2 Presentation complexes as spines of closed 3-manifolds

The presentation complex (or cellular model) $K=K_{\mathcal {P}}$ of a group presentation $\mathcal {P} = \langle {X}\ |\ {R} \rangle $ is the 2-complex with one 0-cell O, a loop at O for each generator $x \in X$ and a 2-cell for each relator (the boundary of that 2-cell spelling the relator). If N is a regular neighborhood of O then $K \cap \partial N$ is a 1-dimensional cell complex called the Whitehead graph or link graph of $\mathcal {P}$ . Thus the Whitehead graph of $\mathcal {P}$ is the graph with $2|X|$ vertices $v_x$ , $v_x'$ ( $x \in X$ ) and an edge $(v_x, v_y)$ (resp. $(v_x', v_y')$ , $(v_x, v_y')$ ) for each occurrence of a cyclic subword $xy^{-1}$ (resp. $x^{-1}y$ , $(xy)^{\pm 1}$ ) in a relator $r \in R$ . The reduced Whitehead graph is the graph obtained from the Whitehead graph by replacing all multiedges between two vertices by a single edge. We say that a group presentation $\mathcal {P}$ is the spine of a closed 3-manifold M if there exists a 3-ball $B^3\subset M$ such that $M-\mathring {B}^3$ collapses onto $K_{\mathcal {P}}$ (where $\mathring {B}^3$ denotes the interior of $B^3$ ). Since $K_{\mathcal {P}}$ is connected, the manifold M is necessarily connected.

Suppose that a group presentation $\mathcal {P}=\langle {X}\ |\ {R} \rangle $ with an equal number of generators and relators is a spine of a closed, oriented 3-manifold. Then (see [Reference Seifert and Threlfall49, Chapter 9], [Reference Neuwirth45], [Reference Sieradski50, p. 125]) there is a $3$ -complex C whose set of faces consists of precisely one pair of oppositely oriented faces $F_r^+,F_r^-$ for each relator $r\in R$ , whose boundaries spell r. Let $M_0$ denote the 3-complex obtained from C by identifying the faces $F_r^+,F_r^-$ ( $r\in R$ ). Then $M_0$ is a closed, connected, oriented pseudo-manifold. The Seifert–Threlfall condition for $M_0$ to be a manifold M is that its Euler characteristic is zero [Reference Seifert and Threlfall49, Theorem I, Section 60]. In this case the cell structure on $M=M_0$ has one vertex, one 3-cell, and 2-skeleton homeomorphic to the presentation complex $K_{\mathcal {P}}$ of $\mathcal {P}$ . The Whitehead graph $\Gamma $ of $\mathcal {P}$ is the link of the single vertex of $K_{\mathcal {P}}$ , and so embeds in the link of the single vertex of M. The link of the vertex of M is the 2-sphere, and so $\Gamma $ has a planar embedding on this sphere. When $\Gamma $ is connected the 3-complex C is a polyhedron $\pi $ bounding a 3-ball. Given a planar embedding of $\Gamma $ there is a one-to-one correspondence between the faces F of this embedding and the vertices $u_F$ of $\pi $ . This correspondence maps a face of degree d to a vertex of degree d of $\pi $ . Moreover, if the vertices in the face read cyclically around the face are $w_1,\ldots ,w_d$ where $w_i\in \{v_x,v_x'\ (x\in X)\}$ then the arcs $e_1,\ldots ,e_d$ incident to $u_F$ , read cyclically, are directed toward $u_F$ if $w_i=v_x$ and away from $u_F$ if $w_i=v_x'$ , and the arc $e_j$ , corresponding to vertex $w_j=v_x$ or $v_x'$ , is labeled x.

3 Cyclic presentations $\mathcal {H}(r,n)$ as spines of type (I.5)

For $n>1,r \geq 1$ let

$$\begin{align*}\mathcal{H}(r,n)=\mathcal{G}_n(x_0x_1\ldots x_{r-1})\end{align*}$$

and let $H(r,n)$ be the group it defines. By [Reference Umar54, Theorems 2 and 3], $H(r,n)$ is finite if and only if $(n,r)=1$ , in which case $H(r,n) \cong \mathbb{Z}_r$ . The Whitehead graph of $\mathcal {H}(r,n)$ is connected if and only if $r>1$ and $(n,r)=1$ , and in this case it is of type (I.5), as shown in Figure 1, where (as also in Figure 5) the edge labels denote their multiplicities, a vertex label i denotes $v_{x_i}$ , and a vertex label $\overline {i}$ denotes $v_{x_i}'$ .

Figure 1

Whitehead graph for $\mathcal {H}(r,n)$ (where $(r,n)=1$ ).

As a warm up to our main result, Theorem A, in this section we show that, when the Whitehead graph is connected (that is, if $n, r>1$ and $(n,r)=1$ ), then $\mathcal {H}(r,n)$ is a spine of a closed, oriented 3-manifold M. The positive presentations of type $\mathfrak {Z}$ , considered in [Reference McDermott42], also have Whitehead graphs of type (I.5) and, subject to certain hypotheses, they are spines of closed, oriented manifolds N [Reference McDermott42, Theorem 8]. The manifolds M can often be distinguished from the manifolds N by comparing their fundamental groups $\pi _1(M)=H(r,n)\cong \mathbb{Z}_r$ and $\pi _1(N)$ . For example, by [Reference McDermott42, Proposition 6], $\pi _1(N)$ is often infinite and [Reference McDermott42, Table 1] provides computational evidence that many of the finite groups $\pi _1(N)$ are non-cyclic.

