We derive the exact S-matrix for the scattering of particular representations of the
centrally-extended
Article funded by SCOAP3
= \left|\phi\right> \otimes \left|\phi\right> \,, \qquad & \left|z\right> = \left|\psi\right> \otimes \left|\psi\right> \,, \\ \left|\zeta\right> = \left|\phi\right> \otimes\left|\psi\right> \,, \qquad & \left|\chi\right> = \left|\psi\right> \otimes \left|\phi\right> \,, \end{split}\end{equation} where $\phi$ is bosonic and $\psi$ is fermionic, such that we expect one of the factors of $\mathfrak{psu}(1|1)$ to act on each of the two entries. Furthermore, as a consequence of the form of the symmetry algebra and the integrability of the theory~\cite{Sorokin-ml-2011rr,Murugan-ml-2012mf} we expect that the S-matrix for $y$, $z$, $\zeta$ and $\chi$ can be constructed as a graded tensor product of an S-matrix for $\phi$ and $\psi$, with each factor S-matrix invariant under the symmetry $\mathfrak{psu}(1|1) \ltimes \mathbb{R}^3$. In this section we will construct the relevant massive representation of $\mathfrak{psu}(1|1) \ltimes \mathbb{R}^3$. This representation has an obvious massless limit, and, by analogy with the construction for $AdS_3 \times S^3 \times T^4$~\cite{BogdanLatest,BogdanLatest.m001}, one may expect the massless modes to also transform in representations of $\mathfrak{psu}(1|1) \ltimes \mathbb{R}^3$ in the light-cone gauge-fixed theory. The massless limit is discussed in detail in section~\ref{massless}. Let us also briefly mention that there is an additional $\U(1)$ outer automorphism symmetry~\cite{amsw} of the S-matrix~\eqref{redu}, under which the $\mathfrak{psu}(1|1)$ factors transform in the vector representation. The origin of this $\U(1)$ symmetry is the $T^6$ compact space that is required for a consistent 10-d superstring theory. Under this symmetry $(\z,\chi)^T$ also transforms as a vector, while the bosons are uncharged. It is worth noting that taking the tensor product of two copies of any S-matrix for $\phi$ and $\psi$ \corr{preserving the value of $(-1)^F$, where $F$ is the fermion number operator,} we find that the $\U(1)$ symmetry is present so long as a certain quadratic relation between the parametrizing functions is satisfied (see appendix~\ref{appu1}). In the case of interest, this quadratic identity turns out to be true just from demanding invariance under the $\mathfrak{psu}(1|1) \ltimes \mathbb{R}^3$ symmetry and satisfaction of the Yang-Baxter equation. The $\U(1)$ does not act in a well-defined way on the individual factor S-matrices and hence for now we will ignore it. We will reconsider it in section~\ref{secu1}, where it will play a role in defining a pseudovacuum, an important first step in the algebraic Bethe ansatz. ]]>
&=\left|\psi\right> \,, & \mathfrak S\left|\psi\right>&=2 \, C \left|\phi\right> \,,& \mathfrak N\left|\phi\right>&=\nu\left|\phi\right> \,, & \mathfrak N\left|\psi\right>&=(\nu-1)\left|\psi\right> \,, \nonumber \\ \mathfrak C\left|\Phi\right>&= C\left|\Phi \right> \ & \forall \, \, \, |\Phi\rangle \, &\in \,\{ |\phi\rangle, \, |\psi\rangle\}, & C, \nu \, &\in \, \mathbb{C}, & C &\neq 0. \label{undefa} \end{align} {\color{black} We have summarized the generator action in figure~\ref{fig:fig1}.}\begin{figure} \centerline{\includegraphics[scale=0.3,viewport = 0 0 16cm 18.1cm]{phipsi-m.pdf}} ]]>
&=2 \, C \left|\phi\right> \,, & \mathfrak S\left|\phi\right>&=\left|\psi\right> \,, & \mathfrak N\left|\phi\right>&=(\nu-1)\left|\phi\right> \,, & \mathfrak N\left|\psi\right>&=\nu\left|\psi\right> \,, \nonumber \\ \mathfrak C\left|\Phi\right>&= C\left|\Phi \right> \ & \forall \, \, \, |\Phi\rangle \, &\in \, \{|\phi\rangle, \, |\psi\rangle\}, & C, \nu \, &\in \, \mathbb{C}, & C &\neq 0. \label{undefb} \end{align} However, if $C=0$, the two modules are not isomorphic and they are no longer irreducible. Rather they become reducible but indecomposable. \looseness=-1 To elucidate further we introduce the 1-dimensional modules $\langle \mu \rangle$, which form the atypical (short) irreps of $\mathfrak{gl}(1|1)$. These irreps are characterized by the vanishing of all generators except $\mathfrak N$, which acts with eigenvalue $\mu$. We then see that for the Kac module, $\langle 0, \nu\rangle$, the fermion $|\psi\rangle$ spans a sub-representation $\langle \nu-1\rangle$, and the indecomposable is denoted as \begin{eqnarray} \langle \nu-1 \rangle \, \longleftarrow \langle \nu \rangle \,. \end{eqnarray} The anti-Kac module $\overline{\langle 0, \nu\rangle}$ is also reducible but indecomposable and is denoted as \begin{eqnarray} \langle \nu-1 \rangle \, \longrightarrow \langle \nu \rangle\,, \end{eqnarray} with the fermion $|\psi\rangle$ once again spanning the sub-representation $\langle \nu \rangle$. This indecomposable is \emph{not} isomorphic to $\langle 0, \nu\rangle$. Let us mention that modding out the indecomposable representations by their sub-representations one obtains the \emph{factor} representations, which in this case are isomorphic to the short 1-dimensional $\langle \mu \rangle$ modules and are spanned by the boson $|\phi\rangle$. If we take the tensor product of two typical modules, we get \begin{eqnarray} \langle C_1, \nu_1 \rangle \otimes \langle C_2, \nu_2 \rangle \, &=& \, \langle C_1 + C_2, \nu_1 + \nu_2 - 1 \rangle \oplus \langle C_1 + C_2, \nu_1 + \nu_2 \rangle \qquad \mbox{if} \, \, \, C_1 + C_2 \neq 0\,, \no \\ \langle C_1, \nu_1 \rangle \otimes \langle - C_1, \nu_2 \rangle \, &=& \, P_{\nu_1 + \nu_2}\,, \end{eqnarray} where $P_\nu$ is the so-called \emph{projective module} \begin{eqnarray} \langle \nu \rangle \, \longrightarrow \, \langle \nu + 1 \rangle \oplus \langle \nu - 1 \rangle \, \longrightarrow \, \langle \nu \rangle\,, \end{eqnarray} on which $\mathfrak C$ acts identically as zero. The rightmost 1-dimensional short sub-module $\langle \nu \rangle$ is known as the \emph{socle} of $P_\nu$. Since $\mathfrak N$ does not appear on the r.h.s.\ of the commutation relations, the algebra $\mathfrak{gl}(1|1)$ has a non-trivial ideal generated by $\mathfrak Q$, $\mathfrak S$ and $\mathfrak C$. This ideal is the superalgebra $\mathfrak{sl}(1|1)$. Furthermore, this algebra is also not simple, as $\mathfrak C$, being central, is a non-trivial ideal. Additionally modding out $\mathfrak C$ gives the algebra $\mathfrak{psl}(1|1)$, which is still not simple, as the two remaining anti-commuting fermionic generators each form a separate ideal. The fact that $\mathfrak{psl}(1|1)$ is not simple sets this algebra outside the classification of the possible central extensions of basic classical Lie superalgebras presented in~\cite{IoharaKoga}. ]]>
&=a\left|\psi\right> \,, & \mathfrak Q\left|\psi\right>&=b \left|\phi\right> \,, & \mathfrak S\left|\phi\right>&=c\left|\psi\right> \,, & \mathfrak S\left|\psi\right>&=d\left|\phi\right> \,, \nonumber \\* \mathfrak C\left|\Phi\right>&= C\left|\Phi \right> \,, & \mathfrak P\left|\Phi\right>&=P \left|\Phi \right> \,, & \mathfrak K\left|\Phi\right>&= K \left|\Phi \right> \,. \label{undef} \end{align} Here $a$, $b$, $c$, $d$, $C$, $P$ and $K$ are the representation parameters that will eventually be functions of the energy and momentum of the states. For the supersymmetry algebra to close the following conditions should be satisfied \begin{equation}\begin{split}\label{abcdCPK} a b = P \,, \qquad c d = K \,, \qquad & a d + b c = 2 C \,. \end{split}\end{equation} This representation corresponds to the typical (long) Kac module $\langle C, \nu\rangle$ discussed in the previous section. {\color{black} We have summarized the generator action in figure~\ref{fig:fig2}.} \begin{figure} \centerline{\includegraphics[scale=0.2,viewport = 0 0 36.1cm 27.1cm]{twopointtwo-m.pdf}} ]]>
0 \,, \end{equation} also as a consequence of the reality conditions~\eqref{rc}. Motivated by the fact that $C$ will later be associated to an energy, we will take it to be positive. {\color{black} However, let us point out that the algebraic analysis we perform in this paper is largely insensitive to this choice, \cortwo{and hence it} does not represent a loss of generality. If we make this positivity assumption, it} immediately follows that both $(C+\frac{m}2)$ and $(C -\frac{m}{2})$ are also positive.} {\color{black} The analogy with the higher dimensional AdS/CFT cases \corp{suggests that we should associate \cortwo{(the absolute value of)} $m$ with the} mass of the scattering particle.} Later it will be useful to solve the set of equations~\eqref{abcdCPK} for $a$, $b$, $c$ and $d$ in terms of $\mm$, $C$, $P$ and $K$ \begin{align}\nonumber a &= \a \,e^{-\frac{i\pi}4} \bigg(C + \frac{\mm}2\bigg)^{\frac12} \,, & b &= \a^{-1} e^{\frac{i\pi}4} \bigg(C + \frac{\mm}2\bigg)^{-\frac12}P \,, \\ c &= \a\, e^{-\frac{i\pi}4} \bigg(C + \frac{\mm}2\bigg)^{-\frac12}K \,, & d &= \a^{-1}e^{\frac{i\pi}4} \bigg(C + \frac{\mm}2\bigg)^{\frac12} \, . \label{ad} \end{align} Here $\a$ is a phase parametrizing the normalization of the fermionic states with respect to the bosonic states and can be a function of the central extensions. To define the action of this symmetry on the two-particle states we need to introduce the coproduct \begin{align} \Delta(\mathfrak Q) & = \mathfrak Q \otimes \id + \mathfrak U \otimes \mathfrak Q \,, & \nonumber & & \Delta(\mathfrak S) & = \mathfrak S \otimes \id + \mathfrak U^{-1} \otimes \mathfrak S \,, \\ \Delta(\mathfrak P) & = \mathfrak P \otimes \id + \mathfrak{U}^2 \otimes \mathfrak P \,, & \Delta(\mathfrak C) & = \mathfrak C \otimes \id + \id \otimes \mathfrak C \,, & \Delta(\mathfrak K) & = \mathfrak K \otimes \id + \mathfrak{U}^{-2} \otimes \mathfrak K \,,\label{coproduct} \end{align} and the opposite coproduct, defined as \begin{equation} \Delta^{\text{op}}(\mathfrak J) = \mathcal{P}\,\Delta(\mathfrak J)\,, \end{equation} where $\mathfrak J$ is an arbitrary abstract generator (prior to considering a representation), and $\mathcal{P}$ defines the graded permutation of the tensor product. \looseness=-1 The coproduct differs from the trivial one by the introduction of a new abelian generator $\mathfrak U$, with $\Delta(\mathfrak U) = \mathfrak U \otimes \mathfrak U$~\cite{tor,tor.m001}. This is done according to a $\mathbb{Z}$-grading of the algebra, whereby the charges $-2,-1,1$ and $2$ are associated to the generators $\mathfrak K$, $\mathfrak S$, $\mathfrak Q$ and $\mathfrak P$ respectively, while $\mathfrak C$ remains uncharged. The action of $\mathfrak U$ on the single-particle states is given by \begin{equation} \mathfrak U\left|\phi\right> = U\left|\phi\right> \,, \qquad\qquad \mathfrak U\left|\psi\right> = U\left|\psi\right> \,.\label{uu} \end{equation} This braiding allows for the existence of a non-trivial S-matrix. One important consequence of the non-trivial braiding~\eqref{coproduct} is that it leads to a constraint between $U$ and the eigenvalues of the central charges. This follows from the requirement that, to admit an S-matrix, the coproduct of any central element should be equal to its opposite.\footnote{If $\Delta(\gen{c})$ is central, then \begin{equation*} \Delta^{op}(\gen{c}) \, R \, = \, R \, \Delta(\gen{c}) = \Delta(\gen{c}) \, R\,, \end{equation*} which, for an invertible R-matrix, necessarily implies $\Delta^{op}(\gen{c})=\Delta(\gen{c})$. This is expressed by saying that the coproduct of $\gen{c}$ is \emph{co-commutative}.\label{foo10}} This implies \begin{equation} \mathfrak P \propto (1 - \mathfrak U^2) \,,\qquad \qquad \mathfrak K \propto (1 - \mathfrak U^{-2}) \,. \label{pp} \end{equation} \looseness=-1 We fix the normalization of $\mathfrak P$ relative to $\mathfrak K$ by taking both constants of proportionality to be equal to $\frac12 \hh$ where the reality conditions~\eqref{rc} require that $\hh$ is real.\footnote{The reality conditions~\eqref{rc} do allow for the introduction of an additional phase into the constants of proportionality, i.e.\ $\frac12 \hh e^{i\varphi}$ and $\frac12 \hh e^{-i\varphi}$. However, this phase does not appear in the S-matrix and thus we set $\varphi=0$.} The parameter $\hh$ is a coupling constant and eventually should be fixed in terms of the string tension, which we will return to in section~\ref{secpert}. Acting on the single-particle states then gives us the relations \begin{equation}\label{pku} P = \frac \hh2 \, (1 - U^2) \,,\qquad \qquad K = \frac \hh2 \, (1 - U^{-2}) \,, \end{equation} where $U$ should satisfy, as a consequence of~\eqref{rc}, the following reality condition \begin{equation}\label{urc} U^* = U^{-1} \,. \end{equation} The relation~\eqref{short} in terms of $C$, $U$ and $\mm$ is then given by \begin{equation}\label{apii} C^2 = \frac{\mm^2-\hh^2(U-U^{-1})^2}4\,. \end{equation} While this is a single equation for three undetermined parameters, we will later still attempt to interpret it as a dispersion relation with $C$, $U$ and $\mm$ defined in terms of just two kinematic variables, the energy and momentum. These precise definitions are not fixed by symmetry considerations, and hence should be found from direct string computations. It is now useful to introduce the Zhukovsky variables $x^\pm$, in terms of which we will write the S-matrix, in place of the central extensions, $C$ and $U$. These are defined as~\cite{beis0,beis1,beis1.m001} \begin{equation} \label{xx} U^2 = \frac{x^+}{x^-} \,, \qquad\qquad 2C + \mm = i \hh (x^- - x^+) \,, \end{equation} In these variables the dispersion relation~\eqref{apii} takes the following familiar form \begin{equation} x^+ + \frac{1}{x^+} -x^- - \frac{1}{x^-} = \frac{2i\mm}{\hh} \,. \label{yy} \end{equation} The representation parameters $a$, $b$, $c$ and $d$ in~\eqref{ad} and~\eqref{bep} are then given by \begin{align} & a =\, \a \,e^{-\frac{i\pi}4} \sqrt[4]{\frac{x^+}{x^-}}\sqrt{\frac{\hh}2}\ \h \,, \qquad && b =\, \a^{-1} e^{-\frac{i\pi}4} \sqrt[4]{\frac{x^-}{x^+}}\sqrt{\frac{\hh}2} \ \frac{\h}{x^-} \,, \no \\ & c =\, \a\, e^{\frac{i\pi}4} \sqrt[4]{\frac{x^+}{x^-}} \sqrt{\frac{\hh}2} \ \frac{\h}{x^+} \,, \qquad && d =\, \a^{-1}e^{\frac{i\pi}4} \sqrt[4]{\frac{x^-}{x^+}} \sqrt{\frac{\hh}2}\ \h \,, \label{da} \end{align} where \begin{equation}\label{ett} \h \equiv \sqrt{i(x^- - x^+)} \,. \end{equation} \looseness=-1 Here we clearly see that the advantage of these variables is that the parameters $a$, $b$, $c$ and $d$ do not depend on $\mm$ and hence, written as a function of $x^\pm$ and $m$, neither will the S-matrix. Finally, let us note that for the reality conditions~\eqref{rc} we have the usual $(x^\pm)^* = x^\mp$. We could also eliminate the central extensions, $C$ and $U$, in terms of two variables that will later be identified with the energy and momentum. Motivated by the $AdS_5 \times S^5$ case we write \begin{equation}\label{cuep} C = \frac{\ee}{2} \,,\qquad \qquad U = e^{\frac i2 \pp} \,, \end{equation} where $\ee$ is the energy and $\pp$ is the spatial momentum. \corr{While the identification of $\ee$ with the energy and $\pp$ with the spatial momentum is {\color{black} at present} only motivated by analogy with the $AdS_5 \times S^5$ case, a posteriori it will be further justified by matching with perturbative results in section~\ref{secpert}.} Solving for $x^\pm$ in terms of $\ee$ and $\pp$ we find \begin{equation}\begin{split} x^\pm = r\, U^{\pm 1} \,, \qquad r= \frac{ \ee + \mm }{2 \hh\,\sin\frac {\pp}2} =\frac{2\hh\,\sin\frac{\pp}2}{\ee - \mm }\,, \qquad U = e^{\frac{i\pp}{2}} \,. \label{4444} \end{split}\end{equation} Using~\eqref{pku} and~\eqref{cuep} we can substitute in for $C$, $P$ and $K$ in terms of the energy and the momentum in~\eqref{apii} to find the following familiar dispersion relation \begin{equation}\label{apiii} \ee^2 = \mm^2 + 4 \, \hh^2 \, \sin^2\frac{\pp}2 \,. \end{equation} It is important to emphasize that here $\mm$ is algebraically a free parameter. However, for~\eqref{apiii} to really be interpreted as a dispersion relation $m$ should be fixed by the spectral analysis of the theory. In terms of the energy and the momentum the representation parameters $a$, $b$, $c$ and $d$~\eqref{ad} are given by \begin{align}\nonumber a =\, & \frac{\a \,e^{\frac{i\pp}{4}-\frac{i\pi}4}}{\sqrt2} \sqrt{\ee+\mm} \,, &\ b =\, & \frac{\a^{-1} e^{-\frac{i\pp}{4}+\frac{i\pi}4}}{\sqrt2} \frac{\hh(1-e^{i\pp})}{\sqrt{\ee+\mm}} \,, \\ c =\, & \frac{\a\, e^{\frac{i\pp}{4}-\frac{i\pi}4}}{\sqrt2} \frac{\hh(1-e^{-i\pp})}{\sqrt{\ee+\mm}} \,, & d =\, & \frac{\a^{-1}e^{-\frac{i\pp}{4}+\frac{i\pi}4}}{\sqrt 2} \sqrt{\ee+\mm} \,. \label{bep} \end{align} In the $AdS_5 \times S^5$ and $AdS_3 \times S^3\times M^4$ models, the choice of the phase factor $\a$ that is appropriate for the light-cone gauge-fixed string theory is \begin{equation}\label{nca} \a = 1\,. \end{equation} As we will see, this is also a natural choice for $\a$ in the $AdS_2 \times S^2$ theory. ]]>
PK$, or equivalently that $M$ is real and non-zero (we will briefly discuss the case when $M$ vanishes at the end of this section), the above normalization implies that $|\tilde w_0\rangle$ has the same norm as $|w_0\rangle$. Therefore, the action of $\alg{Q}$ and $\alg{S}$ on $|w_1\rangle$ and $|\tilde w_1\rangle$ is given by \begin{equation}\begin{split} \alg{Q}|w_1\rangle &= P|w_0\rangle \,, \qquad \alg{Q}|\tilde w_1\rangle = C|w_0\rangle + \frac{M}2|\tilde w_0\rangle \,, \\ \alg{S}|\tilde w_1\rangle &= K|w_0\rangle \,, \qquad \alg{S}|w_1\rangle = C|w_0\rangle - \frac{M}2|\tilde w_0\rangle \,. \end{split}\end{equation} Again it is clear that the action of the central elements on $|\tilde w_0\rangle$ is given by \begin{equation} (\alg{P},\alg{K},\alg{C})|\tilde w_0\rangle = (P,K,C)|\tilde w_0\rangle \,. \end{equation} Finally, the action of $\alg{Q}$ and $\alg{S}$ on $|\tilde w_0\rangle$ is given by \begin{equation} \alg{Q}|\tilde w_0\rangle = \frac{2P}{M}|\tilde w_1\rangle - \frac{2C}{M} |w_1\rangle \,, \qquad \alg{S}|\tilde w_0\rangle = -\frac{2K}{M} |w_1\rangle + \frac{2C}{M} |\tilde w_1\rangle \,. \end{equation} Therefore, in summary, we have constructed the following 4-dimensional representation: \begin{equation*} (\alg{P},\alg{K},\alg{C})|\Phi\rangle = (P,K,C)|\Phi\rangle \,, \qquad \forall \ |\Phi\rangle \, \in \, \{|w_0\rangle,|w_1\rangle,|\tilde w_1\rangle,|\tilde w_0\rangle)\} \,, \end{equation*} \begin{align} \alg{Q}|w_0\rangle &= |w_1\rangle \,, & \alg{S}|w_0\rangle &= |\tilde w_1\rangle \,,\no \\ \alg{Q}|w_1\rangle &= P|w_0\rangle \,, & \alg{S}|\tilde w_1\rangle &= K |w_0\rangle \,,\no \\ \alg{Q}|\tilde w_1\rangle &= C |w_0\rangle + \frac{M}2 |\tilde w_0\rangle \,, & \alg{S}|w_1\rangle &= C |w_0\rangle - \frac{M}2 |\tilde w_0\rangle \,,\no \\ \alg{Q}|\tilde w_0\rangle &= \frac {2P}{M} |\tilde w_1\rangle - \frac {2C}{M} |w_1\rangle \,, & \alg{S}|\tilde w_0\rangle &= - \frac {2K}{M} |w_1\rangle + \frac {2C}{M} |\tilde w_1\rangle \,. \end{align} {\color{black} We have summarized the situation in figure~\ref{fig:fig3}.}\begin{figure} \centerline{\includegraphics[scale=0.3,viewport = 0 0 36.1cm 27.1cm]{twopointthree-m.pdf}} ]]>
= S_1 \left|\phi\phi'\right> + Q_1 \left|\psi\psi'\right> \,, \qquad \mathbb{S}\left|\psi\psi'\right> = S_2 \left|\psi\psi'\right> + Q_2 \left|\phi\phi'\right> \,,\nonumber \\ &&\mathbb{S}\left|\phi\psi'\right> = T_1 \left|\phi\psi'\right> + R_1 \left|\psi\phi'\right> \,, \qquad \label{ans} \mathbb{S}\left|\psi\phi'\right> = T_2 \left|\psi\phi'\right> + R_2 \left|\phi\psi'\right> \,, \end{eqnarray} where $x^\pm$, $m$ are the kinematic variables associated to the first particle and $x'^\pm$, $m'$ to the second particle, that is \begin{equation}\label{yyss} x^+ +\frac{1}{x^+} -x^{-}-\frac{1}{x^-} = \frac{2im}{\hh}\,, \qquad\quad x'^+ +\frac{1}{x'^+} -x'^{-}-\frac{1}{x'^-} = \frac{2im'}{\hh}\,. \end{equation} As a consequence of the discussion in section~\ref{32} this symmetry will only fix the S-matrix up to two arbitrary functions. One of these functions can be found by requiring the S-matrix also satisfies the Yang-Baxter equation along with additional physical requirements. There are four solutions to the Yang-Baxter equation, two of which we ignore as they violate crossing symmetry. The other two are related by a sign. To fix the sign, we demand that in the BMN limit (for details see section~\ref{secpert}) the S-matrix reduces to the identity operator. The functions parametrizing the exact S-matrix~\eqref{ans} are then given by\footnote{Note that here we are choosing the branch so that $\bigg(\frac{x^-}{x^+}\bigg)^{\#} = \bigg(\frac{x^+}{x^-}\bigg)^{-\#}$ for $\#=\frac12,\,\frac14$ and similarly for $x'^\pm$. For $\pp\in[-\pi,\pi]$ this corresponds to taking the branch cut on the negative real axis.} \begin{align}\no S_1 &= \sqrt{\frac{x^+\xpr^-}{x^-\xpr^+}}\frac{x^- - \xpr^+}{x^+ - \xpr^-} \frac{1+s_1}{2} \tPo \,, & S_2 &= \frac{1+s_2}{2} \tPo \,, \\ T_1 &=\sqrt{\frac{\xpr^-}{\xpr^+}} \frac{x^+ - \xpr^+}{x^+ - \xpr^-} \frac{1+t_1}{2}\tPo \,, & T_2 &=\sqrt{\frac{x^+}{x^-}} \frac{x^- - \xpr^-}{x^+ - \xpr^-}\frac{1+t_2}{2} \tPo \,,\label{exacta} \\ \frac{Q_1}{\a\a'} \!&=\! \a\a'\, Q_2 \!=\! - \frac{i}{2} \,\sqrt[4]{\frac{x^-\xpr^+}{x^+\xpr^-}} \frac{\h\h'}{x^+-\xpr^-} \frac f{x^-x'^+}\tPo \,, & \frac{\a'}{\a} R_1 \!&=\! \frac{\a}{\a'}R_2 \!=\! -\frac{i}{2} \, \sqrt[4]{\frac{x^+\xpr^-}{x^-\xpr^+}} \frac{\h\h'}{x^+ \!-\! \xpr^-} \tPo \,,\no \end{align} where \begin{align}\label{exacta1} f = \frac{\sqrt{\frac{x^+}{x^-}}(x^--\frac{1}{x^+}) - \sqrt{\frac{x'^+}{x'^-}}(x'^--\frac{1}{x'^+})}{1 - \frac{1}{x^+x^-x'^+x'^-}} \,, \quad s_1 &= \frac{1- \frac{1}{x^+\xpr^-}}{x^- - \xpr^+} f \,, \quad s_2 = \frac{1-\frac{1}{x^-\xpr^+}}{x^+ - \xpr^-} f \,, \\ t_1 &= \frac{1 - \frac{1}{x^-\xpr^-}}{x^+ - \xpr^+} f \,, \quad t_2 = \frac{1 - \frac{1}{x^+\xpr^+}}{x^- - \xpr^-} f \,. \end{align} $\tPo$ is an overall factor that sits outside the matrix structure and is not fixed by symmetries or the Yang-Baxter equation. Let us emphasize that, as discussed beneath eq.~\eqref{da}, when written in these variables the S-matrix is independent of $m$ and $m'$, which can take any value. The limits $m\to0$ and $m'\to0$ are subtle however, and will be discussed in detail in section~\ref{massless}. Let us also note that if we take $\a$ to be given by~\eqref{nca}, which is the choice suitable for string theory, then $Q_1 = Q_2$ and $R_1 = R_2$. From now on we will take $\a$ to be given by this value. The S-matrix~\eqref{redu} can be thought of as a $4 \times 4$ block diagonal matrix \begin{equation} \left(\begin{array}{cccc} S_1 & Q_1 & 0 & 0 \\ Q_2 & S_2 & 0 & 0 \\ 0& 0& T_1 & R_1 \\ 0&0& T_2 &R_2 \end{array}\right)\,.\end{equation} One can then check that each of the two $2 \times 2$ blocks have equal trace and determinant, \begin{equation}\label{trdeteq} S_1 + S_2 = T_1 + T_2 \,, \qquad S_1 S_2 - Q_1 Q_2 = T_1 T_2 - R_1 R_2 \,. \end{equation} The second of these equations is particularly important as it implies the tensor product of two copies of the S-matrix possesses an additional $\U(1)$ symmetry, which will be discussed further in section~\ref{secu1} and appendix~\ref{appu1}. \corr{For completeness let us note that the two solutions that violate crossing symmetry are given by $f=0$ and $f\to\infty$ (for the latter one should first rescale $\tPo$ by $f^{-1}$ and then take $f \to \infty$). As $\phi$ and $\psi$ are real \cortwo{and the charge conjugation matrix diagonal, which will be demonstrated in the next section (cf.~\eqref{eq:C})}, the two processes \begin{equation} \phi\,\phi \to \psi\,\psi \qquad \text{and} \qquad \phi\,\psi \to \psi\,\phi\,, \end{equation} should be related by a crossing transformation. However, if $f$ vanishes then so does the amplitude for the first of these processes, but not for the second. Similarly, if $f\to\infty$ then the amplitude for the second process vanishes, but not for the first. Consequently, in both cases the two processes cannot be related by a crossing transformation and hence there is a violation of crossing symmetry as claimed. It is interesting to note that taking $f=0$ and $f\to \infty$ we recover the massive S-matrices of the $AdS_3 \times S^3 \times T^4$ light-cone gauge superstring~\cite{Borsato-ml-2013qpa,HT}. The symmetry is enhanced accordingly from $\mathfrak{psu}(1|1)\ltimes \mathbb{R}^3$ to $[\mathfrak{u}(1)\inplus \mathfrak{psu}(1|1)^2] \ltimes\mathfrak{u}(1) \ltimes\mathbb{R}^3$. For the $AdS_3 \times S^3 \times T^4$ light-cone gauge-fixed theory there is no issue with crossing symmetry as the fields are complex. Therefore, the individual S-matrices do not map to themselves under the crossing transformation, rather to a different S-matrix with the crossed particle replaced by its antiparticle. Finally let us also point out that the S-matrix relevant for the $AdS_2 \times S^2 \times T^6$ light-cone gauge superstring, see eqs.~\eqref{redu} and~\eqref{exacta}, is a linear combination, with coefficients depending on $x^\pm$ and $x'^\pm$, of the $f=0$ and $f\to \infty$ S-matrices. It is non-trivial that such a combination exists with unitarity, crossing symmetry and the Yang-Baxter equation all satisfied.} \iffalse This makes the formal embedding of~\eqref{redu} into the $AdS_5 \times S^5$ S-matrix hard to implement. Embedding the algebra $\mathfrak{psu}(1|1) \ltimes \mathbb{R}^3$ into $\mathfrak{psu}(2|2)\ltimes \mathbb{R}^3$ is possible.\footnote{One may for instance use $\alg{Q}_1^3 + \alg{Q}_2^4$ and $\alg{S}_3^1 + \alg{S}_4^2$ of $AdS_5 \times S^5$ as $\alg{Q}$ and $\alg{S}$ of the smaller algebra, see section~\ref{massi}. This would be consistent with the central extensions. We thank B. Stefanski for discussions regarding this point.} To embed the S-matrix, one possibility would be that two bosons of the same type scatter to two fermions of the same type in $AdS_5 \times S^5$ (for instance, $\phi^1\, \phi'^1 \to \psi^3\, \psi'^3$, in the notation of~\cite{beis0}). A scattering process like this one is not present in the $AdS_5 \times S^5$ S-matrix. It is prohibited by the extra $\mathfrak{su}(2)^2$ symmetry which is present in $\mathfrak{psu}(2|2)\ltimes \mathbb{R}^3$ but not in $\mathfrak{psu}(1|1)\ltimes \mathbb{R}^3$. An alternative would be to allow different $AdS_5 \times S^5$ states for the two particles being scattered, for instance $(\phi^1, \psi^3)$ and $(\phi'^2, \psi'^4)$. Reducing the S-matrix of~\cite{beis0} according to this choice would leave us with many terms, but not the one involving \begin{eqnarray} \phi^1 \, \psi'^4 \, \to \psi^3 \,\phi'^2 \,. \end{eqnarray} This process is not present before reduction, yet it would be necessary to reproduce the $AdS_2$ S-matrix. Once again the embedding of S-matrices seems not to be straightforward. \fi ]]>
\vv'$, while for a scattering process with $\s = -\s'=-1$ we have $\vv'>\vv$. Therefore, we may associate $\lim_{m,m'\to 0} f \to 1$ with $\vv>\vv'$ and $\lim_{m,m'\to 0}f\to -1$ with $\vv<\vv'$. This is consistent with the crossing symmetry discussed above as the group velocity is invariant under the crossing transformation. Furthermore, one may expect the $\s=-\s'=+1$ and $\s=-\s'=-1$ S-matrices to be related upon interchanging the arguments. Indeed, the following equation is satisfied for real momenta\footnote{Here we are defining $\mathbb{S} |\Phi^{\vphantom{'}}_a\Phi'_b\ket = \mathcal{S}_{ab}^{cd}(\pp,\pp')|\Phi^{\vphantom{'}}_c\Phi'_d\ket$, $\Phi_0 =\phi$, $\Phi_1=\psi$ and $[a]=a$.} \begin{equation}\label{ha} \mathcal{S}^{\pm\mp}{}_{ab}^{cd}(\pp,\pp')\big|_{f\to \pm1} = (-1)^{[a][b]+[c][d]} \mathcal{S}^{\mp\pm}{}_{ba}^{dc}(\pp',\pp)^*\big|_{f\to \mp 1}\,. \end{equation} The corresponding relation for the $\s=\s'=\pm1$ S-matrices is given by \begin{equation}\label{ha2} \mathcal{S}^{\pm\pm}{}_{ab}^{cd}(\pp,\pp')\big|_{f\to \pm1} = (-1)^{[a][b]+[c][d]} \mathcal{S}^{\pm\pm}{}_{ba}^{dc}(\pp',\pp)^*\big|_{f\to \mp 1}\,. \end{equation} To conclude, let us briefly comment on unitarity. Motivated by the physical interpretation outlined above, one may expect that braiding unitarity for the massless S-matrix will involve one S-matrix with $f \to 1$ and one with $f \to -1$, and indeed, one can explicitly check that braiding unitarity relations can be constructed in this way. They are given by \begin{equation}\begin{split}\label{masles} & (-1)^{[c][d]+[e][f]}\mathcal{S}^{\pm\pm}{}_{ab}^{ef}(\pp,\pp')\big|_{f\to \pm1} \mathcal{S}^{\pm\pm}{}_{fe}^{dc}(\pp',\pp)\big|_{f\to \mp 1}\propto\d_a^c\d_b^d\,. \\ & (-1)^{[c][d]+[e][f]}\mathcal{S}^{\pm\mp}{}_{ab}^{ef}(\pp,\pp')\big|_{f\to \pm1} \mathcal{S}^{\mp\pm}{}_{fe}^{dc}(\pp',\pp)\big|_{f\to \mp 1}\propto\d_a^c\d_b^d\,. \end{split}\end{equation} These relations can also be found by taking the massless limit of the braiding unitarity relation for the massive S-matrix. Finally, one can see that by combining~\eqref{ha},~\eqref{ha2} and~\eqref{masles}, all the four massless S-matrices are also QFT unitary so long as the overall factors satisfy appropriate constraints. ]]>