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This article is cited in 3 scientific papers (total in 3 papers) Canonical geometrization of orientable N. Yu. Erokhovets Steklov Mathematical Institute of Russian Academy of Sciences, Moscow, Russia
Russian version:
Matematicheskii Sbornik, 2022, Volume 213, Number 6, DOI: https://doi.org/10.4213/sm9665 
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    IntroductionToric topology (see [1]) allows one to give explicit examples for deep results in various fields of mathematics. In this paper we consider Thurston’s geometrization conjecture, which was finally proved by Perelman. To formulate the precise result we need some definitions. We follow the exposition in the book [2]. All precise references and additional details can be found there. We also recommend the surveys [3]–[5]. Definition 0.1. A $3$-dimensional geometry is a smooth, simply connected $3$-manifold $X$ equipped with a smooth transitive action of a Lie group $G$ by diffeomorphisms with compact point stabilizers. The group $G$ is assumed to be maximal among all such groups for $X$. A geometric structure on a $3$-manifold $N$ (modelled on $X$) is a diffeomorphism from the interior of $N$ to $X/\pi$, where $\pi$ is a discrete subgroup of $G$ acting freely on $X$. Thurston showed that, up to a certain equivalence, there exist precisely eight $3$-dimensional geometries that model compact $3$-manifolds. These are: the sphere $S^3$, the Euclidean space $\mathbb R^3$, the hyperbolic space $\mathbb{H}^3$, $S^2\times \mathbb R$, $\mathbb H^2\times\mathbb R$, the universal cover $\widetilde{\mathrm{SL}(2,\mathbb{R})}$ of $\mathrm{SL}(2,\mathbb {R})$, and two further geometries, called $\mathrm{Nil}$ and $\mathrm{Sol}$. We say that $N$ has a geometric structure of finite volume if $X/\pi$ has a finite volume. After Perelman’s work the geometrization theorem conjectured by Thurston can be formulated in the following way (see [2], Theorem 1.9.1, and [6], the introduction). Theorem 0.2 (Geometric Decomposition Theorem). Let $N$ be a closed, orientable, irreducible $3$-manifold. There is a (possibly empty) collection of disjointly embedded incompressible surfaces $S_1,\dots,S_m$, which are either tori or Klein bottles, such that each component of $N$ cut along $S_1\cup\dots\cup S_m$ has a geometric structure. Any such collection with the minimum number of components is unique up to isotopy. Definition 0.3. We call the decomposition of $N$ with the minimum number of components in Theorem 0.2 simply the geometric decomposition. In [7] Davis and Januszkiewicz introduced a class of $n$-manifolds called small covers over simple $n$-polytopes. Each manifold of this type is glued of $2^n$ copies of a polytope $P$ and is defined by a mapping $\Lambda$ from the set of facets $\{F_1,\dots, F_m\}$ of $P$ to $\mathbb Z_2^n=(\mathbb Z/2\mathbb Z)^n$ such that for any face the images of facets containing it are linearly independent. In [7], Example 1.21, there is a sketch of the proof that geometric decomposition exists for $3$-manifolds realizable as small covers over simple $3$-polytopes: “$\langle\,{\dots}\,\rangle$ let us consider the $3$-manifolds which arise as small covers of such a $P^3$. If $P^3$ has no triangles or squares as faces, then it follows from Andreev’s Theorem $\langle\,{\dots}\,\rangle$ that it can be realized as a right-angled polytope in the hyperbolic $3$-space. Hence, any small cover carries a hyperbolic structure. If $P^3$ has no triangular faces, $M^3$ can be decomposed into Seifert fibred or hyperbolic pieces glued along tori or Klein bottles arising from square faces. In particular, such an $M^3$ is aspherical. If $P^3$ has triangular faces, then $M^3$ has a decomposition into such pieces glued along projective planes. Of course, all this fits with Thurston’s Conjecture. ” Actually, any mapping $\Lambda\colon \{F_1,\dots, F_m\}\to \mathbb Z_2^r$ such that for any face the images of facets containing it are linearly independent gives a manifold $M(P,\Lambda)$, which we call the manifold defined by the vector colouring $\Lambda$. If $r=n$, then $M(P,\Lambda)$ is a small cover, and if $r=m$ and the images of all facets are linearly independent, then $M(P,\Lambda)$ is a real moment-angle manifold $\mathbb R\mathcal{Z}_P$. In this paper we solve the following problem. Problem 0.4. Find explicitly the geometric decomposition of any orientable $3$-manifold $M(P,\Lambda)$ defined by a vector colouring $\Lambda$ of a simple $3$-polytope $P$. In particular, find it for any orientable small cover and any real moment-angle manifold over a $3$-polytope. The answer in given in Theorem 4.12. It has turned out that to give an explicit geometric decomposition, additional arguments should be used. Namely, one needs the following. 1) Use $3$-belts and $4$-belts in addition to triangular and quadrangular facets (for simplicial polytopes this corresponds to $3$-cycles not bounding triangles and chordless $4$-cycles, where a ‘chord’ is an edge of a polytope that connects non-successive vertices of the cycle). 2) Use the canonical decomposition of a simple $3$-polytope into a connected sum along vertices of flag $3$-polytopes (that is, simple $3$-polytopes different from the simplex and having no $3$-belts) and simplices. This corresponds to the Kneser-Milnor decomposition of oriented $3$-manifolds into connected sums of prime manifolds. 3) Use right-angled (so that all dihedral angles between adjacent facets are ${\pi}/{2}$) hyperbolic $3$-polytopes of finite volume and simple polytopes obtained by cutting off their ideal vertices. The latter family of polytopes was studied in [8], § 10.3, and [9]. We call both them, and also the cube and the $5$-prism, almost Pogorelov polytopes. On the basis of Andreev’s theorems it can be shown that these are exactly simple $3$-polytopes different from the simplex such that they have no $3$-belts and any one of their $4$-belts surrounds a facet. 4) Find a canonical ‘minimal’ decomposition of a flag simple $3$-polytope along $4$-belts into $k$-prisms, $k\geqslant 5$, and almost Pogorelov polytopes. This decomposition is a special case of a more general decomposition of a so-called Coxeter orbifold in [10]. It turns out that there can be many inequivalent ways to cut a flag polytope along $4$-belts into almost Pogorelov polytopes. Such decompositions were used in [8] (see also [11], after Theorem 20.5.2) to prove Singer’s conjecture on the vanishing of the reduced $l^2$-homology except in the middle dimension for closed aspherical manifolds in the special case of $3$-manifolds arising from right-angled Coxeter groups. The paper [10] was also motivated by this topic. The canonical decomposition of a flag $3$-polytope corresponds to the $\mathrm{JSJ}$-decomposition of the oriented irreducible $3$-manifold $M(P,\Lambda)$. The $\mathrm{JSJ}$-tori, which are defined up to isotopy, correspond to the ‘canonical’ $4$-belts of the polytope $P$ and to the quadrangles of almost Pogorelov polytopes which do not arise in cutting along these belts. Each ‘free quadrangle’ of an almost Pogorelov polytope corresponds to a disjoint union of incompressible submanifolds, which are either tori $T^2$ (if the corresponding vector $\Lambda_i$ does not lie in the subspace spanned by the vectors $\Lambda_j$ of adjacent facets) or Klein bottles $K^2$ (if $\Lambda_i$ lies in this subspace), and the $4$-belt around the quadrangle corresponds to the boundaries of tubular neighbourhoods of these submanifolds. The tori $T^2$ are $\mathrm{JSJ}$-tori themselves, and in the case of $K^2$ the boundaries of tubular neighbourhoods are $\mathrm{JSJ}$-tori. 5) Prove that the arising surfaces are incompressible. For this we use a retraction of a real moment-angle complex $\mathcal{Z}_K$ of a simplicial complex $K$ onto a subset corresponding to the full subcomplex $K_I$. This method was communicated to the author by T. E. Panov (see [1], Exercise 4.2.13, and [12], Proposition 2.2). An alternative way of reasoning using an explicit description of the fundamental group can be found in [13]–[15]. For submanifolds corresponding to facets the incompressibility was proved in [14] as Theorem 3.3. Another approach to this result was presented in [8], [16] and [11]; it is based on the fact that a facet submanifold is a totally geodesic hypersurface in $M(P,\Lambda)$ with the structure of a cubical complex, which is nonpositively curved in the sense of Alexandrov and Gromov [17]. As mentioned above, some of the $\mathrm{JSJ}$-tori corresponding to $4$-belts around quadrangles bound orientable manifolds, which are $I$-bundles over $K^2$. These pieces can be endowed with a Euclidean geometry with infinite volume. To reduce the number of geometric pieces and make all of them have finite volume, these tori should be replaced with the corresponding Klein bottles. 6) Define explicitly geometric structures on the arising pieces. This can be done using a construction invented by Vesnin and Mednykh [18]–[23] in the case of right-angled polytopes of finite volume in $\mathbb R^3$, $S^3$, $\mathbb{H}^3$, $S^2\times \mathbb R$ or $\mathbb{H}^2\times \mathbb R$. A vector colouring defines a subgroup in the right-angled Coxeter group generated by the reflections in facets of the right-angled polytope. This subgroup acts freely, and the orbit space is a manifold with geometric structure. In our case each flag simple $3$-polytope is canonically cut along $4$-belts and free quadrangles into parts which are homeomorphic (respecting the structure of faces) to right-angled polytopes of finite volume in these five geometries. This defines geometric structures on the corresponding pieces of $M(P,\Lambda)$. For geometries $\mathbb{H}^3$ and $S^3$ two right-angled polytopes of finite volume are combinatorially equivalent if and only if they are isometric, while for $\mathbb{H}^2\times \mathbb R$, $\mathbb R^3$ and $S^2\times \mathbb R$ this is not true. Prime orientable $3$-manifolds $M(P,\Lambda)$ correspond to 1) the simplex $\Delta^3$: $M(P,\Lambda)$ is either $S^3$ or $\mathbb RP^3$; both of them are irreducible; 2) the $3$-prism $\Delta^2\times I$: $M(P,\Lambda)$ is either $S^2\times S^1$, which is prime, or $\mathbb{R}P^3\#\mathbb{R}P^3$, which is not prime; 3) flag simple $3$-polytopes: $M(P,\Lambda)$ is aspherical (that is, $\pi_i(M(P,\Lambda))=0$ for $i>1$) and therefore irreducible. On orientable manifolds $M(P,\Lambda)$ or their pieces we obtain geometric structures of finite volume modelled on $\bullet$ $S^3$: in this geometry $\Delta^3$ can be realized as a right-angled polytope (a realization is unique up to isometries), and $M(P,\Lambda)$ is either $S^3$ or $\mathbb RP^3$. $\bullet$ $S^2\times \mathbb R$: in this geometry $\Delta^2\times I$ can be realized as a right-angled polytope (a right-angled realization $\Delta^2\subset S^2$ is unique up to isometries, while $I\subset\mathbb R$ is not), and $M(P,\Lambda)$ is either $S^2\times S^1$ or $\mathbb RP^3\#\mathbb RP^3$. $\bullet$ $\mathbb R^3$, $\mathbb H^2\times \mathbb R$, $\mathbb{H}^3$: a flag simple $3$-polytope is either the cube $I^3$ ($4$-prism), or the $k$-prism with $k\geqslant 5$, or a Pogorelov polytope (that is, a flag polytope without $4$-belts), or none of the above.
In § 1 we give definitions and basic facts on manifolds defined by vector colourings of polytopes. In particular, in Proposition 1.12 we give a criterion when the manifold $M(P,\Lambda)$ is orientable (which generalizes the case of small covers considered in [24]), and in Proposition 1.22, for a nonorientable manifold $M(P,\Lambda)$, we construct the orientable double cover $M(P,\widehat{\Lambda})$. We consider the partially ordered set $\mathcal{F}(P)$ of subgroups $H(\Lambda)\subset \mathbb Z_2^m$ acting freely on $\mathbb R\mathcal{Z}_P$, and in Example 1.18, for a $3$-polytope $P$ we present a subgroup of dimension less than $m-3$ which corresponds to a maximal element. In § 2 we develop the technique of $k$-belts of simple $3$-polytopes. We introduce the notions of a nested family of belts and a nested family of curves corresponding to it. In Lemma 2.15 we prove that any nested family of belts admits a nested family of curves, and in Lemma 2.20, that any two nested families of curves corresponding to the same family of belts are isotopic in this class. We consider the operations of cutting a $3$-polytope along a belt and taking a connected sum of two $3$-polytopes along vertices and along facets surrounded by belts. In Corollaries 2.32 and 2.34 we prove that for any nested family of $3$-belts and any nested family of $4$-belts there is a geometric realization of a $3$-polytope $P$ such that disjoint planar cross-sections of $P$ give the corresponding nested families of curves. In § 3 we build the prime decomposition of any orientable $3$-manifold $M(P,\Lambda)$. First, in Proposition 3.6 we express the manifold corresponding to a connected sum of simple $3$-polytopes along vertices as a connected sum of copies of the corresponding manifolds and copies of $S^2\times S^1$. Then in Theorem 3.12 we prove the main result in that section. In § 4 we prove the main results of the article. First, in §§ 4.1 and 4.2 we present general information on a $\mathrm{JSJ}$-decomposition and geometrization. Then in § 4.3, on the basis of this information we present the $\mathrm{JSJ}$-decomposition and the geometric decomposition for any irreducible orientable $3$-manifold $M(P,\Lambda)$. The main result of this article is Theorem 4.12. It is proved in several steps. First, in § 4.3.1 we build a canonical decomposition of a flag $3$-polytope along $4$-belts and prove its uniqueness. Then, in § 4.3.2 we study submanifolds corresponding to belts and facets surrounded by belts. In Propositions 4.23 and 4.27 we describe them explicitly and prove that they are incompressible. In § 4.3.3 we finish the proof of Theorem 4.12. In Proposition 4.31 we describe explicitly the pieces of $M(P,\Lambda)$ corresponding to $k$-prisms, in particular, their Seifert fibred structure. Then we prove that the constructed family of tori is indeed the family of all $\mathrm{JSJ}$-tori. Finally, we prove that $M(P,\Lambda)$ cannot be a $\mathrm{Sol}$-manifold and build an explicit geometrization using Construction 4.11 due to Vesnin and Mednykh. § 1. Manifolds defined by vector colourings of polytopesFor an introduction to polytope theory we recommend [25]. In this paper, by a polytope we mean an $n$-dimensional combinatorial convex polytope (in most cases $n$ is equal to $3$). Sometimes we use implicitly its geometric realization in $\mathbb R^n$, and sometimes we use it explicitly. In the latter case we call the polytope geometric. A polytope is simple if each vertex of it is contained in exactly $n$ facets. Let $\{F_1,\dots,F_m\}$ be the set of all the facets, and $\mathbb Z_2=\mathbb{Z}/2\mathbb{Z}$. Definition 1.1. With each geometric simple polytope $P$ one associates an $n$-dimensional real moment-angle manifold: here $e_1,\dots, e_m$ is a basis in $\mathbb Z_2^m$ as a vector space. The space $\mathbb R\mathcal{Z}_P$ was introduced in [7]. It is convenient to imagine $\mathbb{R}\mathcal{Z}_P$ as a space glued from copies of the polytope $P$ along facets. If we fix an orientation on $P\times 0$, then on the polytope $P\times a$ we define the same orientation if $a$ has an even number of coordinates one, otherwise we define the opposite orientation. A polytope $P\times a$ is glued to the polytope $P\times (a+e_i)$ along the facet $F_i$. At each vertex the polytopes are arranged as coordinate orthants at the origin in $\mathbb R^n$, along each edge they are arranged as the orthants along a coordinate axis, and along a face of dimension $i$, as the orthants along an $i$-dimensional coordinate subspace. Therefore, $\mathbb{R}\mathcal{Z}_P$ has the natural structure of an oriented piecewise linear manifold. There is a natural action of $\mathbb Z_2^m$ on $\mathbb{R}\mathcal{Z}_P$ induced by the action on the second factor. The actions of basis vectors $e_i$ can be viewed as reflections in facets of the polytope. Example 1.2. For a $k$-gon $P_k$ the closed orientable $2$-dimensional manifold $\mathbb R\mathcal{Z}_{P_k}$ is homeomorphic to a sphere with $g$ handles. The genus $g$ can be calculated using the Euler characteristic: Thus, $g=(k-4)2^{k-3}+1$. There is another representation of $\mathbb{R}\mathcal{Z}_P$. By a simplicial complex $K$ on the set of vertices $[m]=\{1,\dots,m\}$ we mean an abstract simplicial complex, that is, a collection of subsets $\sigma\subset[m]$ such that for any $\sigma\in K$ and any $\tau\subset\sigma$ we have $\tau\in K$. The subsets $\sigma\in K$ are called simplices. Let $D^1=[-1,1]$ and $S^0=\{-1,1\}=\partial D^1$. There is a natural action of $\mathbb Z_2^m$ on $\mathbb R\mathcal{Z}_K$, which arises from the identification of the multiplicative group $\{-1,1\}$ with the additive group $\mathbb{Z}_2=\{0,1\}$. This group acts by changing the signs of coordinates. Each simple polytope $P$ corresponds to a simplicial complex
$$
\begin{equation*}
K_P=\biggl\{\sigma\subset [m]\colon \bigcap_{i\in \sigma}F_i\ne\varnothing\biggr\},
\end{equation*}
\notag
$$
which is isomorphic to the boundary complex $\partial P^*$ of the dual simplicial polytope. This homeomorphism implies that the topological type of $\mathbb{R}\mathcal{Z}_P$ does not depend on the geometric realization of $P$. Proof (sketch of a proof of Proposition 1.4). Consider the barycentric embedding $b\colon P\to [0,1]^m$, which is the mapping that is linear on simplices of the barycentric subdivision of $P$, while at the vertices of this subdivision it satisfies Denote by $[0,1]^P$ the image of this embedding. Then the homeomorphism $\mathbb R\mathcal{Z}_P\to \mathbb{R}\mathcal{Z}_{K_P}$ is given by $(x,a)\to a(b(x))$, where $a(b(x))$ is the result of the action of $a\in\mathbb Z_2^m$ on $b(x)\in [0,1]^P\subset \mathbb{R}\mathcal{Z}_{K_P}$. This finishes the proof. We consider manifolds obtained as the orbit spaces of free actions of subgroups $H\subset\mathbb Z_2^m$ on $\mathbb R\mathcal{Z}_P$. Each subgroup of $\mathbb Z_2^m$ is isomorphic to $\mathbb Z_2^{m-r}$ for some $r$ and can be defined to be he kernel of a an epimorphism $\Lambda\colon\mathbb Z_2^m\to\mathbb Z_2^r$. Such a mapping is uniquely defined by the images $\Lambda_i\in\mathbb Z_2^r$ of all vectors $e_i\in\mathbb Z_2^m$ corresponding to facets $F_i$, $i=1,\dots,m$. It can be shown that the action of the subgroup $H(\Lambda)\subset \mathbb Z_2^m$ on $\mathbb R\mathcal{Z}_P$ is free if and only if
$$
\begin{equation}
\begin{aligned} \, &\text{for any vertex $F_{i_1}\cap\dots\cap F_{i_n}$ of $P$} \\ & \text{the vectors $\Lambda_{i_1},\dots,\Lambda_{i_n}$ are linearly independent.} \end{aligned}
\end{equation}
\tag{$*$}
$$
Definition 1.5. We call a mapping $\Lambda\colon \{F_1,\dots,F_m\}\to \mathbb Z_2^r$ such that the images $\Lambda_j$ of the facets $F_j$ span $\mathbb Z_2^r$ and $\Lambda$ satisfies (*) a vector colouring of rank $r$. Remark 1.6. Sometimes, by a vector colouring of rank $r$ we mean a mapping $\Lambda$: $\{F_1,\dots,F_m\}\to \mathbb Z_2^s$, $s\geqslant r$, satisfying (*) and such that $\dim\langle\Lambda_1,\dots,\Lambda_m\rangle=r$. Remark 1.7. The notion of a vector colouring is not new. It was implicitly used, for example, in [19], [20], [7] and [27]. The analogy between mappings $\{F_1,\dots, F_m\}\to \mathbb Z_2^r$ and colourings of facets of polytopes was considered in [28]. In particular, in [28] there is a notion of a linearly independent colouring, which is equivalent to our vector colouring. For $r=n$ this notion was used in [24] already. In the case when all vectors $\Lambda_1,\dots,\Lambda_m$ belong to a set $\{e_1,\dots,e_r\}$ forming a basis in $\mathbb Z_2^r$, the manifolds defined by such a colouring were studied in [29]. Denote by $M(P,\Lambda)$ the orbit space $\mathbb R\mathcal{Z}_P/H(\Lambda)$ of the free action of the subgroup corresponding to a vector colouring of rank $r$. If we identify $\mathbb Z_2^m/\operatorname{Ker}\Lambda$ with $\mathbb Z_2^r$ via $\Lambda$, then
$$
\begin{equation*}
M(P,\Lambda)=P\times \mathbb Z_2^r/{\sim}, \quad\text{where } (p,a)\sim(q,b) \ \Longleftrightarrow\ p=q\text{ and } a-b\in\langle \Lambda_i\colon p\in F_i\rangle.
