论文标题
从一个难题中构建一个无限的困境家族
Constructing an infinite family of quandles from a quandle
论文作者
论文摘要
困境是自我分配的，直接的，势力的代数。具有二进制操作的共轭的组是一个搜索的示例。给定一个Quandle $（q，\ ast）$和一个正整数$ n $，定义$ a \ ast_n b =（\ cdots（a \ ast \ ast \ usterbrace {b）\ ast \ ast \ cdots）\ ast b} _ {n} $，其中$ a，b \ in q $。然后，$（q，\ ast_n）$再次是一个困境。我们提出以下问题。 `查找$（q，\ ast）$，使得序列$ \ {（q，q，\ ast_n）\，：\，n \ in \ mathbb {z}^+ \} $由成对的非iSomorphic quandles组成。我们将$ 2 $ - by- $ 2 $矩阵的通用线性组超过$ \ mathbb {c} $作为偶数的困境。它的（代数）连接的组件，即其共轭类是其子量。我们表明后者是作为Quandles连接的，并且证明了它们的僵化结果，例如二订单$ 3 $的二面套对于大多数人而言并不是一个子问题。然后，我们考虑使用共轭的$ 2 $ 2 $ 2 $矩阵的投影线性组，并证明它解决了上述问题。在这项工作的过程中，我们证明了一个足够且必要的条件使人们成为拉丁语。这将显着降低算法的复杂性，以确定是否是拉丁文。
Quandles are self-distributive, right-invertible, idempotent algebras. A group with conjugation for binary operation is an example of a quandle. Given a quandle $(Q, \ast)$ and a positive integer $n$, define $a\ast_n b = (\cdots (a\ast \underbrace{b)\ast \cdots )\ast b}_{n}$, where $a, b \in Q$. Then, $(Q, \ast_n)$ is again a quandle. We set forth the following problem. ``Find $(Q, \ast)$ such that the sequence $\{(Q, \ast_n)\, :\, n\in \mathbb{Z}^+ \}$ is made up of pairwise non-isomorphic quandles.'' In this article we find such a quandle $(Q, \ast)$. We study the general linear group of $2$-by-$2$ matrices over $\mathbb{C}$ as a quandle under conjugation. Its (algebraically) connected components, that is, its conjugacy classes, are subquandles of it. We show the latter are connected as quandles and prove rigidity results about them such as the dihedral quandle of order $3$ is not a subquandle for most of them. Then we consider the quandle which is the projective linear group of $2$-by-$2$ matrices over $\mathbb{C}$ with conjugation, and prove it solves the problem above. In the course of this work we prove a sufficient and necessary condition for a quandle to be latin. This will reduce significantly the complexity of algorithms for ascertaining if a quandle is latin.