Consider the polyhedron in Figure 2 with face pairing given by identifying the faces $F_i^+,F_i^-$ ( $0\leq i<n)$ . We shall show that the identification of faces results in a 3-complex M whose 2-skeleton is the presentation complex of $\mathcal {H}(r,n)$ , so M has one 0-cell, n 1-cells, n 2-cells, and one 3-cell, so is a manifold by [Reference Seifert and Threlfall49, Theorem I, Section 60].

Figure 2

Face pairing polyhedron for $\mathcal {H}(r,n)$ .

Let $[S,N]_i$ denote the arc of the face $F_i^+$ with initial vertex S and terminal vertex N. All the arcs labeled $x_0$ are contained in the following cycle (of length r):

$$ \begin{align*} [S,N]_0 &\stackrel{F_{n-(r-1)}}{\longrightarrow} [w_{n-(r-1)}^2,w_{n-(r-1)}^1] \stackrel{F_{n-(r-2)}}{\longrightarrow} [w_{n-(r-2)}^3,w_{n-(r-2)}^2]\\ & \stackrel{F_{n-(r-3)}}{\longrightarrow} [w_{n-(r-3)}^4,w_{n-(r-3)}^3] \stackrel{F_{n-(r-4)}}{\longrightarrow} \cdots \stackrel{F_{n-2}}{\longrightarrow} [w_{n-2}^{r-1},w_{n-2}^{r-2}]\\ & \stackrel{F_{n-1}}{\longrightarrow} [N,w_{n-1}^{r-1}] \stackrel{F_{0}}{\longrightarrow} [S,N]_0. \end{align*} $$

Figure 3

Heegaard diagram for the manifold $M(r,n)$ .

Figure 4

Heegaard diagram for the manifold $M(r,n)/\rho $ .

All the vertices that are a vertex (either initial or terminal) of an arc labeled $x_0$ are contained in the (induced) cycle of the initial vertices in the arcs above, so in the resulting complex M these vertices are identified. In particular, vertices $N,S$ are identified, and (by comparing initial vertices) for $1< j < r$ vertices $w_{n-j}^{r-j+1}$ are identified with S. Therefore, for each $0\leq i<n$ , $1< j<r$ the vertex $w_i^j=\theta ^{i+r-j+1}(w_{n-(r-j+1)}^j)$ is identified with S. Therefore all the vertices of the polyhedron are identified. Thus the quotient M has one 3-cell, n 2-cells, n 1-cells, and one $0$ -cell, and since the boundaries of the $2$ -cells spell the relators of $\mathcal {H}(r,n)$ , it follows that M is a closed, oriented 3-manifold, and $\mathcal {H}(r,n)$ is a spine of M.

As in (for example) [Reference Vesnin and Kozlovskaya57, proof of Proposition 2], a Heegaard diagram for M arising from the face pairing polyhedron is given in Figure 3. This diagram has a rotational symmetry $\rho $ of order n cyclically permuting the faces $F_i^+\rightarrow F_{i+r}^+$ and $F_i^-\rightarrow F_{i+r}^-$ . The quotient of M by $\rho $ is a 3-orbifold in which the image of the rotation axis is a singular set. This yields the Heegaard diagram for $M/\rho $ in Figure 4, which is the canonical diagram of the lens space $L(r,1)$ . Hence the manifold M is an n-fold cyclic branched cover of $L(r,1)$ .

Figure 5

Whitehead graph for $\mathcal {G}^{k/l}(n,f)$ (where $\lambda = 2l+k-3$ , n even and $fk\equiv 0$ or $2\bmod n$ ).

4 Cyclic presentations $\mathcal {G}^{k/l}(n,f)$ as spines of type (II.7)

For $n\geq 2$ , $k,l\geq 1$ , $0\leq f<n$ , let $\mathcal {G}^{k/l}(n,f)$ be the cyclic presentation

$$\begin{align*}\mathcal{G}_n((y_0y_f\ldots y_{(l-1)f}) (y_{lf+1}y_{(l+1)f+1}\ldots y_{(l+(k-1))f+1}) (y_2y_{2+f}\ldots y_{2+(l-1)f})^{-1})\end{align*}$$

and let $G^{k/l}(n,f)$ be the group it defines. When $n\geq 4$ , the Whitehead graph of $\mathcal {G}^{k/l}(n,f)$ is planar if and only if n is even and either $fk\equiv 0 \bmod n$ or $fk\equiv 2 \bmod n$ , in which case it is of type (II.7) if $(k,l)\neq (1,1)$ and is of type (II.11) if $k=l=1$ ; see Figure 5. Our main result is the following.

Our motivation for studying these presentations (and, in particular, under the condition $fk\equiv 0 \bmod n$ but not under the condition $fk\equiv 2 \bmod n$ ) stems from the following connection to the Fractional Fibonacci groups. The presentations $\mathcal {G}^{k/l}(n,0)$ are the Fractional Fibonacci group presentations

$$\begin{align*}\mathcal{F}^{k/l}(n)= \mathcal{G}_n(x_0^lx_1^kx_2^{-l})\end{align*}$$

