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arXiv 2012.11927math.RTmath.RA
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周期平凡扩张代数与分数 Calabi-Yau 代数

Periodic trivial extension algebras and fractionally Calabi-Yau algebras

Aaron Chan, Erik Darpö, Osamu Iyama, René Marczinzik

AI总结:

本文证明平凡扩张代数的周期或扭曲周期性等价于原代数为有限整体维数的分数或扭曲分数 Calabi-Yau,并推广至轨道代数,解决多个相关公开问题。

摘要:

我们研究有限维代数 $A$ 的平凡扩张代数 $T(A)$ 的周期性与扭曲周期性。主要结果表明,$T(A)$ 的(扭曲)周期性等价于 $A$ 是具有有限整体维数的(扭曲)分数 Calabi-Yau 代数。我们还将此结果推广到一大类自内射轨道代数。作为一个重要推论,这些结果对 Erdmann-Skowroński 周期性猜想给出了部分回答;该猜想预期周期代数与扭曲周期代数这两个类是重合的。在实际层面,这使我们能够构造大量周期代数与分数 Calabi-Yau 代数的新例子。我们还建立了周期性与丛倾斜理论之间的联系,证明 $T(A)$ 的扭曲周期性等价于对某些 $r,d\ge 1$,$r$ 重平凡扩张代数 $T_r(A)$ 是 $d$-表示有限的。这回答了 Darpö 与 Iyama 的一个问题。作为结果的应用,我们回答了其他一些公开问题。我们构造了具有任意大最小周期的野表示型周期对称代数,回答了 Skowroński 的一个问题。我们还证明扭曲分数 Calabi-Yau 代数类在导出等价下封闭,回答了 Herschend 与 Iyama 的一个问题。

英文摘要:

We study periodicity and twisted periodicity of the trivial extension algebra $T(A)$ of a finite-dimensional algebra $A$. Our main results show that (twisted) periodicity of $T(A)$ is equivalent to $A$ being (twisted) fractionally Calabi-Yau of finite global dimension. We also extend this result to a large class of self-injective orbit algebras. As a significant consequence, these results give a partial answer to the periodicity conjecture of Erdmann-Skowroński, which expects the classes of periodic and twisted periodic algebras to coincide. On the practical side, it allows us to construct a large number of new examples of periodic algebras and fractionally Calabi-Yau algebras. We also establish a connection between periodicity and cluster tilting theory, by showing that twisted periodicity of $T(A)$ is equivalent the $d$-representation-finiteness of the $r$-fold trivial extension algebra $T_r(A)$ for some $r,d\ge 1$. This answers a question by Darpö and Iyama. As applications of our results, we give answers to some other open questions. We construct periodic symmetric algebras of wild representation type with arbitrary large minimal period, answering a question by Skowroński. We also show that the class of twisted fractionally Calabi-Yau algebras is closed under derived equivalence, answering a question by Herschend and Iyama.

机构、总结、中文摘要均由AI分析而来,可能存在错误,请以论文正文为准。

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