text:
\subsection{Proof of theorem~\ref{thm:sym}} \label{appsec:thmproof}

\begin{proof}
The proof of theorem~\ref{thm:sym} proceeds in two steps.  In the first step, one studies the conditional distribution of $\bZ$ given the  output of the algorithm up until iteration $t$, treating $\bZ$ as random and the output as deterministic.  In the non-symmetric AMP studied in \cite[Theorem 1]{RushVenkataramanan}, the relevant measurement matrix has i.i.d.\ gaussian entries and this conditional distribution was  originally studied in \cite{bayati2011dynamics}.  The result for the case of i.i.d.\ gaussian $\bZ$ is concisely stated in \cite[ Lemma 4.2]{RushVenkataramanan}. For the symmetric AMP of \eqref{eq:sym_AMP} that we are interested in, the matrix $\bZ$ is $\textsf{GOE}(n)$ and so this conditioning argument needs to take into account the symmetry of the matrix entries (and consequently the added dependencies).  This has been studied in other works that give asymptotic characterizations of the performance of symmetric AMP, for example in \cite[Lemma 3]{Montanari-Javanmard}, and these results apply directly to our case since this distributional characterization is already non-asymptotic and does not change in our setting.  This then allows us to characterize the conditional distribution of the iterates $\bh^{t+1}$, conditional on the previous output of the algorithm. We give this result in Lemma~\ref{lem:hb_cond} below, but before stating the lemma, we introduce some useful notation.

\newcommand{\norm}[1]{\lVert#1\rVert}
\newcommand{\mscrs}{\mathscr{S}}
\newcommand{\bH}{\mathbf{H}}
\newcommand{\bDel}{\boldsymbol{\Delta}}
\newcommand{\bigM}{\mathbf{M}}

First, denote $\mathbf{m}^0 := g_0(\bh^0, \bX^n), ..., \mathbf{m}^t := g_t(\bh^t, \bX^n)$ where the terms $g_t(\bh^t, \bX^n)$ are those used in the symmetric AMP in \eqref{eq:sym_AMP}.  Then we define $\mathscr{S}_{0}$ to be the sigma-algebra generated by $\{g_0(\bh^0, \bX^n), \bX^n\}$ and
$\mathscr{S}_{t}$ for $ t\geq 1$ to be the sigma-algebra generated by
\begin{equation*}
\bh^1, ..., \bh^{t}, \mathbf{m}^0, ...,  \mathbf{m}^t, \text{ and }  \bX^n.
\end{equation*}
Using \cite[Lemma 3]{Montanari-Javanmard} to characterize the distribution of $\bZ$ conditioned on the sigma algebra $\mathscr{S}_{t}$, we are able to specify the conditional distributions of $\bh^{t+1}$ given $\mathscr{S}_{t}$, by observing that conditioning on $\mscrs_{t}$ for $t \geq 1$ is equivalent to conditioning on the linear constraint\footnote{While conditioning on the linear constraints, we emphasize that only $\bA$ is treated as random.}
\begin{equation*}
\bZ \bigM_{t} = \bY_{t},
\end{equation*}
where $\bigM_t \in \mathbb{R}^{n \times t}$ and $\bH_t \in \mathbb{R}^{n \times t}$ are the matrices
\begin{equation*}
\bigM_{t} = [\mathbf{m}^0  \mid  ...  \mid  \mathbf{m}^{t-1}] \quad \text { and } \quad \bH_{t} = [\bh^1  \mid  ...  \mid  \bh^{t}],
\end{equation*}
and $\bY_t \in \mathbb{R}^{n \times t}$ is the matrix $\bY_{1} = \bH_1$ and $\bY_{t} = \bH_t + [\mathbf{0} |\bigM_{t-1}] \boldsymbol{\textsf{C}_t}$ for $t \geq 2$,  where
$ \boldsymbol{\textsf{C}_t} =  \text{diag}(\textsf{c}_0, ..., \textsf{c}_{t-1})$.
Note that $[\bc_1 \mid \bc_2 \mid ... \mid \bc_k]$ denotes a matrix with columns $\bc_1, ..., \bc_k$.

We use the notation $\mathbf{m}^{t+1}_{\|}$ to denote the projection of $\mathbf{m}^{t+1}$ onto the column space of $\bigM_{t+1}$. Let
 \begin{equation}
 \boldsymbol{\alpha}^{t+1} := (\alpha^{t+1}_0, \alpha^{t+1}_1, \ldots, \alpha^{t+1}_{t})^\intercal \in \mathbb{R}^{t+1},
 \label{eq:vec_alph_gam_conc}
\end{equation}
 be the coefficient vectors of these projections, i.e.,

$\mathbf{m}^{t+1}_{\|} := \sum_{i=0}^{t} \alpha^t_i \mathbf{m}^i$,

meaning $ \boldsymbol{\alpha}^t = (\bigM_{t+1}^\intercal \bigM_{t+1})^{-1}\bigM_{t+1}^\intercal \mathbf{m}^{t+1}.$
 The projections of $\mathbf{m}^{t+1}$onto the orthogonal complement of $\bigM_{t+1}$, is denoted by

$\mathbf{m}^{t+1}_{\perp} := \mathbf{m}^{t+1} - \mathbf{m}^{t+1}_{\|}.$

 Lemma \ref{lem:main_lem}  shows that for large $n$, the entries of $\boldsymbol{\alpha}$  concentrate around constants. In what follows we show that,
 for $t \geq 0$, the vector $ \boldsymbol{\alpha}^{t+1} \in \mathbb{R}^{t+1}$ in \eqref{eq:vec_alph_gam_conc} concentrates to the vector
 \begin{equation}
\hat{\boldsymbol{\alpha}}^{t+1} := \left [ 0 , \ldots , 0 , \frac{\sigma^n_{t+2}}{\sigma^n_{t+1}}\right]^\intercal \in \mathbb{R}^{t+1},
 \label{eq:hatalph_hatgam_def}
\end{equation}
for the state evolution values given in \eqref{eq:symm_SE}.  Similarly,  Lemma \ref{lem:main_lem} will show that for large $n$, the norm $\|\mathbf{m}^{t-1}_{\perp} \|^2/n$ concentrates to a constant $\sigma_{t}^{\perp}$,  defined as $\sigma_1^{\perp} = \sigma_1^n$, and for $t \geq 2,$

 \begin{equation}
\sigma_t^{\perp} := \sigma_t^n \Big(1 - \frac{\sigma_t^n }{\sigma_{t-1}^n}\Big).
 \label{eq:sigperp_defs}
\end{equation}

With the above notation, we find the following result for the symmetric AMP in \eqref{eq:sym_AMP}.

\begin{lemma}[Conditional Distribution Lemma]
For the vectors $\bh^{t+1}$ defined in \eqref{eq:sym_AMP}, the following hold for $t \geq 1$, provided $n >t$ and $\bigM_{t}^\intercal \bigM_t$ has full column rank.

 \begin{equation}
 \begin{split}
 \bh^{1}  \lvert_{\mscrs_{0}}   \stackrel{d}{=}  \sqrt{\sigma^n_1}  \, \bU_0 + \bDel_{0}, \qquad \text{ and } \qquad  \bh^{t+1} \lvert_{\mscrs_{t}} &  \stackrel{d}{=} \hat{\alpha}^{t+1}_t  \, \bh^{t} + \sqrt{\sigma^{\perp}_{t+1}} \, \bU_t + \bDel_{t}, \label{eq:Ha_dist}
 \end{split}
\end{equation}

where $\bU_0, \bU_t \in \mathbb{R}^n$ are random vectors with elements that are marginally standard gaussian random variables that are independent of the corresponding conditioning sigma-algebras. The terms $\hat{\alpha}^t_{i}$ for $i \in \{0,1,..., t\}$ are defined in \eqref{eq:hatalph_hatgam_def} and the terms $\sigma_{t}^{\perp}$ in \eqref{eq:sigperp_defs}.  The deviation terms are $\bDel_{t} = \mathbf{0}$ and for $t >0$,
\begin{align}
&\bDel_{t} =   \sum_{r=1}^{t} (\alpha^{t+1}_r - \hat{\alpha}^{t+1}_r)\bh^r   +   \Big[ \Big( \frac{\| \mathbf{m}^t_{\perp} \|}{\sqrt{n}} - \sqrt{\sigma_{t+1}^{\perp}}\Big)\mathsf{I}  -  \frac{\| \mathbf{m}^t_{\perp} \|}{\sqrt{n}}\mathsf{P}^{\parallel}_{\bigM_{t}} \Big]\bU_t  \nonumber  \\
&+     \bigM_{t-1} (\bigM_{t-1}^\intercal\bigM_{t-1})^{-1} \Big[\bH_{t-1}^\intercal   \mathbf{m}^t_{\perp} -  \bigM_{t-1}^*\Big(\textsf{c}_{t} \mathbf{m}^{t-1} - \sum_{i=1}^{t-1} \textsf{c}_{i} \alpha^t_{i} \mathbf{m}^{i-1} \Big)\Big].\label{eq:Dt1t}
\end{align}
\label{lem:hb_cond}
\end{lemma}

The second step of the proof is inductive on the iteration $t$, showing that if the result in \eqref{eq:sym_bound} holds up to iteration $t-1$  then it will hold at iteration $t$ as well. This is done by showing that the standardized $\ell_2$ norms of the terms in the AMP algorithm in \eqref{eq:sym_AMP}, like $\frac{1}{n}\|\bh^t\|^2$ and $\frac{1}{n}\|\mathbf{m}^t\|^2$, concentrate on deterministic values predicted by the state evolution. This is done by relating these iteration $t$ values to the iteration $t-1$ values through the iteration in \eqref{eq:sym_bound} and appealing to the conditional distributions from Lemma~\ref{lem:hb_cond} and your inductive hypothesis.  While the proof for the symmetric AMP is largely similar to that for the non-symmetric version, there are additional challenges due to the dependencies created by the symmetry of the $\textsf{GOE}(n)$ matrix.  The details of the inductive proof are quite technical and long and are therefore not included here but we state the result for the symmetric AMP in \eqref{eq:sym_AMP}.

For $t \geq 0$, let $\kappa_{-1} = K_{-1} = 1$, and
\begin{equation}
\begin{split}

K_t = C (t+1)^5K_{t-1},  &\quad  \kappa_t  = \frac{\kappa_{t-1}}{  c(t+1)^{11}},
\end{split}
\label{eq:Kkappa_def}
\end{equation}
where $C, c > 0$ are universal constants (not depending on $t$, $n$, or $\e$).  To keep the notation compact,  we  use $K, \kappa, \kappa'$ to denote generic positive universal constants whose values may change through  the  lemma statement.

The result of theorem~\ref{thm:sym} follows from lemma~\ref{lem:main_lem} result \eqref{eq:Hb1} below.

\begin{lemma}
The following statements hold for $1 \leq t < T^*$ and $\epsilon\in (0,1)$. Define
\begin{equation}
\gamma_n^{t+1} := (\nu^n +  \sigma^n_{1})  (\nu^n +  \sigma^n_{1} +  \sigma^n_{2})  \cdots  (\nu^n + \sum_{i=1}^{t+1} \sigma^n_{i}) \times \max\{1, \hat{\textsf{c}}_1\}  \max\{1, \hat{\textsf{c}}_2\} \cdots  \max\{1, \hat{\textsf{c}}_t\},
\end{equation}
where $\nu$ is the variance factor of sub-gaussian $\bX^n$ which equals $\kappa \rho_n$ for $p_{X, n}$ Bernoulli.
\begin{enumerate}
\item
\begin{equation}
\mathbb{P}\Big(\frac{1}{n}\norm{\bDel_{{t}}}^2 \geq \epsilon\Big) \leq  K (t+1)^2 K_{t-1} \exp\Big\{-\frac{\kappa \kappa_{t-1} n \epsilon}{ (t+1)^4  L_{g}^{4t}  \gamma_n^{t}  \max\{1, \hat{\textsf{c}}_t\} }\Big\}. \label{eq:Ba}
\end{equation}

\item
Denote $\mathbb{E}_{\phi} :=  \mathbb{E}\, \phi_h(\sqrt{\sigma^n_{1}} \tilde{Z}_{1}, \ldots ,\sqrt{\sigma^n_{t+1}} \tilde{Z}_{t+1}, X^n)$.  Then for pseudo-Lipschitz function $\phi: \mathbb{R}^{t+1} \rightarrow \mathbb{R}$ with constant $L_{\phi}$ we have that
\begin{equation}
\mathbb{P}\Big(\Big \lvert \frac{1}{n}\sum_{i=1}^n \phi(h^1_i, \ldots,  h^{t+1}_i, X^n_{i}) - \mathbb{E}_{\phi} \Big \lvert \geq \ep\Big) \leq K (t+1)^3 K_{t-1} \exp\Big\{-\frac{ \kappa \kappa_{t-1} n \epsilon^2}{(t+1)^7 L_{\phi}^{2} L_{g}^{4t}  \gamma_n^{t+1} }\Big\}.
  \label{eq:Hb1}
\end{equation}
The random variables $\tilde{Z}_{0}, \ldots, \tilde{Z}_t$ are jointly gaussian with zero mean and covariance given by $\mathbb{E}[\tilde{Z}_r \tilde{Z}_t] = \sqrt{{\sigma^n_t}/{\sigma^n_r}}$ for $r < t$, and are independent of $\bX^n \sim p_{X, n}$.

Denote $\mathbb{E}_{\phi_{Lip}} :=  \mathbb{E}\, \phi_{Lip}(\sqrt{\sigma^n_{1}} \tilde{Z}_{1}, \ldots ,\sqrt{\sigma^n_{t+1}} \tilde{Z}_{t+1}, X^n)$.  Then for Lipschitz function $\phi_{Lip}: \mathbb{R}^{t+1} \rightarrow \mathbb{R}$ with constant $L_{\phi_{Lip}}$ we have that
\begin{align}
&\mathbb{P}\Big(\Big \lvert \frac{1}{n}\sum_{i=1}^n m^0_i \phi_{Lip}(h^1_i, \ldots,  h^{t+1}_i, X^n_{i}) - \widetilde{\sigma} \mathbb{E}_{\phi_{Lip}}\Big \lvert \geq \ep\Big) \nonumber\\
&\qquad \leq  K  (t+1)^3 K_{t-1} \exp\Big\{-\frac{ \kappa \kappa_{t-1} n \epsilon^2}{ (t+1)^7 L_{\phi}^{2} L_{g}^{4t}  \gamma_n^{t+1} }\Big\}.
 \label{eq:Bb2}
\end{align}
The random variables $\tilde{Z}_{0}, \ldots, \tilde{Z}_t$ are as above.

