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?

  • $\begingroup$ books.google.de/books/about/… $\endgroup$ – user5831 Dec 26 '19 at 17:10
  • 1
    $\begingroup$ 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! :-) $\endgroup$ – ramanujan_dirac Dec 27 '19 at 3:07

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.