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Let $\mathfrak{g} = \mathfrak{sl}(2)$ Lie algebra

The truncated current Lie algebra is the Lie algebra defined as $\mathfrak{g}^{(r)} = \mathfrak{g}[t]/t^{r+1} \mathfrak{g}[t] $ where $\mathfrak{g}[t] = \mathfrak{g} \otimes \mathbb{C}[t]$, and the Lie bracket is defined as $[X\otimes t^{m}, Y\otimes t^{n}] = [X,Y] \otimes t^{m+n}$

My question is, when it comes to the Universal enveloping algebra of $\mathfrak{g}^{(r)}$, can we write that $X\otimes t^{m} \cdot Y\otimes t^{n}= X\otimes t^{n} \cdot Y\otimes t^{m}$ ?

I am not sure whether I can write something like $X\otimes t^{n} = t^{n} (X\otimes 1) $, i.e. if I can consider $\mathfrak{g}^{(r)} $ to be a $\mathbb{C}[t]$ -algebra.

Thanks for your answer ,

Dearly

Joniloli
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1 Answers1

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No. If $m\neq n$ and $X$ and $Y$ are linearly independent, then the four elements $a=X\otimes t^m,b=Y\otimes t^n,c=X\otimes t^n,d=Y\otimes t^m$ are linearly independent in $\mathfrak{g}^{(r)}$. It follows from the PBW theorem that $ab\neq cd$ in the universal enveloping algebra (you can extend $\{a,b,c,d\}$ to a basis and then these are distinct basis elements for the universal enveloping algebra given by PBW).

(That said, this assumes you are talking about the universal enveloping algebra of $\mathfrak{g}^{(r)}$ as a Lie algebra over $\mathbb{C}$. If you are instead talking about the universal enveloping algebra as a Lie algebra over the ring $\mathbb{C}[t]$ then that means you treat $t$ as a scalar and can write $X\otimes t^n=t^n(X\otimes 1)$.)

Eric Wofsey
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