introduced in [Reference Maclachlan and Duncan40, Reference Maclachlan and Reid41, Reference Vesnin and Kim55, Reference Vesnin and Kim56], generalizing the Fibonacci group presentations $\mathcal {F}^{1/1}(n)$ . For even n and coprime integers $k,l\geq 1$ , the Fractional Fibonacci group $F^{k/l}(n)$ is a 3-manifold group [Reference Maclachlan and Duncan40, Theorem 4.1], [Reference Maclachlan and Reid41, Section 4], [Reference Vesnin and Kim55, Section 2], [Reference Vesnin and Kim56, Section 2]. If $n\geq 2$ is even then $\mathcal {F}^{1/1}(n)$ was shown to be a spine of a closed, oriented, 3-manifold in [Reference Cavicchioli and Spaggiari16],[Reference Helling, Kim and Mennicke22],[Reference Hilden, Lozano and Montesinos-Amilibia23],[Reference Hilden, Lozano and Montesinos-Amilibia24]. More generally, if $k,l\geq 1$ are coprime and n is even, then setting $s_i=q_i=l$ , $p_i=-r_i=k$ in [Reference Ruini and Spaggiari47, Theorem 6] gives that $\mathcal {F}^{k/l}(n)$ is a spine of a closed, oriented, 3-manifold. If n is odd then $\mathcal {F}^{k/l}(n)$ is not a spine, since its Whitehead graph is non-planar. The question as to when, for odd n, $F^{k/l}(n)$ is a 3-manifold group is considered in [Reference Howie and Williams26, Theorem 3] for the case $k=l=1$ , and in [Reference Chinyere and Williams17, Theorem 6.2] for the case $l=1$ and the case $n=3$ . The abelianization $F^{k/1}(n)^{\mathrm {ab}}$ is obtained in [Reference Maclachlan and Reid41, Lemma 1], [Reference Noferini and Williams46, Corollary 4.3].

The shift extension $E^{k/l}(n)=F^{k/l}(n) \rtimes _\theta \mathbb{Z}_n$ of $F^{k/l}(n)$ by $\mathbb{Z}_n=\langle {t}\ |\ {t^n} \rangle $ has the presentation

(1) $$ \begin{align} E^{k/l}(n)=\langle {x,t}\ |\ {t^n, x^ltx^ktx^{-l}t^{-2}} \rangle, \end{align} $$

where the second relator is obtained by rewriting the relators of $F^{k/l}(n)$ in terms of the substitutions $x_i=t^ixt^{-i}$ (see, for example, [Reference Johnson, Wamsley and Wright31, Theorem 4]). As set out in [Reference Bogley5, Section 2], for $0\leq f<n$ such that $fk\equiv 0 \bmod n$ , there is a retraction $\nu ^f:E^{k/l}(n) \rightarrow \mathbb{Z}_n=\langle {t}\ |\ {t^n} \rangle $ given by $\nu ^f(t)=t$ , $\nu ^f(x)=t^f$ . The kernel, $\mathrm {ker} (\nu ^f)$ , has a presentation with generators $y_i=t^ixt^{-(i+f)}$ ( $0\leq i<n$ ) and relators that are rewrites of conjugates of the second relator of $E^{k/l}(n)$ by powers of t, and so has the cyclic presentation $\mathcal {G}^{k/l}(n,f)$ . In particular (under the condition $fk\equiv 0 \bmod n$ ),

$$ \begin{align*} \mathcal{G}^{k/1}(n,0) &= \mathcal{G}_n(y_0 y_1^{k} y_2^{-1})=\mathcal{F}^{k/1}(n),\\ \mathcal{G}^{1/l}(n,f) &=\mathcal{G}^{1/l}(n,0)=\mathcal{G}_n(y_0^l y_1 y_2^{-l})=\mathcal{F}^{1/l}(n),\\ \mathcal{G}^{5/2}(n,f) &= \mathcal{G}_n( (y_0y_f) (y_{2f+1}y_{3f+1}y_{4f+1}y_1y_{f+1}) (y_2y_{2+f})^{-1} ),\\ \mathcal{G}^{2/5}(n,f) &= \mathcal{G}_n((y_0y_fy_{0}y_{f}y_{0}) (y_{f+1}y_1) ( y_2y_{2+f}y_2y_{2+f}y_2)^{-1}). \end{align*} $$

The shift extension

$$ \begin{align*} G^{k/l}(n,f) \rtimes_\theta \mathbb{Z}_n &= \langle {y,t}\ |\ {t^n, (yt^f)^lt(yt^f)^kt^{-fk+1}(yt^f)^{-l}t^{-2}} \rangle\\ &= \langle {x,t}\ |\ {t^n, x^ltx^kt^{-fk+1}x^{-l}t^{-2}} \rangle \end{align*} $$

(by setting $x=yt^f$ and eliminating y). In the case $fk\equiv 0 \bmod n$ , this coincides with $E^{k/l}(n)$ , which is independent of the value of f. This implies, for example, that the order of $G^{k/l}(n,f)$ is independent of f, the shift dynamics of $G^{k/l}(n,f)$ are identical for all values of f [Reference Bogley5, Lemma 2.2], and that $G^{k/l}(n,f)$ is (non-elementary) word hyperbolic if and only if $F^{k/l}(n)$ is (non-elementary) word hyperbolic. In Theorem 4.5 we similarly show that, for fixed $k,l,n$ , if two groups $G^{k/l}(n,f_1),G^{k/l}(n,f_2)$ sharing a shift extension are fundamental groups of closed, connected, orientable 3-manifolds then either both manifolds are hyperbolic, or neither are. Observe that (under the hypothesis $fk\equiv 0\bmod n$ ), as in Figure 5, the Whitehead graph of $\mathcal {G}^{k/l}(n,f)$ has vertices $v_{y_i},v_{y_i}'$ and edges $(v_{y_i},v_{y_{i+1}})$ , $(v_{y_i}',v_{y_{i+2}}')$ , $(v_{y_i},v_{y_{i+f}}')$ (of multiplicity $2l+k-3$ ), and so the Whitehead graph of $\mathcal {G}^{k/l}(n,f)$ is obtained from that of $\mathcal {F}^{k/l}(n)=\mathcal {G}^{k/l}(n,0)$ by replacing each edge $(v_{y_i},v_{y_{j}}')$ by the edge $(v_{y_i},v_{y_{j+f}}')$ .

