Apologies for the lack of posts here lately; I’ve been meaning to say many things that I simply have not gotten around to doing. I’ve been taking a course on infinite-dimensional Lie algebras this semester. There are a number of important examples here, most of which I’ve never seen before. This post will set down two of the most basic.

1. The Heisenberg algebra

The Heisenberg Lie algebra ${\mathcal{A}}$ is the Lie algebra with generators ${\left\{a_j, j \in \mathbb{Z}\right\}}$ and another generator ${K}$. The commutation relations are

$\displaystyle [a_j, a_k] = \begin{cases} 0 & \text{ if } j + k \neq 0 \\ j K & \text{ if } j = -k . \end{cases} ,$

and we require ${K}$ to be central. This is a graded Lie algebra with ${a_j}$ in degree ${j}$ and ${K}$ in degree zero.

The Heisenberg algebra is a simple example of a nilpotent Lie algebra: in fact, it has the property that its center contains the commutator subalgebra ${\mathbb{C} K}$.

The factor of ${j}$ in the relation for ${[a_j, a_k]}$ is of course a moot point, as we could choose a different basis so that the relation read ${[a_j, a_k] = \delta_{j, -k}}$. (The exception is ${a_0}$: that has to stay central.) However, there is a geometric interpretation of ${\mathcal{A}}$ with the current normalization. We have

$\displaystyle [a_j, a_k] = \mathrm{Res}( t^j d t^k)_{t = 0} K.$

Here ${\mathrm{Res}}$ denotes the residue of the differential form ${t^j dt^k = k t^j t^{k-1} dt}$ at ${t = 0}$. Since only terms of the form ${t^{-1} dt}$ contribute to the residue, this is easy to check.

As a result, we can think of as the Lie algebra of Laurent polynomials plus a one-dimensional space:

$\displaystyle \mathcal{A}= \mathbb{C}[t, t^{-1}] \oplus \mathbb{C}K$

where ${K}$ is central, and where the Lie bracket of Laurent polynomials ${f, g }$ is

$\displaystyle \mathrm{Res}_{t =0 } (f dg) K.$

Note that any exact form has residue zero, so ${\mathrm{Res}_{t = 0}(fdg) = -\mathrm{Res}_{t=0}(g df)}$ (by comparing with ${d(gf)}$). This explains the antisymmetry of the above form. (more…)