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

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?

• books.google.de/books/about/… – user5831 Dec 26 '19 at 17:10
• Thanks! I think Theorem 6.1 of the above mentioned references answers my question. I might write up a short summary as an answer to my own question then! :-) – ramanujan_dirac Dec 27 '19 at 3:07

## 1 Answer

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.