\end{equation*}
\notag
$$
In particular, $M(P,\Lambda)$ has the structure of a piecewise linear manifold glued from $2^r$ copies of $P$. It carries an action of $\mathbb Z_2^r$ such that the orbit space is $P$. Definition 1.8. We call $M(P,\Lambda)$ a manifold defined by the vector colouring $\Lambda$ of the polytope $P$. Example 1.9. For $r=m$ and the mapping $\Lambda_i=e_i$ the manifold $M(P,E)$ is $\mathbb R\mathcal{Z}_P$. For $r=n$ a mapping $\Lambda$ satisfying condition (*) is called a characteristic mapping, and the manifold $M(P,\Lambda)$ is called a small cover of the polytope $P$. Proposition 1.10. For vector colourings $\Lambda_1$ and $\Lambda_2$ of ranks $r_1$ and $r_2$ of a polytope $P$ we have $H(\Lambda_1)\subset H(\Lambda_2)$ if and only if there is a surjection $\Pi\colon\mathbb Z_2^{r_1}\to \mathbb Z_2^{r_2}$ such that $\Pi \circ \Lambda_1=\Lambda_2$. Proof. We have $H(\Lambda_1)\subset H(\Lambda_2)$ if and only if each row of the matrix $\Lambda_2$ with columns $\Lambda_{2,i}$ is a linear combination of rows of $\Lambda_1$. This is equivalent to the existence of a surjection $\Pi\colon\mathbb Z_2^{r_1}\to \mathbb Z_2^{r_2}$ such that $\Pi(\Lambda_{1,i})=\Lambda_{2,i}$ for all $i=1,\dots,m$. The proposition is proved. Corollary 1.11. We have $H(\Lambda_1)=H(\Lambda_2)$ if and only if there is a linear isomorphism $\Pi\colon \mathbb Z_2^{r_1}\to\mathbb Z_2^{r_2}$ such that $\Pi\circ\Lambda_1=\Lambda_2$. If $H(\Lambda_1)\subset H(\Lambda_2)$, then $M(P,\Lambda_2)$ is the orbit space of a freely acting group $H(\Lambda_2)/H(\Lambda_1)$ on $M(P,\Lambda_1)$; in particular, there is a covering $M(P,\Lambda_1)\to M(P,\Lambda_2)$ with fibre $H(\Lambda_2)/H(\Lambda_1)$. The following result is a generalization of the result from [24], which considers only small covers. Proposition 1.12. Let the vectors $\Lambda_{j_1}, \dots, \Lambda_{j_r}$ form a basis in $\mathbb Z_2^r$. Then the manifold $M(P,\Lambda)$ is orientable if and only if each $\Lambda_i$ is a sum of an odd number of these vectors. Proof. For $M(P,\Lambda)=P\times \mathbb Z_2^r/\sim$ to be orientable it is necessary and sufficient to define orientations of all the polytopes $P\times a$, $a\in\mathbb Z_2^r$, in such a way that for any $a$ and $i$ the polytopes $P\times a$ and $P\times (a+\Lambda_i)$ glued along the facet $F_i\times a$ have opposite orientations. If we choose an orientation of $P\times 0$ and go through the facets $F_{j_1},\dots,F_{j_r}$, then using combinations of such steps we arrive at any copy $P\times a$ and define its orientation. It is the same as the orientation of $P\times 0$ if and only if $a$ is the sum of an even number of vectors $\Lambda_{j_l}$. Thus, the orientations of $P\times a$ and $P\times (a+\Lambda_i)$ are different if and only if $\Lambda_i$ is the sum of an odd number of vectors $\Lambda_{j_l}$. This finishes the proof. The Four-Colour Theorem implies the existence of a small cover over any simple $3$-polytope. Namely, if all the facets of a $3$-polytope $P$ are coloured with four colours in such a way that adjacent facets have different colours, then to the first three colours we assign the basis vectors $e_1,e_2,e_3\in\mathbb Z_2^3$, and to the fourth colour the vector $e_1+e_2+e_3$. Since any three of these four vectors are linearly independent, we obtain a characteristic function. Corollary 1.13 (see [24]). A $3$-dimensional small cover $M(P,\Lambda)$ is orientable if and only if its characteristic function corresponds to a colouring with at most four colours. Remark 1.14. For bounded right-angled hyperbolic polytopes the orientability of the manifolds corresponding to colourings with at most four colours was already mentioned in [19] and [20]. The maximum dimension of subgroups in $\mathbb Z_2^m$ acting freely on $\mathbb R\mathcal{Z}_P$ is called the real Buchstaber number $s_{\mathbb R}(P)$ (see [27], [28], [30] and [31]). It is easy to see that $1\leqslant s_{\mathbb R}(P)\leqslant m-n$. The above arguments imply that $s_{\mathbb R}(P)=m-3$ for any simple $3$-polytope. Consider all subgroups in $\mathbb Z_2^m$ acting freely on $\mathbb R\mathcal{Z}_P$. They form a partially ordered set $\mathcal{F}(P)$ with respect to inclusion. In our discussion of this article with V. M. Buchstaber the following natural questions have arisen. Question 1.16. Find the set (or multiset) of integer numbers consisting of ranks of subgroups lying in $\mathcal{F}(P)$. Question 1.17. Find the minimum rank $m(P)$ of subgroups that are maximal elements of $\mathcal{F}(P)$. These questions lead to new combinatorial invariants generalizing the real Buchstaber number. It follows from the definition that $s_{\mathbb R}(P)$ is equal to the maximum of the elements of the (multi)set in Question 1.16. The following example shows that $m(P)$ can be less than $s_{\mathbb R}(P)$. In particular, it contains some additional information on the combinatorics of $P$. Example 1.18 (an orientable manifold corresponding to a maximal subgroup of dim $<m-n$). In Figure 1 we show a Schlegel diagram of a $3$-dimensional associahedron $As^3$. This polytope can be realized as a cube $I^3$ with three disjoint pairwise orthogonal edges cut. A Schlegel diagram is a polytopal complex arising when we project the boundary complex of a polytope onto one of its facets from a point lying outside the polytope close to this facet. Also in Figure 1 we show a vector colouring $\Lambda$ of rank $4$. Proposition 1.12 implies that $M(As^3,\Lambda)$ is an orientable manifold glued of 16 copies of $As^3$. Assume that there exists a surjection $\Pi\colon \mathbb Z_2^4\to\mathbb Z_2^3$ such that $\Pi\Lambda$ satisfies condition (*) again. Since $\Pi(e_1)$, $\Pi(e_2)$ and $\Pi(e_3)$ arise at the same vertex of $As^3$, these vectors form a basis in $\mathbb Z_2^3$. On the other hand $\Pi(e_4)$ must be different from $\Pi(e_1)$, $\Pi(e_2)$, $\Pi(e_3)$ and $\Pi(e_1)+\Pi(e_2)+\Pi(e_3)$, since these vectors correspond to facets adjacent to quadrangles corresponding to $\Pi(e_4)$. Furthermore, $\Pi(e_4)$ is different from $\Pi(e_1)+\Pi(e_2)$, $\Pi(e_2)+\Pi(e_3)$ and $\Pi(e_3)+\Pi(e_1)$, since these vectors are sums of vectors corresponding to facets surrounding a vertex together with one of these quadrangles. Moreover, $\Pi(e_4)\ne 0$ by (*). Thus, no vector in $\mathbb Z_2^3$ is suitable for the role of the image of $e_4$. Therefore, $H(\Lambda)$ is a maximal freely acting subgroup of $\mathbb Z_2^m$ for $As^3$. Remark 1.19. Given a simple $3$-polytope $P$, maximal elements in $\mathcal{F}(P)$ are important due to the following fact (see [2], Theorem 1.9.3). Let $N$ be a closed orientable irreducible $3$-manifold, let $p\colon \widehat{N}\to N$ be a finite cover and $S_1,\dots, S_m$ be tori and Klein bottles from the geometric decomposition of $N$ in Theorem 0.2. Then $\widehat{N}$ is irreducible and the connected components of $p^{-1}(S_1\sqcup\dots\sqcup S_m)$ give the geometric decomposition for $\widehat{N}$. In our case each manifold $M(P,\Lambda)$ is covered by $\mathbb R\mathcal{Z}_P$ and covers $M(P,\Lambda')$, where $H(\Lambda')$ is a maximal element of $\mathcal{F}(P)$. We will not actually use the latter covering, since the case of $\mathbb R\mathcal{Z}_P$ is easier and it is more convenient to study the covering $\mathbb R\mathcal{Z}_P\to M(P,\Lambda)$. Proposition 1.20. Let $\Lambda_{i_1},\dots,\Lambda_{i_r}$ be a basis in $\mathbb Z_2^r$ for a vector colouring $\Lambda$ of rank $r$ of a simple $n$-polytope $P$. The subgroup $H(\Lambda)$ corresponds to a maximal element in $\mathcal{F}(P)$ if and only if for any $j=1,\dots,r$, and any mapping $\Pi_{j,a}\colon {\mathbb Z_2^r\to \mathbb Z_2^{r-1}}$ defined by Proof. If the composition $\Pi_{j,a}\circ \Lambda$ satisfies condition (*), then $H(\Lambda)\subset H(\Pi_{j,a}\circ \Lambda)$ is not maximal. On the other hand, if $H(\Lambda)$ is not maximal in $\mathcal{F}(P)$, then it lies in a freely acting subgroup of dimension $\dim H(\Lambda)+1$. This corresponds to a surjection $\varphi\colon \mathbb Z_2^r\to\mathbb Z_2^{r-1}$ such that $\varphi\circ\Lambda$ satisfies (*). Then $r-1$ of the vectors $\varphi(\Lambda_{i_1}),\dots,\varphi (\Lambda_{i_r})$ form a basis in $\mathbb Z_2^{r-1}$, and there is an automorphism $A$ of $\mathbb Z_2^{r-1}$ such that $A\circ \varphi=\Pi_{j,a}$ for some $j$ and $a$. Then $\Pi_{j,a}\circ \Lambda$ also satisfies (*). The proposition is proved. Construction 1.21 (orientable double cover). Let $M(P,\Lambda)$ be a nonorientable manifold and $\Lambda_{j_1},\dots,\Lambda_{j_r}$ be a basis in $\mathbb Z_2^r$. Below we consider coordinates with respect to this basis. We have an inclusion $\mathbb Z_2^r\subset \mathbb Z_2^{r+1}$ as a subset of elements with the last coordinate equal to zero. Then $\Lambda_{j_1},\dots,\Lambda_{j_r}$, $e_{r+1}=(0,\dots,0,1)$ is a basis in $\mathbb Z_2^{r+1}$, and we have a surjection $\Pi\colon \mathbb Z_2^{r+1}\to\mathbb Z_2^r$ given by $\Pi(\Lambda_{j_s})=\Lambda_{j_s}$, ${s=1,\dots,r}$, and $\Pi(e_{r+1})=\Lambda_{i_0}$, where $\Lambda_{i_0}$ has an even number of nonzero coordinates. Then $\operatorname{Ker} \Pi=\langle e_{r+1}+\Lambda_{i_0}\rangle$. Consider a mapping $\widehat{\Lambda}\colon {\{F_1,\dots,F_m\}\to \mathbb Z_2^{r+1}}$: Proposition 1.22. For a nonorientable manifold $M(P,\Lambda)$ the mapping $\widehat{\Lambda}$ is a vector colouring (of rank $r+1$), and $M(P,\widehat{\Lambda})$ is an orientable double cover. Proof. Let us prove that $\widehat{\Lambda}$ satisfies condition (*). In fact, if $F_{i_1}\cap\dots\cap F_{i_n}$ is a vertex, then $\Lambda_{i_1},\dots,\Lambda_{i_n}$, $e_{r+1}+\Lambda_{i_0}$ are linearly independent. If there is a nontrivial linear dependence $\mu_1\widehat{\Lambda}_{i_1}+\dots+\mu_n\widehat{\Lambda}_{i_n}=0$, then we have where $p_1,\dots,p_l\in\{i_1,\dots, i_n\}$ are the subscripts of vectors with an even number of nonzero coordinates. In particular, $\mu_1=\dots=\mu_n=0$, which is a contradiction.
We claim that in the basis $\widehat{\Lambda}_{j_1}=\Lambda_{j_1},\ \dots,\ \widehat{\Lambda}_{j_r}=\Lambda_{j_r}$, $\widehat{\Lambda}_{i_0}=e_{r+1}$ each vector $\widehat{\Lambda}_i$ has an odd number of nonzero coordinates. Indeed, if $\Lambda_i$ has an odd number of nonzero coordinates in the basis $\Lambda_{j_1},\dots,\Lambda_{j_r}$, then $\widehat{\Lambda}_i=\Lambda_i$ and the claim is proved for this vector. If $\Lambda_i$ has an even number of nonzero coordinates in the basis $\Lambda_{j_1},\dots,\Lambda_{j_r}$, then
$$
\begin{equation*}
\widehat{\Lambda}_i=\Lambda_i+(e_{r+1}+\Lambda_{i_0})=(\Lambda_i+\Lambda_{i_0})+e_{r+1}.
\end{equation*}
\notag
$$
This vector has an even number of nonzero coordinates corresponding to the first $r$ basis vectors, and one coordinate corresponds to $e_{r+1}$. The claim is proved. Thus, $M(P,\widehat{\Lambda})$ is an orientable manifold.