In contrast, for fixed $k,l,n$ , two groups $G^{k/l}(n,f_1), G^{k/l}(n,f_2)$ sharing a shift extension will typically be non-isomorphic. We give examples of this in Example 4.2 and Lemma 4.3. Further, in the case $fk\equiv 0 \bmod n$ , we also have that the presentation $\mathcal {G}^{k/l}(n,f)$ has a planar Whitehead graph. Since the group $G^{k/l}(n,f)$ shares its shift extension with that of $F^{k/l}(n)$ , these properties combined suggest that the geometric properties of $F^{k/l}(n)$ and $\mathcal {F}^{k/l}(n)$ may be inherited by $G^{k/l}(n,f)$ and $\mathcal {G}^{k/l}(n,f)$ . However, while [Reference Ruini and Spaggiari47, Theorem 6] implies that $\mathcal {F}^{k/l}(n)$ is a spine for all coprime $k,l$ and even $n\geq 2$ , Lemma 4.4 will show that, additionally, f must be even for $\mathcal {G}^{k/l}(n,f)$ to be a spine of a closed, oriented, 3-manifold. This demonstrates that, generally, given two retractions $\nu ^{f_1},\nu ^{f_2}:\langle {x,t}\ |\ {t^n,W(x,t)} \rangle \rightarrow \langle {t}\ |\ {t^n} \rangle $ whose kernels have cyclic presentations $\mathcal {P}_1,\mathcal {P}_2$ , as in Remark 4.1, it is not the case that $\mathcal {P}_1$ is a spine of a closed, oriented 3-manifold if and only if $\mathcal {P}_2$ is such a spine. In the other planar case, $fk\equiv 2\bmod n$ , the groups ${G}^{k/l}(n,f)$ do not share a shift extension with ${F}^{k/l}(n)$ , so there is no apriori reason to expect similar geometric properties to hold.

In Lemma 4.3, we use the fact that the order of the abelianization $G^{k/l}(n,f)^{\mathrm {ab}}$ is given by $|\mathrm {Res}(p_f(t),t^n-1)|$ , if this is non-zero, and is infinite otherwise, where

$$\begin{align*}p_f(t)= (1-t^2)(1+t^f+\cdots +t^{(l-1)f}) + t^{lf+1}(1+t^f+\cdots + t^{(k-1)f})\end{align*}$$

is the representer polynomial of $G^{k/l}(n,f)$ , and $\mathrm {Res}(\cdot , \cdot )$ denotes the resultant [Reference Johnson32, p. 82]. As we will only be interested in the absolute values of resultants (and not the sign), to avoid repetitive use of modulus signs we will take $\mathrm {Res}(\cdot , \cdot )$ to mean $|\mathrm {Res}(\cdot , \cdot )|$ .

Proof For any $0\leq f<n$

(2) $$ \begin{align} \mathrm{Res}(p_{f}(t),t^n-1) = \mathrm{Res}(p_{f}(t),t^{n/2}-1)\cdot \mathrm{Res}(p_f(t),t^{n/2}+1). \end{align} $$

The hypotheses imply that l is odd so we have

$$ \begin{align*} \mathrm{Res}(p_{n/2}(t),t^{n/2}-1) &= \mathrm{Res}(l(1-t^2) + kt,t^{n/2}-1)\\ &= \mathrm{Res}(p_0(t),t^{n/2}-1), \end{align*} $$

and, setting $\alpha ,\bar {\alpha }=(k\pm \sqrt {k^2+4l^2})/(2l)$ ,

$$ \begin{align*} \mathrm{Res}(p_{0}(t),t^{n/2}+1) &=\mathrm{Res}\left( l(t-\alpha)(t-\bar{\alpha}), t^{n/2}+1\right)\\ &= l^{n/2} (\alpha^{n/2}+1)(\bar{\alpha}^{n/2}+1)\\ &= l^{n/2}\left((-1)^{n/2}+1+(\alpha^{n/2} + \bar{\alpha}^{n/2})\right). \end{align*} $$

On the other hand,

$$ \begin{align*} \mathrm{Res}(p_{n/2}(t),t^{n/2}+1) &=\mathrm{Res}\left( l (1-t^2) + kt(1+t^{n/2})/2, t^{n/2}+1\right)\\ &=l^{n/2}\cdot 2(1+(-1)^{n/2}). \end{align*} $$

Therefore $\mathrm {Res}(p_0(t),t^{n/2}+1)\neq \mathrm {Res}(p_{n/2}(t),t^{n/2}+1)$ so, by equation (2), $\mathrm {Res}(p_0(t), t^{n}-1)\neq \mathrm {Res}(p_{n/2}(t),t^{n}-1)$ . Hence $|G^{k/l}(n,0)^{\mathrm {ab}}| \neq |G^{k/l}(n,n/2)^{\mathrm {ab}}|$ , and the result follows.

Proof First note that if f is odd, then the hypotheses imply that l is odd. The Whitehead graph of $\mathcal {G}^{k/l}(n,f)$ has a planar embedding, as shown in Figure 5. It has two $n/2$ -gons $v_{y_0}'-v_{y_2}'-\cdots -v_{y_{n-2}}'-v_{y_0}'$ and $v_{y_1}'-v_{y_3}'-\cdots -v_{y_{n-1}}'-v_{y_1}'$ , n 3-gons $v_{y_i}-v_{y_{i+1}}-v_{y_{i+f+1}}'-v_{y_i}$ , n 4-gons $v_{y_i}-v_{y_{i+1}}-v_{y_{i+f+2}}'-v_{y_{i+f}}'-v_{y_i}$ , and $\lambda n$ 2-gons $v_{y_i}-v_{y_{i+f+1}}'-v_{y_i}$ , where $\lambda = 2l+k-3$ . It follows that the reduced Whitehead graph is unique up to self homeomorphism of $S^2$ .

