The determinant identity for the half-space and full-space TASEP kernels

From papers

Let K1K_1 be the product of the first three matrices in the displayed factorization of the TASEP kernel, and let K2K_2 be the operator defined by

K2:=(b10b1b10b1b110b0b110b).K_2:=\begin{pmatrix} {\texttt{b}_*}1_0{\texttt{b}_*}^{-1} & {\texttt{b}_*}1_0{\texttt{b}}^{-1}\partial {\texttt{b}_*}^{-1}1_0{\texttt{b}}\\ 0 & {\texttt{b}}^{-1}1_0{\texttt{b}} \end{pmatrix}.

Here the operators act on the indicated L2L^2 spaces. The determinant identity conjecture. Due to the special structure of b{\texttt{b}_*} and b{\texttt{b}}, one should have

det(I+12K1)L2([0,))×L2([0,))=det(I+K1K2)L2(R)×L2(R).\det\left(I+\tfrac{1}{2}K_1\right)_{{L^2([0,\infty))}\times {L^2([0,\infty))}}=\det\left(I+K_1K_2\right)_{L^2(\mathbb{R})\times L^2(\mathbb{R})}.

This identity would relate the half-space kernel to the full-space TASEP formula. The preceding transformation is only formal because some intermediate operators are not well defined, although the resulting kernel makes sense; the conjectured determinant equality is not resolved in the supplied text.

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Sources & referencesView supporting material

Primary source

Xincheng Zhang, “TASEP in half-space”, arXiv:2409.09974 (2025).

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