January 01, 1970
Let \(f:M\longrightarrowM\) be a diffeomorphism on a compact connected smooth manifold of dimension at least \(5\). It is known that the Nielsen number \(N(f)\) of \(f\) vanishes if and only if \(f\) is isotopic to a fixed point free map. For any \(n\ge 5\), there exists a compact \(n\)-dimensional nilmanifold \(M\) such that every self homeomorphism on \(M\) is isotopic to a fixed point free map. The proof depends on the fact that nilmanifolds are of Jiang-type and the fundamental group \(\pi_1(M)\) has the property \(R_{\infty}\) for certain nilmanifolds \(M\). In this paper we construct the first known examples (an infinite family) of non-aspherical closed manifolds whose fundamental groups have property \(R_{\infty}\) and that are also of Jiang-type. In particular, every self homeomorphism on such a manifold is isotopic to be fixed point free. The main objective of this work is to construct non-aspherical manifolds whose fundamental groups have property \(R_{\infty}\) and that are also of Jiang-type.
A space \(X\) is said to have the fixed point property if every selfmap \(f:X\longrightarrowX\) has a fixed point, i.e., \(f(x)=x\) for some \(x\in X\). For instance, if the Lefschetz number \(L(f)\) is nonzero for every \(f\), then it follows from the celebrated Lefschetz fixed point theorem that \(X\) has the fixed point property. Conversely, \(L(f)=0\) does not guarantee that \(f\) can be deformed to be fixed point free. For compact manifolds of dimension at least three, a classical theorem of Wecken asserts that the vanishing of the more subtle homotopy invariant, namely the Nielsen number \(N(f)\), implies that \(f\) is homotopic to a fixed point free map. From the point of view of dynamics, one is interested in knowing if every self homeomorphism \(f: X\longrightarrowX\) can be deformed to be fixed point free.
In [1], it was shown that for every \(n\ge 4\), there exists an \(n\)-dimensional compact nilmanifold \(X\) such that every self homeomorphism is homotopic to a fixed point free homeomorphism. If \(n\ge 5\) then the homotopy can be made to be isotopy. This result is based upon two important concepts, namely the property \(R_{\infty}\) and the concept of a Jiang-type space. These manifolds were the first examples of an infinite family of Jiang-type spaces whose fundamental groups satisfy the so-called property \(R_{\infty}\).
Let \(\varphi: G\longrightarrowG\) be a group endomorphism. The Reidemeister number \(R(\varphi)\) is the number of \(\varphi\)-twisted conjugacy classes of elements of \(G\) under the equivalence relation \(\alpha \sim \sigma \alpha \varphi(\sigma)^{-1}\). In generalizing the classical Burnside-Frobenius theorem, A. Fel’shtyn and R. Hill [2] conjectured that \(R(\varphi)=\infty\) if \(\varphi\) is injective and \(G\) has exponential growth. This conjecture has been proven to be false in general (see [3]). The study of \(R(\varphi)\) in fact originated in the Nielsen fixed point theory where twisted conjugation arises naturally. A group \(G\) is said to have property \(R_{\infty}\) [4] if \(R(\varphi)=\infty\) for every automorphism \(\varphi\in {\rm Aut}(G)\). For a selfmap \(f:X\longrightarrowX\), the Reidemeister number \(R(f)\) of \(f\) is simply \(R(\varphi)\) where \(\varphi\) is the induced homomorphism on the fundamental group. For manifolds of dimension at least \(3\), the Nielsen number \(N(f)\) is a sharp lower bound for the minimal number of fixed points in the homotopy class of \(f\) and \(N(f)\le R(f)\).
The computation of \(N(f)\) is notoriously difficult. In 1963, Jiang introduced the Jiang subgroup \(J(X) \subset \pi_1(X)\) (also known as the first Gottlieb group) of the fundamental group. When \(J(X)=\pi_1(X)\) then for every selfmap \(f:X\longrightarrowX\), (i) \(N(f)=0\) if \(L(f)=0\) or (ii) \(N(f)=R(f)\) if \(L(f)\ne 0\). We say that a space is of Jiang-type if (i) or (ii) holds for all maps. Examples of Jiang-type spaces include simply-connected spaces, \(H\)-spaces, orientable coset spaces \(G/K\) of compact connected Lie groups \(G\) by closed subgroups \(K\), nilmanifolds, certain solvmanifolds, and \(\mathcal{C}\)-nilpotent spaces where \(\mathcal{C}\) is the class of finite groups. If \(X\) is a Jiang-type space and \(\pi_1(X)\) has property \(R_{\infty}\) then it forces \(N(f)=0\) for every self homeomorphism \(f\) because \(N(f)\) is always finite. Furthermore, if, in addition, \(X\) is a compact manifold with \(\dim X\ge 3\) then every self-homeomorphism is homotopic to be fixed point free. (For more background on Nielsen fixed point theory and related topics, see [5] and [6].)
While many finitely generated groups are known to have property \(R_{\infty}\), the only known examples of manifolds \(M\) that are of Jiang-type with \(\pi_1(M)\) having property \(R_{\infty}\) are certain nilmanifollds and certain solvmanifolds (of type \(\mathcal{N}R\)) which are aspherical. Recently, in her dissertation [7], van den Bussche constructed certain solvmanifolds for which every self homeomorphism has zero Nielsen number, extending certain results of [3] (see also [8]). More precisely, she considered a class of spaces called \(N_0\)-spaces, i.e., the family of compact connected manifolds \(M\) such that every self homotopy equivalence \(f:M\longrightarrowM\) has \(N(f)=0\). Evidently, if \(X\) is a compact connected triangulable manifold of dimension at least \(3\) and \(X\) is also an \(N_0\)-space then using a result of Jezierski [9], it is straightforward to show that every self homeomorphism \(f:X\longrightarrowX\) is homotopic to a map \(g\) with no periodic points of any given period.
The main purpose of this paper is to construct non-aspherical \(N_0\)-spaces and also \(N_0\)-spaces that are of Jiang-type. The goal is to focus on constructing manifolds whose fundamental groups are finitely generated virtually cyclic and have property \(R_{\infty}\). The paper is organized as follows. In section 2, we recall some known results on virtually cyclic groups with property \(R_{\infty}\) and some background on \(N_0\)-spaces, Jiang-type spaces, and \(R_{\infty}\)-spaces. In section 3, we construct \(N_0\)-spaces that are mapping tori of certain Lens spaces. In section 4, we consider coset spaces \(U(n)/K\) of unitary groups by finite subgroups that are known to be of Jiang-type. Then we construct for each \(n\ge 5\), an \(N_0\)-space of the form \(U(n)/K_n\). In section 5, we show that for any finite subgroup \(K\) of \(U(2)\), the fundamental group \(\pi_1(U(2)/K)\) does not have property \(R_{\infty}\). In fact, we show that the space \(U(2)/K\) does not have the property \(R_{\infty}\).
The authors thank the CIRM (Luminy) for its support and hospitality under the "Research in Residence" program, February 25 - March 1, 2024 during which this research was conducted. The first author was partially supported by Projeto Regular Fapesp Nielsen theory for maps between spherical 3-manifolds and some homogeneous spaces no. 2023/16051-5 and the travel for the second author was supported by a grant from Bates College.
In this section, we discuss \(N_0\)-spaces and the relationships among Jiang-type spaces and the \(R_{\infty}\)-property for groups and for spaces. We also recall the \(R_{\infty}\)-property for virtually cyclic groups from [10] .
Let \(X\) be a compact connected space.
Definition 1. We say that \(X\) is an \(R_{\infty}\)-space if \(R(f)= \infty\) for every homotopy equivalence \(f : X \longrightarrowX\).
