论文标题
结成不变的倾斜模块
Tilting modules arising from knot invariants
论文作者
论文摘要
我们在结的雅各布代数上构建倾斜模块。到两个桥结$ l [a_1，\ ldots，a_n] $，我们将Quiver $ Q $与潜力及其Jacobian代数$ a $相关联。我们构建了一个规范的不可分解$ a $ modules $ m（i）$的家庭，每个家族都在不同的特定副本$ q（i）$ q $上支持。预计$ m（i）$中的每一个都可以参数结节的琼斯多项式。我们研究了这些inndecosables $ a $以及集群类别中的这些indecosables的直接总和$ m = \ oplus_im（i）$。 在本文中，我们考虑了两个桥结的特殊情况，由两个参数$ a_1，a_2 $给出。我们表明，模块$ m $是刚性和$τ$ - rigid，我们将$ m $构建为倾斜（和$τ$ - tilting）$ a $ a $ -module $ t $。我们表明，$ t $ of $ t $ of operatorm $ \ operatotorname {end} _at $ of $ t $是$ a $的同构，并且映射$ t \ mapsto a [1] $诱导了群集代数$ \ mathcal \ nathcal {a a}（q）$。这种自动形态为二。此外，我们给出一个突变序列，以实现群集自动形态。特别是，我们表明Quiver $ Q $的突变等效于类型$ t_ {p，q，r} $（具有三个分支的树）类型。如果$（a_1，a_2）=（a_1,2），（1，a_2），$或$（2,3）$，则此Quiver是有限类型的，它是$（a_1，a_2）=（2,4）$或$（3,3,3）$的驯服的。
We construct tilting modules over Jacobian algebras arising from knots. To a two-bridge knot $L[a_1,\ldots,a_n]$, we associate a quiver $Q$ with potential and its Jacobian algebra $A$. We construct a family of canonical indecomposable $A$-modules $M(i)$, each supported on a different specific subquiver $Q(i)$ of $Q$. Each of the $M(i)$ is expected to parametrize the Jones polynomial of the knot. We study the direct sum $M=\oplus_iM(i)$ of these indecomposables inside the module category of $A$ as well as in the cluster category. In this paper we consider the special case where the two-bridge knot is given by two parameters $a_1,a_2$. We show that the module $M$ is rigid and $τ$-rigid, and we construct a completion of $M$ to a tilting (and $τ$-tilting) $A$-module $T$. We show that the endomorphism algebra $\operatorname{End}_AT$ of $T$ is isomorphic to $A$, and that the mapping $T\mapsto A[1]$ induces a cluster automorphism of the cluster algebra $\mathcal{A}(Q)$. This automorphism is of order two. Moreover, we give a mutation sequence that realizes the cluster automorphism. In particular, we show that the quiver $Q$ is mutation equivalent to an acyclic quiver of type $T_{p,q,r}$ (a tree with three branches). This quiver is of finite type if $(a_1,a_2)=(a_1,2), (1,a_2),$ or $(2,3)$, it is tame for $(a_1,a_2)=(2,4)$ or $(3,3)$, and wild otherwise.
