Let $\frak{g}$ be a semisimple Lie algebra and $U(\frak{g})$ its universal enveloping algebra. The adjoint action of $\frak{g}$ on itself extends to an action of $\frak{g}$ on $U(\frak{g})$. How does this action interact with the PBW basis of $U(\frak{g})$? More precisely: are the irreducible submodules of $U(\frak{g})$ spanned by the elements of the PBW basis?
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$\begingroup$ I'm not sure this question is tight enough though I suspect that the answer is no however it is clarified. Two initial problems: 'The PBW basis' isn't unique but depends on a choice of basis for $\mathfrak{g}$ and it isn't clear if you mean all irreducible submodules have such a basis or for some decomposition of the enveloping algebra as a direct sum of irreducibles all the summands have this property. $\endgroup$– Simon WadsleyCommented Aug 16, 2023 at 17:37
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$\begingroup$ I mean the latter. $\endgroup$– Béla FürdőházCommented Aug 16, 2023 at 17:38
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$\begingroup$ What basis of the Lie algebra do you choose? $\endgroup$– Simon WadsleyCommented Aug 16, 2023 at 17:46
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$\begingroup$ A general choice of weight basis. Does the scaling of the basis elements play an important role? $\endgroup$– Béla FürdőházCommented Aug 16, 2023 at 18:34
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Since the sum of all the trivial subrepresentations of $U(\mathfrak{g})$ with the adjoint action is its centre, if you could do this then the centre must have a monomial basis but it does not.
For example even if $\mathfrak{g}=\mathfrak{sl}_2$ with basis $e,h,f$ in that order then the centre consists of polynomials in $$C=ef+fe+\frac 1 2 h^2=2ef+\frac 1 2(h^2-h)$$ which does not live in the monomial basis.
answered Aug 16, 2023 at 21:37