Of course, if \(\pi_1(X)\) has the \(R_{\infty}\)-property then \(X\) is an \(R_{\infty}\)-space. On the other hand, we will show later in this section that the converse does not hold. From [7] we have:
Definition 2. We say that \(X\) is an \(N_0\)-space if \(N(f)=0\) for every homotopy equivalence \(f : X \longrightarrowX\).
For instance, \(X = \mathbb{R}P^3\# \mathbb{R}P^3\) is an \(R_{\infty}\)-space but not an \(N_0\)-space (see [11]).
Recall that a space \(M\) is of Jiang-type if for every map \(f:M\longrightarrowM\), either \(L(f)=0 \Rightarrow N(f)=0\) or \(L(f)\ne 0 \Rightarrow N(f)=R(f)\). Known Jiang-type spaces include \(1\)-connected spaces, \(H\)-spaces, generalized lens spaces, orientable \(G/K\) of compact connected Lie group \(G\) by closed subgroup \(K\), nilmanifolds, \(\mathcal{C}\)-nilpotent spaces where \(\mathcal{C}\) is the class of finite groups.
Evidently, if \(M\) is of Jiang-type and is an \(R_{\infty}\)-space then \(M\) is an \(N_0\)-space.
Finitely generated non-elementary word hyperbolic groups have the \(R_{\infty}\)-property [12]. Non-elementary means not virtually cyclic so which (infinite) virtually cyclic groups have the \(R_{\infty}\)-property?
Let \(G\) be a finitely generated virtually cyclic group so that
\[\label{VC} 0 \longrightarrow\mathbb{Z} \longrightarrowG \longrightarrowF \longrightarrow1\tag{1}\] is short exact with \(F\) a finite group. Then \(G\) falls into one of the following two types. Type I: if 1 is not central; Type II if 1 is central. If \(G\) is of Type I then \(G\) has property \(R_{\infty}\) [10].
Now suppose that \(G\) is of Type II. Then \(G\cong K\rtimes_{\theta} \mathbb{Z}\) for some finite group \(K\) with action \(\theta : \mathbb{Z} \longrightarrow{\rm Aut}(K)\). Necessary and sufficient conditions were given in [10] for \(G\) to have property \(R_{\infty}\). However, the statement is not entirely true and we take the opportunity to correct that. We thank Pieter Senden who pointed out this error.
Proposition 2.5 of [10] states: Let \(\mathbb{Z} = \langle t \rangle\), \(H\) a finite group, \(\theta(t)\in {\rm Aut}(H)\), and \(G = H\rtimes_{\theta} \mathbb{Z}\). Then \(G\) has the \(R_{\infty}\)-property if, and only if, \(\theta(t)\) is not conjugate to its inverse in \({\rm Aut}(H)\).
The implication if \(G\) has the \(R_{\infty}\)-property then \(\theta(t)\) is not conjugate to its inverse in \({\rm Aut}(H)\) holds and the proof given in [10] is correct. However, the proof of the converse has a gap. The correct statement should be as follows.
Proposition 3. Let \(\mathbb{Z} = \langle t \rangle\), \(H\) a finite group, \(\theta(t)\in {\rm Aut}(H)\), and \(G = H\rtimes_{\theta} \mathbb{Z}\). Then \(G\) has the \(R_{\infty}\)-property if, and only if, \([\theta(t)]\) is not conjugate to its inverse in \({\rm Out}(H)\).
Here, \([\varphi]\) is the image of an element \(\varphi \in {\rm Aut}(G)\) under the projection homomorphism \({\rm Aut}(H) \longrightarrow{\rm Out}(H)={\rm Aut}(H)/{\rm Inn}(H)\).
The mistake in the proof of [10] occurred when we assumed that an automorphism \(\Phi: G \longrightarrowG\) is of the form \(\Phi(h,t^r) =(\varphi(h),t^ {kr})\) for \(h\in H\) and \(t^r \in \mathbb{Z}\). In fact, an arbitrary automorphism should be of the form \(\Phi(h,t^r) =(\varphi(h)\lambda^r,t^ {kr})\) for an arbitrary \(\lambda\in H\). With this modification, it is straightforward to adapt the proof of the converse given in the proof of [10]. This modification will complete the proof of Proposition 3
Remark 4. It follows from Proposition 3 that if \({\rm Aut}(H)\) is abelian then \(G\) has \(R_{\infty}\) iff \(\theta (1)\) is non-trivial and \(\theta (1)\) is not of order \(2\).
Example 5. Let \(G=\mathbb{Z}_5 \rtimes_{\theta} \mathbb{Z}, H=\mathbb{Z}_5\) so \({\rm Aut}(H)\cong \mathbb{Z}_4\) where \(\theta : \mathbb{Z} \longrightarrow{\rm Aut}(H)\) is multiplication by \(2\). Since \(\theta(1)\) has order \(4\), it follows that \(G\) has property \(R_{\infty}\). Furthermore, every automorphism \(\varphi :G\longrightarrowG\) induces the identity on \(G/H=\mathbb{Z}\).
While it is the case that \(\pi_1(M)\) has \(R_{\infty}\) implies that \(M\) is an \(R_{\infty}\)-space, the converse is false in general. We now construct spaces \(M\) such that its fundamental group does not have the \(R_{\infty}\)-property but the space has the \(R_{\infty}\)-property, i.e. for any homotopy equivalence, it has the property that the induced homomorphism on the fundamental group has infinite Reidemeister number.
First, we give a general construction of such a space with prescribed property on the action of \(\pi_1\) on \(\pi_2\).
Theorem 6. Let \(F\) be a finite group and \(H=F\rtimes _{\theta} \mathbb{Z}\) with the property that any automorphism of \(H\) induces the identity on the quotient \(H/F\simeq \mathbb{Z}\). There is a space \(X\) which has the property that \(\pi_1(X)\simeq \mathbb{Z}\), \(\pi_2(X)\simeq F\) and the action of \(\pi_1(X)\) on \(\pi_2(X)\) is given by \(\theta\). Furthermore, this space has the \(R_{\infty}\)-property but not its fundamental group.
Proof. Given a group \(G\) and a sequence of \(\mathbb{Z}[G]\)-modules \(A_i, i\geq 2\), it follows from [13] that there is a space \(X\) such that \(\pi_1(X)\simeq G\), \(\pi_i(X)=A_i\) and the structure of \(\mathbb{Z}[\pi_1(X)]\)-module of \(\pi_i(X)\) inherited from the space \(X\) is the given \(\mathbb{Z}[G]\)-module structure of \(A_i\) for all \(i\geq 2\).
Now let \(X\) be such a space where \(\pi_1(X)\simeq \mathbb{Z}\), \(\pi_2(X)\simeq F\) such that the action of \(\pi_1(X)\) on \(\pi_2(X)\) is given by \(\theta\). Let \(f:X\longrightarrowX\) be any homotopy equivalence. The action of \(\pi_1\) on \(\pi_2\) is natural with respect to maps so that \(f_{\#}(\alpha\star \beta))=f_{\#}(\alpha)\star f_{\#}(\beta)\) where \(\alpha\in \pi_1(X), \beta \in \pi_2(X)\) and \(\alpha \star \beta=(\theta(\alpha))(\beta)\). Applying this equality to the homotopy equivalence \(f\), we obtain two homomorphisms \(f_{\#1}\) and \(f_{\#2}\) which define an automorphism of the group \(H\). By assumption, \(f_{\#1}\) is the identity so \(R(f)=R(f_{\#})\ge R(f_{\#1})=\infty\). Since \(\mathbb{Z}\) does not have property \(R_{\infty}\), the result follows. ◻
Let \(h:X\longrightarrowX\) be a homeomorphism and \(T_hX\) be the corresponding mapping torus. It follows that we have a fibration \(X\longrightarrowT_hX\longrightarrowS^1\) over the circle and that \(G\simeq \pi_1(T_hX)\) has a presentation \(G=\langle \pi_1(X), t\mid t\beta t^{-1}=h_{\#}(\beta), \beta \in \pi_1(X)\rangle\).
Example 7. Consider the Lens space \(L(5,2)\). Using [14], there is a homeomorphism \(f:L(5,2)\longrightarrowL(5,2)\) which induces an automorphism as multiplication by \(2\) on \(\pi_1(L(5,2))=\mathbb{Z}_5\) and has degree \(-1\). Now consider the homeomorphism \(h=\Sigma f: X=\Sigma L(5,2) \longrightarrow \Sigma L(5,2)\), the suspension of \(f\). Observe that this space \(X\) is simply connected and \(\pi_2(X)=\pi_2(\Sigma L(5,2))\simeq H_2(\Sigma L(5,2)) \simeq H_1(L(5,2)) \simeq \mathbb{Z}_5\). Now if we apply the mapping construction for the homeomorphism \(h=\Sigma f\), we obtain a space \(T_{h}X\) where \(\pi_1(T_{h}X)\simeq \mathbb{Z}\), \(\pi_2(T_{h}X)\simeq \mathbb{Z}_5\) and the action of \(\pi_1(T_{h}X)\) on \(\pi_2(T_{h}X)\) is multiplication by \(2\). From Example 5, every automorphism of \(\mathbb{Z}_5 \rtimes_{\theta} \mathbb{Z}\) induces the identity on the quotient \(\mathbb{Z}\). It follows from Theorem 6 that the space \(T_hX\) has the \(R_{\infty}\)-property while \(\pi_1(T_hX)\simeq \mathbb{Z}\) does not have.
Let \(X\) be a compact connected manifold with \(F=\pi_1(X)\) finite and \(h:X\longrightarrowX\) be a self homeomorphism. Denote by \(M=T_{h}X\) the corresponding mapping torus. Then the natural projection \(M=T_{h}X\longrightarrowS^1\) is a fibre bundle with fibre \(X\).
Note that since \(\pi_1(S^1)=\mathbb{Z}\) is free, we have \(\pi_1(M)\cong F\rtimes \mathbb{Z}=\langle F, t \mid txt^{-1}=h_{\#}(x), x\in F\rangle\). Moreover, since \(F=\pi_1(X)\) is finite, \(F\) is characteristic in \(\pi_1(M)\). If \(\varphi: \pi_1(M)\longrightarrow\pi_1(M)\) is an automorphism, we have the following commutative diagram \[\label{exact} \begin{CD} 1 @>>> \pi_1(X) @>{i_{\#}}>> \pi_1(M) @>{p_{\#}}>> \mathbb{Z} @>>> 0 \\ @. @V{\varphi'}VV @V{\varphi}VV @V{\overline{\varphi}}VV @. \\ 1 @>>> \pi_1(X) @>{i_{\#}}>> \pi_1(M) @>{p_{\#}}>> \mathbb{Z} @>>> 0 \end{CD}\tag{2}\] where the last vertical homomorphism is \(\pm 1_{\mathbb{Z}}\). Again, since \(F\) is finite, it follows from the diagram 2 that \(R(\varphi)=\infty\) if, and only if, \(\overline{\varphi}=1_{\mathbb{Z}}\).
Lemma 8. Suppose \(F\simeq \mathbb{Z}_p\) and \(h_{\#}(l)=ql\) where \(q^ 2\equiv -1 \; mod (p)\). The group \(\pi_1(M)\) has the \(R_{\infty}\)–property and so \(\bar \varphi:\mathbb{Z}\longrightarrow\mathbb{Z}\) is the identity.
Proof. The result follows from Proposition 3. ◻
Proposition 9. Suppose \(F\simeq \mathbb{Z}_p\) and \(h_{\#}(l)=ql\) where \(q^ 2\equiv -1 \; mod (p)\). Let \(f':M\longrightarrowM\) be a homotopy equivalence. Then \(f'\) is homotopic to a fibre preserving map \(f:M \longrightarrowM\) so that we have the
following commutative diagram
\[\begin{CD} X @>{i}>> M @>{p}>> S^1 \\ @V{ f|_X}VV @V{ f}VV @V{\overline{f}}VV @. \\ X @>{i}>> M @>{p}>> S^1 \end{CD}\] which, in turn, induces the commutative diagram 2 at the fundamental group level where \(\varphi=f_\#\) and \(\varphi'=(f|_{X})_{\#}\) and \(\overline{\varphi}=\overline{f}_\#=1_{\mathbb{Z}.}\)
Proof. Since \(f'\) is a homotopy equivalence, the map \(p\circ f'\) is homotopic to \(p\) because \(\overline{\varphi}\) is the identity. We obtain the map \(f\) using the Covering Homotopy Property of the fibration \(M\longrightarrowS^1\). ◻
The following result is immediate.
Theorem 10. Let \(M\) and \(f\) be as in Proposition 9. The map \(f\) can be deformed to a fixed point free map and \(M\) is an \(N_0\)-space.
Now we search for groups of the form \(G=F\rtimes_{\theta} \mathbb{Z}\) such that any automorphism induces the identity on the quotient \(G/F\). Note that such \(G\) is virtually cyclic and necessarily has property \(R_{\infty}\). Once such a group is given, we look for a homeomorphism \(h :X\longrightarrowX\) such that \(\theta(1)=h_{\#}\). Classical lens spaces are natural candidates for such a space \(X\). First, we recall some known results regarding realizing a degree \(\pm 1\) map by a homeomorphism on three dimensional lens spaces.
Theorem 11. ([14] or [15]) (i) Suppose that \(f\) is a degree \(1\) self-map on \(L(p,q)\). Then \(f\) is homotopic to an orientation-preserving homeomorphism if and only if
\[f_{*}(l)=\pm l, \; if \; p \; does \;not \; divide \;q^2-1, \;and \; f_{*}(l)= \pm l, \;\pm ql,\;if \; p\; divides \;q^2-1,\] for all \(l \in \pi_1(L(p,q))\), where \(f_{*}\) is the endomorphism of \(\pi_1(L(p,q))\) induced by \(f\).
(ii) \(L(p,q)\) admits an orientation-reversing homeomorphism if and only if \[q^2\equiv -1\;\; (mod p).\]
In this case, a degree \(-1\) self-map \(f\) on \(L(p,q)\) is homotopic to an orientation-reversing homeomorphism if and only if \[f_{*}(l) = \pm ql, \;for \;all \;l \in \pi_1(L(p,q)),\] where \(f_{*}\) is the endomorphism of \(\pi_1(L(p,q))\) induced by \(f\).
We now describe how to use Lemma 8 to construct an \(N_0\)-space using the procedure above for a pair \((L(p,q), \theta)\) where \(L(p,q)\) is a lens space and \(\theta\) is an automorphism of the fundamental group \(\mathbb{Z}_p\) of the lens spaces \(L(p,q)\). We also write \((L(p,q), k)\) when the automorphism \(\theta\) is multiplication by \(k\).
Theorem 12. Let \(p\) be a positive integer of the form either \(p_1^{e_1}\) or \(2p_1^{e_1}\) where \(p_1\) is an odd prime of the form \(4k+1\). (a) There exists a positive integer \(q\) such that \(q^2\equiv -1\) mod \(p\). (b) The mapping torus \(T_hX\) is an \(N_0\)-space where \(X\) and \(h\) are given by the pair \((L(p,q),q)\).
