Commutators of Schur polynomials of Lie algebra elements

Question

Question:

Is there a well-known formula for computing the commutators of Schur polynomials when the variables are Lie algebra elements? If the algebra has a particularly simple commutation relation, is there a nice way to work this out?

If it helps, the algebra in question is easily treatable as the $x_i$ and $\frac{\partial}{\partial x_i}$ acting on the left on $\mathbb{C}[x_1, x_2, \ldots]$, which is usually the context the Schur polynomials appear in...I have Schur polynomials $f_n$ and $g_n$ in the variables $\frac{\mu}{k} x_k$ and $-\frac{\mu}{k} \frac{\partial}{\partial x_k}$ respectively for some parameter $\mu$, and want to know about $$ [f_m, g_n] $$ in general, and $$ [f_n, g_n] $$ in particular.

Context:

I have an operator $V_\mu(z)$ mapping between `bosonic Fock spaces' $\mathcal{F}_\eta$, the countable-dimensional vector spaces cyclically generated from highest weight vectors $\left| \eta \right>$ by acting freely with the Heisenberg algebra $\mathcal{H} = \mathrm{span}_\mathbb{C}\{\mathbf{1}, \alpha_n: \; n \in \mathbb{Z}\}$. $$ \mathcal{F}_\eta = \mathcal{U}(\mathcal{H}) \cdot \left| \eta \right> $$ where the positive-indexed operators annihilate the highest weight vector; $$ \alpha_n \cdot \left| \eta \right> = 0 \quad \forall n > 0 \\ \alpha_0 \cdot \left| \eta \right> = \eta \left| \eta \right>\\ \mathbf{1} \cdot \left| \eta \right> = \left| \eta \right> $$ and $\mathcal{H}$ has the following well-known commutation relations: $$ [\alpha_m, \alpha_n] = m \delta_{m, -n} \\ [\mathbf{1}, \alpha_n] = 0 \\ $$

$V_\mu(z): \mathcal{F}_\eta \to \mathcal{F}_{\eta + \mu}$ is a screening operator, i.e. $$ V_\mu(z) = e^{\mu q} z^{\mu \eta} \exp\left( \mu \sum_{n > 0}\frac{\alpha_{-n}}{n}z^n \right) \exp\left(-\mu\sum_{n>0}\frac{\alpha_n}{n}z^{-n}\right) $$ which as you can see has exponentials of sums of independent elements, the generating function of the Schur polynomials. We get $$ V_\mu(z) = e^{\mu q} z^{\mu \eta}\sum_{m \in \mathbb{Z}}\sum_{n \in \mathbb{Z}} f_m g_n z^{m-n} $$ where $f_n$ and $g_n$ are the $n$th elementary Schur polynomial in variables $\frac{\mu}{k} \alpha_{-k}$ and $-\frac{\mu}{k}\alpha_k$ respectively.

I am interested in the behaviour of the Fourier modes of this function, taking $$ V_\mu(z) = \sum_{n \in \mathbb{Z}} V_\mu^{(n)} z^{-n-1} $$ and in particular whether they have nice commutation relations.

Answer 1

No, there will be no nice formula for $[V_{\mu}^{(n)},V_{\lambda}^{(m)}]$ in general.

Compute first

$$ V_{\mu}(z_1) V_{\lambda}(z_2)=p(z_1,z_2) : V_{\mu}(z_1)V_{\lambda}(z_2) :$$ $$ V_{\lambda}(z_2) V_{\mu}(z_1)=q(z_1,z_2) : V_{\mu}(z_1)V_{\lambda}(z_2) :$$ where $: \cdot :$ is the normal ordering and $p(z_1,z_2)$ and $q(z_1,z_2)$ are easily computed. Subtract the two expressions and try to simplify the right hand-side. This will be possible when $p(z_1,z_2)-q(z_1,z_2)$ is a derivative of the formal $\delta$-function in $z_1$ and $z_2$.

Here are some special cases:

For $\lambda \cdot \nu \in 2 \mathbb{N}$, your answer should be zero.

For $\lambda \cdot \nu \in - 2\mathbb{N}$, you do get a derivative of the delta function, but the resulting expression is quickly getting very messy.

For $\lambda \cdot \nu \in 2 \mathbb{Z}+1$, the commutator doesn't give a delta function, so it is more natural to consider the anti-bracket. In particular, if $\lambda=1$ and $\mu=-1$ your operators form a pair of free fermions due to boson-fermion correspondence.

For $\lambda \cdot \mu \notin \mathbb{Z}$, the only hope of getting reasonable expressions is by replacing brackets with "generalized brackets", but that's a whole different story ($\leadsto$ parafermions).