On the noncommutative Bondal–Orlov conjecture for some toric varieties

Abstract

We show that all toric noncommutative crepant resolutions (NCCRs) of affine GIT quotients of “weakly symmetric” unimodular torus representations are derived equivalent. This yields evidence for a non-commutative extension of a well known conjecture by Bondal and Orlov stating that all crepant resolutions of a Gorenstein singularity are derived equivalent. We prove our result by showing that all toric NCCRs of the affine GIT quotient are derived equivalent to a fixed Deligne–Mumford GIT quotient stack associated to a generic character of the torus. This extends a result by Halpern–Leistner and Sam which showed that such GIT quotient stacks are a geometric incarnation of a family of specific toric NCCRs constructed earlier by the authors.

Appendix A: Appendix by Jason P. Bell

By Sect. 3.1, understanding the maximal Cohen–Macaulay cliques reduces to a purely combinatorial problem which is a subject of this section. Let $a_1,\ldots ,a_d\in {\mathbb N}_{>0}$. In addition, we assume that $\mathrm{gcd}(a_i)=1$. As a consequence every sufficiently large natural number can be expressed as a linear combination of the $a_i$ with nonnegative integer coefficients. Let $N\in {\mathbb N}$. We define

$$\begin{aligned} \mathcal {S}_+:=\left\{ N + \sum _i c_i a_i :c_i\ge 0\right\} ,\quad \mathcal {S}_{-}=-\mathcal {S}_{+},\quad \mathcal {S}=\mathcal {S}_+\cup \mathcal {S}_-. \end{aligned}$$

Then $\mathcal {S}$ contains all but finitely many integers.

Let

$$\begin{aligned} \mathfrak {M}=\{{\mathcal M}\subset {\mathbb Z}\mid m,m'\in {\mathcal M}\implies m-m'\not \in \mathcal {S}\}. \end{aligned}$$

Given $i\in \{0,\ldots ,N-1\}$, we define p(i) to be the smallest positive integer p such that $i+pN$ is in $\mathcal {S}_+$. Similarly, q(i) is the largest negative integer q such that $i+qN$ is in $\mathcal {S}_{-}$. Let ${\mathcal M}\in \mathfrak {M}$ and let $j\in \mathbb {Z}{\setminus }{\mathcal M}$. We say that $m\in {\mathcal M}$ blocks j if $j-m\in \mathcal {S}$. Notice that if ${\mathcal M}$ is a maximal element of $\mathfrak {M}$ then for each element in the complement of ${\mathcal M}$ there is necessarily some element of ${\mathcal M}$ that blocks it.

Lemma A.1

Let ${\mathcal M}\in \mathfrak {M}$ be a maximal element of $\mathfrak {M}$ containing 0 and let $i\in \{0,\ldots ,N-1\}$. Then ${\mathcal M}$ contains an element that is congruent to i modulo N.

Proof

Suppose not. Then for each $j\in \{q(i),\ldots ,p(i)\}$ we can choose some integer $m_j\in {\mathcal M}$ such that $m_j$ blocks $i+jN$. Since both $i+p(i)N$ and $i+q(i)N$ are in $\mathcal {S}$, we can take $m_{p(i)}=m_{q(i)}=0$. Now let

$$\begin{aligned} X_{\pm } = \{j\in \{q(i),\ldots ,p(i)\} :m_j - (i+jN) \in \mathcal {S}_{\pm }\}. \end{aligned}$$

Then $X_{+}$ and $X_{-}$ are disjoint and their union is all of $\{q(i),\ldots ,p(i)\}$. Moreover, since $m_{p(i)}=m_{q(i)}=0$ and $q(i)<0<p(i)$, we have $q(i)\in X_{+}$ and $p(i)\in X_{-}$. In particular, there must exist some $j\in \{q(i),\ldots ,p(i)-1\}$ such that $j\in X_{+}$ and $j+1\in X_{-}$. Given such a j, we then have

$$\begin{aligned} m_j - (i+jN)\in \mathcal {S}_{+}\quad \mathrm{and}\quad m_{j+1} - (i+(j+1)N)\in \mathcal {S}_{-}. \end{aligned}$$

So we may write $m_j - (i+jN) = N + k_1$ and $m_{j+1}-(i+(j+1)N) = -N-k_2$, where $k_1$ and $k_2$ are ${\mathbb N}$-linear combinations of the $a_i$. Subtracting these two equalities, we see

$$\begin{aligned} m_j - m_{j+1}+N = \left( m_j - (i+jN) \right) - \left( m_{j+1}-(i+(j+1)N)\right) = 2N + k_1+k_2. \end{aligned}$$

In particular, $m_j-m_{j+1}= N+k_1+k_2\in \mathcal {S}_+$. But this contradicts the fact that $m_j,m_{j+1}\in {\mathcal M}\in \mathfrak {M}$. The result follows. $\square$

Corollary A.2

Let N be a ${\mathbb N}$-linear combination of $a_i$. Let ${\mathcal M}$ be a maximal element in $\mathfrak {M}$. For every $0\le i<N$ there exists exactly one element $m\in {\mathcal M}$ such that $m\equiv i \,(N)$. In particular, all maximal elements of $\mathfrak {M}$ have size N.

Proof

Let ${\mathcal M}$ be a maximal element of $\mathfrak {M}$. By translation we can assume that $0\in {\mathcal M}$. By Lemma A.1 we have that for each $i\in \{0,\ldots ,N-1\}$ there is some element of ${\mathcal M}$ that is congruent to i mod N. Thus $|{\mathcal M}|\ge N$. On the other hand, by definition of $\mathcal {S}_{+}$ and the assumption on N we have $\{N,2N,3N,\ldots \}\subseteq \mathcal {S}$. If ${\mathcal M}$ had size strictly larger than N then there would exist $m,m'\in {\mathcal M}$ with $m> m'$ and m and $m'$ congruent to 0 mod N. But then $m-m'$ is a positive multiple of N and hence in $\mathcal {S}$. $\square$