Proof. (a) When \(p=p_1^{e_1}\) for \(p_1\) a prime of the form \(4k+1\), the existence of \(q\) follows from [16]. In the case of \(p=2p_1^{e_1}\), since \({\rm Aut}(\mathbb{Z}_{2p_1^{e_1}})={\rm Aut}(\mathbb{Z}_{p_1^{e_1}})\), the proof of the result is similar.
(b) Using the integer \(q\) obtained from (a), consider the lens space \(X=L(p,q)\). By Theorem 11(ii) (or [14]), there is a homeomorphism \(h:X\longrightarrowX\) of degree \(-1\) such that \(h_{\#}=\theta\) where \(\theta\) is multiplication by \(q\). Thus, Lemma 8 applies and the result follows from Theorem 10. ◻
Remark 13. We do not consider the lens spaces given by [14] since for such lens spaces there are no maps of degree \(-1\). For the other spherical \(3\)-manifolds (except \(S^3\), [14]), there are no degree \(-1\) maps. By Remark 4, we do not consider degree \(1\) maps on lens spaces. Similarly, by Proposition 3, we also do not consider the spherical \(3\)-manifolds with fundamental group isomorphic to \(T^*_{24}, O^*_{48}\) or \(I^*_{120}\) for they have \({\rm Out} \cong \mathbb{Z}_2\).
Theorem 12 provides an infinite family (each for every odd prime of the form \(4k+1\) and there are an infinite number of such primes (see e.g. [17])) of four dimensional compact connected manifolds for which every self homotopy equivalence is homotopic to be fixed point free. The simplest example for which this theorem applies is \((L(5,2),2)\).
Now we consider examples of lens spaces in higher dimensions using [15] which generalizes the results of [14] in the case of lens spaces.
Recall from [15] the following result which generalizes Theorem 11(i).
Theorem 14. Assume that \(f\) is a degree 1 self-map on \(L(p;q_1,q_2,...,q_n)\). Then, \(f\) is homotopic to an orientation-preserving
homeomorphism if and only if \[f_{*}(l) = ql,\;\; for \;\; all \;\; l\in \pi_1(L(p;q_1,q_2,...,q_n)),\] where \(f_{*}\) is the endomorphism on \(\pi_1(L(p;q_1,q_2,...,q_n))\) induced by \(f\), \(q \in \mathbb{Z}_p\) is
coprime to \(p\) with a permutation \(\sigma\), such that \[q^n\equiv 1 \;\; (mod p), \;\;\; q_i \equiv\pm qq_{\sigma(i)} \;\; (mod p)\] for all \(i = 1,2,...,n.\)
Unlike in dimension \(3\) where we did not consider degree \(1\) maps because of Remark 4, we can construct many higher dimensional lens spaces for which Lemma 8 applies.
For simplicity we will restrict to the case where \(p\) is prime. Given \(n\ge 3\), choose a prime \(p\ge 5\) such that \({\rm Aut}(\mathbb{Z}_p)\) admits a non-trivial automorphism \(\theta\) (given by multiplication by some \(q\), i.e., \(\theta(x)=qx\).) such that \(\theta^2\ne 1\) and \(\theta^n=1\). Then the orbit of \(q_1=1\) under \(\theta\) gives rise to \(q_i, i=1,..., n/d\) and a permutation \(\sigma\) such that \(q_i \equiv\pm qq_{\sigma(i)} \;\; (mod p)\) where \(d\) is the order of \(q\).
For instance, take \(p=5\) and \(q=2\). Observe that \(q^4\equiv 1\) mod \(5\). Based on this, define \(q_1=3\), \(q_2=4\), \(q_3=2\), \(q_4=1\) with the permutation \(\sigma=(1342)\) with \(q_i=i\). The map which realizes this has degree \(1\). By Theorem 14, this map can be realized by a homeomorphism \(h\) with \(h_{\#}\) be given by multiplication by \(q=2\). Note that Lemma 8 is still valid if we replace the condition on \(q^2\) by \(q^2\not\equiv 1\) mod \(p\). Now, we can apply Theorem 10 to obtain an \(N_0\)-space \(T_hX\). Moreover, since \(\dim T_hX\ge 5\), it follows that every self homeomorphism of \(T_hX\) is isotopic to be fixed point free.
Similarly, we can take \(n=3\), \(p=7\) and \(q=4\). Observe that \(q^3\equiv 1\) mod \(7\). Define \(q_1=2\), \(q_2=4\), \(q_3=1\) with the permutation \(\sigma=(132)\). The map
which realizes this has degree \(1\). By Theorem 14, this map can be realized by a homeomorphism \(h\) with \(h_{\#}\) be given by multiplication by \(q=4\). Now, applying Theorem 10 yields an \(N_0\)-space \(T_hX\) and since \(\dim T_hX=6\ge 5\), it follows that every self homeomorphism of \(T_hX\) is isotopic to be
fixed point free. Using the same argument, we can take \(n=6\), \(p=7\) and \(q=4\). Since \(q^3\equiv 1\) mod \(7\), we also have \(q^6\equiv 1\) mod \(7\). Define \(q_1=2\), \(q_2=4\), \(q_3=1\), \(q_4=6\), \(q_5=5\), \(q_6=3\) with the permutation \(\sigma=(132)(465)\). Then we
can obtain an \(N_0\)-space \(T_hX\) of dimension \((2(6)-1)+1=12\) such that every self homeomorphism of \(T_hX\) is
isotopic to be fixed point free.
Note that for any prime \(p\ge 5\) and \(q=2\), we can always find a permutation \(\sigma\) (not necessarily a \((p-1)\)-cycle) on \(\{1,...,p-1\}\) and \(\{q_i\}=\{1,...,p-1\}\) so that the hypotheses in Theorem 14 are satisfied and that \(q^2\not\equiv 1\) mod \(p\).
In fact, we have the following.
Theorem 15. Let \(p\ge 5\) be a prime, \(X=L(p; 1,..., p-1)\) be a \((2(p-1)-1)\)-dimensional lens space, and \(\theta : \mathbb{Z}_p \longrightarrow\mathbb{Z}_p\) be the automorphism given by multiplication by \(q=2\). Then \(M=T_hX\) is an \(N_0\)-space where \(h:X\longrightarrowX\) is a degree \(1\) homeomorphism with \(h_{\#}=\theta\). Moreover, every self homeomorphism of \(M\) is isotopic to be fixed point free.
The next result from [15] generalizes Theorem 11(ii).
Theorem 16. A lens space \(L(p;q_1,q_2,...,q_n)\) admits an orientation-reversing homeomorphism if and only if there exist some \(q \in \mathbb{Z}_p\) coprime to \(p\) with a permutation \(\sigma\), such that \[q^n=-1 \;\;(mod p), \;\;\; q_i \equiv\pm qq_{\sigma(i)} \;\; (mod p),\] for all \(i = 1,2,...,n\). In this case, a degree \(-1\) self-map \(f\) on \(L(p;q_1,q_2,...,q_n)\) homotopic to an orientation-reversing homeomorphism if and only if \[f_{*}(l)=ql, \;\;for \;\;all \;\; l\in \pi_1(L(p;q_1,q_2,...,q_n)),\] where the \(f_{*}\) is the endomorphism of \(\pi_1(L(p;q_1,q_2,...,q_n))\) induced by \(f\), \(q\in \mathbb{Z}_p\) is coprime to \(p\) with a permutation \(\sigma\), such that \[q^n\equiv -1 \;\; (mod p), \;\;\;\;\;\;\;\;\; q_i\equiv\pm qq_{\sigma(i)} \;\; (mod p),\] for all \(i = 1,2,...,n.\).
