) \cap E_1(a)$ . By Corollary 2.6, we can write $b = \lambda x + d$ with $d \in E \cap L_{-1}$ . Moreover, $[a, d] = [a, b]$ , and since the pair $(a,b)$ is special and both a and b are collinear with x, we must have $[a,b] = \lambda x$ for some $\lambda \in k^\times $ . Now write $c = \lambda ^{-1} a$ , and we get $[c, d] = x$ , as required.

Next, observe that $g(c, y) = 0$ and $g(d, y) = 0$ because these pairs are special. Using the Jacobi identity, we get

$$ \begin{align*} [c+d,-[c,y]+[d,y]]&=-2g_{c}(y)c+2g_{d}(y)d+[c,[d,y]]-[d,[c,y]] \\ &= [c,[d,y]]+[d,[y,c]]=-[y,[c,d]]\\ &=-[y,x]=[x,y], \end{align*} $$

as claimed.

Assumption 3.1. We assume in this section that L is a simple Lie algebra generated by its set E of pure extremal elements. We also assume that $k \neq \mathbb F_2$ . Notice that this implies that L does not contain sandwich elements, and by [Reference Cohen and IvanyosCI06, Theorem 28], the graph $({\mathcal {E}}, {\mathcal {E}}_2)$ is connected, so L contains many hyperbolic pairs.

Notation 3.2. Let $(x,y) \in E_2$ be a hyperbolic pair and scale x and y so that $g(x,y)=1$ . Let $L=L_{-2}\oplus L_{-1}\oplus L_0\oplus L_1\oplus L_2$ be the corresponding $5$ -grading as in Proposition 2.4, with $L_{-2} = \langle x \rangle $ and $L_2 = \langle y \rangle $ .

Let $I_{-1}$ be the subspace of $L_{-1}$ spanned by $E\cap L_{-1}$ and let $I_1$ be the subspace of $L_{1}$ spanned by $E\cap L_{1}$ . Notice that the extremal geometry has no lines if and only if $E_{-1}$ and $E_1$ are empty or, equivalently, if and only if $I_{-1} = I_1 = 0$ .

Proof. Denote the subalgebra generated by y and $I_{-1}$ by I. Since $I_{-1}\neq \emptyset $ , there are lines in the extremal geometry. By Proposition 2.16, $x \in [I_{-1},I_{-1}] \leq I$ . By Corollary 2.6, all elements in $E_{-1}(x)$ are contained in I.

Now observe that if $a,b \in I$ , then also $\exp (a)(b) = b + [a,b] + g(a,b)a \in I$ . In particular, the automorphism $\varphi $ introduced in Lemma 2.8 maps I to itself. This implies that also $I_1 \in I$ and all elements of $E_{-1}(y)$ belong to I.

We can now replace y by any $y' \in E_2(x) \cap E_{-1}(y)$ ; observe that both $y'$ and the ‘new’ $I_{-1}$ w.r.t. $y'$ are contained in I. By repeating this procedure, we see that

(5) $$ \begin{align} \text{all elements of } E_2(x)\ \textit{connected} \text{ to } y \text{ are contained in } I. \end{align} $$

If the relation $E_0$ is nonempty, then we can invoke [Reference Cuypers and MeulewaeterCM21, Lemma 2.20]. It tells us that ${\mathcal {E}}_2(x)$ is connected, and we conclude that $E_2(x) \subseteq I$ . If $z \in E_1(x)$ , then by Proposition 2.13(iii), we can find an element $y' \in E_{-1}(z) \cap E_2(x)$ . Again, $y' \in I$ , and we can replace y by $y'$ to see that also the collinear point z is contained in I. Thus, $E_1(x) \subseteq I$ . Applying $\varphi $ , we also get $E_1(y) \subseteq I$ . The only extremal elements not considered yet are those in $C := E_0(x) \cap E_0(y)$ ; notice that $C = E \cap L_0$ by Proposition 2.5. So let $z \in C$ and let $\ell $ be any line through z not completely contained in C. Now observe that C is a subspace of the extremal geometry (i.e., if two collinear elements are contained in C, then the whole line through these elements lies in C). (This follows, for instance, from the fact that $L_0 = N_L(x) \cap N_L(y)$ by Proposition 2.4). Hence, z is the unique point of $\ell $ in C. Since we assume $|k| \geq 3$ (see Assumption 3.1), $\ell $ contains at least $2$ other points, which we already know are contained in I, and we conclude that also $z \in I$ . Since L is generated by E, this shows that indeed $I=L$ in this case.

Assume from now on ${\mathcal {E}}_0=\emptyset $ . In this case, the extremal geometry is a generalized hexagon, and the relations ${\mathcal {E}}_{-1}$ , ${\mathcal {E}}_1$ and ${\mathcal {E}}_2$ coincide with the relations ‘distance $1$ ’, ‘distance $2$ ’, and ‘distance $3$ ’, respectively, in the collinearity graph. (Indeed, we have observed in the proof of Proposition 2.13 that $({\mathcal {E}}, \mathcal {F})$ is a nondegenerate root filtration space, and it has been observed in [Reference CohenCoh12, Example 3.5] that a nondegenerate root filtration space with ${\mathcal {E}}_0 = \emptyset $ is necessarily a generalized hexagon.) Recall that in a generalized hexagon, for any point p and any line $\ell $ , there is a unique point on $\ell $ closest to p, called the projection of p on $\ell $ .

