Is it the only solution $\sinh(u-\xi+\eta)=\sinh(\eta)\rightarrow \eta=\pm\frac{i\pi}{3} \hspace{0.3cm} \& \hspace{0.3cm} u-\xi = \pm\frac{i\pi}{3}$

Question

I am working on a problem that maps the 6 vertex model in statistical physics to alternating sign matrices. The main idea is that there is a one to one correspondence between alternating sign matrices and the 6 vertex model and one can use the partition function of the 6 vertex model to determine the number of permutations of a sign alternating matrix of size $N\times N$.

I am facing some trouble with the following result(click here to get redirected to the paper) (check chapter 7)

$$\sinh(u-\xi+\eta)=\sinh(\eta)\rightarrow \eta=\pm\frac{i\pi}{3} \hspace{0.3cm} \& \hspace{0.3cm} u-\xi = \pm\frac{i\pi}{3}$$

Here, $ u \hspace{0.2cm} \& \hspace{0.2cm} \xi$, are complex variables and $\eta$ is a complex constant.

I believe that this isn't the only solution possible, for example we can chose $\eta=\pm\frac{i\pi}{4} \hspace{0.3cm} \& \hspace{0.3cm} u-\xi = \pm\frac{i\pi}{2}$.

Therefore I am not sure as to why they take only this solution. (It might be because it is the solution that is the simplest to take) But I can't think of any other reason and I would appreciate some help.

this is where my question ends

More details about the problem :

The initial partition function of the 6 vertex model with domain wall boundary condition takes the following form :

$ K_N (u_1,u_2,\ldots,u_N|\xi_1,\xi_2,\ldots,\xi_N) =\frac{\prod_{j,k}^N \sinh(u_j -\xi_k + \eta)}{\prod_{j<k}^N\sinh(u_j-u_k)\sinh(\xi_k - \xi_j)}\times \det_N \left(\frac{\sinh \eta}{\sinh (u_j-\xi_k)\sinh(u_j -\xi_k +\eta)}\right)$

Small note about the 6 vertex model with domain wall boundary conditions :

Basically, it is the partition function of an $N\times N$ square grid with varying arrows on its edges. You assign to each edge between the vertices an arrow (left or right if you are on a horizontal edge, up or down if you are on a vertical edge). The boundary conditions are such that all the arrows on the upper and lower side of the grid are pointing outwards and the ones on the left and right sides of the grid are pointing inwards. In short this is the 6 vertex model with DW boundary conditions.

The partition function allows us to count how many grids of size $N \times N$ we can construct respecting the boundary conditions. (with an additional imposed condition on the vertices, which is that we can only have vertices with number of incoming arrows = number of outgoing arrows).

To be able to do the mapping to the alternating sign matrices (which are matrices with entries $\pm 1$ or $0$ with 2 conditions, First is that the sum of each column or row should equal 1, second is that the signs of non-zero elements should alternate) we have to set some parameters equal to one. (basically the weights of the 6 allowed vertices (which are 3 sets of 2)

$a=\frac{\sinh(u_j-\xi_k+\eta)}{\sinh(u_j-\xi_k)}=1 \hspace{0.25cm} \& \hspace {0.25cm} c=\frac{\sinh\eta}{\sinh(u_j-\xi_k)}=1$

This lead us to the equality at the beginning of my post

Answer 1

$\sinh$ is $2\pi i$-periodic, and has an additional reflection-symmetry, thus $$\sinh x = \sinh y \quad\Longleftrightarrow\quad (x-y = 2\pi i\, k\text{ or } x+y=(2k+1)\pi i \text{ for some } k\in \Bbb Z)$$ where the "or" is inclusive. The periodicy of $\sinh$ and symmetry can be seen easily by it's connexion with $\sin$: $$\sinh ix = i \sin x\qquad\text{and}\qquad i\sinh x = \sin ix$$ together with the combinations of the following two symmetries of $\sin$: $$\sin x = \sin (x+2\pi)\qquad\text{and}\qquad \sin x = \sin (\pi-x)$$

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Applying this to your equation

$\sinh(u-\xi+\eta)=\sinh\eta \tag 1$

we find that

1. $u-\xi$ is an even multiple of $\pi i$, or
2. $u-\xi+2\eta$ is an odd multiple of $\pi i$.

$\eta=\pm\frac{i\pi}{3} \quad \& \quad u-\xi = \pm\frac{i\pi}{3}$

• The $(+,+)$ case is case 2. with $u-\xi+2\eta=+\pi i$.
• The $(-,-)$ case is case 2. with $u-\xi+2\eta=-\pi i$.
• The opposite sign cases $(+,-)$ and $(-,+)$ match neither case 1. nor case 2.

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As an aside, simlar symmetries hold for $\cosh$ due to $$\cosh ix = \cos x\qquad\text{and}\qquad \cosh x = \cos ix$$ namely $$\cos x = \cos (x+2\pi)\qquad\text{and}\qquad \cos x = \cos (-x)$$ which translates to $\cosh$ as $$\cosh x = \cosh y \quad\Longleftrightarrow\quad (x-y = 2\pi i\, k\text{ or } x+y=2\pi ik \text{ for some } k\in \Bbb Z)$$