Figure 6

Face pairing polyhedron for $\mathcal {G}^{k/1}(n,f)$ .

Figure 7

Face pairing polyhedron for $\mathcal {G}^{5/2}(n,f)$ .

Figure 8

Face pairing polyhedron for $\mathcal {G}^{1/l}(n,0)=\mathcal {F}^{1/l}(n)$ .

Figure 9

Face pairing polyhedron for $\mathcal {G}^{2/5}(n,f)$ .

If $\mathcal {G}^{k/l}(n,f)$ is a spine of a closed, oriented, 3-manifold then in a putative corresponding face-pairing polyhedron, there are two degree $n/2$ source vertices, $N,S$ , say, where N has outgoing arcs in cyclic order $y_0,y_2,y_4,\ldots , y_{n-2}$ and S has outgoing arcs in cyclic order $y_1,y_3,y_5,\ldots , y_{n-1}$ . The 2-cells incident to N are therefore, in cyclic order, $F_0^-,F_2^-,F_4^-, \dots , F_{n-2}^-$ , where the boundary of $F_i^-$ reads, anticlockwise, the i’th relator of $\mathcal {G}^{k/l}(n,f)$ . The arc labeled $y_2$ , with initial vertex N, is the first of a path $y_2-y_{2+f}-\cdots -y_{2+(l-1)f}$ . Let v denote the terminal vertex of the last arc in this path. Then v has two incoming arcs labeled $y_{(l-1)f+1},y_{(l-1)f+2}$ , and an outgoing arc labeled $y_{lf+3}$ . Therefore v is a degree 4 vertex, and its remaining arc is outgoing, and labeled $y_{lf+1}$ . The outgoing arcs $y_{lf+1},y_{lf+3}$ then bound the $F_{lf+1}^-$ face. Since $l,f$ are odd, $lf+1$ is even, so the face pairing contains two $F_{lf+1}^-$ faces, a contradiction.

In Figures 6, 7, 8, and 9, we present face-pairing polyhedra with n pairs of faces $F_i^+,F_i^-$ ( $0\leq i<n$ ) of opposite orientation whose boundaries spell the defining relators of the relevant presentation $\mathcal {G}^{k/l}(n,f)$ . In the proof of Theorem A, we show that the identification of faces $F_i^+,F_i^-$ results in a 3-complex that satisfies the Seifert–Threlfall condition, and so is a manifold whose spine is the presentation complex of $\mathcal {G}^{k/l}(n,f)$ . (Note that, in these figures, the condition that f is even is necessary for each relator to appear as the label of a pair of oppositely oriented faces.) The diagrams have different forms depending on whether $l>k$ or $k>l$ . The figures should provide the enthusiastic reader with sufficient information to construct face-pairing polyhedra in the general cases. The values $\{k,l\}=\{2,5\}$ were selected so that $k,l,|k-l|$ are distinct, to make it straightforward to infer paths of lengths $k,l$ , and $|k-l|$ . (For example, in Figure 7 the path from $v_{2f}$ to $u_2$ has length $3=k-l$ , and in Figure 9 the path from $w_f$ to $v_{f-1}$ has length $3=l-k$ .) We expect the Seifert–Threlfall condition to hold in the general case for coprime $k,l$ ; however, the corresponding analysis to check this would be highly technical. Since our goal is to exhibit new cyclic presentations that arise as spines of closed 3-manifolds, whose Whitehead graphs are of type (II.7), rather than to be exhaustive, we have not sought to do this.

Proof of Theorem A

We may assume n is even and $fk\equiv 0$ , for otherwise the Whitehead graph of $\mathcal {G}^{k/l}(n,f)$ is not planar [Reference Howie and Williams27] (noting that $fk\not \equiv 2 \bmod n$ , by hypothesis). Moreover, by Lemma 4.4 we may assume f is even.

Case 1: $(k,l)=(k,1)$ . Consider the face pairing polyhedron depicted in Figure 6 (where the arc labels can be deduced from the position of the unique source in the boundary of each face) with face pairing given by identifying the faces $F_i^+,F_i^-$ ( $0\leq i<n)$ . We shall show that the identification of faces results in a 3-complex M whose $2$ -skeleton is the presentation complex of $\mathcal {G}^{k/l}(n,f)$ , so M has one $0$ -cell, n 1-cells, n 2-cells, and one $3$ -cell, so is a manifold by [Reference Seifert and Threlfall49, Theorem I, Section 60].

All the arcs labeled $x_0$ are contained in the following cycle:

$$ \begin{align*} [N,u_0] &\stackrel{F_0}{\longrightarrow} [u_{1-f},v_0] \stackrel{F_{-1}}{\longrightarrow} [w_{f}^{k-2},u_1] \stackrel{F_{f-1}}{\longrightarrow} [w_{2f}^{k-3},w_{2f}^{k-2}]\\ &\stackrel{F_{2f-1}}{\longrightarrow} [w_{3f}^{k-4},w_{3f}^{k-3}] \stackrel{F_{3f-1}}{\longrightarrow} \cdots \stackrel{F_{(k-3)f-1}}{\longrightarrow} [w_{(k-2)f}^{1},w_{(k-2)f}^{2}]\\ &\stackrel{F_{(k-2)f-1}}{\longrightarrow} [v_{(k-1)f-2},w_{(k-1)f}^{1}] \stackrel{F_{(k-1)f-1}}{\longrightarrow} [v_{(k-1)f-3},w_{0}^1] \stackrel{F_{-2}}{\longrightarrow} [N,u_0]. \end{align*} $$