Let $z\in E_2(x)$ be arbitrary; our goal is to show that $z\in I$ . If $(y,z) \in E_{\leq -1}$ , then it follows from (5) that $z \in I$ . Next, assume that $(y,z) \in E_1$ , so $\langle y\rangle $ and $\langle z\rangle $ are at distance $2$ . Recall that the common neighbor of $\langle y\rangle $ and $\langle z\rangle $ is $\langle [y,z]\rangle $ . By (5), we are done if $[y,z]\in E_2(x)$ , so we may assume that $[y,z] \in E_1(x)$ . Now consider any line through $\langle z \rangle $ not containing $\langle [y,z] \rangle $ , and let $\langle p \rangle $ be the projection of $\langle x \rangle $ on that line, so $\operatorname {\mathrm {dist}}(\langle x \rangle , \langle p \rangle ) = 2$ ; let $q := [p,x]$ . Observe now that

$$\begin{align*}\langle x\rangle,\langle [x,[y,z]]\rangle,\langle [y,z]\rangle,\langle z\rangle, \langle p \rangle, \langle q \rangle,\langle x\rangle\end{align*}$$

forms an ordinary hexagon. Next, let $\langle b \rangle $ the projection of $\langle y \rangle $ on the line $\langle p, q \rangle $ . Notice that $\langle b \rangle $ is different from $\langle p \rangle $ and $\operatorname {\mathrm {dist}}(\langle y \rangle , \langle b \rangle ) = 2$ . If $\langle b \rangle = \langle q \rangle $ , we replace y by another $y'$ collinear with y and $[y,z]$ , which will then have a different projection on the line $\langle p,q \rangle $ , so we may assume without loss of generality that $\langle b \rangle \neq \langle q \rangle $ . Let $a := [b,y]$ .

Since all neighbors of $\langle x \rangle $ and $\langle y \rangle $ are contained in I, we see that $q \in I$ and $a \in I$ , and hence also their common neighbor $\langle b \rangle $ belongs to I. Now the line $\langle p, q \rangle $ already contains two points in I (namely $\langle b \rangle $ and $\langle q \rangle $ ), and hence also $p \in I$ . Finally, since $\langle p \rangle $ and $\langle [y,z] \rangle $ belong to I, their common neighbor $\langle z \rangle $ belongs to I.

Finally, assume that $(y,z) \in E_2$ , so $z\in E_2(x)\cap E_2(y)$ . Let $\langle y\rangle ,\langle a\rangle ,\langle b\rangle ,\langle z\rangle $ be a path of length $3$ . If $\langle a,b\rangle $ is not completely contained in $E_{\leq 1}(x)$ , we can apply the previous paragraph twice to obtain $z\in I$ . So assumeFootnote ² $a,b\in E_1(x)$ (i.e., both a and b are at distance $2$ from x). Since a generalized hexagon does not contain cycles of length $\leq 5$ , this can only happen if the line $\langle a,b\rangle $ contains a point $\langle c\rangle $ collinear with $\langle x\rangle $ .

Since $\operatorname {\mathrm {Exp}}(x)$ acts sharply transitively on the set of points of the line $\langle a, b\rangle $ without $\langle c\rangle $ (see Definition 1.5), we find a unique $\psi \in \operatorname {\mathrm {Exp}}(x)$ such that $\psi (a) \in \langle b \rangle $ . Let $z' := \psi (y)$ , and notice that $\psi $ fixes x. Since $x,y \in I$ , also $z' \in I$ . Since $\psi $ fixes x, we get $z'\in E_2(x)$ . Now $\langle b\rangle = \psi (\langle a\rangle )$ is a neighbor of both $\langle z\rangle $ and of $\langle z'\rangle $ , and thus, $z'\in E_{\leq 1}(z)$ . Since $z'\in I$ , the previous paragraph now implies $z\in I$ .

We have now shown that $E_2(x) \subseteq I$ , and just as in the case where ${\mathcal {E}}_0 \neq \emptyset $ , we conclude that $I = L$ .

Proof. By Lemma 3.3 and the Jacobi identity, I is an ideal of L. Since L is simple, this implies $I=L$ .

Proof. Let X be the subspace generated by x, y and $x^\perp \cap y^\perp $ . Since every line through x contains a point collinear to y, we have $x^\perp \subseteq X$ , and similarly $y^\perp \subseteq X$ . Consider $z\in X$ arbitrary. Let $\ell $ be any line through x. If $x^\perp \cap \ell \neq y^\perp \cap \ell $ , then $\ell \subseteq X$ . So we may assume that $\ell $ contains a unique point $a\in x^\perp \cap y^\perp $ . By the nondegeneracy, we find $b\in x^\perp \cap y^\perp $ such that a and b are not collinear. Now $\ell $ and $by$ are opposite lines, and hence, there exists a point c on $\ell $ distinct from a which is collinear to a point of $by$ distinct from y and b. Hence, c lies on a line m such that $m\cap y^\perp \in by\setminus \{b,y\}$ and thus, $m\cap y^\perp \neq m\cap x^\perp $ . As before, $m\subseteq X$ . In particular, $c\in X$ , so together with $a\in X$ we obtain $z\in X$ .

Proof. By Proposition 3.5(v), $S\cap {\mathcal {E}}_0(y)$ consists of a single point $\langle z\rangle $ , and since $S \subseteq {\mathcal {E}}_{\leq 0}(x)$ , this implies $z \in L_0$ . By Lemma 3.6, $L_S$ is spanned by x, z, and all points $\langle a\rangle $ collinear to both $\langle x\rangle $ and $\langle z\rangle $ . For such a point $\langle a \rangle $ , we have of course $a \in E_{-1}(x)$ . Now $z \in E_{-1}(a) \cap E_0(y)$ , so by Proposition 2.13(iv), we have $(a,y) \not \in E_2$ . Hence, $a \in E_{-1}(x) \cap E_{\leq 1}(y) = L_{-1}$ by Lemma 2.14.

