Given a $\mathbb{k}$-vector space $V$ and an endomorphism $f$ of $V$, we can regard $V$ an an abelian Lie algebra and construct the semidirect product $V \rtimes_f \mathbb{k}$. It is given by the vector space $V \oplus \mathbb{k}$ together with the Lie bracket $$ [ (v, \lambda) , (w, \mu) ] = ( \lambda f(w) - \mu f(v), 0 ) \,. $$
Question. Under what conditions is $V \rtimes_f \mathbb{k}$ isomorphic to $W \rtimes_g \mathbb{k}$?
The question is motivated from the following example: Let $f(\tau)$ be the endomorphism of $\mathbb{k}^2$ given by the matrix $$ f(\tau) \equiv \begin{pmatrix} 1 & \\ & \tau \end{pmatrix} \,. $$ The resulting semidirect products $\mathfrak{g}_\tau := \mathbb{k}^2 \rtimes_{f(\tau)} \mathbb{k}$ with $\tau \in \mathbb{k}$ are a standard example for infinitely many non-isomorphic, three-dimensional Lie algebras: the Lie algebras $\mathfrak{g}_\tau$ and $\mathfrak{g}_\upsilon$ are isomorphic if and only if $\tau = \upsilon$ or $\tau \upsilon = 1$.
Let us abbreviate $V \rtimes_f \mathbb{k}$ as $\mathfrak{g}_f$. So far, I’ve found out the following:
- Both $V$ and $W$ need to have the same dimension, so one may assume for simplicity that $W = V$.
- $\mathfrak{g}_f$ is abelian if and only if $f = 0$. This entails that $\mathfrak{g}_0$ is isomorphic to no other $\mathfrak{g}_f$.
- The center of $\mathfrak{g}_f$ is given by $\ker(f) \oplus 0$, and the derived Lie algebra of $\mathfrak{g}_f$ is given by $\operatorname{im}(f) \oplus 0$. We hence find that $f$ and $g$ need to have the same rank, the same nullity, and the same corank.
- Suppose that the two endomorphisms $f$ and $g$ of $V$ are similar, and let $\varphi$ be a vector space automorphism of $V$ with $g = \varphi f \varphi^{-1}$. We have an isomorphism of Lie algebras $$ \mathfrak{g}_f \to \mathfrak{g}_g \,, \quad (v, \lambda) \mapsto (\varphi(v), \lambda) \,. $$
- For any nonzero scalar $\alpha$ in $\mathbb{k}$ we have an isomorphism of Lie algebras $$ \mathfrak{g}_{\alpha f} \to \mathfrak{g}_f \,, \quad (v, \lambda) \mapsto (v, \alpha \lambda) \,. $$
- For every linear map $\beta$ from $\mathbb{k}$ to $V$ we have an automorphism of Lie algebras $$ \mathfrak{g}_f \to \mathfrak{g}_f \,, \quad (v, \lambda) \mapsto (v + \beta(\lambda), \lambda) \,. $$
If $f$ is surjective, then the above observations suffice to answer the question:
It follows from the surjectivity of $f$ that $f$ has corank $0$, which means that $g$ has corank $0$, whence $g$ is surjective, too. It follows that $[\mathfrak{g}_g, \mathfrak{g}_g]$ has codimension $1$ in $\mathfrak{g}_g$, which means that $g$ is also surjective.
Every isomorphism of Lie algebras $\theta$ from $\mathfrak{g}_f$ to $\mathfrak{g}_g$ must therefore map $V \oplus 0 = [\mathfrak{g}_f, \mathfrak{g}_f]$ to $V \oplus 0 = [\mathfrak{g}_g, \mathfrak{g}_g]$. With respects to the decompositions of vector spaces $\mathfrak{g}_f = V \oplus \mathbb{k}$ and $\mathfrak{g}_g = V \oplus \mathbb{k}$ we can therefore express $\theta$ as an upper triangular matrix $$ \theta = \begin{pmatrix} \varphi & \beta \\ 0 & \alpha \end{pmatrix} $$ where $\varphi$ is a vector space endomorphism of $V$, $\alpha$ is a scalar in $\mathbb{k}$ and $\beta$ is a linear map from $\mathbb{k}$ to $V$. That $\theta$ is a homomophism of Lie algebras means that $\varphi f = \alpha g \varphi$, and the invertibility of $\theta$ is equivalent to the invertibility of both $\varphi$ and $\alpha$. We hence find that $f$ is similar to $\alpha g$, which is a nonzero scalar multiple of $g$.
Together with the above observations this shows that if $f$ is surjective, then $\mathfrak{g}_f$ is isomorphic to $\mathfrak{g}_g$ if and only if $f$ is similar to a nonzero scalar multiple of $g$.