Therefore, in the resulting complex M all the arcs labeled $x_0$ are identified. Moreover, all the vertices that are a vertex (either initial or terminal) of an arc labeled $x_0$ are contained in the (induced) cycle of the initial vertices in the arcs above, so these vertices are identified in M. By applying the shift $\theta ^{2j}$ ( $0\leq j<n/2$ ) to the above, all the arcs labeled $x_{2j}$ are identified and all the vertices of such arcs are identified. That is, the vertices $N,u_{2j+(1-f)},w_{2j+f}^{k-2},w_{2j+2f}^{k-3},\dots , w_{2j+(k-2)f}^1, v_{2j+(k-1)f-2}$ are identified; equivalently, N, $u_{2j+1}, w_{2j}^1,\dots w_{2j}^{k-2}, v_{2j}$ ( $0\leq j<n/2$ ) are identified.

All the arcs labeled $x_1$ are contained in the following cycle:

$$ \begin{align*} [S,u_1] &\stackrel{F_1}{\longrightarrow} [u_{2-f},v_1] \stackrel{F_{0}}{\longrightarrow} [w_{f+1}^{k-2},u_2] \stackrel{F_{f}}{\longrightarrow} [w_{2f+1}^{k-3},w_{2f+1}^{k-2}]\\ &\stackrel{F_{2f}}{\longrightarrow} [w_{3f+1}^{k-4},w_{3f+1}^{k-3}] \stackrel{F_{3f}}{\longrightarrow} \cdots \stackrel{F_{(k-3)f}}{\longrightarrow} [w_{(k-2)f+1}^{1},w_{(k-2)f+1}^{2}]\\ &\stackrel{F_{(k-2)f}}{\longrightarrow} [v_{(k-1)f-1},w_{(k-1)f+1}^{1}] \stackrel{F_{(k-1)f}}{\longrightarrow} [v_{(k-1)f-2},w_{1}^1] \stackrel{F_{-2}}{\longrightarrow} [S,u_1]. \end{align*} $$

Therefore, in M all the arcs labeled $x_1$ are identified. Moreover, all the vertices that are a vertex (either initial or terminal) of an arc labeled $x_1$ are contained in the (induced) cycle of the initial vertices in the arcs above, so these vertices are identified in M. By applying the shift $\theta ^{2j}$ ( $0\leq j<n/2$ ) to the above, all the arcs labeled $x_{2j+1}$ are identified and all the vertices of such arcs are identified. That is, the vertices $S,u_{2j+(2-f)},w_{2j+f+1}^{k-2},w_{2j+2f+1}^{k-3},\dots , w_{2j+(k-2)f+1}^1, v_{2j+(k-1)f-1}$ are identified; that is S, $u_{2j}, w_{2j+1}^1,\dots w_{2j+1}^{k-2}, v_{2j+1}$ ( $0\leq j<n/2$ ) are identified. Moreover, examining the terminal vertices of the first cycle above we see that $u_0$ and $u_1$ are identified. Therefore the vertices induced from the first cycle are identified with the vertices from the second cycle, so all vertices of the polyhedron are identified. Thus M has one 3-cell, n 2-cells, n 1-cells, and one 0-cell, and since the boundaries of the $2$ -cells spell the relators of $\mathcal {G}^{1/l}(n,0)$ , it follows that $\mathcal {G}^{1/l}(n,0)$ is a spine of a closed 3-manifold, as required.

Case 2: $(k,l)=(5,2)$ . Consider the face pairing polyhedron depicted in Figure 7 (where the arc labels can be deduced from the position of the unique source in the boundary of each face) with face pairing given by identifying the faces $F_i^+,F_i^-$ ( $0\leq i<n)$ . All the arcs labeled $x_0$ are contained in the following cycle:

$$ \begin{align*} [N,s_0] &\stackrel{F_0}{\longrightarrow} [u_{1-2f},t_0] \stackrel{F_{-1}}{\longrightarrow} [w_{2f-1}^1,w_{2f-1}^2] \stackrel{F_{2f-1}}{\longrightarrow} [r_{2f-1},v_{4f-1}]\\ &\stackrel{F_{4f-2}}{\longrightarrow} [s_{4f},u_{4f}] \stackrel{F_{4f}}{\longrightarrow} [t_{4f},v_{4f}] \stackrel{F_{4f-1}}{\longrightarrow} [w_{f-1}^2,u_{-f+1}]\\ &\stackrel{F_{f-1}}{\longrightarrow} [v_{3f-1},w_{3f-1}^1] \stackrel{F_{3f-1}}{\longrightarrow} [u_{3f-1},r_{3f-1}] \stackrel{F_{-2}}{\longrightarrow} [N,s_0], \end{align*} $$

and all the arcs labeled $x_1$ are contained in the following cycle:

$$ \begin{align*} [S,s_1] &\stackrel{F_1}{\longrightarrow} [u_{2-2f},t_1] \stackrel{F_{0}}{\longrightarrow} [w_{2f}^1,w_{2f}^2] \stackrel{F_{2f}}{\longrightarrow} [r_{2f},v_{4f}]\\ & \stackrel{F_{4f-1}}{\longrightarrow} [s_{4f+1},u_{4f+1}] \stackrel{F_{4f+1}}{\longrightarrow} [t_{4f+1},v_{4f+1}] \stackrel{F_{4f}}{\longrightarrow} [w_{f}^2,u_{-f+2}]\\ & \stackrel{F_{f}}{\longrightarrow} [v_{3f},w_{3f}^1] \stackrel{F_{3f}}{\longrightarrow} [u_{3f},r_{3f}] \stackrel{F_{-1}}{\longrightarrow} [S,s_1]. \end{align*} $$

The proof then proceeds as in Case 1.