Proof. As before, let $I_{-1}$ and $I_1$ be the subspaces of L linearly spanned by the extremal elements contained in $L_{-1}$ and $L_1$ , respectively. Let

$$\begin{align*}I = \langle x\rangle \oplus I_{-1}\oplus [I_{-1},I_1]\oplus I_1\oplus \langle y \rangle. \end{align*}$$

We will use Corollary 3.4 to show that $I = L$ .

We first show that $[I, y] \leq I$ . By Proposition 2.16, we know that $[x,y]\in [I_{-1},I_1]$ . Next, if $a\in E\cap L_{-1}$ , then $a\in E_1(y)$ , and hence, $[a,y]\in I_{1}$ . Since $[L_{\geq 0}, y] \leq L_2 = \langle y \rangle $ , the claim $[I,y]\leq I$ is now clear.

It remains to show that $[I,I_{-1}]\leq I$ . Obviously, $[\langle x\rangle \oplus I_{-1}\oplus I_1,I_{-1}]\leq I$ . Similarly as before, $[\langle y\rangle ,I_{-1}]\leq I_1\leq I$ , so the only case left to prove is

$$\begin{align*}[[I_{-1},I_1],I_{-1}]\leq I_{-1}. \end{align*}$$

So consider arbitrary extremal elements $a,b\in E\cap L_{-1}$ and $c\in E\cap L_1$ ; our goal is to show that

(6) $$ \begin{align} [[a,c],b]\in I_{-1}. \end{align} $$

Observe that $[a,b] \in L_{-2} = \langle x \rangle $ , so by the Jacobi identity and the fact that $[x,c]$ is contained in $E\cap L_{-1}$ , we have

$$\begin{align*}[[a,c], b] \in I_{-1} \iff [[b,c], a] \in I_{-1}. \end{align*}$$

Also observe that $c\in E_{\leq 0}(a)$ implies $[a,c]=0$ and that $c\in E_{\leq 0}(b)$ implies $[b,c]=0$ . Hence, we may assume from now on that $c\in E_{\geq 1}(a)\cap E_{\geq 1}(b)$ .

Since a is extremal, $[[a,c], a] \in \langle a \rangle $ is again extremal, so (6) obviously holds if $(a,b) \in E_{-2}$ . If $(a,b) \in E_{-1}$ , then by [Reference Cuypers and MeulewaeterCM21, Lemma 6.3], $J := \langle a,b\rangle $ is an inner ideal (i.e., $[J, [J, L]] \leq J$ ); in particular, $[b,[a,c]] \in J \leq I_{-1}$ . Next, if $(a,b) \in E_{0}$ , then a and b are contained in a symplecton S. The subspace $L_S$ spanned by all elements of S is an inner ideal by [Reference Cuypers and MeulewaeterCM21, Lemma 6.5], and hence, $[b,[a,c]]\in L_S$ . Since $[b,[a,c]]\in L_{-1}$ , Lemma 3.7 now implies that $[b,[a,c]] \in \langle \{ \langle z\rangle \in S\mid z\in L_{-1}\} \rangle \leq I_{-1}$ .

Now notice that $(a,b) \in E_2$ cannot occur because both a and b are collinear to x, so we can assume from now on that $(a,b) \in E_1$ . In particular, $[a,b] \in \langle x \rangle \setminus \{ 0 \}$ .

If $c\in E_1(a)$ , then $[a,c]\in E$ , and since the extremal form g is associative, we have $g(b,[a,c])=g([b,a],c)=0$ , since $c\in E_1(x)$ . Hence, $[a,c]\in E_{\leq 1}(b)$ , so either $[[a,c],b]\in E$ or $[[a,c],b]=0$ . In either case, we obtain $[[a,c],b]\in I_{-1}$ . The case $c\in E_1(b)$ is similar.

The only case left to consider is $a\in E_1(b)$ and $c\in E_2(a)\cap E_2(b)$ , and this case requires some more effort. We have $g(a,b)=0$ , and by rescaling, we may assume that $g(a,c) = g(b,c) = 1$ , and we also have $g(b, [a,c]) = 0$ as before.

Let $\lambda ,\mu \in k$ be arbitrary nonzero elements such that

(7) $$ \begin{align} \mu (\lambda - 1) = 1 , \end{align} $$

which exist because $k \neq \mathbb F_2$ , and consider the extremal element

$$\begin{align*}d := \exp(\mu c)\exp(\lambda b)\exp(c)(a) .\end{align*}$$

Then

$$ \begin{align*} d &= \exp(\mu c)\exp(\lambda b) \big(a + [c,a] + c \big) \\ &= \exp(\mu c) \big( \underbrace{\lambda [b,a]}_{\in L_{-2}} + \underbrace{a + \lambda [b, [c,a]] + \lambda^2 b}_{\in L_{-1}} + \underbrace{[c,a] + \lambda [b,c]}_{\in L_0} + \underbrace{c\vphantom{[]}}_{\in L_1} \big). \end{align*} $$

We claim that $d \in L_{\leq -1}$ . This could be checked by a direct computation, but it is easier to reverse the process, by starting from an arbitrary extremal element of the form

$$\begin{align*}e := \lambda [b,a] + z \in \langle x \rangle + L_{-1} , \end{align*}$$

compute $\exp (-\mu c)(e)$ , and verify that with the correct choice of z, this is equal to $\exp (-\mu c)(d)$ . We get

$$\begin{align*}\exp(-\mu c)(e) = \underbrace{\lambda [b,a]}_{\in L_{-2}} + \underbrace{z - \lambda \mu [c,[b,a]]}_{\in L_{-1}} - \underbrace{\mu [c,z]}_{\in L_0} + \underbrace{\mu^2 g(c,z)c\vphantom{[]}}_{\in L_1}. \end{align*}$$