We also have $\Pi(\widehat{\Lambda}_i)=\Lambda_i$ for all $i$ by construction. Thus, $H(\widehat{\Lambda})\subset H(\Lambda)$ is a subgroup of index $2$, and $M(P,\Lambda)$ is the quotient space by a freely acting involution on $M(P,\widehat{\Lambda})$. This involution changes orientation, for otherwise $M(P,\Lambda)$ is orientable. For each polytope in the subdivision of $M(P,\Lambda)$ the manifold $M(P,\widehat{\Lambda})$ has two polytopes with opposite orientations mapped to it by the covering mapping. This is exactly the construction of the orientable double cover. The proposition is proved. § 2. Belts and operations on $3$-polytopes2.1. Belts and families of $3$-polytopesDefinition 2.1. A $k$-belt of a $3$-polytope is a cyclic sequence of $k$ facets with the following property: two facets are adjacent if and only if one of them follows the other and no three facets have a common vertex. A $k$-belt is trivial if it surrounds a facet. By the piecewise linear versions of the Jordan curve theorem and Schoenflies’ theorem each piecewise linear simple closed curve $\gamma$ on the surface of a $3$-polytope $P$ divides the sphere $\partial P$ into two connected components $\partial P\setminus\gamma=C_1\sqcup C_2$ such that for each component $C_i$ its boundary is $\gamma$, and there is a piecewise linear homeomorphism of the closure $\overline{C_i}$ to some polygon mapping $\gamma$ to its boundary. For each $k$-belt $\mathcal{B}$ the set $|\mathcal{B}|=\bigcup_{F_i\in \mathcal{B}}F_i\subset\partial P$ is topologically a cylinder bounded by two simple edge cycles bounding topological discs outside the belt. Definition 2.2. With any $k$-belt $\mathcal{B}$ of a simple $3$-polytope $P$ we associate the piecewise linear simple closed curve $\gamma(\mathcal{B})\subset\operatorname{int}|\mathcal{B}|\subset\partial P$ consisting of segments connecting the midpoints of edges of intersection of each facet of the belt with adjacent facets of the belt. We call $\gamma(\mathcal{B})$ the middle line of $\mathcal{B}$. Definition 2.3. A simple polytope $P$ is called a flag polytope if any set of its pairwise intersecting facets $F_{i_1},\dots,F_{i_k}$ has a nonempty intersection: ${F_{i_1}\cap\cdots\cap F_{i_k}\ne\varnothing}$. It can be shown that a simple $3$-polytope is flag if and only if it is different from the simplex $\Delta^3$ and has no $3$-belts. It can also be shown that any flag simple $3$-polytope has $m\geqslant6$ facets, and if it has six facets, then it is combinatorially equivalent to the cube $I^3$. Results due to Pogorelov [32] and Andreev [33] imply that a simple $3$-polytope $P$ can be realized in the Lobachevsky (hyperbolic) space $\mathbb H^3$ as a bounded polytope with right dihedral angles if and only if $P$ is different from the simplex $\Delta^3$ and has no $3$- and $4$-belts. Moreover, the realization is unique up to isometries. Such polytopes are called Pogorelov polytopes. As mentioned in [7], small covers over bounded right-angled polytopes have a natural hyperbolic structure. The same manifolds were discovered by Vesnin and Mednykh [18]–[20] (see Construction 4.11) as generalizations of a manifold invented by Löbell. Results of Birkhoff [34] imply that the $4$-colour problem can be reduced to colouring facets of Pogorelov polytopes with only trivial $5$-belts. Corollary 2.4. The four-colour theorem is equivalent to the fact that any bounded right-angled hyperbolic $3$-polytope with only trivial $5$-belts admits an orientable small cover. In this paper an important role is played by the family of almost Pogorelov polytopes. It consists of simple $3$-polytopes $P\ne \Delta^3$ without $3$-belts such that any $4$-belt is trivial. The combinatorics and hyperbolic geometry of this family of polytopes was studied in [9]. The following fact can be extracted directly from the definition. Proposition 2.5. The cube $I^3$ ($4$-prism) and the $5$-prism are the unique almost Pogorelov polytopes with adjacent quadrangles. All the other almost Pogorelov polytopes have no adjacent quadrangles and $m\geqslant 9$ facets. Results due to Andreev (see [33] and [35]) imply that almost Pogorelov polytopes correspond to right-angled polytopes of finite volume in $\mathbb H^3$ (‘right-angled’ means that all dihedral angles between facets intersecting in edges, perhaps with vertices at infinity (so-called ideal vertices), are ${\pi}/{2}$). Such polytopes may have $4$-valent vertices at infinity, while all proper vertices have valency $3$. Two right-angled polytopes of finite volume in $\mathbb H^3$ are congruent if and only if they are combinatorially equivalent (see [36], Ch. 5, § 2.1). Proposition 2.6 (see [8], Theorem 10.3.1, and also [9], Theorem 6.5). Cutting off $4$-valent vertices defines a bijection between the classes of congruence of right-angled polytopes of finite volume in $\mathbb{H}^3$ and the almost Pogorelov polytopes without adjacent quadrangles. Moreover, it induces a bijection between the ideal vertices of the right-angled polytope and the quadrangles of the corresponding almost Pogorelov polytope. The $3$-dimensional associahedron $As^3$ is the only almost Pogorelov polytope with $m=9$ (see [9]). It corresponds to a right-angled $3$-gonal bipyramid. The bipyramid has two proper vertices of valency $3$ and three ideal vertices of valency $4$. For recent results on the volumes of right-angled hyperbolic polytopes see [37]–[39]. In [15] the right-angled hyperbolicity of handlebodies with simple facial structure was studied. 2.2. Nested families of belts and curvesIn this subsection we develop a technique enabling one to work with families of belts. Definition 2.7. Let $P$ be a simple $3$-polytope. We call two belts $\mathcal{B}_1$ and $\mathcal{B}_2$ compatible if the middle line $\gamma(\mathcal{B}_2)$ lies in the closure of a connected component of $\partial P\setminus\gamma(\mathcal{B}_1)$. Remark 2.8. This definition can be equivalently reformulated in terms of the dual simplicial polytope $P^*$. A belt $\mathcal{B}$ of $P$ corresponds to a chordless cycle $c(\mathcal{B})$ in $P^*$ that does not bound a facet, where a ‘chord’ is an edge of the polytope connecting nonsuccessive vertices of the cycle. Belts $\mathcal{B}_1$ and $\mathcal{B}_2$ are compatible if and only if $c(\mathcal{B}_2)$ lies in the closure of a connected component of $\partial P^*\setminus c(\mathcal{B}_1)$. Since $P$ and $P^*$ have combinatorially equivalent barycentric subdivisions, $\partial P^*$ can be combinatorially realized in the barycentric subdivision of $\partial P$: the vertex $\{i\}$ of $P^*$ corresponds to the barycentre of the facet $F_i$ of $P$, the edge $\{i,j\}$ to the curve consisting of two segments connecting the centre of the edge $F_i\cap F_j$ with barycentres of $F_i$ and $F_j$, and the triangle $\{i,j,k\}$ to the union of six triangles with the common vertex $F_i\cap F_j\cap F_k$. Then $\gamma(\mathcal{B})$ is isotopic to $c(\mathcal{B})$ in this realization. Lemma 2.9. Two belts $\mathcal{B}_1$ and $\mathcal{B}_2$ are compatible if and only if a component of $\partial P\setminus\gamma(\mathcal{B}_2)$ lies in a component of $\partial P\setminus\gamma(\mathcal{B}_1)$. Proof. Let $\gamma_1=\gamma(\mathcal{B}_1)$ divide $\partial P$ into components $C_1$ and $C_2$, $\gamma_2=\gamma(\mathcal{B}_2)$ divide it into $D_1$ and $D_2$, and let $\gamma_2\subset\overline{C_1}$. The component $C_2$ is path-connected and does not intersect $\gamma_2$, hence $C_2$ lies in $D_1$ or $D_2$, for instance, in $D_2$. Then $D_1\cap \overline{C_2}=\varnothing$. In particular, $D_1\subset C_1$.
On the other hand, if $D_1\subset C_1$, then $\gamma_2=\partial D_1\subset \overline{C_1}$. The lemma is proved. Following the notations of the proof, if $D_1\subset C_1$, then $C_2\cap \overline {D_1}=\varnothing$. In particular, $C_2\subset D_2$. Thus, the notion of compatibility is symmetric with respect to $\gamma_1$ and $\gamma_2$. Lemma 2.10. If two belts $\mathcal{B}_1$ and $\mathcal{B}_2$ are not compatible, then each of them contains two facets lying in the closures of different connected components of the complement to the other belt in $\partial P$. Proof. Let $\gamma_1=\gamma(\mathcal{B}_1)$ divide $\partial P$ into connected components $C_1$ and $C_2$, and let $\gamma_2=\gamma(\mathcal{B}_2)$. If the belts are not compatible, then $\gamma_2$ has points in $C_1$ and $C_2$. Each segment of $\gamma_2$ connects the two midpoints of nonadjacent edges of some facet $F_i$ of $\mathcal{B}_2$. Let such a segment $E_2$ have a point in $C_1$. If $F_i$ does not belong to $\mathcal{B}_1$, then $F_i\subset C_1$, otherwise there is a segment $E_1$ of $\gamma_1$ lying in $F_i$. Since the segments do not coincide (for otherwise $E_2\subset \gamma_1$ cannot have points in $C_1$), either $E_2$ does not intersect $E_1$, or it intersects $E_1$ transversally, or they have a unique common vertex. In all cases $E_2$ has a vertex in $C_1$. This vertex is the midpoint of the edge of intersection of $F_i$ with some facet $F_j\in \mathcal{B}_2$. The facet $F_j$ has a point in $C_1$, is adjacent to $F_i$ and does not succeed it in $\mathcal{B}_1$. Therefore, $F_j\notin \mathcal{B}_1$, and $F_j\subset C_1$. One can also similarly find a facet in $\mathcal{B}_2$ lying in $C_2$. This finishes the proof. Definition 2.11. By a nested family of belts we mean a collection of belts such that any two of them are compatible. Proof. If the $3$-belts $\mathcal{B}_1$ and $\mathcal{B}_2$ are not compatible, then $\mathcal{B}_1$ has two facets in the closures of different connected components of $\partial P\setminus|\mathcal{B}_2|$. Hence these facets do not intersect, which is a contradiction. The corollary is proved. Definition 2.13. A nested family of curves is a family of piecewise linear simple closed curves $\widetilde{\gamma}(\mathcal{B})$ corresponding to the belts $\mathcal{B}$ in a nested family bijectively and satisfying the following conditions: 1) the segments of each curve $\widetilde{\gamma}(\mathcal{B})$ correspond bijectively to the facets of the belt $\mathcal{B}$, and each segment connects two points in the relative interiors of two (different) edges of intersection of the corresponding facet with the adjacent facets of the belt; 2) the curves are pairwise disjoint. For any vertex $v$ of an edge $F_i\cap F_j\subset P$ and a belt $\mathcal{B}$ containing $F_i$ and $F_j$, there is a unique connected component $C_v(\mathcal{B})$ of $\partial P\setminus\gamma(\mathcal{B})$ containing $v$. Proposition 2.14. For any nested family of curves and any edge $E$ of $P$ containing the points $v_p\in\widetilde{\gamma}(\mathcal{B}_p)$ and $v_q\in\widetilde{\gamma}(\mathcal{B}_q)$, the vertex $v$ of $E$ is closer to $v_p$ than to $v_q$ if and only if $C_v(\mathcal{B}_p)\subset C_v(\mathcal{B}_q)$. Proof. Assume that $C_v(\mathcal{B}_p)\subset C_v(\mathcal{B}_q)$ and $v$ is closer to $v_q$. Consider the segments of $\widetilde{\gamma}(\mathcal{B}_p)$ and $\widetilde{\gamma}(\mathcal{B}_q)$ lying on the same facet and starting at $v_p$ and $v_q$. If their other endpoints $v_p'$ and $v_q'$ lie on the same edge, then the vertex $v'\in C_{v'}(\mathcal{B}_p)=C_v(\mathcal{B}_p)\subset C_v(\mathcal{B}_q)=C_{v'}(\mathcal{B}_q)$ of this edge is again closer to $v'_q$ than to $v_p'$ since the curves are disjoint. Then we take the next segments. Since the curves correspond to different compatible belts, after several steps we obtain two segments on one facet that start from the same edge and finish on different edges of this facet. Moreover, for the starting edge its vertex $\widehat{v}\in C_{\widehat{v}}(\mathcal{B}_p)\subset C_{\widehat{v}}(\mathcal{B}_q)$ is again closer to the vertex of $\widetilde{\gamma}(\mathcal{B}_q)$. The terminal edge of the segment of $\widetilde{\gamma}(\mathcal{B}_p)$ lies in $C_{\widehat{v}}(\mathcal{B}_q)$, and the terminal edge of the segment of $\widetilde{\gamma}(\mathcal{B}_q)$ lies in $\partial P\setminus \overline{C_{\widehat{v}}(\mathcal{B}_p)}$. Then these segments must intersect, which is a contradiction. The proposition is proved. Definition 2.16. A nested family of belts is said to be cylindrical if there is an ordering of this family $\mathcal{B}_1,\dots,\mathcal{B}_N$ and a choice of connected components $C(\mathcal{B}_i)$ of $\partial P\setminus \gamma(\mathcal{B}_i)$ such that $C(\mathcal{B}_i)\subset C(\mathcal{B}_{i+1})$ for $i=1,\dots,N-1$. Lemma 2.17. For any nested family of belts and an edge $F_i\cap F_j$ the subfamily of belts containing $F_i$ and $F_j$ is cylindrical. Proof. A belt $\mathcal{B}$ contains $F_i$ and $F_j$ if and only if $\gamma(\mathcal{B})$ contains the midpoint of ${F_i\cap F_j}$. Moreover, the vertices $v$ and $w$ of the edge $F_i\cap F_j$ lie in different connected components of $\partial P\setminus\gamma(\mathcal{B})$. Since $v\notin C_w(\mathcal{B})$, we have $C_v(\mathcal{B})\not\subset C_w(\mathcal{B}')$ for any two belts $\mathcal{B}$ and $\mathcal{B}'$ containing $F_i$ and $F_j$. Since belts are compatible, $C_v(\mathcal{B})\subset C_v(\mathcal{B}')$ or $C_v(\mathcal{B}')\subset C_v(\mathcal{B})$. Then the required ordering of belts is the inclusion ordering of the components $C_v(\mathcal{B})$. Lemma is proved. Construction 2.18. For a nested family of belts and each edge $F_i\cap F_j$ of $P$ take its vertex $v$ and consider the ordering of the belts containing $F_i$ and $F_j$ by inclusion of the components $C_v(\mathcal{B})$. In the interior of the edge $F_i\cap F_j$, in the same order in the direction from $v$ to the other vertex we put disjoint points corresponding to these belts (for example, we divide $F_i\cap F_j$ into equal parts by these points). Then for each belt $\mathcal{B}$ we replace the middle line $\gamma(\mathcal{B})$ by a piecewise linear curve $\widehat{\gamma}(\mathcal{B})$ consisting of the line segments in facets of the belt that connect the selected points on edges of intersection with adjacent facets of the belt. Lemma 2.19. For any two belts $\mathcal{B}_1$ and $\mathcal{B}_2$ in a nested family of belts the curves $\widehat{\gamma}(\mathcal{B}_1)$ and $\widehat{\gamma}(\mathcal{B}_2)$ do not intersect. Proof. Let the curves $\widehat{\gamma}(\mathcal{B}_1)$ and $\widehat{\gamma}(\mathcal{B}_2)$ have a common point. By construction these curves have no common points on edges and do not go through vertices of the polytope. Then their common point is an interior point of some facet $F_i$. In particular, this point is the intersection of some segments $\widehat{E_1}$ and $\widehat{E_2}$ of these curves which lie in $F_i$. If the endpoints of these segments lie on four different edges, then the corresponding curves $\gamma(\mathcal{B}_1)$ and $\gamma(\mathcal{B}_2)$ intersect transversally, which is a contradiction. Thus the endpoints of the segments lie on either three or two edges of $F_i$. For $\gamma(\mathcal{B}_1)$ and $\gamma(\mathcal{B}_2)$ the corresponding segments $E_1$ and $E_2$ ether have a unique common vertex, or coincide. Without loss of generality assume that for a vertex $v$ of the edge of $P$ containing the common endpoint of $E_1$ and $E_2$ we have $C_v(\mathcal{B}_1)\subset C_v(\mathcal{B}_2)$. In the first case the other endpoint of $E_1$ lies in $C_v(\mathcal{B}_2)$. After a shift, the first endpoint of $\widehat{E_1}$ lies closer to $v$ than the corresponding endpoint of $\widehat{E_2}$, while the other endpoints of these segments stay on the same edges as before. Hence $\widehat{E_1}\cap \widehat{E_2}=\varnothing$. In the second case, for both edges of $F_i$ containing the endpoints of $E_1$ and $E_2$ their vertices $v$ and $v'$ lie in $C_{v'}(\mathcal{B}_1)=C_v(\mathcal{B}_1)\subset C_v(\mathcal{B}_2)=C_{v'}(\mathcal{B}_2)$. Therefore, the edges $\widehat{E_1}$ and $\widehat{E_2}$ become disjoint after a shift. Lemma is proved. This finishes the proof of Lemma 2.15. Lemma 2.20. Any two nested families of curves corresponding to the same nested family of belts are isotopic in the class of nested families of curves. Proof. The curves are defined uniquely by the corresponding sequences of points on the edges of the polytope. By Proposition 2.14, on each edge the points corresponding to different belts are arranged in the same order. So we can move one sequence of points on each edge to the other sequence by an isotopy. The lemma is proved. 2.3. Operations on polytopesConstruction 2.21. For any $k$-belt $\mathcal{B}$ of a simple $3$-polytope $P$ one can define the operation of cutting the polytope along the belt. Combinatorially, take a piecewise linear closed curve consisting of segments connecting some interior points of edges of intersection of each facet of a belt with the adjacent facets in the belt, for example, $\gamma(\mathcal{B})$. Then cut the surface of the polytope along this curve to obtain two topological discs partitioned into polygons. Gluing up each disc by a polygon along its boundary we obtain two partitions of a sphere into polygons. Using Steinitz’s theorem it is easy to see (see [26], Lemma 2.28, [40], Proposition B.1, and [41], Proposition 5.1.1) that each partition is combinatorially equivalent to the boundary of a simple polytope. Denote these simple polytopes by $P_1$ and $P_2$. Each polytope is different from the simplex, since no three facets of the belt have a common vertex. Facets of the polytope $P_i$ are of three types:
We say that the facet $F_a$ of $P$ is present in $P_i$ if $\widehat{F_a}$ is a facet of $P_i$. Remark 2.22. Lemma 2.20 implies that the operation of cutting a simple $3$-polytope along a nested family of belts is well defined. Lemma 2.23. Facets $\widehat{F_{a_1}},\dots,\widehat{F_{a_k}}$ in $P_i$ intersect if and only if the corresponding facets $F_{a_1},\dots,F_{a_k}$ in $P$ intersect. Proof. If $F_{a_1},\dots,F_{a_k}$ intersect, consider their intersection in $P$. It is either a vertex or an edge. If the corresponding facets $\widehat{F_{a_1}},\dots,\widehat{F_{a_k}}$ do not intersect, then this vertex or edge lies outside $\overline{C_i(\mathcal{B})}$. This is possible only if all facets $F_{a_1},\dots,F_{a_k}$ belong to $\mathcal{B}$. Then $k=2$ and $F_{a_1}$ and $F_{a_2}$ are successive facets in $\mathcal{B}$. Then $F_{a_1}\cap F_{a_2}$ intersects $\gamma(\mathcal{B})$, which is a contradiction. On the other hand, the construction of $P_i$ does not create new intersections of old facets, that is, if $F_{a_1},\dots,F_{a_k}$ do not intersect, then the corresponding facets $\widehat{F_{a_1}},\dots,\widehat{F_{a_k}}$ in $P_i$ do not intersect either. The lemma is proved. Lemma 2.24. For each polytope $P_i$, $i=1,2$, its $3$-belts correspond bijectively to the $3$-belts of $P$ consisting of facets present in $P_i$ (equivalently, $3$-belts $\mathcal{L}$ of $P$ with $\gamma(\mathcal{L})\subset \overline{C_i(\mathcal{B})}$). This bijection preserves the triviality of belts, except in the case when the belt around the facet $F$ corresponds to a nontrivial $3$-belt $\mathcal{B}$. Proof. First let us prove that $3$-belts of $P$ consisting of facets present in $P_i$ are exactly $3$-belts $\mathcal{L}$ of $P$ with $\gamma(\mathcal{L})\subset \overline{C_i(\mathcal{B})}$. Indeed, if $\mathcal{L}$ consists of facets present in $P_i$, then the midpoints of edges of intersection of its successive facets lie in $\overline{C_i(\mathcal{B})}$. Therefore, $\gamma(\mathcal{L})\subset \overline{C_i(\mathcal{B})}$. Vice versa, if $\gamma(\mathcal{L})\subset \overline{C_i(\mathcal{B})}$, then the midpoints of edges of intersection of successive facets of $\mathcal{L}$ lie in $\overline{C_i(\mathcal{B})}$. This implies that all the facets of $\mathcal{L}$ are present in $P_i$.