Based on this result, one can construct an \(N_0\)-space as follows. Take \(p=11\), \(q=2\) and \(n=5\). Observe that \(2^5\equiv -1\) mod(11). Now we define \(q_1=5\), \(q_2=8\), \(q_3=4\), \(q_4=2\), \(q_5=1\). Observe that, modulo \(11\), \[q_1\equiv 2\cdot q_2; \quad q_2\equiv 2\cdot q_3;\quad q_3\equiv 2\cdot q_4;\quad q_4\equiv 2\cdot q_5;\quad q_5\equiv -2\cdot q_1,\] so the permutation is \(\sigma=(12345)\) . It follows from Theorem 16 that we have a homeomorphism \(h\) of \(X=L(11;5,8,4,2,1)\) of degree \(-1\). Thus, the mapping torus \(T_hX\) is an \(N_0\)-space.
It is straightforward to extend this to the following by repeating the sequence \(q_1,..., q_5\), i.e., \(\{q_i\}=\{q_{j+5k}\mid 1\le j\le 5\}\).
Theorem 17. For any positive integer \(N\), let \(q_1=5; q_2=8; q_3=4; q_4=2; q_5=1\) and \(q_{i+5j}=q_i\) for \(1\le i\le 5\) and \(1\le j\le N-1\). There is an \((2(5N)-1)\)-dimensional \(N_0\)-space \(M=T_hX\) which is the mapping torus of the lens space \(X=L(11;q_1,..., q_{5N})\) where the homeomorphism \(h:X\longrightarrowX\) has degree \(-1\) and \(h_{\#}\) is multiplication by \(2\). Moreover, every self homeomorphism of \(M\) is isotopic to be fixed point free.
While we have constructed many \(N_0\)-spaces as in Theorem 15 and in Theorem 17, we do not know if any of these spaces can be of Jiang-type. Next, we focus on certain Jiang-type spaces with virtually cyclic fundamental groups.
Let \(G\) be a compact connected Lie group; \(K\) a finite subgroup. It follows from [18] (see also [19]) that \(M=G/K\) is of Jiang-type. Moreover, it was shown in [19] that \[0\longrightarrow\pi_1(G) \longrightarrow\pi_1(G/K) \longrightarrowK \longrightarrow1\] is a central extension. In particular, when \(G=U(n)\), \(\pi_1(U(n)/K)\) is a virtually cyclic group of Type II so that Proposition 3 is applicable.
Given a finite subgroup \(K\subset U(n)\) we begin by describing the fundamental group of the homogeneous space \(U(n)/K\) (left cosets). The determinant map \(\det:U(n)\longrightarrowS^1\) provides a fibration \(SU(n)\longrightarrowU(n)\stackrel{\det} \longrightarrowS^1\) where the fibre, the pre-image of \(1\in S^1\), is the subgroup \(SU(n)\). Observe that \(\det(K)\) is a finite subgroup of \(S^1\), therefore it is a finite cyclic group. By considering the coset spaces, we obtain a fibration \[SU(n)/K' \longrightarrowU(n)/K \longrightarrowS^1/\det(K)\] where \(K'=K\cap SU(n)\) and \(S^1/\det(K)\) is homeomorphic to \(S^1\).
The last \(5\)-terms of the long exact sequence in homotopy associated to the above fibration is:
\[1\longrightarrowK' \longrightarrow\pi_1(U(n)/K) \longrightarrow\mathbb{Z}\longrightarrow1.\] Further, this short exact sequence splits (since \(\mathbb{Z}\) is free) and the action
of a generator of \(\mathbb{Z}\) is obtained as follows:
Choose a matrix \(A\in K\) such that \(\det(A)\) is a generator of \(\det(K)\) and let \(\alpha \in K'\). The action of
the generator of \(\mathbb{Z}=\pi_1(S^1/\det(K))\) defined by \(A\) on \(\alpha\) is given the conjugation \(A\circ \alpha\circ
A^{-1}\). In order to decide if the group \(K'\rtimes_{\theta} \mathbb{Z}\) has or not the \(R_{\infty}\)-property we use Propostion 3.
Let \(n=5\). Consider the elements \(A,B\in U(5)\) given by
\[A=\begin{pmatrix} \omega & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & \omega & 0 \\ 0 & \omega & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & \omega \\ 0 & 0 & \omega & 0 & 0 \end{pmatrix}, \qquad B=\begin{pmatrix} 0 & 0 & 0 & 0 & 1\\ 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \end{pmatrix}\]
where \(\omega^{20}=1\).
Let \(K\le U(5)\) be
\[K=\langle A,B \mid A^{20}=B^5=I, A^4B=BA^4, ABA^{-1}=B^2 \rangle.\]
Consider the subgroup \(K'=\langle A^4, B\rangle \cong \mathbb{Z}_5 \oplus \mathbb{Z}_5\). It is easy to see that
\[1\longrightarrowK' \longrightarrowK \longrightarrow\mathbb{Z}_4 \longrightarrow1\] is a short exact sequence of finite groups and \(\mathbb{Z}_4\) acts on \(K'\) via \(\begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix}\).
It follows that
\[1\longrightarrowK' \longrightarrow\pi_1(U(5)/K) \longrightarrow\mathbb{Z} \longrightarrow1\] is a short exact sequence such that \(\pi_1(U(5)/K)\cong K' \rtimes_{\theta} \mathbb{Z}\) with action \(\theta\) given by \(\theta(1) =\begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix}\). Note that \(\theta(-1)=\begin{pmatrix} 1 & 0 \\ 0 & 3 \end{pmatrix}\) is the inverse of \(\theta(1)\).
It is straighforward to check that for any \(V\in GL_2(\mathbb{Z}_5)\), \[\begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix}V \ne V\begin{pmatrix} 1 & 0 \\ 0 & 3 \end{pmatrix}.\]
Since for any \(k\ge 1\), we can embed \(M\in U(n)\) in \(U(n+k)\) as \(\begin{pmatrix} M & 0 \\ 0 & I_k \end{pmatrix}\) where \(I_k\) is the \(k \times k\) identity matrix, we have the following.
Theorem 18. For any \(n\ge 5\), there exists a finite group \(K_n \le U(n)\) such that \(M(n)=U(n)/K_n\) is a Jiang-type, \(R_{\infty}\), \(N_0\)-space. Furthermore, every self homeomorphism of \(M(n)\) is isotopic to be fixed point free. Here \(K_5=K\) as described above.
Now, we give another example of a finite subgroup \(K\le U(7)\) with the desired property but \(K\) is not of the form \(\begin{pmatrix} K_5 & 0 \\ 0 & I_2 \end{pmatrix}\) as in Theorem 18.
Consider the elements \(A,B\in U(7)\) given by
\[A=\begin{pmatrix} \omega & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & \omega & 0 & 0 \\ 0 & \omega & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & \omega & 0 \\ 0 & 0 & \omega & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & \omega \\ 0 & 0 & 0 & \omega & 0 & 0 & 0 \end{pmatrix}, \qquad B=\begin{pmatrix} 0 & 0 & 0 & 0 & 0 & 0 & 1\\ 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 \end{pmatrix}\]
where \(\omega^{21}=1\).
Let \(K\le U(7)\) be
\[K=\langle A,B \mid A^{21}=B^7=I, A^3B=BA^3, ABA^{-1}=B^2 \rangle.\]
Consider the subgroup \(K'=\langle A^3, B\rangle \cong \mathbb{Z}_7 \oplus \mathbb{Z}_7\). It is easy to see that
\[1\longrightarrowK' \longrightarrowK \longrightarrow\mathbb{Z}_3 \longrightarrow1\] is a short exact sequence of finite groups and \(\mathbb{Z}_3\) acts on \(K'\) via \(\begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix}\).