Comparing the $L_{-1}$ -components, we have no other choice than setting

(8) $$ \begin{align} z := a + \lambda^2 b + \lambda [b,[c,a]] + \lambda \mu [c,[b,a]]. \end{align} $$

Now, using the Premet identity (3) (see Definition 1.2) and the fact that $g(c, [b,a]) = 0$ , we get

$$ \begin{align*} [c,z] &= [c,a] + \lambda^2 [c,b] + \lambda [c,[b,[c,a]]] + \lambda \mu [c,[c,[b,a]]] \\ &= [c,a] + \lambda^2 [c,b] - \lambda ([c,b] + [c,a]) \\ &= (1 - \lambda) [c,a] + \lambda (\lambda - 1) [c,b], \end{align*} $$

and hence by (7), $-\mu [c,z] = [c,a] - \lambda [c,b]$ , so the $L_0$ -components coincide. Finally,

$$ \begin{align*} g(c,z) &= g(c,a) + \lambda^2 g(c,b) + \lambda g(c,[b,[c,a]]) + \lambda \mu g(c,[c,[b,a]]) \\ &= 1 + \lambda^2 - \lambda g([c, [c, a]], b) \\ &= 1 + \lambda^2 - 2 \lambda g(c,a) g(c, b) = 1 + \lambda^2 - 2 \lambda = (1 - \lambda)^2 , \end{align*} $$

so again by (7), $\mu ^2 g(c,z)= 1$ , and hence also the $L_1$ -components coincide. Therefore, $e=d$ , and we conclude that the element $e = \lambda [b,a] + z \in \langle x \rangle + L_{-1}$ is an extremal element.

If $z = 0$ , then by (8), $[b,[c,a]]$ can be written as the sum of $3$ extremal elements contained in $L_{-1}$ . Indeed, obviously $a, b\in E$ , but also $[c,[b,a]]\in E$ since $\langle c\rangle $ and $\langle x\rangle =\langle [b,a]\rangle $ form a special pair.

If $z \neq 0$ , then $\langle e,x \rangle $ is a line of the extremal geometry. Since $z \in \langle e,x\rangle $ , it is contained in E. So by (8) again, the element $[b,[c,a]]$ can be written as the sum of $4$ extremal elements contained in $L_{-1}$ .

This shows that $[b,[c,a]] \in I_{-1}$ in all possible cases.

Proof. By Proposition 2.4(ii), we have

$$\begin{align*}[l,[l,[x,y]]]=[l,2l_{-2}+l_{-1}-e]=-3[l_{-2},e]-2[l_{-1},e] \in L_{-1} + L_0.\end{align*}$$

However, since l is extremal, we have $[l, [l, [x,y]]] \in \langle l \rangle $ . By the previous identity, this element must have trivial $L_1$ -component, but since $e \neq 0$ by assumption, this implies that $[l, [l, [x,y]]] = 0$ , so

$$\begin{align*}3[l_{-2}, e] = 2[l_{-1}, e] = 0. \end{align*}$$

Since $\operatorname {\mathrm {char}}(k) \neq 3$ , this implies $[l_{-2},e]=0$ , and hence, $l_{-2}=0$ by Proposition 2.4(iii). If $l_{-1}=0$ , then there is nothing to show, so assume $l_{-1}\neq 0$ . Then

$$\begin{align*}l = l_{-1} + e \in L_{-1} + L_1 , \quad l_{-1} \neq 0, \quad e \neq 0. \end{align*}$$

By Proposition 2.5, we have $l\in E_1(y)$ , and thus, $[l_{-1},y] = [l,y] \in E$ . By Proposition 2.5 again, $[l_{-1},y]\in E_1(x)$ , and thus, $l_{-1}=[x,[l_{-1},y]]\in E$ . Since $l_{-1}$ , e and $l=l_{-1}+e$ are contained in E, Proposition 1.8 implies that $(l_{-1},e)\in E_{-1}$ . Hence, $\langle l_{-1}\rangle $ is the common neighbor of the special pair $(\langle x\rangle , \langle e\rangle )$ , and thus, $l_{-1}\in \langle [x,e]\rangle $ .

Proof. Let $\lambda ,\mu \in k$ be such that $l_{-2}=\lambda x$ and $[l_0,x]=\mu x$ . By Proposition 2.4(ii), we have

$$ \begin{align*} [l,[l,x]] &= [l, \mu x + [y,x]] \\ &= \mu [l,x] + \big[ \lambda x+l_{-1}+l_0+y, [y,x] \big] \\ &= (\mu^2 x+\mu[y,x])+(-2\lambda x-l_{-1}+2y). \end{align*} $$

Since l is extremal and $g(l, x) = g(y, x)=1$ by Proposition 2.4(v), we get

(9) $$ \begin{align} \mu^2 = 4\lambda, \quad 3 l_{-1} = 0, \quad \mu[y,x]=2l_0. \end{align} $$

If $\operatorname {\mathrm {char}}(k)$ is not $2$ or $3$ , then this already implies that $l =\exp (-\tfrac {1}{2} \mu x)(y)$ .

