The authors gave an example showing an error in [Reference Lee and Lee2, Lemma 3.3], and below offer at least a partial correction for that error under the unimodularity assumption. This makes all of the remaining results in [Reference Lee and Lee2] valid.
Consider the three-dimensional solvable non-unimodular Lie algebra $\mathfrak{S}$:
This Lie algebra has a faithful matrix representation as follows:
We can choose an ordered (linear) basis for $\mathfrak{S}$:
They satisfy $[\mathbf{b}_{1},\mathbf{b}_{2}]=\mathbf{0}$, $[\mathbf{b}_{3},\mathbf{b}_{1}]=\mathbf{b}_{1}$ and $[\mathbf{b}_{3},\mathbf{b}_{2}]=\mathbf{b}_{2}$. The connected and simply connected solvable Lie group $S$ associated with the Lie algebra $\mathfrak{S}$ is
Let $g=((x,y),t)$ denote an element of $S$. Because $\text{Ad}(g):\mathfrak{S}\rightarrow \mathfrak{S}$ is given by $\text{Ad}(g)(A)=gAg^{-1}$ for $A\in \mathfrak{S}$, a simple computation shows that the adjoint of $g$ is given by
Let ${\it\varphi}$ be a Lie algebra homomorphism on $\mathfrak{S}$. Since $[\mathfrak{S},\mathfrak{S}]$ is generated by $\mathbf{e}_{1}$ and $\mathbf{e}_{2}$, we have
for some $m_{ij},p,q,m\in \mathbb{R}$. Since ${\it\varphi}$ preserves the bracket operations $[\mathbf{b}_{3},\mathbf{b}_{1}]=\mathbf{b}_{1}$ and $[\mathbf{b}_{3},\mathbf{b}_{2}]=\mathbf{b}_{2}$, it follows easily that
Therefore, with respect to the basis $\{\mathbf{b}_{1},\mathbf{b}_{2},\mathbf{b}_{3}\}$ of $\mathfrak{S}$, ${\it\varphi}$ is one of the following:
Now we can easily check that
This example shows that [Reference Lee and Lee2, Lemma 3.3] is not true in general. We remark also that $S$ is not unimodular, and hence, as can be expected, $\det (\text{Ad}(g))=e^{2t}\neq 1$ for all $t\neq 0$. We prove, however, that the lemma is true under the unimodularity assumption of the connected Lie group. That is, the following theorem.
Theorem 1. Let $S$ be a connected and simply connected solvable Lie group, and let $D:S\rightarrow S$ be a Lie group homomorphism. If $S$ is unimodular, then for any $x\in S$,
Remark 2. It is known that if a Lie group admits a lattice (discrete cocompact subgroup), then it is unimodular. Consequently, the remaining results of [Reference Lee and Lee2] are valid.
Lemma 3. Let $S$ be a connected and simply connected solvable Lie group, and let $D:S\rightarrow S$ be a Lie group homomorphism. Then, for any $x\in S$, $I-D_{\ast }$ is an isomorphism if and only if $I-\text{Ad}(x)D_{\ast }$ is an isomorphism.
Proof. Because $I-\text{Ad}(x^{-1})\text{Ad}(x)D_{\ast }=I-D_{\ast }$, it suffices to show the only if.
Let $G=[S,S]$; then $G$ is nilpotent, and $S/G\cong \mathbb{R}^{k}$ for some $k$. Then we have the following commutative diagram:
This induces the following commutative diagram:
For $x\in S$, we denote by ${\it\tau}_{x}$ the inner automorphism on $S$ whose differential is $\text{Ad}(x)$. This induces an automorphism on $G$, and we still denote it by ${\it\tau}_{x}$ and its differential is $\text{Ad}^{\prime }(x)$. Then we can express $I-D_{\ast }$ and $I-\text{Ad}(x)D_{\ast }$ as
with respect to some linear basis for $\mathfrak{S}$.
Assume that $I-\bar{D}_{\ast }$ is an isomorphism. We claim that $I-D_{\ast }^{\prime }$ is an isomorphism if and only if $I-\text{Ad}^{\prime }(x)D_{\ast }^{\prime }$ is an isomorphism.
Since $I-\bar{D}$ is an isomorphism on $\mathbb{R}^{k}$, $\text{fix}(\bar{D})=\ker (I-\bar{D})$ is a trivial group. For any $x\in S$, we consider the exact sequence of the Reidemeister sets
$\hat{p}^{x}$ is surjective, and $(\hat{p}^{x})^{-1}([\bar{1}])=\text{im}(\hat{i} ^{x})$. If $\hat{i} ^{x}([g_{1}])=\hat{i} ^{x}([g_{2}])$ for some $g_{1},g_{2}\in G$, then by definition there is $y\in S$ such that $g_{2}=yg_{1}({\it\tau}_{x}D(y))^{-1}$. The image in $S/G$ is then $\bar{g}_{2}=\bar{y}\bar{g}_{1}\bar{D}(\bar{y})^{-1}$, which yields that $\bar{y}\in \text{fix}(\bar{D})=\{\bar{1}\}$, and so $y\in G$. This shows that $\hat{i} ^{x}$ is injective for all $x\in S$. Because there is a bijection between the Reidemeister sets ${\mathcal{R}}[D]$ and ${\mathcal{R}}[{\it\tau}_{x}D]$ given by $[g]\mapsto [gx^{-1}]$, it follows that $R(D^{\prime })=R({\it\tau}_{x}D^{\prime })$. On the other hand, by [Reference Dekimpe and Penninckx1, Lemma 3.4], since $I-\bar{D}_{\ast }$ is an isomorphism, $R(\bar{D})<\infty$, and
This proves our claim.
Now assume that $I-D_{\ast }$ is an isomorphism. Then it follows that $I-\bar{D}_{\ast }$ and $I-D_{\ast }^{\prime }$ are isomorphisms. By the above claim, $I-\text{Ad}^{\prime }(x)D_{\ast }^{\prime }$, and hence $I-\text{Ad}(x)D_{\ast }$ are isomorphisms.◻
Proof of Theorem 1.
If $S$ is Abelian, then $\text{Ad}(x)$ is the identity and hence there is nothing to prove. We may assume that $S$ is non-Abelian. Further, by Lemma 3, we may assume that $I-D_{\ast }$ is an isomorphism. Hence, $I-\bar{D}_{\ast }$ and $I-\text{Ad}(x)D_{\ast }$ are isomorphisms for all $x\in S$.
Denote $G=[S,S]$ and ${\rm\Lambda}_{0}=S/G$. Then $G$ is nilpotent, and ${\rm\Lambda}_{0}\cong \mathbb{R}^{k_{0}}$ for some $k_{0}>0$. Consider the lower central series of $G$:
where ${\it\delta}_{i+1}(G)=[G,{\it\delta}_{i}(G)]$. Let ${\rm\Lambda}_{i}={\it\delta}_{i}(G)/{\it\delta}_{i+1}(G)$. Then ${\rm\Lambda}_{i}\cong \mathbb{R}^{k_{i}}$ for some $k_{i}>0$. For each $x\in S$, the conjugation ${\it\tau}_{x}$ by $x$ induces an automorphism on $G$. Since each ${\it\delta}_{i}(G)$ is a characteristic subgroup of $G$, ${\it\tau}_{x}\in \text{Aut}(G)$ restricts to an automorphism on ${\it\delta}_{i}(G)$, and hence on ${\rm\Lambda}_{i}$. Now, if $x\in G$, then we have observed that the induced action on ${\rm\Lambda}_{i}$ is trivial. Consequently, there is a well-defined action of ${\rm\Lambda}_{0}=S/G$ on ${\rm\Lambda}_{i}$. Hence, there is a well-defined action of ${\rm\Lambda}_{0}$ on ${\rm\Lambda}_{i}$. This action can be viewed as a homomorphism ${\it\mu}_{i}:{\rm\Lambda}_{0}\rightarrow \text{Aut}({\rm\Lambda}_{i})$. Note that ${\it\mu}_{0}$ is trivial. Moreover, for any $x\in S$ denoting its image under $S\rightarrow {\rm\Lambda}_{0}$ by $\bar{x}$, the differential of conjugation ${\it\tau}_{x}$ by $x$ can be expressed as a matrix of the form
by choosing a suitable basis of the Lie algebra $\mathfrak{S}$ of $S$.
The homomorphism $D:S\rightarrow S$ induces homomorphisms $D_{i}:{\it\delta}_{i}(G)\rightarrow {\it\delta}_{i}(G)$ and hence homomorphisms $\bar{D}_{i}:{\rm\Lambda}_{i}\rightarrow {\rm\Lambda}_{i}$, so that the following diagram is commutative:
Hence, the differential of $D$ can be expressed as a matrix of the form
with respect to the same basis for $\mathfrak{S}$ chosen as above.
Furthermore, the above commutative diagram produces the following identities:
Let $x\in S$ with $\bar{x}\in {\rm\Lambda}_{0}=\mathbb{R}^{k_{0}}$. Since $I-\bar{D}_{0}:\mathbb{R}^{k_{0}}\rightarrow \mathbb{R}^{k_{0}}$ is invertible, we can choose $\bar{y}\in {\rm\Lambda}_{0}$ so that $(I-\bar{D}_{0})(\bar{y})=\bar{x}$. Now, using the above identities, we observe that
Consequently, we have
This completes the proof of our theorem. ◻