\item Let $L_g > 0$ be the pseudo-Lipschitz constant for the denoiser functions $\{g_t\}_{ t\geq 0}$ and let $X_n \overset{\mathbf{. .}}{=} c$ be shorthand for $$\mathbb{P}(\abs{X_n -c} \geq \epsilon) \leq K (t+1)^3 K_{t-1} \exp\Big\{-\frac{ \kappa \kappa_{t-1} n \epsilon^2}{(t+1)^7 L_{g}^{4t+2} \gamma_n^{t+1} }\Big\}.$$
For all $0 \leq r \leq t$,
\begin{align*}
&\mathbb{P}\Big(\Big \lvert \frac{1}{n}(\bh^{r+1})^* \bh^{t+1}- \sigma^n_{t+1} \Big \lvert \geq \epsilon \Big) \leq K (t+1)^3 K_{t-1} \exp\Big\{-\frac{ \kappa \kappa_{t-1} n \epsilon^2}{(t+1)^7L_{g}^{4t} \gamma_n^{t+1} }\Big\}. \\
& \frac{1}{n}(\mathbf{m}^{0})^* \mathbf{m}^{t+1} \overset{\mathbf{. .}}{=} \widetilde{\sigma}^n \mathbb{E}[g_{t+1}(\sqrt{\sigma^n_{t+1}} \tilde{Z}_{t+1}, X^n)],  \quad

\frac{1}{n}(\mathbf{m}^{r+1})^*\mathbf{m}^{t+1}  \overset{\mathbf{. .}}{=} \sigma^n_{t+2}. \\
&\mathbb{P}\Big(\Big \lvert \textsf{c}_{t+1} - \hat{\textsf{c}}_{t+1} \Big \lvert \geq \ep\Big) \leq K (t+1)^3 K_{t-1} \exp\Big\{-\frac{ \kappa \kappa_{t-1} n \epsilon^2}{(t+1)^7 L_{g}^{4(t+1)} \gamma_n^{t+1} }\Big\}. \\%  \label{eq:Hf1}  \\
& \frac{1}{n}(\bh^{t+1})^*\mathbf{m}^{r+1}  \overset{\mathbf{. .}}{=} \hat{\textsf{c}}_{r+1}  \sigma^n_{t+2}, \qquad

 \frac{1}{n}(\bh^{r+1})^*\mathbf{m}^{t+1} \overset{\mathbf{. .}}{=}  \hat{\textsf{c}}_{t+1} \sigma^n_{t+2}.
\end{align*}

\item
\begin{equation}

 \mathbb{P}\left( \frac{1}{n} \bigM_{t+1}^* \bigM_{t+1} \text{  is singular}\right) \leq (t+1) K_{t-1} \exp\Big\{-\frac{\kappa_{t-1} \kappa n}{{(t+1)^7 L_{g}^{4t+2}\gamma_n^{t+1} }}\Big\}. \label{eq:Msing}
\end{equation}
For $\hat{\alpha}^{t+1}$ defined in \eqref{eq:hatalph_hatgam_def}, when the inverse of $\frac{1}{n} \bigM_{t+1}^* \bigM_{t+1} $ exists, for $1\leq i, j \leq t+1$,
\begin{equation}
\begin{split}
& \mathbb{P}\Big( \Big \lvert \Big[(\frac{1}{n} \bigM_{t+1}^* \bigM_{t+1})^{-1} - (\mathbf{C}^{t+1})^{-1} \Big]_{i,j} \Big \lvert \geq \epsilon \Big) \leq K  K_{t-1} \exp\Big\{-\frac{\kappa \kappa_{t-1}n\epsilon^2}{L_{g}^{4t+2} \gamma_n^{t+1} } \Big\},  \\
&  \mathbb{P}\Big( \lvert\alpha^{t+1}_{i-1} - \hat{\alpha}^{t+1}_{i-1} \lvert \geq \epsilon\Big) \leq K (t+1)^4 K_{t-1}   \exp\Big\{-\frac{\kappa \kappa_{t-1}n\epsilon^2}{ (t+1)^9 L_{g}^{4(t+1)} \gamma_n^{t+1} } \Big\}.
\end{split}
\label{eq:Hg}
\end{equation}
In the above, the matrix $\mathbf{C}^{t+1} \in \mathbb{R}^{(t+1) \times (t+1)}$ has elements
$[\mathbf{C}^{t+1}]_{i,j} =  \sigma^n_{\max\{i,j\}}$ for $1 \leq i,j \leq t+1$.

\item With $\sigma_{t}^{\perp}$ defined in \eqref{eq:sigperp_defs},
\begin{equation}
\mathbb{P}\Big( \Big \lvert\frac{1}{n}\norm{\mathbf{m}^{t+1}_{\perp}}^2 - \sigma_{t+2}^{\perp} \Big \lvert \geq \epsilon\Big) \leq K (t+1)^5 K_{t-1}   \exp\Big\{-\frac{\kappa \kappa_{t-1}n\epsilon^2}{ (t+1)^{11} L_{g}^{4(t+1)} \gamma_n^{t+1} } \Big\}. \label{eq:Hh}
\end{equation}

\end{enumerate}
\label{lem:main_lem}
\end{lemma}

\end{proof}


TITLE: What is the Fixed Subfield of this Group?
QUESTION [4 upvotes]: Suppose $\mathbb{K}$ is a field of characteristic zero. Let $G$ be the subgroup of the Galois group of $\mathbb{K} \subset \mathbb{K}(x)$ (the field of rational functions over $\mathbb{K}$ with indeterminate $x$) generated by the automorphism $x \rightarrow x+1$. Find the fixed subfield of $\mathbb{K}(x)$ corresponding to $G$.
My attempt: During the test, I went on a tangent trying to show that the fixed subfield is $\mathbb{K}$ because a rational function being periodic seemed weird to me; however I could not prove it and I'm sure the Galois group isn't generated by $x \rightarrow x+1$. Alternatively, I tried to find a rational function that is fixed by this automorphism to no avail.
Context: This kind of question was on a HW assignment and a test in my Algebra course (that's an indicator it might be on the Qualifying Exam). Any help would be appreciated, Thank you.
$\underline{Edit:}$ I'm seeing answers that result in the fixed field being $\mathbb{K}$ (as I had guessed). Does this imply that the $Gal(\mathbb{K}(x) / \mathbb{K}) \cong <\beta >$ by Gaois Correspondence? Where $\beta$ takes $x \rightarrow x+1$.

REPLY [2 votes]: Let $\Phi(x)\in \mathbb K(x)$ denote a rational function invariant under $x \mapsto x+1$.  Choose $a\in \mathbb K$ for which $\Phi(a)$ exists (always possible since $\mathbb K$ is infinite) and let $k=\Phi(a)$ and let $\Psi(x)=\Phi(x)-k$.
Then $\Psi(x)\in \mathbb K(x)$ has infinitely many zeroes, as $\Psi(a+n)=0\;\forall n\in \mathbb N$.  It follows that the numerator of $\Psi(x)$ is identically $0$ which implies that $\Phi(x)=k \;\forall x\in \mathbb K$.
(Note:  characteristic $0$ was needed to get infinitely many zeroes.  If the characteristic were $p$ then this argument would only yield $p$ zeroes.)


TITLE: Consider the Fredholm equation of second kind where K satisfies a Lipschitz condition...
QUESTION [1 upvotes]: Prove that there exists a unique solution for all $|\lambda|$ $\lt$ $\frac{1} {M(b-a)}$
Where $f(x)=\phi(x)+ \lambda  \int_{a}^{b}K(x,y;f(y)) dy$
And K satisfies a Lipschitz condition of the form $|K(x,y;r)-K(x,y;s)|\le M|r-s|$
And moreover what would the successive approximations to this solution be given by?
I have spent all day researching and attempting to do this however I am stumped.
Any help greatly appreciated.

REPLY [2 votes]: Consider the Banach space $(C[a,b], ||*||)$, where $||*||$ is the $\sup$ -norm and let $T:C[a,b] \to C[a,b]$ be defined by
$(Tf)(x)=\phi(x)+ \lambda  \int_{a}^{b}K(x,y;f(y)) dy$.
Its your turm to show, that
$||Tf-Tg|| \le \lambda M (b-a) ||f-g||$ for all $f,g \in C[a,b]$.
Hence, if $\frac{1} {M(b-a)}$, Banach's fix point theorem applies !


TITLE: Need explanation on asymptotic running time results for various functions
QUESTION [0 upvotes]: I did not understand few results from the book problem.
Here is the problem:
Indicate, for each pair of expressions (A, B) in the table below, whether A is
O, o, Ω, ω, Θ of B. Assume that k ≥ 1,  > 0, and c > 1 are constants.
And solution:

I am not sure

why c. has all NO. How is this can even be the case?
why e. has YES for Ω and Θ . Asymptotically $c^{lg(n)}$ grow faster than $n^{lg(c)}$, right? Why $n^{lg(c)}$ is not in o($c^{lg(n)}$) ?
why f. has YES for  Ω and  Θ . It also seems that asymptotacally $lg(n^n)$ grows faster than $lg(n!)$ . Why $lg(n!)$ is not in o($lg(n^n)$) ?

REPLY [1 votes]: For c: due to $\sin n$ this function is unbounded, so there doesn't exist $c$ such that any of the order functions hold
For e: consider taking $n = e^t$
For f: use Stirling's formula or Euler-Maclaurin for $\sum_k \log k$, these functions are of the same order.


text:
\begin{document}

\pagestyle{empty}
\begin{flushright}
LAPTH-043/14
\end{flushright}

\vspace{20pt}

\begin{center}
\begin{LARGE}
{\bf $R$ matrices of three-state Hamiltonians\\[1.2ex] solvable by Coordinate Bethe Ansatz}
\end{LARGE}

\vspace{50pt}

\begin{large}

{T. Fonseca${}^{a}$, L. Frappat${}^a$, E. Ragoucy${}^a$ \footnote[1]{tiago.dinis@lapth.cnrs.fr, luc.frappat@lapth.cnrs.fr, eric.ragoucy@lapth.cnrs.fr}}

\end{large}

\vspace{15mm}

${}^a$  {\it Laboratoire de Physique Th\'eorique LAPTh, CNRS and Universit\'e de Savoie,\\
BP 110, 74941 Annecy-le-Vieux Cedex, France}

\end{center}

\vfill

\begin{abstract}
We review some of the strategies that can be implemented to infer an $R$-matrix from the knowledge of its Hamiltonian.
We apply them to the classification achieved in \texttt{arXiv:1306.6303}, on three state $U(1)$-invariant Hamiltonians solvable by CBA,
focusing on models for which the $S$-matrix is not trivial.

For the 19-vertex solutions, we recover the $R$-matrices of the well-known Zamolodchikov--Fateev  and Izergin--Korepin models. We point out that the generalized Bariev Hamiltonian is related to both main and special branches studied by Martins in
\texttt{arXiv:1303.4010}, that we prove to generate the same Hamiltonian.
The 19-vertex SpR model still resists to the analysis, although we are able to state some no-go theorems on its $R$-matrix.

For 17-vertex Hamiltonians, we  produce a new $R$-matrix.
\end{abstract}

\vfill

\newpage
\pagestyle{plain}
\setcounter{page}{1}

\section{Introduction}

In his pioneering work \cite{Bethe}, Hans Bethe introduced the now called ``coordinate Bethe ansatz'' (CBA), allowing him to solve the Heisenberg model \cite{Heis}, i.e. to determine the eigenvalues and eigenvectors of the corresponding Hamiltonian. In the case of the one-dimensional quantum many-body problem with repulsive delta-function interaction, similar considerations led to the introduction of a new equation \cite{McGuire,Yang} that appears as a consistency condition for the factorization of the scattering matrix. It also showed up in Baxter's resolution of the eight vertex model \cite{Bax72}. This equation is now known as the Yang--Baxter equation (YBE) \cite{Yang,Baxt}.

A decade later, a crucial breakthrough arose from the works of the Leningrad school \cite{FST,FT}, see also \cite{KulSklya} and references therein. A new and more algebraic approach called quantum inverse scattering method (QISM) was developed, which constitutes nowadays the well-established framework for the study of quantum integrable systems. A particular object highlighted in this context is the quantum $R$-matrix \cite{D1,D2,J86}, satisfying the Yang--Baxter equation, see e.g. \cite{KuReSk,KuSkl}. Beyond the mathematical playground opened by the existence and properties of this key object, it has become the cornerstone of the study of quantum integrable systems. Indeed, in the context of one-dimensional spin chains, $R$-matrices with spectral parameters constitute the basic ingredient for constructing the monodromy matrix. It allows one also to define the transfer matrix and opens to the Algebraic Bethe Ansatz (ABA) \cite{FT}. The Yang--Baxter equation ensures the transfer matrices to commute for different values of the spectral parameter. The transfer matrix therefore encodes the conserved commuting quantities of the system, and among them the Hamiltonian, usually defined as the logarithmic derivative of the transfer matrix at a specific value of the spectral parameter.

A Hamiltonian being given, it is therefore of great importance to determine whether there exists an $R$-matrix from which this Hamiltonian can be generated.
Unfortunately, there is no systematic procedure to induce the form of the $R$-matrix once an integrable Hamiltonian is given.
In this paper, we review some of the strategies that can be implemented to infer such solutions and we apply them to the classification achieved in \cite{r1d3}, on three state $U(1)$-invariant Hamiltonians solvable by CBA, focusing on models for which the $S$-matrix is not trivial.
We also simplify the presentation of these Hamiltonians, relating the so-called telescoping terms and shifts by identity to Drinfeld twists leaving the Hopf structure unchanged.

For the four 19-vertex solutions, one recovers the $R$-matrices of the well-known Zamolodchikov--Fateev \cite{ZF} and Izergin--Korepin \cite{IK} models. They were solved through CBA and ABA in \cite{LS}. We point out that the third solution, the generalized Bariev Hamiltonian, is related to both main and special branches of \cite{M13}, which generate the same Hamiltonian.
Finally, the 19-vertex SpR model still resists to the analysis, although we are able to state some no-go theorems for it.
For this latter model, we show that no univariate $R$-matrix can be associated to the corresponding Hamiltonian, while the question remains open for bivariate $R$-matrices due to the complexity of the calculations.
At this stage, it should be noticed that the equivalence between the CBA and QISM approaches, to the best of our  knowledge, has not been proved yet.

For 17-vertex Hamiltonians, we recover again the special branch $R$-matrix of \cite{M13}, but at a special point, and show how the 19-vertex solution degenerate to 17-vertex at this special point. We also produce a new $R$-matrix, associated to the Hamiltonian called $17V_2$ in \cite{r1d3}. It could be of some interest to study the physical content of this new Hamiltonian.

 There is also a 14-vertex solution with a non-trivial scattering matrix. Surprisingly enough, we prove that it does not exist any (univariate or bivariate) $R$-matrix for generic values of the parameters of the Hamiltonian. Hence, this case could be of some relevance in the comparison between CBA and ABA methods.

The paper is organized as follows. In section \ref{sect:setup}, we fix the main notations and properties of $R$-matrices. In section \ref{sect:constr}, we review some constructions of the $R$-matrices, namely the Baxterization procedure, the iteration procedure, the resolution by brute force and the spectral curve approach. We emphasize when relevant the behavior of the $R$-matrix whether it is univariate or bivariate. In section \ref{sect:resu}, we expose the results for the Hamiltonians under consideration.

