text:
\section{Introduction}
It is well known that the critical dimension of bosonic (or supersymmetric) string theory is 26 (or 10) and we can check this fact by many methods, including Light-cone quantization and BRST quantization \cite{bib:001, bib:002}.
Recently, Luca Mezincescu and Paul K. Townsend showed that by using light-cone gauge a consistent critical string theory can be constructed also in three dimensions, since there is no Lorentz anomaly in three dimensions.
In fact, in three dimensions the dangerous commutator which breaks Lorentz symmetry, $[\mathcal{J}^{-I}, \mathcal{J}^{-J}] \ (I,J=2, \cdots ,D-1)$, vanishes trivially because there is only one transverse direction \cite{bib:003, bib:004, bib:005, bib:021}.
\begin{eqnarray}
[\mathcal{J}^{-I}, \mathcal{J}^{-I}] \equiv 0 ~~~ \mbox{in 3 dim.}
\label{eq:1-1}
\end{eqnarray}
Moreover they found that the spectrum of a 3d string in light-cone gauge includes "anyons", which have non half-integer spins.

The difference between the critical dimension in the light-cone gauge quantization and that in others might mean the fault of the light-cone quantization or the incompleteness of other quantizations including BRST method.
Mezincescu and Townsend suggest that such difference may be caused by the existence of anyon, although it is still not clear whether this is in the case and how the difference arises.
Beside finding a reason for the difference, it is also important to find other examples which give a different result in the light-cone method and others and to invent other quantization schemes which reproduce results obtained by the light-cone method, especially by the covariant one.

In this paper, we investigate 3d tensionless bosonic closed string.
It has been known by using BRST method that tensionless p-branes have no Lorentz anomaly in any dimensions and that conformal tensionless p-branes have the critical dimension for the space-time conformal symmetry, $D=2$ \cite{bib:013, bib:014, bib:015, bib:006}.
And the mass spectrum also has been investigated in \cite{bib:016, bib:020}.

On the other hand, Isberg, Lindstrom, Sundborg and Theodoridis show that there are some anomalous commutators in the space-time conformal group of light-cone gauge \cite{bib:007, bib:017}.
Specifically, $[\mathcal{K}^{I}, \mathcal{J}^{-J}]$, $[\mathcal{K}^{+}, \mathcal{K}^{-}]$, $[\mathcal{K}^{I}, \mathcal{K}^{-}]$ and $[\mathcal{K}^{-}, \mathcal{J}^{-I}]$ have anomalous terms.
Especially Isberg et al. emphasize in \cite{bib:007, bib:017} that the commutator $[\mathcal{K}^{I}, \mathcal{J}^{-J}]$ has anomalous terms in traceless part of $I$ and $J$ and gives the definition of $\mathcal{K}^{-}$ by trace part:
\begin{eqnarray}
[\mathcal{K}^{I}, \mathcal{J}^{-J}] = -i\delta ^{IJ} \mathcal{K}^{-} +L^{IJ}_{t.l.},
\label{eq:1-2}
\end{eqnarray}
where $L^{IJ}_{t.l.}$ is traceless and proportional to the generator of $\sigma $-dependent special linear group.
In $D\geq 4$, we will find that through Jacobi identities all other anomalous commutators are related to this commutator or the definition of $\mathcal{K}^{-}$.
For example, if we choose $J$ different from $I$ and use the fact that there is no Lorentz anomaly for a tensionless string\footnote{$[\mathcal{J}^{-I}, \mathcal{J}^{-J}]\propto $($\sigma $-translation gauge-fixing constraint). This constraint corresponds to the level-matching condition for tensile string and commutes to all generators.}, we obtain
{\setlength\arraycolsep{1pt}\begin{eqnarray}
[\mathcal{K}^{-}, \mathcal{J}^{-I}] &=& i[ [\mathcal{K}^{J}, \mathcal{J}^{-J}]-L^{JJ}_{t.l.}, \mathcal{J}^{-I}] \nonumber \\
&=& -i[ [\mathcal{J}^{-I}, \mathcal{K}^{J}], \mathcal{J}^{-J}] -i[ [\mathcal{J}^{-J}, \mathcal{J}^{-I}], \mathcal{K}^{J}] -i[L^{JJ}_{t.l.}, \mathcal{J}^{-I}] \nonumber \\
&=& i[L^{JI}_{t.l.}, \mathcal{J}^{-J}]-i[L^{JJ}_{t.l.}, \mathcal{J}^{-I}].
\label{eq:1-3}
\end{eqnarray}}
Similarly we will find that all anomalies derive from $L^{IJ}_{t.l.}$.

In three dimensions, we can readily find that no anomalous terms appear in $[\mathcal{J}^{-I}, \mathcal{K}^{J}]$, that is, $L^{IJ}_{t.l.}=0$.
Using Jacobi identities and $[\mathcal{K}^{+}, \mathcal{K}^{I}]=0$ which is checked easily, we obtain $[\mathcal{K}^{+}, \mathcal{K}^{-}]=0$.
Because we can't set $I\not= J$ in three dimensions, however, $[\mathcal{K}^{-}, \mathcal{J}^{-I}]$ and $[\mathcal{K}^{I}, \mathcal{K}^{-}]$ are not related to $L^{IJ}_{t.l.}$.
We just get the relation between $[\mathcal{K}^{-}, \mathcal{J}^{-I}]$ and $[\mathcal{K}^{I}, \mathcal{K}^{-}]$\footnote{$[\mathcal{K}^{-}, \mathcal{J}^{-I}]=-i[\mathcal{P}^{-}, [\mathcal{K}^{-}, \mathcal{K}^{I}]]$ or $[\mathcal{K}^{I}, \mathcal{K}^{-}]=-i[[\mathcal{J}^{-I}, \mathcal{K}^{-}], \mathcal{K}^{+}]$}.
Therefore it is nontrivial whether $[\mathcal{K}^{-}, \mathcal{J}^{-I}]$ is anomalous and we need to investigate it in order to conclude the absence of space-time conformal anomaly.
Fortunately, because the vanishing of $[\mathcal{K}^{-}, \mathcal{J}^{-I}]$ is equivalent to that of $[\mathcal{K}^{I}, \mathcal{K}^{-}]$, we just have to check $[\mathcal{K}^{-}, \mathcal{J}^{-I}]=0$.
This check is the first content of this paper.

In this paper, we write operators by the Fourier expansion and choose the "reference ordering"\footnote{In the reference ordering all P-modes and $p_{-}$ are to the right of all X-modes and $x^{-}$\cite{bib:007, bib:017}.}.
Then we investigate commutators of the space-time conformal group for Hermitian generators.
In the calculation, we need to introduce some regularization because we find many troublesome divergences.
In this paper, we regularize by removing higher modes of operators than a given cutoff and find $[\mathcal{K}^{-}, \mathcal{J}^{-I}]=0$.

As the second content of this paper, we also investigate the spectrum of a 3d tensionless bosonic closed string.
We get massive states and massless states there.
Because massless states are conformal invariant, we discuss it in more detail.

The content of the paper is as follows: In Section 2, we represent a 3D tensionless closed bosonic string in light-cone gauge.
In section 3, we quantize a 3D tensionless closed bosonic string in light-cone gauge and find that this string theory has the space-time conformal symmetry. In Section 4, we discuss the spectrum of a 3d tensionless string. In Section 5, we end the paper with the conclusion and outlooks. The definition of light-cone coordinate and the algebra of conformal group are collected in appendices.

\section{3D tensionless string in light-cone gauge}
In this section we consider in three dimensions the light-cone quantization of a string without tension, namely a tensionless string.
We follow the method of \cite{bib:003} to quantize a tensionless string.

First we consider a string with tension, namely a tensile string.
A 3D bosonic closed string with tension $T$ is described by Nambu-Goto action:
\begin{eqnarray}
S[\mathbf{X}]=-T\int d\tau \oint \frac{d\sigma }{2\pi } \sqrt{\left( \left( \dot{\mathbf{X}} \cdot \mathbf{X}' \right) ^2 -\dot{\mathbf{X}} ^2 (\mathbf{X}')^2 \right) } ,
\label{eq:2-1}
\end{eqnarray}
where ${\mathbf{X}^{\mu }(\tau ,\sigma );\mu =0,1,2}$ represents an embedding of the world sheet ${(\tau ,\sigma )}$ to 3D Minkowski space with matric $\eta =\mbox{diag}(-1,1,1)$. An overdot indicates a derivative with respect to time parameter $\tau $ and a prime indicates a derivative with respect to string coordinate $\sigma $.
The centerdot or superscript "2" indicate the contraction.
Moreover we assume that the functions are periodic, $X^{\mu }(\tau , \sigma ) = X^{\mu }(\tau , \sigma +2\pi )$.

By using the conjugate momentums $\mathbf{P}_{\mu }$, and the auxiliary fields $V$ and $U$, we can rewrite the action (\ref{eq:2-1}) to the form
\begin{eqnarray}
S[\mathbf{X},\mathbf{P};V,U] = \int d\tau \oint \frac{d\sigma }{2\pi } \left\{ \dot{\mathbf{X}} ^{\mu } \mathbf{P}_{\mu } -\frac{1}{2} V\left[ \mathbf{P}^2 +(T\mathbf{X}') ^2 \right] -U \mathbf{X}'^{\mu }\mathbf{P}_{\mu } \right\} ,
\label{eq:2-2}
\end{eqnarray}
where $V$ and $U$ are the Lagrange multipliers for the Hamiltonian and $S^{1}$-diffeomorphism constraints, respectively.
When one eliminates $\mathbf{P}_{\mu }$, followed by the elimination of $V$ and $U$ in order, the Nambu-Goto action (\ref{eq:2-1}) is reproduced \footnote{The action of a relativistic massive point particle is $S=-m\int d\tau \sqrt{-\dot{x}^{\mu }\dot{x}_{\mu } }$. This can be written as $S=\int d\tau  \left[ \dot{x}^{\mu }p_{\mu } -\frac{1}{2}v\left( p^{2}+m^{2} \right) \right]$, where $p_{\mu }$ is conjugate to $x^{\mu }$ and $v$ is the Lagrange multiplier for the off-shell condition, $p^2 +m^2 =0$. The former action is reproduced by the eliminating $p_{\mu }$ and $v$ in order. }.

The new action (\ref{eq:2-2}) has a local symmetry under the transformations
{\setlength\arraycolsep{1pt}\begin{eqnarray}\begin{split}
\delta \mathbf{X}^{\mu } &= \alpha \mathbf{P}^{\mu } +\beta \mathbf{X}'^{\mu } , \\
\delta \mathbf{P}_{\mu } &= T^2 (\alpha \mathbf{X}'_{\mu })' +(\beta \mathbf{P}_{\mu })' , \\
\delta V &= \dot{\alpha } +U'\alpha -U\alpha ' +V'\beta -V\beta ' , \\
\delta U &= \dot{\beta } +U'\beta -U\beta ' +T^2 (\alpha V'-\alpha 'V) ,
\end{split}\label{eq:2-3}
\end{eqnarray}}
where $\alpha (\tau , \sigma )$ and $\beta (\tau , \sigma )$ are arbitrary functions.
We choose light-cone gauge to fix this local symmetry and investigate whether there is Lorentz anomaly or not. Then we will find that "anyons" appear in the spectrum \cite{bib:003}.

From now on, we set $T$ to be zero to investigate the tensionless string.
The action (\ref{eq:2-2}) with $T=0$ is
\begin{eqnarray}
S[\mathbf{X},\mathbf{P};V,U] = \int d\tau \oint \frac{d\sigma }{2\pi } \left\{ \dot{\mathbf{X}} ^{\mu } \mathbf{P}_{\mu } -\frac{1}{2} V \mathbf{P}^2  -U \mathbf{X}'^{\mu }\mathbf{P}_{\mu } \right\} ,
\label{eq:2-4}
\end{eqnarray}
and the gauge symmetry (\ref{eq:2-3}) becomes
{\setlength\arraycolsep{1pt}\begin{eqnarray}\begin{split}
\delta \mathbf{X}^{\mu } &= \alpha \mathbf{P}^{\mu } +\beta \mathbf{X}'^{\mu } , \\
\delta \mathbf{P}_{\mu } &= (\beta \mathbf{P}_{\mu })' , \\
\delta V &= \dot{\alpha } +U'\alpha -U\alpha ' +V'\beta -V\beta ' , \\
\delta U &= \dot{\beta } +U'\beta -U\beta ' .
\end{split}\label{eq:2-5}
\end{eqnarray}}
In the next subsection we fix the gauge symmetry (\ref{eq:2-6}) with light-cone gauge.

\subsection{Light-cone gauge}
The light-cone components of coordinates and their conjugates , $(X^{+},X^{-},X)$ and $(P_{+},P_{-},P)$, are written with the components in Minkowski base as follows:
\begin{eqnarray}\begin{split}
& X^{\pm } \equiv \frac{1}{\sqrt{2}} (\mathbf{X}^1\pm \mathbf{X}^0), \ X\equiv \mathbf{X}^2, \\
& P_{\pm } \equiv \frac{1}{\sqrt{2}} (\mathbf{P}_1\pm \mathbf{P}_0)=P^{\mp }, \ P\equiv \mathbf{P}_2 .
\end{split}\label{eq:2-6}
\end{eqnarray}
We impose light-cone gauge to fix the gauge symmetry (\ref{eq:2-5}),
\begin{eqnarray}
X^{+} =\tau , \ P_{-}=p_{-}(\tau ) \not= 0 ,
\label{eq:2-7}
\end{eqnarray}
where $p_{-}(\tau )$ is a non-vanishing function of $\tau $.
This gauge choice restricts gauge parameters such that $\alpha =0, \beta =\beta _{0}(\tau )$ and then leaves only the residual global gauge symmetry induced by a constant shift of $\sigma $.
We leave this for a moment to clarify what the constraint is, though we will fix this later.

To obtain the action in light-cone gauge, we decouple the center of mass coordinate which is the average about $\sigma $ from the rest.
Namely, for a given function $F(\tau ,\sigma )$, we decompose it into $f$ and $\bar{F} $:
\begin{eqnarray}\begin{split}
f(\tau ) \equiv  \oint \frac{d\sigma }{2\pi } F(\tau ,\sigma ) , \\
\bar{F} (\tau ,\sigma ) \equiv F(\tau ,\sigma ) -f(\tau ) .
\end{split}\label{eq:2-8}
\end{eqnarray}
Note that $\oint \frac{d\sigma }{2\pi } \bar{F}=0 $.

Using the gauge choice (\ref{eq:2-7}) and the decoupling with regard to $F=X^{-},X,P_{+},P,U$, we find that the Lagrangian (\ref{eq:2-4}) reduces to
{\setlength\arraycolsep{1pt}\begin{eqnarray}\begin{split}
L &= \dot{x} p +\dot{x}^{-} p_{-} +p_{+}+\oint \frac{d\sigma }{2\pi } \dot{\bar{X} }\bar{P} -u\oint \frac{d\sigma }{2\pi } \bar{X}'\bar{P} - \oint \frac{d\sigma }{2\pi } \bar{U} \bar{X}' P \\
& +p_{-} \oint \frac{d\sigma }{2\pi } \left\{ \bar{X}^{-}\bar{U}' -V\left( P_{+} +\frac{1}{2p_{-}} P^2  \right)  \right\} ,
\end{split}\label{eq:2-9}
\end{eqnarray}}
where $\bar{X}^{-} $ is a lagarange multiplier giving the constraint $\bar{U}'=0 $. Together with $\oint \frac{d\sigma }{2\pi } \bar{U}=0 $, we obtain $\bar{U} =0$.
On the contrary the variation of $\bar{U} $ induces the relation
\begin{eqnarray}
p_{-}(\bar{X}^{-} )'=-\bar{X}'P+\oint \frac{d\sigma }{2\pi } \bar{X}'\bar{P} .
\label{eq:2-10}
\end{eqnarray}
which we use to determine $\bar{X}^{-} $.

Moreover the variation of $V$ leads to
\begin{eqnarray}
P_{+} =-\frac{1}{2p_{-}} P^2 .
\label{eq:2-11}
\end{eqnarray}
We regard this equation as expressing $P_{+}$ in terms of other variables.
The center part of $P_{+}$ is the Hamiltonian
\begin{eqnarray}
H\equiv -p_{+} =\frac{1}{2p_{-}} \left( p^2 + \mathcal{M} ^2 \right) ,
\label{eq:2-12}
\end{eqnarray}
and the mass squared is given by
\begin{eqnarray}
\mathcal{M} ^2 =2p_{+}p_{-}-p^2 = \oint \frac{d\sigma }{2\pi } \bar{P} ^2 .
\label{eq:2-13}
\end{eqnarray}
In summary the Lagrangian reduces to
\begin{eqnarray}
L = \dot{x} p +\dot{x}^{-} p_{-} +\oint \frac{d\sigma }{2\pi } \dot{\bar{X} }\bar{P} -H -u\oint \frac{d\sigma }{2\pi } \bar{X}'\bar{P} .
\label{eq:2-14}
\end{eqnarray}

Next we use the residual gauge symmetry induced by $\beta _{0}$ to fix $u=0$ and rewrite the Lagrangian to the form
\begin{eqnarray}
L = \dot{x} p +\dot{x}^{-} p_{-} +\oint \frac{d\sigma }{2\pi } \dot{\bar{X} }\bar{P} -H
\label{eq:2-15}
\end{eqnarray}
with a constraint
\begin{eqnarray}
\oint \frac{d\sigma }{2\pi } \bar{X}'P =0 .
\label{eq:2-16}
\end{eqnarray}

Finally we solve most of the infinite constrains in the action (\ref{eq:2-4}) by eq. (\ref{eq:2-10}) and (\ref{eq:2-11}) and then leave the only constraint (\ref{eq:2-16}).
It is an advantage of using light-cone gauge that we can solve most of the constraints leaving a finite number of simple constarints.
The difference between the light-cone quantization and others originates in whether the number of constraints is finite or infinite and whether we can deal with them adequately.

\subsubsection*{Fourier expansion}
To solve eq. (\ref{eq:2-10}) and (\ref{eq:2-11}) explicitly, we use the Fourier expansion of $X$ and $P$ with respect to $\sigma $:
{\setlength\arraycolsep{1pt}\begin{eqnarray}\begin{split}
X &= \sum _{n=-\infty }^{\infty } X_{n} e^{in\sigma } , ~~ X_{0}=x, \\
P &= \sum _{n=-\infty }^{\infty } P_{n} e^{in\sigma } , ~~ P_{0}=p.
\end{split}\label{eq:2-17}
\end{eqnarray}}
The reality conditions of $X$ and $P$ lead to
{\setlength\arraycolsep{1pt}\begin{eqnarray}
&& (X_{n})^{*}=X_{-n}, \nonumber \\
&& (P_{n})^{*}=P_{-n},
\label{eq:2-17-2}
\end{eqnarray}}
where the asterisk represents the complex conjugate.
From now on the sum without an explicit range specified must be understood to run from minus infinity to infinity \footnote{For example, $\sum _{n} \equiv \sum _{n=-\infty }^{\infty }$, $\sum _{n\not= 0} \equiv \sum _{n=-\infty }^{-1}+\sum _{n=1}^{\infty }$, $\sum _{n>0} \equiv \sum _{n=1}^{\infty }$ and so on.}.
For a tensile string, we usually combine $X_{n}$ and $P_{n}$ as
\begin{eqnarray}\begin{split}
\alpha _{n} &= -i \sqrt{\frac{T}{2}} n X_{n} +\frac{1}{\sqrt{2T}}P_{n}, \\
\tilde{\alpha }_{-n} &= i \sqrt{\frac{T}{2}} n X_{n} +\frac{1}{\sqrt{2T}}P_{n}
\end{split}\label{eq:2-18}
\end{eqnarray}
which express the right-moving or the left-moving respectively.
However, because we have no scale like $T$, it is not clear whether we should introduce some scale to combine Fourier coefficient (\ref{eq:2-17}) in the oscillator form.

Let us rewrite the mass squared (\ref{eq:2-12}) and the constraint (\ref{eq:2-16}).
First we solve eq. (\ref{eq:2-10}) as follows \footnote{The condition $\oint \frac{d\sigma }{2\pi } \bar{X}^{-} =0$ leads to the same constraint as (\ref{eq:2-16}) or (\ref{eq:2-24}).}:
\begin{eqnarray}
\bar{X}^{-} = - \frac{1}{p_{-}} \sum _{n\not= 0} \frac{i}{n} M_{n} e^{in\sigma } ,
\label{eq:2-19}
\end{eqnarray}
where
\begin{eqnarray}
M_{n} \equiv -i \sum _{m} mX_{m}P_{n-m} , ~~~ \mbox{for} ~~ n\not= 0.
\label{eq:2-20}
\end{eqnarray}
Denoting the center part by $x^{-}$, we have $X^{-}=x^{-}+\bar{X}^{-} $.

Next we solve eq. (\ref{eq:2-11})
\begin{eqnarray}
P_{+} = -\frac{1}{2p_{-}} \sum _{n} L_{n} e^{in\sigma } ,
\label{eq:2-21}
\end{eqnarray}
where
\begin{eqnarray}
L_{n} \equiv \sum _{m} P_{m}P_{n-m}.
\label{eq:2-22}
\end{eqnarray}
From the zero mode $L_{0}$, the mass squared reads as
\begin{eqnarray}
\mathcal{M}^2 = 2\sum _{n>0} P_{n}P_{-n}.
\label{eq:2-23}
\end{eqnarray}
Further the constraint (\ref{eq:2-16}) is expressed as
\begin{eqnarray}
0 = \oint \frac{d\sigma }{2\pi } \bar{X}'P = i \sum _{n} nX_{n}P_{-n} \equiv -M_{0},
\label{eq:2-24}
\end{eqnarray}
which corresponds to the level-matching condition for a tensile string.

\subsubsection*{Equations of motion}
Using the Fourier expansion, the Lagrangian (\ref{eq:2-15}) is written in the form
\begin{eqnarray}
L = \dot{x}p+\dot{x}^{-}p_{-}+\sum _{n\not= 0} \dot{X}_{n}P_{-n} -H ,
\label{eq:2-25}
\end{eqnarray}
where $H=\frac{1}{2p_{-}}(p^2+\mathcal{M}^2)$ and the mass squared is given by eq. (\ref{eq:2-24}).
We obtain the equations of motion from this Lagrangian,
\begin{eqnarray}
\dot{p}=\dot{p}_{-}=0, \ \dot{x}=\frac{p}{p_{-}} , \ \dot{x}^{-}=-\frac{H}{p_{-}}
\label{eq:2-26}
\end{eqnarray}
and
\begin{eqnarray}
\dot{X}_{n}=\frac{P_{n}}{p_{-}}, ~~ \dot{P}_{n}=0.
\label{eq:2-27}
\end{eqnarray}
These equations indicate that the center part moves with uniform velocity and the shape of string changes in proportion to $P(\sigma )$,
\begin{eqnarray}
X(\tau ,\sigma ) = X(\tau =0,\sigma ) +\frac{P(\sigma )}{p_{-}}\tau ,
\label{eq:2-28}
\end{eqnarray}
which is expected from the equations of motion before taking the Fourier expansion.
It reflects the fact that the Hamiltonian and the mass squared are independent of $\tau $.

\section{Quantization of a 3D tensionless string and Space-time conformal symmetry}
In this section we first quantize a 3D tensionless string represented in section 2 and next find that this theory has no anomaly for the space-time conformal symmetry.

\subsection{Quantization}
In the quantum theory, the canonical variables in the action (\ref{eq:2-15}) are promoted to operators with the commutation relations
\begin{eqnarray}
[x^{-},p_{-}]=i,\ [X(\sigma ),P(\sigma ')]=2\pi i \delta (\sigma -\sigma ')  \ \ \ \mbox{with others vanishing},
\label{eq:3-1}
\end{eqnarray}
where we set $\hbar =1$. In the mode expansion, the last relation indicates
\begin{eqnarray}
[x,p]=i, \ [X_{n},P_{m}]=i\delta _{n+m,0} ,
\label{eq:3-2}
\end{eqnarray}
where $n,m\in \mathbb{Z}$.
Using these basic relations, we obtain that $L_{n}$ and $M_{n}$ satisfy the following relations:
\begin{eqnarray}\begin{split}
& [X_{n}, M_{m}]=(n+m)X_{n+m} ,~~ [P_{n},M_{m}]=nP_{n+m} ,~~ [X_{n}, L_{m}]=2iP_{n+m} ,~~ [P_{n},L_{m}]=0 \\
& [M_{n},M_{m}]=(n-m)M_{n+m} , ~~ [L_{n},M_{m}]=(n-m)L_{n+m} , ~~ [L_{n}, L_{m}]=0 ,
\end{split}\label{eq:3-3}
\end{eqnarray}
where we use the operator-ordering given in eq. (\ref{eq:2-20}) and (\ref{eq:2-22}) .
We make a remark that $L_{n}$ and $M_{n}$ satisfy  the 2-dimensional Galilean Conformal Algebra (2d GCA) \footnote{Recently this algebra was investigated in terms of a tensionless string \cite{bib:012}}.

The quantum Hamiltonian and the mass suared are then
\begin{eqnarray}\begin{split}
& H=\frac{1}{2p_{-}}(p^2+\mathcal{M}^2), \\
& \mathcal{M}^2=2\sum _{n>0} P_{n}P_{-n},
\end{split}\label{eq:3-4}
\end{eqnarray}
where there is no constant term arising from ambiguity of the operator ordering because $P_{n}$ commute with $P_{-n}$ \footnote{In the case of a tensile string, the mass squared has a constant $a$ arising from the operator ordering ambiguity.
To avoid the Lorentz anomaly, we choose the critical dimension of the string theory and the ordering constant to be $D=26$ and $a=1$, respectively.
However in three dimensions no Lorentz anomaly exists trivially and then $a$ remains as an arbitrary constant \cite{bib:003}.}.
On the contrary, the constraint (\ref{eq:2-24}) has ambiguity of the operator ordering.
This ambiguity is related to the choice of the vacuum.
Here we define $M_{0}$ as in (\ref{eq:2-24}) in order that the action of $M_{0}$ on physical state vanishes.

\subsection{Generators }
In light-cone gauge quantization, the space-time Lorentz and conformal symmetries are not clear.
Therefore we check whether the generators of these symmetries satisfy the expected commutation relations (\ref{eq:a2-7}) and (\ref{eq:a2-11}), respectively.
In $D>3$, we determine the critical dimension of a bosonic string and the ordering constant in the mass squared to preserve the Lorentz invariance in quantum theory.
In three dimensions, there is only one transverse direction and the dangerous commutator (\ref{eq:1-1}) vanishes trivially. Hence we have no Lorentz anomaly.
However the conformal symmetry is not trivial even in three dimensions. We now investigate whether a 3D tensionless closed string has conformal symmetry.

The conformal group is generated by translations, Lorentz rotations, dilatation and specail conformal transformations.
Now we define these generators in the"reference order," which all $P_{n}$ and $p_{-}$ are to the right of $X_{n}$ and $x^{-}$ respectively \cite{bib:007}. We shall call the reference order "R-order" \footnote{We can obtain the R-ordering from the normal ordering in the tensionless limit $T\rightarrow 0$. In detail, the string ground state $|0\rangle _{T}$ of a tensile string with a tension $T$ is annihilated by positive modes of right-moving and left-moving oscillators, $\{ \alpha _{n}, ~ \tilde{\alpha }_{n} ; n>0 \} $. According to eq. (\ref{eq:2-18}), this string ground state in the tensionless limit reduces the vacumm which annihilates all $P_{n}$ for all non-zero $n$. However, when we set $T=0$ from the beginning, there is no reason why this state should be chosen. }.
The definitions of $L_{n}$ and $M_{n}$ in eq. (\ref{eq:2-20}) and (\ref{eq:2-22}) were already into the R-order.
Hereafter, when we want to emphasize operators to be into the R-order, we specify the R-ordered operator by the subscript $R$.

We first define the translation generators as
\begin{eqnarray}
\mathcal{P}_{R}^{\mu } \equiv \oint \frac{d\sigma }{2\pi } P^{\mu }.
\label{eq:3-5}
\end{eqnarray}
In the light-cone base, these are
\begin{eqnarray}
\mathcal{P}_{R} = p, ~~~ \mathcal{P}_{R}^{+}=p_{-},~~~\mathcal{P}_{R}^{-}=p_{+}=-H.
\label{eq:3-6}
\end{eqnarray}
Second, the Lorentz generators are defined as
\begin{eqnarray}
\mathcal{J}_{R}^{\mu } \equiv \epsilon ^{\mu \nu \rho } \oint \frac{d\sigma }{2\pi } X_{\nu }P_{\rho } .
\label{eq:3-7}
\end{eqnarray}
In light-cone base, these are written as
\begin{eqnarray}\begin{split}
& \mathcal{J}_{R} =x^{-}p_{-}+\tau H,~~~\mathcal{J}_{R}^{+}=\tau p-xp_{-}, \\
& \mathcal{J}_{R}^{-} =-x^{-}p -xH +\frac{\Lambda }{p_{-}},
\end{split}\label{eq:3-8}
\end{eqnarray}
where
\begin{eqnarray}
\Lambda = p_{-} \oint \frac{d\sigma }{2\pi } \left[ \bar{X}\bar{P}_{+} -\bar{X}^{-}\bar{P} \right] =\sum _{n\not= 0} \left( -\frac{1}{2} X_{n}L_{-n} +\frac{i}{n} M_{n}P_{-n} \right) .
\label{eq:3-9}
\end{eqnarray}
Next the dilatation generator is defined as
\begin{eqnarray}
\mathcal{D}_{R} = \oint \frac{d\sigma }{2\pi } X^{\mu }P_{\mu } ,
\label{eq:3-10}
\end{eqnarray}
and now expressed as
\begin{eqnarray}
\mathcal{D}_{R} =x^{-}p_{-}-\tau H +\sum _{n} X_{n}P_{-n}.
\label{eq:3-11}
\end{eqnarray}
At last the generators of the special conformal transformations are defined as
\begin{eqnarray}
\mathcal{K}_{R}^{\mu } = \oint \frac{d\sigma }{2\pi } \left[ X^{\mu } \left( X \cdot P \right) - \frac{1}{2} \left( X \cdot X \right) P^{\mu } \right] _{R} ,
\label{eq:3-12}
\end{eqnarray}
where the subscript $R$ in the right-hand side indicates the reordering into R-order \footnote{Because $\mathcal{K}_{R}^{-}$ includes a quadratic term of $X^{-}$, the simply defined $\mathcal{K}^{-}$ is not into R-order as it is.}.
In light-cone base,
{\setlength\arraycolsep{1pt}\begin{eqnarray}\begin{split}
\mathcal{K}_{R}^{+} &= -\frac{1}{2}\sum _{n} X_{n}X_{-n} p_{-} + \tau \sum _{n} X_{n}P_{-n} -\tau ^2 H \\
\mathcal{K}_{R} &= xx^{-}p_{-}+\sum _{n\not= 0} \frac{i}{n}X_{n}M_{-n} +\frac{1}{2}\sum _{n} \sum _{m} X_{n}X_{m}P_{-n-m} +\tau \mathcal{J}^{-} \\
\mathcal{K}_{R}^{-} &= x^{-}x^{-}p_{-} +x^{-}\sum _{n} X_{n}P_{-n} -\frac{i}{p_{-}} \sum _{n}\sum _{m\not= 0}\left( \frac{n}{m^2}+\frac{1}{m} \right) X_{n}M_{m}P_{-n-m} \\
&  +\frac{1}{4p_{-}}\sum _{n}\sum _{m} X_{n}X_{m}L_{-n-m} .
\end{split}\label{eq:3-13}
\end{eqnarray}}

Physical observables should be represented by Hermitian operators. We require the conformal generators to be Hermite.
In our case the R-ordered definitions of the generators in (\ref{eq:3-8})(\ref{eq:3-11})(\ref{eq:3-13}) are simple but not Hermite \footnote{Of course the translation $\mathcal{P}=p$, the Hamiltonian $H$ and the mass squared $\mathcal{M}^2$ are clearly Hermite. Moreover $\Lambda $ is Hermite, and if $\sum _{n}n = 0$ the constraint $M_{0}$ is also Hermite.}.
Then we introduce the Hermitian version $\mathcal{G}$ of the R-ordered generator $\mathcal{G}_{R}$ as follows:
\begin{eqnarray}
\mathcal{G} \equiv \frac{1}{2}\left( \mathcal{G}_{R} + (\mathcal{G}_{R} ) ^{\dagger } \right) .
\label{eq:3-14}
\end{eqnarray}
All Hermitian versions of generators are independent of $\tau $.
Hence, when we deal with Hermitian generators, we can use generators to be set $\tau =0$.
Here note that the differences of the Hermitian version from the R-ordered generators mostly include various divergent terms.
So, if we order generators and commutators into R-order, we need some regularization.

\subsection{Anomaly}
In the previous subsection, we defined the generators of the space-time conformal symmetry.
In three dimensions, there is the only one transverse direction \footnote{So we omit the label $I$ or $J$. $\mathcal{P}^{I}\rightarrow \mathcal{P}$, $\mathcal{K}^{I}\rightarrow \mathcal{K}$, $\mathcal{J}^{\pm I}\rightarrow \mathcal{J}^{\pm }$. And we define $\mathcal{J}^{+-}\rightarrow -\mathcal{J}$.} and the dangerous commutator (\ref{eq:1-1}) vanishes trivially. Therefore all commutation relations of Poincar\'{e} group (\ref{eq:a2-7}) are satisfied.
In this subsection, we investigate whether the anomaly arises in the quantization of a 3D tensionless string, namely whether all commutation relations of conformal symmetry (\ref{eq:a2-12}) are satisfied.

In \cite{bib:007}, J. Isberg et al. showed that in $D>3$ anomalies arise from the commutators, $[\mathcal{K}^{I}, \mathcal{J}^{-J}] $, $[\mathcal{K}^{+}, \mathcal{K}^{-}] $, $[\mathcal{K}^{I}, \mathcal{K}^{-}] $ and $[\mathcal{K}^{-}, \mathcal{J}^{-I}] $.
The first commutator of these has the traceless part $L^{IJ}_{t.l.}$ with respect to $I,J$ as well as the trace part.
The difference of trace part can be absorbed in the redefinition of $\mathcal{K}^{-}$, but the traceless part $L^{IJ}_{t.l.}$ remains as the anomaly.

In three dimensions, this commutator is only one and hence this type of anomaly does not exist.
But we need to calculate other non-trivial commutators to check the space-time conformal symmetry.
In three dimensions, $[\mathcal{K}, \mathcal{J}^{-}]$ corresponds to $[\mathcal{K}^{I}, \mathcal{J}^{-J}] $ in $D>3$.
Though this is slightly different from $\mathcal{K}^{-}=\frac{1}{2}(\mathcal{K}^{-}_{R}+(\mathcal{K}^{-}_{R})^{\dagger })$, we can interpret that this give the redefinition as
\begin{eqnarray}
\hat{\mathcal{K} }^{-} =i[\mathcal{K}, \mathcal{J}^{-}] =\mathcal{K}^{-}+\delta \mathcal{K}^{-} ,
\label{eq:3-15}
\end{eqnarray}
where $\delta \mathcal{K}^{-}$ is a constant times $\frac{1}{p_{-}}$.
Note that in the term with $M_{0}$, we put $M_{0}$ to the right of other operators and set zero on physical Hilbert space if we need.

We still have the three "dangerous" commutators.
Using the Jacobi identity and $[\mathcal{K}, \hat{\mathcal{K}}^{-}]=0$ which has no anomaly, we find that $[\mathcal{K}^{+}, \hat{\mathcal{K}}^{-}]$ has no anomaly, that is, $[\mathcal{K}^{+}, \hat{\mathcal{K}}^{-}]=0$.
On the other hand, it is non-trivial whether the last two "dangerous" commutators vanish and then we have to check it.
Fortunately, because there are the relations between $[\mathcal{K}^{I}, \mathcal{K}^{-}] $ and $[\mathcal{K}^{-}, \mathcal{J}^{-I}] $\footnote{$[\mathcal{K}, \hat{\mathcal{K}}^{-}]=-i[[\mathcal{J}^{-}, \hat{\mathcal{K}}^{-}], \mathcal{K}^{+}]$ and $[\hat{\mathcal{K}}^{-}, \mathcal{J}^{-}]=-i[\mathcal{P}^{-}, [\hat{\mathcal{K}}^{-}, \mathcal{K}]]$.}, all we have to do is the calculation of one commutator.
We calculate the easier one, $[\mathcal{K}^{-}, \mathcal{J}^{-I}] $.

$[\hat{\mathcal{K}}^{-}, \mathcal{J}^{-} ] $ must be zero:
\begin{eqnarray}
[\hat{\mathcal{K}}^{-}, \mathcal{J}^{-} ] =0 .
\label{eq:3-16}
\end{eqnarray}
This commutator can have parts which are in proportion to $(x^{-})^2$ , $x^{-}\frac{1}{p_{-}}$ and $\frac{1}{p_{-}{}^2}$ in the R-ordering and complicated divergences too.
In particular, the part in proportion to $\frac{1}{p_{-}{}^2}$ has quintic, qubic and linear terms with respect to $X_{n}$ and $P_{n}$ and is complicated.
So the calculation of this commutator needs a lot of labor and care.

Though the calculations of other commutators also have divergences, we can deal with them without the concrete regularization.
However, commutators (\ref{eq:3-15}) (\ref{eq:3-16}) have many types of divergences \footnote{For example, $\sum _{n}1$ and $\sum _{n} n$.} and it is too complicated to calculate them correctly.
Moreover we should take care of the shift of dummy variables in the sum and the termwise re-summation\footnote{For operators into some order, $\sum _{n}X_{n}P_{-n}|_{*} = \sum _{n}X_{n+k}P_{-n-k}|_{*} $, where the subscripts mean these terms are into some order. But for the number, $\sum _{n}n \not= \sum _{n}(n+k) \not= \sum _{n}n +k\sum _{n}1$. }.
Then the regularization help us from these divergences.

\subsubsection*{Cutoff regulerization}
We use the cutoff regularization to remove higher modes of $X$ and $P$,
\begin{eqnarray}
X_{n}=P_{n}=0 ~~~\mbox{for} ~~ |n|>N,
\label{eq:3-17}
\end{eqnarray}
where we assume that $N$ is a large integer.
Using this, we obtain
\begin{eqnarray}
M_{n}=-i \sum _{|m|\leq N, |n-m|\leq N} mX_{m}P_{n-m} , ~~~ L_{n}=\sum _{|m|\leq N, |n-m|\leq N} P_{m}P_{n-m} ,
\label{eq:3-18}
\end{eqnarray}
and also find $M_{n}=L_{n}=0$ for $|n|>2N$.

Thanks to the cutoff regularization, summations reduce to finite sum and term-wise treatments are possible.
And the error as to the shift of dummy variables will decrease.
Further, because of the symmetric cutoff with respect to positive and negative modes, we get $\sum _{n}n = \sum _{n} \frac{1}{n} =0$.
Thus we resolve many difficulties for the divergence.
But we have to take care of the informations about the range of the summation.

Then there remains only the lengthy calculation for us.
After the calculation under the cutoff regularization, we take the limit $N\rightarrow \infty $.

\subsection{Check of anomaly free}
Our goal is to check a relation
\begin{eqnarray}
[\hat{\mathcal{K}}^{-}, \mathcal{J}^{-}]=0
\label{eq:3-19}
\end{eqnarray}
under the cutoff regularization.
We can check that most of other commutation relations are satisfied without using the cutoff regularization and that the rest of commutation relations are satisfied by using eq. (\ref{eq:3-19}).

We summary steps of the calculation, instead of representing the process of the lengthy calculation explicitly.
We can deform eq.(\ref{eq:3-19}) to
\begin{eqnarray}
[\hat{\mathcal{K}}^{-}, \mathcal{J}^{-}] = \left[ \frac{1}{2}\left( \mathcal{K}^{-}_{R} +(\mathcal{K}^{-}_{R})^{\dagger } \right) +\delta \mathcal{K}^{-}, \mathcal{J}^{-} \right] =\frac{1}{2} \left( [\mathcal{K}_{R}^{-}, \mathcal{J}^{-} ] -[\mathcal{K}_{R}^{-}, \mathcal{J}^{-} ] ^{\dagger } \right) +[\delta \mathcal{K}^{-}, \mathcal{J}^{-}] .
\label{eq:3-20}
\end{eqnarray}
If we use Hermitian operators from the beginning, we have more terms than in the case of R-ordered operators and many divergent terms.
Then we first use R-ordered operator, $\mathcal{J}^{-}_{R}$, and next consider its Hermite-conjugate and finally add the commutator of the difference $\delta \mathcal{K}^{-}$.

First we calculate
\begin{eqnarray}
[\mathcal{K}_{R}^{-}, \mathcal{J}^{-} ] ,
\label{eq:3-21}
\end{eqnarray}
and order the result into the R-order except for terms with $M_{0}$, which is the right of all other operators.
Here note that $\mathcal{J}^{-} = \frac{1}{2} \left( \mathcal{J}^{-}+(\mathcal{J}^{-})^{\dagger } \right) = \mathcal{J}_{R}^{-} +\frac{1}{2}i\frac{p}{p_{-}}$.
The calculation of eq. (\ref{eq:3-21}) is very lengthy and laborious.
The commutator (\ref{eq:3-21}) can have three parts which are proportional to $(x^{-})^2$, $x^{-}\frac{1}{p_{-}}$ and $\frac{1}{p_{-}{}^{2}}$.
We can check the vanishing of parts proportional to $(x^{-})^2$ and $x^{-}\frac{1}{p_{-}}$ easily.
The part proportional to $\frac{1}{p_{-}{}^2}$ has quintic terms with $XXPPP$-form , cubic terms with $XPP$-form and linear terms with $P$-form.
Maximal quintic terms are independent of the ordering. Therefore we get the same result as the classical case.
In fact, after the lengthy calculation, we find the absence of quintic terms.
However cubic terms and linear terms do not vanish. So we must consider other parts of (\ref{eq:3-20}).

Next we consider the Hermite-conjugate of (\ref{eq:3-21}) and then make the anti-Hermitian version of (\ref{eq:3-21}).
In the anti-Hermitian version (\ref{eq:3-21}), we can deform all cubic terms to the form with $[X,P]$ because all summations have the symmetry which all dummy variables invert simultaneously\footnote{For example, its anti-Hermitian version of $\sum _{|n|\leq N}\sum _{0<|m|\leq N} \frac{n}{m} X_{n}P_{m}P_{-n-m}$ is \\
$\frac{1}{2} \sum _{|n|\leq N}\sum _{0<|m|\leq N} \frac{n}{m} (X_{n}P_{m}P_{-n-m} -P_{n+m}P_{-m}X_{-n}) = \frac{1}{2} \sum _{|n|\leq N}\sum _{0<|m|\leq N} \frac{n}{m} (X_{n}P_{m}P_{-n-m} -P_{-n-m}P_{m}X_{n}) = \frac{1}{2} \sum _{|n|\leq N}\sum _{0<|m|\leq N} \frac{n}{m} [X_{n},P_{m}] P_{-n-m} $. In the case of quintic terms or linear terms, we will get anti-commutator. }.
Therefore all cubic terms reduce to the linear term.
There is this symmetry in the R-ordering, but not always in other orderings, like normal ordering.

In this way, all we have to check is the linear terms.
The sum of subscripts in each terms is zero and hence the linear terms with $P$-form must be the form that a constant times $P_{0}=p$.
Because the calculation of only linear terms is the same as calculating cubic terms, we calculate all terms with $p$.
Though we have more hard work apparently, the calculation becomes easier because we don't have to distinguish the degree of operators.
The part with $p$ of (\ref{eq:3-21}) is \footnote{If we use the following equations, the calculation may become easier:\\
$[\mathcal{K}_{R}^{-}, \mathcal{J}^{-} ] |_{p} = \left( [\mathcal{K}_{R}^{-}|_{p}, \mathcal{J}^{-} ] +[\mathcal{K}_{R}^{-}, \mathcal{J}^{-}|_{p} ] -[\mathcal{K}_{R}^{-}|_{p}, \mathcal{J}^{-}|_{p} ] \right) |_{p}$ and $\mathcal{K}^{-}_{R}|_{p}=-\mathcal{K}_{R}\frac{p}{p_{-}}+\frac{1}{2}\mathcal{K}^{+}\frac{p^2}{p_{-}{}^2}+2x^{-}xp +x\sum _{n} X_{n}P_{-n} \frac{p}{p_{-}}$, $[\mathcal{K}_{R}, \mathcal{J}^{-}] = -i(\mathcal{K}^{-}_{R}+\delta \mathcal{K}^{-}_{R})$ and $[\mathcal{K}^{+}, \mathcal{J}^{-}] =i\mathcal{K}$.}
\begin{eqnarray}
[\mathcal{K}_{R}^{-}, \mathcal{J}^{-} ] |_{p} = \frac{1}{2}i\delta \mathcal{K} \frac{p^2}{p_{-}^{2}} +i\delta \mathcal{K}^{-}_{R} \frac{p}{p_{-}},
\label{eq:3-22}
\end{eqnarray}
where $(\cdots )|_{p}$ represents the part with $p$ of $(\cdots )$ and we define $\delta \mathcal{K} \equiv \mathcal{K}-\mathcal{K}_{R} = x \times (\mbox{constant})$ and $\delta \mathcal{K}^{-}_{R} \equiv i[\mathcal{K}_{R}, \mathcal{J}^{-}] -\mathcal{K}^{-}_{R}$ \footnote{In the representation without regularization, $\delta \mathcal{K} = -\frac{1}{2}i(1+\sum _{n}1)x$ and\\
$\delta \mathcal{K}^{-}_{R} =-\frac{i}{2}x^{-} +\frac{1}{2p_{-}} +\frac{i}{2}(\sum _{l\not= 0} 1)\sum _{n \not= 0} X_{n}P_{-n} \frac{1}{p_{-}} +\frac{i}{2}(\sum _{n} 1) x\frac{p}{p_{-}} +i\sum _{n \not=0} \sum _{m \not= 0} \frac{n^2}{m^2} X_{n}P_{-n} \frac{1}{p_{-}} +2i\sum _{n \not= 0} \frac{1}{n} X_{n}P_{-n} \frac{1}{p_{-}} M_{0}$.}. Furthermore we obtain
\begin{eqnarray}
\frac{1}{2}\left( [\mathcal{K}_{R}^{-}, \mathcal{J}^{-} ] |_{p} - ([\mathcal{K}_{R}^{-}, \mathcal{J}^{-} ] |_{p})^{\dagger } \right) = i\delta \mathcal{K}^{-} \frac{p}{p_{-}} ,
\label{eq:3-23}
\end{eqnarray}
where we use $\delta \mathcal{K}^{-} =\frac{1}{2}(\delta \mathcal{K}^{-}_{R} + (\delta \mathcal{K}^{-}_{R} )^{\dagger })$.

Finally we calculate the commutator of the difference $\delta \mathcal{K}^{-}$.
\begin{eqnarray}
[\delta \mathcal{K}^{-}, \mathcal{J}^{-}] = [\delta \mathcal{K}^{-}, -x^{-}p] = -i \delta \mathcal{K}^{-} \frac{p}{p_{-}},
\label{eq:3-24}
\end{eqnarray}
where we use $\delta \mathcal{K}^{-} = \frac{1}{p_{-}} \times ( \mbox{constant} )$ and $[\frac{1}{p_{-}}, x^{-}] =i\frac{1}{p_{-}{}^2}$.

Adding (\ref{eq:3-23}) and (\ref{eq:3-24}), we can check the commutation relation (\ref{eq:3-19}).
Thus we find that there is no anomaly of the space-time conformal symmetry in a 3D tensionless bosonic closed string.

\section{Spectrum}
In this section, we investigate the spectrum of a 3D tensionless closed string.
We deal the center of mass part and the rest separately and assume that the ground state of the center is $|p,p_{-} \rangle $ like the case of a tensile string. We do not care about the ground state of the center part so much.

On the other hand, we need the discussion about the rest part, which includes non-zero modes.
First we consider the R-ordered vacuum $|0 \rangle _{R}$, which satisfy
\begin{eqnarray}
P_{n} |0 \rangle _{R} =0 ~~~ \mbox{for all non-zero } n ,
\label{eq:4-1}
\end{eqnarray}
as the string ground state.
We can obtain this R-ordered string ground state from the string ground state in the tensionless limit \cite{bib:007}.

When we choose this string ground state, we obtain other states by acting $\{ X_{n}; n \not= 0 \} $ on $|0 \rangle _{R}$.
The fundamental elements of states are
\begin{eqnarray}
X_{n_1} X_{n_2} \cdots X_{n_l} |0\rangle _{R} ,
\label{eq:4-2}
\end{eqnarray}
where $n_{i} ~ (i=1,2,\cdots l ) $ are non-zero integers.
We combine these state to obtain the general states.

Here note that the physical states must satisfy $M_{0}=0$.
Therefore the physical states are the states with the form (\ref{eq:4-2}) which satisfy
\begin{eqnarray}
\sum _{i=1}^{l}n_{i}=0 .
\label{eq:4-3}
\end{eqnarray}
The general physical states satisfy this condition in each terms.
Under this condition, we try to make eigenstates of $\mathcal{M}^2$.
In the case of the R-ordered string ground state, it is convenient to represent $P_{n}$ as
\begin{eqnarray}
P_{n}=-i\frac{\partial }{\partial X_{-n}}, ~~~ [X_{n}, P_{m}] =i\delta _{n+m,0}
\label{eq:4-4}
\end{eqnarray}
Using this representation, $\mathcal{M}^2$ is expressed as
\begin{eqnarray}
\mathcal{M}^2 = -2 \sum _{n>0} \frac{\partial }{\partial X_{-n}} \frac{\partial }{\partial X_{n}} .
\label{eq:4-5}
\end{eqnarray}

By the space-time conformal symmetry, we expect that eigenvalues of $\mathcal{M}^2$ are continuous or zero.
If we find the eigenfunction with the eigenvalue $M^2$, we get an eigenfunction of $\mathcal{M}^2$ with the eigenvalue $\lambda ^{-2} M^2$ by transforming $X_{n}\rightarrow \lambda X_{n}$ for all $n$.
Here we are especially interested in the case of zero eigenvalue, that is massless.

\subsection{Mass eigenstate }
In this subsection, we use the similar method in \cite{bib:016}.
In order to find eigenvalues of $\mathcal{M}^2$ and their eigenfunctions, we rewrite eq.(\ref{eq:4-5}),
\begin{eqnarray}
\mathcal{M}^2 = - \frac{1}{2} \sum _{n>0} \left[ \frac{\partial ^2}{\partial r_{n} {}^2 } +\frac{1}{r_{n}} \frac{\partial }{\partial r_{n}} +\frac{1}{r_{n}{}^2}\frac{\partial ^2}{\partial \theta _{n}{}^2 } \right]
\label{eq:4-6}
\end{eqnarray}
where $r_{n}$ and $\theta _{n}$ are the real operators which are defined by $X_{n}=r_{n}e^{i\theta _{n}}$ and $X_{-n}=r_{n}e^{-i\theta _{n}}$ for all positive $n$.
Because we can separate variables into each $n$, we consider the following differential equation:
\begin{eqnarray}
-\frac{1}{4} \left[ \frac{\partial ^2}{\partial r^2 } +\frac{1}{r} \frac{\partial }{\partial r} +\frac{1}{r^2}\frac{\partial ^2}{\partial \theta ^2 } \right] \psi _{m}(r, \theta ) = m^2 \psi _{m}(r, \theta ) ,
\label{eq:4-7}
\end{eqnarray}
where $r\geq 0$ and $m^2\geq 0$\footnote{Because the eigenvalue of the operator $A A^{\dagger }$ is zero or positive number, we consider the case of $m^2\geq 0$. And we also find that the solutions for $m^2 < 0$ have bad behavior in $r\rightarrow \infty $ and can not be normalized. }.
Here we assume $\psi _{m,s }(r,\theta )=\phi _{m,s }(r) e^{i s \theta }$, where $s$ is integer and $\phi _{m,s}(r)$ is the function which depends on only $r$.
Then we obtain
\begin{eqnarray}
\left[ \frac{d^2}{d r^2 } +\frac{1}{r} \frac{d}{d r} + 4 m^2 -\frac{s^2}{r^2} \right] \phi _{m,s }(r) = 0 .
\label{eq:4-8}
\end{eqnarray}

\subsubsection{$m^2 >0$}
For $m^2 > 0$, we replace $r$ with $\hat{r} =2m r$ and obtain
\begin{eqnarray}
\left[ \frac{d^2}{d \hat{r} ^2 } +\frac{1}{\hat{r} } \frac{d}{d \hat{r} } + 1 -\frac{s ^2}{\hat{r} ^2} \right] \phi _{m,s }\left( \frac{\hat{r} }{2m}\right) = 0 ,
\label{eq:4-9}
\end{eqnarray}
where $m > 0$. The solution of this equation is expressed in terms of Bessel function $J_{|s|}(\hat{r} )$.
In this way, we get the solution $\psi _{m,s }(r,\theta ) = N_{m } J_{|s|}(2mr) e^{i s \theta }$, where $N_{m}$ is the normalization constant \footnote{Note that $J_{\nu }(\hat{r} )=(-1)^{\nu }J_{-\nu }(\hat{r} )$ for integer $\nu $.}.

We rewrite the solutions in terms of $X_{n}$ and $X_{-n}$ with $n>0$:
{\setlength\arraycolsep{1pt}\begin{eqnarray}
\psi _{m,\pm |s|}(r_{n},\theta _{n}) &=& N_{m} J_{|s|}( 2m(X_{n}X_{-n})^{\frac{1}{2}}) \left( \frac{X_{n}}{X_{-n}} \right) ^{\frac{s}{2}} \nonumber \\
&=& N_{m} (m X_{\pm n} )^{|s|} \sum _{l=0}^{\infty } \frac{(-m^2)^l }{l! \ (l+|s|)!} ( X_{n}X_{-n} )^{l} ,
\label{eq:4-10}
\end{eqnarray}}
where we use
\begin{eqnarray}
J_{\nu }(z) = \left( \frac{z}{2} \right) ^{\nu } \sum _{l=0}^{\infty } \frac{(-1)^l}{l! \ \Gamma (l+\nu +1)} \left( \frac{z}{2}\right) ^{2l}.
\label{eq:4-11}
\end{eqnarray}
Thus we find that the eigenfunctions are the combinations of positive integer power with respect to $X_{n}$ and $X_{-n}$ as in eq.(\ref{eq:4-2}).

Now we consider the normalization of wave functions.
Since $m^2$ is not discrete valuable, the wave functions for each $m^2$ can not be normalized to "1" but the wave functions for different values of $m^2$ must be orthogonal.
Here we consider the next scalar product. The scalar product of two wave functions $\psi _{1}(r,\theta )$ and $\psi _{2}(r,\theta )$ is
\begin{eqnarray}
(\psi _{1}, \psi _{2}) = \int _{0}^{\infty } dr \int _{0}^{2\pi } d\theta r \psi _{1}^{*}(r,\theta ) \psi _{2}(r, \theta ) .
\label{eq:4-12}
\end{eqnarray}
Then we obtain for $m>0$ and $m'>0$
\begin{eqnarray}
(\psi _{m,s },\psi _{m',s'}) = \frac{\pi }{2} \frac{|N_{m}|^2}{m} \delta (m-m') \delta _{s,s'}
\label{eq:4-13}
\end{eqnarray}
and find the orthogonality. The detail of the normalization is given in appendix.

\subsubsection{$m^2=0$ }
For $m^2=0$, eq.(\ref{eq:4-8}) becomes
\begin{eqnarray}
\left[ \frac{d^2}{d r^2 } +\frac{1}{r} \frac{d}{d r} -\frac{s^2}{r^2} \right] \phi _{0,s}(r) = 0 .
\label{eq:4-14}
\end{eqnarray}
The solutions of this equation for $s \not= 0$ are $r^{|s|}$ and $r^{-|s|}$. For $s=0$, the solutions are constant and $\log r$.
Because of the orthogonality between two eigenfunctions with $m^2 >0$, $r^{|s|}$ for $s \not= 0$ and constant for $s=0 $ are chosen \cite{bib:016}\footnote{The detail is in appendix.}.
In terms of $X_{n}$ and $X_{-n}$ with $n>0$, the eigenfunctions $\psi _{0,s}(r,\theta )$ are $(X_{n})^{s}$ for $s>0$, $(X_{-n})^{-s}$ for $s<0$ and constant for $s=0$.
If we collect these three cases, we get
\begin{eqnarray}
\psi _{0,\pm |s|}(r,\theta ) \propto r^{|s|}e^{\pm i |s| \theta } = (X_{\pm n})^{|s|} .
\label{eq:4-15}
\end{eqnarray}

\subsubsection{Total eigenfunction}
The eigenfunctions of $\mathcal{M}^2$ are the product
\begin{eqnarray}
\Psi = \prod _{n>0} \psi _{m_{n},s_{n}}(r_{n},\theta _{n})
\label{eq:4-16}
\end{eqnarray}
and their eigenvalues are
\begin{eqnarray}
\mathcal{M}^2 = 2 \sum _{n>0} (m_{n})^2.
\label{eq:4-17}
\end{eqnarray}

\subsection{Physical spectrum}
The physical state must satisfy the constraint $M_{0}=0$.
In terms of $r_{n}$ and $\theta _{n}$, $M_{0}$ becomes
\begin{eqnarray}
M_{0}= -i\sum _{n} n X_{n}P_{-n} = -\sum _{n>0} n \left[ X_{n}\frac{\partial }{\partial X_{n}} -X_{-n}\frac{\partial }{\partial X_{-n}}\right] = i\sum _{n>0} n \frac{\partial }{\partial \theta _{n}}.
\label{eq:4-18}
\end{eqnarray}
Then the constraint in the case of (\ref{eq:4-16}) is
\begin{eqnarray}
\sum _{n>0} n s_{n} =0.
\label{eq:4-19}
\end{eqnarray}
As mentioned in the beginning of this section, the eigenvalues of $\mathcal{M}^2$ are continuous or zero.
Though this fact is expected by the conformal symmetry, it is understood by the explicit solution (\ref{eq:4-10}).
We get continuous eigenvalues if $m_{n}^2>0$ at least for some $n$.
On the other hand, we get massless states only if $m_{n}=0$ for all $n$.
Since we are interested in massless states which are expected to preserve the space-time conformal symmetry,
we do not discuss the continuous spectrum anymore.
Now we investigate the case of massless in detail.

\subsubsection*{Massless states }
If we want to get massless states, we have to choose the solution (\ref{eq:4-15}) for all positive integer $n$.
In other words, we should choose the states (\ref{eq:4-2}) such that their subscripts satisfy
\begin{eqnarray}
X_{n_1} X_{n_2} \cdots X_{n_l} |0\rangle _{R} ~~ \mbox{with } ~~ \sum _{i=1}^{l}n_{i}=0 ~~\mbox{and} ~~ n_{i}+n_{j}\not= 0 ~~ \mbox{for } {}^{\forall }i,j .
\label{eq:4-20}
\end{eqnarray}
One of the simplest examples is
\begin{eqnarray}
X_{2}X_{-1}X_{-1}|0 \rangle _{R} .
\label{eq:4-21}
\end{eqnarray}
Because a linear term of $X_{n}$ is prohibited by the constraint $\sum _{i=1}^{l}n_{i}=0$ and the squared terms like $X_{n}X_{-n}$ do not create massless states which consist of a single term: monomial-type.

By the way, there are massless states consist of not only monomial, but also polynomial.
For example, states like $(X_{2}X_{-2}-X_{1}X_{-1})|0\rangle _{R}$ or $\left( 4(X_{2}X_{-2})(X_{1}X_{-1}) -(X_{2}X_{-2})^2 -(X_{1}X_{-1})^2\right) |0\rangle _{R}$ are also physical states with $\mathcal{M}^2 =0$.

In the 3d Poincar\'{e} group, there are two Lorentz invariant which commute each other: $\mathcal{P}^{2}=-\mathcal{M}^{2}$ and $\mathcal{P} \cdot \mathcal{J} =\Lambda $.
The first invariant is mass squared operator and now we consider the case of $\mathcal{M}^2 =0$.
The second invariant is "spin" operator and represented as
\begin{eqnarray}
\Lambda = -\frac{1}{2} \sum _{n \not= 0} \sum _{m \not= 0, -n} \left( 1+\frac{m}{n}+\frac{n}{m} \right) X_{n+m} \frac{\partial }{\partial X_{n}}\frac{\partial }{\partial X_{m}}
\label{eq:4-22}
\end{eqnarray}
in the X-representation. We find that acting of $\Lambda $ on states decreases the number of $X$ by one.
Because of $[\mathcal{M}^2, \Lambda ]=0$, we can get some polynomial-type massless states by acting $\Lambda $ on monomial-type massless states.
For example,
{\setlength\arraycolsep{1pt}\begin{eqnarray}
\Lambda X_{2}X_{-1}X_{-1}|0\rangle _{R} &=& -3(X_{2}X_{-2}-X_{1}X_{-1})|0\rangle _{R} \nonumber \\
\Lambda X_{2}X_{2}X_{-1}X_{-1}X_{-1}X_{-1}|0\rangle _{R} &\sim & 6(2(X_{1}X_{-1})-3(X_{2}X_{-2}))X_{2}X_{-1}X_{-1}|0\rangle _{R}  \\
\Lambda \Lambda X_{2}X_{2}X_{-1}X_{-1}X_{-1}X_{-1}|0\rangle _{R} &\sim & 54((X_{1}X_{-1})^2+(X_{2}X_{-2})^2-4(X_{2}X_{-2})(X_{1}X_{-1}))|0\rangle _{R}, \nonumber
\label{eq:4-23}
\end{eqnarray}}
where $\sim $ means the omission of monomial-type massless states.
The first line of (\ref{eq:4-23}) is the most simple polynomial-type massless state. The second and third line are polynomial-type massless states with fifth and fourth degree of $X$.
Here note that all massless states with 3rd degree are monomial-type.

Furthermore the inner product of polynomial-type massless states and any massive states, $\psi _{m, s}$, vanish.
Because the number of the power of $r_{n}$ in polynomial-type massless states is greater than that in monomial-type massless states by even number, we will find this vanishing by using the partial integration and the reason that we use in the definition of the delta function (\ref{eq:a3-5}).

In this way, we expect that all polynomial-type massless states are created from monomial-type massless by the action of $\Lambda $ or other operators commutative with $\mathcal{M}^2$ and $M_{0}$, though we don't investigate more in this paper.

Now we consider the theory which consist only of massless states.
Such theory is expected to be space-time conformal invariant.
Here we define the "conformal dimension"\footnote{This definition is the inverse of the natural definition of the conformal dimension.} from the dilatation as
\begin{eqnarray}
\Delta _{R} \equiv i \mathcal{D}_{R} =\sum _{n} X_{n} \frac{\partial }{\partial X_{n}}
\label{eq:4-120}
\end{eqnarray}
and may use this operator to investigate the properties of the massless states.
We find that $\Delta _{R}$ count the number of $X$-type operators by the commutators
\begin{eqnarray}
[\Delta _{R}, X_{n}]=X_{n} .
\label{eq:4-130}
\end{eqnarray}
For example, we can determine that the state (\ref{eq:4-20}) has $\Delta _{R}=l$ and particularly the string ground state $|0 \rangle _{R}$ has $\Delta _{R}=0$.

Though we do not investigate massless states anymore, it is interesting to characterize massless states in terms of the 3d space-time conformal symmetry.

\section{Conclusion and Outlook}
In this paper, we showed that a 3D tensionless bosonic closed string in light-cone gauge have no anomaly of the space-time conformal symmetry under the Hermitian version of the R-ordering and the cutoff regularization.
Further we investigated the spectrum of a 3D tensionless string, particularly massless states.
When we consider the R-ordered string ground state, we obtained the simple expression.

Note that the results we got is in the case of a single string.
We don't understand anything about multi-string theories with interactions or string field theories, as well as in the case of \cite{bib:003, bib:004, bib:005}.
It is interesting to investigate the effects of the interaction.
To study this in detail, we may need the string field theory which reproduces the spectrum of a tensionless string.

A 3D tensionless string has some of prospects.

First, one may image the open string version of our results. In the open string case, we must consider the boundary condition at two endpoints.
Even if we choose any boundary conditions, we can fix the gauge in the same way as our case.
However we need the caution about the Fourier expansion.
In the case of a tensionless (or tensile) closed string, thanks to the periodicity of $\sigma $ (or a combination of $\sigma $ and $\tau $), we can use the Fourier expansion with respect to $\sigma $ (or a combination of $\sigma $ and $\tau $).
But in the case of a open string, the way of the expansion changes according to the boundary condition.
If we choose Neumann boundary condition at two endpoints, the coordinate $X(\tau ,\sigma )$ can be represented with a quasi-periodic function of $\sigma$ or the combination of $\tau $ and $\sigma $.
If we choose Neumann boundary condition at one endpoint and Dirichlet boundary condition at other, $X(\tau ,\sigma )$ can be represented with a quasi-antiperiodic function of $\sigma$ or the combination of $\tau $ and $\sigma $.
If we choose Dirichlet boundary condition at two endpoints, $X(\tau ,\sigma )$ can be represented with a quasi-periodic function of $\sigma$ or the combination of $\tau $ and $\sigma $, but $X$ can have a linear term of $\sigma $.
Therefore, in a tensionless open string except for the case of the Neumann-Neumann condition, we do not know simply whether we obtain the same result as this paper.
Furthermore note that because of the difference of the equation of motion between the tensile case and the tensionless case, the coordinate $X$ in a tensile open string are essentially periodic but in a tensionless open string is a formal expedient to represent the zero point of $X$ or its differential.
\footnote{Here we consider the case of Neumann-Neumann condition as an example. The equation of motion of a tensile string is $\left( \partial ^2_{\tau } -T^2 \partial ^2_{\sigma } \right) X(\tau ,\sigma ) =0$ and its general solution is $X(\tau ,\sigma ) = \frac{1}{2} \left( F(T\tau -\sigma ) + G(T\tau +\sigma ) \right)$. In the case of Neumann-Neumann condition such that $X'(\tau ,0)=X'(\tau ,\pi )=0$, we obtain $F'(\tau )=G'(\tau )$ and $F'(\tau +\pi )=G'(\tau -\pi )$. Using these relations, we get $G(u)=F(u)+a$ and $F(u+2\pi )=F(u)+2\pi \cdot v$ where $a$ and $v$ are constant. In this way, we can write $X(\tau ,\sigma )$ with the quasi-periodic function of the combination of $\tau $ and $\sigma $. On the other hand, the equation of motion of a tensionless string is $\partial ^2_{\tau } X(\tau ,\sigma ) =0$ and its general solution is $X(\tau ,\sigma ) = F(\sigma ) + \tau G(\sigma )$. In the case of Neumann-Neumann condition such that $X'(\tau ,0)=X'(\tau ,\pi )=0$, we get $F'(0)=F'(\pi )=0$ and $G'(0)=G'(\pi )=0$. We can choose $\cos (n\sigma ) $ with integer $n$ as the function that satisfy these relations. The case of other sets of boundary conditions at two endpoints are also discussed in the same way.}
Hence we can not use the Fourier expansion based on the periodicity simply and may need the discussion in another way.
For example, one may deal the functions $X(\sigma )$ or $P(\sigma )$ without using the mode expansions, like the case of \cite{bib:007}.
But it may be difficult to think the concrete regularization.

Next is the relation of the other background metric, which we didn't comment on this in this paper.
We find the invariance of the action(\ref{eq:2-4}) under the space-time Weyl transformation with the transformation of $V$.
Then we expect applications to other background space-times, e.g. $AdS_{3}$, though we must investigate whether this theory represents a tensionless string on non-Minkowski background space-time like a point particle or whether this classical fact is satisfied in quantum theory.

As other possibilities there are the supersymmetric version, the membrane version, the relation of Higher-spin gauge theories and so on.
In the case of a tensionless closed supersymmetric string, we can use the similar way to ours, unlike a tensionless open string.
We expect that the supersymmetric case in light-cone gauge gives the different result from the case of BRST quantization \cite{bib:018, bib:019}.
And it is interesting to consider not only string but also two dimensional object, so-called the membrane.
Moreover we are interested in the relation of Higher-spin gauge theories \cite{bib:009,bib:010}, which may be one of the limits of the string theory.

Of course we are interested in the explanation for the difference between BRST quantization and the light-cone gauge quantization and the covariant method that reproduces results of the light-cone gauge quantization.

\section*{Acknowledgements}
I would like to thank M. Fukuma, K. Sugiyama and S. Shimasaki for useful comments and advices.
The work of KM was supported by the Japan Society for the Promotion of Science (JSPS).

redpajama_set_name: RedPajamaArXiv


text:
John Gill's Commentary of the Whole Bible: 1 Kings 22
INTRODUCTION TO 1 KINGS 22
This chapter relates, that after three years' peace with the king of Syria, Ahab was inclined to go to war with him, to take Ramothgilead out of his hands; and he drew in Jehoshaphat, king of Judah, to join him in it, 1Ki 22:1, but before they went into it, they took advice, Ahab of his four hundred prophets, which Jehoshaphat not being satisfied with, a true prophet of the Lord, Micahah, was sent for, 1Ki 22:5 who, when he came, jeered Ahab with what his prophets had said to him; intimated that he should be killed, and explained it to him how he came to be deceived by his prophets, 1Ki 22:15 upon which he was smitten on the cheek by Zedekiah, one of the false prophets, and imprisoned by the order of Ahab, 1Ki 22:24, after which the two kings went to the battle, and Jehoshaphat was in great danger of his life; but Ahab was wounded, and died, 1Ki 22:29, and the chapter is concluded with an account of the reign of Jehoshaphat king of Judah, 1Ki 22:41, and of Ahaziah king of Israel, 1Ki 22:51.
Earmark 1 Kings 22:1
See 1 Kings 22:1 in… The King James Bible Geneva Bible (1599)
Ver. 1. And they continued three years without war between Syria and Israel. From the time that Benhadad made a covenant with Ahab; not three full years, but part of them: it was threatened by Elijah from the Lord, that Ahab's life should go for Benhadad's, because he had let him, go, 1Ki 22:42, but because of his humiliation, as is thought by Ben Gersom and others, it was respited for those three years; and now an opportunity and occasion would be given for the fulfilment of what was threatened.
Ver. 2. And it came to pass in the third year,… Of the peace, before it was expired:
that Jehoshaphat the king of Judah came down to the king of Israel; to Ahab, from Jerusalem to Samaria, reckoned thirty two miles {m}; either to make peace with him, and put an end to the wars which subsisted between Israel and Judah since the division of the kingdom, 1Ki 22:44 or to contract an affinity with him, by marrying his son to a daughter of Ahab, 2Ki 8:18 or rather after peace was made, and that strengthened by the marriage; and so he went merely to pay a visit, as he judged he might then with great safety; and he and all his retinue were entertained by Ahab in a very sumptuous and liberal manner, 2Ch 18:1.
{m} Bunting's Travels, &c. p. 178. near 40, Rainold. Praelect. 31. col. 266.
Ver. 3. (And the king of Israel said unto his servants,… His nobles, those of his privy council, his ministers of state; or "had said" {n}, some little time before Jehoshaphat came:
know ye that Ramoth in Gilead is ours a city of refuge beyond Jordan, in the tribe of Gad, and so of course must belong to the kingdom of Israel, of which see Jos 20:8
and we be still, and take it not out of the hand of the king of Syria?) neither demand it, nor take any measures to oblige him to deliver it up; representing it as a great omission, and as a piece of negligence and slothfulness, or cowardice.
{n} rmayw "et dixerat", Junius & Tremellius.
Ver. 4 And he said unto Jehoshaphat, wilt thou go with me to battle to Ramothgilead?… This affair being lately canvassed at the council board, and very much on Ahab's mind, he puts this question to Jehoshaphat, his visitor, relation, and ally; wisely considering that his own forces were small, and that to have such an auxiliary might be of great advantage to him:
and Jehoshaphat said to the king of Israel, I am as thou art, my people as thy people, my horses as thy horses; meaning, that he and his soldiers, foot and horse, were at his service.
Ver. 5. And Jehoshaphat said unto the king of Israel, inquire, I pray thee, at the word of the Lord today. Being a pious and religious prince, he did not choose to go into a war at once, without consulting the Lord by his prophets, whether it was his will and pleasure they should engage in it, and should prosper; and he was desirous of having this done immediately, before they proceeded any further.
Ver. 6. Then the king of Israel gathered the prophets together, about four hundred men,… False prophets, as the Targum and Arabic version; and they are called Ahab's prophets, and not the Lord's, 1Ki 22:23 perhaps these were the prophets of the groves, that ate at Jezebel's table, and were preserved when the prophets of Baal were destroyed, since the number agrees with them, see 1Ki 18:19
and said unto them, shall I go against Ramothgilead to battle, or shall I forbear? which would you advise to? signifying he should take their advice:
and they said, go up; for the Lord shall deliver it into the hand of the king: which words are very ambiguous, like the oracles of the Heathens; for they do not express who or what should be delivered up, for the word it is a supplement, nor to what king the delivery should be made; whether the Syrians, and the place they held should be given up to king Ahab, which they would have understood; or whether the Israelites should be delivered up to king Benhadad; so that, whichever had been the case, the credit of their prophecy would be secured. They used the word "Lord", and not Baal, in complaisance to Jehoshaphat, and perhaps as directed by Ahab.
Ver. 7. And Jehoshaphat said, is there not here a prophet of the Lord besides, that we might inquire of him?] He suspected these to be false prophets, though he would not call them so; nor suggest that they were not the prophets of the Lord, because he would not affront Ahab, who had an opinion of them; and therefore asks, if there were no other that went under the character of a prophet of the Lord, that he might inquire of him for his further satisfaction.
Ver. 8. And the king of Israel said unto Jehoshaphat, there is yet one man (Micaiah the son of Imlah), by whom we may inquire of the Lord,… And but one in Samaria; Elijah and Elisha were elsewhere:
but I hate him, for he doth not prophesy of good concerning me, but evil; who is thought to be the same that was several times with him when engaged in the war with the king of Syria, 1Ki 20:13 and each time, excepting the last, he brought him good tidings; but because, in his last message, he told him, that, since he had let Benhadad go, his life should go for his life, and his people for his people, for that he hated him:
and Jehoshaphat said, let not the king say so; which was very modestly, though perhaps too gently, said; suggesting that the prophets of the Lord should be heard, respected, and honoured, let their message be as it would, since they spake not of their own mind and will, but what they were moved unto by the Spirit of God.
Ver. 9. Then the king of Israel called an officer,… An eunuch, as the word is sometimes used, one of pages:
and said, hasten hither Micaiah the son of Imlah; who, as it seems from
1Ki 22:26 was in prison, where perhaps Ahab had cast him for his last prophecy to him, and where he had lain ever since; and this gives a reason why he could so readily send for him, knowing where he was.
Earmark 1 Kings 22:10
See 1 Kings 22:10 in… The King James Bible Geneva Bible (1599)
Ver. 10. And the king of Israel, and Jehoshaphat the king of Judah, sat each on his throne,… In great state and majesty:
having put on their robes; their royal robes, which they wore when they appeared in pomp and grandeur:
in a void place in the entrance of the gate of Samaria; where courts of judicature were held, and there was an open void space for the people to assemble in to hear; the word has the signification of a corn floor, and the Jews suppose they and their attendants sat in a semicircle like the half of a corn floor, after the same manner in which they say the sanhedrim at Jerusalem sat {o}:
and all the prophets prophesied before them; concerning this affair of going to Ramothgilead.
{o} T. Bab. Cholin, fol. 5. 1. Vid. Kimchium in loc.
Ver. 11. And Zedekiah the son of Chenaanah made him horns of iron,… Horns are emblems of power and might, and iron ones of greater strength still; the prophets sometimes made use of visible signs, to represent the things they prophesied of should come to pass, see
Isa 20:2, and the same method this prophet took:
and he saith, thus saith the Lord; imitating the true prophets: with these shall thou push the Syrians until thou hast consumed them: Abarbinel thinks he had in view the blessing of Joseph by Moses, De 33:17 where he is compared to a bullock with horns; and these said to be the ten thousands of Ephraim, and the thousands of Manasseh; and Ahab being of the tribe of Joseph, and ruling in Ephraim and Manasseh, the prophet chose to make use of this emblem for his encouragement.
Ver. 12. And all the prophets prophesied so, saying, go up to Ramothgilead, and prosper,… All encouraged the king to go up against this place, and prophesied of victory, as Zedekiah did:
for the Lord shall deliver it into the king's hand,
See Gill on "1Ki 22:6".
Ver. 13. And the messenger that was gone to call Micaiah spake unto him,… By the way, as they came along together, as Josephus {p} observes:
behold, now, the words of the prophets declare good unto the king with one mouth; they are unanimous that he shall prosper in his undertaking against the Syrians:
let thy word, I pray thee, be like the word of one of them, and speak that which is good; which, as an ignorant man, he might advise to from good will to the prophet, that he might not be branded with singularity, and a spirit of contradiction, and that he might have the favour of the king, and be released from prison, pitying his miserable condition in which he found him.
{p} Antiqu. l. 8. c. 15. sect. 4.
Ver. 14. And Micaiah said, as the Lord liveth,… He swore by the living God, for the confirmation of what he was about to say:
what the Lord saith unto me, that will I speak; truly and faithfully, keeping nothing back, nor adding anything, whether it be good or evil, pleasing or displeasing; it looks as if as yet he had no instruction from the Lord what to say, and yet the vision he later declares seems to have been had by him before, 1Ki 22:17.
Ver. 15. So he came to the king,… Being introduced by the officer:
and the king said unto him, Micaiah, shall we go up against Ramothgilead to battle, or shall we forbear? the same question in the same words that was put to the other prophets, 1Ki 22:6, only there he uses the singular number, here the plural, including Jehoshaphat with him:
and he answered him, go, and prosper: for the Lord shall deliver [it] into the hand of the king; he answered not in the name of the Lord, saying, "thus saith the Lord", nor did he speak his own sense and in his own words, nor seriously, but by way of derision; he took up the words of the prophets, and bantered them; it is as if he should say, the prophets bid you go, and tell you that you shall "prosper", and that the city will be delivered into the king's hand; do as they direct you, and see what the issue will be, no doubt it will be good, since they are all agreed; but he delivered the above words with such gestures, and such a tone, and with a contemptuous smile in his countenance, which showed that he spoke not seriously, but sarcastically; and this the king plainly discovered, as appears by what follows.
Ver. 16. And the king said unto him, how many times shall I adjure thee,… Not that he had as yet adjured him at all, or not till now; but he asks him how often he must be obliged to do it; and now he adjures him once for all, that he might not be forced to repeat it:
that thou tell me nothing but that which is true in the name of the Lord? for he observed he did not speak in the name of the Lord before, and what he said was not in a serious but ludicrous manner, and not to be regarded as truth.
Ver. 17. And he said,… The prophet, in a serious and solemn manner, being adjured by the king:
I saw all Israel scattered on the hills, as sheep that have not a shepherd; the armies of Israel routed, dispersed, and fleeing, some one way and some another, on the mountains of Gilead near Ramoth, weak and helpless, not knowing where to go for safety, having none to direct them; and this was either now instantly represented to his mind, or what had been before in a dream or vision:
and the Lord said, these have no master; these sheep have no shepherd this army hath no general,
[Israel] has [lost its king]: let them return every man to his house in peace, very few slain, Jarchi thinks Ahab only, see 1Ki 22:31 that part of the threatening, 1Ki 20:42 was now to he accomplished, "thy life shall go for his life", but the other part, "and thy people for his people", was to be deferred to another time.
Ver. 18. And the king of Israel said unto Jehoshaphat,… Plainly perceiving that the prophet foretold that he should fall in battle:
did not I tell thee that he would prophesy no good concerning me, but evil? intimating that this proceeded from spite and malice, from ill will to him and hatred of him, and was not from the Lord, and therefore not to be regarded; he had told him three years ago his life should go for letting Benhadad go; but it had not proved true, and no more would this; and Jehoshaphat being an easy man, and too credulous, believed what Ahab said of the character of this prophet, or otherwise it is not to be accounted for that he should go with him to war after such a declaration made.
Ver. 19. And he saith, hear thou therefore the word of the Lord,… Since he had represented what he had said as proceeding from hatred to him, he would make it clear and plain that what he had said was the word of the Lord, and according to his mind; and that what the other prophets had said was owing to a lying spirit in them, which the Lord suffered for his ruin; all which are represented as in a vision, in which things are brought down to the capacities of men, and not as really transacted:
I saw the Lord sitting on his throne; so it was represented to his mind, as if he had seen with his bodily eyes the divine Being in a glorious form, as a king sitting on his throne, to do justice and judgment; as Ahab and Jehoshaphat were now sitting on their thrones, only as a far greater King, even the King of kings, and in a more splendid manner:
and all the host of heaven standing by him on his right hand and on his left the ministering angels ready to do his will.
Ver. 20. And the Lord said, who shall persuade Ahab, that he may go up and fall at Ramothgilead?… Not that it can be supposed that the Lord entered into a consultation with the angels upon this subject; only that it was the decree of God that he should go thither, and fall by the hand of the man whom he had let go, as a just punishment of him:
and one said on this manner, and another said on that manner; not that there was such an altercation among them; it only signifies, that there are various ways and means, by which the purposes and decrees of God may be and are brought about.
Ver. 21. And there came forth a spirit,… Not from the heavenly host on the right hand or the left, for they are pure and holy spirits, and impeccable, and cannot lie or deceive; but the evil spirit, Satan, the father of lies, the old deceiver, who came forth from his own place and his own company:
and stood before the Lord presented himself before him, as Satan did, Job 1:6,
and said, l will persuade him; or prevail upon him; evil spirits love to be employed in doing harm to men, they go about seeking whom they may devour. This could not be the spirit of Naboth, as the Jews say {q}, seeking revenge on Ahab; that was in a state of happiness, could not move from thence, and be capable of sinning.
{q} T. Bab. Sanhedrin, fol. 89. 1. & 102. 2. Targum in 2 Chron. xviii. 20.
Ver. 22. And the Lord said unto him, wherewith?… What way and method did he propose, to persuade Ahab to go up to Ramoth? the Lord is introduced in this visionary narrative as asking this question, not as ignorant of the scheme of the evil spirit, but in order to bring it out, and lead on to the following account:
and he said, I will go forth, and I will be a lying spirit in the mouth of all his prophets; put them on encouraging Ahab to go up, and promising him success, as he had in former battles with the king of Syria, and which might both encourage them to give forth such a prediction, and him to believe it to be true; this proposal was quite agreeable to the character of the devil, as the father of lies:
and he said, thou shalt persuade him, and prevail also; not only make use of this artifice to persuade, but succeed also; the Lord knew that what he should suggest to the prophets, and they should deliver to Ahab, would be agreeable to his inclination, nor would he do anything in the course of his providence to hinder its taking effect:
go forth, and do so; which was giving leave to try his skill in the art of persuasion, in which he knew he would succeed, and bring on the righteous judgment of God upon Ahab; with this compare Joh 13:27.
Ver. 23. Now therefore behold, the Lord hath put a lying spirit in the mouths of all these thy prophets,… That is, suffered the lying spirit to suggest a lie to them, and sent them strong delusions to believe that lie, whose minds were disposed at any rate to flatter Ahab, to whom they told it; which was the way designed to bring him to the ruin appointed for him:
and the Lord hath spoken evil concerning thee: he had decreed it in himself, declared it by Micaiah his prophet, and suffered all those steps to be taken by Satan and the false prophets, to bring him to it.
Ver. 24. But Zedekiah the son of Chenaanah went near,… Stepped in haste and passion from the place where he was:
and smote Micaiah on the cheek; in contempt of him, and to show his indignation at what he said; this he did in open court, before two kings; one he believed would favour and screen him in this lawless action, and the other, out of his own jurisdiction, had not courage and presence of mind to resent it:
and said, which way went the Spirit of the Lord from me to speak unto thee? hereby boasting that he had the Spirit of the Lord, and was directed by him in what he said, and still remained with him, and could not possibly go to Micaiah, and suggest the very reverse; and therefore pertly asks him which way the spirit went, intimating that it was impossible he could steer a course contrary to himself.
Ver. 25. And Micaiah said, behold, thou shalt see in the day when thou shalt go into an inner chamber to hide thyself. Who would either accompany Ahab to the battle, and, upon his being wounded, flee to the first place of secrecy for safety; or, upon the news of his defeat brought to Samaria, would betake himself to a private chamber for security, fearing the enemy would pursue to the very place; or else through fear of the populace, who would attribute the death of the king to the advice of him and the other prophets.
Ver. 26. And the king of Israel said,… To some of his officers:
take Micaiah, and carry him back unto Amon the governor of the city: the chief magistrate under the king; a sort of sheriff, who had the care of malefactors, and of all committed to prison, from whom he was received by the messenger, and now sent back to him:
and to Joash the king's son; who might be over his household, as sometimes the king's son was, 2Ch 26:21 or might be viceroy while the king was without the city, and at the gate of it, and about to go to war.
Ver. 27. And say, thus saith the king, put this fellow in prison,… In the common prison of the city, where he had been before, as it seems; and might be now ordered into a more confined place in it, and what might be called "little ease":
and feed him with bread of affliction, and with water of affliction; with bad bread and foul water, and but little of either; just enough to keep alive, and to continue starving:
until I come in peace; which he seemed confident of, and intimates that then he would punish him more severely, even with death, as a false prophet.
Ver. 28. And Micaiah said, if thou return at all in peace, the Lord hath not spoken by me,… I am content to be reckoned a false prophet, and to be punished as such:
and, he said, hearken, O people, everyone of you; he called aloud unto them to observe what he had predicted, and mark the issue of it, and to bear testimony for him, or against him, as things should be.
Ver. 29. So the king of Israel, and Jehoshaphat the king of Judah, went up to Ramothgilead. Which, according to Bunting {r}, was twenty four miles from Samaria. That Ahab went is no wonder, it was his own motion first, his inclination led to it, his prophets encouraged him, and, in bravado to the prophet of the Lord, was determined upon it; but it may seem much more strange that Jehoshaphat should, after such an account as Micaiah had given, and who, doubtless, could observe a great difference between him and the prophets of Ahab; and yet there is much to be said which might incline him to go, as that there were four hundred prophets all agreed, and who made use of the name of the Lord, and pretended to be true prophets; and though he might suspect them, he could not confute them; and Micaiah, he prophesied evil of Ahab only, and therefore Jehoshaphat might think himself safe in going; and besides, he had given his word to Ahab, and he did not choose to go from it; to which may be added, that Ahab's cause was just, to recover a part of his own dominions.
{r} Travels, &c. p. 178.
Ver. 30. And the king of Israel said unto Jehoshaphat, I will disguise myself, and enter into the battle,… Change his clothes, his royal robes, and put on others, perhaps the habit of a common soldier; having, it may be, been informed by some deserters or spies, of the design of Benhadad against him. Abarbinel thinks the meaning is, that he would clothe himself with a coat of mail, and take to him the each of the instruments of war, and so go into the battle secure; this seems probable from 1Ki 22:34 and this he might do to elude the prophecy of Micaiah:
but put thou on thy robes; his royal robes, or rather keep them on, that he might appear to be the chief commander of the army. There seems to be a good deal of insincerity and treachery in this conduct of Ahab's, whatever honour he might pretend to Jehoshaphat, or safety he might promise him in such a situation; his view seems to be to save himself at the hazard of the life of Jehoshaphat, especially if the Septuagint version could be established, "and put on my clothes"; which is natural enough, but would have been too barefaced:
and the king of Israel disguised himself, and went into the battle; as if he had been a common soldier.
Ver. 31. But the king of Syria commanded his thirty and two captains that had the rule over his chariots,… This was the number of his kings in the first battle with Israel, and of his captains in the second, 1Ki 20:1, and the same number he had now, being very probably not only the number of his chariots, but the division of his army was into so many battalions, under the command of these captains of chariots:
saying, fight neither with small nor great; of those that belonged to Jehoshaphat:
save only with the king of Israel; and his men; for it can hardly be thought that his orders were to fight with none, nor kill any in the battle but Ahab personally; though it is very probable he might give them directions to aim at him chiefly, knowing that, if he was killed or taken, his army would flee or surrender; and he might be desirous of getting him into his hands, as he had been in his; and the rather his spite was against him, as he was the mover of the war.
Ver. 32. And it came to pass, when the captains of the chariots saw Jehoshaphat,… In his royal robes:
that they said, surely it is the king of Israel; for they might not know the persons either of Ahab or him, but judged by his habit:
and they turned aside to fight against him; pressed upon him with all their force, either to take him or slay him:
and Jehoshaphat cried out; with a loud voice, either to the captains to let them know who he was, or to his men to come to his assistance, or rather to God to help and deliver him; since it is said in 2Ch 18:31 that the Lord helped him, and moved or inclined the hearts of the captains to depart from him, as it follows here.
Ver. 33. And it came to pass, when the captains of the chariots perceived that it was not the king of Israel,… Against whom only their orders were to fight:
that they turned back from pursuing him; for upon so great a force coming upon him he could not withstand, he fled.
Ver. 34. And a certain man drew a bow at a venture,… Not aiming at the king of Israel, or knowing whereabout he was. In the Targum on
2Ch 18:33, this man is said to be Naaman, the general of the army of the king of Syria, and so Jarchi here; but though he did this in his simplicity, as the word signifies, without any intention to smite any particular person; yet God directed the arrow to the man he had marked for destruction, and neither his disguise, nor coat of mail, could secure from that:
and smote the king of Israel between the joints of the harness: of which the pieces of armour on him were joined together, the higher and lower parts of it, the breastplate, and what covered the belly; and though these were joined as close as they were capable of joining them in those times, yet the arrow, guided by divine Providence, found its way into his body:
wherefore he said unto the driver of his chariot, turn thine hand; or hands, with which he held the reins, and turn the horses on one side:
and carry me out of the host; where the battle was hottest, to a place more remote and private, that he might have the wound examined, and the blood stopped, and return again, as it seems he did:
for I am wounded; or rather "I am sick" {s}, or ill, as the Targum; somewhat out of order, and therefore chose to retire a little while; not caring it should be known that he was smitten and wounded, lest his soldiers should be disheartened.
{s} ytylxh "aegrotare factus sum", Vatablus; "aegrotus factus sum", Junius & Tremellius.
Ver. 35. And the battle increased that day,… It went on, and did not stop upon Ahab's going out of the host, but was very hot, and both sides fought furiously:
and the king was stayed up in his chariot against the Syrians: the Targum is,
"he strengthened himself, and stood;"
he exerted himself to the uttermost, and stood as long as he could, or could be supported, fighting against the Syrians, to animate his army, and that the Syrians might not have any notion of his being wounded:
and died at even: in his chariot:
and the blood ran out of the wound into the midst of the chariot; or "bosom" {t} of it, the hollow part of it.
{t} qyx la "ad sinum", Montanus; "in sinum", Vatablus.
Ver. 36. And there went a proclamation throughout the host, about the going down of the sun,… Much about the time that Ahab died; and this proclamation by an herald might be made by his order, as he was dying, or by Jehoshaphat, when he understood he was dead:
saying, every man to his city, and every man to his own country; the order was to cease fighting, and make the best of their way as fast as they could to their own homes, since their shepherd and master was dead, which fulfilled the vision of Micaiah, 1Ki 22:17. It seems to have been a drawn battle, at least there is no account of the advantage on either side.
Ver. 37. So the king died, and they brought him to Samaria,… In the chariot he died in:
and they buried the king in Samaria; where his father Omri was buried, 1Ki 16:28.
Ver. 38. And one washed the chariot in the pool of Samaria,… After the body was taken out, very likely the chariot driver, who drove it into the pool, and plunged it into it, as the word signifies, to cleanse it from the blood of his master:
and the dogs licked up his blood; mixed with the water of the pool; the Septuagint adds, "the swine", which is not probable, such creatures not being bred in the land of Israel:
and they washed his armour; his coat of mail, through the joints of which the blood issued, and ran upon it. The word is sometimes used for whores, and is so translated here in the Greek version, and by Munster and Castalio; and both Ben Gersom and Abarbinel say, that women, who were harlots, washed here in his blood, mixed with water; and so Josephus {u} writes, that afterwards it was a custom for whores to wash in this pool; though some say {w} two whores were painted on Ahab's chariot, by the order of Jezebel, to inflame his lust, and these were what were washed; but the word signifies armour, or rather ornaments, clothes, jewels, &c.
and now all this was according to the word of the Lord which he spake; both by Elijah, that as the dogs licked the blood of Naboth, so they should his, as they now did, though not in the same place; nor was it necessary to fulfil the prophecy, See Gill on "1Ki 21:19", though some have thought {x} that his blood, mixed with the water of the pool of Samaria, was carried in a stream down to Jezreel, and there licked by the dogs, where Naboth's was; but chiefly what was spoken by Micaiah is here respected, that thus Ahab fell at Ramothgilead, as he had prophesied, 1Ki 22:17 and his life went for the life of Benhadad, as he had before declared, 1Ki 20:42.
{u} Antiqu. l. 8. c. 15. sect. 6. {w} See Jarchi & Kimchi in loc. {x} Kimchi in loc.
Ver. 39. Now the rest of the acts of Ahab, and all that he did, and the ivory house which he made,… Which, being a very curious and extraordinary thing, is particularly mentioned; though perhaps it might not be made wholly of ivory, but inlaid with it; we read of ivory houses in Am 3:15,
and all the cities that he built, are they not written in the book of the chronicles of the kings of Israel? in which the acts of his predecessors were recorded, see 1Ki 14:19 not the Scripture book of Chronicles, for there none of these things are related.
Ver. 40. So Ahab slept with his fathers, and Ahaziah his son reigned in his stead,] Of whom more is said in the latter part of this chapter, and in the following book.
Ver. 41. And Jehoshaphat the son of Asa began to reign over Judah in the fourth year of Ahab king of Israel. And as Ahab reigned twenty two years, 1Ki 16:29, Jehoshaphat must reign about eighteen years with him, and seven years after him.
Ver. 42. Jehoshaphat was thirty five years old when he began to reign, and he reigned twenty and five years in Jerusalem,… So that he must be sixty years of age when he died:
and his mother's name was Azubah, the daughter of Shilhi; but of what family they were is not said.
Ver. 43. And he walked in all the ways of Asa his father,… Who was a good prince:
he turned not aside from doing that which was right in the eyes of the Lord; in his moral conversation, religious worship, and civil government:
nevertheless, the high places were not taken away, for the people offered and burnt incense yet in the high places; he took away the high places and groves for idolatrous worship, 2Ch 17:6, but not the high places in which sacrifices were offered to the Lord, which ought to have been, especially since the temple was built; and those in the tribes of Judah and Benjamin were altogether inexcusable, being near to the temple, and under no restraint, as those of the ten tribes were; but the people were fond of them, because of their antiquity, and it was difficult for religious princes to remove them, if inclined.
Ver. 44. And Jehoshaphat made peace with the king of Israel. First with Ahab, with whom he contracted an affinity, and joined with him in his expedition to Ramothgilead, and with Ahaziah his successor.
Ver. 45. And all the rest of the acts of Jehoshaphat, and all his might that he showed, and all his wars, are they not written in the book of the Chronicles of the kings of Israel?] Many of them are recorded in the canonical book of Scripture, which bears the name of Chronicles; and more might be in this referred to, out of which might be taken what God has thought fit to transmit to future times; see 2Ch 17:1.
Ver. 46. And the remnant of the Sodomites, which remained in the days of his father Asa, he took out of the land. His father Asa removed many of these filthy creatures, but not all; as many, no doubt, as came within his knowledge, but some remained, whom this his son removed, being of the same disposition with his father, see 1Ki 15:12.
Ver. 47. There was then no king in Edom, a deputy was king. Which had been the case from the times of David, who subdued Edom, and placed garrisons in it, and governors over it, 2Sa 8:14 and continued through the reign of Jehoshaphat, unto the times of his son, under whom the Edomites revolted, and set up a king of their own, 2Ki 8:20, with a view to which this is observed, as also to account for it how Jehoshaphat could build ships in Eziongeber, which was in the land of Edom, of which in the next verse, because the whole country was governed by a viceroy, or deputy, under him.
Ver. 48. Jehoshaphat made ships of Tarshish,… Ships to go to sea, particularly the Indian sea, 1Ki 10:22. Tarshish is used for the sea in general, Ps 48:7, in the Cetib, or text, it is "ten"; in the Keri, or margin, it is "made", which we follow, and may be put together, as in the Tigurine version, and read, "he made ten ships to go by sea":
even to go to Ophir for gold; as Solomon did; of which place see 1Ki 9:28,
but they went not, for the ships were broken at Eziongeber; the port where they were built: as soon as they were launched, or sailed, they were broken to pieces against the rocks near the harbour, which stood up like a man's backbone, whence the port had its name;
See Gill on "1Ki 9:26", and if this was Calzem, as there observed, near to it was a dangerous place for ships, and where many were lost, and is supposed to be the place where Pharaoh and his host were drowned {y}; the reason of this shipwreck was, because Jehoshaphat joined himself with Ahaziah king of Israel, for which he was reproved by the prophet Eliezer, and this was his punishment, 2Ch 20:35.
{y} Vid. Geograph. Nub. Climat. 3. par. 3. in fine.
Ver. 49. Then said Ahaziah the son of Ahab unto Jehoshaphat,… Who very probably had built some more ships on his own, having broke off his partnership with Ahaziah:
let my servants go with thy servants in the ships; since he was refused a part in the ships themselves, he desires leave to send men aboard them to traffic for him abroad:
but Jehoshaphat would not; having been reproved by a prophet of the Lord, and had suffered the loss of his ships by joining with him already.
Ver. 50. And Jehoshaphat slept with his fathers, and was buried with his fathers in the city of David his father,… In the city of Sion, where David, Rehoboam, Abijam, and Asa, were buried:
and Jehoram his son reigned in his stead; who was now thirty two years of age, and he reigned ten years.
Ver. 51. Ahaziah the son of Ahab began to reign over Israel in Samaria the seventeenth year of Jehoshaphat king of Judah,… It is observed, that Jehoshaphat began to reign in the fourth year of Ahab, and Ahab reigned twenty two years, see 1Ki 22:41, and therefore Ahab's son must begin to reign in the eighteenth year of Jehoshaphat; but perhaps he was made king in his father's lifetime, before he went on his expedition to Ramothgilead, or Ahab's reign was not twenty two years complete:
and reigned two years over Israel; not complete, as appears from 2Ki 3:1.
Ver. 52 And he did evil in the sight of the Lord,… Which evil was idolatry:
and walked in the way of his father; his father Ahab, who worshipped Baal:
and in the way of his mother; his mother Jezebel, who was still living, and served Baal and Astarte, the deities of her country:
and in the way of Jeroboam the son of Nebat; who set up and worshipped the golden calves:
who made Israel to sin; by the worship of the same, into which he drew them by his example and authority.
Ver. 53. For he served Baal, and worshipped him,… That is, Ahaziah served him, as his father had done, and his mother still did:
and provoked to anger the Lord God of Israel, according to all that his father had done; of which there is an instance in the first chapter of the following book; for falling through a lattice, and becoming sick upon it, he quickly died, having sent messengers to inquire of the god of Ekron whether he should die or not.

redpajama_set_name: RedPajamaCommonCrawl


The airport in Birmingham, AL, that I fly in and out of, just got a facelift, part of a national effort to update some US airports. Half of it was closed for a major renovation, which was completed last Spring, and now the other half is closed and being ripped apart so that it too can be reinvented. The new terminal is spacious and bright, with new vendors and electrical outlets at many of the seats, and easy listening music playing throughout.


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text:
/** @file
An implementation of the API in qubits.h for a local (non-MPI) environment.
*/

# include <stdlib.h>
# include <stdio.h>
# include <omp.h>
# include "qubits.h"
# include "precision.h"
# include "qubits_internal.h"

void createMultiQubit(MultiQubit *multiQubit, int numQubits, QuESTEnv env)
{
	createMultiQubitCPU(multiQubit, numQubits, env);
}

void destroyMultiQubit(MultiQubit multiQubit, QuESTEnv env)
{
	destroyMultiQubitCPU(multiQubit, env);
}

void initStateVec(MultiQubit *multiQubit){
	initStateVecCPU(multiQubit);
}

void initQuESTEnv(QuESTEnv *env){
        // init MPI environment
	env->rank=0;
	env->numRanks=1;
}

void syncQuESTEnv(QuESTEnv env){
	// MPI Barrier goes here in MPI version.
}

void closeQuESTEnv(QuESTEnv env){
	// MPI finalize goes here in MPI version. Call this function anyway for consistency
}

void reportQuESTEnv(QuESTEnv env){
	printf("EXECUTION ENVIRONMENT:\n");
	printf("Running locally on one node\n");
	printf("Number of ranks is %d\n", env.numRanks);
# ifdef _OPENMP
	printf("OpenMP enabled\n");
	printf("Number of threads available is %d\n", omp_get_max_threads());
# else
	printf("OpenMP disabled\n");
# endif
}

double calcTotalProbability(MultiQubit multiQubit){

  /* IJB - implemented using Kahan summation for greater accuracy at a slight floating
     point operation overhead. For more details see https://en.wikipedia.org/wiki/Kahan_summation_algorithm */
  /* Don't change the bracketing in this routine! */
  REAL pTotal=0;
  REAL y, t, c;
  long long int index;
  long long int numAmpsPerRank = multiQubit.numAmps;

  c = 0.0;
  for (index=0; index<numAmpsPerRank; index++){
    /* Perform pTotal+=multiQubit.stateVec.real[index]*multiQubit.stateVec.real[index]; by Kahan */
   // pTotal+=multiQubit.stateVec.real[index]*multiQubit.stateVec.real[index];

    y = multiQubit.stateVec.real[index]*multiQubit.stateVec.real[index] - c;
    t = pTotal + y;
    c = ( t - pTotal ) - y;
    pTotal = t;

    /* Perform pTotal+=multiQubit.stateVec.imag[index]*multiQubit.stateVec.imag[index]; by Kahan */
    //pTotal+=multiQubit.stateVec.imag[index]*multiQubit.stateVec.imag[index];

    y = multiQubit.stateVec.imag[index]*multiQubit.stateVec.imag[index] - c;
    t = pTotal + y;
    c = ( t - pTotal ) - y;
    pTotal = t;

  }
  return pTotal;
}

void rotateQubit(MultiQubit multiQubit, const int rotQubit, Complex alpha, Complex beta)
{
	// all values required to update state vector lie in this rank
	rotateQubitLocal(multiQubit, rotQubit, alpha, beta);
}

double findProbabilityOfZero(MultiQubit multiQubit,
                const int measureQubit)
{
	double stateProb=0;
	stateProb = findProbabilityOfZeroLocal(multiQubit, measureQubit);
	return stateProb;
}

double measureInZero(MultiQubit multiQubit, const int measureQubit)
{
        double stateProb;
	stateProb = findProbabilityOfZero(multiQubit, measureQubit);
        measureInZeroLocal(multiQubit, measureQubit, stateProb);
        return stateProb;
}

double filterOut111(MultiQubit multiQubit, const int idQubit1, const int idQubit2, const int idQubit3)
{
        double stateProb=0;
        stateProb = probOfFilterOut111(multiQubit, idQubit1, idQubit2, idQubit3);
        filterOut111Local(multiQubit, idQubit1, idQubit2, idQubit3, stateProb);
        return stateProb;
}

double probOfFilterOut111(MultiQubit multiQubit, const int idQubit1, const int idQubit2, const int idQubit3)
{
        double stateProb=0;
        stateProb = probOfFilterOut111Local(multiQubit, idQubit1, idQubit2, idQubit3);
        return stateProb;
}

redpajama_set_name: RedPajamaGithub


See a private or NHS podiatrist at least once a year. You should be eligible for an NHS podiatrist if you have a long term condition such as diabetes. Ask your GP for a referral or find a local podiatrist.
Seek treatment from your GP or podiatrist if blisters or injuries do not heal quickly.


Five years ago the UCI Symphony Orchestra, led by Maestro Stephen Tucker, adopted the Santa Ana High School (SAHS) Symphony Orchestra. The partnership was designed to build awareness of the university experience to make it a tangible goal for the SAHS students; increase access to arts resources; and cultivate engaged young artists, the Trevor School students, who will become involved citizens in their communities not just in the arts.
For the first time, the UCI Symphony Orchestra will perform a full concert in Santa Ana, continuing the intersection of our two communities for the benefit of both, on Nov. 4, 2016 at 8 p.m. Experience the famous Bill Medley Auditorium, whose namesake is the Grammy Award-winning singer and songwriter best known as one-half of The Righteous Brothers and a SAHS alumnus '58, to support both schools at this special concert.