Case 3: $(k,l)=(1,l)$ . Consider the face pairing polyhedron depicted in Figure 8. All the arcs labeled $x_0$ are contained in the following cycle:

$$ \begin{align*} [N,u_0^{1}] &\stackrel{F_0}{\longrightarrow} [w_0,t_0^1] \stackrel{F_{-2}}{\longrightarrow} [u_0^1,u_0^2] \stackrel{F_{0}}{\longrightarrow} [t_0^1,t_0^2] \stackrel{F_{-2}}{\longrightarrow} [u_0^2,u_0^3]\\ &\stackrel{F_{0}}{\longrightarrow} [t_0^2,t_0^3] \stackrel{}{\longrightarrow} \cdots \stackrel{F_{-2}}{\longrightarrow} [u_0^{l-2},u_0^{l-1}] \stackrel{F_{0}}{\longrightarrow} [t_0^{l-2},v_{-1}]\\ &\stackrel{F_{-2}}{\longrightarrow} [u_0^{l-1},w_{-1}] \stackrel{F_{0}}{\longrightarrow} [v_{-1},v_0] \stackrel{F_{-1}}{\longrightarrow} [w_{-2},w_0] \stackrel{F_{-2}}{\longrightarrow} [N,u_0^1], \end{align*} $$

and all the arcs labeled $x_1$ are contained in the following cycle:

$$ \begin{align*} [S,u_1^{1}] &\stackrel{F_1}{\longrightarrow} [w_1,t_1^1] \stackrel{F_{-1}}{\longrightarrow} [u_1^1,u_1^2] \stackrel{F_{1}}{\longrightarrow} [t_1^1,t_1^2] \stackrel{F_{-1}}{\longrightarrow} [u_1^2,u_1^3]\\ &\stackrel{F_{1}}{\longrightarrow} [t_1^2,t_1^3] \stackrel{}{\longrightarrow} \cdots \stackrel{F_{-1}}{\longrightarrow} [u_1^{l-2},u_1^{l-1}] \stackrel{F_{1}}{\longrightarrow} [t_1^{l-2},v_{0}]\\ &\stackrel{F_{-1}}{\longrightarrow} [u_1^{l-1},w_{0}] \stackrel{F_{1}}{\longrightarrow} [v_{0},v_1] \stackrel{F_{0}}{\longrightarrow} [w_{-1},w_1] \stackrel{F_{-1}}{\longrightarrow} [S,u_1^1]. \end{align*} $$

The proof then proceeds as in Case 1.

Case 4: $(k,l)=(2,5)$ . Consider the face pairing polyhedron depicted in Figure 9. All the arcs labeled $x_0$ are contained in the following cycle:

$$ \begin{align*} [N,u_0^{1}] &\stackrel{F_0}{\longrightarrow} [w_0,t_0^1] \stackrel{F_{-2}}{\longrightarrow} [u_{0}^2,u_{0}^3] \stackrel{F_{0}}{\longrightarrow} [t_0^2,v_{-1}] \stackrel{F_{-2}}{\longrightarrow} [u_0^4,w_{f-1}]\\ &\stackrel{F_{0}}{\longrightarrow} [s_{-1},v_0] \stackrel{F_{-1}}{\longrightarrow} [r_{f-2},w_f] \stackrel{F_{f-2}}{\longrightarrow} [u_f^{1},u_{f}^2] \stackrel{F_{f}}{\longrightarrow} [t_f^{1},t_{f}^2]\\ & \stackrel{F_{f-2}}{\longrightarrow} [u_{f}^3,u_f^4] \stackrel{F_{f}}{\longrightarrow} [v_{f-1},s_{f-1}] \stackrel{F_{f-1}}{\longrightarrow} [w_{2f-2},r_{2f-2}] \stackrel{F_{2f-2}}{\longrightarrow} [N,u_{2f}^1], \end{align*} $$

and all the arcs labeled $x_1$ are contained in the following cycle:

$$ \begin{align*} [S,u_1^{1}] &\stackrel{F_1}{\longrightarrow} [w_1,t_1^1] \stackrel{F_{-1}}{\longrightarrow} [u_{1}^2,u_{1}^3] \stackrel{F_{1}}{\longrightarrow} [t_1^2,v_{0}] \stackrel{F_{-1}}{\longrightarrow} [u_1^4,w_{f}]\\ &\stackrel{F_{1}}{\longrightarrow} [s_{0},v_1] \stackrel{F_{0}}{\longrightarrow} [r_{f-1},w_{f+1}] \stackrel{F_{f-1}}{\longrightarrow} [u_{f+1}^{1},u_{f+1}^2] \stackrel{F_{f+1}}{\longrightarrow} [t_{f+1}^{1},t_{f+1}^2]\\ & \stackrel{F_{f-1}}{\longrightarrow} [u_{f+1}^3,u_{f+1}^4] \stackrel{F_{f+1}}{\longrightarrow} [v_{f},s_{f}] \stackrel{F_{f}}{\longrightarrow} [w_{2f-1},r_{2f-1}] \stackrel{F_{2f-1}}{\longrightarrow} [S,u_{2f+1}^1]. \end{align*} $$

The proof then proceeds as in Case 1.

We now consider the structures of the manifolds $M^{k/l}(n,f)$ of Theorem A.

