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Apr 22, 2021 at 6:08 comment added Vladimir Dotsenko @YCor : for the right hand side, the most classical description of that algebra is, of course, as coinvariants of the Weyl group action on the Cartan subalgebra, that is $\mathbb{C}[\mathfrak{h}]/(\mathbb{C}[\mathfrak{h}]^W_+)$, and the grading by the integer lattice in the Cartan subalgebra seems to disappear in the quotient. That said, it is indeed a completely separate question, I think, since the algebra structures on the right and on the left are quite drastically different from the very beginning.
Apr 21, 2021 at 15:16 comment added YCor I don't know if it's of interest here, but such a nilpotent Lie algebra admits a natural grading in a free abelian group of rank $\mathrm{rank}(\mathfrak{g})$, coming from the action of the maximal torus normalizing $\mathfrak{n}$. So the whole cohomology algebra inherits this grading, and one can wonder whether it makes sense at the level of the right hand de Rham cohomology space.
Apr 21, 2021 at 15:06 history edited Vladimir Dotsenko CC BY-SA 4.0
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Apr 18, 2021 at 20:49 comment added user108998 Thanks, don't know why I didn't see that!
Apr 18, 2021 at 16:19 comment added Vladimir Dotsenko @EBz because elements of cohomological degree one anticommute and elements of cohomological degree two commute?
Apr 18, 2021 at 12:05 comment added user108998 Why does the fact that the grading scales by 2 imply that the multiplicative structure cannot be preserved?
Apr 18, 2021 at 7:01 history edited Vladimir Dotsenko CC BY-SA 4.0
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Apr 18, 2021 at 6:52 history edited YCor CC BY-SA 4.0
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Apr 18, 2021 at 6:51 history edited Vladimir Dotsenko CC BY-SA 4.0
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Apr 18, 2021 at 6:44 history asked Vladimir Dotsenko CC BY-SA 4.0