Assume now $\operatorname {\mathrm {char}}(k)=3$ . We claim that also in this case, $l_{-1} = 0$ ; it then follows again that $l=\exp (-\tfrac {1}{2} \mu x)(y)$ . To show the claim, suppose that $l_{-1} \neq 0$ and let $z := \exp (\tfrac {1}{2} \mu x)(l)$ . Then z is an extremal element, and we compute that $z = l_{-1} + y$ . We apply the same technique as in the last paragraph of the proof of Lemma 3.9: By Proposition 2.5, $z \in E_1(y)$ , and thus, $[l_{-1},y] = [z,y] \in E$ . By Proposition 2.5 again, $[l_{-1}, y]\in E_1(x)$ , and thus, $l_{-1} = [x, [l_{-1}, y]] \in E$ . Since $l_{-1}$ , y and $z = l_{-1}+y$ are contained in E, Proposition 1.8 implies that $(l_{-1}, y) \in E_{-1}$ , contradicting Proposition 2.5.

Assume, finally, that $\operatorname {\mathrm {char}}(k)=2$ . By (9), we have $l_{-1}=0$ , so $l \in L_{-2} + L_0 + L_2$ . By Proposition 2.16, there exist $c,d \in E\cap L_{-1}$ such that $[c, d] = x$ . By Lemma 2.15, we get for any $e\in E\cap L_{-1}$ that $g(l,e)=0$ , and since $[y,e]\neq 0$ , this implies that the pair $(l,e)$ is special; in particular, $[l,e] = [l_0,e] + [y,e] \in L_{-1} + L_1$ is extremal. By Lemma 3.9 applied on $[l,c]$ and $[l,d]$ , we find $\lambda ,\mu \in k$ such that $[l_0, c] = \lambda [x, [y, c]] = \lambda c$ and $[l_0, d] = \mu [x, [y, d]] = \mu d$ . Let $\alpha := \exp (\lambda x)$ . Then

$$\begin{align*}\alpha([l,c])=[y,c], \quad \alpha([l,d])=[y,d] + (\lambda + \mu) d. \end{align*}$$

By the Premet identity (2), we have

$$ \begin{align*} &[[l,c],[l,d]] = g_l ([c,d])l + g_l (d)[l,c] + g_{l}(c)[l,d] = l \quad \text{and} \\ &[[y,c],[y,d]] = g_y ([c,d])y + g_y (d)[l,c] + g_{y}(c)[l,d] = y. \end{align*} $$

Hence,

$$ \begin{align*} \alpha(l) &= [\alpha([l,c]),\alpha([l,d])]=[[y,c], [y,d] + (\lambda + \mu)d] \\ &= y + (\lambda + \mu) [[y, c], d]. \end{align*} $$

Now note that by the Premet identity (3), we have

$$\begin{align*}[y, [d, [y, c]]] = g_y([d,c]) y + g_y(c) [y,d] + g_y(d) [y,c] = y, \end{align*}$$

and hence,

$$\begin{align*}[\alpha(l), y] = (\lambda + \mu) y , \end{align*}$$

but then since $\operatorname {\mathrm {char}}(k) = 2$ and $\alpha (l)$ is an extremal element, we get

$$\begin{align*}0 = [\alpha(l), [\alpha(l), y]] = (\lambda + \mu)^2 y , \end{align*}$$

so $\lambda = \mu $ and therefore $\alpha (l) = y$ and so $l = \alpha (y)$ (since $\alpha ^2 = 1$ ).

Proof.

1. (i) Let $a,b\in E \cap L_{\geq 1}$ ; notice that $g(a,b)=0$ . Since L is generated by $L_{\geq -1}$ , it suffices to check that both sides coincide when applied on an element $l \in L_{\geq -1}$ . We have

   (12) $$ \begin{align} \exp(a)\exp(b)(l) &= \exp(a)\big( l + [b,l] + g(b,l)b \big) \nonumber \\ &= l + [a+b, l] + [a,[b,l]] \nonumber \\ &+ g([a,b],l)a + g(b,l)[a,b] + g(a,l)a + g(b,l)b. \end{align} $$

   Using the Premet identity (3), we have

   (13) $$ \begin{align} [[a,b],[a+b,l]] &= [[a,b],[a,l]]-[[b,a],[b,l]] \nonumber \\ &= g_a([b,l])a+g_a(l)[a,b]-g_a(b)[a,l]\nonumber\\ & -g_b([a,l])b-g_b(l)[b,a]+g_b(a)[b,l]\nonumber\\ &= g([a,b],l)a + g([a,b],l)b + (g(a,l)+g(b,l))[a,b]. \end{align} $$

   Since $[a,b]\in L_2$ , the automorphism $\exp ([a,b])$ fixes all elements of $L_{\geq 1}$ . Hence, by (12) with the roles of a and b reversed, (13) and the Jacobi identity, we see that

   $$\begin{align*}\exp(a)\exp(b)(l)=\exp([a,b])\exp(b)\exp(a)(l), \end{align*}$$
   as claimed.

2. (ii) We again check that both sides coincide when applied on an element $l \in L_{\geq -1}$ . Comparing (12) with the expression for $\exp (a+b)(l)$ and using the fact that $[a,b] = 0$ and that b is a multiple of y, we see that this is equivalent with

   $$\begin{align*}[a, [y, l]] = g(a,l)y + g(y,l)a. \end{align*}$$

   Now both sides are equal to $0$ if $l \in L_{\geq 0}$ . Furthermore, if $l \in L_{-2}$ , then we can write $l = \lambda x$ with $\lambda \in k$ to see that both sides are equal to $\lambda a$ , and if $l \in L_{-1}$ , then the equality follows from Lemma 2.7.

