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The answer to this question may be well-known, but I failed to locate it in any obvious source. From the results of Bott (Ann. Math. 66 (1957), 203-248) and Kostant (Ann. Math. 74 (1961), 329-387), it follows that for the maximal nilpotent subalgebra $\mathfrak{n}$ of a complex semisimple Lie algebra $\mathfrak{g}$, one has an isomorphism of graded vector spaces $$ H^\bullet(\mathfrak{n},\mathbb{C})\cong H^{2\bullet}(G/B,\mathbb{C}) $$ of the Lie algebra cohomology of $\mathfrak{n}$ with constant coefficients and the cohomology of the corresponding flag variety $G/B$. However, this isomorphism multiplies degrees by two, and so cannot have anything to do with the multiplicative structure of the cohomology. My question is:

What is the (super-commutative) algebra structure of $H^\bullet(\mathfrak{n},\mathbb{C})$?

Even though I feel like it might be useful to state this question in full generality, I should probably say that I am mostly interested in the case $\mathfrak{g}=\mathfrak{sl}_{n}$, so that $\mathfrak{n}$ is the Lie algebra of all strictly upper-triangular matrices, and I ideally want the description to be by generators and relations.

Perhaps to give an illustration of what is going on, I should give the answer for upper triangular matrices of size three. In this case, the algebra has two generators $x,y$ of degree one and two generators $a,b$ of degree two, and the relations between them (besides the implied super-commutativity $fg=(-1)^{|f||g|} gf$ for all possible choices of $f,g$ among them) are: $$ xy=0, a^2=ab=b^2=0, xa+yb=xb=ya=0. $$

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  • $\begingroup$ Why does the fact that the grading scales by 2 imply that the multiplicative structure cannot be preserved? $\endgroup$
    – user108998
    Commented Apr 18, 2021 at 12:05
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    $\begingroup$ @EBz because elements of cohomological degree one anticommute and elements of cohomological degree two commute? $\endgroup$
    – Vladimir Dotsenko
    Commented Apr 18, 2021 at 16:19
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    $\begingroup$ Thanks, don't know why I didn't see that! $\endgroup$
    – user108998
    Commented Apr 18, 2021 at 20:49
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    $\begingroup$ 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. $\endgroup$
    – YCor
    Commented Apr 21, 2021 at 15:16
  • $\begingroup$ @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. $\endgroup$
    – Vladimir Dotsenko
    Commented Apr 22, 2021 at 6:08

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