The significance of Theorem 4.5 is that the manifolds $M^{k/l}(n,0)$ are known to be hyperbolic in many cases, as we now describe. By [Reference Vesnin and Kim56, Corollary 2.1], if $n\geq 6$ is even then $M^{k/l}(n,0)$ is hyperbolic for all but finitely many pairs of coprime integers $k,l$ , and it is conjectured [Reference Vesnin and Kim56, p. 658] that $n=6$ and $k=l=1$ is the only non-hyperbolic case (which is an affine Riemannian manifold by [Reference Helling, Kim and Mennicke22, Proposition 6]). Moreover, $M^{k/l}(n,0)$ is hyperbolic in each of the following cases: $k\geq 2,l=1$ , $n\geq 6$ and even [Reference Maclachlan and Reid41, Theorem 3]; $k=1,l\geq 2$ , $n\geq 6$ and even [Reference Vesnin and Kim56, Corollary 3.5]; $k=l=1$ , $n\geq 8$ and even [Reference Helling, Kim and Mennicke22, Theorem C]. If $n=4$ and $k=1$ then $M^{k/l}(n,0)=M^{1/l}(4,0)$ is the lens space $L(4l^2+1,2l)$ and $G^{1/l}(4,0)\cong \mathbb{Z}_{4l^2+1}$ by [Reference Vesnin and Kim56, Corollary 3.4]. For $k\geq 2$ the manifolds $M^{k/1}(4,0)$ are not hyperbolic and are described in [Reference Maclachlan and Reid41, pp. 170–171]. In the next theorem we consider the manifolds $M^{k/l}(4,f)$ where $k\geq 2$ .

Proof We show that the subgroup A of $E=E^{k/l}(4)$ generated by $x^k, tx^kt^{-1}$ is free abelian of rank 2. Then, since $\nu ^f(x^k)=1$ and $\nu ^f(tx^kt^{-1})=1$ , $A=\mathbb{Z}\oplus \mathbb{Z}$ is a subgroup of $\mathrm {ker}(\nu ^f)=G^{k/l}(4,f)$ , and hence $M^{k/l}(4,f)$ is not hyperbolic by [Reference Callahan and Reid6, Theorem 3.3].

Let $G=\mathrm {ker}(\nu ^0)$ , so $G=F^{k/l}(4)$ and is generated by $x_i=t^ixt^{-i}$ , and let $G_1=\langle {x_0,x_2}\ |\ {x_0^kx_2^k} \rangle $ , $G_2=\langle {x_1,x_3}\ |\ {x_1^kx_3^k} \rangle $ , $H=\langle {a,b}\ |\ {ab=ba} \rangle \cong \mathbb{Z}^2$ . Let $\phi _1:H\rightarrow G_1$ , $\phi _2:H\rightarrow G_2$ , be given by $\phi _1(a)=x_0^k$ , $\phi _1(b)=x_0^{-l}x_2^l$ , $\phi _2(a)=x_3^{-l}x_1^l$ , $\phi _2(b)=x_1^k$ . We shall show that $\phi _1,\phi _2$ are injections. This implies that G can be expressed as an amalgamated free product $G=G_1*_H G_2$ , and $\phi _1(a)=x_0^k,\phi _1(b)=x_0^{-l}x_2^l$ generate a free abelian subgroup of G (compare [Reference Maclachlan and Reid41, p. 171], which deals with the case $l=1$ ). That is, $x_0^k$ and $x_1^k$ generate a free abelian subgroup of G and so $x^k$ and $tx^kt^{-1}$ generate a free abelian subgroup of E, as required.

We show that $\phi _1$ is an injection, the argument for $\phi _2$ being similar. Let $\alpha : G_1\rightarrow \mathbb{Z}_k$ be given by $\alpha (x_0)=1,\alpha (x_2)=1\in \mathbb{Z}_k$ . Then $K=\mathrm {ker}(\alpha )$ is generated by $z=x_0^k$ , $g_i=x_0^ix_2x_0^{-(i+1)}$ ( $0\leq i\leq k-2$ ), $g_{k-1}=x_0^{k-1}x_2$ , and has a presentation

$$\begin{align*}\langle {z,g_0,\ldots , g_{k-1}}\ |\ {zg_0g_1\ldots g_{k-2}g_{k-1}, zg_1g_2\ldots g_{k-1}g_{0}, \ldots , zg_{k-1}g_0\ldots g_{k-3}g_{k-2}} \rangle. \end{align*}$$

Now let $\theta :K\rightarrow \mathbb{Z}^k$ be an epimorphism given by $\theta (g_0)=(1,0,\ldots ,0)$ , $\theta (g_1)=(0,1,\ldots ,0), \ldots , \theta (g_{k-1})=(0,0,\ldots ,1)$ , and $\theta (z)=(-1,-1,\ldots , -1)$ . Let $l=qk+r$ where $q\geq 0$ , $0\leq r<k$ and note that $r\neq 0$ , since $\gcd (k,l)=1$ . Then $x_0^{-l}x_2^l= (x_0^k)^{-q}x_0^{-r}(x_2^k)^qx_2^r=(x_0^k)^{-q}x_0^{-r}(x_0^{-k})^qx_2^r=z^{-2q-1} g_{k-r}g_{k-(r-1)}\ldots g_{k-1}$ and $x_2^k=z$ . Therefore

$$ \begin{align*} \theta (x_0^{-l}x_2^l)&=(2q+1)(1,1,\ldots ,1)+(0,\ldots ,0,1,\ldots ,1)\\ &=(2q+1,\ldots ,2q+1,2q+2,\ldots ,2q+2) \end{align*} $$

and $\theta (x_2^k)=(-1,-1,\ldots , -1)$ so $\theta (x_0^{-l}x_2^l), \theta (x_2^k)$ generate a free abelian subgroup of rank 2 in $\mathbb{Z}^k$ . Moreover, in $G_1$ , $x_0^{-l}x_2^l$ and $x_2^k$ commute and so generate a free abelian subgroup of $G_1$ , and hence $\phi _1$ is injective.