\section{General set-up}
\label{sect:setup}

\subsection{Notations}

We consider $U(1)$-invariant Hamiltonians $H$ acting on a spin chain of length $L$ with nearest neighbour interactions and assuming periodic conditions for the chain (sites $L+1$ and 1 are identified), that is
\beq
H=\sum_{j=1}^L H_{j,j+1}.
\label{eqH1}
\eeq
The $U(1)$ symmetry of $H$ is generated by the $S^z$ component of the total spin:
\beq
[H\,,\,S^z]=0\,,\quad S^z=\sum_{\ell=1}^L \fs^z_\ell
\mb{with} \fs^z|j\rangle =j\,|j\rangle.
\eeq
We are interested in models where the two-site Hamiltonian $H_{j,j+1}$ describes a three-state system, in other words $H_{j,j+1}$ acts in $\CC^3$ as a vector space, with basis vectors $|0\rangle = (1,0,0)$, $|1\rangle = (0,1,0)$, $|2\rangle = (0,0,1)$. If $E_{ij}$ denote the elementary $3 \times 3$ matrices with entry 1 in position $(i,j)$ and zero elsewhere, the two-site Hamiltonian under consideration reads
\begin{align}
H_{12} = {} & \sum_{i_1,i_2,j_1,j_2 \in \{0,1,2\}} h_{i_1\;i_2}^{j_1\;j_2} \, E_{i_1,j_1} \otimes E_{i_2,j_2} \nonumber \\
= {} & p E_{01} \otimes E_{10} + q E_{10} \otimes E_{01} + t_1 E_{21} \otimes E_{01} + s_1 E_{12} \otimes E_{10} + t_2 E_{01} \otimes E_{21} + s_2 E_{10} \otimes E_{12} \nonumber \\
&+ t_3 E_{12} \otimes E_{21} + s_3 E_{21} \otimes E_{12} + t_p E_{02} \otimes E_{20} + s_p E_{20} \otimes E_{02} + \sum_{i,j} v_{ij} E_{ii} \otimes E_{jj}\,.
\label{eqH2}
\end{align}

As already mentionned, Hamiltonians of the form \eqref{eqH1}-\eqref{eqH2} and
solvable by CBA have been classified in \cite{r1d3}. The next question is then whether these Hamiltonians can be related to an $R$-matrix, using the transfer matrix formalism.
More precisely, let $\check R(x,y)$ be a (general bivariate) solution of the braided Yang--Baxter equation (YBE):
\begin{equation}\label{YBE}
\check R_{12}(y,z) \check R_{23}(x,z) \check R_{12}(x,y) = \check R_{23}(x,y) \check R_{12}(x,z) \check R_{23}(y,z) \,.
\end{equation}
We will assume that the $R$-matrix is unitary
\beq\label{unitarite}
\check R(x,y) \check R(y,x) \propto \II,
\eeq
and regular
\beq\label{regularite}
\check R(x,x) = \II.
\eeq
The transfer matrix associated to this $R$-matrix reads
\beq
t_{<12...L>}(x|y_1,...,y_L) = Tr_0 R_{01}(x,y_1)\,R_{02}(x,y_2)\,\cdots R_{0L}(x,y_L).
\eeq
The regularity and unitary conditions on $R$ ensure the existence of local interaction
\beq
H=\left.\frac{d}{dx} \ln t_{<12...L>}(x|y,y,...,y)\right|_{x=y}=\sum_{j=1}^L H_{j,j+1},
\eeq
the local Hamiltonian being given by
\begin{equation}\label{H=dR}
H_{12} = \left. \partial_x \check R_{12}(x,y) \right|_{x=y} \,.
\end{equation}

 In most cases, the $R$-matrix will be \emph{univariate}: \emph{multiplicative}, $\check R(x,y)=\check R(x/y)$, or \emph{additive}, $\check R(x,y)=\check R(x-y)$. However, we will also encounter cases where the $R$-matrix is genuinely \emph{bivariate}, see below.
We will also use the non braided $R$-matrix defined by $R = P \check R$ where $P$ is the permutation operator.

\subsection{Transformations on $R$-matrices and corresponding Hamiltonians}
\label{subsect:transfo}

Before computing the $R$-matrix, one can note that there exists transformations on $H$ that lead to physically equivalent models.
We shall show that most of the transformations can be re-interpreted as some particular Drinfeld twists that preserve the Hopf structure (hence the Yang--Baxter equation). We remind the form of a Drinfeld twist (once represented):
\beq
\check R(x,y) \to \check R^F(x,y) = F(x,y)\, \check R(x,y)\, F^{-1}(y,x)
\mb{where} F(x,y)\in \text{End}(\CC^3\otimes \CC^3)[x,y]
\eeq
The following particular Drinfeld twists can be related to such transformations.

\paragraph{Factorized twist / gauge transformation.}

Let $F = g\otimes g$, $g\in \text{End}(\CC^3)$. It is known that the transformed $\check R^F$-matrix satisfies YBE. The two-site Hamiltonian is then transformed as
\begin{equation}
H_{j,j+1} \to g \otimes g \; H_{j,j+1} \; g^{-1}\otimes g^{-1} \,.
\end{equation}

\paragraph{Grading twist / rescaling of parameters.}

Let $F = g \otimes g^{-1}$ with $g=\exp({\alpha \fs^z})$. It can be checked that since $[g\otimes g, \check R] = 0$, the  $\check R^F$-matrix satisfies  YBE.
The two-site Hamiltonian is then transformed as
\begin{equation}
H_{j,j+1} \to e^{\alpha \fs^z} \otimes e^{-\alpha \fs^z} \; H_{j,j+1} \; e^{-\alpha \fs^z} \otimes e^{\alpha \fs^z} \,.
\label{eq:gradation}
\end{equation}

\paragraph{Conjugation / telescopic terms.}

Let $F(x,y) = g(x)\otimes g(y)$, where $g(x)=\exp(-xA)$ and $A$ is a diagonal matrix. It is easy to see that the  $\check R^F$-matrix satisfies YBE. The two-site Hamiltonian is then transformed as:
\begin{equation}
H_{j,j+1} \to H_{j,j+1} + A \otimes \II - \II \otimes A  \,,
\end{equation}
which generates a generic telescopic term.

 \paragraph{Normalization / shifts by identity.}
Rescaling the $\check R$-matrix by a function $f(x,y)=\exp(\alpha(x-y))$ amounts to shift the Hamiltonian as
\beq
H_{j,j+1} \to H_{j,j+1} + \alpha\,\II\otimes\II
\eeq

Finally, let us note that since $H$ and $S^z$ commute, one can also shift $H$ by $S^z$.
However, this transformation is not related to a transformation on $R$ that preserves the Hopf structure.
Hence, in the course of reconstructing an $R$-matrix from $H$, one has to deal with $H+\beta\,S^z$
and tune the parameter $\beta$.

\section{Methods for constructing $R$-matrix}
\label{sect:constr}
There is no general constructive procedure to obtain an $R$-matrix from an integrable Hamiltonian. However, there are some techniques that may (or may not) work, depending on the considered Hamiltonian. We briefly review them, adding some properties for some of them.

\subsection{Baxterization}

The method of Baxterization has been proposed by V.F.R. Jones \cite{Jones}. It allows one to obtain solutions of the Yang--Baxter equation with spectral parameter from representations of the braid group, in particular in the Hecke, Temperly--Lieb and Birman--Murakami--Wenzl cases \cite{Isaev,KMN,Jim86}. Note that beyond this procedure, many authors tried to generalize or produce other suitable formulae that may lead to solutions of the YBE \cite{CGX,ZGB,YQL}.

Consider the braid group $\mathcal{B}_N$ generated by generators $T_i$ ($i=1,\ldots,N-1$), their inverses $T_i^{-1}$ and the relations (see~\cite{CP}):
\begin{equation}
\begin{split}
& T_i \, T_{i+1} \, T_i = T_{i+1} \, T_i \, T_{i+1} \;, \\
& T_i \, T_j = T_j \, T_i \quad \mbox{for} \ |i-j| > 1.
\end{split}
\label{BG}
\end{equation}
We set $Z_i=T_i-T_i^{-1}-\xi$ for some constant $\xi$.

\paragraph{Hecke case:} When the $T_i$'s satisfy the supplementary quadratic relations $Z_i = 0$, then
\begin{equation}
\check R_{i,i+1}(z) = zT_i - z^{-1}T_i^{-1}
\end{equation}
is unitary and satisfies the Yang--Baxter equation with multiplicative spectral parameter $z$.

\paragraph{Temperly--Lieb case:} When $\ft_i=T_i+1$ satisfy the supplementary relations
\begin{equation}
 \ft_i \, \ft_{i\pm1} \, \ft_i = \ft_i \mb{and}
 \ft_i\, \ft_i  = 2a\,\ft_i,
\label{TL}
\end{equation}
for some parameter $a$, then
\begin{equation}
\check R_{i,i+1}(z) = \ft_i -a +\frac{z+1}{z-1}\sqrt{a^2-1}
\end{equation}
is unitary and satisfies the Yang--Baxter equation with multiplicative spectral parameter $z$.

\paragraph{Birman--Murakami--Wenzl case:} When $Z_i$ and $T_i$ satisfy the following supplementary relations

\begin{equation}
 Z_i \, T_{i-1}^{\pm 1} \, Z_i = \xi a^{\mp 1} Z_i \,, \quad Z_i \, T_{i}=T_i \, Z_i = - a\, Z_i
 \end{equation}
for some constant $a$ (BMW algebra), both matrices
\begin{equation}
\check R^{(\pm)}_{i,i+1}(z) =  T_i + \frac{a_\pm z^2}{1-a_\pm z^2}\, Z_i + z  \frac{q-1/q}{z-1/z}\, \II \mb{with} \xi = q-\frac1q
\mb{and} a_\pm= \mp aq^{\pm1}
\end{equation}
are unitary and satisfy the Yang--Baxter equation with spectral parameter $z$.

\bigskip

In all cases, eq. \eqref{H=dR} leads to $H=T_i$.
Hence the Baxterization procedure is the simplest way to get an $R$-matrix from an Hamiltonian, but it works only for specific Hamiltonians satisfying the braid group relations and Hecke, Temperly--Lieb or BMW algebra relations.

\subsection{Iteration procedure \textsl{\`a la Idzumi et al.}}

In this section we review Idzumi's method \cite{ITA} to construct solutions of the Yang--Baxter equation and derive some interesting properties that can help for simplifying the problem. Idzumi and collaborators used this method to build 19-vertex solutions, but for univariate $R$-matrices only. They found however four new Hamiltonians, that were solved in \cite{KMZ} through TQ relations.
Unfortunately the method is not so efficient for bivariate $R$-matrices.
We will give below a generalization that works for the latter case, but some freedom is left that cannot be resolved without any further assumption on $R$. Moreover, in both cases (univariate or bivariate), the $R$-matrix is obtained as a series that may be difficult to handle.

\subsubsection{Recursion formulae for multiplicative $R$-matrix}
In what follows, we consider the case of $R$-matrices with multiplicative spectral parameters, although the original paper deals with additive ones.

Let $\check R(u)$ be a solution of the multiplicative Yang--Baxter equation:
\begin{equation}
\label{eq:YBE_multiplicative}
\check R_{12} (u) \check R_{23} (uv) \check R_{12} (v) =
\check R_{23} (v) \check R_{12} (uv) \check R_{23} (u) \,,
\end{equation}
and suppose that $\check R (u)$ is analytical around $u = 1$:
\begin{equation}
\check R (u) = \sum_{i=0}^\infty \check R^{(i)} (u-1)^i \,.
\end{equation}
We demand regularity for $R$ and thus $\check R^{(0)} = \II \otimes \II$.
Furthermore, the Hamiltonian being defined by the first derivative, one has $\check R^{(1)} = H$.

\begin{thm}\cite{ITA}
Let $\check R(u)$ be an analytical solution of the multiplicative Yang--Baxter equation \eqref{eq:YBE_multiplicative}.
An Hamiltonian $H$ being given, the full matrix $\check R(u)$ such that $\check R^{(1)} = H$ can be reconstructed, up to an arbitrary normalization factor.
\label{thm:idzumi}
\end{thm}
\emph{Proof}.
We perform a Taylor expansion of \eqref{eq:YBE_multiplicative} and select the coefficient of $(u-1)^k (v-1)$, for $k \geq 1$.

Using $\check R^{(0)} = \II \otimes \II$, we get:
\begin{align}
(k+1) \left( \check R_{12}^{(k+1)} - \check R_{23}^{(k+1)}\right)
= {}& \sum_{j=0}^k \left( \check R_{12}^{(j)} \check R_{23}^{(k-j)} H_{12} - H_{23} \check R_{12}^{(k-j)} \check R_{23}^{(j)} \right) \nonumber \\
& + \sum_{j=0}^k (k-j) \left( \check R_{12}^{(j)} \check R_{23}^{(k-j)} - \check R_{12}^{(k-j)} \check R_{23}^{(j)}\right) \nonumber \\
& + \sum_{j=1}^k (k-j+1) \left( \check R_{12}^{(j)} \check R_{23}^{(k-j+1)} - \check R_{12}^{(k-j+1)} \check R_{23}^{(j)}\right).
\end{align}
Only terms of order less than $k$ appear in the right hand side. Thus we write this equation as
\begin{equation}
\label{eq:rec_multiplicative}
\check R_{12}^{(k+1)} - \check R_{23}^{(k+1)} = Q^{(k)} \,,
\end{equation}
where $Q^{(k)}$ only depends on lower terms, so that the system is triangular, expressed on matrices in $\text{End}(\CC^3\otimes\CC^3\otimes\CC^3)$.
Any solution of the Yang--Baxter equation~\eqref{eq:YBE_multiplicative} can be normalized such that $\check R_{aa}^{aa}(u) = 1$,
that is $\check R_{aa}^{aa,(k+1)} = 0$ for all $k \in \mathbb N$, for some given $a$. This corresponds to the arbitrary
normalization factor.
Then, looking at the different entries, i.e. computing $\bra{abc} (\ldots) \ket{def}$, where $(\ldots)$ represents
eq.~\eqref{eq:rec_multiplicative}, one can deduce the entries of the matrix at level $k+1$ from the ones at level $k$:
\bea
\check R_{ab}^{ab,(k+1)} &=& -\bra{aab} Q^{(k)} \ket{aab} \,,\quad \forall a,b
\\
\check R_{bd}^{ce,(k+1)} &=& - \bra{abd} Q^{(k)} \ket{ace} \,,\quad d \neq e
\\
\check R_{bd}^{ce,(k+1)} &=& \bra{bda} Q^{(k)} \ket{cea} \,,\quad b \neq c
\eea

\finproof

\subsubsection{Specific case of $U(1)$-invariant models}
Having established that the computation of the matrix $\check R$ can be achieved through recursion formulae in a rather general framework, we now restrict ourselves to the specific case of 19-vertex models.
\begin{prop}
\label{prop:ice-type}
Let $H$ be a spin preserving Hamiltonian:
\[
[H, \fs^z \otimes \II + \II \otimes \fs^z] = 0 \mb{i.e.} H_{ab}^{cd} = 0 \quad \textrm{if} \quad a+b \neq c+d
\mbox{ (ice rule condition)}\,.
\]
If $H$ is obtained from a multiplicative $R$-matrix,  this $R$-matrix must also preserves the spin
\[
[ R(u), \fs^z \otimes \II + \II \otimes \fs^z] = 0 \,.
\]
\end{prop}

\emph{Proof}.
We proceed by induction. Suppose that the ice-rule property is satisfied by the matrix $\check R^{(k)}$ at order $k$ and use the relation \eqref{eq:rec_multiplicative}.
Computing an entry $\check R_{ab}^{cd,(k+1)}$ where $c+d \neq a+b$ amounts to compute
\[
\check R_{bd}^{ce,(k+1)} = - \bra{abd} Q^{(k)} \ket{ace}
\mb{and} \check R_{bd}^{ce,(k+1)} = - \bra{bda} Q^{(k)} \ket{cea} \,.
\]
In any case, the right hand side vanishes because $Q^{(k)}$ preserves the spin.
Moreover the property is obviously true for $k=1$, hence $\check R^{(k)}$ satisfies the ice-rule property for all $k$. Therefore the proposition is proved.
\finproof

So, if we start with an ice-type Hamiltonian, we only need to use the part of the recursion such that $a+b = c+d$, hence avoiding most of the computations.

