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Elliptic free-fermion model with OS boundary and elliptic Pfaffians

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Abstract

We introduce and study a class of partition functions of an elliptic free-fermionic face model. We study the partition functions with a triangular boundary using the off-diagonal K-matrix at the boundary (OS boundary), which was introduced by Kuperberg as a class of variants of the domain wall boundary partition functions. We find explicit forms of the partition functions with OS boundary using elliptic Pfaffians. We find two expressions based on two versions of Korepin’s method, and we obtain an identity between two elliptic Pfaffians as a corollary.

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Acknowledgements

The author thanks the referees for careful reading, various invaluable comments and suggestions to improve the paper. This work was partially supported by Grant-in-Aid for Scientific Research (C) No. 18K03205 and No. 16K05468.

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  1. Faculty of Marine Technology, Tokyo University of Marine Science and Technology, Etchujima 2-1-6, Koto-Ku, Tokyo, 135-8533, Japan

    Kohei Motegi

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Appendix: An elementary proof of (4.35) for the case \(n=2\)

Appendix: An elementary proof of (4.35) for the case \(n=2\)

In this Appendix, we check (4.35) for the case \(n=2\) by elementary manipulations. In this case, one can see from the definition of Pfaffians (2.1) that proving (4.35) is equivalent to showing the following identity

$$\begin{aligned}&\frac{[u_2-u_1][u_1+u_2+h][u_4-u_3][u_3+u_4+h]}{[u_2-u_1+1/2][u_4-u_3+1/2]} \nonumber \\&\qquad \times [u_3+u_1+1/2][u_4+u_1+1/2][u_3+u_2+1/2][u_4+u_2+1/2] \nonumber \\&\qquad -\frac{[u_3-u_1][u_1+u_3+h][u_4-u_2][u_2+u_4+h]}{[u_3-u_1+1/2][u_4-u_2+1/2]} \nonumber \\&\qquad \times [u_2+u_1+1/2][u_4+u_1+1/2][u_3+u_2+1/2][u_4+u_3+1/2] \nonumber \\&\qquad +\frac{[u_4-u_1][u_1+u_4+h][u_3-u_2][u_2+u_3+h]}{[u_4-u_1+1/2][u_3-u_2+1/2]} \nonumber \\&\qquad \times [u_2+u_1+1/2][u_3+u_1+1/2][u_4+u_2+1/2][u_4+u_3+1/2] \nonumber \\&\quad =\frac{[u_2-u_1][u_1+u_2+h][u_4-u_3][u_3+u_4+h][u_3+u_1][u_4+u_1][u_3+u_2][u_4+u_2]}{[u_2-u_1+1/2][u_4-u_3+1/2]}\nonumber \\&\qquad -\frac{[u_3-u_1][u_1+u_3+h][u_4-u_2][u_2+u_4+h][u_2+u_1][u_4+u_1][u_3+u_2][u_4+u_3]}{[u_3-u_1+1/2][u_4-u_2+1/2]} \nonumber \\&\qquad +\frac{[u_4-u_1][u_1+u_4+h][u_3-u_2][u_2+u_3+h][u_2+u_1][u_3+u_1][u_4+u_2][u_4+u_3]}{[u_4-u_1+1/2][u_3-u_2+1/2]}. \end{aligned}$$
(A.1)

Let us show this using the addition formula for theta functions (2.8) repeatedly. The difference between the left hand side and the right-hand side of (A.1) can be expressed as

$$\begin{aligned}&\frac{[u_2-u_1][u_1+u_2+h][u_4-u_3][u_3+u_4+h]}{[u_2-u_1+1/2][u_4-u_3+1/2]} \nonumber \\&\quad \times ([u_3+u_1+1/2][u_4+u_1+1/2][u_3+u_2+1/2][u_4+u_2+1/2] \nonumber \\&\quad -[u_3+u_1][u_4+u_1][u_3+u_2][u_4+u_2]) \nonumber \\&\quad -\frac{[u_3-u_1][u_1+u_3+h][u_4-u_2][u_2+u_4+h]}{[u_3-u_1+1/2][u_4-u_2+1/2]} \nonumber \\&\quad \times ([u_2+u_1+1/2][u_4+u_1+1/2][u_3+u_2+1/2][u_4+u_3+1/2] \nonumber \\&\quad -[u_2+u_1][u_4+u_1][u_3+u_2][u_4+u_3]) \nonumber \\&\quad +\frac{[u_4-u_1][u_1+u_4+h][u_3-u_2][u_2+u_3+h]}{[u_4-u_1+1/2][u_3-u_2+1/2]}\nonumber \\&\quad \times ([u_2+u_1+1/2][u_3+u_1+1/2][u_4+u_2+1/2][u_4+u_3+1/2] \nonumber \\&\quad -[u_2+u_1][u_3+u_1][u_4+u_2][u_4+u_3]). \end{aligned}$$
(A.2)