It follows that
\[1\longrightarrowK' \longrightarrow\pi_1(U(7)/K) \longrightarrow\mathbb{Z} \longrightarrow1\] is a short exact sequence such that \(\pi_1(U(7)/K)\cong K' \rtimes_{\theta} \mathbb{Z}\) with action \(\theta\) given by \(\theta(1) =\begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix}\). Note that \([\theta(1)]^2=\begin{pmatrix} 1 & 0 \\ 0 & 4 \end{pmatrix}\) is the inverse of \(\theta(1)\).
We can check that there exists NO \(V\in GL_2(\mathbb{Z}_7)\) such that \[\begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix}V=V\begin{pmatrix} 1 & 0 \\ 0 & 4 \end{pmatrix}.\]
We have constructed for each \(n\ge 5\) a finite subgroup \(K_n\) such that \(\pi_1(M(n))\) has the \(R_{\infty}\)-property where \(M(n)=U(n)/K_n\). It is therefore natural to ask whether such examples can be found when \(n=2,3,\) or \(4\). In this section we show that for any finite subgroup \(K\subset U(2)\), the fundamental group of the homogeneous space \(U(2)/K\) does not have the \(R_{\infty}\)-property. Moreover, we show that the space \(U(2)/K\) is not an \(R_{\infty}\)-space.
First, we briefly describe the classification by Du Val [20] and relate the subgroup \(K'\) with the notations of finite subgroups used by Du Val.
For any \(A\in U(2)\), the corresponding transformation \(T_A\) sends \(x \longmapsto\varepsilonx \kappa\) for some \(\varepsilon\in S^1\) (unit complex numbers) and \(\kappa \in Sp(1)\) (unit quaternions), for every \(x\in \mathbb{C}^2\). Equivalently, \[\label{DV} A=\begin{pmatrix} \varepsilon& 0 \\ 0 & \varepsilon\end{pmatrix} \begin{pmatrix} a & -\bar c \\ c & \bar a \end{pmatrix}\tag{3}\] where \(a,c \in \mathbb{C}\), \(a \bar a + c \bar c=1\), \(\varepsilon\in S^1\), and \(\kappa=a+cj \in Sp(1)\). Thus for every \(A\), we can associate to it a pair \((\varepsilon, \kappa) \in S^1 \times Sp(1)\). Note that the pair \((\varepsilon, \kappa)\) is not unique since we can also associate to \(A\) the pair \((-\varepsilon, -\kappa)\). In [20], it was shown that any finite subgroup \(K\) of \(U(2)\) is determined by four groups as follows. Given \(K\), let \(R=\{r \mid \exists \ell \in S^1, (\ell, r)\in K\}\), \(R_K=\{r \mid (1,r)\in K\}\). Similarly, let \(L=\{\ell \mid \exists r\in Sp(1), (\ell, r)\in K\}\) and \(L_K=\{\ell \mid (\ell, 1)\in K\}\). Moreover, \(R/R_K \cong L/L_K\). Note that different isomorphisms \(\overline{\varphi}: L/L_K \longrightarrowR/R_K\) may yield different finite subgroups \(K\). According to [20], every finite subgroup is denoted by \((L/L_K, R/R_K)\) together with an isomorphism \(\overline{\varphi}\) between \(L/L_K\) and \(R/R_K\) and the identification \((\varepsilon, \kappa)\sim (-\varepsilon, -\kappa)\). In other words, every finite subgroup is given by \[\label{finiteK} K=\{(\varepsilon, \kappa)\in L\times R \mid \overline{\varphi}(\overline{\varepsilon})=\overline{\kappa}\}/\sim.\tag{4}\]
Now, given a finite subgroup \(K \subset U(2)\), we recall that \(K'\) is the kernel of the determinant map \(\det : K \longrightarrowS^1\). Suppose \(A\in K\). It follows from 3 that \(A\in K'\) iff \(\varepsilon^2=1\) or \(A\) corresponds to \((1,\kappa)\). In other words, \(K'\) is the subgroup \(\{(1,r) \in K\}\) and hence \(K' \cong R_K\).
For the list of all finite groups of \(U(2)\), we will use the classification given in [21] and [22]. From there the finite groups of unitary transformations in the plane are:
\((\textswab{C}_{2m}/\textswab{C}_f; \textswab{C}_{2n}/\textswab{C}_g)_{d},\) of order \(gm=fn\)
\(\langle p,2,2\rangle_m=(\textswab{C}_{2m}/\textswab{C}_{2m}; {\boldsymbol{D}}_{p}/{\boldsymbol{D}}_p),\) \(4mp\)
\((\textswab{C}_{4m}/\textswab{C}_{2m}; \langle p,2,2\rangle= {\boldsymbol{D}}_{p}/\textswab{C}_{2p}) ,\) \(4mp\)
\((\textswab{C}_{4m}/\textswab{C}_{m}; \langle p,2,2\rangle={\boldsymbol{D}}_{p}/\textswab{C}_{p}), \;m \;and \;\;p\;\;odd,\) \(2mp\)
\((\textswab{C}_{4m}/\textswab{C}_{2m}; \langle 2p,2,2\rangle={\boldsymbol{D}}_{2p}/ \langle p,2,2\rangle= {\boldsymbol{D}}_{p}),\) \(8mp\)
\(\langle 3,3,2\rangle_m=(\textswab{C}_{2m}/\textswab{C}_{2m}; {\boldsymbol{T}}/{\boldsymbol{T}}),\) \(24m\)
\((\textswab{C}_{6m}/\textswab{C}_{2m}; \langle 3,3,2\rangle/\langle 2,2,2\rangle={\boldsymbol{T}}/{\boldsymbol{Q}}_8),\) \(24m\)
\(\langle 4,3,2 \rangle_m=(\textswab{C}_{2m}/\textswab{C}_{2m}; {\boldsymbol{O}}/{\boldsymbol{O}}),\) \(48m\)
\((\textswab{C}_{4m}/\textswab{C}_{2m}; \langle 4,3,2\rangle/\langle 3,3,2 \rangle={\boldsymbol{O}}/{\boldsymbol{T}}),\) \(48m\)
\(\langle 5,3,2 \rangle_m=(\textswab{C}_{2m}/\textswab{C}_{2m}; {\boldsymbol{I}}/{\boldsymbol{I}}),\) \(120m\)
For the case (1) the numbers \(f,g,m,n,d\) are positive integers such that \(f\equiv g(mod(2)\), \(gm=nf\), \(f\) divides \(2m\), \(d\) is relatively prime to \(2m/f\), and \(1\leq d<m/f\). Then there is a group \((\textswab{C}_{2m}/\textswab{C}_f; \textswab{C}_{2n}/\textswab{C}_g)_{d},\) consisting of the \(gm=fn\) transformations \[\label{abelian-K39} e^{\mu \pi i/m}xe^{d\nu \pi i/n}\tag{5}\] where \(\mu=0,1,\cdots, \lambda m-1\); \(\nu=0,1,\cdots,n-1\); \(\mu\equiv \nu(mod \;\;\lambda m/f)\), and \(\lambda=\) 1 or 2 according as \(f\) and \(g\) are odd or even (see [21].
