Split extension with group action

Question

I don't understand a step of the definition of the Sol geometry. It says that Sol is the split extension of $\mathbb R$ by $\mathbb R^2$ with the group action of $\mathbb R$ on $\mathbb R^2$ defined as $(t,(x,y))\mapsto (e^t x, e^{-t}y)$. But the split extension is defined as the short exact sequence $0\to \mathbb R^2 \to Sol \to \mathbb R \to 0$, which implies that there is a isomorphism $\mathbb R \cong Sol/\mathbb{R}^2$. Why the group action is defined then between these spaces?

Answer 1

There is a close relation between split extensions of groups and groups acting on groups.

First, for any extension of groups $1 \mapsto K \mapsto G \mapsto Q \mapsto 1$ and for any splitting $\sigma : Q \to G$, there is an associated action of $Q$ on $K$, which to each $q \in Q$ associates an automorphism of $K$ given by the following formula:

• For each $q \in Q$ and $k \in K$ we have $q \cdot k = \sigma(q)^{-1} \, k \, \sigma(q)$

The idea here is that $K$ is identified with a normal subgroup of $Q$ using the injection $K \hookrightarrow Q$, and so $\sigma(q)^{-1} \, k \, \sigma(q)$ is guaranteed to be an element of $K$. One can check that for each $q$ this defines an isomorphism of $K$, and one can check the "action equation" $q_1 \cdot (q_2 \cdot k) = (q_1 q_2) \cdot k$.

In the other direction, for any groups $Q$ and $K$ and any action of $Q$ on $K$ there is an associated group called the semidirect product. As a set, the semidirect product is just the product set $K \times Q$. The group action on this set is given by the formula $$(k_1,q_1) * (k_2,q_2) = (k_1 (q_1 \cdot k_2), q_1 q_2) $$ One can check that this defines a group structure on the product set, and one can check that the inclusion $K \to K \times Q$ given by $k \mapsto (k,1)$ and the projection $K \times Q \mapsto Q$ given by $(k,q) \mapsto q$ define a split extension.

These two operations (from split extension to group action, and from group action to split extension) are inverses of each other, in some appropriate sense.

Added: To answer an additional question in the comments, the algebraic structures of $K$ and $Q$ do not determine the group operation on the extension group. But with one additional piece of information, namely a homomorphism $Q \mapsto \text{Aut}(K)$, the extension group structure is determined.

In the first part of my answer where the extension is given to you, the homomorphism $Q \mapsto \text{Aut}(K)$ is given to you by the choice of a section $\sigma : Q \to G$: it is the homomorphism that takes $q \in Q$ to the element of $\text{Aut}(K)$ given by the formula $k \mapsto \sigma(q)^{-1} k \sigma(q)$.

In the second part of my answer, the homomorphism $Q \mapsto \text{Aut}(K)$ was given in the phrase "for any action of $Q$ on $K$": an action of $Q$ on $K$ is exactly the same thing as a homomorphism $Q \mapsto \text{Aut}(K)$.

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Up to this point, I've been working with arbitrary groups, and I've just used concatenation to denote the group operation, i.e. the group operation applied to $q_1$ and $q_2$ yields $q_1 q_2$.

So, on to Sol, and now the group operations on $\mathbb R^2$ and $\mathbb R$ are addition. As a set, Sol is just the product $\mathbb R^2 \times \mathbb R$. But the group action is given by the formula $$\bigl((x_1,y_1),t_1\bigr) * \bigl((x_2,y_2),t_2\bigr) = \bigl((x_1 + e^{t_1} x_2,y_1 + e^{-t_1} y_2),t_1 + t_2\bigr) $$