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This article is cited in 1 scientific paper (total in 1 paper) On possible symmetry groups of 27-vertex triangulations of manifolds like the octonionic projective plane A. A. Gaifullinabcd a Steklov Mathematical Institute of Russian Academy of Sciences, Moscow, Russia b Skolkovo Institute of Science and Technology, Moscow, Russia c Lomonosov Moscow State University, Moscow, Russia d Institute for Information Transmission Problems of the Russian Academy of Sciences (Kharkevich Institute), Moscow, Russia
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     § 1. IntroductionThroughout this paper we denote by $\mathrm{C}_n$, $\mathrm{D}_n$, $\mathrm{S}_n$, and $\mathrm{A}_n$ the cyclic group of order $n$, the dihedral group of order $2n$, the symmetric group of degree $n$, and the alternating group of degree $n$, respectively. In 1987 Brehm and Kühnel [14] obtained the following result. We are not going to give a definition of a PL Morse function here. In the present paper we need only the following properties of manifolds $M$ like projective planes; see [21].
In view of Theorem 1.1, for $d=2,4,8,16$ it is interesting to study combinatorial $d$-manifolds with exactly $3d/2+3$ vertices that are not homeomorphic to a sphere, and their symmetry groups. The known examples of such combinatorial manifolds are as follows.
For interesting connections of $(3d/2+3)$-vertex triangulations of $d$-manifolds like projective planes with the Freudenthal magic square and Severi varieties; see [18]. The four $27$-vertex combinatorial $16$-manifolds like the octonionic projective plane with the symmetry group $G_{351}$ were found using a specially designed computer program; see [24]. This program lists all $27$-vertex combinatorial $16$-manifolds like the octonionic projective plane with a given symmetry group $G$. It works rather well for groups $G$ with $|G|\gtrsim 250$. However, its running time grows uncontrollably for groups of smaller order. The fact that the program found examples of combinatorial manifolds for $G=G_{351}$ and did not find them for a number of groups $G$ of close orders received no reasonable explanation. In the present work we clarify this issue. Our main result is as follows. Theorem 1.2. Suppose that $K$ is a $27$-vertex combinatorial $16$-manifold that is not homeomorphic to the sphere $S^{16}$. Then the symmetry group $\operatorname{Sym}(K)$ is conjugate to one of the $26$ subgroups of $\mathrm{S}_{27}$ listed in Table 1. Table 1.Possible symmetry groups of $27$-vertex triangulations of $16$-manifolds like the octonionic projective plane
Remark 1.3. Let us make several comments concerning Table 1. Remark 1.4. So far, examples of $27$-vertex combinatorial $16$-manifold like the octonionic projective plane are known only for $6$ of the $26$ potentially possible symmetry groups from Table 1, namely, for the symmetry groups $G_{351}=\mathrm{C}_3^3\rtimes \mathrm{C}_{13}$, $\mathrm{C}_3^3$, $\mathrm{C}_{13}$, $\mathrm{C}_3^2$, $\mathrm{C}_3$ and $1$. All these examples were constructed by this author in [22]; the numbers of such examples are indicated in the last column of Table 1. For the other $20$ groups the question of the existence of a $27$-vertex combinatorial $16$-manifold like the octonionic projective plane with such a symmetry group is open. The group $G_{351}$ is the only symmetry group for which we know a complete list of all (four) $27$-vertex combinatorial $16$-manifold like the octonionic projective plane. For the other five symmetry groups it is unknown whether the examples obtained in [22] exhaust all possible combinatorial manifolds. Note also that it is still unknown whether the combinatorial manifolds constructed in [22] are homeomorphic to $\mathbb{OP}^2$ or they are some other manifolds like the octonionic projective plane. Remark 1.5. A partial explanation for the group $G_{351}$ arising as the symmetry group of a $27$-vertex combinatorial $16$-manifold like the octonionic projective plane is that $G_{351}$ can nicely be realized as a subgroup of the isometry group $\operatorname{Isom}(\mathbb{OP}^2)$, where $\mathbb{OP}^2$ is endowed with the Fubini–Study metric; see [22], Remark 1.7, for more detail. Namely, $\operatorname{Isom}(\mathbb{OP}^2)$ is isomorphic to the exceptional simply connected compact Lie group $F_4$, and Alekseevskii [2] constructed an important subgroup of $F_4$ isomorphic to $\mathrm{C}_3^3\rtimes\mathrm{SL}(3,\mathbb{F}_3)$, which contains $G_{351}$ since $\mathrm{SL}(3,\mathbb{F}_3)$ contains an element of order $13$. On the other hand the group $F_4$ has many different finite subgroups (see [19]) and, as Theorem 1.2 shows, not all of them are realized as the symmetry groups of $27$-vertex triangulations of manifolds like the octonionic projective plane. So the connection between the symmetry groups of such triangulations and finite subgroups of $F_4$ is yet to be clarified. All the above can be extended to the case of $\mathbb{Z}$-homology manifolds: see Definition 2.7 below. We start with the following analogue of the Brehm–Kühnel theorem. Theorem 1.6 (Novik [34]). Suppose that $K$ is a $\mathbb{Z}$-homology $d$-manifold with $n$ vertices. Remark 1.7. This theorem was not stated explicitly in [34]. Let us explain how to extract it from there. First, Lemma 5.5 in [34] implies assertion (1) of Theorem 1.6, and, in addition, implies that in the case of even $d=2m$ and $n=3m+3$ one has where the $\beta_i$ are the Betti numbers of $K$ with coefficients in any field of characteristic two, or in any other field, provided that, in addition, $K$ is orientable. Since any $\mathbb{Z}$-homology $2$-manifold is a combinatorial manifold, in the case $d=2$ the required assertion follows immediately from the uniqueness of a $6$-vertex triangulation of a $2$-manifold not homeomorphic to $S^2$. So we may assume that $d\geqslant 4$. Then by (1.2) we have $\beta_1=0$ with coefficients in the field $\mathbb{F}_2$. Hence the fundamental group $\pi_1(K)$ admits no nontrivial homomorphisms to $\mathbb{F}_2$. It follows that $K$ is orientable, and so (1.2) holds with coefficients in any field. Then Poincaré duality implies that $H_0(K;\mathbb{Z})\cong H_{2m}(K;\mathbb{Z})\cong\mathbb{Z}$, the group $H_m(K;\mathbb{Z})$ is free abelian, and $H_i(K;\mathbb{Z})=0$ for $i\notin\{0,m,2m\}$. Now, Theorem 5.6 in [34] implies that $\beta_m\leqslant 1$, so the group $H_m(K;\mathbb{Z})$ is either trivial or isomorphic to $\mathbb{Z}$. In the former case we have $H_*(K;\mathbb{Z})\cong H_*(S^d;\mathbb{Z})$, and in the latter case, using Poincaré duality we arrive at the isomorphism (1.1). Finally, it is a well-known corollary of Adams’ Hopf invariant one theorem (see [1]) that such isomorphism can exist only for $d\in \{4,8,16\}$. A generalization of Theorem 1.2 for homology manifolds is as follows. Theorem 1.8. Suppose that $K$ is a $27$-vertex $\mathbb{Z}$-homology $16$-manifold such that the graded group $H_*(K;\mathbb{Z})$ is not isomorphic to $H_*(S^{16};\mathbb{Z})$. Then the symmetry group $\operatorname{Sym}(K)$ is conjugate to one of the $26$ subgroups of $\mathrm{S}_{27}$ listed in Table 1. Of course, Theorem 1.2 follows from Theorem 1.8. A natural conjecture (cf. [4], § 20, Conjecture 0) is that any $(3d/2+3)$-vertex $\mathbb{Z}$-homology $d$-manifold is a combinatorial manifold. This conjecture has been proved for $d=2$ and $d=4$: see [4], §§ 19 and 21, but it is completely open for $d=8$ and $d=16$. So it is not clear whether Theorem 1.8 is really stronger than Theorem 1.2. Our approach to the symmetry groups $\operatorname{Sym}(K)$ (and thus to the proof of Theorem 1.8) is to consider elements $g\in\operatorname{Sym}(K)$ of different orders and examine the topology and combinatorics of the corresponding fixed point sets $K^{\langle g\rangle}$. (In what follows we denote by $\langle g\rangle$ the cyclic subgroup generated by $g$.) We obtain results about fixed-point sets by using two sets of ideas:
Combining these two approaches we obtain a series of conditions on the orders of elements and subgroups of $\operatorname{Sym}(K)$: see Proposition 5.1. The next part of our proof is purely group theoretic: we prove that any group satisfying the conditions obtained is in Table 1; see Proposition 5.2. More precisely, at this stage we obtain one extra possibility for $\operatorname{Sym}(G)$, namely, the group $\mathrm{PSU}(3,2)$ of order $72$ acting with three orbits of length $9$. Then this extra possibility is excluded by a more detailed study of the fixed-point complexes for subgroups isomorphic to $\mathrm{C}_4$. All of our proofs in this paper are not computer assisted, except for the proof of the fact that the symmetry group of a $15$-vertex $\mathbb{F}_2$-homology $8$-manifold that satisfies complementarity cannot contain an element of order $4$ (see Proposition 6.1). The proof of this fact uses a modification of the software created by this author for finding $27$-vertex triangulations of $16$-manifolds like the octonionic projective plane (see [24]). Remark 1.9. It is natural to ask to what extent the results and methods of this work transfer to the case of $15$-vertex triangulations of $\mathbb{HP}^2$ or manifolds like the quaternion projective plane. One would expect that in this case the conditions on the symmetry group would be even more restrictive. Nevertheless, owing to the difference between some number-theoretic properties of the integers $15$ and $27$, it has turned out that this is not the case. An attempt to apply the methods of this work to the $8$-dimensional case has led to the discovery of a number of new $15$-vertex triangulations of $\mathbb{HP}^2$ with various symmetry groups and a partial classification of them; see [23]. The present paper is organized as follows. Sections 2, 3 and 4 provide preliminary information on simplicial complexes, transformation groups and finite groups, respectively. Section 5 contains the scheme of the proof of Theorem 1.8. In that section we formulate Proposition 5.1 (the properties of $\operatorname{Sym}(K)$), Proposition 5.2 (list of groups satisfying these properties) and Proposition 5.3 (excluding the group $\mathrm{PSU}(3,2)$), which in combination imply Theorem 1.8. In § 6 we obtain an auxiliary result on the absence of elements of order $4$ in the symmetry group of a $15$-vertex $\mathbb{F}_2$-homology $8$-manifold that satisfies complementarity. Further on we prove Propositions 5.1, 5.2 and 5.3 in §§ 7, 8 and 9, respectively. AcknowledgementsThe author is grateful to Denis Gorodkov, Vasilii Rozhdestvenskii and Constantin Shramov for useful discussions. The author would like to thank particularly Andrey Vasil’ev, who pointed out a way to simplify radically the proof of Proposition 5.2 by using the results of Higman and Suzuki on groups in which every element has a prime power order. § 2. Preliminaries on simplicial complexes2.1. Simplicial complexes and group actionsDefinition 2.1. An (abstract) simplicial complex on a vertex set $V$ is a set $K$ of finite subsets of $V$ such that The dimension of a simplex $\sigma$ is the cardinality of $\sigma$ minus one. We always assume that a simplicial complex has no ghost vertices, that is, all one-element subsets of $V$ are simplices of $K$. In this paper we work with finite simplicial complexes only. Throughout the paper we denote by $\chi(K)$ the Euler characteristic of $K$. We denote by $\Delta^k$ the standard $k$-dimensional simplex and by $\partial\Delta^k$ the boundary of it. The geometric realization $|K|$ is the set of all linear combinations such that
The set $|K|$ is always endowed with the topology of a CW-complex. Given a simplex $\sigma\in K$, the point is called the barycentre of $\sigma$.We say that a finite group $G$ acts simplicially on $K$ if $G$ acts on the vertex set $V$ so that every element of $G$ takes simplices of $K$ to simplices of $K$ and nonsimplices of $K$ to nonsimplices of $K$. A simplicial action of $G$ on $K$ induces a piecewise linear action of $G$ on the geometric realization $|K|$. We denote the set of all $G$-fixed points in $|K|$ by $|K|^G$. The following proposition is standard. Proposition 2.2. Suppose that $K$ is a finite simplicial complex on a vertex set $V$ with a simplicial action of a finite group $G$. Let $\sigma_1,\dots,\sigma_s$ be all $G$-orbits in $V$ that are simplices of $K$. Then the fixed-point set $|K|^G$ coincides with the geometric realization of the simplicial complex $K^G$ with vertices $b(\sigma_1),\dots,b(\sigma_s)$ such that $\bigl\{b(\sigma_{i_1}),\dots,b(\sigma_{i_t})\bigr\}\in K^G$ if and only if $\sigma_{i_1}\cup\cdots\cup\sigma_{i_t}\in K$. The simplicial complex $K^G$ will be called the fixed-point complex of the action. 2.2. Pseudomanifolds, combinatorial manifolds and homology manifoldsDefinition 2.3. A finite simplicial complex $K$ is called a weak $d$-pseudomanifold if it satisfies the following two conditions: A weak $d$-pseudomanifold $K$ is called a $d$-pseudomanifold if, in addition, it satisfies the condition A $d$-pseudomanifold is said to be orientable if the $d$-simplices of it can be endowed with compatible orientations, which means that the orientations of $\sigma$ and $\tau$ should induce opposite orientations of $\sigma\cap\tau$ whenever $\dim(\sigma\cap\tau)=d-1$. We will need the following easy proposition. Proposition 2.4. Suppose that a nontrivial finite group $G$ acts faithfully and simplicially on a $d$-pseudomanifold $K$. Then $\dim K^G$ is strictly smaller than $d$. Proof. Obviously, $\dim K^G\leqslant d$. Assume that $\dim K^G=d$. Then there exists a $d$-simplex $\sigma\in K$ such that every element of $G$ fixes every vertex of $\sigma$. It is easy to see that if a $d$-simplex $\sigma$ has this property, then any $d$-simplex $\tau\in K$ such that $\dim(\sigma\cap\tau)=d-1$ also has the same property. Since $K$ is strongly connected, it follows that every element of $G$ fixes every vertex of $K$ and so the action of $G$ on $K$ is not faithful.
The proposition is proved. Definition 2.5. The link of a simplex $\sigma$ of a simplicial complex $K$ is the simplicial complex Definition 2.7. Suppose that $R$ is either the ring $\mathbb{Z}$ or a finite field. A finite simplicial complex $K$ is called an $R$-homology $d$-manifold if The following assertions are straightforward.