The facet $F$ does not lie in any $3$-belt. For otherwise this belt has the form $(F,\widehat{F_a},\widehat{F_b})$, where $F_a, F_b\in\mathcal{B}$. In this case $F_a\cap F_b\ne\varnothing$. Hence $F_a$ and $F_b$ are successive facets of $\mathcal{B}$. In particular, $F_a\cap F_b$ is an edge with a point on $\gamma(\mathcal{B})$, which corresponds to the vertex $F\cap\widehat{F_a}\cap\widehat{F_b}$. The correspondence $\widehat{\mathcal{L}}=(\widehat{F_a},\widehat{F_b},\widehat{F_c})\leftrightarrow (F_a,F_b,F_c)=\mathcal{L}$ between the $3$-belts of $P_i$ and the $3$-belts of $P$ consisting of facets present in $P_i$ follows from Lemma 2.23. If the belt $\widehat{\mathcal{L}}$ surrounds a facet $\widehat{F_d}$, then $\mathcal{L}$ surrounds $F_d$. If $\widehat{\mathcal{L}}$ surrounds the triangle $F$, then this belt corresponds to a $3$-belt $\mathcal{B}$, which can be either trivial or nontrivial. If the $3$-belt $\mathcal{L}$ consisting of facets present in $P_i$ surrounds a facet $F_d$ of $P$, then either $\mathcal{B}=\mathcal{L}$ is a trivial belt, or $F_d$ is also present in $P_i$. Moreover, in the latter case $F_d$ does not belong to $\mathcal{B}$, since a belt cannot contain triangles; therefore, $\widehat{\mathcal{L}}$ surrounds $\widehat{F_d}$. The lemma is proved. Corollary 2.25. Any $3$-belt of $P$ different from $\mathcal{B}$ corresponds to a $3$-belt of exactly one polytope $P_1$ or $P_2$. Proof. Since the $3$-belts form a nested family, any $3$-belt of $P$ consists of facets which are simultaneously present in one of the polytopes $P_1$ and $P_2$. If all its facets are present in both $P_1$ and $P_2$, then it coincides with $\mathcal{B}$. The corollary is proved. Construction 2.26. Given simple $3$-polytopes $P_1$ and $P_2$ with some chosen facets $F_i$ and $F_j$ surrounded by $k$-belts with the same $k$, one can take a connected sum of $P_1$ and $P_2$ along $k$-gons. The result is a simple $3$-polytope $P$ such that $\partial P$ is combinatorially obtained from $\partial P_1$ and $\partial P_2$ by deleting the interiors of the chosen facets, gluing the resulting discs partitioned into polygons along their boundary $k$-cycles, and then deleting the arising $k$-cycle. Steinitz’s theorem implies (see [40], Proposition B.1, and [41], Proposition 5.1.1) that $P$ is indeed a simple $3$-polytope. This operation depends on the identification of the boundaries of the chosen facets. It is inverse to the operation of cutting a polytope along a belt. Corollary 2.27. Any $k$-belt $\mathcal{B}$ of a simple $3$-polytope $P$ corresponds to a decomposition of $P$ into a connected sum of some polytopes $P_1$ and $P_2$ along $k$-gons. Definition 2.28. A connected sum of simple $3$-polytopes $P$ and $Q$ along vertices $v\in P$ and $w\in Q$ is a simple $3$-polytope $P\#_{v,w}Q$ which is obtained by cutting off the vertices $v$ and $w$ and taking the connected sum along the arising triangles. The operation of connected sum along vertices gives a $3$-belt on the resulting polytope. Remark 2.29. Steinitz’s theorem implies that for a $3$-belt $\mathcal{B}$ in Construction 2.21 each polytope $P_i$ admits a shrinking of the triangle obtained from the belt into a point so that a new $3$-polytope $Q_i$ is produced. Then $P$ is a connected sum of $Q_1$ and $Q_2$ along vertices. Topologically, the polytopes $Q_1$ and $Q_2$ are obtained from $P$ by cutting it along the middle line of the belt and shrinking the arising triangles to points. In particular, each triangle of a polytope $P\ne\Delta^3$ corresponds to a connected sum of a polytope $Q$ with this triangle shrunk, with a simplex along vertices. Corollary 2.30. Let a simple $3$-polytope $P$ be a connected sum of $Q_1$ and $Q_2$ along vertices. Then for each polytope $Q_i$, $i=1,2$, its $3$-belts correspond bijectively to the $3$-belts of $P$ different from the arising $3$-belt and consisting of facets present in $Q_i$. Proof. This follows from Lemma 2.24 and Corollary 2.25. Namely, since the facet $F$ cannot belong to any $3$-belt of $P_i$, shrinking this facet to a point transforms $3$-belts of $P_i$ different from the belt around $F$ into $3$-belts of $Q_i$. On the other hand the polytope $P_i$ is obtained from $Q_i$ by cutting off a vertex. This operation evidently transforms $3$-belts into $3$-belts. The corollary is proved. Now consider a geometric realization of a connected sum of two simple $3$-polytopes along triangles and quadrangles. We need it below. For triangles the result is simpler. Proposition 2.31. Let $P$ and $P'$ be geometric simple $3$-polytopes, different from the simplex, and $F_i$ and $F_j'$ be their triangular facets. Fix an identification of $F_i$ and $F_j'$ as combinatorial polygons. Then, up to projective transformations, $P$ and $P'$ can be glued along congruent facets $F_i$ and $F_j'$ following the combinatorial identification to produce a geometric $3$-polytope $Q$ with a $3$-belt instead of these facets. Proof. Consider the three edges of $P$ intersecting $F_i$ at vertices. Each pair of these edges lies in a plane. Therefore, either these edges are parallel, or the lines containing them intersect at a common point. In the latter case the point does not belong to $P$, for otherwise $P=\Delta^3$. Then there is a projective transformation mapping this point to a point at infinity and taking $P$ to a bounded polytope. In the new polytope the three edges are parallel. By an affine transformation we can make them orthogonal to $F_i$ and transform this facet into a regular triangle. Applying to $P'$ similar transformations, and, perhaps, a reflection we can glue the resulting polytopes along congruent facets by following the combinatorial identification. The proposition is proved. Corollary 2.32. For any family $\mathcal{F}$ of $3$-belts of a simple $3$-polytope $P$ there is a geometric realization of $P$ and a family of planes corresponding to belts bijectively such that the intersections of planes with $\partial P$ form a nested family of curves corresponding to $\mathcal{F}$. Proof. By Corollary 2.12, $\mathcal{F}$ is a nested family of belts. By Lemmas 2.15 and 2.20 this family defines a decomposition of $P$ into a connected sum of polytopes along triangles. These operations can be encoded by a tree, where vertices correspond to polytopes and edges to pairs of identified triangles. Take any geometric realizations of these polytopes. Then starting from any vertex of the tree as a root and applying projective transformations, using Proposition 2.31 we can glue all the polytopes one by one with respect to the combinatorial identifications of their triangles. The resulting polytope is combinatorially equivalent to $P$, the identified triangles correspond to planar cross-sections of it, and these cross-sections are disjoint. The corollary is proved. For a connected sum along quadrangles the situation is a bit more complicated. Let $P$ be a simple geometric $3$-polytope in $\mathbb R^3\subset\mathbb RP^3$, and $F_i$ be its quadrangular facet surrounded by a $4$-belt $(F_j,F_k,F_l,F_r)$. Then no three of the four planes $\pi_j=\operatorname{aff}(F_j)$, $\pi_k=\operatorname{aff}(F_k)$, $\pi_l=\operatorname{aff}(F_l)$ and $\pi_r=\operatorname{aff}(F_r)$ contain the same line in $\mathbb RP^3$. There are two possibilities: either these planes contain a common point of $\mathbb RP^3$, or not. In the first case, up to a projective transformation we can assume that this point lies at infinity. Then these four planes in $\mathbb R^3$ bound an infinite cylinder over $F_i$, which, after the addition of one point in $\mathbb R P^3$, is divided by $F_i$ into two $4$-pyramids. Let us call the pyramid containing $P$ the inner pyramid associated with $F_i$, and the other one the outer pyramid. The addition of the outer pyramid to $P$ transforms it into a polytope obtained from $P$ combinatorially by shrinking $F_i$ to a point. In the second case the four planes are in general position in $\mathbb RP^3$ and, up to a projective transformation, these planes bound a simplex in $\mathbb R^3$ containing $P$. The plane $\pi_i=\operatorname{aff}(F_i)$ divides this simplex into two polytopes which are combinatorially equivalent to $3$-prisms. These prisms have a common quadrangle $F_i$ and are ‘twisted’, that is, the bases of each prism are glued to a pair of quadrangles of the other along edges. Let us call the prism containing $P$ the inner prism associated to $F_i$ and the other $3$-prism the outer prism. There are two possibilities: $F_j$ and $F_l$ correspond to the bases of the inner prism, or to those of the outer prism. In the first case the addition of the outer prism to the polytope transforms $P$ into a polytope with $F_i$ shrunk combinatorially to an edge ‘parallel’ to $F_i\cap F_j$ and $F_i\cap F_l$, and in the second case such an edge is parallel to $F_i\cap F_k$ and $F_i\cap F_r$. Each of the three possibilities is a projective invariant of the geometric polytope $P$. Proposition 2.33. Let $P$ and $P'$ be geometric simple $3$-polytopes, and $F_i$ and $F_j'$ be their quadrangular facets surrounded by belts. Fix an identification of $F_i$ and $F_j'$ as combinatorial polygons. Then, up to projective transformations, $P$ and $P'$ can be glued along congruent facets $F_i$ and $F_j'$ following the combinatorial identification to produce a geometric simple $3$-polytope $Q$ having a $4$-belt instead of these facets if and only if both $F_i$ and $F_j'$ correspond either to $4$-pyramids or to $3$-prisms so that the inner $3$-prisms are ‘twisted’ under the identification of their quadrangles. Proof. From the above arguments we see that if after projective transformations we can glue the polytopes along congruent facets, then either both facets $F_i$ and $F_j'$ correspond to $4$-pyramids, and the inner $4$-pyramid of one polytope is identified with the outer $4$-pyramid of the other, or both facets correspond to $3$-prisms and the inner $3$-prism of one polytope is identified with the outer $3$-prism of the other polytope. In particular, the inner $3$-prisms are twisted under the identification of their facets. On the other hand, if $F_i$ and $F_j'$ correspond to $4$-pyramids, then there are projective transformations making congruent squares orthogonal to the adjacent facets of these polytopes, and we can glue the polytopes following the combinatorial identification of facets, perhaps after a reflection of one of the polytopes. If $F_i$ and $F_j'$ correspond to twisted $3$-prisms, then using the classical facts that any two combinatorial $3$-prisms are projectively equivalent (see [25], Exercise 6.24) and the group of symmetries of a right triangular prism with regular bases acts transitively on the pairs (a quadrangle, an edge of it orthogonal to the bases) we can find a projective transformation mapping the inner $3$-prism associated with $F_i$ to the outer $3$-prism associated with $F_j'$ and respecting the combinatorial identification of these facets. This finishes the proof. Corollary 2.34. For any nested family $\mathcal{F}$ of $4$-belts of a simple $3$-polytope $P$ there is a geometric realization of $P$ and a family of planes bijectively corresponding to belts such that the intersections of planes with $\partial P$ form a nested family of curves corresponding to $\mathcal{F}$. Proof. By Lemmas 2.15 and 2.20 the family $\mathcal{F}$ defines uniquely a decomposition of $P$ into a connected sum of polytopes along quadrangles. These operations can be encoded by a tree $T$, where vertices correspond to polytopes and edges to pairs of identified quadrangles. Choose any vertex of $T$ as a root and take any geometric realization of the corresponding polytope $Q$. Since $Q$ is simple, there is an arbitrarily small deformation of its defining inequalities such that the new realization has the same combinatorial type and all planes containing facets are in general position, that is, no point in $\mathbb R^3$ lies in more than three of these planes. Then any quadrangle of $Q$ corresponds to an inner $3$-prism of $Q$. Consider the edge of $T$ corresponding to an identification of a quadrangle $F_i$ of $Q$ with a quadrangle $F_j'$ of another polytope $R$. According to Proposition 2.33, to make this identification geometric after projective transformations one needs to realize $R$ in such a way that the inner $3$-prism of $R$ is twisted under this identification. Again, the realization can be deformed to be in general position. Since $F_j'$ is surrounded by a belt, Steinitz’s theorem implies (see [9], Theorem 11.4) that both polytopes arising from $R$ by shrinking $F_j'$ to an edge in the two possible ways exist. Cutting this edge we obtain two realizations of $R$ with the inner $3$-prisms corresponding to $F_j'$ twisted under the identification of their quadrangles. So one of these realizations fits our condition. Repeating this argument for each new edge of $T$, in the end we obtain a polytope combinatorially equivalent to $P$. In this polytope the identified quadrangles correspond to pairwise disjoint planar cross-sections. The corollary is proved.
§ 3. Prime decomposition for $3$-manifolds defined by vector colourings3.1. General informationAs mentioned above, we follow the exposition of $3$-manifold theory presented in [2]. It is important, that in dimension $3$ there is no difference between continuous, smooth and piecewise linear categories. Namely, it is known that any (not necessarily compact) topological $3$-manifold admits a unique piecewise linear structure and a unique smooth structure. A topological action of a finite group on a closed $3$-manifold $N$ is smoothable if and only if it is simplicial in some triangulation of $N$. Definition 3.1. By a topological $2$-submanifold of a closed (that is, compact without boundary) topological $3$-manifold $N$ we mean a subset $S$ of $N$ such that for any point $p\in N$ there exists a homeomorphism $j\colon U\to V$ from an open neighbourhood of $p$ in $N$ to an open subset of $\mathbb R^3$ such that $j(U\cap S)=V\cap (\{0\}\times \mathbb R^2)$. It is essential that one can replace topological $2$-submanifolds by $\mathrm{PL}$-submanifolds and by smooth submanifolds. In particular, if $N$ is a topological closed $3$-manifold and $S$ is a topological $2$-submanifold of $N$, then there exists a smooth (or $\mathrm{PL}$) structure on $N$ and a smoothly (or $\mathrm{PL}$) embedded surface $S'$ in $N$ that is isotopic to $S$ (see [2], § 1.1). Definition 3.2. We denote a tubular neighbourhood of the submanifold $S$ of $N$ by $\nu(S)$. Given a surface $S$ in $N$ we refer to $N\setminus\nu(S)$ as $N$ cut along $S$. In the case of manifolds defined by vector colourings there are natural structures of piecewise linear manifolds and submanifolds. Definition 3.3. Let $N_1$ and $N_2$ be oriented $3$-manifolds and let $B_i\subset N_i$, $i=1,2$, be embedded closed $3$-balls with boundaries which are $2$-submanifolds homeomorphic to $S^2$. For $i = 1,2$ we endow $\partial B_i$ with the orientation induced from the orientation as a boundary component of $N_i\setminus\operatorname{int} B_i$. We pick an orientation-reversing homeomorphism $f \colon \partial B_1 \to \partial B_2$ and refer to $N_1\#N_2 \colon= (N_1\setminus \operatorname{int}B_i)\cup_f (N_2\setminus \operatorname{int}B_2)$ as the connected sum of $N_1$ and $N_2$. The homeomorphism type of $N_1\#N_2$ does not depend on the choice of $f$. Definition 3.4. An orientable $3$-manifold $N$ is called prime if it cannot be decomposed into a nontrivial connected sum of two manifolds, so that, if $N = N_1\#N_2$, then $N_1 = S^3$ or $N_2 = S^3$. A $3$-manifold $N$ is called irreducible if every one of its $2$-submanifolds homeomorphic to a $2$-sphere bounds a subset in $N$ homeomorphic to a closed $3$-ball. An orientable irreducible $3$-manifold is evidently prime. It can be proved that $S^2\times S^1$ is prime, but not irreducible. Conversely, if $N$ is a closed orientable prime $3$-manifold, then either $N$ is irreducible or $N = S^2\times S^1$. It is a consequence of the generalized Schoenflies Theorem that $S^3$ is irreducible. If $p\colon\widetilde{M}\to M$ is a covering space and $\widetilde{M}$ is irreducible, then $M$ is also irreducible (see [42], Proposition 1.6). In particular, $\mathbb RP^3$ is an irreducible manifold. The following theorem is due to Kneser (1929), Haken (1961) and Milnor (1962). Theorem 3.5 (Prime Decomposition Theorem; see [2], Theorem 1.2.1). Let $N$ be a closed oriented $3$-manifold. Then there exist oriented closed prime $3$-manifolds $N_1,\dots,N_m$ such that $N=N_1\#\dotsb\#N_m$. Moreover, if $N=N_1\#\dotsb\#N_m$, and $N=N_1'\#\dotsb\#N_n'$, with oriented prime $3$-manifolds $N_i$ and $N_i'$, then $m=n$ and (possibly after reordering) there exist orientation-preserving homeomorphisms ${N_i\to N_i'}$. 3.2. The case of manifolds defined by vector colouringsDenote by $X^{\#k}$ the connected sum $\underbrace{X\#\dotsb\#X}_k$. Proposition 3.6. Let $P$ be a connected sum of two simple $3$-polytopes $P_1$ and $P_2$ along vertices, and $M(P,\Lambda)$ be an orientable manifold defined by a vector colouring $\Lambda$ of rank $r$. Then there is a homeomorphism where $\Lambda_1$ and $\Lambda_2$ are induced vector colourings of ranks $r_1$ and $r_2$ respectively. Remark 3.7. In the nonorientable case the statement of Proposition 3.6 is not valid. The reason is that the vectors $\Lambda_i$, $\Lambda_j$ and $\Lambda_k$ corresponding to the facets $F_i$, $F_j$ and $F_k$ of a $3$-belt can be linearly dependent. Then the restrictions of $\Lambda$ to $P_1$ and $P_2$ do not satisfy condition (*). For example, consider $\Delta^2\times I=\Delta^3\#\Delta^3$ and set $\Lambda$ to be $(1,0,0)$ on the bases of the $3$-prism, and $(0,1,0)$, $(0,0,1)$ and $(0,1,1)$ on side facets. Proof of Proposition 3.6. Consider the $3$-belt $\mathcal{B}=(F_i,F_j,F_k)$ of $P$ which corresponds to the connected sum $P=P_1\#P_2$. Since the facets of the belt are pairwise adjacent, the vectors $\Lambda_i$, $\Lambda_j$ and $\Lambda_k$ are pairwise different. If $\Lambda_i+\Lambda_j+\Lambda_k=0$, then $\Lambda_k=\Lambda_i+\Lambda_j$, and $M(P,\Lambda)$ is not orientable, which is a contradiction. Thus, $\Lambda_i$, $\Lambda_j$ and $\Lambda_k$ are linearly independent. In particular, the induced mappings $\Lambda_1$ and $\Lambda_2$ for $P_1$ and $P_2$ satisfy condition (*) and are vector colourings.
