Hodge theory for Lie algebra (co)homology - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T03:06:01Z http://mathoverflow.net/feeds/question/79022 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/79022/hodge-theory-for-lie-algebra-cohomology Hodge theory for Lie algebra (co)homology robot 2011-10-24T20:49:22Z 2011-10-26T14:57:19Z <p>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: $C^k(\mathfrak{n},V) = C_k(\mathfrak{n^*},V) = \Lambda^k \mathfrak{n^*}\otimes V$.</p> <p>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. </p> <p>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 $H^*(\mathfrak{n},V)$.</p> <p>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: $\delta d u = 0 \implies du=0\quad \&amp;\quad d\delta u = 0 \implies \delta u = 0$.</p> <blockquote> <p><strong>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?</strong></p> </blockquote> <p>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.</p>