3. (iii) By (i), we already know that $[E_+(x,y),E_+(x,y)]\leq \operatorname {\mathrm {Exp}}(y)$ . By Proposition 2.16, the element y can be written as $y = [c,d]$ for certain $c,d \in E \cap L_1$ . By (i) again, we see that $\exp ([c,d]) \in [E_+(x,y),E_+(x,y)]$ , and by rescaling, we get $\operatorname {\mathrm {Exp}}(y) \leq [E_+(x,y),E_+(x,y)]$ , as required. The second statement follows immediately from (i) since $[a,b] = 0$ if $b \in L_2$ .

4. (iv) Let $l\in L_1$ . By Theorem 3.8, we can write l as the sum of a finite number of extremal elements in $L_1$ :

   $$\begin{align*}l = e_1+\dots+e_n \quad \text{with } e_i\in E\cap L_1. \end{align*}$$

   We will prove, by induction on n, that the automorphism

   $$\begin{align*}\alpha := \exp(e_1) \dotsm \exp(e_n) \in E_+(x,y) \end{align*}$$
   satisfies (10) and (11) for the appropriate choices of maps $q_\alpha $ , $n_\alpha $ and $v_\alpha $ . (Notice that $\alpha $ depends on the choice of the elements $e_1,\dots ,e_n$ as well as on their ordering.)

   If $n=1$ , then we set

   $$\begin{align*}q_{\alpha}(m) := g_l(m)l, \quad n_{\alpha}(m) := 0, \quad v_{\alpha}(m) := 0 , \end{align*}$$
   for all $m \in L$ . By Definition 1.5, we see that (11) is satisfied. By Lemma 2.15, $q_\alpha (L_i) = 0$ unless $i = -1$ , so (10) holds as well.

   Now assume $n>1$ , let $e := e_1$ and let $l'=e_2+\dots +e_n$ , so $l = e + l'$ . Let $\alpha ' := \exp (e_2) \dotsm \exp (e_n)$ , so $\alpha := \exp (e)\alpha '$ . By the induction hypothesis, there exist maps $q_{\alpha '}, n_{\alpha '}$ and $v_{\alpha '}$ satisfying (10) and (11). We set

   (14) $$ \begin{align}e q_{\alpha}(m) &:= q_{\alpha'}(m)+g_{e}(m)e+[e,[l',m]],\end{align} $$
   (15) $$ \begin{align} n_{\alpha}(m) &:= n_{\alpha'}(m)+g_{e}([l',m])e+[e,q_{\alpha'}(m)], \end{align} $$
   (16) $$ \begin{align} v_{\alpha}(m) &:= v_{\alpha'}(m)+[e,n_{\alpha'}(m)],\qquad\qquad \end{align} $$
   for all $m\in L$ . By Lemma 2.15 and the fact that (10) holds for $\alpha '$ , we get
   $$ \begin{align*} &[e,v_{\alpha'}(m)]\in L_{\geq (1+4-2)}=L_{\geq 3}=0 \quad \text{and} \\ &g_{e}(q_{\alpha'}(m))=g_{e}(n_{\alpha'}(m))=g_{e}(v_{\alpha'}(m))=0 \end{align*} $$
   for all $m \in L$ , and by expanding
   $$\begin{align*}\alpha(m) = \exp(e)(\alpha'(m)) = \alpha'(m) + [e, \alpha'(m)] + g_e(\alpha'(m)) e, \end{align*}$$
   using the fact that (11) holds for $\alpha '$ , we now see that (11) is satisfied for $\alpha $ . By Lemma 2.15 again, we see that also (10) holds for $\alpha $ .

5. (v) Let $l \in L_1$ and let $\beta \in \operatorname {\mathrm {Aut}}(L)$ be any automorphism satisfying (10) and (11). Let $\alpha $ be the automorphism $\alpha = \exp (e_1) \dotsm \exp (e_n)$ as constructed in (iv) with respect to some choice $l = e_1 + \dots + e_n$ , and observe that by the same construction,

   $$\begin{align*}\alpha^{-1} = \exp(-e_n) \dotsm \exp(-e_1) \end{align*}$$
   then satisfies (10) and (11) for the element $-l = -e_n - \dots - e_1$ .

   Now $\alpha $ and $\beta $ coincide on $L_{\geq 1}$ . However, any element of $E_+(x,y)$ preserves the filtration $(L_{\geq i})$ , so in particular, $\alpha ^{-1}(L_{\geq 1}) = L_{\geq 1}$ . Hence, $\alpha \alpha ^{-1}$ and $\beta \alpha ^{-1}$ coincide on $L_{\geq 1}$ , so $\gamma := \beta \alpha ^{-1}$ fixes $L_{\geq 1}$ .

   Moreover, by (11) for $\beta $ and $\alpha ^{-1}$ , we have

   $$\begin{align*}\gamma(x) = \beta \alpha^{-1}(x) \in \beta(x - [l,x] + L_{\geq 0}) = x + L_{\geq 0}, \end{align*}$$
   so $\gamma (x)$ is an extremal element with trivial $L_{-1}$ -component. By Lemma 3.10, $\gamma (x) = \exp (z)(x)$ for some unique $z = \lambda y \in L_2$ . Notice that also $\exp (z)$ fixes $L_{\geq 1}$ . Since L is generated by $L_1$ and x, we conclude that $\gamma = \exp (z)$ .