2.3. ComplementarityThe following property plays a key role in our considerations. Definition 2.8. We say that a finite simplicial complex $K$ on a vertex set $V$ satisfies complementarity if for each subset $\sigma\subseteq V$ exactly one of the two subsets $\sigma$ and $V\setminus \sigma$ is a simplex of $K$. The following theorem by Arnoux and Marin (see [4], § 20), in combination with Theorem 1.6, implies that all $(3d/2+3)$-vertex $\mathbb{Z}$-homology $d$-manifolds with $H_*(K;\mathbb{F}_2)\not\cong H_*(S^d;\mathbb{F}_2)$ satisfy complementarity. Theorem 2.9 (Arnoux and Marin). Suppose that $K$ is a simplicial complex such that the cohomology ring $H^*(K;\mathbb{F}_2)$ contains a subring isomorphic to $\mathbb{F}_2[a]/(a^3)$, where $m=\deg a$ is even. Then $K$ has at least $3m+3$ vertices. Moreover, if $K$ has exactly $3m+3$ vertices, then $K$ satisfies complementarity. The next theorem is due to Datta [20], except for the case of a $6$-pseudomanifold with $12$ vertices, which was excluded by Bagchi and Datta [6]. Theorem 2.10 (Datta and Bagchi). Suppose that $K$ is a $d$-pseudomanifold that satisfies complementarity, where $d>0$. Then one of the following assertions hold: Remark 2.11. Throughout the paper we also encounter another type of simplicial complexes that satisfy complementarity, namely, the disjoint unions $\mathrm{pt}\sqcup\partial\Delta^k$, where $\mathrm{pt}$ is a point and $\partial\Delta^k$ is the boundary of a $k$-dimensional simplex. It is not hard to show that these are the only disconnected simplicial complexes that satisfy complementarity. In particular, for $k=1$ we arrive at a disjoint union of three points. We will also need the following easy observation by Bagchi and Datta; see [6]. Proposition 2.12 (Bagchi and Datta). Suppose that $K$ is a finite simplicial complex that satisfies complementarity. Then the Euler characteristic $\chi(K)$ is odd. Moreover, if the number of vertices of $K$ is even, then $\chi(K)=1$. It follows from Poincaré duality that the Euler characteristic of any odd-dimensional $\mathbb{F}_p$-homology manifold is equal to zero. Complementarity is closely related to neighbourliness. Definition 2.14. A simplicial complex $K$ on a vertex set $V$ is said to be $k$-neighbourly if every $k$-element subset of $V$ is a simplex of $K$. The next proposition follows immediately from the definition. Proposition 2.15. Suppose that $K$ is an $n$-vertex $d$-dimensional simplicial complex that satisfies complementarity. Then $K$ is $(n-d-2)$-neighbourly. Finally, we need the following result on how the complementarity property is inherited after going over to the fixed-point complex. Proposition 2.16. Suppose that $K$ is a finite simplicial complex that satisfies complementarity, $V$ is the vertex set of $K$ and $G$ is a finite group acting simplicially on $K$. Let $r$ be the number of $G$-orbits in $V$. Moreover, $K^G$ is empty or a simplex if and only if at least one of the $H$-orbits in $V$ is not a simplex of $K$. Proof. Let $\sigma_1,\dots,\sigma_r$ be all $G$-orbits in $V$. First assume that $r=1$. Since $\varnothing\in K$, we have $\sigma_1=V\notin K$ by complementarity. So $K^G$ is empty by Proposition 2.2.
Second assume that $r=2$. By complementarity exactly one of the two $G$-orbits $\sigma_1$ and $\sigma_2$ is a simplex of $K$. So $K^G$ is a point. Third assume that $r\geqslant 3$. Then we have two cases. Case 1: one of the $G$-orbits in $V$, say $\sigma_1$, is not a simplex of $K$. Then $K^G$ is a simplicial complex with $r-1$ vertices $b(\sigma_2),\dots,b(\sigma_r)$. Since $\sigma_1\notin K$, by complementarity we have $\sigma_2\cup\cdots\cup\sigma_r\in K$. So $K^G$ is an $(r-2)$-simplex with vertices $b(\sigma_2),\dots,b(\sigma_r)$. Case 2: all $G$-orbits in $V$ are simplices of $K$. Then $K^G$ is a simplicial complex with $r$ vertices $b(\sigma_1),\dots,b(\sigma_r)$. The complementarity property of $K^G$ follows immediately from the complementarity property of $K$. 2.4. Kühnel’s triangulation $ {\mathbb{CP}}^2_9$In this subsection we recall an explicit description of Kühnel’s triangulation $\mathbb{CP}^2_9$ in terms of an affine plane over a field of three elements. This construction is due to Bagchi and Datta; see [5]. Let $\mathcal{P}$ be an affine plane over the field $\mathbb{F}_3$. Then $|\mathcal{P}|=9$. Fix a decomposition where $\ell_0$, $\ell_1$ and $\ell_2$ are mutually parallel lines, and fix a cyclic order of these three lines. We conveniently consider the indices $0$, $1$ and $2$ as elements of $\mathbb{F}_3$. The three lines $\ell_0$, $\ell_1$ and $\ell_2$ are called special, and all other lines in $\mathcal{P}$ are nonspecial.The affine plane $\mathcal{P}$ will serve as the set of vertices of $\mathbb{CP}^2_9$. The $4$-dimensional simplices of $\mathbb{CP}^2_9$ are exactly the following $36$ five-element subsets of $\mathcal{P}$:
The simplicial complex $\mathbb{CP}^2_9$ consisting of these $36$ four-simplices and all their faces is a combinatorial manifold PL-homeomorphic to $\mathbb{CP}^2_9$. Moreover, $\mathbb{CP}^2_9$ satisfies complementarity. Remark 2.17. In [5] the $9$ simplices of the second type are $(\ell_t\setminus\{u\})\cup\ell_{t+1}$, where $t\in\mathbb{F}_3$ and $u\in\ell_t$. So our convention on the cyclic order of the three special lines is opposite to the one used in [5]. It follows immediately from the construction that the simplicial complex $\mathbb{CP}^2_9$ is invariant under all affine transformations of $\mathcal{P}$ that take special lines to special lines and preserve their cyclic order. In fact, such affine transformations form the whole group $\operatorname{Sym}(\mathbb{CP}^2_9)$, which has order $54$. If we introduce affine coordinates $(x,y)$ in $\mathcal{P}$ so that the special lines are the lines $\{y=0\}$, $\{y=1\}$ and $\{y=2\}$ in this cyclic order, then the group $\operatorname{Sym}(\mathbb{CP}^2_9)$ consists of all transformations of the form
$$
\begin{equation}
\begin{pmatrix} x\\ y \end{pmatrix} \mapsto \begin{pmatrix} \pm 1 & c\\ 0 & 1 \end{pmatrix} \begin{pmatrix} x\\ y \end{pmatrix}+ \begin{pmatrix} a\\ b \end{pmatrix}, \qquad a,b,c\in\mathbb{F}_3,
\end{equation}
\tag{2.1}
$$
which can be rewritten as
$$
\begin{equation*}
\begin{pmatrix} x\\ y\\1 \end{pmatrix} \mapsto \begin{pmatrix} \pm 1 & c & a\\ 0 & 1 & b\\ 0&0&1 \end{pmatrix} \begin{pmatrix} x\\ y \\ 1 \end{pmatrix}.
\end{equation*}
\notag
$$
It follows that $\operatorname{Sym}(\mathbb{CP}^2_9)\cong\mathrm{He}_3\rtimes\mathrm{C}_2$. By Sylow’s theorems all elements of order $2$ in $\operatorname{Sym}(\mathbb{CP}^2_9)$ are conjugate to each other and therefore are conjugate to the transformation The following result was obtained in [15], § 7.2.Proposition 2.18 (Brehm and Kühnel). For any element $g\!\in\!\operatorname{Sym}(\mathbb{CP}^2_9)$ of order $2$, the fixed-point complex $(\mathbb{CP}^2_9)^{\langle g\rangle}$ is isomorphic to $\mathbb{RP}^2_6$. Also, we will need the following proposition. Proposition 2.19. Suppose that $m$ is a $3$-element subset of $\mathcal{P}$. Then the following two assertions are equivalent: Proof. It can be easily checked that any $3$-element subset $m\subset\mathcal{P}$ can be taken by a transformation of the form (2.1) to one of the following five subsets: so it is enough to check the claim of the proposition for each of these five subsets, which can be done by direct enumeration.
The proposition is proved. § 3. Preliminaries on transformation groupsThroughout this section $K$ is a finite simplicial complex and $p$ is a prime. We need two classical results from the theory of transformation groups. In the following theorem assertions (1) and (2) are due to Smith (see [35], Theorems 1, 5 and 6) and assertion (3) is due to Bredon (see [10], Theorem 7.7); see also [9], Ch. V, for more detail. Theorem 3.1 (Smith and Bredon). Suppose that $K$ is an $\mathbb{F}_p$-homology manifold, and $G$ is a finite $p$-group that acts faithfully and simplicially on $K$. Then To formulate the second result we conveniently use the following standard notation. We write $K\sim_p\mathbb{P}^h(m)$ if
$$
\begin{equation}
H^*(K;\mathbb{F}_p)\cong\mathbb{F}_p[a]/(a^{h+1}), \qquad \deg a = m.
\end{equation}
\tag{3.1}
$$
The following theorem is due to Bredon [11]; see also [12], Theorem 4.1, and [13], Ch. VII, Theorem 3.1. Theorem 3.2 (Bredon). Suppose that $K\sim_p\mathbb{P}^h(m)$ and the group $\mathrm{C}_p$ acts simplicially on $K$. Then either $K^{\mathrm{C}_p}=\varnothing$ or for some $F_i\sim_p\mathbb{P}^{h_i}(m_i)$, where $m_i\leqslant m$, $h_i\geqslant 0$ and $\sum_{i=1}^s(h_i+1)=h+1$. Remark 3.3. We are especially interested in complexes $K\sim_p\mathbb{P}^2(m)$. Recall that Adams’ Hopf invariant one theorem (see [1]) implies that if $p=2$, then $m\in\{1,2,4,8\}$. We will conveniently write $\mathbb{RP}^2$, $\mathbb{CP}^2$, $\mathbb{HP}^2$, and $\mathbb{OP}^2$ instead of $\mathbb{P}^2(1)$, $\mathbb{P}^2(2)$, $\mathbb{P}^2(4)$, and $\mathbb{P}^2(8)$, respectively. On the contrary, if $p$ is odd, then a space $K\sim_p\mathbb{P}^2(m)$ exists for any even $m$; see [13], Ch. VII, § 4. Remark 3.4. We have formulated Theorems 3.1 and 3.2 for simplicial actions on finite simplicial complexes, since this is the only case of interest for us. However, in all papers and books cited above these theorems are formulated in much greater generality, namely, for continuous actions on locally compact and paracompact Hausdorff topological spaces of finite cohomology dimension. In general, care must be taken with regard to the choice of the definition of a homology (or cohomology) manifold and the cohomology theory used. Namely, Alexander–Spanier, Čech or sheaf theoretic cohomology are used in various versions of Theorems 3.1 and 3.2. Nevertheless, for finite simplicial complexes (and their open subsets) all these cohomology theories coincide with singular cohomology. A general definition of $\mathbb{F}_p$-cohomology manifold suitable for Theorem 3.1 can be found in [9], Ch. I. It is based on the concept of local Betti numbers around a point, which goes back to Alexandroff [3] and Čech [17]. The fact that for finite simplicial complexes the general definition of $\mathbb{F}_p$-cohomology manifold is equivalent to Definition 2.7 follows from two simple observations. First, for a finite simplicial complex $K$, the ‘clever’ definition of local Betti numbers around a point $x$ used in [9] is equivalent to the ‘naive’ definition § 4. Preliminaries on finite groups4.1. Sylow theoremsSuppose that $G$ is a finite group and $p$ is a prime number. Recall that if $p$ divides $|G|$, then a Sylow $p$-subgroup is a maximal $p$-subgroup of $G$. We need the following classical Sylow theorems:
4.2. Small groupsIn this paper we need a number of classification results on finite groups. We start with the following standard assertion. Proposition 4.1. Any group of order $2p$, where $p$ is an odd prime number, is isomorphic either to $\mathrm{C}_{2p}$ or to $\mathrm{D}_p$. Further, we need some specific classification results on groups of small order. For the convenience of the reader, we have collected all these results in Proposition 4.2 below. Note that we are only interested in groups of order up to $351$, so we definitely do not require any hard results on the classification of finite simple groups. The classification of finite groups of order up to $2000$ was obtained by Besche, Eick and O’Brien [7]. The databases of small groups are accessible in GAP (see [8]) or Magma (see [16], Ch. 62). All assertions of the following proposition can immediately be extracted from any of these databases. On the other hand, each of these assertions can easily be proved by hand using standard group-theoretic methods. Proposition 4.2. Up to isomorphism, Remark 4.3. Recall the definition of the group $\mathrm{PSU}(n,p)$, where $p$ is a prime number. (One can similarly define the groups $\mathrm{PSU}(n,p^r)$ with $r>1$, but we do not need them.) Let $\mathbb{F}_{p^2}$ be a field of $p^2$ elements and $F\colon x\mapsto x^p$ be the Frobenius automorphism of it. Then the unitary group $\mathrm{U}(n,p)$ is the subgroup of $\mathrm{GL}(n,\mathbb{F}_{p^2})$ consisting of all matrices $U$ that satisfy $F(U)^t\,U=I$, where $I$ is the identity matrix. The special unitary group is the group Finally, the projective special unitary group $\mathrm{PSU}(n,p)$ is the quotient of the group $\mathrm{SU}(n,p)$ by its centre, which consists of diagonal matrices. In this paper we deal only with the group $\mathrm{PSU}(3,2)$. It can be checked directly that it has order $72$ and is isomorphic to the semidirect product $\mathrm{C}_3^2\rtimes \mathrm{Q}_8$, where $\mathrm{Q}_8$ acts faithfully on $\mathrm{C}_3^2$. So a Sylow $2$-subgroup of $\mathrm{PSU}(3,2)$ is isomorphic to $\mathrm{Q}_8$. Moreover, it is easy to see that $\mathrm{PSU}(3,2)$ has nine Sylow $2$-subgroups. 4.3. Finite groups in which every element has prime power orderWe need two classical theorems on finite groups in which every element $\ne 1$ has prime power order. The first is due to Suzuki (see [36], Theorem 16); it classifies simple groups with this property. The second theorem is due to Higman (see [28], Theorem 1); it describes the structure of solvable groups with this property. Theorem 4.4 (Suzuki). Up to isomorphism, there are exactly eight finite noncyclic simple groups all of whose elements $\ne 1$ have prime power order, namely, the groups $\mathrm{PSL}(2,\mathbb{F}_q)$ for $q=5$, $7$, $8$, $9$ and $17$, $\mathrm{PSL}(3,\mathbb{F}_4)$ and the Suzuki groups of Lie type $\mathrm{Sz}(8)$ and $\mathrm{Sz}(32)$. The groups $\mathrm{PSL}(2,\mathbb{F}_q)$ for $q=5$, $7$, $8$, $9$ and $17$ have orders $60$, $168$, $504$, $360$ and $2448$, respectively. The order of $\mathrm{PSL}(3,\mathbb{F}_4)$ is $20160$. The Suzuki groups $\mathrm{Sz}(q)$ have orders $q^2(q^2+1)(q-1)$. So the order of each group in the list from Theorem 4.4 is divisible by at least one of the numbers $5$, $7$ and $17$. Corollary 4.6. Let $G$ be a finite group all of whose elements $\ne 1$ have prime power order. Suppose that $|G|$ is not divisible by any of the numbers $5$, $7$ and $17$. Then $G$ is solvable. Proof. All composition factors of $G$ are simple finite groups in which every element $\ne 1$ has a prime power order. Then it follows from Theorem 4.4 that all of them are cyclic.
