Example of C-rigid polytopes which are not B-rigid

TL;DR

This paper investigates the relationship between B-rigidity and C-rigidity in simple polytopes, demonstrating that B-rigidity does not necessarily imply C-rigidity, thus clarifying their distinction.

Contribution

It provides the first known example of C-rigid polytopes that are not B-rigid, showing these concepts are not equivalent.

Findings

B-rigid simple polytopes are not always C-rigid

C-rigidity is characterized by cohomology rings of quasitoric manifolds

B-rigidity is characterized by Tor-algebra of the polytope

Abstract

A simple polytope $P$ is said to be \emph{B-rigid} if its combinatorial structure is characterized by its Tor-algebra, and is said to be \emph{C-rigid} if its combinatorial structure is characterized by the cohomology ring of a quasitoric manifold over $P$. It is known that a B-rigid simple polytope is C-rigid. In this paper, we, further, show that the B-rigidity is not equivalent to the C-rigidity.

Equations

$$(t_1,\dots ,t_n)\cdot (x_0,x_1,\dots ,x_n)=(x_0,t_1{x}_1,\dots ,t_n{x}_n)$$

$$0\to R^{-m}\stackrel{d}{\to }\cdots \stackrel{d}{\to }{R}^{-1}\stackrel{d}{\to }\mathcal{A}\stackrel{d}{\to }\mathbb{Q}\to 0.$$

$$\mathrm{Tor}(P):=H(R{\otimes }_{\mathcal{A}}\mathbb{Q}(P))=H(\Lambda [u_1,\dots ,u_m]\otimes \mathbb{Q}(P)).$$

$$I(\subset [n+3])\text{mbox}isasimplexofK⟺O\in \mathrm{conv}\{\varphi (F_i)\mid i\in [n+3]\setminus I\}.$$

$${\beta }^{-1,2j}(P)=|\{i\in [2k+1]\mid |e_i|=j\}=|\{i\in [2k+1]\mid a_i+a_{i+1}+\cdots +a_{i+k-1}=j\}|,$$

$${\mathcal{Z}}_P\simeq {\#}_{i=1}^{2k-1}{S}^{2{\varphi }_i-1}\times S^{2{\phi }_{i+k-1}-2},$$

$$\{a+b,b+c,c+d,d+e,e+a\}=\{a^{\prime }+b^{\prime },b^{\prime }+c^{\prime },c^{\prime }+d^{\prime },d^{\prime }+e^{\prime },e^{\prime }+a^{\prime }\}.$$

$$a^{\prime }+b^{\prime }=A,b^{\prime }+c^{\prime }=B,c^{\prime }+d^{\prime }=C,d^{\prime }+e^{\prime }=D,e^{\prime }+a^{\prime }=E,$$

$$\mathbb{Z}[v_1,\dots ,v_m]/I_P+J,$$

$$H^{\ast }(M;{\mathbb{Z}}_2)={\mathbb{Z}}_2[v_{1_1},v_{1_2},v_{1_3},v_{3_1},v_{3_2},v_5,v_2,v_4]/I_{\mathcal{P}}+J\cong {\mathbb{Z}}_2[x,y,z]/I_A$$

$$\begin{array}[]{|l|l|}\hline\cr A_{1}&(x^{3}y+yz^{3},y^{3}+yz^{2},y^{2}z+z^{3},xz,x^{4})\\ \hline\cr A_{2},A_{3},A_{5}&(x^{3}y+x^{2}y^{2}+y^{2}z^{2},y^{3}+yz^{2},y^{2}z+z^{3},xz,x^{4}+x^{3}y)\\ \hline\cr A_{4},A_{6},A_{7}&(x^{3}y+y^{3}z,y^{3}+yz^{2},y^{2}z+z^{3},xz,x^{4}+x^{2}y^{2})\\ \hline\cr A_{8}&(x^{3}y+x^{2}y^{2}+y^{4},y^{3}+yz^{2},y^{2}z+z^{3},xz,x^{4}+y^{4})\\ \hline\cr A_{9},A_{11}&(x^{3}y+yz^{3},xy^{2}+y^{3}+yz^{2},y^{2}z+z^{3},xz,x^{4})\\ \hline\cr A_{10},A_{12}&(x^{3}y+xy^{3}+y^{4},xy^{2}+y^{z}+yz^{2},yz^{2}+z^{3},xz,x^{4}+y^{4})\\ \hline\cr A_{13}&(x^{3}y+yz^{3},x^{2}y+y^{3}+yz^{2},y^{2}z+z^{3},xz,x^{4})\\ \hline\cr A_{14}&(x^{2}y^{2}+xy^{3}+y^{4},x^{2}y+yz^{2},z^{3},xz,x^{4}+y^{4})\\ \hline\cr A_{15},A_{16},A_{18}&(x^{4}+y^{4}+y^{3}z,x^{2}y+yz^{2},z^{3},xz,x^{4}+x^{2}y^{2}+xy^{3})\\ \hline\cr A_{17},A_{19},A_{20}&(x^{4}+y^{4}+y^{2}z^{2},x^{2}y+yz^{2},z^{3},xz,x^{4}+x^{2}y^{2}+xy^{3})\\ \hline\cr A_{21}&(x^{4}+y^{4}+y^{3}z+y^{2}z^{2},x^{2}y+yz^{2},z^{3},xz,x^{4}+x^{2}y^{2}+xy^{3})\\ \hline\cr\end{array}$$

$$H^{\ast }(N;{\mathbb{Z}}_2)={\mathbb{Z}}_2[v_{1_1},v_{1_2},v_{2_1},v_{3_1},v_{3_2},v_5,v_{2_2},v_4]/I_{\mathcal{P}}+J\cong {\mathbb{Z}}_2[x,y,z]/I_B$$