6. (vi) Let $\alpha \in E_+(x,y)$ . We claim that $\alpha $ can be written as a product of elements of the form $\exp (a)$ with $a \in E \cap L_1$ . Indeed, by definition, it is the product of elements $\exp (l)$ with $l \in E \cap L_{\geq 1}$ , but each such l can be written as $l = a+b$ with $a \in E \cap L_1$ and $b \in L_2$ . By (ii), $\exp (l) = \exp (a) \exp (b)$ , and by the proof of (iii), $\exp (b) = [\exp (c), \exp (d)]$ for certain $c,d \in E \cap L_1$ ; this proves our claim. Hence, $\alpha = \exp (e_1) \dotsm \exp (e_n)$ for certain $e_1,\dots ,e_n \in E \cap L_1$ . Now let $m := e_1 + \dots + e_n$ ; then by (iv), $\alpha $ is an m-exponential automorphism. This shows that any automorphism in $E_+(x,y)$ is m-exponential for some $m \in L_1$ ; the other inclusion ‘ $\supseteq $ ’ holds by (v).

   If $\alpha $ is any l-exponential automorphism, then in particular, $\alpha \in E_+(x,y)$ , so by the previous paragraph, $\alpha $ is m-exponential for $m = e_1 + \dots + e_n$ . Then by (11), the $L_{-1}$ -component of $\alpha (x)$ is equal to both $[m, x]$ and $[l, x]$ , hence $m = l$ , and we have shown the required decomposition.

7. (vii) By (vi), we can write $\alpha = \exp (e_1) \dotsm \exp (e_n)$ and $l = e_1 + \dots + e_n$ , where $e_i \in E \cap L_1$ . Now let $\beta := \exp (\lambda e_1) \dotsm \exp (\lambda e_n) \in E_+(x,y)$ , with corresponding element $\lambda l = \lambda e_1 + \dots + \lambda e_n \in L_1$ and corresponding maps $q_\beta , n_\beta , v_\beta \colon L \to L$ . Observe now that if we replace each $e_i$ by $\lambda e_i$ in the recursive formulas (14) to (16) in the proof of (iv), then this yields

   $$\begin{align*}q_\beta(m) = \lambda^2 q_\alpha(m), \quad n_\beta(m) = \lambda^3 n_\alpha(m), \quad v_\beta(m) = \lambda^4 v_\alpha(m), \end{align*}$$
   for all $m \in L$ . We conclude that $\beta = \alpha _\lambda $ and hence $\alpha _\lambda \in E_+(x,y)$ , which is then a $(\lambda l)$ -exponential automorphism.

8. (viii) By (vi), $\alpha = \exp (e_1) \dotsm \exp (e_n)$ and $\beta = \exp (f_1) \dotsm \exp (f_m)$ with $e_i, f_i \in E \cap L_1$ and with $l = e_1 + \dots + e_n$ and $l' = f_1 + \dots + f_m$ . Hence, $\alpha \beta = \exp (e_1) \dotsm \exp (e_n) \exp (f_1) \dotsm \exp (f_m)$ , so by (iv), $\alpha \beta $ is an $(l + l')$ -exponential automorphism.

9. (ix) Assume $\operatorname {\mathrm {char}}(k)\neq 2$ . We first show the existence of such an automorphism. As in the proof of part (iv), we write $l = e_1 + \dots e_n$ with $e := e_1$ and $l' := e_2 + \dots + e_n$ , and we proceed by induction on n.

   If $n=1$ , then we choose $\alpha = \exp (e)$ , which has $q_{\alpha }(m) = g_l(m)l = \tfrac {1}{2} [l,[l,m]]$ , as required. Now assume $n>1$ . By the induction hypothesis, there exists an $l'$ -exponential automorphism $\alpha '$ with $q_{\alpha '}(m)=\tfrac {1}{2} [l',[l',m]]$ for all $m \in L$ , and we first set $\beta := \exp (e) \alpha '$ , so that we can invoke the formulas (14) to (16). In particular, we have

   $$\begin{align*}q_{\beta}(m)=\tfrac{1}{2} [l',[l',m]]+\tfrac{1}{2} [e,[e,m]]+[e,[l',m]] \end{align*}$$
   for all $m \in L$ . Now set $\alpha := \exp (-\tfrac {1}{2} [e,l'])\beta $ , and notice that $[e, l'] \in L_2$ , so $\alpha $ is again an l-exponential automorphism. We get, using the Jacobi identity,
   $$ \begin{align*} q_{\alpha}(m) &=\tfrac{1}{2} [l',[l',m]]+\tfrac{1}{2} [e,[e,m]] + [e,[l',m]] +\tfrac{1}{2} [m,[e,l']] \\ &=\tfrac{1}{2} [l'+e,[l'+e,m]]=\tfrac{1}{2} [l,[l,m]], \end{align*} $$
   so $\alpha $ satisfies the required assumptions, proving the existence.

   To prove uniqueness, assume that $\alpha $ and $\beta $ are two l-exponential automorphisms with $q_\alpha = q_\beta $ ; then in particular, $q_\alpha (x) = q_\beta (x)$ . By (v), we have $\beta = \exp (z) \alpha $ for some $z = \lambda y \in L_2$ . By expanding $\exp (z)$ – see also Remark 3.14 below – we get $q_\beta (x) = q_\alpha (x) + [z,x]$ , but $[z,x] = \lambda [y,x]$ , which is $0$ only for $\lambda = 0$ , and hence $\beta = \alpha $ .