The corollary is proved. Theorem 4.7 (Higman). Let $G$ be a finite solvable group all of whose elements $\ne 1$ have prime power order. Then either $G$ is a $p$-group or $G$ satisfies the following: Remark 4.8. All assertions of Theorem 4.7, except for the divisibility $q^n\mid p^k-1$, are contained in Theorem 1 in [28], while this divisibility is implicitly contained in the proof of that theorem. For the convenience of the reader, let us prove that $q^n\mid p^k-1$. Let $Q$ be a Sylow $q$-subgroup of $G$. Then $Q$ acts by conjugation on $P$. The restriction of this action to the set $P\setminus\{1\}$ is free. Indeed, if $yxy^{-1}=x$ for some $x\in P\setminus\{1\}$ and $y\in Q\setminus\{1\}$, then the order of $xy$ is the product of the orders of $x$ and $y$ and therefore not a prime power. Thus, $|Q|=q^n$ divides $|P\setminus\{1\}|=p^k-1$. Remark 4.9. Theorem 1 in [28] also contains further results on the structure of $G/P$, which are not important for us. § 5. Scheme of proof of Theorem 1.8The proof of Theorem 1.8 consists of three steps. First, using results by Smith, Bredon, Datta and Bagchi (Theorems 3.1, 3.2 and 2.10) we prove the following. Proposition 5.1. Suppose that $K$ is a $27$-vertex $\mathbb{Z}$-homology $16$-manifold such that the graded group $H_*(K;\mathbb{Z})$ is not isomorphic to $H_*(S^{16};\mathbb{Z})$. Let $V$ be the vertex set of $K$. Then the symmetry group $G=\operatorname{Sym}(K)$ has the following properties: The second part of the proof of Theorem 1.8 is purely group theoretic. Namely, we classify all subgroups of $\mathrm{S}_{27}$ that have the properties from Proposition 5.1. Proposition 5.2. Up to conjugation, there are exactly $27$ subgroups $G\subset \mathrm{S}_{27}$ that satisfy properties (a)–(g) from Proposition 5.1; these are the $26$ subgroups from Table 1 and also the group $\mathrm{PSU}(3,2)$ that acts on $V=\{1,\dots,27\}$ with three orbits of length $9$. Finally, we need to exclude the group $\mathrm{PSU}(3,2)$. Proposition 5.3. The group $\mathrm{PSU}(3,2)$ cannot serve as the symmetry group of a $27$-vertex $\mathbb{Z}$-homology $16$-manifold $K$ with graded homology group $H_*(K;\mathbb{Z})$ not isomorphic to that of the $16$-sphere. Theorem 1.8 follows immediately from Propositions 5.1, 5.2 and 5.3. We prove these three propositions in §§ 7, 8 and 9, respectively. § 6. Absence of elements of order 4 in the symmetry groups of 15-vertex homology 8-manifoldsThe aim of this section is to obtain the following result, which will be used in the proofs of properties (a) and (b) from Proposition 5.1. Proposition 6.1. Suppose that $K$ is a $15$-vertex $\mathbb{F}_2$-homology $8$-manifold that satisfies complementarity. Then the symmetry group $\operatorname{Sym}(K)$ contains no elements of order $4$. We start with the following proposition. Proposition 6.2. Suppose that $K$ is a $15$-vertex $\mathbb{F}_2$-homology $8$-manifold that satisfies complementarity. Then $K$ has exactly $490$ top-dimensional simplices. Proof. Brehm and Kühnel [15], § 1, proved this proposition in the case of a combinatorial $8$-manifold. Let us show that their proof can be extended literally to the case of a homology manifold.
Let $f_i$ denote the number of $i$-simplices of $K$. We conveniently put $f_{-1}=1$. By Proposition 2.15 the simplicial complex $K$ is $5$-neighbourly. Hence
$$
\begin{equation}
f_i=\binom{15}{i+1}, \qquad -1\leqslant i\leqslant 4.
\end{equation}
\tag{6.1}
$$
By a theorem of Klee [29] the numbers $f_i$ satisfy the Dehn–Sommerville equations
$$
\begin{equation}
h_{9-i}-h_i=(-1)^i\binom{9}{i}\bigl(\chi(K)-2\bigr),\qquad 0\leqslant i\leqslant 9,
\end{equation}
\tag{6.2}
$$
where $(h_0,\dots,h_9)$ is the $h$-vector of $K$ defined by
$$
\begin{equation}
\sum_{i=0}^9h_it^{9-i}=\sum_{i=0}^9f_{i-1}(t-1)^{9-i},
\end{equation}
\tag{6.3}
$$
and $\chi(K)$ is the Euler characteristic of $K$, that is, Brehm and Kühnel [15], § 1, showed that the system of equations (6.1)–(6.4) has a unique solution and, for this solution, $f_8=490$.
The proposition is proved. Proposition 6.1 is the only assertion in the present paper whose proof is computer assisted. In [22] this author proposed an algorithm that, given numbers $d$, $n$ and $N$ and a subgroup $G\subset S_n$, produces the list of all $G$-invariant weak $d$-pseudomanifolds $K$ with $n$ vertices and at least $N$ maximal simplices such that $K$ satisfies ‘half of the complementarity condition’: for each subset $\sigma\subset V$ at most one of the two subsets $\sigma$ and $V\setminus\sigma$ is a simplex of $K$. (Here $V$ is the set of vertices of $K$.) This algorithm was implemented as a C++ program; see [24]. It completes in reasonable time, provided that the symmetry group $G$ is large enough. We could try to obtain Proposition 6.1 by running this program for and subgroups $G$ of $S_{15}$ that are isomorphic to $\mathrm{C}_4$. (We would have to take for $G$ a representative of every conjugacy class of such subgroups.) If the result were an empty list, Proposition 6.1 would follow. However, when run on such data, the program did not finish in a reasonable amount of time. Perhaps such calculations can still be performed using a high-performance computer, but we prefer to simplify the computational task by proving the following proposition. Proposition 6.3. Suppose that $K$ is a $15$-vertex $\mathbb{F}_2$-homology $8$-manifold that satisfies complementarity. Assume that the symmetry group $\operatorname{Sym}(K)$ contains a subgroup $H\cong\mathrm{C}_4$. Then renumbering the vertices of $K$ we can achieve that the group $H$ is generated by the permutation and $K$ contains the following $12$ eight-dimensional simplices: Remark 6.4. Of course, once we have achieved that $K$ contains the $12$ simplices listed in Proposition 6.3, we obtain automatically that $K$ also contains all simplices obtained from those $12$ simplices by the action of the group $H$. Once we will have proved Proposition 6.3, we will be able to simplify our computational task considerably. First, we will need to check only one subgroup of $\mathrm{S}_{15}$ isomorphic to $\mathrm{C}_4$, namely, the subgroup generated by the permutation $A$. Second, which is more important, we will be able to modify the algorithm by including the $12$ eight-dimensional simplices from Proposition 6.3 (and all simplices in their $H$-orbits) in the simplicial complex $K$ from the very beginning. This reduces the running time significantly since the inclusion of a simplex in $K$ immediately prohibits the inclusion of a number of other simplices in $K$; see [22], Remark 4.3. Being implemented in this way, the program completes in a reasonable amount of time (about $3$ hours on one processor core of clock frequency 3.3 GHz), and yields the following result. Proposition 6.5. There are no $15$-vertex weak $8$-pseudomanifolds $K$ that satisfy the following conditions: Proposition 6.1 follows immediately from Propositions 6.3 and 6.5. In the rest of this section we prove Proposition 6.3. We start with the following auxiliary assertion. Proposition 6.6. Suppose that $K$ is a $15$-vertex $\mathbb{F}_2$-homology $8$-manifold that satisfies complementarity and $g\in\operatorname{Sym}(K)$ is an element of order $2$. Then $K^{\langle g\rangle}\cong\mathbb{CP}^2_9$. Proof. Since $K$ satisfies complementarity, Proposition 2.15 implies that $K$ is $5$-neighbourly. Hence the union of any two $\langle g\rangle$-orbits is a simplex of $K$. It follows that $K^{\langle g\rangle}$ is nonempty and connected. Then by Theorem 3.1 the complex $K^{\langle g\rangle}$ is an $\mathbb{F}_2$-homology manifold and thus a pseudomanifold. Therefore, $K^{\langle g\rangle}$ is not a simplex of positive dimension. Since $g$ acts on the vertices of $K$ with at least eight orbits, Proposition 2.16 implies that $K^{\langle g\rangle}$ satisfies complementarity and has at least eight vertices. On the other hand, since $g\ne 1$, the complex $K^{\langle g\rangle}$ has at most $14$ vertices. From Theorem 2.10 we see that either $K^{\langle g\rangle}\cong\mathbb{CP}^2_9$ or $K^{\langle g\rangle}$ is a $14$-vertex $7$-dimensional pseudo-manifold that satisfies complementarity. The latter is impossible by Corollary 2.13, since $K^{\langle g\rangle}$ is an $\mathbb{F}_2$-homology manifold.
The proposition is proved. Proof of Proposition 6.3. Let $V=\{1,\dots,15\}$ be the vertex set of $K$. By Proposition 2.15 the simplicial complex $K$ is $5$-neighbourly.
Let $N\subset H$ be the subgroup isomorphic to $\mathrm{C}_2$. By Proposition 6.6 we have $K^N\cong\mathbb{CP}^2_9$. Hence, $N$ acts on $V$ with nine orbits, that is, with six orbits of length $2$ and three fixed points. We identify $K^N$ with $\mathbb{CP}^2_9$ along some isomorphism and identify the vertex set of $\mathbb{CP}^2_9$ with the affine plane $\mathcal{P}$ over $\mathbb{F}_3$ as in § 2.4. These identifications provide a bijection $\varphi\colon V/N\to \mathcal{P}$. Then some six points in $\mathcal{P}$ correspond to $N$-orbits of length $2$, and the other three points in $\mathcal{P}$ correspond to the $N$-fixed vertices of $K$. We denote by $m\subset \mathcal{P}$ the $3$-element subset consisting of the points that correspond to $N$-fixed vertices of $K$. Every $4$-element subset $\sigma\subset \mathcal{P}$ that contains $m$ corresponds to a $5$-element subset of $V$, which is a simplex of $K$ since $K$ is $5$-neighbourly. Hence $\sigma$ is a simplex of $\mathbb{CP}^2_9$. From Proposition 2.19 it follows that $m$ is a nonspecial line. Now we can choose affine coordinates $(x,y)$ in $\mathcal{P}$ so that:
The action of $H$ on $K$ induces an action of the quotient group $H/N\cong\mathrm{C}_2$ on $K^N=\mathbb{CP}^2_9$. We denote by $s$ the automorphism of $\mathbb{CP}^2_9$ given by the generator of $H/N$; then $s^2=\mathrm{id}$. The group $\operatorname{Sym}(\mathbb{CP}^2_9)$ consists of affine transformations of $\mathcal{P}$ that take special lines to special lines and preserve their cyclic order; see § 2.4. In particular, $s$ is such an affine transformation. Moreover, since the set of $N$-fixed points in $V$ is $H$-invariant, we see that the line $m$ is invariant under $s$. Finally, since the action of $H$ on $V$ is faithful, the set $V$ contains an $H$-orbit of length $4$; hence $s\ne\mathrm{id}$. It follows easily that Renumbering the vertices of $K$ we can achieve that the bijection $\varphi\colon V/N\to\mathcal{P}$ is given by Table 2. Then the group $N$ is generated by the permutation Since a generator $A$ of $H$ satisfies $A^2=B$ and induces the involution $s$ on $\mathcal{P}$, we see that
$$
\begin{equation*}
A=(1\ 3\ 2\ 4)^{\pm 1}(5\ 7\ 6\ 8)^{\pm 1}(9\ 11\ 10\ 12)^{\pm 1}
\end{equation*}
\notag
$$
for some signs $\pm$. Swapping the vertices in the pairs $\{3,4\}$, $\{7,8\}$ and $\{11,12\}$ we can achieve that all signs are pluses. Table 2.The bijection between the points $(x,y)\in\mathcal{P}$ and the $N$-orbits in $V$
By the construction of the simplicial complex $\mathbb{CP}^2_9$, the following $12$ five-element subsets of $\mathcal{P}$ are $4$-simplices of $\mathbb{CP}^2_9$:
$$
\begin{equation*}
\begin{aligned} \, &\bigl\{(0,2),(1,0),(2,0),(1,1),(2,1)\bigr\},& &\bigl\{(0,0),(1,0),(2,0),(1,1),(2,1)\bigr\},\\ &\bigl\{(0,0),(1,1),(2,1),(1,2),(2,2)\bigr\},& &\bigl\{(0,1),(1,1),(2,1),(1,2),(2,2)\bigr\},\\ &\bigl\{(0,1),(1,0),(2,0),(1,2),(2,2)\bigr\},& &\bigl\{(0,2),(1,0),(2,0),(1,2),(2,2)\bigr\},\\ &\bigl\{(0,1),(1,0),(1,1),(1,2),(2,0)\bigr\},& &\bigl\{(0,2),(1,0),(1,1),(1,2),(2,0)\bigr\},\\ &\bigl\{(0,2),(1,0),(1,1),(1,2),(2,1)\bigr\},& &\bigl\{(0,0),(1,0),(1,1),(1,2),(2,1)\bigr\},\\ &\bigl\{(0,0),(1,0),(1,1),(1,2),(2,2)\bigr\},& &\bigl\{(0,1),(1,0),(1,1),(1,2),(2,2)\bigr\}. \end{aligned}
\end{equation*}
\notag
$$
So the corresponding $12$ subsets of $V/N$ are simplices of $K^N$. Hence the preimages of these $12$ subsets under the projection $V\to V/N$ are simplices of $K$. It is easy to check that they are exactly the $12$ simplices listed in Proposition 6.3. Proposition 6.3 is proved. § 7. Proof of Proposition 5.1Throughout this section, $K$ is a $27$-vertex $\mathbb{Z}$-homology $16$-manifold such that the graded group $H_*(K;\mathbb{Z})$ is not isomorphic to $H_*(S^{16};\mathbb{Z})$, $V$ is the vertex set of $K$ and $G=\operatorname{Sym}(K)$ is the symmetry group of $K$. We identify $V$ with the set $\{1,\dots,27\}$. Then $G$ is a subgroup of $\mathrm{S}_{27}$. In this section we study the group $G$ and prove all properties from Proposition 5.1. Namely, property (a) is Proposition 7.22, property (b) follows from Propositions 7.4 and 7.14, property (c) is Propositions 7.2 and 7.6, property (d) is Corollary 7.12, property (e) is Proposition 7.16, property (f) is Corollary 7.19, and property (g) is Proposition 7.20. From Theorems 1.6 and 2.9 it follows that
We will make constant use of these two properties. 7.1. Subgroups of order $2^k$We start with the following standard lemma. Lemma 7.1. Suppose that $H$ is a finite $p$-group that acts non-trivially on a finite set $W$. Then there exists an index $p$ normal subgroup $N\vartriangleleft H$ such that at least one $H$-orbit in $W$ splits into $p$ different $N$-orbits. Proof. It is well known that any index $p$ subgroup of $H$ is normal and any proper subgroup of $H$ is contained in an index $p$ subgroup. Let $\alpha\subseteq W$ be an $H$-orbit of length greater than $1$ and $S\subset H$ be the stabilizer of an element $w\in \alpha$. Then $|S|< |H|$. We take for $N$ an index $p$ subgroup of $H$ that contains $S$. Then the orbit $\alpha$ splits into $p$ different $N$-orbits.