$$\begin{array}[]{|l|l|}\hline\cr B_{1}&(x^{2}y^{2}+y^{2}z^{2},y^{4}+y^{2}z^{2},y^{2}z+z^{3},xz,x^{3})\\ \hline\cr B_{2},B_{3}&(x^{2}y^{2}+xy^{3}+y^{3}z,y^{4}+y^{2}z^{2},y^{2}z+z^{3},xz,x^{3}+x^{2}y)\\ \hline\cr B_{4}&(x^{2}y^{2}+y^{4},y^{4}+y^{2}z^{2},y^{2}z+z^{3},xz,x^{3}+xy^{2})\\ \hline\cr B_{5},B_{7},B_{18}&(x^{2}y^{2}+y^{2}z^{2},xy^{3}+y^{4}+y^{2}z^{2},y^{2}z+z^{3},xz,x^{3})\\ \hline\cr B_{6},B_{8}&(x^{2}y^{2}+y^{4},xy^{3}+y^{3}z+y^{2}z^{2},yz^{2}+z^{3},xz,x^{3}+xy^{2})\\ \hline\cr B_{9},B_{19},B_{20}&(x^{2}y^{2}+y^{2}z^{2},y^{4},y^{2}z+z^{3},xz,x^{3})\\ \hline\cr B_{10}&(x^{2}y^{2}+y^{4},x^{2}y^{2}+y^{2}z^{2},z^{3},xz,x^{3}+xy^{2})\\ \hline\cr B_{11},B_{12},B_{14}&(x^{2}y^{2}+y^{4}+y^{3}z,x^{2}y^{2}+y^{2}z^{2},z^{3},xz,x^{3}+xy^{2})\\ \hline\cr B_{13},B_{15},B_{16}&(y^{4},x^{2}y^{2}+y^{2}z^{2},z^{3},xz,x^{3}+xy^{2})\\ \hline\cr B_{17}&(y^{4}+y^{3}z,x^{2}y^{2}+y^{2}z^{2},z^{3},xz,x^{3}+xy^{2})\\ \hline\cr B_{21}&(x^{2}y^{2}+y^{2}z^{2},xy^{3}+y^{4},y^{2}z+z^{3},xz,x^{3})\\ \hline\cr\end{array}$$

Full text

The first named author was supported by Basic Science Research Program through the National Research Foundation of Korea(NRF) funded by the Ministry of Science, ICT & Future Planning(NRF-2016R1D1A1A09917654).

1. Introduction

Let $P$ be a simple polytope of dimension $n$. A closed, smooth manifold of dimension $2n$ is called a quasitoric manifold over $P$ if it admits a locally standard action of the $n$-dimensional torus $T^{n}$ whose orbit space can be identified with $P$ (see [8] for more details). One typical example of quasitoric manifolds is a complex projective space $\mathbb{C}P^{n}=\mathbb{C}^{n+1}\setminus\{O\}/\sim$ where $(x_{0},x_{1},\ldots,x_{n})\sim(rx_{0},rx_{1},\ldots,rx_{n})$ for all non-zero real numbers $r$. One can see that a $T^{n}$-action on $\mathbb{C}^{n+1}\setminus\{O\}$ defined by

[TABLE]

induces a locally standard $T^{n}$-action on $\mathbb{C}P^{n}$, and its orbit space is an $n$-dimensional simplex $\Delta^{n}$. Hence, $\mathbb{C}P^{n}$ is a quasitoric manifold over $\Delta^{n}$. Here is one naive question: can $\mathbb{C}P^{n}$ be a quasitoric manifold over a different simple polytope $Q$ other than $\Delta^{n}$? The answer is “no”. It is well-known that for a quasitoric manifold $M$ over $P$, the Betti numbers of $M$ coincide with the $h$-numbers of $P$. Since $\Delta^{n}$ is the only simple polytope whose $h$-vector is $(1,1,\ldots,1)$ which is the sequence of the Betti numbers of $\mathbb{C}P^{n}$, only $\Delta^{n}$ can be an orbit space of $\mathbb{C}P^{n}$.

This phenomenon raises one fundamental question which asks how much combinatorial information of $P$ is decided by the topology of a quasitoric manifold $M$ over $P$. Masuda and Suh [14] dealt with it as the property of simple polytope. Throughout this paper, $H^{\ast}(X)$ denotes the integral cohomology ring of a topological space $X$.

Definition 1.1**.**

A simple polytope $P$ is said to be (toric) cohomologically rigid (or C-rigid) if there is no simple polytope $Q\not\approx P$ such that $H^{\ast}(M)\cong H^{\ast}(N)$ for some quasitoric manifolds $M$, $N$ over $P$, $Q$, respectively.

We note that there are many simple polytopes which do not support any quasitoric manifold. Since the “C-rigidity” requires the existence of quasitoric manifold over a given polytope, there is no canonical way to define the C-rigidity of such polytopes. Thus there has been confusion; some literatures such as [7] define that such polytope is not C-rigid, but some literatures such as [3] define that such polytopes are C-rigid as their conventions. However, it does not matter because the C-rigidity of $P$ should be considered only when $P$ supports a quasitoric manifold.