Using the addition formula for theta functions (2.8) (and (2.7)), one finds

$$\begin{aligned}&{[}u_3+u_1+1/2][u_4+u_1+1/2][u_3+u_2+1/2][u_4+u_2+1/2] \nonumber \\&\qquad -[u_3+u_1][u_4+u_1][u_3+u_2][u_4+u_2] \nonumber \\&\quad =[1/2][u_1+u_2+u_3+u_4+1/2][u_2-u_1+1/2][u_4-u_3+1/2], \end{aligned}$$
(A.3)
$$\begin{aligned}&{[}u_2+u_1+1/2][u_4+u_1+1/2][u_3+u_2+1/2][u_4+u_3+1/2] \nonumber \\&\qquad -[u_2+u_1][u_4+u_1][u_3+u_2][u_4+u_3]\nonumber \\&\quad =[1/2][u_1+u_2+u_3+u_4+1/2][u_3-u_1+1/2][u_4-u_2+1/2], \end{aligned}$$
(A.4)
$$\begin{aligned}&{[}u_2+u_1+1/2][u_3+u_1+1/2][u_4+u_2+1/2][u_4+u_3+1/2] \nonumber \\&\qquad -[u_2+u_1][u_3+u_1][u_4+u_2][u_4+u_3] \nonumber \\&\quad =[1/2][u_1+u_2+u_3+u_4+1/2][u_3-u_2+1/2][u_4-u_1+1/2]. \end{aligned}$$
(A.5)

Using identities (A.3), (A.4) and (A.5), (A.2) reduces to

$$\begin{aligned}&\frac{[u_2-u_1][u_1+u_2+h][u_4-u_3][u_3+u_4+h]}{[u_2-u_1+1/2][u_4-u_3+1/2]}\nonumber \\&\qquad \times [1/2][u_1+u_2+u_3+u_4+1/2][u_2-u_1+1/2][u_4-u_3+1/2]\nonumber \\&\qquad -\frac{[u_3-u_1][u_1+u_3+h][u_4-u_2][u_2+u_4+h]}{[u_3-u_1+1/2][u_4-u_2+1/2]}\nonumber \\&\qquad \times [1/2][u_1+u_2+u_3+u_4+1/2][u_3-u_1+1/2][u_4-u_2+1/2]\nonumber \\&\qquad +\frac{[u_4-u_1][u_1+u_4+h][u_3-u_2][u_2+u_3+h]}{[u_4-u_1+1/2][u_3-u_2+1/2]}\nonumber \\&\qquad \times [1/2][u_1+u_2+u_3+u_4+1/2][u_3-u_2+1/2][u_4-u_1+1/2] \nonumber \\&\quad =[1/2][u_1+u_2+u_3+u_4+1/2] ([u_2-u_1][u_1+u_2+h][u_4-u_3][u_3+u_4+h]\nonumber \\&\qquad -[u_3-u_1][u_1+u_3+h][u_4-u_2][u_2+u_4+h]\nonumber \\&\qquad +[u_4-u_1][u_1+u_4+h][u_3-u_2][u_2+u_3+h]). \end{aligned}$$
(A.6)

One can apply addition formula (2.8) again to get

$$\begin{aligned}&{[}u_2-u_1][u_1+u_2+h][u_4-u_3][u_3+u_4+h]\nonumber \\&\quad -[u_3-u_1][u_1+u_3+h][u_4-u_2][u_2+u_4+h] \nonumber \\&\quad +[u_4-u_1][u_1+u_4+h][u_3-u_2][u_2+u_3+h]=0, \end{aligned}$$
(A.7)

and we find the right-hand side of (A.6) becomes zero. Hence (4.35) for the case \(n=2\) is proved.

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Motegi, K. Elliptic free-fermion model with OS boundary and elliptic Pfaffians. Lett Math Phys 109, 923–943 (2019). https://doi.org/10.1007/s11005-018-1130-8

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