**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.



Commutators of Schur polynomials of Lie algebra elements
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). 