Let $\pi_i$ be the linear span of the vectors $\Lambda_j\in V=\mathbb Z_2^r$ corresponding to facets of the polytope $P_i$. Then $\pi_1+\pi_2=V$ and $r_i=\dim \pi_i$. The $3$-dimensional subspace $\pi_{\mathcal{B}}=\langle\Lambda_i,\Lambda_j,\Lambda_k\rangle$ corresponding to $\mathcal{B}$ lies in $\pi_1\cap\pi_2$. In $M(P,\Lambda)$ the belt $\mathcal{B}$ corresponds to $2^r$ copies of the triangle bounded in $P$ by the middle line of $\mathcal{B}$. These triangles form $2^{r-3}$ disjoint $2$-spheres. Each sphere is glued of eight triangles corresponding to polytopes $P\times a$ with $a$ lying in the same coset $x+\pi_\mathcal{B}\in V/\pi_{\mathcal{B}}$. For $P_1$ and $P_2$ each vertex involved in the connected sum corresponds to a tetrahedron — the convex hull of this vertex and the midpoints of the three edges incident to it. In $M(P_i,\Lambda_i)$ this corresponds to a disjoint set of $2^{r_i-3}$ balls corresponding to polytopes $P_i\times a$ with $a$ lying in the same coset in $\pi_i/\pi_{\mathcal{B}}$. Each ball is glued of eight tetrahedra. In $M(P,\Lambda)$ we have $2^{r-r_i}$ copies of the manifold $N_i$ obtained from $M(P_i,\Lambda_i)$ by the deletion of these balls. Each copy is glued of parts of polytopes $P_i\times a$ with $a$ lying in the same coset in $V/\pi_i$ and is bounded by $2^{r_i-3}$ spheres. The boundary spheres of the same copy of $N_i$ that correspond to the same coset in $\pi_i/(\pi_1\cap\pi_2)$ are also boundary components of the same copy of $N_{2-i}$. Spheres corresponding to different cosets in $\pi_i/(\pi_1\cap\pi_2)$ are boundary components of different copies of $N_{2-i}$. Indeed, if a copy of $N_i$ is glued of parts of polytopes $P_i\times a$ with $a\in x+\pi_i$ and two spheres are glued of triangles lying in polytopes $P_i\times b$, where $b\in x+(y+\pi_{\mathcal{B}})$ for the first sphere and $b\in x+(z+ \pi_{\mathcal{B}})$ for the second, and if $y-z\notin \pi_1\cap\pi_2$, then $(x+y)-(x+z)=y-z\notin\pi_{2-i}$, for otherwise $y-z\in \pi_1\cap \pi_2$. If we choose one copy of $N_1$ and one sphere for each coset $\pi_1/(\pi_1\cap\pi_2)$ and also choose one copy of $N_2$ and one sphere for each coset in $\pi_2/(\pi_1\cap\pi_2)$ including the particular sphere chosen for $N_1$ and corresponding to $N_2$, then these spheres correspond to a connected sum $M(P_1,\Lambda_1)^{\#2^{r-r_1}}\#M(P_2,\Lambda_2)^{\#2^{r-r_2}}$. There are $2^{r-r_1}+2^{r-r_2}-1$ such spheres. Any other sphere corresponds to an addition of a handle, which is equivalent to taking the connected sum with $S^2\times S^1$. There are $2^{r-3}-2^{r-r_1}-2^{r-r_2}+1$ such spheres. This finishes the proof. Corollary 3.8. Let $P$ be a connected sum of two simple $3$-polytopes $P_1$ and $P_2$ along vertices. Then there is a homeomorphism where $m_i$ is the number of facets of the polytope $P_i$, $i=1,2$. Corollary 3.9. Let $P$ be a connected sum of two simple $3$-polytopes $P_1$ and $P_2$ along vertices. And let $M(P,\Lambda)$ be an orientable small cover. Then there is a homeomorphism where $\Lambda_1$ and $\Lambda_2$ are the characteristic functions for $P_1$ and $P_2$ obtained as restrictions of $\Lambda$. Example 3.10. For $P=\Delta^3$, up to a change of basis in the image, there are two possible vector colourings: one of rank $4$ and one of rank $3$. The first of them corresponds to the real moment-angle manifold $\mathbb{R}\mathcal{Z}_{\Delta^3}=S^3$, and the second to the unique small cover, namely, $\mathbb RP^3$. Both of them are irreducible. Example 3.11. Let us classify the orientable manifolds $M(P,\Lambda)$ for $P=\Delta^2\times I$. It is necessary that the vectors $\Lambda_i$, $\Lambda_j$ and $\Lambda_k$ corresponding to side facets of the prism, are linearly independent. Let $\Lambda_l$ and $\Lambda_r$ correspond to the bases. If $\Lambda$ has rank $5$, then $M(P,\Lambda)=\mathbb R\mathcal{Z}_{\Delta^2\times I}=S^2\times S^1$ is a prime manifold. If $\Lambda$ has rank $4$, then up to a reflection of $\Delta^2\times I$ we can assume that the vectors $\Lambda_i$, $\Lambda_j$, $\Lambda_k$, and $\Lambda_l$ form a basis. Then up to a rotation of $\Delta^3\times I$ we have the following cases: $\Lambda_r=\Lambda_l$, $\Lambda_r=\Lambda_i+\Lambda_j+\Lambda_k$ and $\Lambda_r=\Lambda_i+\Lambda_j+\Lambda_l$. Example 3.10 and Proposition 3.6 imply that in these cases $M(P,\Lambda)$ is homeomorphic to $S^3\#S^3\#(S^2\times S^1)=S^2\times S^1$, $S^3\#(\mathbb RP^3)^{\#2}\#(S^2\times S^1)^{\#0}=\mathbb RP^3\#\mathbb RP^3$ and $S^3\#S^3\#(S^2\times S^1)=S^2\times S^1$, respectively. If $\Lambda$ has rank $3$, then $\Lambda_r=\Lambda_l=\Lambda_i+\Lambda_j+\Lambda_k$, and $M(P,\Lambda)=\mathbb RP^3\mathbin{\#}\mathbb RP^3\mathbin{\#}(S^2\times S^1)^{\#0}=\mathbb RP^3\#\mathbb RP^3$. Theorem 3.12. 1. Any simple $3$-polytope $P$ can be decomposed into a connected sum of simplices and flag $3$-polytopes along vertices: Such a decomposition is unique in the following sense: if there is another decomposition $P=Q_1'\#\dotsb\#Q_{N'}'$, then $N=N'$ and after a reordering of the polytopes there are combinatorial equivalences $Q_i\simeq Q_i'$ which are compatible with the correspondence between the polytopes, vertices chosen and facets at these vertices under the connected sums. 2. The orientable manifold $M(\Delta^3,\Lambda)$ is either $S^3$ or $\mathbb RP^3$. It is irreducible. The orientable manifold $M(\Delta^2\times I,\Lambda)$ is either $S^2\times S^1$, which is prime, or $\mathbb{R}P^3\#\mathbb{R}P^3$. For a simple $3$-polytope $P\ne \Delta^3, \Delta^2\times I$ the decompositions from part 1 and Proposition 3.6 give the prime decomposition of any orientable manifold $M(P,\Lambda)$ into a connected sum of copies of $S^2\times S^1$, copies of $\mathbb RP^3$ corresponding to simplices with the induced vector colourings of rank $3$, and copies of the manifolds $M(Q_i,\Lambda_i)$ defined by flag $3$-polytopes $Q_i$ with induced vector colourings $\Lambda_i$. The latter manifolds are aspherical; therefore, they are irreducible. Proof. The nested set of curves $\widehat{\gamma}(\mathcal{B})$ corresponding to the family of all $3$-belts gives a canonical decomposition of $P$ into a connected sum of polytopes $P_i$ along triangles and polytopes $Q_i$ along vertices. By Corollary 2.30 the polytopes $Q_i$ have no $3$-belts. Hence each of them is either a flag polytope or the simplex.
Let us prove the uniqueness of the decomposition. Let $P$ be represented as a connected sum along vertices of polytopes $Q_i'$ without $3$-belts. Each operation of connected sum along vertices produces a $3$-belt on the resulting polytope. Let us choose a piecewise linear curve which satisfies condition 1) in Definition 2.13 on this belt. After all operations we obtain a nested family of curves. The corresponding belts on $P$ form a nested family. If there is a $3$-belt of $P$ which is not in this family, then by Corollary 2.30 it corresponds to a $3$-belt on some polytope $Q_i'$, which is a contradiction. Thus, all the $3$-belts of $P$ lie in this family. By Lemma 2.20 the nested family of curves corresponding to the polytopes $Q_i'$ is isotopic to the nested family of curves corresponding to $Q_i$. Since polytopes in each family are combinatorially obtained by cutting $P$ along the curves from the corresponding family and shrinking these curves to points, we see that there is a bijection between the sets of polytopes $\{Q_i\}$ and $\{Q_i'\}$ which preserves the combinatorial equivalence and the correspondence between vertices and facets containing them. This finishes the proof of the first part of the theorem. The case of manifolds $M(\Delta,\Lambda)$ and $M(\Delta^2\times I,\Lambda)$ is covered by Examples 3.10 and 3.11. Consider the decomposition given by part 1 and Proposition 3.6. Each manifold in this decomposition is either $M(\Delta,\Lambda)$, which is irreducible, or ${S^2\times S^1}$, which is prime, or $M(Q_i,\Lambda_i)$ for a flag polytope $Q_i$. For a flag polytope $Q_i$ the manifold $\mathbb R\mathcal{Z}_{Q_i}$ is aspherical. This follows from Theorem 2.2.5 in [16] (see also [11], Proposition 1.2.3). It is known that any aspherical closed orientable $3$-manifold is irreducible. For example, Lemma 3.2 in [43] implies that if a closed aspherical $3$-manifold $M$ is decomposed into a connected sum $M_1\#M_2$, then one of the summands is homotopy equivalent to $S^3$. Then the Poincaré conjecture proved by Perelman implies that this summand is homeomorphic to $S^3$. Since each manifold $M(Q_i,\Lambda_i)$ is covered by $\mathbb R\mathcal{Z}_{Q_i}$, it is also aspherical and irreducible. The theorem is proved. Corollary 3.13. The orientable manifold $M(P,\Lambda)$ over a simple $3$-polytope $P$ is prime if and only if either $P=\Delta^3$ or $P$ is flag, or $P=\Delta^2\times I$ and $\Lambda$ either has rank $5$, or it has rank $4$ and no vector of the base lies in the subspace spanned by the vectors of side facets. § 4. $\mathrm{JSJ}$-decomposition and geometrization4.1. General information on a $\mathrm{JSJ}$-decompositionA Seifert fibred manifold is a compact $3$-manifold $N$ together with a decomposition of $N$ into disjoint simple closed curves (called Seifert fibres) such that each Seifert fibre has a tubular neighbourhood that forms a standard fibred torus. The standard fibred torus corresponding to a pair of coprime integers $(a,b)$ with $a>0$ is the mapping torus of the automorphism of a disc given by the rotation by an angle of ${2\pi b}/{a}$, equipped with the natural fibering by circles. If $a=1$, then the middle Seifert fibre is called regular, otherwise it is called singular. A $3$-manifold $N$ is atoroidal, if any map $T\to N$ from a torus to $N$ which induces a monomorphism $\pi_1(T)\to \pi_1(N)$ can be homotoped to the boundary of $N$. There exist orientable, irreducible $3$-manifolds that cannot be cut into atoroidal components in a unique way (for example, the $3$-torus). Nonetheless, any compact, orientable, irreducible $3$-manifold with empty or toroidal boundary admits a canonical decomposition along tori into atoroidal and Seifert fibred manifolds. Definition 4.1. A $2$-submanifold $\Sigma$ of a $3$-dimensional manifold $N$ is called incompressible if none of its connected components $\Sigma_i$ is homeomorphic to $S^2$ or $D^2$, and each induced mapping $\pi_1(\Sigma_i)\to \pi_1(N)$ is injective. The following theorem was first announced by Waldhausen (1969) and proved independently by Jaco and Shalen (1979) and Johannson (1975, 1979). Theorem 4.2 ($\mathrm{JSJ}$-Decomposition Theorem; see [2], Theorem 1.6.1). Let $N$ be a compact, orientable, irreducible $3$-manifold with empty or toroidal boundary. Then there exists a (possibly empty) collection of disjointly embedded incompressible tori $T_1,\dots,T_m$ such that each component of $N$ cut along $T_1,\dots,T_m$ is atoroidal or Seifert fibred. Any such collection with a minimal number of tori is unique up to isotopy. Definition 4.3. The tori in the minimal collection in Theorem 4.2 are called $\mathrm{JSJ}$-tori. The decomposition of $N$ into atoroidal and Seifert fibred components given by these tori is called the $\mathrm{JSJ}$-decomposition. Notation 4.4. Denote by $K^2$ the Klein bottle, and by $K^2\widetilde{\times} I$ the unique orientable total space of (a necessarily twisted) $I$-bundle over $K^2$. The following theorem says, in particular, that ‘sufficiently complicated’ Seifert fibred manifolds carry a unique Seifert fibred structure. Theorem 4.5 (see [2], Theorem 1.5.2). Let $N$ be an orientable $3$-manifold that admits the Seifert fibred structure. If $N$ has a nontrivial boundary, then the Seifert fibred structure of $N$ is unique up to isotopy if and only if $N$ is not homeomorphic to one of the following: $S^1\times D^2$, $T^2\times I$ and $K^2\widetilde{\times}I$. For an additively written abelian group $G$ we denote by $PG$ the set of equivalence classes of the equivalence relation $g\sim-g$ on $G\setminus\{0\}$, and for $g\in G\setminus\{0\}$ we denote by $[g]$ its equivalence class in $PG$. Given a Seifert fibred manifold $N$ with a choice of a Seifert fibred structure and a boundary torus $T$, the regular fibre determines an element $f(N,T)\in PH_1(T)$ in a natural way. It follows from Theorem 4.5 that $f(N,T)$ is independent of the choice of the Seifert fibred structure, unless $N$ is one of $S^1\times D^2$, $T^2\times I$ or $K^2\widetilde{\times} I$. For later use we record the following. Lemma 4.6 (see [2], Lemma 1.5.3). Let $N_1$ and $N_2$ be two Seifert fibred manifolds (where $N_1=N_2$ is allowed) and let $N_1\cup_{T_1=T_2}N_2$ be the result of gluing $N_1$ and $N_2$ along boundary tori $T_1$ and $T_2$. Then the following are equivalent: 1) there exists a Seifert fibred structure on $N_1\cup_{T_1=T_2}N_2$ that restricts (up to isotopy) to the Seifert fibred structures on $N_1$ and $N_2$; 2) $f(N_1,T_1)=f(N_2,T_2)\in PH_1(T_1)=PH_1(T_2)$. There is a criterion for showing that a collection of tori in $N$ is in fact the $\mathrm{JSJ}$-tori. Proposition 4.7 (see [2], Proposition 1.6.2). Let $N$ be a compact, orientable, irreducible $3$-manifold with empty or toroidal boundary. Let $T_1,\dots,T_m$ be disjointly embedded incompressible tori in the $3$-manifold $N$. Then the $T_i$ are the $\mathrm{JSJ}$-tori of $N$ if and only if and only if the following hold. 1. The components $M_1,\dots,M_n$ of $N$ cut along $T_1\cup\dots\cup T_m$ are Seifert fibred or atoroidal. 2. For any choice of Seifert fibred structures on all Seifert fibred components among $M_1,\dots,M_n$ the following two conditions hold:
3. If one of the $M_i$ equals $T^2\times I$, then $m=n=1$. To distinguish between Seifert fibred and atoroidal components we will use the following result. Theorem 4.8 (see [3], Theorem 2.10). A complete hyperbolic manifold of finite volume cannot be a Seifert fibred manifold. 4.2. A connection between the $\mathrm{JSJ}$-decomposition and geometrizationA connection between the $\mathrm{JSJ}$-decomposition and the Geometric Decomposition Theorem (Theorem 0.2) is described in the following way. We refer to the union of tori and Klein bottles in Theorem 0.2 as the geometric decomposition surface. Proposition 4.9 (see [2], Proposition 1.9.2). Let $N$ be a closed orientable, irreducible $3$-manifold. 1. If $N$ is a $\mathrm{Sol}$-manifold, then $N$ is geometric, that is, the geometric decomposition surface is empty. On the other hand $N$ has one $\mathrm{JSJ}$-torus, namely, if $N$ is a torus bundle, then the fibre is the $\mathrm{JSJ}$-torus, and if $N$ is a twisted double of $K^2\widetilde{\times}I$, then the $\mathrm{JSJ}$-torus is given by the boundary of $K^2\widetilde{\times}I$. 2. Suppose that $N$ is not a $\mathrm{Sol}$-manifold, and denote by $T_1,\dots,T_m$ the $\mathrm{JSJ}$-tori of $N$. We assume that they are ordered in such a way that the tori $T_1,\dots,T_n$ do not bound copies of $K^2\widetilde{\times}I$ and that for $i=n+1,\dots, m$, each $T_i$ cobounds a copy $K_i\widetilde{\times}I$ of its own. Then the geometric decomposition surface of $N$ is given by Conversely, if $T_1\cup\dots\cup T_n\cup K_{n+1}\cup\dots\cup K_m$ is the geometric decomposition surface such that $T_1,\dots,T_n$ are tori and $K_{n+1},\dots,K_m$ are Klein bottles, then the $\mathrm{JSJ}$-tori are given by where $\nu(K_i)$ is a tubular neighbourhood of $K_i$.Remark 4.10. We need not give details on $\mathrm{Sol}$-manifolds to use part 1 of Proposition 4.9. The only facts we will need is that for a torus bundle the complement to the $\mathrm{JSJ}$-torus is homeomorphic to the interior of $T^2\times I$, and for the twisted double of $K^2\widetilde{\times}I$ the complement of the $\mathrm{JSJ}$-torus consists of two copies of the interior of $K^2\widetilde{\times}I$ (see [2], § 1.5). The formulations of all of the previous results and the appropriate references can be found in [2]. 4.3. Applications to manifolds defined by vector colouringsNow, using the general results described above we find the $\mathrm{JSJ}$-decomposition and the geometric decomposition of any orientable manifold $M(P,\Lambda)$ defined by a vector colouring of a simple $3$-polytope. It turns out that they correspond to a canonical decomposition of a flag $3$-polytope along $4$-belts into $k$-prisms, which correspond to Seifert fibred pieces, and almost Pogorelov polytopes, which produce hyperbolic parts. The $\mathrm{JSJ}$-tori, which are defined uniquely up to isotopy, correspond to these canonical belts, and also to the ‘free quadrangles’ of almost Pogorelov polytopes; these quadrangles are not surrounded by canonical $4$-belts. An explicit geometrization of the pieces may be given by a construction invented by Vesnin and Mednykh in the series of papers [18]–[23]. Construction 4.11. Let $P$ be a right-angled polytope of finite volume in $\mathbb R^3$, $S^3$, $\mathbb{H}^3$, $S^2\times \mathbb R$ or $\mathbb{H}^2\times \mathbb R$, where for a product of spaces we mean that $P$ is a product of right-angled polytopes of finite volume in these spaces. For $\mathbb R^3$ such a polytope is combinatorially a cube, for $S^3$ it is a simplex, for $\mathbb{H}^3$ an almost Pogorelov polytope without adjacent quadrangles with all quadrangles shrunk to points, for $S^2\times \mathbb R$ the polytope is a product of a right-angled triangle and a line segment, and for $\mathbb{H}^2\times \mathbb R$ it is a product of a right-angled polygon of finite area and a line segment. The polygon may have vertices at infinity (ideal vertices), while all of its proper vertices have right angles between the corresponding edges. The polytope $P$ corresponds to a right-angled Coxeter group