One important step on the theory of rigidity of simple polytopes is to find some combinatorial invariants of $P$ determined by the cohomology ring of a quasitoric manifold over $P$. Let $P$ be an $n$-dimensional simple polytope with $m$ facets and $\mathcal{A}=\mathbb{Q}[v_{1},\ldots,v_{m}]$ the polynomial ring over $\mathbb{Q}$ with $\deg v_{i}=2$ for all $i$. We consider $\mathbb{Q}$ as an $\mathcal{A}$-module via the map $\mathcal{A}\rightarrow\mathbb{Q}$ sending each $v_{i}$ to [math]. The Stanley-Reisner ring of $P$ is the quotient ring $\mathbb{Q}(P)=\mathbb{Q}[v_{1},\ldots,v_{m}]/I_{P},$ where $I_{P}$, the Stanley-Reisner ideal of $P$, is the homogeneous ideal generated by all square-free monomials $v_{i_{1}}\cdots v_{i_{r}}$ such that $F_{i_{1}}\cap\cdots\cap F_{i_{r}}=\emptyset$. We also regard $\mathbb{Q}(P)$ as an $\mathcal{A}$-module.

Let $\Lambda[u_{1},\ldots,u_{m}]$ be an exterior algebra. Then, we have a differential bigraded algebra $R=\Lambda[u_{1},\ldots,u_{m}]\otimes\mathcal{A}$ with map $d\colon R\rightarrow R$, where $\operatorname{bideg}u_{i}=(-1,2)$, $\operatorname{bideg}v_{i}=(0,2)$, $du_{i}=v_{i}$, $dv_{i}=0$ and $\otimes$ is a tensor product over $\mathbb{Q}$. Then, $R$ is a free $\mathcal{A}$-module. Let $R^{-i}=\Lambda^{i}[u_{1},\ldots,u_{m}]\otimes\mathcal{A}$, where $\Lambda^{i}[u_{1},\ldots,u_{m}]$ is the submodule of $\Lambda[u_{1},\ldots,u_{m}]$ spanned by monomials of length $i$. Then, we have the free resolution of $\mathbb{Q}$, known as the Koszul resolution, as follows:

[TABLE]

By taking $\otimes_{\mathcal{A}}\mathbb{Q}(P)$ to the Koszul resolution, we obtain the Tor-module of $P$

[TABLE]

Furthermore, it has the natural algebra structure induced from the Koszul resolution, so it is called the Tor-algebra $\operatorname{Tor}(P)$ of $P$. Since $\operatorname{Tor}(P)$ has the bigraded structure, the bigraded Betti number $\beta^{-i,2j}(P)$ is also defined for $P$.

It is shown in [7, Lemma 3.7] that, for two quasitoric manifolds $M$ and $N$ over $P$ and $Q$, respectively, if $H^{\ast}(M)\cong H^{\ast}(N)$ as graded rings, then $\operatorname{Tor}(P)\cong\operatorname{Tor}(Q)$ as bigraded rings. This fact stimulates to consider the hierarchy of rigidities of simple polytopes (see [6] and [2]).

• •

$P$ is combinatorially rigid (or A-rigid) if there is no $n$-dimensional simple polytope $Q\not\approx P$ such that $\beta^{-i,2j}(P)=\beta^{-i,2j}(Q)$ for all $i,j$.

• •

$P$ is B-rigid if there is no $n$-dimensional simple polytope $Q\not\approx P$ such that $\operatorname{Tor}(P)\cong\operatorname{Tor}(Q)$ as bigraded rings.

In addition, we have the following implications.

1. (1)

If $P$ is A-rigid, then $P$ is B-rigid. 2. (2)

If $P$ supports a quasitoric manifold and $P$ is B-rigid, then $P$ is C-rigid.

It is natural to ask whether the converse of the above statements hold or not. In [5], the first named author found the counterexample of the reverse implication of (1); there is a $3$-dimensional B-rigid simple polytope with $11$ facets which is not A-rigid. However, the reverse implication of (2) has been open (see the remark in Section 3 of [3]).

In this paper, we shall provide a counterexample of the reverse implication of (2), that is, there is a C-rigid simple polytope which is not B-rigid while it supports a quasitoric manifold. More precisely, in Section 3, we provide two distinct simple polytopes $\mathcal{P}$ and $\mathcal{Q}$ of dimension $5$ having $8$ facets satisfying the following:

1. (a)

both $\mathcal{P}$ and $\mathcal{Q}$ support quasitoric manifolds, 2. (b)

$\operatorname{Tor}(\mathcal{P})\cong\operatorname{Tor}(\mathcal{Q})$ as bigraded rings, 3. (c)

there is no other polytope whose Tor-algebra is isomorphic that of $\mathcal{P}$, and 4. (d)

no two quasitoric manifolds $M$ over $\mathcal{P}$ and $N$ over $\mathcal{Q}$ have the isomorphic cohomology rings, that is, $H^{\ast}(M)\not\cong H^{\ast}(N)$ as graded rings.

It proves that the A-, B-, and C-rigidities are not equivalent to each other under the condition that $P$ supports a quasitoric manifold.

This paper is mainly based on a part of the Ph.D. thesis of the second named author [15] supervised by the first named author.

2. Simple polytopes with a few facets

In this section, we recall some useful facts on $n$-dimensional simple polytopes with $n+3$ facets. The structure of $n$-simple polytope $P$ with $n+3$ facets is well-known (see, for example, [12]): $P$ can be obtained by a sequence of wedge operations from either a cube or the dual of cyclic polytope $C^{\ast}_{2k-2,2k+1}$ for some $k\geq 2$. If $P$ is obtained from a cube by a sequence of wedge operations, then $P$ is the product of three simplices. It is shown in [7, Theorem 5.3] that every product of simplices is A-rigid, and so is $P$. Hence, in the remain of the section, we only consider the case where $P$ is obtained from a cyclic polytope. It is convenient to represent $P$ by the Gale diagram in $\mathbb{R}^{2}$.