The lemma is proved. Proposition 7.2. Suppose that $H\subset G$ is a $2$-subgroup. Proof. First assume that $|H|\leqslant 4$. Since $K$ is $9$-neighbourly, the union of any two $H$-orbits is a simplex of $K$. Therefore, $K^H$ is nonempty and connected, and, moreover, the number of vertices of $K^H$ is equal to the number of $H$-orbits in $V$. From Theorem 3.1 it follows that $K^H$ is a connected $\mathbb{F}_2$-homology $d$-manifold for some $d$ and therefore a $d$-pseudomanifold. By Proposition 2.4 we have ${d<16}$. By Theorem 3.2 and Remark 3.3 we have $K^H\sim_2\mathbb{P}^2(m)$ for some $m\in\{1,2,4,8\}$. Since the fundamental cycle of $K^H$ represents a nontrivial homology class in $H_d(K^H;\mathbb{F}_2)$, we obtain that the homology dimension of $K^H$ coincides with the geometric dimension, that is, ${d=2m}$. So $d\in\{2,4,8\}$. Since all $H$-orbits in $V$ are simplices of $K$, Proposition 2.16 implies that $K^H$ satisfies complementarity. From Theorem 2.10 we obtain that $K^H$ is isomorphic to $\mathbb{RP}^2_6$, or $\mathbb{CP}^2_9$, or is an $8$-pseudomanifold with at least $15$ vertices.
Assume that $|H|=2$. Then $H$ acts on $V$ with at least $14$ orbits, so $K^H$ has at least $14$ vertices. Hence, $K^H$ is neither $\mathbb{RP}^2_6$ nor $\mathbb{CP}^2_9$. Therefore, $K^H$ is an $\mathbb{F}_2$-homology $8$-manifold with $n\geqslant 15$ vertices and $K^H\sim_2\mathbb{HP}^2$. Since $K^H$ satisfies complementarity, from Proposition 2.15 we obtain that $K^H$ is $(n-10)$-neighbourly. If $n$ were greater than $15$, we would obtain that $K^H$ is $6$-neighbourly, so that $H^4(K^H;\mathbb{F}_2)\,{=}\,0$, which would contradict the equivalence $K^H\,{\sim_2}\,\mathbb{HP}^2$. Thus, ${n\,{=}\,15}$. Assume that $|H|=4$. Then $K^H$ has at least seven vertices, so $K^H$ is not isomorphic to $\mathbb{RP}^2_6$. Hence $K^H$ is either isomorphic to $\mathbb{CP}^2_9$, or it is a $8$-pseudomanifold with at least $15$ vertices. By Lemma 7.1 there is a subgroup $N\vartriangleleft H$ such that $N\cong\mathrm{C}_2$ and at least one $H$-orbit in $V$ splits into two $N$-orbits. Then $K^H=(K^{N})^{H/N}$ and the group $H/N\cong\mathrm{C}_2$ acts nontrivially on $K^{N}$. By the above $K^{N}$ is a $15$-vertex connected $\mathbb{F}_2$-homology $8$-manifold. Then, by Proposition 2.4, $\dim K^H<\dim K^{N}=8$. Hence $K^H\cong\mathbb{CP}^2_9$. Therefore, $H$ acts on $V$ with nine orbits. Now assume that $|H|=8$. By Lemma 7.1 there exists a subgroup $N\vartriangleleft H$ of order $4$ such that at least one $H$-orbit in $V$ splits into two $N$-orbits. Then the quotient group $H/N$ acts nontrivially on $K^N$. By the above we have $K^{N}\cong\mathbb{CP}^2_9$. So by Proposition 2.18 we have
$$
\begin{equation*}
K^H=(K^{N})^{H/N}\cong(\mathbb{CP}^2_9)^{\mathrm{C}_2}\cong\mathbb{RP}^2_6.
\end{equation*}
\notag
$$
Since $K$ is $9$-neighbourly, every $H$-orbit in $V$ is a simplex of $K$. Therefore, the number of $H$-orbits is equal to the number of vertices of $K^H$, so $H$ acts on $V$ with six orbits.
Proposition 7.2 is proved. Proposition 7.4. The group $G$ does not contain a subgroup isomorphic to $\mathrm{C}_8$ or $\mathrm{C}_2\times\mathrm{C}_4$. Proof. Assume that $H\subset G$ is a subgroup isomorphic to $\mathrm{C}_8$ or $\mathrm{C}_2\times\mathrm{C}_4$. Then $H$ contains a subgroup $N\cong\mathrm{C}_2$ such that $H/N\cong\mathrm{C}_4$. By Proposition 7.2 we have that $K^N$ is a $15$-vertex $\mathbb{F}_2$-homology $8$-manifold satisfying complementarity. If the action of $H/N$ on $K^N$ were not faithful, then we would obtain a subgroup $Q\subset H$ of order $4$ such that $Q\supset N$ and $K^Q=K^N$, which would yield a contradiction with Proposition 7.2. So the action of $H/N$ on $K^N$ is faithful, that is, the symmetry group $\operatorname{Sym}(K^N)$ contains the subgroup $H/N\cong\mathrm{C}_4$. We obtain a contradiction with Proposition 6.1.
The proposition is proved. Suppose that $H\subset G$ is a subgroup of order $8$. From Proposition 7.4 and assertion (1) of Proposition 4.2 it follows that $H$ is isomorphic to $\mathrm{C}_2^3$, $\mathrm{D}_4$, or $\mathrm{Q}_8$. By Proposition 7.2 the group $H$ acts on $V$ with $6$ orbits. The length of every orbit is a power of two and does not exceed $8$. It is easy to see that there are exactly two ways to decompose $27$ into the sum of six powers of two that do not exceed $8$, namely, The following proposition says that the latter partition never occurs if $H$ is isomorphic to either $\mathrm{C}_2^3$ or $\mathrm{Q}_8$.Proposition 7.5. Suppose that $H$ is a subgroup of $G$ that is isomorphic to either $\mathrm{C}_2^3$ or $\mathrm{Q}_8$. Then $H$ acts on $V$ with $3$ orbits of length $8$ and $3$ fixed points. Proof. Assume the converse, that is, $H$ acts on $V$ with two orbits of length $8$, two orbits of length $4$, one orbit of length $2$ and one fixed point. Let $\alpha$ be one of the two $H$-orbits of length $4$. Consider the stabilizer $H_v$ of a vertex $v\in\alpha$. We have $H_v\cong\mathrm{C}_2$. Since $H$ is isomorphic to either $\mathrm{C}_2^3$ or $\mathrm{Q}_8$, it follows that $H_v$ is a normal subgroup of $H$. Hence $H_v$ fixes every vertex in $\alpha$, and we obtain a contradiction with Corollary 7.3.
The proposition is proved. Proof. Assume the converse, that is, $G$ contains a subgroup $H$ with $|H|=16$. From Proposition 7.4 and assertion (2) of Proposition 4.2 it follows that $H\cong\mathrm{C}_2^4$. By Lemma 7.1 there exists a subgroup $N\subset H$ of order $8$ such that at least one $H$-orbit in $V$ splits into two $N$-orbits. Then
$$
\begin{equation*}
K^H=(K^{N})^{H/N}\cong(\mathbb{RP}^2_6)^{\mathrm{C}_2},
\end{equation*}
\notag
$$
where the action of $\mathrm{C}_2$ on $\mathbb{RP}^2_6$ is nontrivial. All subgroups isomorphic to $\mathrm{C}_2$ of the group $\operatorname{Sym}(\mathbb{RP}^2_6)\cong\mathrm{A}_5$ are conjugate to each other. It is easy to check that $(\mathbb{RP}^2_6)^{\mathrm{C}_2}\cong\mathrm{pt}\sqcup\partial\Delta^2$; cf. [22], Table 9.1. Hence $K^H\cong\mathrm{pt}\sqcup\partial\Delta^2$. So from Proposition 2.16 it follows that all $H$-orbits are simplices of $K$. Thus, the number of $H$-orbits is equal to the number of vertices of $K^H$, that is, $H$ acts on $V$ with four orbits. The number of ones in the binary representation of $27$ is four, so $27$ can be written as the sum of four powers of two in a unique way, namely, It follows that $V$ splits into four $H$-orbits $\alpha_1$, $\alpha_2$, $\alpha_3$ and $\alpha_4$ of length $16$, $8$, $2$ and $1$, respectively. Let $S\subset H$ be the stabilizer of a vertex $v\in\alpha_3$. Then $S\cong\mathrm{C}_2^3$. Take any subgroup $Q\subset H$ such that $Q\cong\mathrm{C}_2^3$ and $Q\ne S$. Then $\alpha_3$ is a $Q$-orbit of length $2$, and we obtain a contradiction with Proposition 7.5.
The proposition is proved. 7.2. Elements of odd prime orderProposition 7.7. Suppose that $H\subset G$ is a subgroup that is isomorphic to $\mathrm{C}_p$, where $p$ is an odd prime number. Then every $H$-orbit $\sigma\subset V$ is a simplex of $K$ and one of the following assertions holds for the fixed-point simplicial complex $K^H$: Proof. The group $H=\mathrm{C}_p$ cannot act on a $27$-element set with one or two orbits. So from Proposition 2.16 it follows that $K^H$ is nonempty and is either a simplex of positive dimension or satisfies complementarity. However, by Theorem 3.1 any connected component of $K^H$ is an $\mathbb{F}_p$-homology manifold. Hence $K^H$ cannot be a simplex of positive dimension. Therefore, $K^H$ satisfies complementarity. Moreover, from Proposition 2.16 it follows that every $H$-orbit $\sigma\subset V$ is a simplex of $K$.
Let $F_1,\dots,F_s$ be the connected components of $K^H$. Each vertex $w$ of $K^H$ is the barycentre of some simplex $\sigma_{w}\in K$ that is an $H$-orbit. For every $i=1,\dots,s$ let $V_i$ be the union of all $H$-orbits $\sigma_w$, where $w$ ranges over the vertices of $F_i$. The subsets $V_1,\dots,V_s$ are certainly pairwise disjoint. Since all $H$-orbits in $V$ are simplices of $K$, we obtain $V_1\cup\cdots\cup V_s=V$. Moreover, $V_i\cup V_j$ is not a simplex of $K$ unless $i=j$. Note that $s\leqslant 3$. Indeed, otherwise both $V_1\cup V_2$ and $V_3\cup \cdots\cup V_s$ would be non-simplices of $K$, which would contradict complementarity. Consider three cases. Case 1: $s=3$. Since $V_1\cup V_2$, $V_1\cup V_3$ and $V_2\cup V_3$ are non-simplices of $K$, complementarity implies that $V_i\in K$ for $i=1,2,3$. So the connected components $F_i$ are simplices. On the other hand, by Theorem 3.1 the connected components $F_i$ are $\mathbb{F}_p$-homology manifolds. It follows that, in fact, $F_1$, $F_2$ and $F_3$ are points. So $K^H\cong \mathrm{pt}\sqcup\mathrm{pt}\sqcup\mathrm{pt}=\mathrm{pt}\sqcup\partial\Delta^1$. Case 2: $s=2$. By complementarity exactly one of the two sets $V_1$ and $V_2$ is a simplex of $K$. We may assume that $V_1\in K$ and $V_2\notin K$. Then $F_1$ is a simplex. On the other hand $F_1$ is an $\mathbb{F}_p$-homology manifold by Theorem 3.1. So $F_1=\mathrm{pt}$. Since $K^H$ satisfies complementarity, it follows that $F_2\cong\partial \Delta^k$ for some $k\geqslant 2$. Case 3: $s=1$. By Theorem 3.1 the simplicial complex $K^H$ is a connected orientable $\mathbb{F}_p$-homology manifold of even dimension $d=2m$. Hence $K^H$ is an orientable pseudo-manifold of dimension $2m$. By Theorem 3.2 we have
$$
\begin{equation}
H^*\bigl(K^H;\mathbb{F}_p\bigr)\cong\mathbb{F}_p[a]/(a^3),\qquad \deg a =m.
\end{equation}
\tag{7.1}
$$
Since the product of odd-dimensional cohomology classes is anticommutative and $p$ is odd, the isomorphism (7.1) can hold only if $m$ is even. Also, $m>0$ since $K^H$ is connected. Therefore, $d$ is positive and divisible by $4$. Moreover, by Proposition 2.4 we have $d<16$. Thus, $d$ is $4$, $8$ or $12$.
If $d=4$, then by Theorem 2.10 we have $K^H\cong\mathbb{CP}^2_9$. Suppose that $d$ is either $8$ or $12$. Let $n$ be the number of vertices of $K^H$ and $W$ be the vertex set of $K^H$. By Theorem 2.10 we have On the other hand, from (7.1) it follows that $K^H$ is not $m$-connected. Hence there exists an $(m+2)$-element subset $\eta\subset W$ such that $\eta\notin K^H$. By complementarity $W\setminus\eta\in K^H$. Therefore, $|W\setminus\eta|\leqslant d+1$. Thus, Combining inequalities (7.2) and (7.3), we obtain that $n=15$ if $d=8$ and ${19\leqslant n\leqslant 21}$ if $d=12$.Finally, it follows from (7.1) that $\chi\bigl(K^H\bigr)=3$. If $n$ were $20$, then by Proposition 2.12 we would obtain $\chi\bigl(K^H\bigr)=1$. Thus, $(d,n)\ne (12,20)$. Proposition 7.7 is proved. 7.3. The absence of elements of prime order $p\ne 2,3,11,13$Since $G$ is a subgroup of $\mathrm{S}_{27}$, it follows that $G$ does not contain elements of prime order $p>27$. Proof. Suppose that $p$ is either $19$ or $23$. Assume that there is an element $g\in G$ of order $p$. Then the group $H=\langle g\rangle\cong \mathrm{C}_p$ acts on the vertex set $V$ with one orbit $\alpha$ of length $p$ and $27-p$ fixed points. Since $\dim K=16$ and $p>17$, we have $\alpha\notin K$, which contradicts Proposition 7.7.
The proposition is proved. Proof. Assume that there is an element $g\in G$ of order $17$. Then the group $H=\langle g\rangle\cong \mathrm{C}_{17}$ acts on $V$ with one orbit $\alpha$ of length $17$ and $10$ fixed points. By Proposition 7.7 we have $\alpha\in K$. Let $u_0$ be a vertex of $\alpha$. Then $\beta=\alpha\setminus\{u_0\}$ is a $15$-simplex of $K$. Since $K$ is a $16$-pseudomanifold, it follows that $\beta$ is contained in exactly two $16$-simplices of $K$. One of these two simplices is $\alpha$. The other has the form where $v\notin\alpha$. The group $H$ acts transitively on $\alpha$ and fixes $v$. Since $K$ is $H$-invariant, we see that for all $u\in\alpha$. Thus, $K$ contains the boundary of the $17$-simplex $\alpha\cup\{v\}$, which is impossible since $K$ is a $16$-pseudomanifold with $27$ vertices.