$$
\begin{equation*}
\langle \rho_1,\dots,\rho_m\rangle/(\rho_1^2,\dots,\rho_m^2,\ \rho_i\rho_j=\rho_j\rho_i \text{ if }F_i\cap F_j\ne\varnothing),
\end{equation*}
\notag
$$
where we do not take points at infinity in the intersection of facets into account. For example, for the product of a right-angled polygon with a vertex at infinity and a line segment the products of sides containing the ideal vertex and the segment are facets with an empty intersection. The Coxeter group is isomorphic to the subgroup $G(P)$ of isometries of the ambient space generated by the reflections in planes containing facets of $P$, where $\rho_i$ corresponds to the reflection in the plane containing $F_i$. The group $G(P)$ acts discretely on the ambient space, and the polytope is a fundamental domain, that is the polytopes $\{gP\}_{g\in G(P)}$ fill the whole space and their interiors do not intersect. Moreover, $P$ is the orbit space, and for any point in $P$ its stabilizer is generated by reflections in facets containing this point (see [36], Ch. 5, Theorem 1.2). Given a mapping $\Lambda\colon\{F_1,\dots,F_m\}\to\mathbb Z_2^r\setminus\{0\}$ such that for any edge $F_i\cap F_j$ (perhaps with some vertices at infinity) and any proper vertex $F_i\cap F_j\cap F_k$ the images of the corresponding facets are linearly independent and $\operatorname{Im} \Lambda$ spans $\mathbb Z_2^r$, we define an epimorphism $\varphi_{\Lambda}\colon {G(P)\to \mathbb Z_2^r}$ by the rule $\varphi_{\Lambda}(\rho_i)=\Lambda(F_i)$. The subgroup $\operatorname{Ker} \varphi_{\Lambda}$ acts freely on the ambient space (apart from the points at infinity), and the quotient space $N(P,\Lambda)$ is a finite volume manifold glued of $2^r$ copies of the polytope. This manifold has automatically the geometric structure modelled on the ambient space. If $P$ is a compact polytope, then the mapping $\Lambda(F_i)=e_i\in\mathbb Z_2^m$ produces $N(P,\Lambda)=\mathbb R\mathcal{Z}_P$, and a vector colouring $\Lambda$ produces $N(P,\Lambda)=M(P,\Lambda)$. The inverse homeomorphism is given by the mapping $(p,t)\to \varphi^{-1}_{\Lambda}(t)\cdot p$. Recall that the notions of a nested family of belts, a nested family of curves, and cutting a polytope along a belt or a nested family of belts were introduced in Definitions 2.11 and 2.13, and Construction 2.21 together with Remark 2.22 respectively. Using these notions we formulate the main result of the article. Theorem 4.12. Let $P$ be a simple flag $3$-polytope different from $I^3$. Then the following hold. 1) There is a unique nested family of $4$-belts such that cutting $P$ along all these belts gives (a) almost Pogorelov polytopes without adjacent quadrangles, and (b) $k$-prisms, $k\geqslant 5$, where any two adjacent prisms ‘are twisted’, that is, if two prisms have quadrangles corresponding to the same belt in $P$, then their bases correspond to different facets of this belt. We call the belts in the above family canonical. We call a quadrangle of an almost Pogorelov polytope free if it does not correspond to a canonical belt. Each free quadrangle corresponds to a quadrangular facet of $P$, which we call a free quadrangle of $P$. 2) For any orientable manifold $M(P,\Lambda)$ the family of canonical belts gives the $\mathrm{JSJ}$-decomposition and the geometric decomposition in the following way. (a) There is a geometric realization of $P$ in $\mathbb R^3$ such that planar cross-sections of the canonical belts are disjoint quadrangles bounded by curves forming a nested family. (b) The deletion of these quadrangles and all free quadrangles from $P$ decomposes $P$ into a disjoint union of parts $Q_i$, $i=1,2,\dots$ . Each $Q_i$ is homeomorphic (preserving the face structure) to a right-angled polytope of finite volume in either $\mathbb {H}^2\times \mathbb R$ (if $Q_i$ corresponds to a prism) or $\mathbb{H}^3$ (if $Q_i$ corresponds to an almost Pogorelov polytope) and has an induced vector colouring $\Lambda_{Q_i}$. (c) The $\mathrm{JSJ}$-tori of $M(P,\Lambda)$ are the connected components of the preimages of
(d) The tori of types (i) and (ii) and the Klein bottles corresponding to tori of type (iii) give the geometric decomposition of $M(P,\Lambda)$. Each piece of the complement to these surfaces in $M(P,\Lambda)$ is homeomorphic to a manifold $N(Q_i,\Lambda_i)$ and has the geometric structure of finite volume given by Construction 4.11. Remark 4.13. There are two extreme cases when the set of canonical belts is empty. The first case is when $P$ is a $k$-prism, $k\geqslant 4$. Then the orientable manifold $M(P,\Lambda)$ is a closed Seifert fibred manifold, which has the geometric structure already. For $k=4$ it is modelled on $\mathbb R^3$, and for $k\geqslant 5$ on $\mathbb H^2\times\mathbb R$. The set of $\mathrm{JSJ}$-tori in part 2) is empty in this case. The second case is when $P$ is an almost Pogorelov polytope without adjacent quadrangles. If $P$ is a Pogorelov polytope, then the orientable manifold $M(P,\Lambda)$ is a closed hyperbolic manifold, and the set of $\mathrm{JSJ}$-tori in part 2) is also empty. If $P$ has quadrangles, then each quadrangle in part 2) corresponds to some $\mathrm{JSJ}$-tori, and $M(P,\Lambda)$ is divided into homeomorphic pieces with hyperbolic structure of finite volume. Remark 4.14. Part 1) of this theorem is a special case of a more general decomposition of so-called Coxeter orbifolds in [10]. Here one should take a right-angled Coxeter orbifold. Remark 4.15. Part 1) is equivalent to the fact that $P$ can uniquely be decomposed into a connected sum of $k$-prisms, $k\geqslant5$, and almost Pogorelov polytopes without adjacent quadrangles in such a way that adjacent prisms are twisted. Indeed, cutting along a $4$-belt is the inverse operation to taking a connected sum along quadrangles. Therefore, part 1) gives the described decomposition automatically. On the other hand, if we take a collection of $k$-prisms, $k\geqslant 5$, and almost Pogorelov polytopes without adjacent quadrangles and start taking connected sums of these polytopes along quadrangles following the twisted prisms rule, then the $4$-belt arising at each step cannot contain quadrangles. In particular, the arising $4$-belts form a nested family and each quadrangle of the new polytope corresponds to a quadrangle of a unique summand. This is clear in the case of two polytopes. Assume that we have proved this fact for each polytope obtained as a connected sum of fewer than $k$ polytopes from the collection. Consider the case of $k$ polytopes. The last connected sum involves two polytopes $Q_1$ and $Q_2$ each of which is either a polytope from the collection or a connected sum of fewer than $k$ such polytopes. By induction the quadrangles chosen on $Q_1$ and $Q_2$ correspond to uniquely defined polytopes $P_1$ and $P_2$ in the collection. If the $4$-belt arising in the last connected sum contains a quadrangle, then $P_1$ and $P_2$ are prisms. Since they should be twisted, we obtain a contradiction. We prove Theorem 4.12 in several steps. 4.3.1. The decomposition of a flag polytopeWe start with a lemma. Lemma 4.16. Let the simple $3$-polytope $P$ be a connected sum of polytopes $P_1$ and $P_2$ along $k$-gons surrounded by belts, $k\geqslant 4$. Then $P$ is flag if and only if $P_1$ and $P_2$ are flag. In the case of $4$-belts the situation is more complicated than for $3$-belts. It is possible that the middle lines of $4$-belts intersect transversally. For example, for any $k$-prism, $k\geqslant 4$, there are such $4$-belts (in particular, the belts around adjacent quadrangles). As we will see below, this is almost the general ‘bad’ case. An algorithm for decomposition. First we find algorithmically the decomposition in Remark 4.15 to part 1) of Theorem 4.12, and then we prove that it is unique. Step 1. We cut off all sequences of adjacent quadrangles. Proof. These three quadrangles are surrounded by a sequence of three pairwise adjacent facets. Since $P$ has no $3$-belts, they should have a common vertex. Hence $P=I^3$. The lemma is proved. Assume that $P\ne I^3$. Then each of its vertices belongs to at most two quadrangles. Let there be two adjacent quadrangles, $F_i$ and $F_j$. If a quadrangle $F_q$ is adjacent to one of them, then their edge of intersection should be opposite to $F_i\cap F_j$ in this facet. Thus, moving in each direction we obtain a sequence of quadrangles with ‘parallel’ edges of intersection. This sequence either becomes a ring, or terminates at some moment of time. In the first case $P$ is a $k$-prism for $k\geqslant 4$. In the second case it can be shown that the sequence of quadrangles is surrounded by a $4$-belt. Let us cut the polytope along this belt to obtain a $k$-prism, $k\geqslant 5$, and a flag polytope $P_1$. Apply the same procedure to $P_1$. Since the belts we cut along do not contain quadrangles, all of them correspond to a nested family of $4$-belts in $P$. In the end we obtain a polytope $Q$ without adjacent quadrangles or a $k$-prism, $k\geqslant 5$. Each step decreases the number of facets at least by one. If $Q$ has only trivial $4$-belts or $Q$ is a $k$-prism, we move to Step 3. Otherwise, we move to Step 2. Step 2. We cut a flag $3$-polytope without adjacent quadrangles along a nontrivial $4$-belt containing no quadrangles. Lemma 4.18. Let $Q$ be a flag $3$-polytope without adjacent quadrangles. Then either $Q$ has only trivial $4$-belts, or it has a nontrivial $4$-belt containing no quadrangles. Proof. Let $\mathcal{B}=(F_i,F_j,F_k,F_l)$ be a nontrivial $4$-belt. Then it contains at most two quadrangles. Let $F_i$ be one of them. In the closure of each component of the complement $\partial Q\setminus|\mathcal{B}|$ the quadrangle $F_i$ is adjacent to some facets $F_p$ and $F_q$ such that $(F_j,F_p,F_l,F_q)$ is a $4$-belt (each facet of a flag $3$-polytope is surrounded by a belt). If $F_p\cap F_k\ne\varnothing$, then $F_p\cap F_k\cap F_j$ and $F_p\cap F_k\cap F_l$ are vertices, and $\mathcal{B}$ is a belt around $F_p$, which is a contradiction. Similarly, $F_q\cap F_k=\varnothing$. Thus, $\mathcal{B}_1=(F_p,F_j,F_k,F_l)$ is a $4$-belt and $F_p$, $F_j$ and $F_l$ are not quadrangles. The belt $\mathcal{B}_1$ is nontrivial, for otherwise $\mathcal{B}$ surrounds two adjacent quadrangles. Similarly, $\mathcal{B}_2=(F_q,F_j,F_k,F_l)$ is a nontrivial $4$-belt. If $F_k$ is a quadrangle, then we can apply to it a similar argument to obtain nontrivial belts containing no quadrangles. The lemma is proved. Thus, if $Q$ is not an almost Pogorelov polytope nor a $k$-prism, then we can cut it along a $4$-belt containing no quadrangles to obtain two flag polytopes $Q_1$ and $Q_2$ with fewer facets. If one of the polytopes $Q_i$ is neither an almost Pogorelov polytope nor a $k$-prism, we move to Step $1$ for this polytope. If $P$ has already been cut into almost Pogorelov polytopes and $k$-prisms, we move to Step $3$. Step 3. Now we have cut a polytope $P$ along a nested set of $4$-belts into almost Pogorelov polytopes without adjacent quadrangles and $k$-prisms, $k\geqslant 5$ (the case $k=4$ is impossible, since it corresponds to cutting along a trivial $4$-belt). This corresponds to a representation of $P$ as a connected sum of the corresponding polytopes along quadrangles. Assume that two prisms in this decomposition are glued along quadrangles. If their bases are glued along edges, then we obtain a prism again. We delete the corresponding belt from the nested family. Repeating this argument, in the end we arrive at the case when any two adjacent prisms are ‘twisted’. Here the algorithm stops. Uniqueness of the decomposition. Now let us prove that the decomposition in part 1) of Theorem 4.12 is unique. Similarly to the case of $3$-belts we can represent the decomposition combinatorially as follows. For $4$-belts we take the curves $\widehat{\gamma}(\mathcal{B})$ from Construction 2.18. Then we cut $\partial P$ along these curves and glue up each curve by a quadrangle to obtain the boundaries of the polytopes $R_i$. Again, for a facet $F_a$ of $P$ we denote by $\widehat{F_a}$ the corresponding facet of $R_i$. The facet $\widehat{F_a}$ can be considered as a part of $F_a$, and $F_a$ is assembled from such parts. Lemma 4.19. Let $\mathcal{F}$ be a nested family of $4$-belts on a flag $3$-polytope $P\ne I^3$ such that cutting along all these belts gives polytopes $R_i$, which are either $k$-prisms, $k\geqslant 5$, or almost Pogorelov polytopes without adjacent quadrangles, and any two adjacent prisms ‘are twisted’. Then 1) any $4$-belt in $\mathcal{F}$ contains no quadrangles; 2) any $4$-belt of $P$ either belongs to $\mathcal{F}$ or is compatible with all belts in $\mathcal{F}$. In the latter case it consists of facets which are simultaneously present in the same polytope $R_i$, where they form a $4$-belt. Proof. If a belt $\mathcal{B}\in\mathcal{F}$ contains a quadrangle $F_i$, then $\widehat{\gamma}(\mathcal{B})\cap F_i=E$ is a line segment and any nonempty intersection $\widehat{\gamma}(\mathcal{B}')\cap F_i\ne\varnothing$ for $\mathcal{B}'\in\mathcal{F}\setminus\{\mathcal{B}\}$ is a line segment ‘parallel’ to $E$. Thus, if $F_i$ is present in some polytope $R_j$, then the corresponding facet $\widehat{F_i}$ is also a quadrangle. Let $\mathcal{B}$ separate the polytopes $R_p$ and $R_q$. Then in each of these polytopes $F_i$ corresponds to a quadrangle adjacent to the quadrangle arising from $\mathcal{B}$. Then both $R_p$ and $R_q$ are prisms. Since they are twisted, $\widehat{F}_i$ is a base for one of these polytopes. Hence it is not a quadrangle, which is a contradiction.
