It seems I am feeling my way down a blind alley. I'm trying to understand this Heisenberg vertex operator in the concrete case of 3 dimensional Heisenberg group.
It would seem that generally vertex algebra can be defined this way for a polynomial ring.
Given say $V = \mathbb{C}[x_{-1}, x_{-2}, . . . .]$, a vacuum,a translation and vertex operation can be defined.
A simple form for the translation and vertex operation are the below:
data
$V = \mathbb{C}[x_{-1}, x_{-2}, . . . .]$
Let a vacuum exist, $\mathbb{1}$
$T(x_{-n}) =(-n) . x_{-n -1}$
$Y(x,z)$ = ${\Sigma}_n x_n z^{-n -1}$
Then
$a, b \in V $, $a_n b = Y(a,z)b$
This seems to be a simple form of this idea.
If anything I've said above is true, then maybe I can ask my question.
In my own notation (which seems to be natural as I have checked and it is close to what others are using), I am referring to the VOA as (V, 1, T, Y) following a definition I stitched up based on the above.
That is what are V, 1, T, and Y for the Heisenberg vertex operator algebra in the simplest possible form?
The 3-dimensional Heisenberg Lie algebra seems to be formed from its have 3 basis elements.
$X = \pmatrix{ 0 &1&0 \\ 0 &0&0 \\ 0&0&0}$ $Y = \pmatrix{ 0 &0&0 \\ 0 &0&1 \\ 0 &0&0}$ $Z = \pmatrix{ 0 &0&1 \\ 0 &0&0 \\ 0&0&0}$
Page 18 through 20 of this reference construct the vertex operator algebra but it is a bit too abstract and I don't believe I have understood it because I am having trouble translating it.
Also, see page 13 of this
My understanding of the matter is something like this:
Take h, the lie algebra . . Look at the basis and add $h_c$ a central term then this is supposed to be your V (or H in some notation ). Then the 'data' proceed from here.
How does one construct the concrete Heisenberg vertex operator algebra for the case of 3d Heisenberg group and associated lie algebra?