The proposition is proved. Proof. Assume that there is an element $g\in G$ of order $p$, where $p$ is either $5$ or $7$. Let $H=\langle g\rangle$ and
$$
\begin{equation*}
V=\alpha_1\sqcup\cdots\sqcup\alpha_s\sqcup\{v_{1},\dots,v_t\},
\end{equation*}
\notag
$$
where $\alpha_1,\dots,\alpha_s$ are the $H$-orbits with $|\alpha_i|=p$ and $v_{1},\dots,v_t$ are the $H$-fixed points. We have $s\geqslant 1$ since $g\ne 1$, and $t\geqslant 1$ since $p$ is not a divisor of $27$. Then Since $K$ is $9$-neighbourly, we see that $\alpha_i\cup\{v_j\}\in K$ and so $\{b(\alpha_i),v_j\}\in K^H$ for all $i$ and $j$. Hence $K^H$ is a connected simplicial complex with $s+t$ vertices By Proposition 7.7, either $K^H\cong\mathbb{CP}^2_9$, or $K^H$ is an $\mathbb{F}_p$-homology manifold satisfying assertion (3) in Proposition 7.7. Consider two cases.
Case 1: $K^H\cong\mathbb{CP}^2_9$. Then $s+t=9$. So it follows from equation (7.4) that Hence $p=7$, $s=3$ and $t=6$. The combinatorial manifold $\mathbb{CP}^2_9$ is $3$-neighbourly, so and therefore We arrive at a contradiction sinceCase 2: $K^H$ is an $n$-vertex $\mathbb{F}_p$-homology $d$-manifold satisfying assertion (3) from Proposition 7.7. Then
$$
\begin{equation}
(d,n)=(8,15),\quad (d,n)=(12, 19),\quad\text{or}\quad(d,n)=(12,21).
\end{equation}
\tag{7.5}
$$
Combining the equality $s+t=n$ with (7.4), we obtain Since $n\geqslant 15$ and $p\geqslant 5$, we have $s\leqslant 3$. Since $K^H$ satisfies complementarity and $n\geqslant d+7$, it follows from Proposition 2.15 that $K^H$ is $5$-neighbourly. Hence
$$
\begin{equation*}
\sigma=\{b(\alpha_1),\dots,b(\alpha_s),v_1,\dots,v_{5-s}\}\in K^H.
\end{equation*}
\notag
$$
Since $K^H$ is a $d$-pseudomanifold, $\sigma$ is contained in a $d$-simplex $\tau\in K^H$. We have
$$
\begin{equation*}
\tau =\bigl\{b(\alpha_1),\dots,b(\alpha_s),v_{i_1},\dots,v_{i_{d+1-s}}\bigr\}
\end{equation*}
\notag
$$
for some $i_1<\cdots<i_{d+1-s}$. Then
$$
\begin{equation*}
\alpha_1\cup\cdots\cup\alpha_s\cup \bigl\{v_{i_1},\dots,v_{i_{d+1-s}}\bigr\}\in K.
\end{equation*}
\notag
$$
Since every simplex of $K$ has at most $17$ vertices, we obtain Combining this inequality with (7.6), we see that which contradicts (7.5).
Proposition 7.10 is proved. 7.4. Elements of order $3$Proposition 7.11. Suppose that an element $g\in G$ has order $3$. Then $g$ acts on $V$ without fixed points and $K^{\langle g\rangle}\cong\mathbb{CP}^2_9$. Proof. Put $H=\langle g\rangle$. Since $K$ is $9$-neighbourly, we see that the union of any two $H$-orbits is a simplex of $K$. Therefore any two vertices are joined by an edge in $K^H$. So $K^H$ is connected. If we prove that $g$ acts on $V$ without fixed points, then it will follow that $K^H$ has nine vertices; hence Proposition 7.7 will imply that $K^H\cong\mathbb{CP}^2_9$.
Let us prove that $g$ acts on $V$ without fixed points. Assume the converse, that is, let $V$ contain an $H$-fixed point and then at least three $H$-fixed points. The number of vertices of $K^H$ is equal to the number of $H$-orbits in $V$ and therefore is greater than $9$. From Proposition 7.7 it follows that $K^H$ is an $n$-vertex $d$-pseudomanifold satisfying complementarity, where $(d,n)$ is $(8,15)$ or $(12,19)$ or $(12,21)$. Consider two cases. Case 1: $d=8$. Then $n=15$. Hence $H$ acts on $V$ with exactly $15$ orbits, so with six orbits $\alpha_1,\dots,\alpha_6$ of length $3$ and nine fixed points. Since $K^H$ satisfies complementarity, it follows from Proposition 2.15 that $K^H$ is $5$-neighbourly. Therefore, the set $\sigma=\{b(\alpha_1),\dots,b(\alpha_5)\}$ is a simplex of $K^H$. But $K^H$ is an $8$-pseudomanifold, so $\sigma$ must lie in an $8$-simplex. This means that the union of the five orbits $\alpha_1,\dots,\alpha_5$ together with some other four orbits is a simplex of $K$. However, such union has cardinality at least $19$, which yields a contradiction since $\dim K=16$. Case 2: $d=12$. Then $n$ is either $19$ or $21$. Hence $H$ acts on $V$ either with four orbits $\alpha_1$, $\alpha_2$, $\alpha_3$ and $\alpha_4$ of length $3$ and $15$ fixed points or with three orbits $\alpha_1$, $\alpha_2$ and $\alpha_3$ of length $3$ and $18$ fixed points. Since $K^H$ satisfies complementarity, it follows from Proposition 2.15 that $K^H$ is $5$-neighbourly. Therefore, the set $\sigma=\{b(\alpha_1),b(\alpha_2),b(\alpha_3)\}$ is a simplex of $K^H$. But $K^H$ is a $12$-pseudomanifold, so $\sigma$ must be contained in a $12$-simplex. This means that the union of the three orbits $\alpha_1$, $\alpha_2$ and $\alpha_3$ with some other ten $H$-orbits is a simplex of $K$. However, such a union has cardinality at least $19$, which again gives a contradiction, since $\dim K=16$. The proposition is proved. Proposition 7.14. The group $G$ does not contain a subgroup isomorphic to $\mathrm{C}_{27}$ or $\mathrm{C}_3\times\mathrm{C}_9$. Proof. Assume that $H\subset G$ is a subgroup isomorphic to either $\mathrm{C}_{27}$ or $\mathrm{C}_3\times\mathrm{C}_9$. Then $H$ contains a subgroup $N\cong\mathrm{C}_3$ such that $H/N\cong \mathrm{C}_9$. By Proposition 7.11 we have $K^N\cong\mathbb{CP}^2_9$. By Corollary 7.12 the group $H$ acts on $V$ freely and therefore transitively. Hence the group $H/N\cong\mathrm{C}_9$ acts on $K^N$ simplicially and transitively on the vertices. We arrive at a contradiction, since the Sylow $3$-subgroup of the group $\operatorname{Sym}(\mathbb{CP}^2_9)$ is isomorphic to the Heisenberg group $\mathrm{He}_3=3^{1+2}_+$ (see [31]), which contains no elements of order $9$.
The proposition is proved. 7.5. Absence of elements of order $6$ Proof. Assume that $G$ contains an element of order $6$. Then $G$ contains a subgroup
$$
\begin{equation*}
H\cong\mathrm{C}_6\cong\mathrm{C}_2\times\mathrm{C}_3.
\end{equation*}
\notag
$$
Using Propositions 7.11 we obtain
$$
\begin{equation*}
K^H=(K^{\mathrm{C}_3})^{\mathrm{C}_2}\cong (\mathbb{CP}^2_9)^{\mathrm{C}_2}.
\end{equation*}
\notag
$$
If the action of $\mathrm{C}_2$ on $\mathbb{CP}^2_9$ is trivial, then $K^H\cong\mathbb{CP}^2_9$, and if the action of $\mathrm{C}_2$ on $\mathbb{CP}^2_9$ is nontrivial, then $K^H\cong\mathbb{RP}^2_6$ (by Proposition 2.18). So $K^H$ has either nine or six vertices. On the other hand, by Proposition 7.2 the factor $\mathrm{C}_2\subset H$ acts on $V$ with three fixed points and twelve orbits of length $2$. Further, the action of the factor $\mathrm{C}_3$ on $V$ is free and commutes with the action of the factor $\mathrm{C}_2$. It follows that $\mathrm{C}_3$ acts freely on the $15$-element set $V/\mathrm{C}_2$. Hence $|V/H|\!=\!5$. Therefore, $K^H$ has at most five vertices. The contradiction obtained shows that $G$ contains no elements of order $6$.
The proposition is proved. 7.6. Elements of order $11$Proposition 7.16. Suppose that an element $g\in G$ has order $11$. Then $g$ acts on $V$ with five fixed points and two orbits of length $11$ and $K^{\langle g\rangle}\cong\mathrm{pt}\sqcup\partial\Delta^5$, where the separate point $\mathrm{pt}$ is the barycentre of one of the two orbits of length $11$. Proof. Let $H=\langle g\rangle\cong\mathrm{C}_{11}$. First we need to exclude the possibility that $H$ acts on $V$ with one orbit $\alpha$ of length $11$ and $16$ fixed points. Assume that this is the case. By Proposition 7.7 we have $\alpha\in K$. Then the simplicial complex $K^H$ has $17$ vertices, namely, the barycentre $b(\alpha)$ and the $16$ fixed vertices of $K$. Since $K$ is $9$-neighbourly, any two vertices in $V\setminus\alpha$ are connected by an edge in $K$ and therefore in $K^H$. Moreover, since $K$ is a $16$-pseudomanifold, the $10$-simplex $\alpha$ is contained in a $16$-simplex of $K$. Hence $b(\alpha)$ is contained in a $6$-simplex of $K^H$. Consequently, $K^H$ is connected. However, then it follows from Proposition 7.7 that $K^H$ must have $9$, $15$, $19$ or $21$ vertices, so we arrive at a contradiction.
Thus, $H$ acts on $V$ with two orbits $\alpha$ and $\beta$ of length $11$ and five fixed points. By Proposition 7.7 we have $\alpha\in K$ and $\beta\in K$. Then the simplicial complex $K^H$ has seven vertices, namely, the barycentres $b(\alpha)$ and $b(\beta)$ and the five fixed points $v_1,\dots,v_5$ in $V$. From Proposition 7.7 it follows that $K^H\cong\mathrm{pt}\sqcup\partial\Delta^5$. Any two vertices $v_i$ and $v_j$ are connected by an edge in $K$ and therefore in $K^H$. Thus, the separate point $\mathrm{pt}$ is either $b(\alpha)$ or $b(\beta)$. The proposition is proved. Proposition 7.17. Suppose that $H\subset G$ is a subgroup that is isomorphic to one of the groups $\mathrm{C}_{22}$, $\mathrm{D}_{11}$, or $\mathrm{C}_{33}$. Then $V$ contains two $H$-orbits of length $11$. Proof. In each of the three cases the group $H$ contains a normal subgroup $N\cong\mathrm{C}_{11}$. The quotient group $H/N$ is isomorphic to either $\mathrm{C}_2$ or $\mathrm{C}_3$. By Proposition 7.16 the group $N$ acts on $V$ with two orbits $\alpha$ and $\beta$ of length $11$ and five fixed points, and $K^N\cong\mathrm{pt}\sqcup\partial\Delta^5$, where the separate point $\mathrm{pt}$ is the barycentre of either $\alpha$ or $\beta$. We may assume that $\mathrm{pt}=b(\alpha)$. The action of $H$ on $K$ induces an action of $H/N$ on $K^N$. Of course, the separate point $\mathrm{pt}=b(\alpha)$ is fixed under this action, and so the point $b(\beta)$ is also fixed. Consequently, $\alpha$ and $\beta$ are $H$-orbits.
The proposition is proved. Proof. Assume that an element $g\in G$ has order $22$ or $33$, and let $H=\langle g\rangle$. Then the element $g^{11}$ has order $2$ or $3$. By Proposition 7.17 the set $V$ contains two $H$-orbits $\alpha$ and $\beta$ of length $11$. Since neither $2$ nor $3$ divides $11$, the element $g^{11}$ fixes at least one point in each of the $H$-orbits $\alpha$ and $\beta$. Since $H$ is commutative, $g^{11}$ fixes every point in $\alpha$ and every point in $\beta$, which contradicts either Corollary 7.3 or Proposition 7.11.
The corollary is proved. Corollary 7.19. Suppose that $H\subset G$ is a subgroup isomorphic to $\mathrm{D}_{11}$. Then $H$ acts on $V$ with two orbits of length $11$, two orbits of length $2$ and one fixed point. Proof. By Proposition 7.17 the set $V$ contains two $H$-orbits $\alpha$ and $\beta$ of length $11$. Let $v_1,\dots,v_5$ be the five points of $V$ not belonging to $\alpha\cup\beta$. Let $g\in H$ be an element of order $2$. Then $g$ fixes an odd number of points in $\alpha$ and an odd number of points in $\beta$. On the other hand, by Corollary 7.3 the element $g$ fixes exactly three points in $V$. Therefore, $g$ fixes exactly one of the points $v_1,\dots,v_5$. We may assume that $g$ fixes $v_5$ and swaps the points in each of the pairs $\{v_1,v_2\}$ and $\{v_3,v_4\}$. Since the length of any $H$-orbit must be a divisor of $22$, we conclude that $\{v_1,v_2\}$, $\{v_3,v_4\}$ and $\{v_5\}$ are $H$-orbits.
The corollary is proved. 7.7. Elements of order $13$Proposition 7.20. Suppose that an element $g\in G$ has order $13$. Then $g$ acts on $V$ with two orbits of length $13$ and one fixed point, and $K^{\langle g\rangle}$ is the disjoint union of three points. Proof. Let $H=\langle g\rangle\cong\mathrm{C}_{13}$. We need to exclude the possibility that $H$ acts on $V$ with one orbit $\alpha$ of length $13$ and $14$ fixed points. Assume that this is the case. By Proposition 7.7 we have $\alpha\in K$. Then the simplicial complex $K^H$ has $15$ vertices, namely, the barycentre $b(\alpha)$ and the $14$ fixed points. Since $K$ is $9$-neighbourly, any two vertices in $V\setminus\alpha$ are connected by an edge in $K$ and thus in $K^H$. In addition, since $K$ is a $16$-pseudomanifold, the $12$-simplex $\alpha$ is contained in a $16$-simplex of $K$. Hence $b(\alpha)$ is contained in a $4$-simplex of $K^H$. Consequently, $K^H$ is connected. Then from Proposition 7.7 it follows that $K^H$ is an $\mathbb{F}_{13}$-homology $8$-manifold and therefore an $8$-pseudomanifold. On the other hand any nine vertices in $V\setminus\alpha$ form a simplex of $K$ and therefore a simplex of $K^H$. Hence the complex $K^H$ contains a subcomplex isomorphic to the $8$-skeleton of a $13$-simplex. Therefore, $K^H$ is not an $8$-pseudomanifold, which yields a contradiction.
