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Let $G$ be a simple Lie group and $P$ its parabolic subgroup such that on the level of Lie algebras we have $\mathfrak{p} = \mathfrak{g}_0 \oplus \mathfrak{n}$. The dual $\mathfrak{n}^{*}$ is identified with the nilradical of the opposite Lie algebra via the Killing form of $\mathfrak{g}$. Now for any representation $V$ of $\mathfrak{g}$ one can define Lie algebra cohomology of $\mathfrak{n}$ with values in $V$ and Lie algebra homology of $\mathfrak{n}^*$ with values in $V$. The cochain and chain spaces are the same: $C^k(\mathfrak{n},V) = C_k(\mathfrak{n}^*,V) = \Lambda^k \mathfrak{n^*}\otimes V$.

All details can be found in the paper "Lie Algebra Cohomology and the Generalized Borel-Weil Theorem" by Kostant, where it is proved that for finite dimensional $V$, the Lie algebra homology differential $\delta$ is adjoint to the Lie algebra codifferential $d$ with respect to some invariant positive definite inner product on (co)chain spaces.

From the fact that these operators are adjoint follows that there is a sort of Hodge theory on the (co)chain complex for the operator $\square = d\delta + \delta d$. In other words, there is a unique representative in each (co)homology class that is harmonic; i.e. it lies in the kernel of $\square$. The induced isomorphism of (co)homology groups with $\ker \square$ lies at the heart of the famous Kostant's theorem on the structure of $H^*(\mathfrak{n},V)$.

One actually doesn't need $d$ and $\delta$ to be adjoint wrt an invariant inner product in order to get such a result. As Kostant actually proved in the paper, it is sufficient to prove that these operators are disjoint: $\delta d u = 0 \implies du=0\quad \&\quad d\delta u = 0 \implies \delta u = 0$.

Let $G$ be a simple complex Lie group and let $P$ be a parabolic subgroup. For which representations $V$ of $(\mathfrak{g},P)$ are the Lie algebra differential and codifferential acting on $C(\mathfrak{n},V)$ disjoint?

Remark: (Co)Chain complexes with values in Segal-Shale-Weil representation are an example that these operators are not disjoint in general. This in particular means that there is no $\mathfrak{sp}$-invariant Hermitian product on the Segal-Shale-Weil representation.

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    $\begingroup$ What is your choice for the identification of $C_*(n,V)$ with $C^*(n,V)$? $\endgroup$ – Vladimir Dotsenko Oct 25 '11 at 23:52
  • $\begingroup$ I am sorry for the confusion. The Lie algebra (co)differential is for $C^*(\mathfrak{n},V)$ while the Lie algebra differential is for $C_*(\mathfrak{n^*},V)$. $\endgroup$ – Vít Tuček Oct 26 '11 at 12:47
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    $\begingroup$ OK, now it has just got worse. Before it was $C_*(n,V)$ and $C^*(n,V)$, the Chevalley--Eilenberg complexes computing the respective (co)homology. I am afraid I do not understand the notation $C_*(n^*,V)$. What I am asking, however, is that you obviously want to identify those two as vector spaces, since for $d$ and $\delta$ to be disjoint they should at least act on the same space. My question is, what is that identification? $\endgroup$ – Vladimir Dotsenko Oct 26 '11 at 14:20
  • $\begingroup$ I've tried to address your comments in the question. I hope it is clear now. $\endgroup$ – Vít Tuček Oct 26 '11 at 15:01
  • $\begingroup$ @Vladimir: I think that r0b0t is using doing the following. By $\mathfrak n$ I understand to mean the Lie algebra of "upper triangular" matrices. Then $\mathfrak n^*$ denotes the Lie algebra of "lower triangular" matrices, and the two Lie algebras are in duality via the Killing form. This is part of a general story. The identification $\mathfrak n^*=$lower triangulars induces on $\mathfrak n$ the structure of a Lie bialgebra, which is (more or less) equivalent to using the identification $C_\bullet(\mathfrak n)=C^\bullet(\mathfrak n^*)$ to make it into a BV algebra. $\endgroup$ – Theo Johnson-Freyd Oct 26 '11 at 16:20

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