9
$\begingroup$

There is an exact sequence

$$0 \to H^2(\mathfrak{g}, k) \to H^1(\mathfrak{g}, \mathfrak{g}^*) \to H^0(\mathfrak{g}, S^2\mathfrak{g}) \xrightarrow{d} H^3(\mathfrak{g}, k) \to H^2(\mathfrak{g}, \mathfrak{g}^*) \to H^1(\mathfrak{g}, S^2\mathfrak{g}),$$

where $\mathfrak{g}$ is a Lie algebra over a field $k$ and $H^i(\mathfrak{g}, M)$ are Lie algebra cohomology. It is interesting because of a map $d$, a Koszul homomorphism, which sends an invariant symmetric bilinear form $\left< \cdot, \cdot \right>$ to a 3-cocycle $\left<[\cdot, \cdot], \cdot\right>$ and is an isomorphism for a semi-simple Lie algebra.

The question is whether it is a part (namely, lower-degree terms) of some spectral sequence? If not, what is a natural way to obtain it? An unclear proof may be found in Neeb, Wagemann The second cohomology of current algebras of general Lie algebras, proposition 7.2.

$\endgroup$
2
  • $\begingroup$ You mean $H^0(g,S^2g)$ and $H^1(g,S^2g)$, right? Otherwise I don't know how to interpret this. $\endgroup$ Commented Feb 21, 2017 at 17:49
  • $\begingroup$ If so, one would expect it to arise from en.wikipedia.org/wiki/Five-term_exact_sequence, but the indicies don't quite line up $\endgroup$
    – Drew Heard
    Commented Feb 21, 2017 at 18:01

1 Answer 1

5
$\begingroup$

It's part of the Pirashvili exact sequence, relating Lie algebra and Leibniz cohomologies of Lie algebras. This is discussed (in homology terms) in the end of p2 of this paper of mine on Koszul's homomorphism (arxiv link). Pirashvili's paper is freely accessible here on Numdam; it's also written in terms of homology but the cohomological statement follows.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .