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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
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