Let us prove that a $4$-belt on $P$ different from the belts in $\mathcal{F}$ cannot be present in different polytopes $R_i$ and $R_j$. Indeed, any two polytopes in the decomposition are separated on $P$ by some of the curves $\widehat{\gamma}(\mathcal{B})$, $\mathcal{B}\in\mathcal{F}$. If a facet is present in both polytopes, then it belongs to $\mathcal{B}$. Hence the belt coincides with $\mathcal{B}$. Now consider any $4$-belt $\mathcal{B}=(F_i,F_j,F_k,F_l)\notin\mathcal{F}$. If $\mathcal{B}$ is compatible with each belt in $\mathcal{F}$, then using Construction 2.18 and Lemma 2.19 we can move each vertex of the middle line $\gamma(\mathcal{B})$ in the interior of the corresponding edge of $P$ in such a way that the new curve $\widehat{\gamma}(\mathcal{B})$ does not intersect the curves $\widehat{\gamma}(\mathcal{B}')$ for $\mathcal{B}'\in\mathcal{F}$. In particular, it lies in the interior of a part of $\partial P$ corresponding to some $R_i$. Then all the facets of $\mathcal{B}$ are present in $R_i$. Lemma 2.23 implies that two facets $F_a$ and $F_b$ present in $R_i$ are adjacent in $P$ if and only if the facets $\widehat{F_a}$ and $\widehat{F_b}$ are adjacent in $R_i$. Hence the corresponding facets of $R_i$ also form a $4$-belt. Assume that $\mathcal{B}$ is not compatible with some belt $\mathcal{B}'\in\mathcal{F}$ connecting polytopes $R_p$ and $R_q$. Then Lemma 2.10 implies that after some relabelling of the facets, $F_j$ lies in the closure of the connected component of $\partial P\setminus|\mathcal{B}'|$ corresponding to $R_p$, and $F_l$ lies in the closure of the component corresponding to $R_q$. Then $F_i$ and $F_k$ should belong to $\mathcal{B}'$. We claim that in this case at least one of the facets $F_j$ and $F_l$ is disjoint from one of the facets $F_i$ and $F_k$. Indeed, consider the polytope $R_p$. If it is not a prism, then it is an almost Pogorelov polytope without adjacent quadrangles. Then there is exactly one quadrangle in $R_p$ adjacent to both $\widehat{F_i}$ and $\widehat{F_k}$, namely, the quadrangle $F$ corresponding to the belt $\mathcal{B}'$. If there is another quadrangle $F'$ with this property, then $(\widehat{F_i},F,\widehat{F_k},F')$ is a $4$-belt. Since $R_p$ has only trivial $4$-belts, this belt must surround a quadrangle adjacent to both $F$ and $F'$, which is a contradiction. Therefore, each of the parts of $\partial P$ separated from $R_p$ by curves $\widehat{\gamma}(\mathcal{B}'')$, where $\mathcal{B}''\in \mathcal{F}\setminus\{\mathcal{B}'\}$ are $4$-belts present in $R_p$, intersects at most one of the facets $F_i$ and $F_k$. Thus, if $F_j$ is not present in $R_p$, it cannot intersect both $F_i$ and $F_k$. If it is present in $R_p$, then $\widehat{F_j}$ is adjacent to both $\widehat{F_i}$ and $\widehat{F_k}$ in this polytope. The sequence $(F,\widehat{F_i},\widehat{F_j},\widehat{F_k})$ cannot be a $4$-belt, for otherwise this belt surrounds a quadrangle adjacent to $F$. Thus, $\widehat{F_j}\cap F\ne\varnothing$. Then $F_j$ belongs to $\mathcal{B}'$, which is a contradiction. So $F_j$ cannot intersect both $F_i$ and $F_k$ if $R_p$ is not a prism. Similarly, $F_l$ cannot intersect both $F_i$ and $F_k$ if $R_q$ is not a prism. If both $R_p$ and $R_q$ are prisms, then in one of them, say, $R_p$, $\widehat{F_i}$ and $\widehat{F_k}$ are quadrangles, and in the other, $R_q$, they are bases. In this case there is only one quadrangle adjacent to both $\widehat{F_i}$ and $\widehat{F_k}$ in $R_p$, namely, the quadrangle $F$ corresponding to $\mathcal{B}'$. Therefore, each of the parts of $\partial P$ separated from $R_p$ by curves $\widehat{\gamma}(\mathcal{B}'')$, where the $\mathcal{B}''\in \mathcal{F}\setminus\{\mathcal{B}'\}$ are $4$-belts present in $R_p$, intersects at most one of the facets $F_i$ and $F_k$. Hence, if $F_j$ is not present in $R_p$, it cannot intersect both $F_i$ and $F_k$. If it is present in $R_p$, then $\widehat{F_j}$ is adjacent to both $\widehat{F_i}$ and $\widehat{F_k}$ in this polytope. Then $\widehat{F_j}$ is a base and $F_j$ belongs to $\mathcal{B}'$, which is a contradiction. The lemma is proved. Now let $\mathcal{F}'$ be another nested family of $4$-belts such that cutting along all of these belts gives polytopes $R_i'$ which are either $k$-prisms, $k\geqslant 5$, or almost Pogorelov polytopes without adjacent quadrangles, and any two adjacent prisms are ‘twisted’. By Lemma 4.19 the union $\mathcal{F}'\cup \mathcal{F}$ is a nested family of belts again. Using Construction 2.18 we can choose a nested family of curves corresponding to $\mathcal{F}'\cup\mathcal{F}$ as an extension of such a family for $\mathcal{F}$. For each belt $\mathcal{B}\in \mathcal{F}'\setminus\mathcal{F}$ the curve $\widehat{\gamma}(\mathcal{B})$ lies on some polytope $R_i$ and corresponds to a $4$-belt $\widehat{\mathcal{B}}$ on it. Assume that $\widehat{\mathcal{B}}$ is a trivial belt. Then it cannot surround a quadrangle corresponding to belts in $\mathcal{F}$. Thus, it surrounds some quadrangle $\widehat{F_j}\in R_i$. This quadrangle is adjacent only to facets in $\widehat{\mathcal{B}}$, and all of them have the form $\widehat{F_k}$ for some facets $F_k$ of $P$. Hence $F_j$ does not intersect the curves $\widehat{\gamma}(\mathcal{B}')$, $\mathcal{B}'\in \mathcal{F}$. Therefore, $F_j$, as well as $\widehat{F_j}$, is a quadrangle. Then the component of $\partial P\setminus\gamma(\mathcal{B})$ containing $F_j$ cannot contain curves $\gamma(\mathcal{B}')$ for any $4$-belt $\mathcal{B}'$ of $P$. In particular, when we cut along belts in $\mathcal{F}'$, one of the polytopes $R_j'$ is a $4$-prism, which is a contradiction. Thus, $\widehat{\mathcal{B}}$ is a nontrivial belt. Hence the polytope $R_i$ is a prism and $\widehat{\mathcal{B}}$ contains two bases and two quadrangles. Consider all belts $\mathcal{B}\in\mathcal{F}'\setminus\mathcal{F}$ corresponding to nontrivial belts in $R_i$. If we cut $R_i$ along the corresponding curves $\widehat{\gamma}(\mathcal{B})$, we obtain a set of $k$-prisms with $k\geqslant 5$ such that any two adjacent prisms are not twisted. Taking such cuts for all prisms $R_i$ we obtain a set of prisms. If we add all almost Pogorelov polytopes without adjacent quadrangles $R_j$, we obtain the family of polytopes arising from $P$ when we cut it along all curves $\widehat{\gamma}(\mathcal{B})$, $\widehat{\gamma}(\mathcal{B})\in\mathcal{F}\cup\mathcal{F'}$. Now if we do the same, starting with $\mathcal{F}'$ and then considering $\mathcal{F}\setminus\mathcal{F}'$, then we obtain the same family of polytopes. Consider now two adjacent prisms corresponding to the above polytope $R_i$. When we cut $P$ in the second way, they arise from different prisms $R_p'$ and $R_q'$, and therefore they must be twisted. This is a contradiction. Thus, $\mathcal{F}=\mathcal{F}'$. 4.3.2. Incompressible surfaces corresponding to belts and facetsFor a subset $\omega\subset[m]$ the simplicial complex $K_{\omega}=\{\sigma\,{\in}\, K\colon \sigma\,{\subset}\,\omega\}$ is called a full (or induced) subcomplex. In $\mathbb{R}\mathcal{Z}_K$ the simplices of a full subcomplex correspond to the subset of the form $\mathbb R\mathcal{Z}_{K_{\omega}}\times \mathbb Z_2^{m-|\omega|}$. This is a disjoint union of $2^{m-|\omega|}$ copies of $\mathbb R\mathcal{Z}_{K_{\omega}}$. The following result was communicated to the author by Panov. Lemma 4.20 (see [1], Exercise 4.2.13, and [12], Proposition 2.2). For any subset $\omega\subset[m]$ and any $a\in\mathbb Z_2^{m-|\omega|}$ there is a retraction $\mathbb R\mathcal{Z}_K\to\mathbb R\mathcal{Z}_{K_{\omega}}\times a$. Proof. The retraction has the form This mapping is induced by the projection of the cube $(D^1)^m$ onto its face ${(D^1)^{|\omega|}\times a}$. It acts by the formula and is continuous by construction. It is also identical on $\mathbb R\mathcal{Z}_{K_{\omega}}\times a$. The lemma is proved. Corollary 4.21. For any subset $\omega\subset[m]$ and any $a\in\mathbb Z_2^{m-|\omega|}$ the mapping of fundamental groups $\pi_1(\mathbb R\mathcal{Z}_{K_\omega}\times a)\to \pi_1(\mathbb R\mathcal{Z}_K)$ is injective. Any $k$-belt $\mathcal{B}$ of a simple $3$-polytope $P$ corresponds to the set $\omega(\mathcal{B})=\{i\colon F_i\in \mathcal{B}\}$. The induced subcomplex $K_{\omega(\mathcal{B})}$ in $K=\partial P^*$ is a simple cycle, which is a boundary of a $k$-gon. This $k$-gon can be realized as a piecewise linear subset $P_k(\mathcal{B})$ of $P$ consisting of triangles with vertices at the barycentre of $P$, the barycentre of a facet of the belt, and the barycentre of its edge of intersection with an adjacent facet of the belt. Definition 4.22. For a vector colouring of rank $r$ of a simple $3$-polytope $P$ and a $k$-belt $\mathcal{B}$ set $\pi_{\mathcal{B}}=\langle \Lambda_j\colon F_j\in\mathcal{B}\rangle$ and $r(\mathcal{B})=\dim \pi_{\mathcal{B}}$. Then there is an induced vector colouring $\Lambda_{\mathcal{B}}$ of $P_k(\mathcal{B})$ of rank $r(\mathcal{B})$, $2\leqslant r(\mathcal{B})\leqslant r$. Proposition 4.23. Let $M(P,\Lambda)$ be a manifold defined by a vector colouring $\Lambda$ of rank $r$ of a simple $3$-polytope $P$ and $\mathcal{B}$ be a $k$-belt, $k\geqslant 4$. Then the copies of the $k$-gon $P_k(\mathcal{B})\subset P$ in $M(P,\Lambda)$ form a disjoint union of $2^{r-r(\mathcal{B})}$ incompressible $2$-submanifolds $M_x(\mathcal{B})$, $x\in \mathbb Z_2^r/\pi_\mathcal{B}$, with product neighbourhoods. Each submanifold $M_x(\mathcal{B})$ is homeomorphic to $M(P_k(\mathcal{B}),\Lambda_{\mathcal{B}})$, which is either $(T^2)^{\# g}$, $g=1+(k- 4)2^{r(\mathcal{B})-3}$, or $(\mathbb RP^2)^{\#l}$, $l=2+(k- 4)2^{r(\mathcal{B})-2}$. Remark. For small covers another approach to the incompressibility of these submanifolds, which is based on an explicit representation of fundamental groups, can be found in [13]–[15]. Proof of Proposition 4.23. Under the homeomorphism $\mathbb{R}\mathcal{Z}_{K_P}\simeq\mathbb R\mathcal{Z}_P$ from the proof of Proposition 1.4 the subspace $\mathbb{R}\mathcal{Z}_{K_{\omega(\mathcal{B})}}\times a\subset \mathbb{R}\mathcal{Z}_{K_P}$ is mapped to the $2$-submanifold $\mathbb{R}\mathcal{Z}_{P_k(\mathcal{B})}(a)$ in $\mathbb{R}\mathcal{Z}_P$ with a product neighbourhood. This submanifold is homeomorphic to $\mathbb{R}\mathcal{Z}_{P_k(\mathcal{B})}$ and is glued of parts of polytopes $P\times b$ with $b$ lying in the coset in $\mathbb Z_2^m/\mathbb Z_2^{\mathcal{B}}$ corresponding to $a\in\mathbb Z_2^{m-|\omega(\mathcal{B})|}$, where $\mathbb Z_2^{\mathcal{B}}=\langle e_i\colon F_i\in\mathcal{B}\rangle\subset \mathbb Z_2^m$.
The copies of $P_k(\mathcal{B})$ in $M(P,\Lambda)$ are glued in a disjoint union of $2$-submanifolds $M_x(\mathcal{B})$ homeomorphic to $M(P_k(\mathcal{B}),\Lambda_{\mathcal{B}})$. By construction each $2$-submanifold $M_x(\mathcal{B})$ has a product neighbourhood and is glued of $k$-gons $P_k(\mathcal{B})\times b$, where $b$ belong to the coset $x+\pi_{\mathcal{B}}$ in $\mathbb Z_2^r/\pi_{\mathcal{B}}$. In particular, there are $2^{r-r(\mathcal{B})}$ such submanifolds. The submanifold $M_x(\mathcal{B})$ is the orbit space of a free action of the subgroup
$$
\begin{equation*}
H(\mathcal{B})=H(\Lambda)\cap \mathbb Z_2^{\mathcal{B}}=\operatorname{Ker}[\Lambda_{\mathcal{B}}\colon \mathbb Z_2^{\mathcal{B}}\to\mathbb Z_2^r]
\end{equation*}
\notag
$$
on $\mathbb{R}\mathcal{Z}_{P_k(\mathcal{B})}(a)\subset\mathbb{R}\mathcal{Z}_P$ for any $a\in \widehat{\Lambda}^{-1}(x+\pi_{\mathcal{B}})$, where $\Lambda\colon\mathbb Z_2^m\to\mathbb Z_2^r$, and $\widehat{\Lambda}\colon \mathbb Z_2^m/\mathbb Z_2^{\mathcal{B}}\to \mathbb Z_2^r/\pi_{\mathcal{B}}$. In particular, there is a covering $\mathbb{R}\mathcal{Z}_{P_k(\mathcal{B})}(a)\to M_x(\mathcal{B})$ with $|H(\mathcal{B})|=2^{k-r(\mathcal{B})}$ sheets.
According to Example 1.2, each manifold $\mathbb R\mathcal{Z}_{P_k(\mathcal{B})}(a)$ is homeomorphic to a sphere with $g$ handles, $g=(k-4)2^{k-3}+1$. For an orientable manifold $M(P,\Lambda)$ each submanifold $M_x(\mathcal{B})$ is also orientable, since it has a product neighbourhood. Thus, it is also a sphere with handles. Let $g'$ be its genus. From the covering $\mathbb{R}\mathcal{Z}_{P_k(\mathcal{B})}(a)\to M_x(\mathcal{B})$ we see that $\chi(\mathbb R\mathcal{Z}_{P_k(\mathcal{B})})=|H(\mathcal{B})|\chi(M_x(\mathcal{B}))$. Therefore, $2-2g=|H(\mathcal{B})|(2-2g')$ and
$$
\begin{equation*}
g'=1+\frac{g-1}{|H(\mathcal{B})|}=1+(k-4) \frac{2^{k-3}}{2^{k-r(\mathcal{B})}}=1+(k-4)2^{r(\mathcal{B})-3}.
\end{equation*}
\notag
$$
If $M(P,\Lambda)$ is nonorientable, then $M_x(\mathcal{B})$ can be either orientable or not. In the latter case it has the form $(\mathbb RP^2)^{\#l}$, where $2-2g=|H(\mathcal{B})|(2-l)$, and
$$
\begin{equation*}
l=2+2\frac{g-1}{|H(\mathcal{B})|}=2+2(k-4) \frac{2^{k-3}}{2^{k-r(\mathcal{B})}}=2+(k-4)2^{r(\mathcal{B})-2}.
\end{equation*}
\notag
$$
Now we prove the incompressibility of the submanifolds $M_x(\mathcal{B})$. Corollary 4.21 implies that each submanifold $\mathbb{R}\mathcal{Z}_{P_k(\mathcal{B})}(a)\subset \mathbb R\mathcal{Z}_P$ is incompressible. If $\Lambda$ has rank $m$, then $M_x(\mathcal{B})=\mathbb{R}\mathcal{Z}_{P_k(\mathcal{B})}(x)$. Otherwise consider the commutative diagrams of mappings and the induced homomorphisms of groups: It is a classical fact that for a free action of a finite group $G$ on a Hausdorff topological space $X$ the projection $p\colon X\to X/G$ is a normal covering, $p_*\colon\pi_1(X)\to\pi_1(X/G)$ is injective, and $G\simeq \pi_1(X/G)/p_*\pi_1(X)$ (see [44], Proposition 1.40). This concerns the mappings $q_1$ and $q_2$. In particular, $(q_1)_*$ and $(q_2)_*$ are injective, their images are normal subgroups, and
$$
\begin{equation*}
\begin{gathered} \, \pi_1(M_x(\mathcal{B}))/\operatorname{Im}(q_1)_*\simeq H(\mathcal{B})\simeq \mathbb Z_2^{k-r(\mathcal{B})}, \\ \pi_1(M(P,\Lambda))/ \operatorname{Im}(q_2)_*\simeq H(\Lambda)\simeq \mathbb Z_2^{m-r}. \end{gathered}
\end{equation*}
\notag
$$
Furthermore, $i_*$ is injective by Corollary 4.21. Since $j_*(q_1)_*$ is injective, $\operatorname{Ker}j_*\cap \operatorname{Im}(q_1)_*=\{1\}$. Then there is an injection $\operatorname{Ker}j_*\to \pi_1(M_x(\mathcal{B}))/\operatorname{Im}(q_1)_*\simeq H(\mathcal{B})$. If $k\geqslant 4$, then $\mathbb R\mathcal{Z}_{P_k(\mathcal{B})}(a)$ is a sphere with at least one handle. In particular, ${\chi(\mathbb R\mathcal{Z}_{P_k(\mathcal{B})}(a))\leqslant 0}$. Since $q_1$ is a covering, $\chi(M_x(\mathcal{B}))\leqslant 0$. It is a classical fact that a $2$-dimensional closed manifold $X$ with $\chi(X)\leqslant 0$ is a $K(\pi,1)$-space, that is, $\pi_i(X)=0$ for $i\geqslant 2$. It is also known that any element of $\pi_1(K(\pi,1))$, where $K(\pi,1)$ is a finite cell complex, has an infinite order (see [44], Proposition 2.45). Since $\operatorname{Ker}j_*$ consists of elements of finite order, it must be equal to $\{1\}$. This finishes the proof of Proposition 4.23. Example 4.25. Consider a $4$-belt $\mathcal{B}=(F_i,F_j,F_k,F_l)$ of a simple $3$-polytope $P$. For the quadrangle $P_4(\mathcal{B})\simeq I\times I$ the real moment-angle manifold $\mathbb{R}\mathcal{Z}_{P_4}$ is a torus $T^2$ glued from 16 copies of $P_4$. In $\mathbb{R}\mathcal{Z}_P$ it corresponds to $2^{m-4}$ disjoint incompressible tori. For $M(P,\Lambda)$ defined by a vector colouring of rank $r$ the following possibilities exist. 1. $\operatorname{rk}\Lambda_{\mathcal{B}}=4$. Then $M_x(\mathcal{B})\simeq \mathbb{R}\mathcal{Z}_{P_4}\simeq T^2$. 2. $\operatorname{rk}\Lambda_{\mathcal{B}}=3$. Then either any three vectors form a basis or three subsequent vectors span a $2$-dimensional subspace, and the fourth does dot lie in this subspace. In the second case either two opposite vectors coincide, or not. Thus, up to a symmetry of the quadrangle we can assume that $\Lambda_i$, $\Lambda_j$, $\Lambda_k$ is a basis and $\Lambda_l\in \{\Lambda_i+\Lambda_j+\Lambda_k,\Lambda_j,\Lambda_j+\Lambda_k\}$. The first two cases give $M_x(\mathcal{B})=T^2$, and the last case gives $M_x(\mathcal{B})=K^2$. 3. $\operatorname{rk}\Lambda_{\mathcal{B}}=2$. Then $\Lambda_i$ and $\Lambda_j$ form a basis, and either $\{\Lambda_k,\Lambda_l\}=\{\Lambda_i,\Lambda_j\}$ or $\Lambda_i+\Lambda_j\in\{\Lambda_k,\Lambda_l\}$. In the first case $\Lambda_k=\Lambda_i$, $\Lambda_l=\Lambda_j$ and $M_x(\mathcal{B})=T^2$, and in the second case, up to a symmetry of the quadrangle, we can assume that $\Lambda_k=\Lambda_i$ and $\Lambda_l=\Lambda_i+\Lambda_j$. Then $M_x(\mathcal{B})=K^2$. We also need to consider $2$-submanifolds of $M(P,\Lambda)$ corresponding to facets of a simple $3$-polytope $P$ surrounded by belts. Definition 4.26. Let $F_i$ be a $k$-gonal facet of a simple $3$-polytope $P$ surrounded by a $k$-belt $\mathcal{B}$. Denote $\pi_{F_i}=\langle \Lambda_i\rangle+\pi_{\mathcal{B}}$, and $r(F_i)=\dim \pi_{F_i}$. Proposition 4.27. Let $M(P,\Lambda)$ be a manifold defined by a vector colouring $\Lambda$ of rank $r$ of a simple $3$-polytope $P$, and $F_i$ be a facet surrounded by a $k$-belt $\mathcal{B}$ with $k\geqslant 4$. Then the copies of the $k$-gon $F_i$ in $M(P,\Lambda)$ form a disjoint union of $2^{r-r(F_i)}$ incompressible $2$-submanifolds $M_x(F_i)$, $x\in \mathbb Z_2^r/\pi_{F_i}$. If $\Lambda_i\notin \pi_{\mathcal{B}}$, then each manifold has a product neighbourhood with two boundary components $M_x(\mathcal{B})$ and $M_{x+\Lambda_i}(\mathcal{B})$ isotopic to $M_x(F_i)$. If $\Lambda_i\in \pi_{\mathcal{B}}$, then each submanifold has a nontrivial tubular neighbourhood with boundary $M_x(\mathcal{B})$. This neighbourhood is homeomorphic to a mapping cylinder of a quotient map of a free action of an involution on $M_x(\mathcal{B})$ corresponding to the vector $\Lambda_i$. Remark 4.28. In the case of small covers over a $3$-polytope the incompressibility of facet submanifolds follows from Theorem 3.3 in [14]. Another approach to this result is presented in [8], [16] and [11]; it is based on the fact that $M_x(F_i)$ is a totally geodesic hypersurface in $M(P,\Lambda)$ with the structure of a cubical complex which is nonpositively curved in the sense of Alexandrov and Gromov [17]. Proof of Proposition 4.27. In $M(P,\Lambda)$ the copies of $F_i$ are glued in a disjoint union of $2^{r-r(F_i)}$ manifolds $M_x(F)$. Each manifold is glued of copies of $F_i$ corresponding to polytopes $P\times b$ with $b$ lying in the coset $x+\pi_{F_i}\in \mathbb Z_2^r/\pi_{F_i}$. There is a piecewise linear homeomorphism between the part of the polytope $P$ between $P_k(\mathcal{B})$ and $F_i$ and $F_i\times I$. This homeomorphism gives an isotopy between $P_k(\mathcal{B})$ and $F_i$.