   Finally, let $\gamma := e_+(-l)e_+(l)$ . By (viii), $\gamma $ is a $0$ -exponential automorphism, so by (v), we have $\gamma = \exp (z)$ for some $z = \lambda y \in L_2$ again. However,

   $$ \begin{align*} q_\gamma(x) &= q_{e_+(-l)}(x) + q_{e_+(l)}(x) + [-l, [l, x]] \\ &= \tfrac{1}{2}[-l, [-l, x]] + \tfrac{1}{2}[l, [l, x]] + [-l, [l, x]] = 0 , \end{align*} $$
   so $\lambda = 0$ , and hence, $\gamma = \operatorname {\mathrm {id}}$ .

Proof. Consider $l\in L_1$ arbitrary. To simplify the notation, we will identify each $m \in L$ with $m \otimes 1 \in L \otimes k'$ .

By Theorem 3.13(v), we find an l-exponential automorphism $\alpha $ of $L\otimes k'$ . Since $[x,y]\neq 0$ , we can find a basis $\mathcal B$ for $L_0$ containing $[x,y]$ . Now, by (10), we have $q_{\alpha }(x)\in L_0 \otimes k'$ , so we can write

$$\begin{align*}q_{\alpha}(x) = [x,y]\otimes \lambda+b_1\otimes\lambda_1+\dots +b_n\otimes\lambda_n \end{align*}$$

with $b_1,\dots ,b_n\in \mathcal B\setminus \{[x,y]\}$ and $\lambda ,\lambda _1,\dots ,\lambda _n\in k'$ .

Let $\beta := \exp (\lambda y)\alpha $ and notice that $\beta $ is again an l-exponential automorphism of $L \otimes k'$ . By Remark 3.14, we have $q_{\beta }=q_{\alpha }+\operatorname {\mathrm {ad}}_{\lambda y}$ , and hence,

$$\begin{align*}q_{\beta}(x) = q_\alpha(x) + [y,x] \otimes \lambda =b_1\otimes\lambda_1+\dots+ b_n\otimes\lambda_n. \end{align*}$$

Now each $\sigma \in \operatorname {\mathrm {Gal}}(k'/k)$ acts on the Lie algebra $L \otimes k'$ by sending each $m \otimes \lambda $ to $m\otimes \lambda ^\sigma $ . Since $\sigma (l) = l$ , we have, for each $m'\in L\otimes k'$ , that

$$ \begin{align*} (\sigma\circ \beta\circ \sigma^{-1})(m') &= m' + [l,m'] + (\sigma \circ q_{\beta} \circ \sigma^{-1})(m') \\ &\quad+ (\sigma \circ n_{\beta_{l}} \circ \sigma^{-1})(m') + (\sigma \circ v_{\beta_{l}} \circ \sigma^{-1})(m') , \end{align*} $$

so $\gamma := \sigma \circ \beta \circ \sigma ^{-1}$ is again an l-exponential automorphism of $L \otimes k'$ , with corresponding maps $q_{\gamma }=\sigma \circ q_{\beta }\circ \sigma ^{-1}$ , $n_{\gamma }=\sigma \circ n_{\beta }\circ \sigma ^{-1}$ and $v_{\gamma }=\sigma \circ v_{\beta }\circ \sigma ^{-1}$ . Hence, Theorem 3.13(v) implies that there exists $\mu \in k'$ such that $\gamma =\exp (\mu y)\beta $ , and by Remark 3.14 again, we have $q_\gamma = q_\beta + \operatorname {\mathrm {ad}}_{\mu y}$ . In particular,

$$ \begin{align*} b_1 \otimes \lambda_1^\sigma + \dots + b_n \otimes \lambda_n^\sigma &= q_{\gamma}(x) = [y,x] \otimes \mu +q_{\beta}(x) \\ &= [y,x] \otimes \mu + b_1 \otimes \lambda_1 + \dots + b_m \otimes \lambda_n. \end{align*} $$

Since $b_1,\dots , b_n$ and $[x,y]$ are linearly independent by construction, this implies $\mu =0$ , and thus, $\gamma = \beta $ . Since $\sigma \in \operatorname {\mathrm {Gal}}(k'/k)$ was arbitrary, we get $\beta =\sigma \circ \beta \circ \sigma ^{-1}$ for all $\sigma \in \operatorname {\mathrm {Gal}}(k'/k)$ , and therefore,

$$\begin{align*}q_{\beta}=\sigma\circ q_{\beta}\circ \sigma^{-1}, \quad n_{\beta}=\sigma\circ n_{\beta}\circ \sigma^{-1}, \quad v_{\beta}=\sigma\circ v_{\beta}\circ \sigma^{-1}, \end{align*}$$

for all $\sigma \in \operatorname {\mathrm {Gal}}(k'/k)$ . Since $k'/k$ is a Galois extension and thus $\operatorname {\mathrm {Fix}}(\operatorname {\mathrm {Gal}}(k'/k))=k$ , this implies that the maps $q_\beta $ , $n_\beta $ and $v_\beta $ stabilize L, proving that the restriction of $\beta $ to L is an automorphism of L. Since $\beta $ is an l-exponential automorphism, we are done.

Assumption 3.18. For the rest of this section, we assume that L is a simple Lie algebra over k with extremal elements x and y with $g(x,y)=1$ and that $k'/k$ is a Galois extension such that $L\otimes k'$ is a simple Lie algebra generated by its pure extremal elements, with $\mathcal F(L\otimes k')\neq \emptyset $ and $|k'|\geq 3$ .

Proof. Notice that by Theorem 3.13(viii), both $\alpha \beta $ and $\exp ([l,l'])\beta \alpha $ are $(l+l')$ -exponential automorphisms. By the uniqueness statement of Remark 3.14, it remains to show that $q_{\alpha \beta }(x) = q_{\exp ([l,l'])\beta \alpha }(x)$ . By Remark 3.14(ii), we have