Thus, $H$ acts on $V$ with two orbits of length $13$ and one fixed point. From Proposition 7.7 it follows that $K^H$ is the disjoint union of three points. The proposition is proved. Proof. Assume that an element $g\in G$ has order $26$ or $39$. Then the order of the element $g^{13}$ is either $2$ or $3$. Let $H=\langle g\rangle$, and let $N\subset H$ be the subgroup isomorphic to $\mathrm{C}_{13}$. By Proposition 7.20 the group $N$ acts on $V$ with two orbits $\alpha$ and $\beta$ of length $13$ and one fixed point $v$. Hence either both $\alpha$ and $\beta$ are $H$-orbits or $\alpha\cup\beta$ is an $H$-orbit. In the former case, as in the proof of Corollary 7.18, $g^{13}$ fixes every point in $\alpha$ and every point in $\beta$. In the latter case, the element $g^{13}$ swaps $\alpha$ and $\beta$ and therefore fixes exactly one point $v$ in $V$. In both cases we obtain a contradiction with either Corollary 7.3 or Proposition 7.11.
The proposition is proved. 7.8. End of the proof of property (a) Proof. From Propositions 7.8, 7.9, and 7.10 it follows that the only prime numbers that can occur among the orders of elements of $G$ are $2$, $3$, $11$ and $13$. Let us show that $G$ does not contain elements of orders $pq$, where $p$ and $q$ are different primes from $\{2,3,11,13\}$. For all pairs $\{p,q\}$, except $\{11,13\}$, this follows from Propositions 7.15 and 7.21 and Corollary 7.18. So we need to show that $G$ contains no elements of order $143$. If $g$ were such an element, then the permutation $g$ would be a product of two disjoint cycles of lengths $11$ and $13$, respectively. Hence $g^{13}$ would be a cycle of length $11$, which is impossible by Proposition 7.16. Therefore, the order of every element of $G$ is a prime power $p^k$, where $p\in\{2,3,11,13\}$. Further, by Propositions 7.4 and 7.14 the group $G$ does not contain elements of order $8$ or $27$. Also, $G$ contains no elements of order $121$ or $169$ since there are no such elements in $\mathrm{S}_{27}$. The proposition follows.
§ 8. Proof of Proposition 5.2The aim of this section is to prove Proposition 5.2. This is a purely group theoretic result, so we can completely forget about triangulations. It is easy to check that all subgroups $G\subset\mathrm{S}_{27}$ listed in Table 1 and also the group $\mathrm{PSU}(3,2)=\mathrm{C}_3^2\rtimes\mathrm{Q}_8$ acting on $\{1,\dots,27\}$ with three orbits of length $9$ satisfy properties (a)–(g) from Proposition 5.1. So we only need to prove that every subgroup $G$ satisfying these properties is in the list. Remark 8.1. The first part of the proof below (up to and including Proposition 8.4) follows the approach proposed to the author by Andrey Vasil’ev. A key feature of this approach is the use of the solvability of the group $G$. The author’s original proof was based on counting elements of different orders using Sylow’s theorems; it was more cumbersome. Throughout this section we assume that $G\subset \mathrm{S}_{27}$ is a nontrivial subgroup that satisfies properties (a)–(g) from Proposition 5.1. We put $V=\{1,\dots,27\}$. Proof. From property (a) it follows that $2$, $3$, $11$ and $13$ are the only prime numbers that can enter the prime factorization of $|G|$. By property (c) the exponent of $2$ in the prime factorization of $|G|$ does not exceed $3$. By property (d) a Sylow $3$-subgroup of $G$ acts freely on $27$ points. Hence the exponents of $3$ in the prime factorization of $|G|$ also does not exceed $3$. Now, let $p$ be either $11$ or $13$. The exponent of $p$ in $|\mathrm{S}_{27}|=27!$ is equal to $2$, so any subgroup $H\subset\mathrm{S}_{27}$ of order $p^2$ is a Sylow $p$-subgroup of $\mathrm{S}_{27}$. By Sylow’s theorems all Sylow $p$-subgroups are conjugate to one another. Hence any such subgroup $H$ is generated by two disjoint $p$-cycles. So if $|G|$ were divisible by $p^2$, then $G$ would contain a $p$-cycle, that is, a permutation of order $p$ that acts on $V$ with one orbit of length $p$ and ${27-p}$ fixed points. Then we would obtain a contradiction with property (e) or (g). Consequently, the exponents of $11$ and $13$ in the prime factorization of $|G|$ do not exceed $1$. Thus, $|G|$ divides $2^3\cdot 3^3\cdot 11\cdot 13$.
By property (a) all elements of $G$ have prime power orders. Since $|G|$ is not divisible by any of the primes $5$, $7$ and $17$, it follows from Corollary 4.6 that $G$ is solvable. The proposition is proved. Remark 8.3. Instead of using Theorem 4.4 due to Suzuki, the solvability of $G$ can be proved in another way. Indeed, the list of all noncyclic finite simple groups of order $\leqslant 2^3\cdot 3^3\cdot 11\cdot 13=30888$ is well known and consists of only $23$ groups. (Note that, in contrast to the general problem of the classification of finite simple groups, listing the finite simple groups of such a small order does not involve any hard results; cf. [27].) It is easy to check that for none of these simple groups its order divides $30888$. Therefore, any group whose order divides $30888$ is solvable. Proposition 8.4. The order of $G$ is one of the numbers $2$, $3$, $4$, $6$, $8$, $9$, $11$, $12$, $13$, $18$, $22$, $24$, $26$, $27$, $36$, $39$, $52$, $54$, $72$ or $351$. Proof. By property (a) and Proposition 8.2, $G$ is a solvable group all of whose elements $\ne 1$ have prime power order and, moreover, $|G|$ divides $2^3\cdot 3^3\cdot 11\cdot 13$. If $G$ is a $p$-group, then its order is either $2^m$ or $3^m$, where $1\leqslant m\leqslant 3$, or $11$ or $13$. Suppose that $G$ is not a $p$-group. Then from Theorem 4.7 it follows that
Consider four cases. Case 1: $p=2$. Then $2^k-1$ is $1$, $3$ or $7$. Since $q^n$ divides $2^k-1$, we have $k=2$, $q=3$ and $n=1$. Hence $|G|$ is either $12$ or $24$. Case 2: $p=3$. Then $3^k-1$ is $2$, $8$ or $26$. Since $q^n$ divides $3^k-1$, we have the following two subcases. Subcase 2a: $q=2$ and $(k,n)$ is one of the pairs $(1,1)$, $(2,1)$, $(2,2)$, $(2,3)$ or $(3,1)$. Then $m=k$, since $2^{m-k}$ must divide $q-1=1$. Hence $|G|$ is $6$, $18$, $36$, $72$ or $54$. Subcase 2b: $q=13$, $k=3$ and $n=1$. Then $|G|=351$. Case 3: $p=11$. Then $k=m=1$, so $q^n$ divides $p-1=10$. Hence $q=2$ and $n=1$. Therefore, $|G|=22$. Case 4: $p=13$. Then $k=m=1$, so $q^n$ divides $p-1=12$. Hence $q^n$ is $2$, $3$ or $4$. Therefore, $|G|$ is $26$, $39$ or $52$. The proposition is proved. Proof. First, assume that $g$ has order $2$. Then by property (c) the group $\langle g\rangle$ acts on $V$ with exactly $15$ orbits. Hence $g$ fixes exactly three points.
Second, assume that $g$ has order $4$. By property (c) the group $\langle g\rangle$ acts on $V$ with exactly nine orbits. There are two possibilities:
However, in the latter case we obtain that the element $g^2$ fixes seven points in $V$, so the latter case is impossible. The proposition is proved. Proof of Proposition 5.2. Let us consider in turn all possibilities for $|G|$ from Proposition 8.4.
Case 1: $|G|=2^k$, where $1\leqslant k\leqslant 3$. By properties (a) and (b) and assertion (1) of Proposition 4.2 we obtain that $G$ is isomorphic to one of the groups $\mathrm{C}_2$, $\mathrm{C}_2^2$, $\mathrm{C}_4$, $\mathrm{C}_2^3$, $\mathrm{D}_4$ and $\mathrm{Q}_8$. If $G$ is isomorphic to either $\mathrm{C}_2$ or $\mathrm{C}_4$, then from Proposition 8.5 it follows that the group $G$ fixes exactly three points in $V$ and acts freely on the other $24$ points. These two actions are present in Table 1. Suppose that $G\cong\mathrm{C}_2^2$. By Proposition 8.5 every nontrivial element $g$ of $G$ fixes exactly three points in $V$. It follows easily that up to isomorphism exactly two actions of $\mathrm{C}_2^2$ have this property:
Both actions are presented in Table 1. Finally, suppose that $|G|=8$. Then $G$ is isomorphic to $\mathrm{C}_2^3$, $\mathrm{D}_4$, or $\mathrm{Q}_8$. By property (c) the group $G$ acts on $V$ with exactly six orbits. Moreover, every $G$-orbit has length $1$, $2$, $4$ or $8$. As we have already mentioned in § 7.1, there are exactly two ways to decompose $27$ into a sum of six powers of two not exceeding $8$, namely, The actions of the three possible groups $G$ with three orbits of length $8$ and three fixed points are presented in Table 1.Now we consider the case of $G$ acting on $V$ with two orbits of length $8$, two orbits $\alpha$ and $\beta$ of length $4$, one orbit $\gamma$ of length $2$ and one fixed point $v_0$. As in the proof of Proposition 7.5, we see that if $G$ were isomorphic to $\mathrm{C}_2^3$ or $\mathrm{Q}_8$, then $G$ would contain a nontrivial element that fixes every point of $\alpha$, which would contradict Proposition 8.5. Therefore, $G\cong\mathrm{D}_4$. From Proposition 8.5 it follows that each subgroup $\mathrm{C}_2\subset\mathrm{D}_4$ fixes exactly two points in $\alpha\cup\beta\cup\gamma$. The group $\mathrm{D}_4$ has exactly five subgroups isomorphic to $\mathrm{C}_2$, one of which is the centre $Z$. If $Z$ fixed a point in an orbit of length $4$, then it would fix every point in this orbit, so we would arrive at a contradiction with Proposition 8.5 again. Hence $Z$ fixes the two points of $\gamma$. Then each of the eight points in $\alpha\cup\beta$ must be fixed by some noncentral subgroup $\mathrm{C}_2\subset\mathrm{D}_4$. Therefore, each of the four noncentral subgroups $\mathrm{C}_2\subset\mathrm{D}_4$ fixes exactly two points in $\alpha\cup\beta$. These two points lie in the same $G$-orbit (either $\alpha$ or $\beta$), since they must be swapped by the generator of $Z$. Thus, the stabilizers $G_u$ and $G_v$ of any points $u\in\alpha$ and $v\in\beta$ are not conjugate to each other. Moreover, the stabilizer $G_w$ of any point $w\in \gamma$ is a subgroup of order $4$ not containing a noncentral subgroup $\mathrm{C}_2\subset\mathrm{D}_4$. Therefore, $G_w=\mathrm{C}_4$. The action obtained is presented in Table 1. Case 2: $|G|=3^k$, where $1\leqslant k\leqslant 3$. From properties (a) and (b) and assertion (3) of Proposition 4.2 we obtain that $G$ is isomorphic to one of the groups $\mathrm{C}_3$, $\mathrm{C}_3^2$, $\mathrm{C}_9$, $\mathrm{C}_3^3$, $\mathrm{He}_3$ and $3_-^{1+2}$. Moreover, by property (d) the group $G$ acts freely on $V$. All these actions are presented in Table 1. Case 3: $|G|=2^k\cdot 3$, where $1\leqslant k\leqslant 3$. Since $G$ contains no elements of order $6$, from assertion (4) of Proposition 4.2 it follows that $G$ is isomorphic to one of the three groups $\mathrm{S}_3$, $\mathrm{A}_4$ and $\mathrm{S}_4$. By property (d) the lengths of all $G$-orbits in $V$ are divisible by $3$. On the other hand the lengths of all $G$-orbits must divide $|G|$. So the only possible orbit lengths are $3$, $6$, $12$ and $24$. If $G\cong \mathrm{S}_3$, then every $G$-orbit in $V$ has length $3$ or $6$. Any element of order $2$ in $G$ acts without fixed points on every $G$-orbit of length $6$ and fixes exactly one point in every $G$-orbit of length $3$. From Proposition 8.5 it follows that there are exactly three $G$-orbits of length $3$ and so exactly three $G$-orbits of length $6$. This action is presented in Table 1. If $G\cong\mathrm{A}_4$, then every $G$-orbit in $V$ has length $3$, $6$ or $12$. Since $|V|$ is odd, at least one $G$-orbit has length $3$. Suppose that $g\in G$ is an element of order $2$. The action of $G$ on each $G$-orbit $\alpha$ of length $3$ corresponds to the surjective homomorphism $\mathrm{A}_4\to\mathrm{C}_3$, so $g$ fixes every point of $\alpha$. Moreover, if $V$ contained a $G$-orbit $\beta$ of length $6$, then the action on $\beta$ would be isomorphic to the left action of $\mathrm{A}_4$ on the left cosets of $\mathrm{C}_2$ in $\mathrm{A}_4$. Then $g$ would fix exactly two points of $\beta$. By Proposition 8.5 the element $g$ fixes exactly three points in $V$. Therefore, $G$ acts on $V$ with one orbit of length $3$ and two orbits of length $12$. This action is presented in Table 1. Finally, suppose that $G\cong\mathrm{S}_4$. Since $|V|$ is odd, at least one $G$-orbit $\alpha$ has length $3$. The action of $G$ on $\alpha$ corresponds to the surjective homomorphism ${\mathrm{S}_4\!\to\!\mathrm{S}_3}$, so any element $g\in G$ of order $4$ swaps a pair of points in $\alpha$. It follows that the number of fixed points of $g^2$ in $V$ is strictly greater than the number of fixed points of $g$ in $V$, which contradicts Proposition 8.5. So this case is impossible. Case 4: $|G|=2^k\cdot 9$, where $1\leqslant k\leqslant 3$. Since $G$ contains no elements of order $6$ or $8$, it follows from assertions (5)–(7) of Proposition 4.2 that $G$ is one of the four groups $\mathrm{D}_9$, $\mathrm{C}_3\rtimes\mathrm{S}_3$, $\mathrm{C}_3^2\rtimes\mathrm{C}_4$ (where $\mathrm{C}_4$ acts faithfully on $\mathrm{C}_3^2$) and $\mathrm{PSU}(3,2)$. From property (d) it follows that $G$ acts on $V$ either with three orbits of length $9$ or with two orbits of length $18$ and $9$, respectively. Consider the two subcases. Subcase 4a: $G$ acts on $V$ with three orbits of length $9$. Since all Sylow $2$-subgroups of $G$ are conjugate to one another, $G$ has a unique up to isomorphism transitive action on nine points. The corresponding actions of $\mathrm{D}_9$, $\mathrm{C}_3\rtimes\mathrm{S}_3$, and ${\mathrm{C}_3^2\rtimes\mathrm{C}_4}$ on $27$ points with three orbits of length $9$ are presented in Table 1, and the action of $\mathrm{PSU}(3,2)$ is mentioned separately in Proposition 5.2. Subcase 4b: $G$ acts on $V$ with two orbits $\alpha$ and $\beta$ of length $18$ and $9$, respectively. First assume that $G$ is isomorphic to either $\mathrm{D}_9$ or $\mathrm{C}_3\rtimes\mathrm{S}_3$. It can be checked directly that any element $g\in G$ of order $2$ fixes exactly one point in $\beta$. Moreover, $G$ acts regularly transitively on $\alpha$. Hence $g$ fixes exactly one point in $V$, which contradicts Proposition 8.5. Second assume that $G$ is isomorphic to $\mathrm{C}_3^2\rtimes\mathrm{C}_4$ or $\mathrm{PSU}(3,2)$. It is easy to check that