There are two possibilities: either $\Lambda_i\notin \pi_{\mathcal{B}}$, or $\Lambda_i\in \pi_{\mathcal{B}}$. In the first case $M_x(F_i)$ has a product tubular neighbourhood with boundary consisting of two manifolds $M_x(\mathcal{B})$ and $M_{x+\Lambda_i}(\mathcal{B})$. In particular, all the three manifolds are isotopic and homeomorphic to $M(F_i,\Lambda_{\mathcal{B}})$, and $M_x(F_i)$ is incompressible by Proposition 4.23. In the second case $M_x(F_i)$ has a nontrivial tubular neighbourhood with boundary $M_x(\mathcal{B})$. The manifold $M_x(F_i)$ is homeomorphic to the quotient space of $M(F_i,\Lambda_{\mathcal{B}})$ by the action of the involution given by $\Lambda_i\in\pi_{\mathcal{B}}\simeq \mathbb Z_2^{r(\mathcal{B})}$. The action if free since $\Lambda_i\notin \langle\Lambda_p,\Lambda_q\rangle$ for any vertex $F_i\cap F_p\cap F_q$ of $F_i$. Then $M_x(F_i)$ is homeomorphic to the manifold $M(F_i,\Lambda_{F_i})$, where $\Lambda_{F_i}$ is the vector colouring given by the composition $\{F_j\colon F_j\in\mathcal{B}\}\to\pi_{\mathcal{B}}\to \pi_{\mathcal{B}}/\langle\Lambda_i\rangle$. The structure of $I$-bundle of the tubular neighbourhood of $M_x(F_i)$ gives a homeomorphism of this neighbourhood and the mapping cylinder of the quotient mapping $M(F_i,\Lambda_{\mathcal{B}})\to M(F_i,\Lambda_{F_i})$ described above. It corresponds to the mapping from the boundary $M_x(\mathcal{B})$ to the zero section $M_x(F_i)$ of the tubular neighbourhood. This mapping is homotopic to the identical mapping and is a $2$-sheeted covering. Thus, the inclusion $i\colon M_x(\mathcal{B})\to M(P,\Lambda)$ can be decomposed up to homotopy into the composition of the $2$-sheeted covering $c\colon M_x(\mathcal{B})\to M_x(F_i)$ and the inclusion $j\colon M_x(F_i)\to M(P,\Lambda)$. We have the composition of the homomorphisms
$$
\begin{equation*}
\pi_1(M_x(\mathcal{B}))\xrightarrow{c_*} \pi_1(M_x(F_i))\xrightarrow{j_*} \pi_1(M(P,\Lambda)).
\end{equation*}
\notag
$$
The first mapping is injective due to the covering space property, and the composition is injective by Proposition 4.23. Moreover, $\pi_1(M_x(F_i))/\operatorname{Im}c_*\simeq \mathbb Z_2$. Thus, as in the case of belts, we obtain $\operatorname{Ker} j_*\cap \operatorname{Im}c_*=\{1\}$. In particular, there is an embedding $\operatorname{Ker} j_*\to \pi_1(M_x(F_i))/\operatorname{Im}c_*\simeq \mathbb Z_2$. As discussed in the proof of Proposition 4.23, this implies that $\operatorname{Ker} j_*$ is trivial if $\chi(M_x(F_i))\leqslant 0$. The last condition is satisfied since $M_x(F_i)$ is covered by $M_x(\mathcal{B})$, which has a nonpositive Euler characteristic for $k\geqslant 4$. The proposition is proved. Example 4.29. If $F_i$ is a quadrangle, then taking Example 4.25 into account we have the following possibilities. 1. $\Lambda_i\notin \pi_{\mathcal{B}}$. Then $M_x(F_i)\simeq M_x(\mathcal{B})\simeq M_{x+\Lambda_i}(F_i)$ is either $T^2$ or $K_2$. If $M(P,\Lambda)$ is orientable, then only the case of $T^2$ is possible. 2. $\Lambda_i\notin \pi_{\mathcal{B}}$. Then $M_x(F_i)$ has a tubular neighbourhood with the boundary $M_x(\mathcal{B})$, and there is a $2$-sheeted covering $M_x(\mathcal{B})\to M_x(F_i)$. The following possibilities exist: $(M_x(\mathcal{B}), M_x(F_i))\in \{(T^2,T^2), (T^2,K^2), (K^2,K^2)\}$. If $M(P,\Lambda)$ is orientable, then $M_x(\mathcal{B})\simeq T^2$ since it has a product neighbourhood. The vector $\Lambda_i$ is a linear combination of an odd number of vectors from the base of the set $\{\Lambda_j\colon F_j\in \mathcal{B}\}$. Therefore, the corresponding involution on $M_x(\mathcal{B})$ changes the orientation, and $M_x(F_i)\simeq K^2$. 4.3.3. The $\mathrm{JSJ}$-decomposition and an explicit geometrizationTo pass to the $\mathrm{JSJ}$-decomposition of the orientable manifold $M(P,\Lambda)$, first we realize geometrically the combinatorial decomposition of item 1) of Theorem 4.12 using Corollary 2.34. Each planar cross-section $\widehat{P}_4(\mathcal{B})$ of $P$ corresponding to a belt $\mathcal{B}$ is isotopic to the piecewise linear polygon $P_4(\mathcal{B})$. Hence the corresponding submanifolds $\widehat{M}_x(\mathcal{B})$ and $M_x(\mathcal{B})$ are isotopic in $M(P,\Lambda)$. Thus, we obtain a disjoint family of incompressible tori $\widehat{M}_x(\mathcal{B})$ corresponding to $4$-belts $\mathcal{B}$ in the nested family, incompressible tori $M_x(F_i)$ corresponding to ‘free’ quadrangles $F_i$ of almost Pogorelov polytopes with $r(F_i)>r(\mathcal{B}(F_i))$, and incompressible tori $\widehat{M}_x(\mathcal{B})$ that are the boundaries of tubular neighbourhoods of the incompressible submanifolds $M_x(F_i)\simeq K^2$ corresponding to ‘free’ quadrangles $F_i$ of almost Pogorelov polytopes with $r(F_i)=r(\mathcal{B}(F_i))$. For the $4$-belt around a quadrangle the cross-section $\widehat{P}_4(\mathcal{B})$ can be obtained by a slight shift of the plane containing the facet. Now from the flag polytope $P\ne I^3$ we delete all cross-sections $\widehat{P}_4(\mathcal{B})$ corresponding to belts in the decomposition in part 1) of Theorem 4.12 and delete all the ‘free’ quadrangles of almost Pogorelov polytopes from the decomposition. Then each part $Q_i$ corresponding to an almost Pogorelov polytope $R_i$ without adjacent quadrangles is homeomorphic (preserving the face structure) to this polytope with all quadrangles deleted and to a right-angled polytope of finite volume in $\mathbb H^3$ with ideal points deleted. The realization of a right-angled polytope of finite volume in $\mathbb H^3$ is unique up to isometries of $\mathbb H^3$. Each part $Q_j$ corresponding to a $k$-prism $R_j$, $k\geqslant 5$, is homeomorphic (preserving the face structure) to this prism with some disjoint set of quadrangles deleted, and also to a direct product of a polygon with some set of vertices deleted and a line segment. The polygon can be realized as a right-angled polygon of finite area in $\mathbb H^2$ with deleted points lying at infinity. This realization is not unique. In the orientable manifold $M(P,\Lambda)$ this partition of the polytope corresponds to the partition of the manifold into pieces produced by the deletion of the incompressible tori corresponding to belts from part 1) and the incompressible tori and Klein bottles corresponding to ‘free’ quadrangles of almost Pogorelov polytopes. Each piece is glued from right-angled polytopes of finite volume in $\mathbb H^2\times \mathbb R$ and $\mathbb H^3$. The vector colouring $\Lambda$ of rank $r$ induces a vector colouring $\Lambda_{Q_i}$ of each piece $Q_i$, which we consider as a right-angled polytope of finite volume in $\mathbb H^3$ or $\mathbb H^2\times \mathbb R$. Definition 4.30. Denote by $\pi_{Q_i}\subset \mathbb Z_2^r$ the linear span of vectors $\Lambda_j$ corresponding to facets of $Q_i$, and let $r_i=\dim \pi_{Q_i}$. In the orientable manifold $M(P, \Lambda)$ each peace $M_x(Q_i)$ corresponding to $Q_i$ is homeomorphic to $N(Q_i,\Lambda_{Q_i})$ and is glued of polytopes $Q_i\times b$ with $b$ lying in the coset $x+\pi_{Q_i}\in\mathbb Z_2^r/\pi_{Q_i}$. There are $2^{r-r_i}$ such pieces. The closure of the open set $M_x(Q_i)$ in $M(P,\Lambda)$ contains a torus $\widehat{M}_y(\mathcal{B})$ from the above collection if and only if $\widehat{P}_4(\mathcal{B})$ lies in the boundary of $Q_i$ and $y+\pi_{\mathcal{B}}\subset x+\pi_{Q_i}$. It contains a torus or a Klein bottle $M_y(F_j)$ from the above collection if and only if $F_j$ is a free quadrangle of an almost Pogorelov polytope $R_i$ and $y+\pi_{\mathcal{B}}\subset x+\pi_{Q_i}$. This gives an explicit geometrization of the pieces of the manifold $M(P,\Lambda)$. Now our goal is to prove that this is actually the geometric decomposition from Theorem 0.2. For this we use Proposition 4.9. First, using Proposition 4.7 we prove that the collection of incompressible tori from Theorem 4.12 gives the $\mathrm{JSJ}$-decomposition, and then we prove that $M(P,\Lambda)$ is not a $\mathrm{Sol}$-manifold. Proposition 4.31. For an orientable manifold $M(P,\Lambda)$ over a flag $3$-polytope $P$ different from a $k$-prism, the closure of each piece $M_x(Q_i)$ corresponding to a $k$-prism, $k\geqslant 5$, is a Seifert fibred manifold with toric boundary, and it is not homeomorphic to $D^2\times S^1$, $T^2\times I$ or $K^2\widetilde{\times} I$. Moreover, for any two pieces $M_x(Q_i)$ and $M_y(Q_j)$ with common boundary torus $T$ we have $f(T,\overline{M_x(Q_i)})\ne f(T,\overline{M_y(Q_j)})$. Remark 4.32. If $P$ is a $k$-prism, $k\geqslant 4$, then the same argument as in the proof of Proposition 4.31 gives the structure of a closed Seifert fibred manifold on the orientable manifold $M(P,\Lambda)$. Proof of Proposition 4.31. First consider the case when $\Lambda=E$, and ${M(P,E)\!=\!\mathbb R\mathcal{Z}_P}$.
Consider the $k$-gon $L$ which is a base of the prism $S=L\times I$. It has $l$ edges corresponding to the canonical $4$-belts of $P$ (we call these edges ‘singular’). Copies of the base $L$ of $S$ in $\mathbb{R}\mathcal{Z}_P$ are glued into several copies of an oriented $2$-dimensional submanifold $N$, which is homeomorphic to a sphere $S^2_{g,s}$ with $g$ handles and $s$ holes. Denote by $L'$ the polygon obtained by the contraction of $l$ singular edges of $L$, and set $k'=k-l$. It is easy to see that $k'\geqslant 3$. Denote by $\widehat{N}\simeq S^2_g$ the corresponding manifold without holes. Then $\widehat{N}\simeq\mathbb{R}\mathcal{Z}_{L'}$. In particular, $g=1+(k'-4)2^{k'-3}$ according to Example 1.2. Each singular edge of $L$ corresponds to a disjoint set of $2^{k'-2}$ holes with boundary of each hole consisting of four edges. Thus, there are $s=l2^{k'-2}$ holes. Then the $k$-prism corresponds to a disjoint union of pieces homeomorphic to $S^2_{g,s}\times S^1$ in $\mathbb{R}\mathcal{Z}_P$. We have $l\geqslant 2$ for $k'=3$, and $l\geqslant 1$ for $k'\geqslant 4$. Therefore, $s\geqslant 4$. Lemma 4.33. The manifold $S^2_{g,s}\times S^1$ with $g\geqslant 0$ and $s\geqslant 3$ cannot cover a manifold homotopy equivalent to $D^2\times S^1$, $T^2\times I$ or $K^2\widetilde{\times} I$. Proof. Set $X=S^2_{g,s}\times S^1$. The manifold $S^2_{g,s}$ is homotopy equivalent to a wedge of $p=2g+s-1\geqslant 2$ circles. In particular, $\pi_1(X)=F_p\times \mathbb Z$, where $F_p$ is a free group on $p$ generators, and there is a subgroup in $\pi_1(X)$ isomorphic to $F_2$. If $\varphi\colon X\to Y$ is a covering, then $\varphi_*$ is an injection, and $\pi_1(Y)$ contains a subgroup $G$ isomorphic to $F_2$. Then $Y$ is not homotopy equivalent to $D^2\times S^1$ or $T^2\times I$, since their fundamental groups are abelian. If $Y$ is homotopy equivalent to $K^2\widetilde{\times} I$, then $\pi_1(Y)\simeq \pi_1(K^2)$ and we identify these groups. Consider the $2$-sheeted covering $\psi\colon T^2\to K^2$. It corresponds to the short exact sequence ${1\to\pi_1(T^2)\to \pi_1(K^2)\to \mathbb Z_2\to 1}$. Since $G$ is not abelian, it is not contained in $\psi_*(\pi_1(T^2))$, and there is a short exact sequence $1\to\operatorname{Im}\psi_*\cap G\to G\to \mathbb Z_2\to 1$, which shows that the subgroup $\operatorname{Im}\psi_*\cap G$ has index $2$ in $G$. Then it corresponds to a $2$-sheeted covering $\zeta\colon Z\to S^1\vee S^1$, where $Z$ is connected. In this case $Z$ is homotopy equivalent to a wedge of three circles, and $\operatorname{Im}\psi_*\cap G\simeq F_3$. But the group $\operatorname{Im}\psi_*\cap G$ is abelian. This is a contradiction, which finishes the proof of Lemma 4.33. Thus, each piece of $\mathbb R\mathcal{Z}_P$ corresponding to a $k$-prism, is not homeomorphic to $D^2\times S^1$, $T^2\times I$ or $K^2\widetilde{\times}I$ and has a unique Seifert fibred structure by Theorem 4.5. If two pieces $M_1$ and $M_2$ have a common boundary torus $T$, they correspond to different prisms. Since the prisms are twisted, the elements $f(T,M_1)$ and $f(T,M_2)$ in $PH_1(T^2)$ correspond to the classes of the vectors $(1,0)$ and $(0,1)$, and are different. Now consider the case of an arbitrary vector colouring $\Lambda$. Each manifold $\overline{M_x(Q_i)}$ corresponding to a $k$-prism is homeomorphic to the quotient space of $S^2_{g,s}\times S^1$ by a freely acting group isomorphic to $H(\Lambda)\cap \mathbb Z_2^{Q_i}$, where $\mathbb Z_2^{Q_i}$ is the linear span of the basis vectors $e_j\in\mathbb Z_2^m$ corresponding to facets of $Q_i$ (recall that $Q_i$ is obtained from the prism $L\times I$ by the deletion of the quadrangles corresponding to canonical belts). The action moves fibres to fibres and induces the following Seifert fibred structure on $\overline{M_x(Q_i)}$. Fibres are formed again by copies of $z\times I\subset L\times I$. Denote by $F_p$ and $F_q$ the facets of $P$ containing bases of the prism. Then the copies of $z\times I\subset L\times I$ in $\overline{M_x(Q_i)}$ form circles $C(z)$. 1. If the point $z$ lies in the interior of $L$ or the interior of a singular edge, then there is a neighbourhood $U(z)$ such that the copies of $U(z)\times I$ are glued into a direct product $U(z)\times S^1$, where $S^1$ is glued of $2^{\dim \langle\Lambda_p,\Lambda_q\rangle}$ copies of $I$. Hence the fibre $C(z)$ is regular. 2. If $z$ lies in the interior of a regular edge of $L$ or it is a common vertex of a regular edge and a singular edge, then the regular edge corresponds to some facet $F_a$ of $P$ with $\Lambda_a\notin\{\Lambda_p,\Lambda_q\}$. Since $M(P,\Lambda)$ is orientable, either $\Lambda_p=\Lambda_q$ or $\Lambda_p\ne\Lambda_q$ and $\Lambda_a\notin\langle \Lambda_p,\Lambda_q\rangle$. In both cases there is a neighbourhood $U(z)$ such that the copies of $U(z)\times I$ are glued into a direct product $U(z)\times S^1$, where $S^1$ is glued of $2^{\dim \langle\Lambda_p,\Lambda_q\rangle}$ copies of $I$. Hence the fibre $C(z)$ is regular. 3. If $z$ is a common vertex of two regular edges of $L$, then these edges correspond to facets $F_a$ and $F_b$ of $P$, with $\Lambda_p,\Lambda_q\notin \langle \Lambda_a,\Lambda_b\rangle$. If $\Lambda_p=\Lambda_q$, then again $C(z)$ has a trivial tubular neighbourhood. If $\Lambda_p\ne\Lambda_q$, then $\Lambda_q=\Lambda_a+\Lambda_b+\Lambda_p$, since $M(P,\Lambda)$ is orientable. In this case $C(z)$ has a tubular neighbourhood that forms a standard fibred torus corresponding to the rotation by $\pi$. Lemma 4.33 implies that $\overline{M_x(Q_i)}$ is not homeomorphic to $D^2\times S^1$, $T^2\times I$ or $K^2\widetilde{\times} I$. Therefore, it has a unique Seifert fibred structure by Theorem 4.5. If two pieces $M_x(Q_i)$ and $M_y(Q_j)$ have a common boundary torus $T$, they correspond to different prisms. Since the prisms are twisted, Example 4.25 shows that the elements $f(T,M_1)$ and $f(T,M_2)$ in $PH_1(T^2)$ are different. This finishes the proof of Proposition 4.31. Now we are ready to finish the proof of Theorem 4.12. By Theorem 4.8 pieces corresponding to almost Pogorelov polytopes without adjacent quadrangles do not have the structure of a Seifert fibred manifold, and all the Seifert fibred pieces of $M(P,\Lambda)$ correspond to $k$-prisms and to Klein bottles arising from some free quadrangles of almost Pogorelov polytopes. The latter pieces have a common boundary only with hyperbolic pieces. Thus, Propositions 4.31 and Proposition 4.7 imply that we have indeed a $\mathrm{JSJ}$-decomposition. To use Proposition 4.9 we need to prove that an orientable $M(P,\Lambda)$ cannot be a $\mathrm{Sol}$-manifold. As mentioned in this proposition, a $\mathrm{Sol}$-manifold has one $\mathrm{JSJ}$-torus, moreover, either both components of the complement of this torus are homeomorphic to the interior of $K^2\widetilde{\times}I$, or the complement to the torus is homeomorphic to the interior of $T^2\times I$. For $M(P,\Lambda)$ the hyperbolic pieces are not homeomorphic to the interiors of $K^2\widetilde{\times}I$ or $T^2\times I$, since they do not admit a Seifert fibred structure. We have also proved that the pieces corresponding to $k$-prisms are not homotopy equivalent to $K^2\widetilde{\times}I$ or $T^2\times I$. Only a piece corresponding to a Klein bottle arising from a free quadrangle can be homeomorphic to the interior of $K^2\widetilde{\times}I$, but not to $T^2\times I$. Then the adjacent piece with the same boundary component is hyperbolic, which is a contradiction. This finishes the proof of Theorem 4.12. Corollary 4.34. For orientable manifolds $M(P,\Lambda)$ defined by vector colourings of simple $3$-polytopes the following five of the eight Thurston geometries arise: $\bullet$ $S^3$ for the simplex $\Delta^3$; $\bullet$ $S^2\times \mathbb R$ for the $3$-prism $\Delta^2\times I$; $\bullet$ $\mathbb R^3$ for the cube $I^3$; $\bullet$ $\mathbb{H}^2\times\mathbb R$ for the $k$-prisms, $k\geqslant 5$, and pieces corresponding to them; $\bullet$ $\mathbb H^3$ for Pogorelov polytopes and pieces corresponding to almost Pogorelov polytopes. AcknowledgementsThe author is grateful to Victor Buchstaber for his encouraging support and attention to this work, to Taras Panov for his valuable advice, and to Vladimir Shastin and Dmitry Gugnin for fruitful discussions.
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