One can wonder whether it is possible to go further in the simplification of the $R$-matrix when considering Hamiltonians that have more zero entries, i.e. if some zeros are preserved when computing $\check R(u)$ from $H$. One can show the following:
\begin{prop}
Let $H$ be a spin preserving Hamiltonian, that satisfies the additional constraint $H_{ce}^{bd} = 0$ for all $b \ne c$ for fixed $c$. Then
\[
\check R_{ce}^{bd} (u)= 0 \qquad \textrm{for all }b \neq c \,.
\]
\end{prop}

\emph{Proof}. Suppose that the property is true for $\check R^{(k)}$ at order $k$. We use the recursion relation \eqref{eq:rec_multiplicative} to compute the next term:
\[
\check R_{ce}^{bd,(k+1)} = \bra{bda} Q^{(k)} \ket{cea} \,.
\]
The right hand side is composed by terms like
$\bra{bda} \check R_{12}^{(j)} \check R_{23}^{(i)} H_{12} \ket{cea}$,
$\bra{bda} H_{23} \check R_{12}^{(i)} \check R_{23}^{(j)} \ket{cea}$,
$\bra{bda} \check R_{12}^{(j)} \check R_{23}^{(i)} \ket{cea}$,
where $i,j \leq k$. All of them vanish trivially.
The property being obviously true for $k=1$, the result is proved.
\finproof

This property is valid for any entry, for instance
\begin{align*}
\textrm{For }b \textrm{ fixed}, \quad H_{ce}^{bd} = 0 \quad \textrm{for all }c \neq b \quad \Rightarrow \quad \check R_{ce}^{bd} (u) = 0 \quad  \textrm{for all }c \neq b  \,.
\end{align*}
In particular, one can deduce:
\begin{coro}
 The multiplicative matrices $\check R(u)$ that may lead to the 14-vertex Hamiltonians of Ref. \cite{r1d3}  can have at most $15$ non-zero entries.
\end{coro}

\subsubsection{Recursion formulae for bivariate $R$ matrices}

We focus now on a more general case by considering bivariate $R$ matrices, i.e. that depend on two spectral parameters.
We start by reviewing some basic results.
\begin{lemma}
\label{prop:RR=Id}
Let $\check R(x,y)$ be a solution of the braided Yang--Baxter equation \eqref{YBE}, that is regular. Then $\check R(x,y)$ is unitary:
\beq\label{unit:general}
\check R(x,y) \check R(y,x) = \lambda(x,y) \,\II \otimes \II \,,
\eeq
where $\lambda$ is some symmetric scalar function.
\end{lemma}
\emph{Proof}. Equation \eqref{YBE} taken at $z=x$ gives
\begin{equation}
\label{eq:pre_identity}
\check R_{12} (y,x)\check R_{12} (x,y) =
\check R_{23} (x,y)\check R_{23} (y,x) \,.
\end{equation}
that is $M_{12} (y,x) \otimes \II_3 = \II_1 \otimes M_{23} (x,y)$ with $M(x,y) = \check R (x,y) \check R (y,x)$.
This equality shows that $M_{23}(x,y)$ is proportional to identity in space 3, while $M_{12}(y,x)$ is proportional to identity in space 1. This implies that $M(x,y)=\lambda(x,y)\,\II\otimes\II$, where $\lambda(x,y)$ is some scalar function that must be symmetric.
\finproof
\begin{lemma}
If $\check R(x,y)$ is a regular solution of the braided Yang--Baxter equation \eqref{YBE}, then the two Hamiltonians
\beq
H=\prt_x \check R(x,y)\Big|_{x=y} \mb{and} \wt H=\prt_y \check R(x,y)\Big|_{x=y}
\eeq
differ only by some term proportional to identity.
\end{lemma}
\emph{Proof}. Obvious by differentiating the unitary condition \eqref{unit:general} and using regularity. \finproof

\bigskip

Now, we would like to implement Idzumi's method in the case of bivariate $R$-matrices and try to build the full solution for the $R$ matrix starting from the minimum possible knowledge.
Consider the braided Yang--Baxter equation \eqref{YBE} and expand the $R$-matrix as
\begin{align} \label{eq:R_2_variables_taylor}
\check R(x,y) = \sum_{i=0}^\infty \sum_{j=0}^\infty \check R^{(ij)} x^i y^j \,.
\end{align}
\begin{thm}
Let $\check R$ be an analytical solution of the braided Yang--Baxter equation~\eqref{YBE}, with Taylor series expansion of the form~\eqref{eq:R_2_variables_taylor}. Suppose that $\check R(x,x) = \II \otimes \II$.
Then, one can reconstruct the full solution $\check R(x,y)$ once the matrix $\check R (x,0)$ is given.
\end{thm}

\emph{Proof}.
Using the identity equation $\check R (x,0) \check R (0,x) = \II \otimes \II$, one can compute $\check R (0,x)$, hence $\check R^{(i0)}$ and $\check R^{(0i)}$ are known for all $i \in \mathbb N$. Recall that $\check R^{(00)} = \II \otimes \II$. \\
Consider the braided Yang--Baxter equation~\eqref{YBE}, and take the coefficient of $x^m y^0 z^n$:
\begin{equation}
\sum_{ij} \check R_{12}^{(0,j)} \check R_{23}^{(m-i,n-j)} \check R_{12}^{(i,0)}
=
\sum_{ij} \check R_{23}^{(i,0)} \check R_{12}^{(m-i,n-j)} \check R_{23}^{(0,j)} \,.
\end{equation}
This last equation can be rewritten as a recursion relation for $\check R^{(m,n)}$:
\begin{align}
\check R^{(m,n)}_{12} - \check R^{(m,n)}_{23} &=
\sum_{i=1}^m \sum_{j=1}^n \left(\check R_{12}^{(0,j)} \check R_{23}^{(m-i,n-j)} \check R_{12}^{(i,0)} -
\check R_{23}^{(i,0)} \check R_{12}^{(m-i,n-j)} \check R_{23}^{(0,j)} \right) \nonumber \\
& \quad +
\sum_{i=1}^m \left(\check R_{23}^{(m-i,n)} \check R_{12}^{(i,0)} -
\check R_{23}^{(i,0)} \check R_{12}^{(m-i,n)} \right) \nonumber \\
& \quad +
\sum_{j=1}^n \left(\check R_{12}^{(0,j)} \check R_{23}^{(m,n-j)} -
\check R_{12}^{(m,n-j)} \check R_{23}^{(0,j)} \right) \,.
\end{align}
which is of the form
\begin{equation}
\check R^{(m,n)}_{12} - \check R^{(m,n)}_{23} = Q^{(m,n)} \,,
\label{eq:recursion_double}
\end{equation}
where $Q^{(m,n)}$ depends only on terms of smaller order in $m,n$.
Notice that we need $m$ and $n$ to be non-zero, otherwise the equation is trivial. \\
The rest of the proof is similar to the one variable case.
\finproof

\begin{coro}
If $\check R (x,0)$ preserves the spin, that is
$[ \check R (x,0) , \fs^z \otimes \II + \II \otimes \fs^z ] =0$,
then it is also true for the full solution:
$[ \check R (x,y) , \fs^z \otimes \II + \II \otimes \fs^z ] =0$.
\end{coro}
We need to check that the recursion relation preserves this property. However, relation \eqref{eq:recursion_double} is exactly of the same type that the one that appears in the one variable case, and the proof follows.

\bigskip

To be complete, notice that the fact that $\check R (x,x) = \II \otimes \II$ also leads to the equation:
\[
\sum_{i=0}^n \check R^{(i,n-i)} = \delta_{n,0}\, \II \otimes \II \,.
\]

\subsection{Resolution of YBE by brute force for multiplicative $R$-matrices}

Although the resolution of the Yang--Baxter equation seems to be an impossible task in the general case, the property of $U(1)$-invariance implies strong constraints on the resulting equations, and allows one in some cases to compute directly the $R$-matrix when it is univariate. Let us start with a regular $R$-matrix of the form
\beq
R(u)=
\begin{pmatrix}
f_{11}(u) & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & f_{22}(u) & 0 & f_{24}(u) & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & f_{33}(u) & 0 & f_{35}(u) & 0 & f_{37}(u) & 0 & 0 \\
0 & f_{42}(u) & 0 & f_{44}(u) & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & f_{53}(u) & 0 & f_{55}(u) & 0 & f_{57}(u) & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & f_{66}(u) & 0 & f_{68}(u) & 0 \\
0 & 0 & f_{73}(u) & 0 & f_{75}(u) & 0 & f_{77}(u) & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & f_{86}(u) & 0 & f_{88}(u) & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & f_{99}(u)
\end{pmatrix}
\label{eq:Rmat19}
\eeq
where the functions $f_{ij}(u)$ are to be determined. \\
We impose the Yang--Baxter equation and $\check{R}'(1)=H$ for a given Hamiltonian $H$, and we set
\beq
p_{ij} = f_{ij}'(1).
\eeq
 This leads to a set of equations that should be satisfied by the functions $f_{ij}$. Among these equations, we start from the relation
\begin{equation}
f_{44}(u) f_{42}(uv) f_{22}(v) = f_{11}(u) f_{42}(uv) f_{11}(v) - f_{42}(u) f_{11}(uv) f_{42}(v) \,.
\end{equation}
Since the right hand side is symmetric in the exchange $u \leftrightarrow v$, one gets the consistency condition
$p_{22} \, f_{44}(u) = p_{44} \, f_{22}(u)$.

In the same way, one obtains the consistency condition $p_{66} \, f_{88}(u) = p_{88} \, f_{66}(u)$.

Moreover, one has also a relation of the type $F(u)\,F(v)=F(uv)$ for $F(u) = f_{24}(u)/f_{42}(u)$ and for $F(u) = f_{68}(u)/f_{86}(u)$, therefore
\begin{equation}
\frac{f_{24}(u)}{u^{p_{24}}} = \frac{f_{42}(u)}{u^{p_{42}}}
\qquad \text{and} \qquad
\frac{f_{68}(u)}{u^{p_{68}}} = \frac{f_{86}(u)}{u^{p_{86}}} \,.
\end{equation}
The remaining equations imply the two following relations:
\beq \frac{f_{35}(u)}{f_{53}(u)}\,v^{p_{42}-p_{24}}=\frac{f_{35}(uv)}{f_{53}(uv)}
\mb{and} \frac{f_{75}(u)}{f_{57}(u)}\,v^{p_{24}-p_{42}}=\frac{f_{75}(uv)}{f_{57}(uv)},
\eeq
from which one obtains $f_{53}(u) = \lambda_{53} \, f_{35}(u) \, u^{p_{42}-p_{24}}$ and $f_{57}(u) = \lambda_{57} \, f_{75}(u) \, u^{p_{24}-p_{42}}$, where $\lambda_{53}$ and $\lambda_{57}$ are (at that point) arbitrary parameters.

Note that $f_{35}(u)$ and $f_{75}(u)$ cannot be identically zero since the parameters of the Hamiltonian are restricted to $t_1 \ne 0$, $t_2 \ne 0$.
It is then convenient to introduce equations that depend only on one variable. To this end, we derive the YBE with respect to the $v$ variable and we set $v=1$. Similarly, we derive the YBE with respect to the $u$ variable and we set $u=1$. In this way, we obtain a set of differential equations satisfied by the functions $f_{ij}(u)$. We note however that among these equations, a particular subset ${\cal P}$ is constituted by polynomial functional equations.

Two types of models emerge:
\begin{itemize}
\item[$\diamond$]
In the first case, the parameters of the Hamiltonian satisfy $s_1=s_2=0$, from which it follows $\lambda_{53} = \lambda_{57} = 0$. Plugging this constraints in the YBE leads immediately to the following relations:
\begin{equation}
\frac{f_{37}(u)}{u^{p_{37}}} = \frac{f_{73}(u)}{u^{p_{73}}} \,,\quad p_{77} \, f_{44}(u) = p_{77} \, f_{44}(u) \,,\quad p_{88} \, f_{77}(u) = p_{77} \, f_{88}(u)
\end{equation}
At this stage, it is necessary to implement the different models by specifying for each case the entries of the Hamiltonian (i.e. the $p_{ij}$), at least for the off-diagonal part, and solve case by case the remaining equations.
\item[$\diamond$]
In the second case, $s_1$ and/or $s_2$ are nonzero and one should have $p_{24}-p_{42} = p_{68}-p_{86}$.
We focus now on the polynomial equations, once $f_{33}$ and $f_{77}$ have been expressed in terms of the other functions. In the case where $p_{22}p_{88}=p_{66}p_{44}$ (which is satisfied by the ZF, IK and SpR models), we can determine quite easily simple relations for the functions $f_{11}$, $f_{44}$, $f_{88}$, $f_{99}$, $f_{35}$, $f_{75}$, and in particular $f_{42} = f_{86}$. One deduces that $p_{22}^2p_{77} = p_{44}^2p_{33}$ (case of ZF or IK models for which one can proceed further) or $p_{44}^2 f_{11}(u) = p_{88}^2 f_{99}(u)$ (SpR model).
\end{itemize}
This method is exhaustive although rather heavy. However it remains tractable for univariate $R$-matrices in the general case.
In the bivariate case, it becomes too intricate and one needs to restrict with some symmetry assumptions.