So each of the two groups $G$ has a unique up to isomorphism transitive action on $18$ points, which is thus realized on $\alpha$. One can easily check that in both cases the group $G$ contains an element $g$ of order $4$ that acts on $\alpha$ with an orbit of length $2$. Therefore, the number of fixed points of $g^2$ is strictly greater than the number of fixed points of $g$, which contradicts Proposition 8.5 again. So this subcase is impossible. Case 5: $|G|$ is either $11$ or $22$. Since $G$ contains no elements of order $22$, it follows from Proposition 4.1 that $G$ is isomorphic to either $\mathrm{C}_{11}$ or $\mathrm{D}_{11}$. If $G\cong\mathrm{C}_{11}$, then by property (e) it acts on $V$ with two orbits of length $11$ and five fixed points. If ${G\cong\mathrm{D}_{11}}$, then by property (f) it acts on $V$ with two orbits of length $11$, two orbits of length $2$ and one fixed point. Both these actions are presented in Table 1. Case 6: $|G|=2^k\cdot 13$, where $0\leqslant k\leqslant 2$. Since $G$ contains no elements of order $26$, it follows from Proposition 4.1 and assertion (8) of Proposition 4.2 that $G$ is isomorphic to one of the three groups $\mathrm{C}_{13}$, $\mathrm{D}_{13}$ and $\mathrm{C}_{13}\rtimes\mathrm{C}_4$ (where $\mathrm{C}_4$ acts faithfully on $\mathrm{C}_{13}$). Let $H$ be a Sylow $13$-subgroup of $G$; then $H\cong\mathrm{C}_{13}$. By property (g) the subgroup $H$ acts on $V$ with three orbits of lengths $13$, $13$ and $1$. Every $G$-orbit is the union of several $H$-orbits, and the length of every $G$-orbit divides $|G|$. Therefore, the lengths of $G$-orbits in $V$ are either $13$, $13$ and $1$, or $26$ and $1$. Consider two subcases. Subcase 6a: $G$ acts on $V$ with three orbits of length $13$, $13$ and $1$. Each of the three possible groups $G$ has a unique up to isomorphism transitive action on $13$ points. The corresponding three actions on $27$ points with orbit lengths $13$, $13$ and $1$ are presented in Table 1. Subcase 6b: $G$ acts on $V$ with two orbits of length $26$ and $1$. Then $G$ is isomorphic to either $\mathrm{D}_{13}$ or $\mathrm{C}_{13}\rtimes\mathrm{C}_4$. If $G\cong\mathrm{D}_{13}$, then $G$ acts regularly transitively on the orbit of length $26$, so every element $g\in G$ of order $2$ fixes only one point in $V$, which contradicts Proposition 8.5. If $G\cong\mathrm{C}_{13}\rtimes\mathrm{C}_4$, then $G$ contains an element $g$ of order $4$. The stabilizer $G_v$ of any point $v$ in the orbit of length $26$ has order $2$; hence $g\notin G_v$. Therefore, $g$ fixes only one point in $V$, which contradicts Proposition 8.5 again. So this subcase is impossible. Case 7: $|G|=39$. From property (d) it follows that the length of every $G$-orbit in $V$ is divisible by $3$. Since the length of every $G$-orbit divides $39$, it follows that all $G$-orbits in $V$ have length $3$. We arrive at a contradiction since $G$ contains an element of order $13$. So this case is impossible. Case 8: $|G|=54$. We are going to prove that this case is also impossible. Let $N$ be the number of pairs $(g,v)\in G\times V$ such that $g$ has order $2$ and $g(v)=v$. To arrive at a contradiction we compute $N$ in two ways and show that the results are different. Firstly, from property (d) it follows that $G$ acts transitively on $V$. Hence the stabilizer $G_v$ of every point $v\in V$ has order $2$. Therefore, $N=|V|=27$. Secondly, from property (a) it follows that the order of every element of $G$ is either $2$ or a power of $3$. The Sylow $3$-subgroup $H$ of $G$ has index $2$ and therefore is normal. It follows that any $3$-subgroup of $G$ is contained in $H$, so $G$ has exactly $26$ elements $g\ne 1$ whose order is a power of $3$. Therefore, $G$ contains exactly $27$ elements $g$ that have order $2$. By Proposition 8.5 each such element fixes exactly three points in $V$. Thus, $N=3\cdot 27=81$. This is a contradiction. Case 9: $|G|=351$. Since $G$ contains no elements of order $39$, it follows from assertion (9) of Proposition 4.2 that $G$ is isomorphic to the semidirect product ${\mathrm{C}_3^3\rtimes\mathrm{C}_{13}}$. It is easy to check that all subgroups of order $13$ of $\mathrm{C}_3^3\rtimes\mathrm{C}_{13}$ are conjugate to each other, so this group admits a unique (up to isomorphism) action on $27$ points. This action is in Table 1. We have looked through all the cases. Proposition 5.2 follows. § 9. Absence of triangulations with symmetry group $\mathrm{PSU}(3,2)$In this section we prove Proposition 5.3. We assume that $K$ is a $27$-vertex $\mathbb{Z}$-homology $16$-manifold such that $H_*(K;\mathbb{Z})\!\not\cong\! H_*(S^{16};\mathbb{Z})$ and $\operatorname{Sym}(K)\!\cong\!\mathrm{PSU}(3,2)$, and our goal is to arrive at a contradiction. We denote the group $\operatorname{Sym}(K)$ by $G$. Throughout this section a special role is played by subgroups $H\subset G$ that are isomorphic to $\mathrm{C}_4$. We denote the set of all such subgroups $H$ by $\mathcal{H}$. By Remark 4.3 a Sylow $2$-subgroup of $G$ is isomorphic to $\mathrm{Q}_8$, so $\mathcal{H}$ is nonempty. In fact, it is easy to check that the set $\mathcal{H}$ consists of three conjugacy classes of $9$ subgroups each, but we do not need this fact. Recall that it follows from Theorems 1.6 and 2.9 that $K$ satisfies complementarity. Then by Proposition 2.15 we obtain that $K$ is $9$-neighbourly. Let $V$ denote the vertex set of $K$. By Propositions 5.1 and 5.2 the group $G$ acts on $V$ with three orbits of length $9$. We denote these orbits by $\alpha_0$, $\alpha_1$ and $\alpha_2$, where we conveniently consider the indices $0$, $1$ and $2$ as elements of the field $\mathbb{F}_3$. Since $K$ is $9$-neighbourly, we have that $\alpha_0$, $\alpha_1$ and $\alpha_2$ are simplices of $K$. By Proposition 8.5 each subgroup $H\in\mathcal{H}$ acts on $V$ with six orbits of length $4$ and three fixed points. Then every $G$-orbit $\alpha_t$ contains one $H$-fixed point and two $H$-orbits of length $4$. We identify the set of vertices of $\mathbb{CP}^2_9$ with an affine plane $\mathcal{P}$ over $\mathbb{F}_3$ as in § 2.4. Proposition 9.1. The following assertions hold (possibly, after reversing the cyclic order of the orbits $\alpha_0$, $\alpha_1$ and $\alpha_2$). Table 3.The bijection $\varphi_H$ between the points $(x,y)\in\mathcal{P}$ and the $H$-orbits in $V$
Remark 9.2. We conveniently assume that the isomorphisms $\varphi_H$ for different $H$ lead to different copies of $\mathbb{CP}^2_9$, so the coordinates $(x,y)$ for different $H$ are not related. Of course, the fixed vertices $u_t$ and the orbits $\beta_t$ and $\gamma_t$ depend on the choice of $H$. For the sake of simplicity we do not reflect this dependence in the notation. Proof of Proposition 9.1. Let $H\in\mathcal{H}$. For each $t\in\mathbb{F}_3$ the $G$-orbit $\alpha_t$ consists of the $H$-fixed point $u_t$ and two $H$-orbits $\beta_t$ and $\gamma_t$ of length $4$: Note that we have the freedom to choose which of the two orbits is $\beta_t$ and which is $\gamma_t$. In what follows we take the opportunity to swap these orbits.
By Proposition 7.2 we have an isomorphism $\varphi_H\colon K^H\to\mathbb{CP}^2_9$. The isomorphism $\varphi_H$ provides a bijection $V/H\to\mathcal{P}$, which we also denote by $\varphi_H$. Under this bijection some six points in $\mathcal{P}$ correspond to $H$-orbits of length $4$, and the other three points in $\mathcal{P}$ correspond to the $H$-fixed vertices of $K$. We denote by $m\subset \mathcal{P}$ the $3$-element subset consisting of the points corresponding to $H$-fixed vertices of $K$. Then every $4$-element subset $\sigma\subset \mathcal{P}$ that contains $m$ corresponds to a $7$-element subset of $V$, which is a simplex of $K$ since $K$ is $9$-neighbourly. Hence $\sigma$ is a simplex of $\mathbb{CP}^2_9$. From Proposition 2.19 it follows that $m$ is a nonspecial line. Hence we can choose affine coordinates $(x,y)$ in $\mathcal{P}$ so that
Swapping the orbits $\alpha_1$ and $\alpha_2$ (if necessary), we can achieve that We denote the stabilizer $G_{u_0}$ by $S$; then $H\subset S$. By Remark 4.3 we have $S\cong\mathrm{Q}_8$. By Sylow’s theorems all subgroups of $G$ of order $8$ are conjugate to each other. So every $G$-orbit $\alpha_t$ contains a point with stabilizer $S$. This point is $u_t$ since this is the unique point fixed by $H$. So we have $G_{u_t}=S$ for all $t\in\mathbb{F}_3$. Since $S\cong\mathrm{Q}_8$, we see that $H$ is a normal subgroup of $S$. The action of $S$ on $K$ induces an action of the quotient group $S/H\cong\mathrm{C}_2$ on $K^H$, which we identify with $\mathbb{CP}^2_9$ along $\varphi_H$. We denote the involutive automorphism of $\mathbb{CP}^2_9$ given by the generator of $S/H$ by $s$. From Proposition 7.5 it follows that the group $S$ acts on $V$ with three fixed points $u_t$ and three orbits $\beta_t\cup\gamma_t$ of length $8$. Hence for each $t\in\mathbb{F}_3$ the automorphism $s$ fixes the point $\varphi_H(u_t)=(0,t)$ and swaps the points $\varphi_H(\beta_t)$ and $\varphi_H(\gamma_t)$. On the other hand, since $s\in\operatorname{Sym}(\mathbb{CP}^2_9)$, we see that $s$ is an affine transformation of $\mathcal{P}$ that takes special lines to special lines preserving their cyclic order. It follows easily that for all $(x,y)\in\mathcal{P}$. Therefore, for each $t\in\mathbb{F}_3$ the points $\varphi_H(\beta_t)$ and $\varphi_H(\gamma_t)$ have the same $y$-coordinate $\nu(t)$. Then $\nu$ is a permutation of the elements of $\mathbb{F}_3$. In addition, swapping the orbits in pairs $\{\beta_t,\gamma_t\}$, we can achieve that
$$
\begin{equation*}
\varphi_H(\beta_t)=\bigl(1,\nu(t)\bigr) \quad\text{and}\quad \varphi_H(\gamma_t)=\bigl(2,\nu(t)\bigr)
\end{equation*}
\notag
$$
for all $t\in\mathbb{F}_3$. We would like to prove that the permutation $\nu$ is in fact trivial. To do this, we enumerate all nontrivial permutations and arrive at a contradiction in each case. There are three cases. Case 1: $\nu$ is a transposition. We may assume that $\nu=(1\ 2)$ since the cases of the two other transpositions are similar. Then the bijection $\varphi_H\colon V/H\to\mathcal{P}$ is as in Table 4 on the left. By the construction of $\mathbb{CP}^2_9$ the following two sets are simplices of $\mathbb{CP}^2_9$:
$$
\begin{equation}
\begin{aligned} \, &\bigl\{(0,1),(1,0),(2,0),(1,2),(2,2)\bigr\}, \\ &\bigl\{(0,1),(1,1),(2,1),(1,2),(2,2)\bigr\}. \end{aligned}
\end{equation}
\tag{9.4}
$$
Hence the corresponding unions of $H$-orbits are simplices of $K$. These unions of $H$-orbits are
$$
\begin{equation*}
\{u_1\}\cup\beta_0\cup\gamma_0\cup\beta_1\cup\gamma_1 =\alpha_1\cup\bigl(\alpha_0\setminus\{u_0\}\bigr)
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
\{u_1\}\cup\beta_2\cup\gamma_2\cup\beta_1\cup\gamma_1 =\alpha_1\cup\bigl(\alpha_2\setminus\{u_2\}\bigr),
\end{equation*}
\notag
$$