The group denoted by \(\langle p,q,r\rangle_{m}\) has the following presentation (see [21]). :
\[\langle A, B, C, Z \;| \;
A^p=B^q=C^r=ABC=Z^m; \;\;[Z,A]=[Z,,B]=[Z,C]=1 \rangle.\] When \(m=1\) we denote simply by \(\langle p,q,r\rangle\). We have the following identifications: \(\langle p,2, 2\rangle={\boldsymbol{D}}_p; \langle 2p,2,2\rangle={\boldsymbol{D}}_{2p} \;(dicyclic \; groups); \langle 2,2,2\rangle={\boldsymbol{Q}}_8\) (\(quaternionic \;group \;of \;order \;
eight\)); \(\langle 3,3,2\rangle={\boldsymbol{T}}; \langle 4,3,2\rangle={\boldsymbol{O}}; \langle 5,3,2\rangle={\boldsymbol{I}}.\) From the description of the finite subgroups given by the classification above, we
promptly read the subgroup \(K'\). More precisely the subgroup \(K'\) is as follows:
finite abelian subgroup
\({\boldsymbol{D}}_p\)
\(\textswab{C}_{2p}\)
\(\textswab{C}_{p}\)
\(\langle p,2,2\rangle={\boldsymbol{D}}_p\)
\(\langle 3,3,2 \rangle={\boldsymbol{T}}\)
\(\langle 2,2,2\rangle= \;quaternionic \;group \;of \;order \; eight={\boldsymbol{Q}}_8\)
\(\langle 4,3,2 \rangle={\boldsymbol{O}}=\langle X, P,Q, R \;\;| \;\;X^3=1, \;\;P^2=Q^ 2=R^2, \;\; XPX^{-1}=Q, \;\;XQX^{-1}=PQ, \;\; RXR^{-1}=X^{-1}, \;\; RPR^{-1}=QP, \;\; RQR^{-1}= Q^{-1}\rangle\)
\(\langle 3,3,2\rangle={\boldsymbol{T}}\)
\(\langle 5,3,2 \rangle={\boldsymbol{I}}\)
The automorphism \(\theta\) of \(K'\) which appears in the group \(K'\rtimes_{\theta}\mathbb{Z}\) is obtained from the conjugation by an element of the group \(K\) of the form \((e,r)\). Here, \(e\) is a generator of the cyclic group \(L\).
Based upon the classification above, we divide the cases into four subfamilies.
Let \(P_1\) be the family of finite subgroups given by (2), (5), (7) and (9).
Let \(P_2\) be the family of finite subgroups given by (1).
Let \(P_3\) be the family of finite subgroups given by (3), (3’) and (4).
Let \(P_4\) be the family of finite subgroups given by (6) and (8).
For this subfamily we make use of the fact that the finite subgroup is also given by a presentation. We will use this presentation to describe \(K'\) and \(\theta\). The particular case of \(K={\boldsymbol{I}}\), case \((9)\), in fact there is no group of the form \({\boldsymbol{I}}\rtimes \mathbb{Z}\) which has the \(R_{\infty}\)-property. This follows from the fact that \({\rm Out}({\boldsymbol{I}})\) is isomorphic to \(\mathbb{Z}_2\), so the result follows from Proposition 3.
Nevertheless we will give a proof which works for all four cases in the subfamily \(P_1\). The subgroups in question are isomorphic to the group \(\langle r,s,t \rangle_m\), which by [21] is the group defined by the following presentation \[\langle A,B,C,Z \;| \; A^r=B^s=C^t=ABC=Z^m \;\;and \; \;Z \;\; commutes \;with \;\;A,B,C \rangle.\] The subgroups \(K'\) are \(\langle p,2,2 \rangle={\boldsymbol{D}}_p, \langle 3,2,2\rangle={\boldsymbol{T}}, \langle 4,3,2 \rangle={\boldsymbol{O}}, \langle 5,3,2\rangle={\boldsymbol{I}}\), respectively. The action \(\theta\) is the conjugation by \(Z\), so we have a central extention. Therefore for each \(K\) in this subfamily, \(\pi_1(U(2)/K) \cong K'\times \mathbb{Z}\) which does not have the \(R_{\infty}\)-property.
By the description of the finite groups in (1), \(K\cong \mathbb{Z}_m \times \mathbb{Z}_g\) (or \(\mathbb{Z}_{2m} \times \mathbb{Z}_{g/2}\)) is Abelian. Therefore the automorphism of \(K'\) is the identity and thus \(\pi_1(U(2)/K) \cong K' \times \mathbb{Z}\) which does not have the \(R_{\infty}\)-property.
For the case \((3)\), the automorphism \(\theta(1)\) is given by \(\theta(1)(x)=x^ {-1}\) where \(\langle x\rangle=\textswab{C}_{2m}\). Since \(\theta(1)\) has order \(2\), it follows from Proposition 3 that \(\textswab{C}_{2m}\rtimes_{\theta}\mathbb{Z}\) does not have the \(R_{\infty}\)-property. The case of \((3' )\) is similar where the two possibilities for \(\theta(1)\) are in fact the same, i.e. \(\theta(1)(x)=x^ {-1}\). For the case \((4)\), let us consider the presentation \({\boldsymbol{D}}_{2p}=\langle x,y|x^{2p}=y^ 2 \;\; yxy^ {-1}=x^ {-1}\rangle\) and the corresponding presentation for \({\boldsymbol{D}}_{p}=\langle z,y | z^{p}=y^2 \;\; yzy^ {-1}=z^ {-1}\rangle\), as well as the inclusion \({\boldsymbol{D}}_{p} \hookrightarrow {\boldsymbol{D}}_{2p}\) by sending \(z\longmapstox^ 2\) and \(y\longmapstoy\). Thus, up to an inner automorphism of \({\boldsymbol{D}}_{p}\), the automorphism \(\theta(1)\) is given by \(z\longmapstoxzx^ {-1}=z\), \(y \longmapstoxyx^ {-1}=xyx^ {-1}y^ {-1}y=x^2y=zy\). The inverse \(\theta(-1)\) is given by \(z\longmapstoz\), \(y \longmapstoz^{-1}y\). Note that the two automorphisms \(\theta(1)\) and \(\theta(-1)\) are conjugated by the automorphism which is conjugation by \(z^ {p-1}\).
If \(K'={\boldsymbol{Q}}_8\) in (6) then \({\rm Out}(K')\cong S_3\), the symmetric group on \(3\) letters. Now \([\theta(1)]\) is of order either \(2\) or \(3\) in \({\rm Out}(K')\) where \([\theta(1)]\) denotes the image of \(\theta(1)\) in \({\rm Out}(K')\). In \(S_3\), we can conclude that \(\theta(1)\) and \(\theta(-1)\) are conjugate. If \(K'={\boldsymbol{T}}\) in (8), then \({\rm Out}(K')\cong \mathbb{Z}_2\) so that \([\theta(1)]=[\theta(-1)]\). Hence, \(K' \rtimes \mathbb{Z}\) do not have property \(R_{\infty}\). Alternatively, one can find automorphisms \(\psi_1:{\boldsymbol{Q}}_8 \longrightarrow{\boldsymbol{Q}}_8\) and \(\psi_2:{\boldsymbol{T}} \longrightarrow{\boldsymbol{T}}\) such that \(\psi_i\circ\psi_i=Id\) for \(i=1,2\). Then it follows that \(\psi_i\circ \theta(1)\circ\psi_i^{-1}=\theta(-1)\) for \(i=1,2\). Hence, \(K' \rtimes \mathbb{Z}\) do not have property \(R_{\infty}\).
We now summarize our discussion with the following.
Theorem 19. For any finite subgroup \(K\) of \(U(2)\), \(\pi_1(U(2)/K)\) does not have property \(R_{\infty}\).
As we have seen in section 2 that there are \(R_{\infty}\)-spaces \(M\) where \(\pi_1(M)\) does not have property \(R_{\infty}\). Here we study the converse question. That is, we ask whether the space \(U(2)/K\) has the \(R_{\infty}\)-property for any finite subgroup \(K\). In order to do that, we consider the locally trivial fibration \(SU(2)/K' \longrightarrowU(2)/K \longrightarrowS^1\) over \(S^1\) induced by the determinant map \(U(2) \longrightarrowS^1\). We will explicitly define a fibre-preserving self homeomorphism \(\eta\) on \(U(2)/K\) which in turn induces the map \(r:S^1\longrightarrowS^1\) of degree \(-1\) given by \(r(z)=\bar z\) (the complex conjugation). Using the commutative diagram 2 , we conclude that \(R(\eta)<\infty\) so \(U(2)/K\) is not an \(R_{\infty}\)-space.