For $k\geq 2$, let $P_{2k+1}$ be a regular $(2k+1)$-gon in $\mathbb{R}^{2}$ with center at the origin $O$ with the vertex set $[2k+1]=\{1,2,\ldots,2k+1\}$ in counterclockwise order. For a given surjective map $\phi\colon\mathfrak{F}=\{F_{1},\ldots,F_{n+3}\}\rightarrow[2k+1]$, we construct a simplicial complex $K$ on $[n+3]$ by

[TABLE]

It is known that $K$ is a boundary complex of some simple $n$-polytope $P$ with the facet set $\mathfrak{F}$. One observes that the combinatorial structure of $P$ only depends on $|\phi^{-1}(1)|,\ldots,|\phi^{-1}(2k+1)|$ up to rotating and reflecting of $P_{2k+1}$. Hence, an $n$-dimensional simple polytope $P$ with $n+3$ facets, which is not a product of three simplices, is representable on $P_{2k+1}$ with the assigned numbers $[a_{1},\ldots,a_{2k+1}]$ where $a_{1}+\cdots+a_{2k+1}=n+3$ up to rotation and reflection.

Now, let us compute the bigraded Betti numbers of $P$ represented on $P_{2k+1}$ with assigned numbers $[a_{1},\ldots,a_{2k+1}]$. We firstly remark that every simple polytope has the following duality on its bigraded Betti numbers.

Proposition 2.1**.**

Let $Q$ be an $n$-dimensional simple polytope with $m$ facets. Then

1. (1)

$\beta^{-i,2j}(Q)=0$* if $i<0$ or $i>m-n$,* 2. (2)

$\beta^{0,0}(Q)=\beta^{-(m-n),2m}(Q)=1$, and 3. (3)

$\beta^{-i,2j}(Q)=\beta^{-(m-n)+i,2(m-j)}(Q)$.

By the above proposition, we have $\beta^{-1,2j}(P)=\beta^{-2,2(n+3-j)}(P)$, and, hence, $\{\beta^{-1,2j}(P)\mid j\geq 2\}$ completely determines all bigraded Betti numbers of $P$. It should be noted that $\beta^{-1,2j}$ is equal to the number of degree $2j$ monomial elements in a minimal basis of the Stanley-Reisner ideal $I_{P}$. Thus it is enough to find the minimal monomial basis of $I_{P}$. By definitions of $I_{P}$ and a Gale-diagram, we have the following proposition.

Proposition 2.2**.**

Let $P$ be an $n$-dimensional simple polytope with $n+3$ facets $F_{1},\ldots,F_{n+3}$. Let $v_{1},\ldots,v_{n+3}$ be indeterminates corresponding to the facets of $P$ and $\phi$ the corresponding map of $P$. Then, the followings are equivalent:

1. (1)

$v_{i_{1}}\cdots v_{i_{\ell}}$* is a monomial generator of $I_{P}$,* 2. (2)

$F_{i_{1}}\cap\cdots\cap F_{i_{\ell}}=\emptyset$, 3. (3)

$O\notin\operatorname{conv}\{\phi(F_{j})\mid j\in[n+3]\setminus\{i_{1},\ldots,i_{\ell}\}\}$**

Moreover the monomial $v_{i_{1}}\cdots v_{i_{\ell}}$ is an element of the minimal monomial basis if and only if $v_{i_{1}}\cdots v_{i_{\ell}}\in I_{P}$ and $F_{J}=\bigcap_{j\in J}F_{j}\neq\emptyset$ for any $J\subsetneq\{i_{1},\ldots,i_{\ell}\}$. Now consider the collection $\mathcal{S}$ of subsets of $[n+3]$ which satisfies (3) in Proposition 2.2, then $\mathcal{S}$ has a natural poset structure by an inclusion. Note that $\mathcal{S}$ is nonempty and finite. So we can choose the set $\mathcal{T}$ of the minimal elements of $\mathcal{S}$. Then $\mathcal{T}$ supports the minimal monomial basis of $I_{P}$. If $P$ is represented on $P_{2k+1}$ with assigned numbers $[a_{1},a_{2},\ldots,a_{2k+1}]$, then we can assign each $i\in[2k+1]$ by each element $e_{i}=\{i_{1},\ldots,i_{\ell}\}\subset[n+3]$ of $\mathcal{T}$ such that $\phi(F_{j})\in\{i,i+1,\ldots,i+k-1\}$ modulo $2k+1$ for all $j\in e_{i}$ and $\phi(F_{j^{\prime}})\not\in\{i,\ldots,i+k-1\}$ modulo $2k+1$ for all $j^{\prime}\not\in e_{i}$. We, therefore, have

[TABLE]

where $a_{2k+2}=a_{1},a_{2k+3}=a_{2},\ldots,a_{3k}=a_{k-1}$.

On the other hand, Erokhovets [9] showed that the Tor-algebra $\operatorname{Tor}(P)$ is completely determined by the bigraded Betti numbers of $P$. Indeed, $\operatorname{Tor}(P)$ is isomorphic to the cohomology ring of the moment-angle complex $\mathcal{Z}_{P}$ (see [4]), and it is known in [13] that

[TABLE]

where $\phi_{i}=a_{i}+a_{i+1}+\cdots+a_{i+k-2}$ and $\varphi_{i}=a_{i}+a_{i+1}+\cdots+a_{i+k-1}$ for all $i\in[2k+1]$, and indices are taken module $2k+1$. We remark that the A-rigidity of $P$ is equivalent to the B-rigidity, that is, if $P$ is B-rigid, then $P$ is A-rigid as well. However, $P$ is rarely to be A-rigid (and B-rigid) as seen in [1].

In particular, we, therefore, note that two simple polytopes represented on $P_{5}$ with assigned numbers $[a,b,c,d,e]$ and $[a^{\prime},b^{\prime},c^{\prime},d^{\prime},e^{\prime}]$ have isomorphic Tor-algebras if and only if $\{a+b,b+c,c+d,d+e,e+a\}=\{a^{\prime}+b^{\prime},b^{\prime}+c^{\prime},c^{\prime}+d^{\prime},d^{\prime}+e^{\prime},e^{\prime}+a^{\prime}\}$ as multi-sets. We further have the following criterion to find all polytopes whose Tor-algebras are isomorphic to that of a given polytope represented on $P_{5}$.

Proposition 2.3**.**

Let $P$ be the simple $n$-polytope with $n+3$ facets represented on $P_{5}$ with assigned number $[a,b,c,d,e]$. We also let $Q$ be another simple $n$-polytope with $n+3$ facets represented on $P_{5}$ with assigned number $[a^{\prime},b^{\prime},c^{\prime},d^{\prime},e^{\prime}]$. Then, the following are equivalent:

1. (1)

$\operatorname{Tor}(P)\cong\operatorname{Tor}(Q)$* as bigraded rings,* 2. (2)

$P$* and $Q$ have the same bigraded Betti numbers, and* 3. (3)

$[a^{\prime},b^{\prime},c^{\prime},d^{\prime},e^{\prime}]$* appears as a cyclic subgraph with $5$ vertices of the Peterson graph with assigned numbers as in Figure 1.*

Proof.

It is enough to show that (2) and (3) are equivalent. We first show that the set of the sum of the adjacent 2 vertices of each subgraph with 5 vertices in the above Peterson graph is $\{a+b,b+c,c+d,d+e,e+a\}$. We consider 4 cases for the sequence of vertices as follow:

• (1)

$[a,b,c,d,e]$ or reverse order,

• (2)

kinds of $[a,e,d,a+b-d,c+d-a]$ or reverse order,

• (3)

kinds of $[a,e,b+c-e,e+a-c,c+d-a]$ or reverse order, and

• (4)

$[a+b-d,d+e-b,b+c-e,e+a-c,c+d-a]$ or reverse order.

One can easily check that for each case we have $\{a+b,b+c,c+d,d+e,e+a\}$ for the set of the sum of the adjacent 2 vertices, and, hence, we have $24$ sequences from the Peterson graph.

We then show that there are $24$ sequences $[a^{\prime},b^{\prime},c^{\prime},d^{\prime},e^{\prime}]$ satisfying

[TABLE]

Consider the system of equations

[TABLE]

where $\{A,B,C,D,E\}=\{a+b,b+c,c+d,d+e,e+a\}$. Then, we have $5!=120$ systems, and there, thus, can be at most $24$ sequences $[a^{\prime},b^{\prime},c^{\prime},d^{\prime},e^{\prime}]$ up to rotation. Therefore, each sequence $[a^{\prime},b^{\prime},c^{\prime},d^{\prime},e^{\prime}]$ satisfying $\{a+b,b+c,c+d,d+e,e+a\}=\{a^{\prime}+b^{\prime},b^{\prime}+c^{\prime},c^{\prime}+d^{\prime},d^{\prime}+e^{\prime},e^{\prime}+a^{\prime}\}$ is uniquely obtained from a cyclic subgraph with $5$ vertices of the Peterson graph with assigned numbers as in Figure 1. ∎

Example 2.4**.**

We consider the simple polytope $\mathcal{P}$ represented on $P_{5}$ with assigned numbers $[3,1,2,1,1]$. Then the corresponding diagram as in Proposition 2.3 is the following.

\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$3$$1$$2$$1$$1[math]1$$2$$3$$2

One can easily see that there are only two types of cyclic subgraphs with $5$ vertices whose assigned numbers are all positive integers: $[3,1,2,1,1]$ and $[2,2,2,1,1]$ up to rotating and reflecting elements. Let $\mathcal{Q}$ be the simple polytope represented on $P_{5}$ with assigned numbers $[2,2,1,1,1]$. Hence, $\mathcal{Q}$ is the only polytope whose Tor-algebra is isomorphic to that of $\mathcal{P}$ with $\mathcal{P}\not\approx\mathcal{Q}$.

3. Main theorem

In this section, we give an example of C-rigid simple polytope which is not B-rigid, hence we show that the C-rigidity of a given simple polytope $P$ does not guarantee the B-rigidity of $P$.

More precisely, we will show that two simple polytopes $\mathcal{P}$ and $\mathcal{Q}$ represented on $P_{5}$ with assigned numbers $[3,1,2,1,1]$ and $[2,2,2,1,1]$ as in Figure 2, respectively, satisfy the following;

1. (a)

both $\mathcal{P}$ and $\mathcal{Q}$ support quasitoric manifolds, 2. (b)

$\operatorname{Tor}(\mathcal{P})\cong\operatorname{Tor}(\mathcal{Q})$ as bigraded rings, 3. (c)

there is no other polytope whose Tor-algebra is isomorphic that of $\mathcal{P}$, and 4. (d)

no two quasitoric manifolds $M$ over $\mathcal{P}$ and $N$ over $\mathcal{Q}$ have the isomorphic cohomology rings, that is, $H^{\ast}(M)\not\cong H^{\ast}(N)$ as graded rings.

If $\mathcal{P}$ and $\mathcal{Q}$ hold the above (a)–(d), then both $\mathcal{P}$ and $\mathcal{Q}$ are indeed our desired examples. The statement (b) implies that $\mathcal{P}$ and $\mathcal{Q}$ are not B-rigid, and both (c) and (d) imply the C-rigidity of $\mathcal{P}$ and $\mathcal{Q}$ under the condition (a) which confirms the existence of supporting quasitoric manifolds.

The statement (a) immediately follows the following theorem due to Erokhovets [9] (cf. [11]).

Theorem 3.1**.**

Let $P$ be a simple polytope represented on $P_{2k+1}$. Then, $P$ supports a quasitoric manifold if and only if $k\leq 3$.

