$GL(\infty)$ group action through the boson-fermion correspondence

Question

Every point of the Sato Grassmannian can be used to generate a tau function of the KP hierarchy. In addition, the Sato Grassmannian can be seen as a subset of the "second quantized fermion Fock space" $\mathcal{F}$ of the Clifford algebra. The infinite dimensional group $GL(\infty)$ has a representation on the Fock space $\mathcal{F}$. An element $A_{ij} \in GL(\infty)$ describes an operator, $$U = \exp(\sum_{ij}A_{ij}\psi_i \psi_j^*)$$ that acts transitively on the space of tau functions. The boson-fermion correspondence gives a description of this picture in terms of the Heisenberg algebra and it's Fock space. The precise isomorphism is given by, $$\sigma: \mathcal{F}^{f} \to \mathcal{F}^{b}, \quad \sigma(:\psi(z)\psi^*(z):)\sigma^{-1}=\alpha(z)$$ where $\alpha(z)=\sum \alpha_n z^{-n-1}$ is the Heisenberg field and the superscripts in the domain and range denote the fermionic and bosonic Fock space respectively. $::$ denotes normal ordering of the modes.

Question Can $U$ be written in terms of modes of the Heisenberg algebra $\alpha_n$ using the boson-fermion correspondence? What does this look like?

Answer 1

The full question is beyond my understanding, but let me suggest a substitution to your operator that can in some cases simplify handing the infinite matrices. If you introduce a new index $k$, set $k=i+j$ and consider the sequence $s_k = \sum_{i\in [0,k]}A_{i(k-i)}\psi_i\psi^*_{k-i}$, then each element $s_k$ is a finite partial sum since $k$ is finite and you should be able to decompose $U = \exp(\sum_k s_k)$, which can be easier to manipulate, because you'd only have one dimension where indices go to infinity, and evaluating the partial sums $s_k$ can often utilize symmetries inherent in the matrices.