\subsection{Spectral curves \textsl{\`a la Martins}}

In \cite{M13}, M.J. Martins revisited the problem of $U(1)$-invariant three-state vertex models, when the Boltzmann weights $W_{\alpha\beta}^{\gamma\delta}$ configurations break the parity-time reversal symmetry, i.e. $W_{\alpha\beta}^{\gamma\delta} \ne W_{\delta\gamma}^{\beta\alpha}$. More precisely, the Lax operator being given by
\begin{equation}
L_i(z) = \sum_{\alpha,\beta,\gamma,\delta} W_{\alpha\beta}^{\gamma\delta} \, e_{\alpha\gamma}^{(q)} \otimes e_{\beta\delta}^{(i)}
\end{equation}
where the index $q$ denotes the quantum space and $i$ the auxiliary spaces ($i=1,...,N$), one investigates the solutions of a general two parametric Yang--Baxter equation
\begin{equation}
R(x,y)\, L_1(x)\,  L_2(y) = L_2(y)\, L_1(x)\, R(x,y) \,.
\label{eq:YB_RLL}
\end{equation}
Under some symmetry requirements for the $L$-matrix, it is possible in a first step to present a general formula for the $R$-matrix entries in terms of those of the $L$-matrix, and in a second step to express the $L$-matrix in terms of only two functions $a(z)$ and $b(z)$.

At this point, two  solutions naturally emerge, denoted \emph{main branch (MB)} and \emph{special branch (SB)}. For each of these branches, the two functions $a(z)$ and $b(z)$ satisfy some polynomial relation that defines an elliptic curve, whose degree depends on the considered branch
\footnote{There are some misprints in \cite{M13} in the elliptic curve equations.}:
\bea
0 &=& (\alpha\beta-1)\big(a^4+a^2b^2+b^4\big)^2
+ (2-\alpha\beta+\alpha^2+\beta^2)\big(\beta(a^4+a^2b^2+b^4)+ab\big)ab
\nonu
&&-(\alpha\beta-2)a^4+\beta^2a^2b^2-(2-\alpha\beta+\beta^2)b^4-2\beta ab-1,
\qquad (MB)\label{MB},\\
0 &=& (a^2+jb^2)\big(a^4+a^2b^2+b^4+\Lambda_4ab\big)
+ b^2-a^2,\qquad (SB)\label{SB},
\eea
where $a\equiv a(z)$, $b\equiv b(z)$, $j^2-j+1=0$, and $\alpha$, $\beta$, $\Lambda_4$ are free parameters.

We will show below that the branches share the same Hamiltonian, if one generalizes the point where the Hamiltonian is defined (see below). In that sense, it is enough to consider the special branch, and study what is called the generalized Bariev model in \cite{r1d3}. Let us also stress that the branches are the most natural, but not the full set of choices one can do in solving the equations: a detailed study of these equations could lead to new (marginal) integrable models.

Relaxing the symmetry constraints used in \cite{M13}, one can try to use this method to compute solutions of the Yang--Baxter equation~\eqref{eq:YB_RLL}. This is a system of cubic equations containing in total $129$ equations (the others being zero by conservation of spin), each equation being linear in the entries of $R$ and quadratic in the entries of $L$. By choosing a suitable set of equations, one can obtain the entries of $R$ in terms of the entries of $L$, up to a multiplicative constant set by normalizing some given entry of $R$ to $1$.

The remaining equations can be seen as complementary equations for the entries of $L$. Plugging the computed entries of $R$ in these equations, such complementary equations can become very involved, and so, it is crucial to chose wisely the set of equations to use for determining $R$.

These final equations depend on all sorts of entries of $L(x)$ and $L(y)$. Untangling them may happen naturally, otherwise one can try to write the equation $f(L(x),L(y))=0$ as:
\begin{equation}
  f(L(x),L(y)) = P(L(x)) Q(L(y)) - P(L(y)) Q(L(x)) = 0 \,.
\end{equation}
When this is possible, one can then write $P(L(x)) = c\, Q(L(x))$, hence the emergence of elliptic curves depending on some constants in the computation.
Notice that we should make sure that $Q(L(x))$ does not vanish.

\section{Hamiltonians and $R$-matrices}
\label{sect:resu}

We consider here only Hamiltonians that lead to non-trivial scattering matrices ($S\neq \pm 1$). Indeed, when the scattering matrix is trivial, the Bethe equations become also trivial, and the CBA obviously fails to provide a complete spectrum of the Hamiltonian.
Note however that there are some cases where the scattering is trivial, but one can nevertheless construct a $R$-matrix. The status of the corresponding Hamiltonians concerning integrability remains unclear, since CBA does not provide the complete spectrum, but one still gets a transfer matrix $t(u)$ that commutes for different values of the spectral parameter.

The Hamiltonians of \cite{r1d3} that have non-trivial scattering matrix are the four 19-vertex, two 17-vertex and one 14-vertex. Before
presenting them, we simplify them using a twist procedure.

\subsection{19-vertex Hamiltonians}

\subsubsection{Zamolodchikov--Fateev Hamiltonian}

We found in \cite{r1d3} an expression for the Zamolodchikov--Fateev Hamiltonian $H_{\textrm{ZF}}$ that contained a supplementary parameter $\tau_p$ with respect to the original one \cite{ZF}. For $\tau_p=-1$, this Hamiltonian is related to the one based on $\cU_q(B_1^{(1)})$ given in \cite{J86} by
\begin{equation}
H_{\cU_q(B_1^{(1)})}(1/k^2) = H_{\textrm{ZF}}(k)\vert_{\tau_p=-1} + (\II\otimes e_{22}-e_{22}\otimes\II) + 2(\II\otimes e_{33}-e_{33}\otimes\II) \,,
\label{eq:HZF}
\end{equation}
where the $R$-matrix of $\cU_q(B_1^{(1)})$ is normalized by $R_{11}^{11}=1$. \\
In fact, one can remove this parameter by considering the transformation
\begin{equation}
\wt H_{\textrm{ZF}} = F_{12}(\tau_p) H_{\textrm{ZF}} F_{12}^{-1}(\tau_p)
\label{eq:conjZF}
\end{equation}
where $F_{12}(\tau_p) = g \, \exp({\alpha\, \fs^z}) \otimes g \, \exp({-\alpha\, \fs^z})$ with $g=\diag\,(1,\tau_p^{-1/2},1)$ and $e^\alpha=\tau_p^{1/2}$. The explicit expression of $\wt H_{\textrm{ZF}}$ is formally obtained by setting $\tau_p=1$ in $H_{\textrm{ZF}}$. \\
It follows that $\wt H_{\textrm{ZF}}(k)$ is related to $H_{\cU_q(B_1^{(1)})}(1/k^2)$, up to the telescopic terms of \eqref{eq:HZF}, by the transformation of the type \eqref{eq:conjZF} with $F_{12}(-1)=\diag(1,-1,-1,1,-1,-1,-1,1,1)$.
Therefore the $R$-matrix from which one can deduce $\wt H_{\textrm{ZF}}$ is obtained by twisting the $R$-matrix of $\cU_q(B_1^{(1)})$ by this $F_{12}(-1)$. It can be checked that the obtained $R$-matrix satisfies the Yang-Baxter equation.
The explicit expression of the $R$-matrix is as follows:
\begin{equation}
R(u) = \left( \begin {array}{ccccccccc}
1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & b(u) & 0 & c_-(u) & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & f(u) & 0 & d_-(u) & 0 & h_-(u) & 0 & 0 \\

0 & c_+(u) & 0 & b(u) & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & d_+(u) & 0 & g(u) & 0 & d_-(u) & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & b(u) & 0 & c_-(u) & 0 \\

0 & 0 & h_+(u) & 0 & d_+(u) & 0 & f(u) & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & c_+(u) & 0 & b(u) & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1
\end{array}\right)
\label{eq:RmatZF}
\end{equation}
the entries of the $R$-matrix being given by
\begin{align}
& b(u)=-\frac{(u^2-1)k^2}{k^4u^2-1} \,,\qquad c_\pm(u)=u^{\pm 1}\,\frac{u(k^4-1)}{k^4u^2-1} \,,\qquad f(u)=\frac{(u^2-k^2)(u^2-1)k^2}{(k^4u^2-1)(k^2u^2-1)} \\
& d_\pm(u)=-u^{\pm 1}\,\frac{uk(k^4-1)(u^2-1)}{(k^4u^2-1)(k^2u^2-1)} \,,\qquad h_\pm(u)=u^{\pm 2}\,\frac{u^2(k^4-1)(k^2-1)}{(k^4u^2-1)(k^2u^2-1)} \\
& g(u)=\frac{k^4u^4+u^2(k^2+1)(k^2+k-1)(k^2-k-1)+k^2}{(k^4u^2-1)(k^2u^2-1)}
\end{align}

\subsubsection{Izergin--Korepin Hamiltonian}

We found in \cite{r1d3} an expression for the Izergin--Korepin Hamiltonian $H_{\textrm{IK}}$ that contained a supplementary parameter $\tau'_p$, see formula (5.7) therein. This parameter can be removed by considering the transformation
\begin{equation}
\wt H_{\textrm{IK}} = F_{12}(\tau'_p) H_{\textrm{IK}} F_{12}^{-1}(\tau'_p)
\label{eq:conjIK}
\end{equation}
where $F_{12}(\tau'_p) = \exp({\alpha\, \fs^z}) \otimes \exp({-\alpha\, \fs^z})$ with $e^\alpha={\tau'_p}^{1/2}$. The explicit expression of $\wt H_{\textrm{IK}}$ is formally obtained by setting $\tau'_p=1$ in $H_{\textrm{IK}}$. One gets the Hamiltonian based on the $R$-matrix of $\cU_q(A_2^{(2)})$ given in \cite{J86} and normalized such that $R_{11}^{11}=1$. One recovers also the Izergin--Korepin $R$-matrix of the Shabat--Mikhailov model \cite{IK}, after exchanging the roles of the states $|0\rangle$ and $|1\rangle$, and taking into account into the $R$-matrix the existence of telescopic terms and a gauge transformation between the corresponding Hamiltonians, see section \ref{subsect:transfo}.

The $R$-matrix has the same shape as in \eqref{eq:RmatZF} but its entries are now given by
\begin{align}
& b(u)=\frac{k(u-1)}{u-k^2} \,,\quad d_-(u)=k^2\,d(u) \,, \quad d_+(u)=-u\,d(u) \,, \quad d(u)=\frac{k^{1/2}(1-k^2)(u-1)}{(u+k^3)(u-k^2)} \\
& c_-(u)=\frac{1-k^2}{u-k^2} \,,\qquad c_+(u)=uc_-(u) \,,\qquad f(u)=\frac{k^2(u+k)(u-1)}{(u+k^3)(u-k^2)} \\
& g(u)=\frac{k(u+k^3)(u-1)+u(k^3+1)(1-k^2)}{(u+k^3)(u-k^2)} \\
& h_-(u)=\frac{(u+k^3+k^2(u-1))(1-k^2)}{(u+k^3)(u-k^2)} \,,\qquad h(u)_+=\frac{(u+k^3-k(u-1))(1-k^2)}{(u+k^3)(u-k^2)}
\end{align}

\subsubsection{Generalized Bariev Hamiltonian} \label{sect:model25}

As in the previous cases, one can simplify further the Hamiltonian found in \cite{r1d3}, see formula (5.13) therein. First of all, we
perform the following change of variables:
\begin{equation}
\phi = \frac{\xi}{J_0\tau_p\mu^{1/2}} \qquad \text{and} \qquad \psi = \frac{\xi\theta}{J_0^3\tau_p\mu^{3/2}}
\end{equation}
and define $\upsilon$ as
\begin{equation}
-4\xi\upsilon = \phi^2 - \phi\psi + \psi^2 \,,
\label{eq:def_delta}
\end{equation}
where $J=-1/J_0^2$ and $\xi$ is some normalization constant introduced for later convenience (although it may be set to one for the moment, it is useful for taking some limits, see below). Then, using the transformation \eqref{eq:gradation} with $\alpha=(-J^2\mu)^{-1/4}$, one gets an Hamiltonian that depends only on $\phi$, $\psi$ and $\xi$:
\begin{equation}
H_{\textrm{GB}} =
\begin{pmatrix}
-\upsilon & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & \phi & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & -\upsilon - J^2\xi & 0 & \phi & 0 & \xi & 0 & 0 \\
0 & \psi & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & -J^2 (\psi-\frac{\xi^2}{\phi}) & 0 & \upsilon - \xi & 0 & \psi-\frac{\xi^2}{\phi} & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & \psi & 0 \\
0 & 0 & \xi & 0 & -J \phi & 0 & -\upsilon - J \xi & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & \phi & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -\upsilon
\end{pmatrix} \,.
\label{htilde24}
\end{equation}
As stated in \cite{r1d3}, this Hamiltonian is a generalization of the one found by Alcaraz and Bariev \cite{AB}, further corrected in \cite{AN}.

\paragraph{Relation with the Hamiltonian of the main branch at the special point $H_{\textrm{MB}_0}$.}
The Hamiltonian $H_{\textrm{GB}}$ can be related to the one $H_{\textrm{MB}_0}$ obtained in \cite{M13} in the case of the main branch. More precisely, the two-site Hamiltonian $H_{\textrm{MB}_0}$ is defined as
\begin{equation}
H_{\textrm{MB}_0} = P \; \frac{\partial}{\partial x} L(x)\Big\vert_{x=x_0}
\end{equation}
where the point $x_0$ is chosen such that $a(x_0)=1$, $b(x_0)=0$ (constraints satisfied by the corresponding elliptic curve). In order to compute explicitly $H_{\textrm{MB}_0}$, one differentiates the elliptic curve $\gamma(a(x),b(x))=0$ and obtains $b^\prime(x_0)$ as a function of $a^\prime(x_0)$:
\begin{equation}
b^\prime(x_0) = \frac{-4 \alpha}{\alpha^2-\alpha \beta + \beta^2} \, a^\prime(x_0) \,,
\end{equation}
where $\alpha$ and $\beta$ are the two constants entering in the definition of the curve, see \eqref{MB}. \\
The Hamiltonian $H_{\textrm{MB}_0}$ then takes the form \cite{M13}:
\begin{equation}
\label{eq:H_MB_0}
H_{\textrm{MB}_0} =
\begin{pmatrix}
1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & -\beta \rho & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 1+J_0^2 \rho & 0 & -J_0^2 \eta \rho & 0 & \rho & 0 & 0 \\
0 & - \alpha \rho & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & - J_0^2 \eta \rho & 0 & -1 - \rho & 0 & -\eta \rho & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & - \alpha \rho & 0 \\
0 & 0 &  \rho & 0 & -\eta \rho & 0 & 1+J_0^{-2}\rho & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & -\beta \rho & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1
\end{pmatrix}\ ,
\end{equation}
where
\begin{equation}
\rho = \frac{4}{\alpha^2 - \alpha \beta + \beta^2} \qquad \text{and} \qquad \eta = -J_0^{-1} \sqrt{\alpha\beta-1} \,.
\end{equation}
Then the Hamiltonian $H_{\textrm{GB}}$ can be related to $H_{\textrm{MB}_0}$ by the gauge transformation
\beq
H_{\textrm{GB}} = (F \otimes F)\, H_{\textrm{MB}_0} \,(F \otimes F)^{-1}
\eeq
where
\beq
F = \diag\left(1,\sqrt{\frac{J_0^2\eta}{\beta}},1\right)
\label{eq:twistbariev}
\eeq
and the correspondence between the parameters is
\beq
\upsilon = -1 \;, \qquad \alpha = \frac{\psi}{\xi} \;, \qquad \beta = \frac{\phi}{\xi} \;, \qquad \rho = \xi
\eeq
Therefore the $R$-matrix from which the Hamiltonian $H_{\textrm{GB}}$ can be derived is obtained by twisting by $F$, formula \eqref{eq:twistbariev}, the $R$-matrix corresponding to the main branch of ref. \cite{M13}.

\paragraph{Relation with the Hamiltonian of the special branch at a generic point $H_{\textrm{SB}}$.}
As mentioned in section \ref{sect:setup}, when the $R$ matrix satisfies $\check R(x,x)=\II$, one can also define the Hamiltonian for bivariate $R$ matrices by $H = \left. \partial_x \check R(x,y) \right|_{x=y}$ keeping the spectral parameter $x$ free\footnote{One can check by direct calculation that the $R$-matrices in \cite{M13} are regular, both for MB and SB.}. In this way, one introduce a new parameter $a(x)$ (or equivalently $b(x)$) that was set to 1 in the construction of \cite{M13} (since $a(x_0)=1$).
We now present the corresponding Hamiltonians, both for the main branch and for the special branch.

In the case of the special branch, defining $H_{\textrm{SB}} = \left. \partial_x \check R(x,y) \right|_{x=y}$ where $\check R$ is given by the formulae (119-121) of \cite{M13}, one gets
\begin{align}
H_{\textrm{SB}} &=
h_a (E_{00} \otimes E_{00}+E_{22} \otimes E_{22}) + h_{\bar h} E_{00} \otimes E_{22} + h_g E_{11} \otimes E_{11} + h_h E_{22} \otimes E_{00} \nonumber \\
& + h_f (E_{02} \otimes E_{20}+E_{20} \otimes E_{02}) + h_b (E_{10} \otimes E_{01}+E_{12} \otimes E_{21}) + h_{\bar b} (E_{01} \otimes E_{10}+E_{21} \otimes E_{12}) \nonumber \\
& + h_d (E_{12} \otimes E_{10}+E_{21} \otimes E_{01}) + h_{\bar d} (E_{01} \otimes E_{21}+E_{10} \otimes E_{12}) \,,
\label{eq:HSB}
\end{align}
and we set $h_f=\xi$, $h_b=\psi$, $h_{\bar b}=\phi$, $h_a=-\upsilon$ (expressed in terms of the two functions $a(x)$ and $b(x)$). From the expressions of the entries of the $R$-matrix, one obtains immediately $h_h = -\upsilon - J\xi$ and $h_{\bar h} = -\upsilon - J^2\xi$. Using then a symbolic computation program, it can be checked that $H_g = \upsilon + \xi$ up to the curve $\gamma$.
Consider then $F$, a diagonal matrix of the kind $g \otimes g$ where $g=\diag(1,\sqrt{\zeta},1)$, and compute $F H_{\textrm{SB}} F^{-1}$. The entries that are changed are those corresponding to $h_d$ and $h_{\bar d}$ with a factor $\zeta$ for the $_{02}^{11}$ and $_{20}^{11}$ entries and a factor $\zeta^{-1}$ for the $_{11}^{02}$ and $_{11}^{20}$ entries. In order to get further in the identification with the generalized Bariev Hamiltonian, one has to impose
\begin{equation}
\zeta^{-1} h_{\bar d} = \phi \,, \qquad
\zeta^{-1} h_d = - J \phi \,, \qquad
\zeta h_{\bar d} = -J^2 (\psi - \frac{\xi^2}{\phi}) \,, \qquad
\zeta h_d = \psi - \frac{\xi^2}{\phi} \,.
\end{equation}
Hence we set $\displaystyle \zeta = \frac{h_{\bar d}}{\phi}$ and the last equation becomes
\[
h_d h_{\bar d} = \psi \phi - \xi^2 \qquad \textrm{up to the curve}\  \gamma\ ,
\]
which can be checked by a symbolic computation program.
\\
Finally, equation~\eqref{eq:def_delta} can also be checked, up to the curve $\gamma$.
The factor $\xi$ is just a normalization and therefore there are only two genuine parameters $\psi/\epsilon$ and $\phi/\epsilon$, which will depend on the two parameter of the model $\zeta$ and $\Lambda_4$ (one can prove that they are nonzero and independent by computing the Wronskian).

\paragraph{Relation with the Hamiltonian of the main branch at a generic point $H_{\textrm{MB}}$.}
In the case of the main branch, defining $H_{\textrm{MB}} = \left. \partial_x \check R(x,y) \right|_{x=y}$ where $\check R$ is given by the formulae (115-118) of \cite{M13}, one gets for the two-site Hamiltonian a matrix similar to \eqref{eq:HSB}, where $H_c = 0$ and all other entries are very complicated (the smallest one $H_a$ occupying five lines in Mathematica output and the biggest one $H_g$ occupying more than six pages). The reasoning follows the same lines as in the special branch case, and finally the Hamiltonian $H_{\textrm{GB}}$ also appears as a twist of the Hamiltonian $H_{\textrm{MB}}$.

It follows that the same Hamiltonian $H_{\textrm{GB}}$ can be derived in three different ways by considering either the $R$-matrix of the main branch (at the special point $x_0$ or at a generic point) or the $R$-matrix of the special branch at a generic point.
Hence the main branch and the special branch of \cite{M13} share the same conserved quantities.

Therefore the $R$-matrix of the generalized Bariev model is the one given in \cite{M13}, where one can restrict oneself to the special branch case. Hence, the $R$-matrix takes the form
\begin{align}
R(x,y) &=
r_a (E_{00} \otimes E_{00}+E_{22} \otimes E_{22}) + r_b (E_{00} \otimes E_{11}+E_{22} \otimes E_{11}) + r_{\bar b} (E_{11} \otimes E_{00}+E_{11} \otimes E_{22}) \nonumber \\
& + r_f (E_{00} \otimes E_{22}+E_{22} \otimes E_{00}) + r_g E_{11} \otimes E_{11} \nonumber \\
& + r_c (E_{01} \otimes E_{10}+E_{10} \otimes E_{01}+E_{12} \otimes E_{21}+E_{21} \otimes E_{12}) + r_h E_{02} \otimes E_{20} + r_{\bar h} E_{20} \otimes E_{02} \nonumber \\
& + r_d (E_{01} \otimes E_{21}+E_{12} \otimes E_{10}) + r_{\bar d} (E_{10} \otimes E_{12}+E_{21} \otimes E_{01}) \,,
\label{eq:Rmartins}
\end{align}
where the entries are given in terms of two functions $a(z)$ and $b(z)$ that satisfy equation \eqref{SB}:
\begin{align}
r_a(x,y) &= \frac{a(x)a(y)}{a(y)^2+jb(y)^2} + \frac{jb(x)b(y)}{a(x)^2+jb(x)^2} \\
r_b(x,y) &= \frac{b(x)a(y)}{a(y)^2+jb(y)^2} - \frac{a(x)b(y)}{a(x)^2+jb(x)^2} \\
r_{\bar b}(x,y) &= \frac{jb(x)a(y)}{a(x)^2+jb(x)^2} - \frac{ja(x)b(y)}{a(y)^2+jb(y)^2} \\
r_d(x,y) &=
j\frac{b(x)a(y)(a(y)^2+jb(y)^2)-a(x)b(y)(a(x)^2+jb(x)^2)}{ja(x)a(y)+b(x)b(y)(a(x)^2+jb(x)^2)(a(y)^2+jb(y)^2)}
\\
r_f(x,y) &= r_d(x,y)\;\frac{b(x)a(y)(a(x)^2+jb(x)^2)(a(y)^2+jb(y)^2)-j^2a(x)b(y)}{(a(x)^2+jb(x)^2)(a(y)^2+jb(y)^2)} \\
r_g(x,y) &= -r_d(x,y)\;\frac{r_f(y,x)+r_a(y,x)}{r_d(y,x)} \;, \qquad r_{\bar d}(x,y) = jr_d(x,y) \\
r_h(x,y) &= r_a(x,y) + j^{-1}r_f(x,y) \;, \qquad r_{\bar h}(x,y) = r_a(x,y) + jr_f(x,y)
\end{align}
the $R$-matrix being normalized such that $r_c(x,y)=1$.

\subsubsection{Generalized SpR Hamiltonian\label{sect:SpR}}

Performing the change of variable $\displaystyle\theta\tau_p^2=\theta_0=\frac{qt_p^2}{p^3}$, the transformation \eqref{eq:gradation} acting on $H_{red}$ (see eq. (5.19) of \cite{r1d3}) with $\alpha={\tau_p}^{-1/2}$ leads to an Hamiltonian which does not depend any longer on $\tau_p$, with $\delta_0=\tau_3^2-\tau_3+1$. The explicit expression of $\wt H_{\textrm{SpR}}$ is formally obtained by setting $\tau_p=1$, $\theta\to\theta_0$, $\delta\to\delta_0+\theta_0$ in $H_{red}$. \\

The method of finding the $R$-matrix by brute force when considering multiplicative spectral parameter implementation can be used in the case of the SpR model. Referring to the ``algorithm'' explained above, one is led to the following constraint: $p_{44}^2 f_{11}(u) = p_{88}^2 f_{99}(u)$, that is here $f_{11}(u) = \tau_3^2 f_{99}(u)$. Plugging this last equation in the polynomial set ${\cal P}$ leads to impose specific values of the parameters of the Hamiltonian, namely $\tau_3^2=1$ and $\theta_0=\tau_3^2-\tau_3+1$. One concludes that the SpR model does not admit a $R$-matrix with only one multiplicative spectral parameter for generic values of the parameters $\theta_0$ and $\tau_3$.

It is of course tempting to test the case of bivariate $R$-matrices. Unfortunately, the SpR Hamiltonian shows very few symmetry and it is too intricate to deal with the general case. Although the resolution of this case remains open, it implies that if a (bivariate) $R$-matrix exists for this Hamiltonian, it looks certainly very intricate.

Moreover, we checked that the spectrum (and indeed the multiplicities) of the Hamiltonian was obtained from CBA in the case of a chain of length two.

It leaves the question open whether "it always exists an $R$-matrix when an Hamiltonian is solvable by CBA"? or in other words "does CBA implies ABA?"

\subsection{17-vertex Hamiltonians}

\subsubsection{Generalized ``Special Branch'' Hamiltonian}

Performing the transformation \eqref{eq:gradation} acting on $\wt H$ (eq. (5.25) of \cite{r1d3} suitably normalized) with $\alpha=(-JQ)^{-1/4}$ leads to an Hamiltonian which does not depend any longer on $Q$:

\begin{align}
H_{17} = &-\Lambda (E_{00} \otimes E_{00} + E_{00} \otimes E_{22} - E_{11} \otimes E_{11} + E_{22} \otimes E_{00} + E_{22} \otimes E_{22}) \nonumber \\
&- J (E_{01} \otimes E_{10} + E_{01} \otimes E_{21} + E_{10} \otimes E_{12} + E_{21} \otimes E_{21}) \nonumber \\
&+ E_{10} \otimes E_{01} + E_{12} \otimes E_{10} + E_{12} \otimes E_{21} + E_{21} \otimes E_{01}
\end{align}
which is directly related to the Hamiltonian $H_{\textrm{SB0}}$ of \cite{M13}, where $\displaystyle\Lambda = \frac{\Upsilon}{4J\sqrt{-Q}}$\,. \\
Note that the Hamiltonian $H_{17}$ can be obtained as a limit of the Hamiltonian $H_{\textrm{GB}}$. Indeed, if we take the limit $\xi=0$ in the Hamiltonian $H_{\textrm{GB}}$ and set
\begin{equation*}
\upsilon = \Lambda \;,\quad \phi = -J^2 \;,\quad \psi = 1 \;,
\end{equation*}
which satisfy equation~\eqref{eq:def_delta}, we obtain the Hamiltonian $H_{17}$ for the value $J \to J^2$.

\subsubsection{$17V_2$ Hamiltonian}

Performing the change of variable $\displaystyle\theta\tau_p^2=\theta_0=\frac{qt_p^2}{p^3}$, the transformation \eqref{eq:gradation} acting on $H_{red}$ (eq. (5.33) of \cite{r1d3}) with $\alpha={\tau_p}^{-1/2}$ leads to an Hamiltonian $\wt H_{\textrm{17V$_2$}}$ which does not depend any longer on $\tau_p$. The explicit expression of $\wt H_{\textrm{17V$_2$}}$ is formally obtained by setting $\tau_p=1$ in $H_{red}$.

It can be checked that this Hamiltonian satisfies the Hecke relations, and therefore one can derive the $R$-matrix using a Baxterization procedure. Explicitly one obtains:
\begin{equation}
R(z) = \begin{pmatrix}
z-{\frac {\theta_0}{z}} & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & (z-\frac{1}{z})\theta_0 & 0 & \frac {1-\theta_0}{z} & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & (z-\frac{1}{z})\theta_0 & 0 & \frac{1}{z}-z & 0 & \frac {1-\theta_0}{z} & 0 & 0 \\
0 & z (1-\theta_0) & 0 & z-\frac{1}{z} & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & z-{\frac {\theta_0}{z}} & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & (z-\frac{1}{z})\theta_0 & 0 & \frac {1-\theta_0}{z} & 0 \\
0 & 0 & z (1-\theta_0) & 0 & z-\frac{1}{z} & 0 & z-\frac{1}{z} & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & z (1-\theta_0) & 0 & z-\frac{1}{z} & 0\\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & z-{\frac {\theta_0}{z}}
\end {pmatrix}
\label{R17V2}
\end{equation}
To the best of our knowledge, this $R$-matrix is new.

\subsection{$14V$ model}

The transformation \eqref{eq:gradation} acting on $H_{red}$ (eq. (5.35) of \cite{r1d3}) with $\alpha={\tau_p}^{-1/2}$ leads to an Hamiltonian which does not depend any longer on $\tau_p$:

\begin{equation}
H_{14V} = E_{01} \otimes E_{10} + E_{01} \otimes E_{21} - E_{21} \otimes E_{01} + E_{12} \otimes E_{21} + E_{02} \otimes E_{20} + \sum_{i,j} v_{ij} E_{ii} \otimes E_{jj}
\end{equation}
where the non vanishing $v_{ij}$ are $v_{02}=v_{20}=1/2$, $v_{11}=1$, $v_{12}=3/2$, $v_{21}=\xi-3/2$, $v_{22}=\xi$.

Although this Hamiltonian is very simple, one cannot construct a suitable $R$-matrix. More precisely, the following result can be proved by a brute force calculation: it is not possible to find a univariate or a bivariate $R$-matrix satisfying the YBE unless one chooses $\xi=2$. In that case, the obtained Hamiltonian $H_{14V}\vert_{\xi=2}$ is a particular case of the Hamiltonian $H_{17V_2}$ when $\theta_0=0$, hence the corresponding $R$-matrix is given by \eqref{R17V2} with $\theta_0=0$.

Thus, we have a model whose Hamiltonian is not built from an $R$-matrix, but on which one can perform (at least partially) the CBA. Note that  the pseudo-vacuum and the pseudo-plump are here both necessary (see appendix). However direct calculations for $L=2$ and $L=3$ show that a third reference state is needed for completeness. It \emph{could} be an example of a model solvable by CBA but not by ABA. However, a more detailed study on completeness of the CBA is needed before reaching such a conclusion.

\section*{Acknowledgements}

T.F. was supported by ANR Project
DIADEMS (Programme Blanc ANR SIMI1 2010-BLAN-0120-02).

\appendix

\section{CBA for the second reference state\label{CBA-e1}}

Given a $U(1)$-invariant Hamiltonian $H$, see eqs. \eqref{eqH1} and \eqref{eqH2}, we are interested here in deriving the CBA when considering the second reference state. Let us recall briefly the ansatz for the first reference state (pseudo-vacuum).

\subsection{CBA on the pseudo-vacuum}
 Since the $S^z$ component of the total spin commutes with the Hamiltonian, one can decompose the space of states into subspaces ${\cal V}_M$ with given $S^z$-eigenvalue $M$. The subspace ${\cal V}_0$ ($M=0$) is one-dimensional with basis given by the eigenvector $|\Omega\rangle = \bigotimes_{i=1}^L |0\rangle$, called the pseudo-vacuum, corresponding to the eigenvalue $Lv_{00}$. A basis of states in ${\cal V}_M$ with a given number $M$ of pseudo-excitations is obtained by acting with the raising operator on the pseudo-vacuum such that
\begin{equation}
\vert x_1,\dots,x_M \rangle = \underbrace{|0\rangle \otimes \dots \otimes |0\rangle}_{x_1-1} \otimes |m_1\rangle \otimes \underbrace{|0\rangle \otimes \dots \otimes |0\rangle}_{x_{m_1+1}-x_{m_1}-1} \otimes |m_2\rangle \otimes \underbrace{|0\rangle \otimes \dots \otimes |0\rangle}_{x_{m_1+m_2+1}-x_{m_1+m_2}-1} \otimes |m_3\rangle \otimes \dots
\label{eq:elemstate}
\end{equation}
where $1\leq x_1\leq x_2\leq...\leq x_M\leq L$. The $x_j$'s are the locations of the pseudo-excitations along the chain, and $m_k \in \{1,2\}$ such that $\sum m_k = M$. For $j=1+m_1+\dots+m_{k-1}$, one has $m_k=2$ if $x_{j+1}=x_j$ and $m_k=1$ otherwise. \\

An eigenstate $\Psi_M$ for the Hamiltonian $H$ in ${\cal V}_M$ is given by suitable linear combinations of the elementary states \eqref{eq:elemstate} with coefficients $a(x_1,\dots,x_M)$, which are complex-valued functions to be determined:
\begin{equation}\label{eq:psiM}
\Psi_M = \sum_{1 \le x_1 \le \dots \le x_M \le L} a(x_1,\dots,x_M) \vert x_1,\dots,x_M \rangle,
\end{equation}
and we assume a plane wave decomposition for the functions $a(x_1,\dots,x_M)$:
\begin{equation}
a(x_1,\dots,x_M) = \sum_{\sigma \in {\mathfrak S}_M} A_\sigma^{(j_1,\dots,j_P)}(k_1,\dots,k_M) \exp\left(\sum_{n=1}^M ik_{\sigma(n)} x_n \right) = \sum_{\sigma \in {\mathfrak S}_M} A_\sigma^{(j_1,\dots,j_P)}(\vec{k}) e^{i\vec{k_\sigma}\cdot\vec{x}}.
\label{eq:planewave}
\end{equation}
Here ${\mathfrak S}_M$ is the permutation group of $M$ elements and $A_\sigma^{(j_1,\dots,j_P)}(k_1,\dots,k_M)$ are functions on the symmetric group algebra depending on the Bethe roots $k_n$ to be determined by the so-called Bethe ansatz equations. The indices $(j_1,\dots,j_P)$ correspond to double excitations, i.e. indices such that $x_{j_k+1}=x_{j_k}$ for $k=1,\dots,P$.

The energy of the eigenstate $\Psi_M$ is then given by
\begin{equation}
E_M = Lv_{00} + M(v_{01}+v_{10}-2v_{00}) + \sum_{n=1}^M (q\,e^{ik_{n}}+p\,e^{-ik_{n}})
\label{eq:energie0}
\end{equation}
and the Bethe equations that determine the Bethe roots $k_n$ are (see proposition 3.1 of \cite{r1d3} for an explicit expression of the $S$-matrix)
\begin{equation}
e^{ik_{j}L} = \prod_{n \ne j} S(k_n,k_j)\,, \qquad j=1,...,M.
\label{eq:BAE}
\end{equation}

\subsection{CBA on the pseudo-plump}
There exists another one-dimensional subspace, ${\cal V}_{2L}$, with eigenvector $|\widetilde\Omega\rangle = \bigotimes_{i=1}^L |2\rangle$ corresponding to the eigenvalue $Lv_{22}$. This eigenvector is another possible reference state, that we call the pseudo-plump by opposition to the pseudo-vacuum, on which one can develop the CBA method. The existence of the pseudo-plump is just the reflection of the charge conjugation transformation, $h_{i_1\;i_2}^{j_1\;j_2} \to h_{2-i_1\;2-i_2}^{2-j_1\;2-j_2}$, or in terms of the parameters of the Hamiltonian:
\begin{equation}
v_{ij} \leftrightarrow v_{2-i,2-j} \,, \;\; p\leftrightarrow s_3 \,, \;\; q\leftrightarrow t_3 \,, \;\; t_1\leftrightarrow t_2 \,, \;\; s_1\leftrightarrow s_2 \,, \;\; t_p\leftrightarrow s_p.
\label{eq:transfoC}
\end{equation}
In most cases, there is no need to perform this second CBA, because all eigenstates can be obtained from the pseudo-vacuum. However, there are cases where both CBA are needed to get a complete set.

We choose now as a basis of states in ${\cal V}_N$ with a given number $N$ of pseudo-holes (note that $N=2L-M$), the states obtained by acting with the lowering operator on the pseudo-plump $|\widetilde\Omega\rangle$ such that
\begin{equation}
\vert y_1,\dots,y_N \rangle = \underbrace{|2\rangle \otimes \dots \otimes |2\rangle}_{y_1-1} \otimes |m_1\rangle \otimes \underbrace{|2\rangle \otimes \dots \otimes |2\rangle}_{y_{m_1+1}-y_{m_1}-1} \otimes |m_2\rangle \otimes \underbrace{|2\rangle \otimes \dots \otimes |2\rangle}_{y_{m_1+m_2+1}-y_{m_1+m_2}-1} \otimes |m_3\rangle \otimes \dots
\label{eq:elemstate2}
\end{equation}
where $1\leq y_1\leq y_2\leq...\leq y_N\leq L$. The $y_j$'s are the locations of the pseudo-holes along the chain, and $m_k \in \{1,2\}$ such that $\sum m_k = N$. For $j=1+m_1+\dots+m_{k-1}$, one has $m_k=2$ if $y_{j+1}=y_j$ and $m_k=1$ otherwise. \\

An eigenstate $\Psi_N$ for the Hamiltonian $H$ in ${\cal V}_N$ is given by suitable linear combinations of the elementary states \eqref{eq:elemstate2} with coefficients $b(y_1,\dots,y_N)$, which are complex-valued functions to be determined:
\begin{equation}\label{eq:psiN}
\Psi_N = \sum_{1 \le y_1 \le \dots \le y_N \le L} b(y_1,\dots,y_N) \vert y_1,\dots,y_N \rangle,
\end{equation}
and we assume again a plane wave decomposition for the functions $b(y_1,\dots,y_N)$ (the notations are similar to the ones used for the CBA based on $|\Omega\rangle$):
\begin{equation}
b(y_1,\dots,y_N) = \sum_{\sigma \in {\mathfrak S}_N} B_\sigma^{(j_1,\dots,j_P)}(k_1,\dots,k_N) \exp\left(\sum_{n=1}^N ik_{\sigma(n)} y_n \right) = \sum_{\sigma \in {\mathfrak S}_N} B_\sigma^{(j_1,\dots,j_P)}(\vec{k}) e^{i\vec{k_\sigma}\cdot\vec{y}}.
\label{eq:planewave2}
\end{equation}
The energy of the eigenstate $\Psi_N$ is then given by
\begin{equation}
\widetilde E_N = Lv_{22} + N(v_{21}+v_{12}-2v_{22}) + \sum_{n=1}^M (t_3\,e^{ik_{n}}+s_3\,e^{-ik_{n}})
\label{eq:energie2}
\end{equation}
By consistency, one has of course $\{ \widetilde E_{2L-M} \} = \{ E_M \}$ for a given value of $M$. \\
The Bethe equations has the same shape
\begin{equation}
e^{ik_{j}L} = \prod_{n \ne j} \widetilde S(k_n,k_j)\,, \qquad j=1,...,M.
\end{equation}
where now the $S$-matrix is the image of the original one by the transformation \eqref{eq:transfoC}.

\pagebreak[4]

config: arxiv


TITLE: Extending bounded functions to unbounded ones
QUESTION [1 upvotes]: In a proof I read - which I will omit here since it does not contribute to the question - something has to be proven for a general function $h(x): IR \rightarrow IR^+$ where $h$ is a Borel function.
At the beginning, the proof was restricted to a bounded function $h(x)$.
After the the proof had been finished, it was stated that the assumption of boundedness of $h$ can easily be relaxed by $\min(n,h(x))$, where one should note that $\lim_{n\rightarrow \infty}\min(n,h(x))= h(x)$.
However, I don't manage to see why $\min(n,h(x))$ relaxes the assumption of the bounded function $h$? For me, $|\min(n,h(x))| \leq h(x)$ holds, where $h(x)$ is still bounded, so the boundedness remains?
Thanks in advance for your help! :-)

REPLY [2 votes]: Without seeing what result is being proven about $h$, I can not be certain, but here is a common approach:

For every $n\in\mathbb{N}$, Apply the previous result to the bounded function $x\mapsto \textrm{min}(n,h(x))$.

Consider the sequence of functions, $(\textrm{min}\{n,h(x)\})_{n\in\mathbb{N}}$ which converges to $h(x)$. Every element of the sequence satisfies the property implied by your proof.

If whatever property in your proof is preserved "in the limit" then you are done. More precisely, if the set of functions satisfying your property is closed, then you are done. It should be noted that there are plenty of properties which are not preserved "in the limit" (e.g. boundedness)!


TITLE: sign of odd central moments of binomial distribution
QUESTION [2 upvotes]: I am interested in the sign of odd, central moments of a binomial distribution. From DOI. 10.1137/070700024 I have the formulae:
$ E\left[\left(X-\mu\right)^d\right]= \sum_{i=0}^n\binom{n}{i}\left(-p\right)^i\sum_{l=0}^{i}\left(-1\right)^l\binom{i}{l}\left(l-\mu\right)^d$
where i can see that in the odd case the sign of $\left(l-\mu\right)^n$ changes if $l > pn$. But I get no control about all the other stuff, not even the inner sum.
Another way to approach that, also from this paper mentioned above, would be to use the identity:
$E\left[\left(X-\mu\right)^d\right]= \sum_{i=0}^n\binom{n}{i}\left(-\mu\right)^{d-i}E\left[X^d\right]$
and take advantage of the fact that:
$\left(np\right)^d \leq E\left[X^d\right] \leq n^d$
I would highly appreciate to derive the sign of an odd, central moment in dependence of $p$ and $n$, where $p \neq 0.5$.

REPLY [3 votes]: Let $X$ have the binomial distribution with parameters $n,p$. Then for any natural $d$
$$E(X-np)^d=E\Big(\sum_1^n Y_i\Big)^d,$$
where $Y,Y_1,\dots,Y_n$ are iid random variables such that $P(Y=q)=p=1-P(Y=-p)$, with $q:=1-p$. Let $d$ be a natural number.
Suppose now that $p\le1/2$. Then for all natural $k$
$$m_k:=EY^k=pq(q^{k-1}-(-p)^{k-1})\ge0.$$
Also, $E\Big(\sum_1^n Y_i\Big)^d$ is a linear combination with positive coefficients of terms of the form $m_{k_1}\cdots m_{k_j}\ge0$ with natural $k_1,\dots,k_j$.
Thus, $E(X-np)^d\ge0$ if $p\le1/2$. By symmetry, $E(X-np)^d\le0$ if $p\ge1/2$ and $d$ is odd.
Working slightly harder, we can see that, if $d$ is an odd number $\ge3$, then
$$
E(X-np)^d\begin{cases}
>0&\text{ if }0<p<1/2,\\
<0&\text{ if }1/2<p<1,\\
=0&\text{ if }p=1/2.
\end{cases}
$$


TITLE: Is there a function that returns the degree of a polynomial?
QUESTION [12 upvotes]: I'm playing around with some equations in Geogebra and the degree of the polynomials I am using has become a variable. Is there an exact method for determining the degree of a polynomial?
I am currently approximating the degree of $f(x)$ by using:
$$\text{Degree}(f(x)) = \lfloor\log_{1000}(f(1000))\rfloor$$
This has been a decent enough approximation, but it got me wondering whether or not there exists a precise method? One that could be applied to non-integer degree polynomials.

REPLY [23 votes]: According to the GeoGebra wiki, the function you're looking for is Degree[<Polynomial]. If you have a polynomial in several variables, you can also use Degree[<Polynomial>,<Variable>] to get the degree of the polynomial in the specified variable.


TITLE: Easier way to prove a really simple theorem
QUESTION [2 upvotes]: While I was in class I noticed that if we take a number $n$ and we subtract from $n$ his digits we always get a multiple of $9$. More specifically:
Any number can be expressed as $10p + b$ (for example $42$ is $4\cdot 10 + 2$); the digits of $n$ will be $p$ and $b$ so:
$$10p + b-(p+b)=9p$$
Q.E.D
I'm asking if there is another way to prove this little theorem for example without using the $10p + b-(p+b)$ notation

REPLY [1 votes]: Note that $$10≡1\mod 9.$$
Therefore for any $n \in N$ $$10^n≡1\mod 9.$$  Take any integer of the form $$a_n...a_2a_1a_0=a_n10^n+...+a_210^2+a_110+a_0.$$ Then $$a_n...a_2a_1a_0-(a_n+...+a_2+a_1+a_0)\\=a_n(10^n-1)+...+a_2(10^2-1)+a_1(10-1)$$ is divisible by $9.$
For similar result:
(1) Take any integer with more than $1$ digit.
(2) Compute the sum of digits in even places and odd places separately.
(3) Obtain the difference of them.
(4) Subtract and add it to the initial number.
At least one of these number is divisible by $11.$


text:
\paragraph{Linear model and ordinary least squares estimator.}

Consider a collection $\Fc$ of functions $f : \Xc \to \reals$ on a probability space $(\Xc, \Ac, P)$. We think of $f(x)$ as the real-valued response given an input $x \in \Xc$. Given an independent random sample $X_1, \ldots, X_n$ from $P$ together with the associated responses $f(X_1), \ldots, f(X_n)$ for every $f \in \Fc$, we wish to learn the values $f(x)$ of the response functions $f \in \Fc$ for new but yet unobserved inputs $x \in \Xc$. To this end, we map the input space $\Xc$ into a feature space $\Rd$ via a feature map $h : \Xc \to \Rd : x \mapsto h(x) = (h_1(x), \ldots, h_d(x))^\T$. One of the feature functions $h_j$ could be the constant function $1$, corresponding to an intercept. The response functions are modelled as linear functionals of the mapped features with coefficients estimated by ordinary least squares. The approximation to the response function $f \in \Fc$ is thus
\[
	\forall x \in \Xc, \qquad
	\hat{f}(x) = h(x)^\T \hat{\beta}_f
	\quad \text{ where } \quad \hat{\beta}_f = \argmin_{b \in \Rd} \sum_{i=1}^n \{f(X_i) - h(X_i)^\T b\}^2.
\]
We wish to control the learning error $\hat{f} - f$ uniformly in $f \in \Fc$.

Sharing the same inputs and the same feature map across multiple response functions brings computational gains. Classical least-squares theory yields
\[
	\forall f \in \Fc, \qquad
	\hat{\beta}_f
	= G_n^{-1} \frac{1}{n} \sum_{i=1}^n h(X_i) f(X_i)
	\quad \text{ where } \quad G_n = \frac{1}{n} \sum_{i=1}^n h(X_i) h(X_i)^\T.
\]
In case the empirical Gram matrix $G_n$ is not invertible, a pseudo-inverse is used instead. It follows that the predicted responses are linear in the observed responses:
\[
	\forall f \in \Fc, x \in \Xc, \qquad
	\hat{f}(x) = \frac{1}{n} \sum_{i=1}^n w(x, X_i) f(X_i)
	\quad \text{ where } \quad w(x, X_i) = h(x)^\T G_n^{-1} h(X_i).
\]
The weights $w(x, X_i)$ do not depend on the response function $f$. This invariance is a computational advantage if multiple response functions $f \in \Fc$ are to be learned simultaneously.

\paragraph{Worst-case excess prediction risk.}

The response functions are modelled as linear functionals of the mapped features via
\begin{equation}
\label{eq:linmod}
	\forall f \in \Fc, \forall x \in \Xc, \qquad
	f(x) = h(x)^{\T} \beta_f + \eps_f(x).
\end{equation}
The coefficient vector $\beta_f$ is defined as the minimizer over $b \in \Rd$ of the prediction risk
\[
	\forall f \in \Fc, \forall b \in \Rd, \qquad
	L_f(b) = \expec[\{f(X) - h(X)^\T b\}^2].
\]
Here, the expectation is taken with respect to a random feature $X$ with distribution $P$ and it is assumed that $f$ and $h$ have finite second moments. Let $G = \expec[h(X) h(X)^\T]$ denote the $d \times d$ Gram matrix of the feature map, assumed to be positive definite. Classical least squares theory yields
\[
	\forall f \in \Fc, \qquad
	\beta_f = \argmin_{b \in \Rd} L_f(b) = G^{-1} \expec[h(X) f(X)].
\]

Since $\eps_f(X) = f(X) - h(X)^\T \beta_f$ is orthogonal to $h(X)$, i.e., $\expec[h(X) \eps_f(X)] = 0$, the \textit{excess risk} associated to any other coefficient vector $b \in \Rd$ is
\[
	L_f(b) - L_f(\beta_f) = \expec[\{h(X)^\T(b - \beta_f)\}^2] = (b - \beta_f)^\T G (b - \beta_f).
\]

The optimal coefficient vector $\beta_f$ is unknown, so we estimate it by the least-squares estimator $\hat{\beta}_f$. The expected squared error made by the approximated response function for a new input distributed according to $P$ is
\begin{equation}
\label{eq:Lfdecomp}
	\int_{\Xc} \{\hat{f}(x) - f(x)\}^2 \, \diff P(x)
	=
	L_f(\hat{\beta}_f)
	=
	L_f(\beta_f) + (\hat{\beta}_f - \beta_f)^\T \, G \, (\hat{\beta}_f - \beta_f).
\end{equation}
The right-hand side of \eqref{eq:Lfdecomp} decomposes the expected squared error into two parts.
\begin{itemize}
	\item The first term, $L_f(\beta_f)$, is deterministic. It represents the \emph{modelling error} stemming from the linear model in~\eqref{eq:linmod}. Given the model, this term is incompressible and does not depend on the learning algorithm nor on the training data.
	\item The second term on the right-hand side in \eqref{eq:Lfdecomp} is random. It represents the \emph{learning error} due to the estimation step and the randomness of the training data.
\end{itemize}

It is on the learning error that we focus our analysis, with the particularity that we consider the error uniformly in the response function. Our object of interest is thus the \textit{worst-case excess prediction risk}
\begin{equation}
\label{eq:Lfworst}
	\sup_{f \in \Fc} \{L_f(\hat{\beta}_f) - L_f(\beta_f)\}
	=
	\sup_{f \in \Fc} (\hat{\beta}_f - \beta_f)^\T \, G \, (\hat{\beta}_f - \beta_f).
\end{equation}
We are interested in the rate at which this supremum tends to zero as the sample size $n$ and the dimension $d$ of the feature map tend to infinity.

It is to be emphasized that our setting is that of a random design. The excess prediction risk $L_f(\hat{\beta}_f) - L_f(\beta_f)$ is a nonnegative random variable that is constructed out of the random training sample $X_1, \ldots, X_n$. As is clear from \eqref{eq:Lfdecomp}, it incorporates the risk associated to a new and yet unobserved input $x \in \Xc$, averaged over $P$. The expression in \eqref{eq:Lfworst} is thus a supremum over potentially infinitely many random variables, each variable being built on the same inputs.

\paragraph{Excess prediction risk of a single response.}

Fix a response function $f \in \Fc$. Let $P_n$ denote the empirical distribution of the training sample $X_1, \ldots, X_n$, assigning probability $1/n$ to each observed input. For a real-valued, vector-valued or matrix-valued function $g$ on $\Xc$, expectations with respect to the unknown sampling distribution $P$ and the empirical distribution $P_n$ are denoted respectively by
\begin{align*}
	P(g) &= \expec[g(X)] = \int_{\Xc} g(x) \, \diff P(x), &
	P_n(g) &= \frac{1}{n} \sum_{i=1}^n g(X_i).
\end{align*}
Both operators are linear: for instance, $P(Ag) = A P(g)$ and $P_n(A g) = A P_n(g)$ if $A$ is a linear map between Euclidean spaces of suitable dimension. Using this operator notation, we get
\begin{align*}
	G &= P(h h^\T), & \beta_f &= G^{-1} P(h f), \\
	G_n &= P_n(h h^\T), & \hat{\beta}_f &= G_n^{-1} P_n(h f).
\end{align*}
Since $f = h^\T \beta_f + \eps_f$, the estimated coefficient vector is
\[
	\hat{\beta}_f
	= G_n^{-1} P_n[h (h^\T \beta_f + \eps_f)]
	= \beta_f + G_n^{-1} P_n(h \eps_f).
\]
The excess prediction risk is thus
\begin{equation}
\label{eq:Lfrepres0}
	L_f(\hat{\beta}_f) - L_f(\beta_f)
	=
	P_n(h \eps_f)^\T G_n^{-1} G G_n^{-1} P_n(h \eps_f).
\end{equation}

The predicted response functions $\hat{f}$ are linear combinations of the components $h_1, \ldots, h_d$ of the feature map $h$. They only depend on the feature map $h$ through the linear span of the functions $h_1, \ldots, h_d$. Therefore, the predicted responses remain unchanged if we compose the feature map with an invertible linear transformation $A$ of $\Rd$. Let $G^{\half}$ be the unique symmetric square root matrix of $G$ and let $G^{-\half}$ be its inverse. The whitened feature map is $\hbar = G^{-\half} h : \Xc \to \Rd$. If the random input $X$ has distribution $P$, then $\expec[ \hbar(X) \hbar(X)^\T ] = P(\hbar \hbar^\T) = I_d$, the $d \times d$ identity matrix. The empirical Gram matrix of the whitened feature map is
\[
	P_n(\hbar \hbar^\T) = G^{-\half} P_n(h h^\T) G^{-\half} = G^{-\half} G_n G^{-\half}.
\]
Since $h = G^{\half} \hbar$ and $P_n(\hbar \hbar^\T)^{-1} = G^{\half} G_n^{-1} G^{\half}$, the excess prediction risk in \eqref{eq:Lfrepres0} becomes
\begin{equation}
\label{eq:Lfrepres}
	L_f(\hat{\beta}_f) - L_f(\beta_f)
	=
	\abs{ P_n(\hbar \hbar^\T)^{-1} P_n(\hbar \eps_f) }_2^2,
\end{equation}
where $\abs{y}_2 = (y^\T y)^{\half}$ denotes the Euclidean norm of a vector $y \in \Rd$.

Since $P(\hbar \hbar^\T) = I_d$, it is reasonable to expect that $P_n(\hbar \hbar^\T)^{-1}$ is approximately equal to $I_d$, at least if $n$ is large and $d$ is not too large compared to $n$; see Lemma~\ref{lem:Gninv} below for a precise statement. In that case, the excess prediction risk in \eqref{eq:Lfrepres} is approximately equal to $\abs{ P_n(\hbar \eps_f) }_2^2$. The expectation of the latter random variable can be easily calculated: since $P(\hbar \eps_f) = 0$, the terms with $i \ne j$ in the double sum below vanish and we find
\begin{equation}
\label{eq:gamma}
	\expec\left[ \abs{ P_n(\hbar \eps_f) }_2^2 \right]
	= \frac{1}{n^2} \sum_{i=1}^n \sum_{j=1}^n \expec[\eps_f(X_i) \hbar(X_i)^\T \hbar(X_j) \eps_f(X_j)]
	= \frac{1}{n} P(\hbar^\T \hbar \eps_f^2).
\end{equation}
The excess prediction risk for a single response function $f$ can thus be expected to have an order of magnitude equal to $n^{-1} P(\hbar^\T \hbar \eps_f^{2})$. In comparison, in the fixed-design case, where the training sample is considered as non-random, the expected excess risk is equal to $\sigma_f^2 d / n$, where $\sigma_f^2$ is the error variance \citep{hsu+etal:2014}. Eq.~\eqref{eq:gamma} motivates why in Theorem~\ref{thm:main}, the convergence rate for the worst-case prediction risk over the whole response family $\Fc$ involves the quantity $\supf P(\hbar^\T \hbar \eps_f^2)$.


TITLE: A random walk on the positive integers with P[+1 Step] <= P[-1 Step] and a reflecting boundary at the origin
QUESTION [2 upvotes]: Consider a one-dimensional random walk on the positive integers with:
(1) - A fully reflecting boundary at "x = 0".
(2) - Initiation of the walker at "x = 1".
(3) - A P[+1 Step] / P[RHS Step] probability for the walker less than or equal to the P[-1 Step] / P[LHS Step] probability.
What's the average distance of the walker from the origin and the probability distribution of the walker?

REPLY [1 votes]: Added:
As Douglas Zare points out, there is no limiting distribution.
There are two different limits as time runs over even or odd values.
The distribution I describe below is the average of these two, and hence
my $\pi_n$ means the long run average probability that the walk finds
itself in state $n$.
More formally, I mean that
$$\pi_n=\lim_{t\to\infty} {1\over 2} \left(\mathbb{P}(X(t)=n)+\mathbb{P}(X(t+1)=n)\right)$$
where $X(t)$ is the position of the walk at time $t$.

Suppose that the walk has been running for a long time and
has reached its equilibrium. We need the probability of a
negative step (which I'll call $q$) to be strictly greater
than probability of a positive step (which I'll call $p$),
otherwise the equilibrium doesn't exist.
The steady state probabilities are
$$\pi_n=\cases{{q-p\over 2q}   & $n=0$\\[8pt]
               {q-p\over 2pq} \left({p\over q}\right)^n & $n\geq 1$}$$
Therefore, the average position of the walk (in the long run) is
$$\sum_{n=0}^\infty n \pi_n ={1\over 2(q-p)}.$$
When $p$ and $q$ are close together, this number is very large.
But as $q$ gets close to 1, the walk spends most of its time
jumping back and forth between positions 0 and 1, and its average
position is close to $1/2$.


TITLE: Cup Product Structure on the Projective Space
QUESTION [10 upvotes]: I am reading about cup products and am stuck on this exercise in Hatcher (3.2.5). Taking as given that $H^*(\mathbb{R}P^\infty,\mathbb{Z}_2)\simeq\mathbb{Z}_2[\alpha]$, how does one show $H^*(\mathbb{R}P^\infty,\mathbb{Z}_4)\simeq \mathbb{Z}_4[\alpha,\beta]/(2\alpha,2\beta,\alpha^2)$ with $\deg(\alpha)=1,\deg(\beta)=2$?
The cohomology groups are $\mathbb{Z}_4$ in position 0 and $\mathbb{Z}_2$ in every other position. I want to find the cup product structure. To do that, I need to find a cochain map $f:C^\bullet(\mathbb{R}P^\infty,\mathbb{Z}_4)\rightarrow C^\bullet(\mathbb{R}P^\infty,\mathbb{Z}_2)$ induced by the ring map $\mathbb{Z}_4\rightarrow \mathbb{Z}_2$.
The two cochain complexes look like
$
\cdots {\leftarrow}\mathbb{Z}_4\stackrel{0}{\leftarrow} \mathbb{Z}_4\stackrel{2}{\leftarrow} \mathbb{Z}_4\stackrel{0}{\leftarrow} \mathbb{Z}_4\leftarrow 0
$
and
$
\cdots {\leftarrow}\mathbb{Z}_2\stackrel{0}{\leftarrow} \mathbb{Z}_2\stackrel{0}{\leftarrow} \mathbb{Z}_2\stackrel{0}{\leftarrow} \mathbb{Z}_2\leftarrow 0
$
Here are my questions:
$\bullet$ what is the induced cochain map?
$\bullet$ Can someone tell me why $\alpha\cup \alpha=0$?
$\bullet$ Why does $\alpha\cup \beta$ generate in degree 3?
Is there a quick way to answer these geometrically?

REPLY [2 votes]: Looking at complexes we see that the induced map of cohomology groups is an isomorphism in even degrees and zero in odd degrees (so the notation is slightly misleading: $\alpha$ maps to $0$ and not to $\alpha$). So $\alpha^2=0$ (since $f(\alpha^2)=f(\alpha)^2=0)$ and powers of $\beta$ generate even-dimensional cohomology.
The remaining question is why $\alpha\beta^n$ is non-zero. From Gysin exact sequence (see Hatcher 4.D) for the sphere bundle $S^1\to S^\infty\to\mathbb RP^\infty$,
$$
\ldots\to H(S^\infty)=0\to H^k(\mathbb RP^\infty)\to H^{k+2}(\mathbb RP^\infty)\to H(S^\infty)=0\to\ldots,
$$
we see that multiplication by $\beta$ gives (for $k>0$) an isomorphism $H^k(\mathbb RP^\infty)\to H^{k+2}(\mathbb RP^\infty)$. QED. (But certainly Hatcher had some other argument in mind...)