The statements (b) and (c) are already showed in Example 2.4. In the remain of the section, let us show the statement (d) which would be the most difficult part.

Theorem 3.2**.**

The polytopes $\mathcal{P}$ and $\mathcal{Q}$ do not support quasitoric manifolds having the isomorphic $\mathbb{Z}$-cohomology rings.

Proof.

Let $P$ be an $n$-dimensional simple polytope with the set of facets $\mathfrak{F}=\{F_{1},\ldots,F_{m}\}$. We recall the general fact on quasitoric manifolds due to [8] that each quasitoric manifold $M$ over $P$ is assigned by the characteristic map $\lambda\colon\mathfrak{F}\to\mathbb{Z}^{n}$ satisfying that whenever $F_{i_{1}}\cap\cdots\cap F_{i_{\ell}}\neq\emptyset$, the set $\{\lambda(F_{i_{1}}),\ldots,\lambda(F_{i_{\ell}})\}$ of integral vectors is unimodular. We denote $\lambda=(\Lambda_{i,j})_{1\leq i\leq n,1\leq j\leq m}$ by an $n\times m$ integer matrix $\Lambda$ such that the $i$th column of $\Lambda$ is $\lambda(F_{i})$. It should be noted that we may assume that the first $n$ columns of $\Lambda$ form an identity matrix of size $n$.

Furthermore, the cohomology of $M$ associated to $\lambda$ is isomorphic to

[TABLE]

where $\deg v_{i}=2$ for all $i=1,\ldots,m$, $I_{P}$ is the ideal generated by the square free monomials $v_{i_{1}}\cdots v_{i_{r}}$ such that $F_{i_{1}}\cap\cdots\cap F_{i_{r}}=\emptyset$, and $J$ is the ideal generated by $n$ linear terms $\Lambda_{i,1}v_{1}+\cdots+\Lambda_{i,m}v_{m}$ for $1\leq i\leq n$.

Now, let us suppose there are quasitoric manifolds $M$ over $\mathcal{P}$ and $N$ over $\mathcal{Q}$ such that $H^{\ast}(M)\cong H^{\ast}(N)$ as graded rings. Then, $H^{\ast}(M;\mathbb{Z}_{2})=H^{\ast}(M)\otimes_{\mathbb{Z}}\mathbb{Z}_{2}\cong H^{\ast}(N)\otimes_{\mathbb{Z}}\mathbb{Z}_{2}=H^{\ast}(N;\mathbb{Z}_{2})$ as graded rings. We remark that $H^{\ast}(M;\mathbb{Z}_{2})$ is determined by $\mathbb{Z}_{2}$-characteristic map \lambda^{\mathbb{R}}\colon\mathcal{F}\stackrel{{\scriptstyle\lambda}}{{\to}}\mathbb{Z}^{n}\stackrel{{\scriptstyle\text{mod 2}}}{{\longrightarrow}}\mathbb{Z}_{2}^{n}. A $\mathbb{Z}_{2}$-characteristic map also can be represented by an $n\times m$ $\mathbb{Z}_{2}$-matrix $\Lambda^{\mathbb{R}}$, called the $\mathbb{Z}_{2}$-characteristic matrix, whose $i$th column is $\lambda^{\mathbb{R}}(F_{i})$. It should also be remarked that the first $n$ columns of $\Lambda^{\mathbb{R}}$ form an identity $\mathbb{Z}_{2}$-matrix. From now on, we shall show that $H^{\ast}(M;\mathbb{Z}_{2})$ is not isomorphic to $H^{\ast}(N;\mathbb{Z}_{2})$ as graded rings for any pairs of $\mathbb{Z}_{2}$-characteristic maps over $\mathcal{P}$ and over $\mathcal{Q}$, which contradicts to the assumption, so we prove the theorem.

Here is the list of $\mathbb{Z}_{2}$-characteristic matrices over $\mathcal{P}$. We give a re-labeling and an order of the facet set of $\mathcal{P}$ as $\mathfrak{F}=(F_{1_{1}},F_{1_{2}},F_{1_{3}},F_{3_{1}},F_{3_{2}},F_{5},F_{2},F_{4})$ such that the face structure of $\mathcal{P}$ is determined by a surjective map $\phi\colon\mathfrak{F}\to[5]$ defined by $\phi(F_{1_{j}})=1$ for $j=1,2,3$, $\phi(F_{3_{j}})=3$ for $j=1,2$, and $\phi(F_{i})=i$ for $i=2,4,5$, where $[5]$ is the vertex set of a regular pentagon $P_{5}$ as a Gale-diagram. In this order, we consider all possible $\mathbb{Z}_{2}$-characteristic matrices over $\mathcal{P}$. The following is the list of $A_{i}$’s where $\Lambda^{\mathbb{R}}=(I_{5}|A_{i})$ is a $\mathbb{Z}_{2}$-characteristic matrix over $\mathcal{P}$:

A_{1}=\left(\begin{array}[]{ccc}1&0&1\\ 1&0&1\\ 1&0&1\\ 0&1&1\\ 0&1&1\\ \end{array}\right) A_{2}=\left(\begin{array}[]{ccc}1&0&1\\ 1&0&1\\ 1&1&0\\ 0&1&1\\ 0&1&1\\ \end{array}\right) A_{3}=\left(\begin{array}[]{ccc}1&0&1\\ 1&1&0\\ 1&0&1\\ 0&1&1\\ 0&1&1\\ \end{array}\right) A_{4}=\left(\begin{array}[]{ccc}1&0&1\\ 1&1&0\\ 1&1&0\\ 0&1&1\\ 0&1&1\\ \end{array}\right) A_{5}=\left(\begin{array}[]{ccc}1&1&0\\ 1&0&1\\ 1&0&1\\ 0&1&1\\ 0&1&1\\ \end{array}\right) A_{6}=\left(\begin{array}[]{ccc}1&1&0\\ 1&0&1\\ 1&1&0\\ 0&1&1\\ 0&1&1\\ \end{array}\right) A_{7}=\left(\begin{array}[]{ccc}1&1&0\\ 1&1&0\\ 1&0&1\\ 0&1&1\\ 0&1&1\\ \end{array}\right) A_{8}=\left(\begin{array}[]{ccc}1&1&0\\ 1&1&0\\ 1&1&0\\ 0&1&1\\ 0&1&1\\ \end{array}\right) A_{9}=\left(\begin{array}[]{ccc}1&0&1\\ 1&0&1\\ 1&0&1\\ 0&1&1\\ 1&1&1\\ \end{array}\right) A_{10}=\left(\begin{array}[]{ccc}1&1&0\\ 1&1&0\\ 1&1&0\\ 0&1&1\\ 1&0&1\\ \end{array}\right) A_{11}=\left(\begin{array}[]{ccc}1&0&1\\ 1&0&1\\ 1&0&1\\ 1&1&1\\ 0&1&1\\ \end{array}\right) A_{12}=\left(\begin{array}[]{ccc}1&1&0\\ 1&1&0\\ 1&1&0\\ 1&0&1\\ 0&1&1\\ \end{array}\right) A_{13}=\left(\begin{array}[]{ccc}1&0&1\\ 1&0&1\\ 1&0&1\\ 1&1&1\\ 1&1&1\\ \end{array}\right) A_{14}=\left(\begin{array}[]{ccc}1&1&0\\ 1&1&0\\ 1&1&0\\ 1&0&1\\ 1&0&1\\ \end{array}\right) A_{15}=\left(\begin{array}[]{ccc}1&1&0\\ 1&1&0\\ 1&1&1\\ 1&0&1\\ 1&0&1\\ \end{array}\right) A_{16}=\left(\begin{array}[]{ccc}1&1&0\\ 1&1&1\\ 1&1&0\\ 1&0&1\\ 1&0&1\\ \end{array}\right) A_{17}=\left(\begin{array}[]{ccc}1&1&0\\ 1&1&1\\ 1&1&1\\ 1&0&1\\ 1&0&1\\ \end{array}\right) A_{18}=\left(\begin{array}[]{ccc}1&1&1\\ 1&1&0\\ 1&1&0\\ 1&0&1\\ 1&0&1\\ \end{array}\right) A_{19}=\left(\begin{array}[]{ccc}1&1&1\\ 1&1&0\\ 1&1&1\\ 1&0&1\\ 1&0&1\\ \end{array}\right) A_{20}=\left(\begin{array}[]{ccc}1&1&1\\ 1&1&1\\ 1&1&0\\ 1&0&1\\ 1&0&1\\ \end{array}\right) A_{21}=\left(\begin{array}[]{ccc}1&1&1\\ 1&1&1\\ 1&1&1\\ 1&0&1\\ 1&0&1\\ \end{array}\right)

Here is the list of $\mathbb{Z}_{2}$-characteristic matrices over $\mathcal{Q}$. We give a re-labeling and an order of the facet set of $\mathcal{Q}$ as $\mathfrak{F}=(F_{1_{1}},F_{1_{2}},F_{2_{1}},F_{3_{1}},F_{3_{2}},F_{5},F_{2_{2}},F_{4})$ such that the face structure of $\mathcal{Q}$ is determined by a surjective map $\phi\colon\mathfrak{F}\to[5]$ defined by $\phi(F_{i_{j}})=1$ for $i=1,2,3$ and $j=1,2$, and $\phi(F_{i})=i$ for $i=4,5$. In this order, we consider all possible $\mathbb{Z}_{2}$-characteristic matrices over $\mathcal{Q}$. The following is the list of $B_{i}$’s where $(I_{5}|B_{i})$ is a $\mathbb{Z}_{2}$-characteristic matrix over $\mathcal{Q}$:

B_{1}=\left(\begin{array}[]{ccc}1&0&1\\ 1&0&1\\ 0&1&0\\ 0&1&1\\ 0&1&1\\ \end{array}\right) B_{2}=\left(\begin{array}[]{ccc}1&0&1\\ 1&1&0\\ 0&1&0\\ 0&1&1\\ 0&1&1\\ \end{array}\right) B_{3}=\left(\begin{array}[]{ccc}1&1&0\\ 1&0&1\\ 0&1&0\\ 0&1&1\\ 0&1&1\\ \end{array}\right) B_{4}=\left(\begin{array}[]{ccc}1&1&0\\ 1&1&0\\ 0&1&0\\ 0&1&1\\ 0&1&1\\ \end{array}\right) B_{5}=\left(\begin{array}[]{ccc}1&0&1\\ 1&0&1\\ 0&1&0\\ 0&1&1\\ 1&1&1\\ \end{array}\right) B_{6}=\left(\begin{array}[]{ccc}1&1&0\\ 1&1&0\\ 0&1&0\\ 0&1&1\\ 1&0&1\\ \end{array}\right) B_{7}=\left(\begin{array}[]{ccc}1&0&1\\ 1&0&1\\ 0&1&0\\ 1&1&1\\ 0&1&1\\ \end{array}\right) B_{8}=\left(\begin{array}[]{ccc}1&1&0\\ 1&1&0\\ 0&1&0\\ 1&0&1\\ 0&1&1\\ \end{array}\right) B_{9}=\left(\begin{array}[]{ccc}1&0&1\\ 1&0&1\\ 0&1&0\\ 1&1&1\\ 1&1&1\\ \end{array}\right) B_{10}=\left(\begin{array}[]{ccc}1&1&0\\ 1&1&0\\ 0&1&0\\ 1&0&1\\ 1&0&1\\ \end{array}\right) B_{11}=\left(\begin{array}[]{ccc}1&1&0\\ 1&1&0\\ 0&1&1\\ 1&0&1\\ 1&0&1\\ \end{array}\right) B_{12}=\left(\begin{array}[]{ccc}1&1&0\\ 1&1&1\\ 0&1&0\\ 1&0&1\\ 1&0&1\\ \end{array}\right) B_{13}=\left(\begin{array}[]{ccc}1&1&0\\ 1&1&1\\ 0&1&1\\ 1&0&1\\ 1&0&1\\ \end{array}\right) B_{14}=\left(\begin{array}[]{ccc}1&1&1\\ 1&1&0\\ 0&1&0\\ 1&0&1\\ 1&0&1\\ \end{array}\right) B_{15}=\left(\begin{array}[]{ccc}1&1&1\\ 1&1&0\\ 0&1&1\\ 1&0&1\\ 1&0&1\\ \end{array}\right) B_{16}=\left(\begin{array}[]{ccc}1&1&1\\ 1&1&1\\ 0&1&0\\ 1&0&1\\ 1&0&1\\ \end{array}\right) B_{17}=\left(\begin{array}[]{ccc}1&1&1\\ 1&1&1\\ 0&1&1\\ 1&0&1\\ 1&0&1\\ \end{array}\right) B_{18}=\left(\begin{array}[]{ccc}1&0&1\\ 1&0&1\\ 1&1&0\\ 0&1&1\\ 0&1&1\\ \end{array}\right) B_{19}=\left(\begin{array}[]{ccc}1&0&1\\ 1&0&1\\ 1&1&0\\ 0&1&1\\ 1&1&1\\ \end{array}\right) B_{20}=\left(\begin{array}[]{ccc}1&0&1\\ 1&0&1\\ 1&1&0\\ 1&1&1\\ 0&1&1\\ \end{array}\right) B_{21}=\left(\begin{array}[]{ccc}1&0&1\\ 1&0&1\\ 1&1&0\\ 1&1&1\\ 1&1&1\\ \end{array}\right)

The above lists can be found by hand-computation or using computer algorithm such as [10, Algorithm 4.1].

We remark that the $\mathbb{Z}_{2}$-cohomology rings of $M$ over $\mathcal{P}$ associated to $A_{i}$ can be written by

[TABLE]

where $x=v_{5}$, $y=v_{2}$ and $z=v_{4}$.

The following table is the list of generators of $I_{A}$ with respect to each $A_{i}$.

[TABLE]

Similarly, the $\mathbb{Z}_{2}$-cohomology rings of $N$ over $\mathcal{Q}$ associated to $B_{i}$ can also be written by

[TABLE]

where $x=v_{5}$, $y=v_{2_{2}}$ and $z=v_{4}$. The following table is the list of generators of $I_{B}$ with respect to each $B_{i}$.

[TABLE]

Let $V$ be the vector space generated by $x,y,z$ over $\mathbb{Z}_{2}$. In order to show that there is no pair $(M,N)$ such that $H^{\ast}(M;\mathbb{Z}_{2})\cong H^{\ast}(N;\mathbb{Z}_{2})$, we shall check that there is no linear bijective map $V$ to $V$ which sends $I_{A}$ to $I_{B}$ for any $A_{i}$ and $B_{j}$.

For $\gamma\in\{x,y,z,x+y,x+z,y+z,x+y+z\}$, we define the codimension of $\gamma$, denoted by $\operatorname{codim}\gamma$, as the minimum order of $\delta$ such that $\gamma\delta\in I$ and the order of $\gamma$, denoted by $\operatorname{ord}\gamma$, as the minimum of $k$ such that $\gamma^{k}\in I$.

Here are the lists of codimensions and orders of all linear terms corresponding to each $A_{i}$.

[TABLE]

[TABLE]

Here are the lists of codimensions and orders of all linear terms corresponding to each $B_{i}$.

[TABLE]

[TABLE]

Note that the multi-sets of codimensions and orders of all elements in $V$ must be invariant under a linear map which sends $I_{A}$ to $I_{B}$. Let us firstly consider codimensions. Since codimensions of only $x$ and $z$ are all 1 in $I_{A}$ and $I_{B}$, the linear map should sends $\{x,z\}$ to $\{x,z\}$. Then, $x+z\mapsto x+z$ via the linear map.

For $A_{1},A_{2},A_{4},A_{8},A_{9},A_{13}$, since the codimension of $x+z$ in $I$ is different from that in $I_{B}$ for any $B_{i}$, there is not a such linear map between $I_{A}$ and $I_{B}$. For $A_{10}$, while the orders of $x$, $z$, and $x+z$ are all 6 in $I_{A}$, there is no $B_{i}$ satisfying the orders of $x$, $z$, and $x+z$ are all 6. Thus, there is not a such linear map between $I_{A}$ and $I_{B}$. For $A_{14}$, $A_{17}$ and $A_{21}$, while the orders of $x$ and $z$ are 6, 3, respectively, there is no $B_{i}$ satisfying the set of orders of $x$ and $z$ is $\{6,3\}$ in $I_{B}$. Thus, there is not a such linear map between $I_{A}$ and $I_{B}$. Finally, for $A_{15}$, the order of each element in $V$ is either $3$ or $5$ in $I_{A}$. However, for any $B_{i}$, there is at least one element in $V$ whose order is $6$ in $I_{B}$. Therefore, there is no such linear map between $I_{A}$ and $I_{B}$ for any pair of $A_{i}$ and $B_{j}$, which proves the theorem. ∎