respectively. Therefore, $\alpha_0\setminus\{u_0\}$ and $\alpha_2\setminus\{u_2\}$ are simplices of $\operatorname{link}(\alpha_1,K)$. But the simplex $\alpha_1$ is $G$-invariant, so the simplicial complex $\operatorname{link}(\alpha_1,K)$ is also $G$-invariant. Since $G$ acts transitively on each of the sets $\alpha_0$ and $\alpha_2$, it follows that $\operatorname{link}(\alpha_1,K)$ contains the simplices $\alpha_0\setminus\{v\}$ for all $v\in\alpha_0$ and the simplices $\alpha_2\setminus\{v\}$ for all $v\in\alpha_2$. In other words, $\operatorname{link}(\alpha_1,K)$ contains simultaneously the boundaries of the $8$-simplices $\alpha_0$ and $\alpha_2$. This yields a contradiction since $K$ is a $\mathbb{Z}$-homology $16$-manifold, so that $\operatorname{link}(\alpha_1,K)$ must be a connected $\mathbb{Z}$-homology $7$-manifold with the homology of a $7$-sphere. Table 4.The wrong bijections $\varphi_H$
Case 2: $\nu(t)=t+1$ for all $t\in\mathbb{F}_3$. Then the bijection $\varphi_H\colon V/H\to\mathcal{P}$ is as in the middle of Table 4. We arrive at a contradiction literally as in the previous case. Indeed, the sets (9.4) are simplices of $\mathbb{CP}^2_9$ again. Therefore, $\alpha_0\setminus\{u_0\}$ and $\alpha_2\setminus\{u_2\}$ are simplices of $\operatorname{link}(\alpha_1,K)$, and so $\partial\alpha_0$ and $\partial\alpha_2$ are subcomplexes of $\operatorname{link}(\alpha_1,K)$, which is impossible. Case 3: $\nu(t)=t-1$ for all $t\in\mathbb{F}_3$. Then the bijection $\varphi_H\colon V/H\to\mathcal{P}$ is as in Table 4, on the right. This case is a bit different. The set is a simplex of $\mathbb{CP}^2_9$. Hence
$$
\begin{equation*}
\{u_1\}\cup\beta_1\cup\gamma_1\cup\beta_0\cup\gamma_0 =\alpha_1\cup\bigl(\alpha_0\setminus\{u_0\}\bigr)
\end{equation*}
\notag
$$
is a simplex of $K$. As in Case 1, this implies that $\operatorname{link}(\alpha_1,K)$ contains a subcomplex $\partial\alpha_0$. Since $\operatorname{link}(\alpha_1,K)$ must be a connected $\mathbb{Z}$-homology $7$-manifold, it follows that, in fact, $\operatorname{link}(\alpha_1,K)=\partial\alpha_0$. On the other hand is a simplex of $\mathbb{CP}^2_9$ and so
$$
\begin{equation*}
\{u_1\}\cup\beta_0\cup\beta_1\cup\beta_2\cup\gamma_1 =\alpha_1\cup\bigl(\beta_0\cup\beta_2\bigr)
\end{equation*}
\notag
$$
is a simplex of $K$. Therefore, $\beta_0\cup\beta_2$ is a simplex of $\operatorname{link}(\alpha_1,K)$, and we arrive at a contradiction since $\beta_0\cup\beta_2$ is not contained in $\alpha_0$. Thus, $\nu$ is a trivial permutation. Hence the bijection $\varphi_H\colon V/H\to\mathcal{P}$ is as in Table 3, and we obtain assertion (2). For each $t\in\mathbb{F}_3$, we have that and are simplices of $\mathbb{CP}^2_9$, while and are not. So the unions of the $H$-orbits corresponding to elements of the sets (9.5) and (9.6) are simplices of $K$ and the unions of the $H$-orbits corresponding to elements of the sets (9.7) and (9.8) are not. For (9.6), (9.7) and (9.8) we obtain exactly the statements (9.1), (9.2) and (9.3) from assertion (3), respectively. For (9.5) we obtain As above, since $G$ acts transitively on $\alpha_{t+1}$, it follows that $\operatorname{link}(\alpha_t,K)\supseteq \partial\alpha_{t+1}$. Since $\operatorname{link}(\alpha_t,K)$ is a connected $\mathbb{Z}$-homology $7$-manifold, we obtain $\operatorname{link}(\alpha_t,K)=\partial\alpha_{t+1}$, which is assertion (1).Proposition 9.1 is proved. Remark 9.3. Though we have considered a subgroup $H\in\mathcal{H}$ to prove assertion (1), the resulting cyclic order of the orbits $\alpha_0$, $\alpha_1$ and $\alpha_2$ is certainly independent of $H$, since the assertion cannot be true simultaneously for the two opposite orders. In what follows we always number the orbits $\alpha_0$, $\alpha_1$ and $\alpha_2$ so that the assertions of Proposition 9.1 are true for all $H$. The group $G$ has nine Sylow $2$-subgroups, each isomorphic to $\mathrm{Q}_8$; see Remark 4.3. By Sylow’s theorems all of them are conjugate to one another. Hence for each of the three orbits $\alpha_0$, $\alpha_1$ and $\alpha_2$ the stabilizers of the nine vertices in this orbit are exactly the nine different subgroups of $G$ isomorphic to $\mathrm{Q}_8$. Therefore, there is a unique permutation $\theta$ of the set $V$ such that:
It is obvious that $\theta^3=1$. It follows immediately from the construction that the permutation $\theta$ commutes with the action of $G$ on $V$. So for each subgroup $H\in\mathcal{H}$ the permutation $\theta$ takes $H$-orbits to $H$-orbits. Let us use the notation from Proposition 9.1 for $H$-orbits. Then Proposition 9.4. For each $t\in\mathbb{F}_3$ and each vertex $v\in \alpha_t$ the set is a simplex of $K$. Proof. The stabilizer $G_v$ of $v$ is isomorphic to $\mathrm{Q}_8$ and therefore contains a subgroup $H\in\mathcal{H}$ (in fact, it contains exactly three such subgroups). Then $v$ is the $H$-fixed point in $\alpha_t$, that is, the point $u_t$ for the subgroup $H$. The required assertion now follows from assertion (3) of Proposition 9.1. For each subgroup $H\in\mathcal{H}$ and each $t\in\mathbb{F}_3$ the permutation $\theta$ takes the pair of $H$-orbits $\{\beta_{t+1},\gamma_{t+1}\}$ to the pair of $H$-orbits $\{\beta_{t+2},\gamma_{t+2}\}$, where we use the notation for $H$-orbits from Proposition 9.1. Nevertheless, it is by no means true that $\theta$ always takes $\beta_{t+1}$ to $\beta_{t+2}$ and $\gamma_{t+1}$ to $\gamma_{t+2}$. We conveniently introduce signs $\eta_{H,t}=\pm 1$ indexed by pairs $(H,t)\in\mathcal{H}\times\mathbb{F}_3$ in the following way:
$$
\begin{equation*}
\eta_{H,t}= \begin{cases} 1&\text{if }\theta(\beta_{t+1})=\beta_{t+2} \text{ and }\theta(\gamma_{t+1})=\gamma_{t+2}, \\ -1&\text{if }\theta(\beta_{t+1})=\gamma_{t+2} \text{ and } \theta(\gamma_{t+1})=\beta_{t+2}. \end{cases}
\end{equation*}
\notag
$$
Since $\theta^3=1$, we obtain the following statement. Remark 9.6. Actually, the orbits $\beta_t$ and $\gamma_t$ are not determined uniquely by the conditions from assertion (2) of Proposition 9.1. Namely, we are free to change the sign of the coordinate $x$ on the plane $\mathcal{P}$ and swap simultaneously the orbits in every pair $\{\beta_t,\gamma_t\}$, where $t\in\mathbb{F}_3$. Nevertheless, this operation does not change the signs $\eta_{H,t}$, so they are well defined. Further, if a group $H$ acts on a set $Z$ and $z_1,z_2\in Z$, then we put
$$
\begin{equation*}
\varepsilon_{H}(z_1,z_2)= \begin{cases} 1&\text{if }z_1 \text{ and }z_2 \text{ lie in the same }H\text{-orbit}, \\ -1&\text{if }z_1\text{ and }z_2 \text{ lie in different }H\text{-orbits}. \end{cases}
\end{equation*}
\notag
$$
The definitions of $\eta_{H,t}$ and $\varepsilon_H(z_1,z_2)$ immediately imply the following assertion. Proposition 9.7. Suppose that $H\in\mathcal{H}$. Denote the $H$-orbits in $V$ as in Proposition 9.1. Suppose that $t\in\mathbb{F}_3$, $v\in\alpha_{t+1}\setminus\{u_{t+1}\}$ and $w\in\alpha_{t+2}\setminus\{u_{t+2}\}$. Then the following hold: We also need one easy observation concerning actions of the quaternion group $\mathrm{Q}_8$. Lemma 9.8. Suppose that the group $\mathrm{Q}_8$ acts transitively and freely on an $8$-element set $Z$. Let $H_1$, $H_2$ and $H_3$ be the three subgroups of $\mathrm{Q}_8$ that are isomorphic to $\mathrm{C}_4$. Then for all $z_1,z_2\in Z$. Proof. Let $g\in \mathrm{Q}_8$ be the element that takes $z_1$ to $z_2$. The assertion of the lemma follows immediately from the fact that $g$ belongs either to all the three subgroups $H_1$, $H_2$ and $H_3$ or to exactly one of them. Now suppose that $t\in\mathbb{F}_3$, $v\in\alpha_{t+1}$ and $w\in\alpha_{t+2}$. By assertion (1) of Proposition 9.1 we have that $\sigma=\alpha_{t+1}\cup(\alpha_{t+2}\setminus\{w\})$ is a $16$-simplex of $K$. Then
$$
\begin{equation*}
\rho=\sigma\setminus\{v\}=(\alpha_{t+1}\setminus\{v\})\cup(\alpha_{t+2}\setminus\{w\})
\end{equation*}
\notag
$$
is a $15$-simplex of $K$. Since $K$ is a $16$-pseudomanifold, the simplex $\rho$ is contained in exactly one $16$-simplex $\tau\in K$ different from $\sigma$. Since $\operatorname{link}(\alpha_{t+2},K)=\partial\alpha_t$, we see that $\tau\ne (\alpha_{t+1}\setminus\{v\})\cup\alpha_{t+2}$. So there is a unique vertex $u\in\alpha_t$ such that
$$
\begin{equation*}
\tau=(\alpha_{t+1}\setminus\{v\})\cup(\alpha_{t+2}\setminus\{w\})\cup\{u\}
\end{equation*}
\notag
$$
is a simplex of $K$. We obtain a well-defined map
$$
\begin{equation*}
f_t\colon \alpha_{t+1}\times\alpha_{t+2}\to\alpha_t, \qquad (v,w)\mapsto u.
\end{equation*}
\notag
$$
The map $f_t$ is certainly $G$-equivariant with respect to the diagonal action of $G$ on the set $\alpha_{t+1}\times\alpha_{t+2}$. Since $G$ acts transitively on $\alpha_t$, all preimages $f_t^{-1}(u)$, where $u\in\alpha_t$, consist of the same number of elements. Hence $|f_t^{-1}(u)|=9$ for all $u$. Lemma 9.9. (1) If $w=\theta(v)$, then $f_t(v,w)=\theta^2(v)$. (2) If $w\ne\theta(v)$, then $f_t(v,w)$ is neither $\theta^2(v)$ nor $\theta(w)$. Proof. Assertion (1) is a reformulation of Proposition 9.4. Let us prove assertion (2). Assume that $f_t(v,w)=\theta^2(v)$. Then
$$
\begin{equation*}
\tau=(\alpha_{t+1}\setminus\{v\})\cup(\alpha_{t+2}\setminus\{w\})\cup\{\theta^2(v)\}
\end{equation*}
\notag
$$
is a simplex of $K$. Applying Proposition 9.4 to the vertex $v$ we obtain that the set
$$
\begin{equation*}
\kappa=\{v\}\cup\bigl(\alpha_{t+2}\setminus\{\theta(v)\}\bigr)\cup\bigl(\alpha_{t}\setminus\{\theta^2(v)\}\bigr)
\end{equation*}
\notag
$$
is a simplex of $K$. Since $\tau\cup\kappa=V$, we arrive at a contradiction with complementarity. In a similar way, if $f_t(v,w)=\theta(w)$, then we arrive at a contradiction by applying Proposition 9.4 to the vertex $w$.
The lemma is proved. Proposition 9.10. There exists a subgroup $H\in\mathcal{H}$, an element $t\in\mathbb{F}_3$, and vertices $v\in\alpha_{t+1}\setminus\{u_{t+1}\}$ and $w\in\alpha_{t+2}\setminus\{u_{t+2}\}$ such that (Here, as in Proposition 9.1, we denote by $u_0$, $u_1$ and $u_2$ the $H$-fixed points in $\alpha_0$, $\alpha_1$ and $\alpha_2$, respectively.) Proof. Fix an arbitrary vertex $u_0\in \alpha_0$ and put $u_1=\theta(u_0)$ and $u_2=\theta^2(u_0)$. By the definition of $\theta$ we have $G_{u_0}=G_{u_1}=G_{u_2}$; we denote this stabilizer by $S$. Then ${S\cong\mathrm{Q}_8}$. Hence $S$ contains exactly three subgroups that belong to $\mathcal{H}$. We denote these three subgroups by $H_1$, $H_2$ and $H_3$.
As mentioned above, each preimage $f_t^{-1}(u_t)$ consists of nine elements. By Lemma 9.9 one of these elements is $(u_{t+1},u_{t+2})$, and the other eight elements $(v,w)\in f_t^{-1}(u_t)$ satisfy $v\ne u_{t+1}$ and $w\ne u_{t+2}$. We choose one of these eight preimages and denote it by $(v_{t+1},w_{t+2})$. Thus, we obtain six vertices $v_0$, $v_1$, $v_2$, $w_0$, $w_1$ and $w_2$, which have the following properties: For each $t\in\mathbb{F}_3$ the group $S$ acts transitively and freely on the eight-element set $\alpha_{t+2}\setminus\{u_{t+2}\}$. From Lemma 9.8 we see that
$$
\begin{equation*}
\varepsilon_{H_1}\bigl(\theta(v_{t+1}),w_{t+2}\bigr)\varepsilon_{H_2}\bigl(\theta(v_{t+1}),w_{t+2}\bigr)\varepsilon_{H_3}\bigl(\theta(v_{t+1}),w_{t+2}\bigr)=1,\qquad t\in\mathbb{F}_3.
\end{equation*}
\notag
$$
Taking the product of these three equations and the equations from Proposition 9.5 for the three subgroups $H_1$, $H_2$ and $H_3$ we obtain
$$
\begin{equation}
\prod_{i=1}^3\prod_{t\in\mathbb{F}_3}\eta_{H_i,t}\varepsilon_{H_i}\bigl(\theta(v_{t+1}),w_{t+2}\bigr)=1.
\end{equation}
\tag{9.9}
$$
Therefore, at least one multiplier $\eta_{H_i,t}\varepsilon_{H_i}\bigl(\theta(v_{t+1}),w_{t+2}\bigr)$ is equal to $1$. Then the required assertions hold for the $4$-tuple $(H_i,t,v_{t+1},w_{t+2})$.
The proposition is proved. End of the proof of Proposition 5.3. Let us show how to arrive at a contradiction by using Propositions 9.1, 9.7 and 9.10. Let $(H, t, v, w)$ be the $4$-tuple from Proposition 9.10. We denote the $H$-fixed points and the $H$-orbits in $V$ as in Proposition 9.1. By assertion (1) of Proposition 9.10 we have $f_t(v,w)=u_t$. Hence
$$
\begin{equation*}
\tau=\{u_t\}\cup\bigl(\alpha_{t+1}\setminus\{v\}\bigr)\cup\bigl(\alpha_{t+2}\setminus\{w\}\bigr)
\end{equation*}
\notag
$$
is a simplex of $K$. From assertion (2) of Proposition 9.10 and Proposition 9.7 it follows that either $v\in\beta_{t+1}$ and $w\in\beta_{t+2}$, or $v\in\gamma_{t+1}$ and $w\in\gamma_{t+2}$. In the former case $\tau$ contains the subset and in the latter case $\tau$ contains the subset
$$
\begin{equation*}
\lambda=\{u_0,u_1,u_2\}\cup\beta_{t+1}\cup\beta_{t+2}.
\end{equation*}
\notag
$$
We arrive at a contradiction, since by assertion (3) of Proposition 9.1 neither $\lambda$ nor $\mu$ is a simplex of $K$. This completes the proof of Proposition 5.3.
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\Bibitem{Gai24} |
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