Let $G$ be a commutative Lie group and let $T \in \operatorname{Aut}(G)$ be an automorphism of the Lie group. We put $x * y=T x+(1-T) y$ for $x, y \in G$, and then $(G, T)$ is a smooth quandle. We call $(G, T)$ an Alexander quandle and denote it by $G_{T}$ , if $T$ is a multiplication by a scalar $a$, we simply denote it by $G_{a} .$ I need to show that $G_{T}$ is transitive if and only if the endomorphism $1-T$ on $G$ is surjective. I get stuck in this question.
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We say that $X$ is transitive if the action of $\operatorname{Inn}(X)$ on $X$ is transitive, with $\operatorname{Inn}(X)$ is the sub group of the $Aut(X)$ generated by $s_y$ $(y \in X)$ and called the inner antonorphisma group.
Return to my question, we take the action of $\operatorname{Inn}(X)$ on $G$ defince by
$$
\begin{aligned}
Inn(G) \times G & \longrightarrow G \\
(y, x)&\longrightarrow y* x=s_y(x)=T{y}+(1-T) x
\end{aligned}
$$
such that
$$\begin{aligned} \text { $s_y$ : }\;\;\; G & \longrightarrow G \\ x & \longrightarrow s_y(x)=y * x \end{aligned}$$
if $G$ is transitive then $H$ is transitive, for all $x, y$ for there is $z$ on $\operatorname{Inn}(X)$ such that $$x=y * z=s_{y}(z)=T{y}+(1-T) z$$ then $$(1-T)z=x-T y $$
then $(1-T)$ is surjective.
Conversaly: we have $T \in Aut(G)$ and $\forall x, y \in G$, $x-Ty \;\in G$, and $1 -T$ surjective then $\exists z \in G$ sush that $$ (1-T) z=x-T y $$ then $$ x=y * z . \quad \forall x, y \in G $$ Hence G is transitive.
Imade Allal
- 63
(Q1) For any $x \in X, x * x=x .$
(Q2) For any $y \in X$, the map $s_{y}: X \ni x \mapsto x * y \in X$ is a bijection.
(Q3) For any $x, y, z \in X,(x * y) * z=(x * z) (y z)$.
– Imade Allal Feb 26 '22 at 10:34