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The quantum algebra $$U_q(\mathfrak{sl}_2)$$ and its equitable presentation. (English) Zbl 1090.17004
In the paper under review the authors prove and derive some corollaries from the following amazing result:
Theorem: Let $$\mathbb{K}$$ denote a field and $$q$$ a nonzero scalar in $$\mathbb{K}$$ such that $$q^2\neq 1$$. Let further $$U_q(sl_2)$$ be the unital associative $$\mathbb{K}$$-algebra, generated by $$k^{\pm 1}$$, $$e$$ and $$f$$, subject to the relations $$kk^{-1}=k^{-1}k=1$$, $$ke=q^2ek$$, $$kf=q^{-2}fk$$, and $$ef-fe=\frac{k-k^{-1}}{q-q^{-1}}$$. Then $$U_q(sl_2)$$ is isomorphic to the unital associative $$\mathbb{K}$$-algebra, generated by $$x^{\pm 1}$$, $$y$$, and $$z$$, and subject to the following relations: $xx^{-1}=x^{-1}x=1,\quad \frac{qxy-q^{-1}yx}{q-q^{-1}}=1,\quad \frac{qyz-q^{-1}zy}{q-q^{-1}}=1,\quad \frac{qzx-q^{-1}xz}{q-q^{-1}}=1.$ The only asymmetry of the latter presentation is the axiom that the element $$x$$ is invertible. In general it is shown that neither $$y$$ nor $$z$$ are invertible: the authors construct an infinite-dimensional $$U_q(sl_2)$$-module, where these elements have a non-trivial kernel. At the same time, it is shown that the action of both $$y$$ and $$z$$ on each finite-dimensional $$U_q(sl_2)$$-module is invertible provided that $$q$$ is not a root of $$1$$ and $$\mathrm{char}(\mathbb{K}) \neq 2$$. Moreover, the authors construct a special linear operator, which conjugates the action of $$x$$ to that of $$y$$, the action of $$y$$ to that of $$z$$, and the action of $$z$$ to that of $$x$$ on every finite-dimensional $$U_q(sl_2)$$-module.

##### MSC:
 17B37 Quantum groups (quantized enveloping algebras) and related deformations
##### Keywords:
quantum algebra; presentation; module
Full Text:
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