1 Correction to: The Journal of Geometric Analysis (2017) 27:56–64 https://doi.org/10.1007/s12220-015-9672-9

A few years ago Profs. Yurii Nikonorov and Ming Xu found a gap [1] in my paper [2] on homogeneity and bounded isometries. In lines 2 and 3 of the third paragraph of the proof of [2, Theorem 2.5] it was only shown that Ad(sk) is bounded, and from that it should have been shown that Ad(s) is bounded. Now Prof. Nikonorov modified my original proof, completing the argument, and agreed to its publication as a correction. See below. The statement of the theorem is unchanged; the modified argument uses a result [3] of Moskowitz.

Theorem

([2, Theorem 2.5]). Let (M, d) be a metric space on which an exponential solvable Lie group S acts effectively and transitively by isometries. Let \(G = I(M,d)\). Then G is a Lie group, any isotropy subgroup K is compact, and \(G = SK\). If \(g\in G\) is a bounded isometry then g is a central element in S.

Proof

(Yurii Nikonorov [4]). M carries a differentiable manifold structure for which \(s \mapsto s(x_0)\) is a diffeomorphism \(S\cong M\). In the compact–open topology \(G = I(M, d)\) is locally compact and its its action on M is proper [5]. In particular, if \(x_0 \in M\) then the isotropy subgroup \(K = \{k\in G\, \mid \, k(x_0) = x_0\}\) is compact. Further [6, Corollary in §6.3] G is a Lie group. Now S and K are closed subgroups, \(G = SK\), and \(M = G/K\). The action of G on \(M = SK/K = S\) is \((sk) : s' \mapsto s\cdot k s' k^{-1}\). We now assume that \(x_0=1\in S\).

Express \(g = sk \in G\) with \(s\in S\) and \(k\in K\). Suppose that g is a bounded isometry of (M, d). So there is a compact set \(C\subset S\) such that \(\tau (g)(s') s'^{-1}= s[(ks'k^{-1})s'^{-1}]\in C\) for every \(s'\in S\). Then \((ks'k^{-1})s'^{-1} \in s^{-1}C\) so the automorphism \(a: S \rightarrow S\), \(a(s')=ks'k^{-1}\) is an automorphism of bounded displacement. By Corollary 1.3 in [3], we see that a is trivial. Since the isotropy representation is faithful, it follows that \(k=1\) and \(g=s\in S\).

Express \(g = s = \exp (\xi )\) where \(\xi \in \mathfrak {s}:={\text {Lie}}(S)\). Decompose \(\mathfrak {s}\) as a vector space direct sum \(\mathfrak {n}+\mathfrak {a}\), where \(\mathfrak {n}\) is the nilradical of \(\mathfrak {s}\). In a basis respecting that direct sum, \({{\,\mathrm{ad}\,}}(\xi )|_{\mathfrak {s}}\) has matrix of the form \(\left( {\begin{matrix} 0 &{} h \\ 0 &{} 0 \\ \end{matrix}} \right) \). Let N denote the unipotent radical of S, so \(\mathfrak {n}:={\text {Lie}}(N)\). Then the (1, 1) and (2, 1) blocks vanish because g centralizes N by [7, Théorème 1], and the (2, 2) block vanishes because \([\mathfrak {s},\mathfrak {s}]\subset \mathfrak {n}\). Note \(\left( {\begin{matrix} 0 &{} h \\ 0 &{} 0 \\ \end{matrix}} \right) ^2 = 0\), so \({{\,\mathrm{Ad}\,}}(g)|_{\mathfrak {s}}- I = {{\,\mathrm{ad}\,}}(\xi )|_{\mathfrak {s}}\). It cannot have relatively compact image unless \(h = 0\). Thus \({{\,\mathrm{Ad}\,}}(g)|_{\mathfrak {s}}= I\), and \(g\in (Z_G(S) \cap S)=Z_S\) as asserted. \(\square \)