Case (1)
Let \(\kappa \in Sp(1)\) be given by \(\kappa =a+bi+cj+dk\) where \(a,b,c,d \in \mathbb{R}\) and \(\hat{\kappa}=a-bi-dj-ck\). Consider the map \(\Theta : U(2) \longrightarrowU(2)\) given by \((\varepsilon, \kappa) \longrightarrow(\varepsilon^{-1}, \hat{\kappa})\). It is straightforward to check that the map \(\kappa \longmapsto\hat{\kappa}\) is an automorphism of \(Sp(1)\) and thus \(\Theta\) is an automorphism of \(U(2)\). From the description of these subgroups which are finite abelian subgroups of \(U(2)\), every element of \(K\) is of the form 5 . It follows that if \((\varepsilon, \kappa)\in K\) then \(\kappa=a+bi+0j+0k\) where \(a^2+b^2=1\) so that \(\hat{\kappa}=a-bi=\kappa^{-1}\). Thus, \((\varepsilon^{-1}, \hat{\kappa})=(\varepsilon^{-1}, {\kappa}^{-1}) \in K\). In other words, \(K\) is invariant under \(\Theta\). It is easy to see that \(K'\) is also invariant under \(\Theta\). The quotient homeomorphism \(\eta : U(2)/K \longrightarrowU(2)/K\) induces the map \(r:S^1\longrightarrowS^1\) given by the complex conjugation \(r(z)=\bar z\).
Cases (2) - (5), (7) - (9)
In each of these cases, the index \([R:R_K]\) is either \(1\) or \(2\). Suppose \(K\) is given by 4 and \((\varepsilon, \kappa)\in K\). Since \([R:R_K]\in \{1,2\}\), it follows that \((1,\kappa^2)\in K\). Since \((\varepsilon^{-1}, \kappa^{-1})\in K\), \((\varepsilon^{-1},\kappa)=(1,\kappa^2)\cdot (\varepsilon^{-1}, \kappa^{-1})\in K\). Now, the map \(F:U(2) \longrightarrowU(2)\) given by \((\varepsilon, \kappa) \longrightarrow(\varepsilon^{-1}, \kappa)\) is a group isomorphism which leaves \(K\) and \(K'\) invariant. Thus, the quotient homeomorphism \(\eta\) on \(U(2)/K\) induces the complex conjugation \(r(z)=\overline{z}\) on \(S^1\).
Case (6)
Let \(\Theta :U(2)\longrightarrowU(2)\) be the automorphism as in Case (1). The map \(\kappa \longmapsto\hat{\kappa}\) is an automorphism of \(Sp(1)\) whose restriction to \({\boldsymbol{Q}}_8\) is given by \(\psi: i \longmapsto-i\), \(j \longmapsto-k\), \(k \longmapsto-j\). Note that \(R={\boldsymbol{T}}=\{\pm1, \pm i, \pm j, \pm k, (1/2)(\pm 1 \pm i \pm j \pm k)\} \subset Sp(1)\). If \(\kappa \in R_K={\boldsymbol{Q}}_8\) then \(\hat{\kappa} \in {\boldsymbol{Q}}_8\). Suppose \(\kappa \in {\boldsymbol{T}}-{\boldsymbol{Q}}_8\), then \(\kappa=\pm a\pm bi \pm cj \pm dk\) where \(|a|=|b|=|c|=|d|=1/2\). Since \(|c|=|d|\), we conclude that \(\hat{\kappa} =\kappa^{-1}\). Thus \(K\) is invariant under \(\Theta\). Note that in this case \(K'={\boldsymbol{Q}}_8\) so the map \(\Theta\) also keeps \(K'\) invariant and it induces a self homeomorphism \(\eta\) on \(U(2)/K\) whose restriction on \(X=SU(2)/K'\), denoted by \(h\), induces on the fundamental group the automorphism \(\psi\). Moreover, \(\eta\) induces on \(S^1\) the map \(r(z)=\overline{z}\).
Case (3’)
Similar to Case (6), for any \(\kappa =a+bi+cj+dk \in Sp(1)\), we define \(\bar \kappa=a+bi-cj-dk\). Equivalently, if \(\kappa =\alpha + \beta j, \alpha, \beta \in \mathbb{C}\), then \(\bar \kappa = \alpha -\beta j\). The map \(\varphi: \kappa \longmapsto\bar \kappa\) is an automorphism of \(Sp(1)\) so that \(\Phi: U(2) \longrightarrowU(2)\) given by \((\varepsilon, \kappa) \longmapsto(\varepsilon^{-1}, \bar \kappa)\) is an automorphism. To obtain our conclusion as before, it suffices to show that the subgroups \(K\) and \(K'\) are invariant under \(\Phi\). The group \({\boldsymbol{D}}_p\) of order \(4p\) can be embedded in \(Sp(1)\) and is generated by \(u=e^{\pi i/p}\) and \(w=j\) so that \({\boldsymbol{D}}_p=\langle u,w \mid u^{p}=w^2, wuw^{-1}=u^{-1} \rangle\). Now the subgroup \(R_K=\langle u^2\rangle\). Suppose \((\varepsilon, \kappa)\in K\). Since \(\kappa \in R\), \(\kappa=u^sw^t\). Note that \(\varphi(u)=\varphi(e^{\pi i/p})=e^{\pi i/p}=u\) so \(K'\) is invariant under \(\Phi\). It follows that \[\begin{align} \bar \kappa&=\varphi(u^sw^t)=u^s\varphi(w^t) \\ &=u^s(-j)^{t} \end{align}\] which implies that \[\bar \kappa=\begin{cases} \kappa , \text{~for t=0}; \\ u^{s}(-j)=\kappa^{-1} , \text{~for t=1}; \\ \kappa , \text{~for t=2}; \\ u^{s}(j)=\kappa^{-1} , \text{~for t=3}; \\ \end{cases}\] It follows that \((\varepsilon^{-1}, \bar \kappa)=(\varepsilon^{-1}, \kappa^{-1}) \in K\) when \(t\) is odd. When \(t\) is even, \(\kappa^2=u^{2s}\in R_K\) so \((1, \kappa^2)\in K\). In this case, \((\varepsilon^{-1}, \bar \kappa)=(\varepsilon^{-1}, \kappa)=(\varepsilon^{-1}, \kappa^{-1})\cdot (1,\kappa^2)\in K\). Therefore, \(K\) is invariant under \(\Phi\).
Remark 20. In all of the cases except (6) and (3’), we constructed a homeomorphism \(f:U(2)/K \longrightarrowU(2)/K\) with \(R(f)<\infty\) without specifying how we embed \(K\) inside \(U(2)\). For the other two cases, we use the fact that any two isomorphic finite subgroups of \(Sp(1)\) are conjugate in \(Sp(1)\) (see e.g. [20] or [21]) so that the result above is independent of the embedding \(K\hookrightarrow U(2)\) since \(U(2)/K\) is unique.
Thus, we have shown the following
Theorem 21. For any finite subgroup \(K\) of \(U(2)\), the space \(U(2)/K\) does not have property \(R_{\infty}\).
Remark 22. Although Theorem 21 implies Theorem 19, we used different arguments which may be useful in determining the \(R_{\infty}\)-property of \(\pi_1(U(n)/K)\) and of \(U(n)/K\) respectively for \(n=3,4\).