3
$\begingroup$

Algebra that I'm going to describe pop-up in my research, it looks completely elementary, but I don't know any appropriate references.

Let $k$ be an algebraically closed field of characteristic zero. Let us define algebra $A$ as tensor algebra on four variables $k\langle x_1, x_2, y_1, y_2\rangle$ with the following relations $x_1y_1=1$, $x_2y_1=1$, $x_1y_2=1$ and $x_2y_2=1$.

I'm interested in irreducible representations of this algebra. Perhaps, there is a way to rewrite generator and relations in a way that allows to reduce this algebra to some known algebras. Any advice will be appreciated.

Improve this question
$\endgroup$
1
  • $\begingroup$ So, a basis is $y_{i_1} y_{i_2} \cdots y_{i_r} x_{j_1} x_{j_2} \cdots x_{j_s}$, for any binary sequences $(i_1, \ldots, i_r)$, $(j_1, \ldots, j_s)$. $\endgroup$
    – David E Speyer
    Commented Oct 7, 2014 at 16:39

1 Answer 1

Reset to default
-2
$\begingroup$

Since $x_1 y_1=x_2 y_1$, you have $x_1 = x_2$. Similarily $y_1=y_2$. So you algebra is (isomorphic to) the algebra in two variables $k⟨x,y⟩$, with the relation $xy=1$.

This is the algebra $k⟨x,x^{-1}⟩$, which is the algebra of finite power series $\sum_n^m a_i x^i=a_{-n} x^{-n} + ... + a_{-1} x^{-1} + a_0 + a_1 x + ... + a_m x^{-m}$.

Improve this answer
$\endgroup$
1
  • 2
    $\begingroup$ This is not true. You cannot cancel. In fact consider the representation of $A$ on $k^\mathbb{N}$ where $x_1,x_2$ act as left shift and $y_1,y_2$ act as $y_1 (a_0,a_1,...) := (1,a_0,a_1,...)$ and $y_2 (a_0,a_1,...) := (2,a_1,a_2,...)$ respectively. $\endgroup$
    – Johannes Hahn
    Commented Oct 7, 2014 at 14:09

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .