\section{ Introduction}

The conjectured duality between
the type IIB superstring theory on the AdS$_5\times$S$^5$ space
(AdS superstring)
and
$D=4,~{\cal N}=4$ Yang-Mills theory
\cite{M,GKP,W}
has been driven not only
studies of variety of background theories
but also studies of basic aspects such as integrability.
The approach of the pp-wave background superstring theory \cite{MTpp}
was explored
by Berenstein, Maldacena and Nastase \cite{BMN} and
developed in, for example
\cite{GKP2,FT}.
For further development Mandal, Suryanarayan and Wadia
pointed out the relevance with the integrability \cite{MSW},
and Bethe anzatz approach was explored
 by Minahan and Zarembo \cite{MZ} and in for example \cite{B,DNW,BFST}.
The integrability
is a powerful aspect expected in the large N QCD \cite{Lipatov}
and shown to exist in
 the IIB superstring theory on the AdS$_5\times$S$^5$ space
by Bena, Polchinski and Roiban \cite{BPR}.
The integrability provides hidden symmetry generated by
an infinite number of
conserved ``non-local" charges \cite{LP,BIZZ}
as well as
an infinite number of  conserved ``local" charges
\cite{pol2} which are related by a spectral parameter
at different points.
Related aspects on the integrability of the AdS superstring were
   discussed in \cite{new18}.

Recently the conformal symmetry of AdS superstrings
was conjectured due to  the $\kappa$ symmetry
\cite{polyakov}.
The classical conformal symmetry of the AdS superstring theory
also  leads to an infinite number of conserved Virasoro operators.
The naive questions are
how the conformal generator  is related to the infinite number of conserved
``local" currents, and how many independent conserved currents exist.
For principal chiral models the stress-energy tensor
is written by trace of the square of the conserved flat current;
for  reviews see refs.
\cite{EHMM,MSW}.
For the AdS superstring theory
the Wess-Zumino term and the $\kappa$ symmetry make a difference.
Recently issues related to
the integrability and the conformal symmetry
of the AdS superstring theory have been discussed
\cite{MP,Mh,AAT}.
In this paper we will obtain the expression of the conformal generator,
which is the stress-energy tensor relating to
 the lowest spin ``local" current,
  and we calculate the higher spin  ``local"  currents
to clarify independent components.

The AdS space contains the Ramond/Ramond flux which causes
difficulty of
the standard Neveu-Schwarz-Ramond
(NSR) formulation of the superstring theory.
The AdS superstring was described in the Green-Schwarz (GS) formalism
by Metsaev and Tseytlin based on the coset
PSU(2,2$\mid$4)/[SO(4,1)$\times$SO(5)]
\cite{MT}.
Later  Roiban and Siegel reformulate it in terms of the unconstrained
GL(4$\mid$4) supermatrix coordinate based on an alternative
coset GL(4$\mid$4)/[Sp(4)$\times$GL(1)]$^2$  \cite{RS}.
In this formalism the local Lorentz is gauged,
and it turns out that this treatment is essential for
separation into  $+/-$ modes (right/left moving modes) easier.
Furthermore the fermionic constraint including the first class and second class
is necessary for
separation of the fermionic modes into $+/-$ modes.
As the first step toward the CFT formulation of the AdS superstring,
the affine Sugawara construction \cite{Halpern},
the Virasoro algebra and the algebra of currents carrying the
space-time indices are also listed.

The organization of this paper is the following;
in the next section the notation is introduced.
In section 3 we analyze the superparticle in the AdS$_5\times$S$^5$ space,
and the relation between
the reparametrization constraint and the conserved right invariant (RI) current
is given.
In section 4 we analyze the superstring in the AdS$_5\times$S$^5$ space,
and the infinite number of conserved currents are presented
both from the conformal point of view and
from the integrability point of view.
We show that the stress-energy tensor
is written by
the ``supertrace" of the square of the RI current
 as the lowest spin ``local" current.
Then we  calculate higher spin ``local" currents
to clarify independent components of the ``local" currents.

\par\vskip 6mm
\section{ GL(4${\mid}$4) covariant coset}

We review the Roiban-Siegel formulation of the  AdS$_5\times$S$^5$ coset
\cite{RS} and follow the notation in \cite{HKAdS}.
The coset GL(4$\mid$4)/[GL(1)$\times$Sp(4)]$^2$ is used instead of
PSU(2,2$\mid$4)/[SO(4,1)$\times$SO(5)] for the linear realization of the
global symmetry after Wick rotations and introducing the auxiliary variables.
A coset element  $Z_M{}^A$
 is an unconstrained matrix defined on a world-volume
 carrying indices $M=(m,\bar{m}),~A=(a,\bar{a})$ with
 $m,\bar{m},a,\bar{a}=1,\cdots,4$.
The left invariant (LI) current, $L^L$, is invariant under the left action
$Z_M{}^A~\to~\Lambda_M{}^NZ_N{}^A
$ with
 a global parameter GL(4$\mid$4)$\ni \Lambda$
\begin{eqnarray}
(J^L)_A{}^B=(Z^{-1}d Z)_A{}^B~~.
\end{eqnarray}
The LI current satisfies the flatness condition by definition
\begin{eqnarray}
dJ^L=-J^LJ^L~~~.
\end{eqnarray}
The right  invariant (RI) current, $J^R$, is invariant under the right action
$Z_M{}^A~\to~Z_M{}^B\lambda_B{}^A
$ with a local parameter
[Sp(4)$\otimes$GL(1)]$^2$ $\ni\lambda$
\begin{eqnarray}
(J^R)_M{}^N=({\cal D}ZZ^{-1})_M{}^N~~,~~({\cal D}Z)_M{}^A\equiv
dZ_M{}^A+Z_M{}^BA_B{}^A
\end{eqnarray}
with
\begin{eqnarray}
A~\to~\lambda A\lambda^{-1}+(d\lambda) \lambda^{-1}~~,
\end{eqnarray}
and
\begin{eqnarray}
dJ^R=J^RJ^R+Z(dA-AA)Z^{-1}~~~.\label{dAAA}
\end{eqnarray}
Originally $A$ is bosonic  [Sp(4)$\otimes$GL(1)]$^2$
$\ni$$A$, but we will show that
the fermionic constraint i.e. $\kappa$ symmetry
gives fermionic components of $A$.

The conjugate momenta are introduced
\begin{eqnarray}
\{
Z_M{}^A,\Pi_B{}^N
\}=(-)^A\delta_B^A\delta_M^N~~~
\end{eqnarray}
as the graded Poisson bracket and
 $\{q,p\}=-(-)^{qp}\{p,q\}$.
There are also two types of differential operators;
 the global symmetry generator (left action generator), $G_M{}^N$,
and
the  supercovariant derivatives (right action generator), $D_A{}^B$,
\begin{eqnarray}
G_M{}^N=Z_M{}^A\Pi_A{}^N~~,~~D_A{}^B=\Pi_A{}^M Z_M{}^B~~~.\label{DGDG}
\end{eqnarray}
In our coset approach $8\times 8=64$ variables for $Z_M{}^A$ are introduced
and auxiliary variables are eliminated by the following constraints
corresponding to the stability group [Sp(4)$\times $GL(1)]$^2$,
 \begin{eqnarray}
({\bf D})_{(ab)}=(\bar{\bf D})_{(\bar{a}\bar{b})}={\rm tr}~{\bf D}=
{\rm tr}~\bar{\bf D}\equiv 0~~~\label{DSp4GL1}~~~,
\end{eqnarray}
where the bosonic components are denoted by  boldfaced characters as
 ${\bf D}_{ab}\equiv D_{ab}$ and $\bar{\bf D}_{\bar{a}\bar{b}}\equiv
 D_{\bar{a}\bar{b}}$ of \bref{DGDG}.
The number of the coset constraints is $10+10+1+1=22$,
so the number of the coset parameters is $64-22=42$
where $10$ bosons and $32$ fermions.
The $[Sp(4)]^2$ invariant metric is anti-symmetric
and a matrix is decomposed into
 trace part, anti-symmetric-traceless part and the symmetric part,
denoted by
\begin{eqnarray}
{\bf M}_{ab}=-\frac{1}{4}\Omega_{ab}{\bf M}^c{}_c+{\bf M}_{\langle ab\rangle}+{\bf M}_{(ab)}\equiv - \frac{1}{4}\Omega~{\rm tr}{\bf M}
+\langle {\bf M}\rangle+({\bf M})~~~,
\end{eqnarray}
with $M_{(ab)}=\frac{1}{2}(M_{ab}+M_{ba})$,
and similar notation for the barred sector.

Both $G_M{}^N$ and $D_A{}^B$  in \bref{DGDG} satisfy GL(4$\mid$4) algebra.
If we focus on the AdS superalgebra part,
the global symmetry generators $G_M{}^N$ satisfies the global AdS superalgebra
\begin{eqnarray}
\left\{Q_{A\alpha},Q_{B,\beta}\right\}&=&-2\left[
\tau_3{}_{AB}P_{\alpha\beta} +\epsilon_{AB}
M_{\alpha\beta}\right]\label{QQPM}\\
Q_{1\alpha}&=&G_{m\bar{m}}+G_{\bar{m}m
\nn\\
Q_{2\alpha}&=&G_{m\bar{m}}-G_{\bar{m}m
\nn\\
P_{\alpha\beta}
&=&{G}_{\langle mn\rangle}\Omega_{\bar{m}\bar{n}}
-G_{\langle\bar{m}\bar{n}\rangle}\Omega_{mn}\cdots {\rm total~ momentum}\nn\\
M_{\alpha\beta}&=&-G_{(mn)}\Omega_{\bar{m}\bar{n}}
+G_{(\bar{m}\bar{n})}\Omega_{mn}\cdots{\rm total~ Lorentz}\nn~~~.
\end{eqnarray}
The right hand side of \bref{QQPM} can not be diagonalized
by the real SO(2) rotation of $Q_A$'s
because of the total Lorentz charge term with $\epsilon_{AB}$.
On the other hand the local AdS supersymmetry algebra is given by
\begin{eqnarray}
\left\{d_{A\alpha},d_{B,\beta}\right\}&=&2\left[
\tau_3{}
_{AB}
\tilde{p}_{\alpha\beta} +
\epsilon_{AB}m_{\alpha\beta}\right]\\
d_{1\alpha}&=&{D}_{a\bar{a}}+{\bar{D}}_{\bar{a}a}
\nn
\\
d_{2\alpha}&=&{D}_{a\bar{a}}-{\bar{D}}_{\bar{a}a}
\nn
\\
\tilde{p}_{\alpha\beta}&=&
{\bf D}
_{\langle ab\rangle}\Omega_{\bar{a}\bar{b}}
-{\bar{\bf D}}
_{\langle \bar{a}\bar{b}\rangle}\Omega_{ab}
\cdots {\rm local~LI~momentum}\nn
\\m_{\alpha\beta}&=&
-{\bf D}_{(ab)}\Omega_{\bar{a}\bar{b}}
+\bar{\bf D}
_{(\bar{a}\bar{b})}\Omega_{a{b}}\cdots{\rm local~Lorentz}\nn
~~~.
\end{eqnarray}
In our coset approach the local Lorentz generator
is a constraint \bref{DSp4GL1},
so  the
local supercovariant derivative $d_{A\alpha}$'s
can be separated into;
\begin{eqnarray}
\left\{d_{1\alpha},d_{2\beta}\right\}=
2 m_{\alpha\beta}\equiv 0~,~~
 \left\{d_{1\alpha},d_{1\beta}\right\}=
2 \tilde{p}_{\alpha\beta}~~,~~
\left\{d_{2\alpha},d_{2\beta}\right\}=
-2\tilde{p}_{\alpha\beta}
  \end{eqnarray}
Although the global superalgebra can not be separated into
irreducible algebras in the AdS background,
the local superalgebra can be separated into
irreducible sets on the GL(4${\mid}$4) covariant coset approach.
This property allows simpler description of the AdS superstring
as the flat case at least in the classical mechanics level.

\par\vskip 6mm
\section{ AdS Superparticle}

We begin with the action for a superparticle in the AdS$_5\times$S$^5$
\begin{eqnarray}
&S=\displaystyle\int  d\tau~\displaystyle\frac{1}{2e}
\left\{-{\bf J}_\tau^{\langle ab\rangle}{\bf J}_{\tau, \langle ab\rangle}
+\bar{\bf J}_\tau^{\langle \bar{a}\bar{b}\rangle}\bar{\bf J}_{\tau, \langle \bar{a}\bar{b}\rangle}
\right\}&~~~.\label{RS}
\end{eqnarray}
Here we omit the upper-subscript $L$ for the LI currents and their components are denoted as
\begin{eqnarray}
(J^L_{~\mu})_A{}^B
=\left(
\begin{array}{cc}
{\bf J}_{\mu,}{}_a{}^b&j_{\mu,}{}_a{}^{\bar{b}}\\\bar{j}_{\mu,}{}_{\bar{a}}{}^b&\bar{\bf J}_{\mu,}{}_{\bar{a}}{}^{\bar{b}}
\end{array}
\right)~~~.
\end{eqnarray}
From the definition of the canonical conjugates,
$
\Pi_A{}^M={\delta S}/{\delta \partial_\tau Z_M{}^A} (-)^A
$,
we have
the following primary constraints \cite{HKAdS}
\begin{eqnarray}
{\cal A}_{\rm P}=\frac{1}{2}{\rm tr}\left[
\langle{\bf D}\rangle^2-
\langle\bar{\bf D}\rangle^2\right]=0~~,~~
D_{a\bar{b}}=\bar{D}_{\bar{a}{b}}=0~~~\label{ApDD}
\end{eqnarray}
with
\begin{eqnarray}
D_A{}^B=\left(
\begin{array}{cc}{\bf D}_a{}^b&D_a{}^{\bar{b}}\\
\bar{D}_{\bar{a}}{}^b&\bar{\bf D}_{\bar{a}}{}^{\bar{b}}
\end{array}
\right)~~~.
\end{eqnarray}

The Hamiltonian is chosen as
\begin{eqnarray}
{\cal H}=-{\cal A}_{\rm P}=- \frac{1}{2}{\rm tr}\left[
\langle{\bf D}\rangle^2-
\langle\bar{\bf D}\rangle^2\right]
\end{eqnarray}
and the $\tau$-derivative is determined by
the Poisson bracket with ${\cal H}$,
 $\partial_\tau{\cal O}=\{{\cal O},{\cal H}\}$.
The fact that a half of the fermionic constraints  is second class
requires the Dirac bracket in general.
Fortunately the Dirac bracket with the Hamiltonian is equal to
its Poisson bracket because the fermionic
constrains are ${\cal H}$ invariant.

The LI current is calculated as
\begin{eqnarray}
J^L_\tau=Z^{-1}\partial_\tau Z=\left(
\begin{array}{cc}
\langle{\bf D}\rangle&0\\0&\langle \bar{\bf D}\rangle
\end{array}
\right)~~,~~\partial_\tau J^L=0~~~.
\end{eqnarray}
The RI current, generating the global GL(4$\mid$4) symmetry,
is given as
\begin{eqnarray}
J^R_\tau\equiv Z\Pi=Z\left(J^L_\tau+A_\tau\right)Z^{-1}~~,~~
A_\tau=
\left(
\begin{array}{cc}
({\bf D})-\frac{1}{4}\Omega {\rm tr}{\bf D}&D
\\\bar{D}&(\bar{\bf D})-\frac{1}{4}\Omega {\rm tr}\bar{\bf D}
\end{array}
\right)~~~.\label{JRSP}
\end{eqnarray}
Although the stability group does not contain fermionic components originally,
the fermionic components of the gauge connection $A$ in \bref{JRSP} is induced.
It is noted that ``$A$" is the gauge connection
distinguishing from
 the reparametrization
constraint ``${\cal A}$".
The RI current is conserved, since the Hamiltonian is written
by LI currents which are manifestly global symmetry invariant
\begin{eqnarray}
\partial_\tau J^R=0~~~.
\end{eqnarray}

The $\kappa$ symmetry generators are half of the fermionic constraints
by projecting out with the null vector as
\begin{eqnarray}
{\cal B}_{\rm P}{}_a{}^{\bar{b}}=\langle{\bf D}\rangle_a{}^b D_b{}^{\bar{b}}
+D_a{}^{\bar{a}}\langle\bar{\bf D}\rangle_{\bar{a}}{}^{\bar{b}}~~,~~
\bar{\cal B}_{\rm P}{}_{\bar{a}}{}^{{b}}=\langle\bar{\bf D}\rangle_{\bar{a}}{}^{\bar{b}}\bar{D}_{\bar{b}}{}^{{b}}
+\bar{D}_{\bar{a}}{}^{{a}}\langle{\bf D}\rangle_{a}{}^{b}~~~.
\end{eqnarray}
If we construct the closed algebra including these $\kappa$ generators
with keeping
 the bilinear of the fermionic constraints,
the $\tau$-reparametrization constraint, ${\cal A}_{\rm P}$, is modified to \cite{HKAdS}
\begin{eqnarray}
\tilde{\cal A}_{\rm P}&=&\frac{1}{2}{\rm tr}\left[
\langle{\bf D}\rangle^2-
\langle\bar{\bf D}\rangle^2
+2D\bar{D}
\right]~~~~~.
\end{eqnarray}
This expression appears in the Poisson bracket of
${\cal B}$ with $\bar{\cal B}$,
when we keep the bilinear of fermionic constrains.
The  RR flux is responsible for the last term ``$D\bar{D}$".
The term which is bilinear of the constraints
does not change the Poisson bracket
since its bracket with an arbitrary variable
gives terms proportional to the constraints
which are zero on the constrained surface.
In another word ${\cal A}_{\rm P}$ has an ambiguity of bilinear of the constraints,
and the $\kappa$ invariance fixes it.
On the original coset constrained surface \bref{DSp4GL1}
it is also rewritten as
\begin{eqnarray}
\tilde{\cal A}_{\rm P}=\frac{1}{2}~{\rm Str}~[D_A{}^B]^2=\frac{1}{2}~{\rm Str}~[J^R_\tau]^2~~~.
\end{eqnarray}
This is zero-mode contribution of the classical Virasoro constraint
for a superstring
in the AdS$_5\times$S$^5$ background.

\par\vskip 6mm
\section{ AdS Superstring}
\subsection{Conserved currents}

We take the action for a superstring in the AdS$_5\times$S$^5$ given by
\begin{eqnarray}
&S=\displaystyle\int  d^2\sigma~\frac{1}{2}\left\{
-\sqrt{-g}g^{\mu\nu}({\bf J}_\mu^{\langle ab\rangle}{\bf J}_{\nu, \langle ab\rangle}
-\bar{\bf J}_\mu^{\langle \bar{a}\bar{b}\rangle}\bar{\bf J}_{\nu, \langle \bar{a}\bar{b}\rangle}
)
+\frac{k}{2}\epsilon^{\mu\nu}(
E^{1/2}j_\mu^{a\bar{b}}j_{\nu, a\bar{b}}
-
E^{-1/2}\bar{j}_\mu^{\bar{a}{b}}\bar{j}_{\nu, \bar{a}{b}}
)\right\}&\nn\\\label{RS}
\end{eqnarray}
where ``$k$" represents the WZ term contribution with $k=1$ and
 $E={\rm sdet} Z_M{}^A$.
The consistent $\tau$ and $\sigma$ reparametrization generators \cite{HKAdS} are
\begin{eqnarray}
{\cal A}_\perp&=&{\cal A}_{0\perp} +k~{\rm tr}\left[-E^{1/4}Fj_\sigma+E^{-1/4}\bar{F}\bar{j}_\sigma\right]\nn\\
{\cal A}_\parallel&=&{\cal A}_{0\parallel} +k~{\rm tr}\left[E^{-1/4}F\bar{j}_\sigma-E^{1/4}\bar{F}{j}_\sigma\right]\label{Aperp}
\end{eqnarray}
with the following primary constraints
\begin{eqnarray}
{\cal A}_{0\perp}&=&\frac{1}{2}{\rm tr}\left[
(\langle{\bf D}\rangle^2+\langle{\bf J}_\sigma\rangle^2)-
(\langle\bar{\bf D}\rangle^2+\langle\bar{\bf J}_\sigma\rangle^2)
\right]=0\nn\\
{\cal A}_{0\parallel}&=&{\rm tr}\left[
\langle{\bf D}\rangle\langle{\bf J}_\sigma\rangle-
\langle\bar{\bf D}\rangle\langle\bar{\bf J}_\sigma\rangle
\right]=0\\
F_{a\bar{b}}&=&E^{1/4}D_{a\bar{b}}
+\frac{k}{2}E^{-1/4}(\bar{j}_\sigma)_{\bar{b}a}=0\nn\\
\bar{F}_{\bar{a}{b}}&=&E^{-1/4}\bar{D}_{\bar{a}{b}}+\frac{k}{2}E^{1/4}({j}_\sigma)_{{b}\bar{a}}=0~~~.\label{FermionicF}
\end{eqnarray}
Their Poisson brackets are
\begin{eqnarray}
\left\{{\cal A}_\perp(\sigma),{\cal A}_\perp(\sigma')\right\}&=&
2{\cal A}_\parallel(\sigma)\partial_\sigma\delta(\sigma-\sigma')+
\partial_\sigma{\cal A}_\parallel(\sigma)\delta(\sigma-\sigma')\nn\\
\left\{{\cal A}_\perp(\sigma),{\cal A}_\parallel(\sigma')\right\}&=&
2{\cal A}_\perp(\sigma)\partial_\sigma\delta(\sigma-\sigma')+
\partial_\sigma{\cal A}_\perp(\sigma)\delta(\sigma-\sigma')\\
\left\{{\cal A}_\parallel(\sigma),{\cal A}_\parallel(\sigma')\right\}&=&
2{\cal A}_\parallel(\sigma)\partial_\sigma\delta(\sigma-\sigma')+
\partial_\sigma{\cal A}_\parallel(\sigma)\delta(\sigma-\sigma')\nn~~~.
\end{eqnarray}

The Hamiltonian is chosen as
\begin{eqnarray}
{\cal H}&=&-\int d\sigma {\cal A}_\perp\label{HamiltonianSUST}\\&=&
-\int d\sigma {\rm tr}\left[
\frac{1}{2}\left\{
\langle{\bf D}\rangle^2+\langle{\bf J}_\sigma\rangle^2-
\langle\bar{\bf D}\rangle^2-\langle\bar{\bf J}_\sigma\rangle^2\right\}
+\left(kE^{-1/2}\bar{D}\bar{j}_\sigma-k E^{1/2}Dj_\sigma
+j_\sigma\bar{j}_\sigma\right)
\right]
~~~.\nn
\end{eqnarray}
From now on $E=1$ gauge is taken using the local GL(1) invariance.
The global GL(1) symmetry is broken by the WZ term.

Using the Hamiltonian in  \bref{HamiltonianSUST}
the $\tau$-derivative of ${\cal A}_{\perp}$ and ${\cal A}_\parallel$ are given as
\begin{eqnarray}
\partial_\tau {\cal A}_\perp=\partial_\sigma {\cal A}_\parallel~~,~~
\partial_\tau {\cal A}_\parallel=\partial_\sigma {\cal A}_\perp~~~.\label{dAdA}
\end{eqnarray}
Although the coset parameter $Z_M{}^A$ does not satisfy
the world-sheet free wave equation, it is essential to introduce
the world-sheet lightcone
coordinate
\begin{eqnarray}
\sigma^\pm=\tau \pm \sigma~~,~~
\partial_\pm=\frac{1}{2}(\partial_\tau \pm \partial_\sigma)~~~.
\end{eqnarray}
The differential equations \bref{dAdA} are rewritten as
\begin{eqnarray}
\partial_- {\cal A}_+=0~~,~~
\partial_+ {\cal A}_-=0~~,~~
{\cal A}_\pm={\cal A}_\perp\pm {\cal A}_\parallel~~~,
\end{eqnarray}
so the infinite number of the conserved
currents are
\begin{eqnarray}
\partial_- \left[f(\sigma^+){\cal A}_+\right]=0~~,~~
\partial_+ \left[f(\sigma^-){\cal A}_-\right]=0~~\label{conformal}
\end{eqnarray}
with an arbitrary function $f$.
Then there exist infinite number of conserved  charges
\begin{eqnarray}
\partial_- \left[\displaystyle\int d\sigma~
f(\sigma^+){\cal A}_+\right]=0~~,~~
\partial_+ \left[\displaystyle\int d\sigma~
f(\sigma^-){\cal A}_-\right]=0~~~.
\end{eqnarray}

On the other hand the integrability of the superstring
will provide the infinite number of ``local" charges as well as the
``non-local" charges written down in \cite{HY}.
The LI currents
is given by
\begin{eqnarray}
\left\{\begin{array}{ccl}
J^L_\tau&=&\left(
\begin{array}{cc}
\langle{\bf D}\rangle&-k\bar{j}_\sigma\\-kj_\sigma&\langle \bar{\bf D}\rangle
\end{array}
\right)
=\left(
\begin{array}{cc}
\langle{\bf D}\rangle&
2D-2F\\2\bar{D}-2\bar{F}&\langle \bar{\bf D}\rangle
\end{array}
\right)
\approx
\left(
\begin{array}{cc}
\langle{\bf D}\rangle&
2D\\2\bar{D}&\langle \bar{\bf D}\rangle
\end{array}
\right)\nn\\
J^L_\sigma&=&
\left(
\begin{array}{cc}
{\bf J}_{\sigma}
&j_{\sigma}\\
\bar{j}_{\sigma}&
\bar{\bf J}_{\sigma}
\end{array}
\right)
\end{array}\right.~~~
\end{eqnarray}
where the $\tau$ component is determined by \bref{HamiltonianSUST}.
The LI currents satisfy the flatness condition but
does not satisfy the conservation law.
The RI currents are obtained in \cite{HY} as
\begin{eqnarray}
\left\{\begin{array}{ccl}
J^R_\tau&=&ZDZ^{-1}=Z(J^L_\tau+A_\tau)Z^{-1}\label{RISUST}\\
J^R_\sigma&=&Z(J^L_\sigma+A_\sigma)Z^{-1}~~,~~
J^L_\sigma+A_\sigma=
\left(
\begin{array}{cc}
\langle{\bf J}_{\sigma}\rangle
&\bar{F}+\frac{1}{2}j_{\sigma}\\
F+\frac{1}{2}\bar{j}_{\sigma}&
\langle\bar{\bf J}_{\sigma}\rangle
\end{array}
\right)
\end{array}\right.
\end{eqnarray}
where the gauge connection $A_\mu$ is
\begin{eqnarray}
\left\{\begin{array}{ccl}
A_\tau&=&\left(
\begin{array}{cc}
({\bf D})-\frac{1}{4}\Omega {\rm tr}{\bf D}&-D\\-\bar{D}&(\bar{\bf D})-\frac{1}{4}\Omega {\rm tr}\bar{\bf D}
\end{array}
\right)\nn\\
A_\sigma&=&
\left(
\begin{array}{cc}
-({\bf J}_\sigma)+\frac{1}{4}\Omega {\rm tr}{\bf J}_\sigma&
\bar{F}-\frac{1}{2}j_\sigma
\\F-\frac{1}{2}\bar{j}_\sigma
&-(\bar{\bf J}_\sigma)+\frac{1}{4}\Omega {\rm tr}\bar{\bf J}_\sigma
\end{array}
\right)
\end{array}\right.~~~.
\end{eqnarray}
The fermionic components of $A_\mu$ appear again.
In this paper
the fermionic constraints, $F$ and $\bar{F}$,
in
the fermionic components of $A_\sigma$  are kept
while they were absent in our previous paper \cite{HY}
depending on the treatment of the constraint bilinear terms.
Then the integrability of the superstring
leads to the current conservation and the flatness condition
for the RI current;
\begin{eqnarray}
\partial_\tau J^R_\tau=\partial_\sigma J^R_\sigma~~,~~
\partial_\tau J^R_\sigma-\partial_\sigma J^R_\tau=
2\left[J^R_\tau, J^R_\sigma\right]~~~\label{dJRdJR}~~~.
\end{eqnarray}
They are rewritten as
\begin{eqnarray}
\partial_-J^R_+=\left[J^R_-,J^R_+\right]~~,~~
\partial_+J^R_-=\left[J^R_+,J^R_-\right]~~,~~
J^R_\pm=J^R_\tau\pm J^R_\sigma~~~.\label{JRpm}
\end{eqnarray}
Taking the supertrace, denoting ``Str", leads to
the infinite number of conserved ``local"
currents because $J^R_\mu$ are supermatrices,
\begin{eqnarray}
\partial_-~{\rm Str}\left[(J^R_+)^n\right]=0~~,~~
\partial_+~{\rm Str}\left[(J^R_-)^n\right]=0~~,n=1,2,\cdots~~~.\label{JRn}
\label{integrablity}
\end{eqnarray}
It gives the infinite number of conserved ``local" charges
\begin{eqnarray}
\partial_\tau \left[\displaystyle\int d\sigma~
 f(\sigma^+){\rm Str}(J^R_+)^n
\right]=0~~,~~
\partial_\tau \left[\displaystyle\int d\sigma~
 f(\sigma^-){\rm Str}(J^R_-)^n
\right]=0~~~.\end{eqnarray}

In this way classical 2-dimensional conformal symmetry and
integrability of AdS superstring lead to two infinite sets of
 conserved currents,
\bref{conformal} and \bref{integrablity}.
In next sections the relation between them is examined.

\subsection{Stress-energy tensor ($n=2$)}

The ``$+/-$" (right/left moving) modes of the RI currents
on the original coset constrained space
\bref{DSp4GL1}
are written as
\begin{eqnarray}
J^R_\pm=
Z\left(
\begin{array}{cc}
\langle{\bf D}_\pm \rangle&D\pm (\bar{F}+\frac{1}{2}j_\sigma)
\\
\bar{D}\pm(F+\frac{1}{2}\bar{j}_\sigma)&\langle\bar{\bf D}_\pm\rangle
\end{array}
\right)Z^{-1}=
Z\left(
\begin{array}{cc}
\langle{\bf D}_\pm\rangle&d_\pm+\frac{1}{2}j_\pm
\\
\pm(d_\pm -\frac{1}{2}j_\pm)
&\langle\bar{\bf D}_\pm\rangle
\end{array}
\right)Z^{-1}\nn\\
\end{eqnarray}
  with
\begin{eqnarray}
{\bf D}_\pm={\bf D}\pm{\bf J}_\sigma~~,~~
\bar{\bf D}_\pm=\bar{\bf D}\pm\bar{\bf J}_\sigma~~,~~d_\pm=F\pm\bar{F}~~,~~
j_\pm=j_\tau\pm j_\sigma=-\bar{j}_\sigma\pm j_\sigma
\label{pmpm}
\end{eqnarray}
carrying the LI currents indices, $AB$.
This is supertraceless, Str$J^R_\pm=0$, so $n=1$ case of
\bref{JRn} gives just trivial equation.

Let us look at the $n=2$ case of \bref{JRn},
 ${\rm Str}\left[(J^R_\pm)^2\right]$.
Then the ``+" sector  is written as
\begin{eqnarray}
\frac{1}{2}{\rm Str}\left[(J^R_+)^2\right]&=&
\frac{1}{2}{\rm Str}\left[
\left(
\begin{array}{cc}
\langle{\bf D}_+\rangle&d_++\frac{1}{2}j_+
\\
d_+-\frac{1}{2}j_+
&\langle\bar{\bf D}_+\rangle
\end{array}
\right)^2
\right]\nn\\
&=&\frac{1}{2}{\rm tr}\left[
\langle{\bf D}_+\rangle^2-\langle\bar{\bf D}_+\rangle^2
+2(d_++\frac{1}{2}j_+
)(d_+-\frac{1}{2}j_+
)
\right]\nn\\
&=&{\rm tr}\left[\frac{1}{2}\left(
\langle{\bf D}_+\rangle^2-\langle\bar{\bf D}_+\rangle^2\right)
+j_+d_+\right]~~~.\label{419}
\end{eqnarray}
The ``$-$" sector is
\begin{eqnarray}
\frac{1}{2}{\rm Str}\left[(J^R_-)^2\right]&=&
{\rm tr}\left[\frac{1}{2}\left(
\langle{\bf D}_-\rangle^2-\langle\bar{\bf D}_-\rangle^2\right)
-j_-d_-
\right]~~~.\label{420}
\end{eqnarray}

On the other hand the conformal symmetry generator ${\cal A}_\pm$ is
rewritten from the relation \bref{Aperp} and \bref{pmpm} as
\begin{eqnarray}
{\cal A}_\pm&=&
{\rm tr}
\left[\frac{1}{2}\left(
\langle{\bf D}_\pm \rangle^2-\langle\bar{\bf D}_\pm \rangle^2\right)
\pm j_\pm d_\pm
\right]~=~
\frac{1}{2}{\rm Str}\left[(J^R_\pm)^2\right]
~~~.\label{421}
\end{eqnarray}
If we take care of the square of the fermionic constraints,
the closure of the first class constraint set
including the $\kappa$ symmetry
generators,
\begin{eqnarray}
{\cal B}_\pm&=&
\langle{\bf D}_\pm\rangle d_\pm
+d_\pm\langle\bar{\bf D}_\pm\rangle\label{kappaSUST}
~~
\end{eqnarray}
determines the ambiguity of  bilinear of the constraints as
\begin{eqnarray}
\tilde{\cal A}_\pm=
{\rm tr}\left[\frac{1}{2}\left(
\langle{\bf D}_\pm \rangle^2-\langle\bar{\bf D}_\pm \rangle^2\right)
\pm(\frac{1}{2}d_\mp +j_\pm)d_\pm
\right]=
{\cal A}_\pm+{\rm tr}F\bar{F}~~
\end{eqnarray}
obtained in \cite{HKAdS} as a generator of the ${\cal ABCD}$
constraint set
known to exist for a superstring in a flat space
 \cite{WSmech,ABCD}.
Then the stress-energy tensor is
\begin{eqnarray}
T_{\pm\pm}\equiv\tilde{\cal A}_\pm
\approx
{\cal A}_\pm
={\rm Str}J^R_\pm J^R_\pm
~~~.\label{423}
\end{eqnarray}
This is $\kappa$ symmetric stress-energy tensor
in a supersymmetric generalization of
 Sugawara form.

\subsection{Supercovariant derivative algebra}

Existence of the conformal invariance should present the
irreducible coset components of supercovariant derivatives
\cite{HKAdS};
\begin{eqnarray}
\langle{\bf D}_\pm\rangle&=&\langle{\bf D}\rangle\pm\langle{\bf J}_\sigma\rangle
~~,~~
\langle\bar{\bf D}_\pm\rangle~=~\langle\bar{\bf D}\rangle\pm\langle\bar{\bf J}_\sigma\rangle\nn\\
d_\pm&=&F\pm\bar{F}=(D\pm\frac{1}{2}j_\sigma)\pm(\bar{D}\pm\frac{1}{2}\bar{j}_\sigma)
\nn~~~.
\end{eqnarray}

On the  constraint surface \bref{DSp4GL1} and \bref{FermionicF}
the $+/-$ sector supercovariant derivatives are separated as
\begin{eqnarray}
\left\{\langle{\bf D}_+\rangle_{ab}(\sigma),
\langle{\bf D}_-\rangle_{cd}(\sigma')\right\}
&=&2\Omega_{\langle c|\langle b}({\bf D})_{a\rangle|d\rangle}
\delta(\sigma-\sigma')
\equiv 0\nn\\
\left\{\langle{\bf D}_+\rangle_{ab}(\sigma),
d_{-,c\bar{d}}(\sigma')\right\}
&=&\Omega_{c\langle b}d_{+,a\rangle \bar{d}}
\delta(\sigma-\sigma')=
\Omega_{c\langle b}(F+\bar{F})_{a\rangle \bar{d}}
\delta(\sigma-\sigma')
\approx 0\nn\\
\left\{d_{+,a\bar{b}}(\sigma),
d_{-,c\bar{d}}(\sigma')\right\}
&=&2\left[
\Omega_{ac}(\bar{\bf D})_{\bar{b}\bar{d}}
+\Omega_{\bar{b}\bar{d}}({\bf D})_{ac}
\right]
\delta(\sigma-\sigma')
\equiv 0\nn
\end{eqnarray}
with
analogous relation for  the barred sector, $\langle\bar{\bf D}_\pm\rangle$.

The ``+" sector supercovariant derivative algebra is
\begin{eqnarray}
\left\{\langle{\bf D}_+\rangle_{ab}(\sigma),
\langle{\bf D}_+\rangle_{cd}(\sigma')\right\}
&=&2\Omega_{\langle c|\langle b}\Omega_{a\rangle|d\rangle}
\delta'(\sigma-\sigma')+4\Omega_{\langle c|\langle b}
({\bf J}_\sigma)_{a\rangle|d\rangle}
\delta(\sigma-\sigma')\nn\\
&\equiv& 2\Omega_{\langle c|\langle b} \nabla_{a\rangle|d\rangle}
\delta(\sigma-\sigma')
\nn\\
\left\{d_{+,a\bar{b}}(\sigma),
d_{+,c\bar{d}}(\sigma')\right\}
&=&2\left[
\Omega_{\bar{b}\bar{d}}\langle{\bf D}_+\rangle_{ac}
-\Omega_{ac}\langle\bar{\bf D}_+\rangle_{\bar{b}\bar{d}}
\right]
\delta(\sigma-\sigma')
\nn\\
\left\{\langle{\bf D}_+\rangle_{ab}(\sigma),
d_{+,c\bar{d}}(\sigma')\right\}
&=&
\Omega_{c\langle b}(d_-+2j_+)_{a\rangle \bar{d}}
\delta(\sigma-\sigma')\approx
2\Omega_{c\langle b}\omega_{+,a\rangle \bar{d}}
\delta(\sigma-\sigma')
\nn\\
\left\{d_{+,a\bar{b}}(\sigma),
\omega_{+,c\bar{d}}(\sigma')\right\}
&=&-2\Omega_{\bar{b}\bar{d}}\Omega_{ac}\delta'(\sigma-\sigma')
2\left[
-\Omega_{\bar{b}\bar{d}}({\bf J}_\sigma)_{ac}
-\Omega_{ac}(\bar{\bf J}_\sigma)_{\bar{b}\bar{d}}\right]\delta(\sigma-\sigma')\nn\\
&\equiv& -2\nabla_{\bar{b}\bar{d};ac}\delta(\sigma-\sigma')\nn\\
\label{sucovder}\\
\left\{\langle{\bf D}_+\rangle_{ab}(\sigma),
\omega_{+,c\bar{d}}(\sigma')\right\}
&=&
\Omega_{c\langle b}\omega_{-,a\rangle \bar{d}}
\delta(\sigma-\sigma')
\nn\\
\left\{\omega_{+,a\bar{b}}(\sigma),
\omega_{+,c\bar{d}}(\sigma')\right\}
&=&0~~~\nn
\end{eqnarray}
where
\begin{eqnarray}
\omega_\pm&=&j_\pm=-\bar{j}_\sigma\pm j_\sigma~~~.
\end{eqnarray}
This is comparable with the flat case where
the non-local term, $\partial_\sigma\delta(\sigma-\sigma')$, is replaced by the
local Lorentz (~[Sp(4)]$^2$~) covariant non-local term,
$\nabla_\sigma\delta(\sigma-\sigma')$.
For the fifth Poisson bracket,
 $\left\{\langle{\bf D}_+\rangle,
\omega\right\}
$,
it is zero for the flat case but it is not for the AdS case.
For a superstring in a flat space
the consistency of the $\kappa$ symmetry constraint
requires
the first class constraint set,
namely ``${\cal ABCD}$" constraint,
which are bilinear of the supercovariant derivatives
 \cite{WSmech,ABCD}.
 For the AdS case the situation is completely the same,
despite of this anomalous term \cite{HKAdS}.

\par\vskip 6mm

\subsection{``Local" currents ($n\geq 3$)}

Next let us look at $n\geq 3 $ cases
of
the infinite number of conserved ``local" current \bref{JRn}.
For simplicity we focus on the ``+" sector and replace
$``+"$ by $``~\hat{~}~"$, as
$J_+ \to \hat{J}$.
The first  three powers of the RI current,
$(J^R)^n$ with $n=1,2,3$, are listed as below:
\begin{eqnarray}
\left[Z^{-1}\hat{J}^R Z\right]_{AB}&=&\left(
\begin{array}{cc}
\langle\hat{\bf D}\rangle_{\langle ab\rangle}&
(\hat{d}+\frac{1}{2}\hat{j})_{a\bar{b}}
\\
\pm(\hat{d} -\frac{1}{2}\hat{j})_{b\bar{a}}
&\langle\hat{\bar{\bf D}}\rangle_{\bar{a}\bar{b}}
\end{array}
\right)
\end{eqnarray}
\begin{eqnarray}
\left[Z^{-1}(\hat{J}^R)^2 Z\right]_{AB}&=&-\frac{1}{4}
\left(
\begin{array}{cc}
\Omega_{ab}~{\rm tr}(
\langle\hat{\bf D}\rangle^2
+\hat{j}\hat{d})
&\\&
\Omega_{\bar{a}\bar{b}}~{\rm tr}(
\langle\hat{\bar{\bf D}}\rangle^2
-\hat{j}\hat{d})
\end{array}
\right)\\
&&+
\left(
\begin{array}{cc}
(\hat{d}^2-\frac{1}{4}\hat{j}^2)_{(ab)}
+\langle \hat{j}\hat{d}\rangle_{\langle ab \rangle}
&\hat{\cal B}_{a\bar{b}}
+\frac{1}{2}(\langle \hat{\bf D}\rangle \hat{j}
+\hat{j}\langle \hat{\bar{\bf D}}\rangle
)_{a\bar{b}}
\\
\hat{\cal B}_{b\bar{a}}
-\frac{1}{2}(\langle \hat{\bf D}\rangle \hat{j}
+\hat{j}\langle \hat{\bar{\bf D}}\rangle
)_{b\bar{a}}
&(\hat{d}^2-\frac{1}{4}\hat{j}^2)_{(\bar{a}\bar{b})}
-\langle\hat{j}\hat{d}\rangle_{\langle \bar{a}\bar{b} \rangle}
\end{array}
\right)\nn
\end{eqnarray}
\begin{eqnarray}
&&\left[Z^{-1}(\hat{J}^R)^3 Z\right]_{AB}~=~
\frac{1}{4}
\left(
\begin{array}{cc}
\Omega_{ab}~{\rm tr}\left[\hat{\cal B}\hat{j}-
(\langle\hat{\bf D}\rangle \hat{d} )\hat{j}
\right]
&\\&
-\Omega_{\bar{a}\bar{b}}~{\rm tr}\left[\hat{\cal B}\hat{j}
-(\hat{d}\langle\hat{\bar{\bf D}}\rangle)\hat{j}
\right]
\end{array}
\right)\\
&&-
\left(
\begin{array}{cc}
\left[\frac{1}{4}{\rm tr}(\langle\hat{\bf D}\rangle^2+\hat{j}\hat{d})~
\langle\hat{\bf D}\rangle
-\langle\hat{\bf D}\rangle (\hat{j} \hat{d})
+\hat{\cal B}\hat{j}
\right]_{\langle ab\rangle}
&
\frac{1}{4}{\rm tr}(\langle\hat{\bf D}\rangle^2
-\langle\hat{\bar{\bf D}}\rangle^2)~(\hat{d}+\frac{1}{2}\hat{j})_{a\bar{b}}
\\
\frac{1}{4}{\rm tr}(\langle\hat{\bf D}\rangle^2
-\langle\hat{\bar{\bf D}}\rangle^2)~(\hat{d}-\frac{1}{2}\hat{j})_{b\bar{a}}
&
\left[\frac{1}{4}{\rm tr}(\langle\hat{\bar{\bf D}}\rangle^2
+\hat{j}\hat{d})~
\langle\hat{\bar{\bf D}}\rangle
-(\hat{j} \hat{d})
\langle\hat{\bar{\bf D}}\rangle -\hat{\cal B}\hat{j}
\right]_{\langle \bar{a}\bar{b}\rangle}
\end{array}
\right)\nn\\
&&+
\left(
\begin{array}{cc}
\left[
2(\hat{d}^2-\frac{1}{4}\hat{j}^2)\langle\hat{\bf D}\rangle
+\hat{d}\langle\hat{\bar{\bf D}}\rangle\hat{d}
-\frac{1}{4}\hat{j}\langle\hat{\bar{\bf D}}\rangle \hat{j}
\right]_{(ab)}&
-\frac{1}{4}{\rm tr}(\hat{j}\hat{d})
(\hat{d}+\frac{1}{2}\hat{j})_{a\bar{b}}
+
\left[\langle\hat{{\bf D}}\rangle
(\hat{d}+\frac{1}{2}\hat{j})\langle\hat{\bar{\bf D}}\rangle
\right]_{a\bar{b}}
\\
\frac{1}{4}{\rm tr}(\hat{j}\hat{d})
(\hat{d}-\frac{1}{2}\hat{j})_{b\bar{a}}
+
\left[\langle
\hat{{\bf D}}\rangle
(\hat{d}-\frac{1}{2}\hat{j})\langle\hat{\bar{\bf D}}\rangle
\right]_{b\bar{a}}
&
\left[
2(\hat{d}^2-\frac{1}{4}\hat{j}^2)\langle\hat{\bar{\bf D}}\rangle
+\hat{d}\langle\hat{{\bf D}}\rangle\hat{d}
-\frac{1}{4}\hat{j}\langle\hat{{\bf D}}\rangle \hat{j}
\right]_{(\bar{a}\bar{b})}
\end{array}
\right)\nn\\
&&+
\left(
\begin{array}{cc}
&
\left[\left\{
(\hat{d}^2-\frac{1}{4}\hat{j}^2)
+\langle\hat{j}\hat{d}
\rangle
\right\}(\hat{d}+\frac{1}{2}\hat{j})
\right]_{a\bar{b}}
\\
\left[\left\{
(\hat{d}^2-\frac{1}{4}\hat{j}^2)
-\langle\hat{j}\hat{d}
\rangle
\right\}(\hat{d}-\frac{1}{2}\hat{j})
\right]_{\bar{a}b}
&
\end{array}
\right)\nn
\end{eqnarray}
In this computation
 5-dimensional $\gamma$-matrix relations are used,
for example
${\bf V}^{\langle ab\rangle}{\bf U}_{\langle bc\rangle}
+{\bf U}^{\langle ab\rangle}{\bf V}_{\langle bc\rangle}
=\frac{1}{2}\delta^a_c~{\rm tr}{\bf V}{\bf U}$ for bosonic
vectors ${\bf V},~{\bf U}$.

The conserved ``local" current with $n=3$ becomes
\begin{eqnarray}
{\rm Str}
(\hat{J}^R)^3 &=&
{\rm tr}\left[2\hat{\cal B}\hat{j}
-(\langle\hat{\bf D}\rangle \hat{d} )\hat{j}
-(\hat{d}\langle\hat{\bar{\bf D}}\rangle)\hat{j}
\right]
=~{\rm tr}
~(\hat{\cal B}\hat{j}) \label{430}
\end{eqnarray}
where $\hat{\cal B}$ is the $\kappa$ generating constraint
 \bref{kappaSUST}.
 The conserved ``local" current with $n=4$ becomes
\begin{eqnarray}
{\rm Str}
(\hat{J}^R)^4 &=&
-\frac{1}{2}
{\rm tr}\left(
\langle\hat{\bf D}\rangle^2+
\langle\hat{\bar{\bf D}}\rangle^2
\right)
\hat{\cal A}+\left(~\cdots~\right){\rm tr}
~(\hat{\cal B}\hat{j})~~.
\end{eqnarray}
The conserved ``local" current with $n=5,6$ are given as;
 Str$(\hat{J}^R)^5=$(
$\hat{\cal B}$ dependent terms),
 Str$(\hat{J}^R)^6=$(
$\hat{\cal A}$ and $\hat{\cal B}$ dependent terms).
In general for even $n=2m$ its bosonic part is given as
\begin{eqnarray}
{\rm Str}
(\hat{J}^R)^{2m} \mid_{\rm bosonic}&=&
\left({\rm tr}\langle\hat{\bf D}\rangle^2\right)^m-
\left({\rm tr}\langle\hat{\bar{\bf D}}\rangle^2\right)^m\nn\\
&=&{\rm tr}\left(\langle\hat{\bf D}\rangle^2-\langle\hat{\bar{\bf D}}\rangle^2
\right)~\left\{\left(
{\rm tr}\langle\hat{\bf D}\rangle^2\right)^{m-1}
+\cdots +
\left({\rm tr}\langle\hat{\bar{\bf D}}\rangle^2\right)^{m-1}
\right\}\nn\\
&\Rightarrow& (\cdots)\hat{\cal A}+\left(~\cdots~\right){\rm tr}
~(\hat{\cal B}\hat{j})
\end{eqnarray}
where the last equality is guaranteed by the $\kappa$ invariance.
It is also pointed out that
the conserved supertraces of multilinears in the currents factorize in traces of lower number of currents and that for an even number of currents one of the factors
is the stress tensor in \cite{BPR}.
For odd $n=2m+1$ its bosonic part is given as
\begin{eqnarray}
{\rm Str}
(\hat{J}^R)^{2m+1} \mid_{\rm bosonic}~=~0
~\Rightarrow~\left(~\cdots~\right){\rm tr}
~(\hat{\cal B}\hat{j})
\end{eqnarray}
where the possible fermionic variable dependence is
a term proportional to $\hat{\cal B}$
 guaranteed by the $\kappa$ invariance.

In this way, after taking supertrace the even
$n$-th power of $J^R$
reduces terms proportional to ${\cal A}$ and ${\cal B}$,
and the odd   $n$-th power of $J^R$
reduces a term proportional to ${\cal B}$ only.
In this paper
${\cal CD}$ constraints in the ${\cal ABCD}$ first class constraint set
are not introduced for simpler argument,
and set to zero because they are bilinears of constraints.

\par\vskip 6mm
\section{ Conclusion and discussions}

We obtained
the expression of
the conserved ``local" currents
derived from the integrability of a superstring
in the AdS$_5\times$S$^5$ background.
The infinite number of the conserved ``local" currents
 are written by the supertrace of the $n$-th power of the RI currents.
 The lowest nontrivial case, $n=2$, is nothing but the stress-energy tensor
  which is also Virasoro constraint,
  Str$(J^R)^2_\pm$ in \bref{419} and \bref{420}.
  For even $n$ the ``local" current reduces to terms proportional to the
  Virasoro constraint and the $\kappa$ symmetry constraint.
 For odd $n$ it reduces to a term   proportional to the
 $\kappa$ symmetry constraint.
In another word  the integrability
reduces to
the ${\cal AB}({\cal CD})$ first class constraint set
where ${\cal A}$ is the Virasoro generator
and ${\cal B}$ is the $\kappa$ symmetry generator.
The ${\cal ABCD}$ first class constraint set
is the local symmetry generator of superstrings both on the flat space
and on the AdS space.
It is natural that the physical degrees of freedom of
a superstring is common locally,
  independently of  flat or AdS backgrounds.

 It seems that the combination of the  ${\cal B}_\pm j_\pm$
 in \bref{430}
plays the role of the world-sheet supersymmetry operator
in a sense of the grading of the conformal generator.
However it is not
straightforward
to construct the worldsheet supersymmetry operator.
As in the flat case where the lightcone gauge makes
the relation between
the GS fermion and the NSR fermion more transparent,
the $\kappa$ gauge fixing will be a clue to
make a connection to the world-sheet supersymmetry.
We leave this problem in addition to the quantization
problem for future investigations.

\par\vskip 6mm

\noindent{\bf Acknowledgments}

The author thanks to K. Kamimura, S. Mizoguchi and K. Yoshida for fruitful discussions.

\par\vskip 6mm


\section{Introduction}

Two problems that have attracted much attention in the quantum
information community are the MUB and SIC problems for Hilbert spaces of
finite dimension $N$. In the MUB problem \cite{Ivanovic, Wootters1}
one looks for $N+1$ orthonormal bases that are mutually unbiased, in the sense
that

\begin{equation} |\langle e_m|f_n\rangle|^2 = \frac{1}{N} \ , \hspace{8mm}
0 \leq m,n \leq N-1 \ , \end{equation}

\noindent whenever the vector $|e_m\rangle$ belongs to one basis and the
vector $|f_n\rangle$ to another. In the SIC problem \cite{Zauner, Renes}
one looks for a symmetric and informationally complete POVM, which translates
to the problem of finding $N^2$ unit vectors $|\psi_i\rangle$ such that

\begin{equation} |\langle \psi_i|\psi_j\rangle |^2 = \frac{1}{N+1}
\ , \hspace{8mm} 0 \leq i,j \leq N^2 - 1 \ , \end{equation}

\noindent whenever $i \neq j$. These problems are hard. For the MUB problem
an elegant solution exists whenever $N$ is a power of a prime \cite{Wootters2}.
For the SIC problem quite ad hoc looking analytic solutions are known for
eighteen different dimensions; these are described (and in some cases derived)
by Scott and Grassl, who also give full references to the earlier literature
\cite{Grassl}. The belief in the community is that a complete set of $N+1$
MUB does not exist for general $N$, while the SICs do.

Since the problems are so easy to state, it is not surprising that they have
been posed independently in many different branches of science. One purpose
of this article is to describe what nineteenth century geometers had to say
about them. A story told by Eddington \cite{Eddington}
is relevant here:

\

\noindent {\small Some years ago I worked out the structure of this group of operators
in connection with Dirac's theory of the electron. I afterwards learned that
a great deal of what I had written was to be found in a treatise on Kummer's
quartic surface. There happens to be a model of Kummer's quartic surface in
my lecture-room, at which I had sometimes glanced with curiosity, wondering
what it was all about. The last thing that entered my head was that I had
written (somewhat belatedly) a paper on its structure. Perhaps the author
of the treatise would have been equally surprised to learn that he was
dealing with the behaviour of an electron.}

\

\noindent We will see what Eddington saw as we proceed. Meanwhile, let us
observe that when $N$ is a prime
the MUB are the eigenbases of the $N+1$ cyclic subgroups of the Heisenberg
group, while there is a conjecture (enjoying very considerable numerical
support \cite{Grassl}) that the SICs can always be chosen to be special orbits of
this group. When $N$ is a power of a prime the solution of the MUB problem
shifts a little, since the MUBs now consist of eigenvectors of the cyclic subgroups
of the Heisenberg group defined over a finite field rather than over the
ring of integers modulo $N$. Concerning SICs that are orbits under the
Heisenberg group there is a link to the MUB problem: If the dimension
$N$ is a prime the SIC Bloch vectors, when projected onto any one of the
MUB eigenvalue simplices, have the same length for all the
$N+1$ MUB \cite{Khat, ADF}.

In mathematics elliptic curves provide the natural home for the Heisenberg
group, so it seems natural to investigate if elliptic curves can be used
to illuminate the MUB and SIC problems. In dimensions 3 \cite{Hughston} and
4 they certainly can, as we will see, but in higher dimensions I am not so
sure. There will be some comments and formulas that I could not find in
the books and papers I studied, but keeping
Eddington's example in mind I do not claim originality for them.

\section{Two pieces of background information}
We had better define the Heisenberg group properly. A defining non-unitary
representation is given by the upper triangular matrices

\begin{equation} g(\gamma, \alpha, \beta) =
\left( \begin{array}{ccc} 1 & \alpha &  \gamma \\ 0 & 1 & \beta \\
0 & 0 & 1 \end{array} \right) \ . \end{equation}

\noindent Here the matrix elements belong to some ring. In the original
Weyl-Heisenberg group \cite{Weyl} they are real numbers, but here we are
more interested in the case that they belong to the ring of integers
modulo $N$. We denote the resulting group by $H(N)$. It is generated
by two elements $X$ and $Z$ obeying

\begin{equation} ZX = qXZ \ , \hspace{8mm} X^N = Z^N = {\bf 1} \ ,
\hspace{8mm} q = e^{\frac{2\pi i}{N}} \ . \end{equation}

\noindent For $N = 2$ we can use the Pauli
matrices to set $X = \sigma_X$, $Z = \sigma_Z$, which makes it possible
to remember the notation. We will consider the group projectively, so for
our purposes it can often be regarded as a group of order $N^2$.

Because $q$ is a primitive $N$th root of unity the unitary representation
in which $Z$ is diagonal is unique up to permutations \cite{Weyl}.
It is known as the clock and shift representation. If the components of
any vector are denoted $x_a$ the action is given by

\begin{equation} \begin{array}{lll} X: & & x_0 \rightarrow x_{N-1} \rightarrow
x_{N-2} \rightarrow \dots \rightarrow x_1 \rightarrow x_0 \\
\ \label{group} \\ Z: & & x_a \rightarrow q^ax_a \end{array} \ ,
\hspace{8mm} 0 \leq a \leq N-1 \ . \end{equation}

\noindent The unitary automorphism group of the Heisenberg group plays
prominent roles in quantum information theory \cite{Fivel, Gottesman},
and is often called the Clifford group. In the older literature the
Heisenberg group is sometimes called the Clifford collineation group,
and the Clifford group is called the Clifford transform group \cite{Horadam}.
Although we will discuss it in detail for the case $N = 3$ later on, we will
mostly be concerned with automorphisms of order 2. In the clock and shift
representation such an automorphism acts according to

\begin{equation} A: \ \ \  x_a \leftrightarrow x_{-a} \ . \label{A} \ .
\end{equation}

\noindent Adding this generator leads us to consider an extended group which is
twice as large as $H(N)$. In quantum information language the involution $A$ is
generated by one of Wootters' phase point operators \cite{Wootters1}.
Finally there is the curious conjecture \cite{Zauner} that
the SIC vectors are always left invariant by a unitary automorphism of the
Heisenberg group having order 3. No one knows why this should be so,
but it does appear to be true \cite{Marcus, Grassl}, and in four dimensions
we will see exactly how it happens.

What is special about the case when $N$ is prime is that $H(N)$ then admits
$N+1$ cyclic subgroups of order $N$, forming a flower with $N+1$ petals
with only the unit element in common. Correspondingly there are $N+1$
eigenbases, and they necessarily form a complete set of
MUB \cite{Vatan}. In prime power dimensions $N = p^k$ the known complete set
of MUB is the
set of eigenbases of the cyclic subgroups of a Heisenberg group defined
over a Galois field. The only case we will discuss is when $N = 4$, for
which the Galois Heisenberg group is the tensor product $H(2) \otimes H(2)$.

Another piece of background information is that SICs and MUBs look
natural in Bloch space, which is the $N^2-1$ dimensional
space of Hermitean operators of trace 1, considered as a vector space with
the trace inner product and with the maximally mixed state at the origin.
Density matrices form a convex body in Bloch space. A SIC is simply a regular
simplex in Bloch space, inscribed into this convex body. But it is not easy
to rotate the simplex while keeping the body of density matrices fixed,
because the symmetry group of this body is only $SU(N)/Z_N$, a rather
small subgroup of $SO(N^2-1)$ as soon as $N > 2$. This is why the SIC
problem is hard. An orthonormal basis is a regular simplex with only
$N$ corners, spanning some $(N-1)$-plane through the origin in Bloch space.
Two bases are mutually unbiased if the corresponding $(N-1)$-planes are
totally orthogonal, from which it immediately follows that no more than
$N+1$ MUB can exist.

Any pure state corresponds to a Bloch vector of a definite length. Given a
complete set of MUB we can project this vector onto the $N+1$ different
$(N-1)$-planes defined by the MUB. Should it happen that these projected
vectors all have the same length the vector is as it were unbiased with
respect to the MUB, and is then---for some reason---called a
Minimum Uncertainty State \cite{Wootters3, Appleby}. The condition on a
state vector to be unbiased in this sense is easily worked out using the
Euclidean metric on Bloch space in conjunction with Pythagoras' theorem.
Choose any one of the MUB as the computational basis, and express the
Hilbert space components of a unit vector with respect to that basis as

\begin{equation} x_a = \sqrt{p_a}e^{i\mu_a} \ , \hspace{8mm}
\sum_{n = 0}^{N-1} p_a = 1 \ . \label{octant} \end{equation}

\noindent If the corresponding Bloch vector projected onto the $(N-1)$-plane spanned by
the computational basis has the length appropriate to a Minimum Uncertainty
State it must be true that

\begin{equation} \sum_{a = 0}^{N-1}p_a^2 = \frac{2}{N+1} \ . \label{MUS}
\end{equation}

\noindent This is simple enough, but there is the complication that this has
to be done for all the $N+1$ MUB, which will give an additional set of $N$
constraints on the phases ensuring that the vector has the appropriate
length when projected to the other MUB planes. We spare the reader from
the details, but we repeat that all Heisenberg covariant SIC vectors are
Minimum Uncertainty States whenever $N$ is a prime. Examining the
proof of this interesting statement shows that something similar
is true also when no complete set of MUB is available: In any eigenbasis
of a cyclic subgroup of $H(N)$ of order $N$ eq. (\ref{MUS}) will hold
for any vector belonging to a Heisenberg covariant SIC \cite{Khat, ADF}.
This is true regardless of how many bases of this kind there are.

\section{The syzygetic Hesse pencil}
We now descend to the complex projective plane, and begin by introducing the
language used by nineteenth century geometers. Points are represented by ket
or column vectors in ${\bf C}^3$, or more precisely by one-dimensional
subspaces, while lines are represented by two-dimensional subspaces. Using
the scalar product in Hilbert space we can equally well represent the
lines by bra or row vectors orthogonal to the subspaces they represent,
so that the relation

\begin{equation} \langle Y|X\rangle = 0 \label{ett} \end{equation}

\noindent means that the point $X$ lies on the line $Y$. The two-dimensional
subspace representing the line consists of all vectors whose scalar product with
the bra vector $\langle Y|$ vanishes. Since there is a one-to-one correspondence
$|X\rangle \leftrightarrow \langle X|$ between bras and kets there is also a
one-to-one correspondence between points and lines. Clearly eq.
(\ref{ett}) implies that

\begin{equation} \langle X|Y\rangle = 0 \ , \end{equation}

\noindent which says that the point $Y$ lies on the line $X$. This is known as
the duality between points and lines in the projective plane.

We will study complex plane curves defined by homogeneous polynomials
in three variables. Linear polynomials define two-dimensional
subspaces, that is to say two real-dimensional subsets of the complex plane,
and by the above they define projective lines. Intrinsically they are
spheres, namely Bloch spheres, because ${\bf CP}^1 = {\bf S}^2$.
Quadratic polynomials or quadrics define conic sections, and over the
complex numbers the intrinsic geometry of a conic section is again that of a
sphere. The set of spin coherent states is an example \cite{BH}. To
the next order in complication we choose a cubic polynomial. We require the curve
to transform into itself under the Heisenberg group in the clock and shift
representation (\ref{group}). Up to an irrelevant overall constant the most
general solution for the cubic is then

\begin{equation} P = x^3 + y^3 + z^3 + txyz \ . \label{cubic} \end{equation}

\noindent Here $t$ is a complex number parametrising what is known as the
syzygetic Hesse pencil of cubics. Intrinsically each cubic is a torus rather
than a sphere. We observe that the polynomial is automatically invariant
under the additional involution $A$ given above in (\ref{A}).

Hesse \cite{Hesse}, and before him Pl\"ucker \cite{Plucker}, studied this family
of curves in detail. Their first object was to determine the inflection points.
They are given by those points on the curve for which the determinant of its
matrix of second derivatives---its Hessian---vanishes. In the present case this
is a cubic polynomial as well; in fact

\begin{equation} H = \det{\partial_i\partial_jP} =
(6^3 + 2t^3)xyz - 6t^2(x^3 + y^3 + z^3) \ . \end{equation}

\noindent This is again a member of the Hesse pencil of cubics. In astronomy
a ``syzygy'' occurs when three planets lie on a line, so we can
begin to appreciate why the pencil is called ``syzygetic''. The inflection
points are given by $P = H = 0$. By B\'ezout's theorem two cubics in the
complex projective plane intersect in nine points, hence there are nine
inflection points. They coincide for all cubics in the pencil, and are
given by

\begin{equation} \left[ \begin{array}{ccccccccc} 0 & 0 & 0 & -1 & - q & - q^2 &
1 & 1 & 1 \\ 1 & 1 & 1 & 0 & 0 & 0 & -1 & - q & - q^2 \\ -1 & - q & - q^2 &
1 & 1 & 1 & 0 & 0 & 0 \end{array} \right] \ . \label{points} \end{equation}

\noindent This is recognisable as a set of nine
SIC vectors covariant under the Heisenberg group \cite{Zauner, Renes}. We can
normalise our vectors if we want to, but in the spirit of projective geometry
we choose not to.

There are four singular members of the Hesse pencil, defined by values of the
parameter $t$ such that there are non-zero solutions to $P = P_{,x} = P_{,y}
= P_{,z} = 0$. These values are

\begin{equation} t = \infty \hspace{5mm} \mbox{and} \hspace{5mm} t^3 = -
3^3 \ . \end{equation}

\noindent If $t = \infty$ the polynomial reduces to $xyz = 0$. In
this case the singular cubic consists of three projective lines that make
up a triangle. The remaining three singular cases will give rise to three
other triangles. Therefore the syzygetic pencil singles
out 4 special triangles in the projective plane, given by their 12 vertices

\begin{eqnarray} \triangle^{(0)} = \left[ \begin{array}{ccc} 1 & 0 & 0 \\
0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right] \ , \hspace{6mm}
\triangle^{(1)} = \left[ \begin{array}{ccc} 1 & q^2 & q^2 \\ q^2 & 1 & q^2 \\
q^2 & q^2 & 1 \end{array} \right] \ ,  \hspace{12mm} \nonumber \\
\label{MUB3} \\
\triangle^{(2)} = \left[ \begin{array}{ccc} 1 & q & q \\ q & 1 & q \\
q & q & 1 \end{array} \right] \ , \hspace{8mm}
\triangle^{(\infty )} = \left[ \begin{array}{ccc} 1 & 1 & 1 \\ 1 & q & q^2 \\
1 & q^2 & q \end{array} \right] \ , \hspace{12mm} \nonumber \end{eqnarray}

\noindent where $q = e^{2\pi i/3}$. The columns, labelled consecutively by
$0,1,2$, can indeed be regarded as 12 points or by duality as 12 lines.
The four triangles are referred to as the inflection triangles.

What gives the triangles their name is the remarkable fact that the nine
inflection points lie by threes on their twelve edges. Hesse calls this
a ``{\it sch\"onen Lehrsatz}'', and attributes it to Pl\"ucker \cite{Plucker}.
It is not hard
to verify. After a small calculation one finds that the orthogonalities
between the columns in the four triangles and the vectors representing
the inflection points are as follows:

\

{\tiny \begin{tabular}{|c||ccc|ccc|ccc|ccc|} \hline
\  & $\triangle_0^{(0)}$ & $\triangle_1^{(0)}$& $\triangle_2^{(0)}$ & $\triangle_0^{(1)}$
& $\triangle_1^{(1)}$ & $\triangle_2^{(1)}$ & $\triangle_0^{(2)}$ & $\triangle_1^{(2)}$ &
$\triangle_2^{(2)}$ & $\triangle_0^{(\infty )}$ & $\triangle_1^{(\infty )}$ &
$\triangle_2^{(\infty )}$ \\ \hline \hline
$X_0$ & $\bullet$ & \ & \ & $\bullet$ & \ & \ & $\bullet$ & \ & \ & $\bullet$ & \ & \ \\
\ & \ & \ & \ & \ & \ & \ & \ & \ & \ & \ & \ & \ \\
$X_1$ & $\bullet$ & \ & \ & \ & \ & $\bullet$ & \ & $\bullet$ & \ & \ & $\bullet$ & \ \\
\ & \ & \ & \ & \ & \ & \ & \ & \ & \ & \ & \ & \ \\
$X_2$ & $\bullet$ & \ & \ & \ & $\bullet$ & \ & \ & \ & $\bullet$ & \ & \ & $\bullet$ \\
\ & \ & \ & \ & \ & \ & \ & \ & \ & \ & \ & \ & \ \\
$Y_0$ & \ & $\bullet$ & \ & \ & $\bullet$ & \ & \ & $\bullet$ & \ & $\bullet$ & \ & \ \\
\ & \ & \ & \ & \ & \ & \ & \ & \ & \ & \ & \ & \ \\
$Y_1$ & \ & $\bullet$ & \ & $\bullet$ & \ & \ & \ & \ & $\bullet$ & \ & $\bullet$ & \ \\
\ & \ & \ & \ & \ & \ & \ & \ & \ & \ & \ & \ & \ \\
$Y_2$ & \ & $\bullet$ & \ & \ & \ & $\bullet$ & $\bullet$ & \ & \ & \ & \ & $\bullet$ \\
\ & \ & \ & \ & \ & \ & \ & \ & \ & \ & \ & \ & \ \\
$Z_0$ & \ & \ & $\bullet$ & \ & \ & $\bullet$ & \ & \ & $\bullet$ & $\bullet$ & \ & \ \\
\ & \ & \ & \ & \ & \ & \ & \ & \ & \ & \ & \ & \ \\
$Z_1$ & \ & \ & $\bullet$ & \ & $\bullet$ & \ & $\bullet$ & \ & \ & \ & $\bullet$ & \ \\
\ & \ & \ & \ & \ & \ & \ & \ & \ & \ & \ & \ & \ \\
$Z_2$ & \ & \ & $\bullet$ & $\bullet$ & \ & \ & \ & $\bullet$ & \ & \ & \ & $\bullet$ \\
\hline
\end{tabular}}

\

\

\noindent Thus we have

\begin{equation} \langle \Delta_0^{(0)} |X_0\rangle =
\langle \Delta_0^{(0)} |X_1\rangle = \langle \Delta_0^{(0)} |X_2\rangle =
0 \end{equation}

\noindent and so on. Recalling the interpretation of the vanishing scalar
products we see by
inspection of the table that Hesse's beautiful theorem is true.

We have verified that there exists a configuration of 9 points and 12 lines
such that each point belongs to four lines, and each line goes through three points.
This is denoted $(9_4, 12_3)$, and is known as the Hesse configuration. Using the
duality between points and lines we have also proved the existence of the
configuration $(12_3, 9_4)$. From an abstract point of view such a configuration
is a combinatorial object known as a finite affine plane \cite{Dolgachev}. In the
language of quantum information theory the inflection triangles form a complete
set of four MUB, while the inflection points form a SIC.

We can now expand on our discussion of group theory in section 2.
First, every plane cubic can be
regarded as a commutative group in a natural way. This is not surprising,
given that the curve is intrinsically a torus---that is a group manifold.
The idea relies on B\'ezout's theorem, which this time assures us that any
line intersects the cubic in three points---two of which coincide
if the line is a tangent, and all of which coincide if the line is a line of
inflection. An arbitrary point on the cubic is taken to be the identity element,
and denoted $O$. To add two arbitrary points $A$ and $B$ on the
cubic, draw the line between them and locate its third intersection point $P$
with the cubic. Then draw the line between $O$ and $P$ and again locate the
third intersection point $C$. By definition then $A + B = C$. All the group
axioms are obeyed, although it is non-trivial to prove associativity.

Now choose the origin to sit at one of the inflection points. With Hesse's
construction in hand one sees that the nine inflection points form a
group of order nine, which is precisely the projective Heisenberg group.
This is also the torsion group of the curve, meaning that it contains all
group elements of finite order. Because they are group elements of order
3 the inflection points are also called 3-torsion points.

Next we ask for the group of transformations transforming the cubics of the
Hesse pencil among themselves. Recall that the parameter $t$ in the Hesse
cubic (\ref{cubic}) can serve as a complex coordinate on a sphere. The
four singular members of the pencil defines a tetrahedron
on that sphere. Transformations within the pencil act as M\"obius transformations
on the complex number $t$. Moreover they must permute the singular members
of the Hesse pencil among themselves. This means that they form a well known
subgroup of $SO(3)$, namely the symmetry group $A_4$ of the regular
tetrahedron. It enjoys the isomorphism

\begin{equation} A_4 \sim PSL(2, {\bf F}_3) \ , \end{equation}

\noindent where ${\bf F}_3$ is the field of integers modulo 3. The group
$SL(2, {\bf F}_3)$ consists of unimodular two by two matrices with
integer entries taken modulo three; here only its projective part enters
because the subgroup generated by the matrix $-{\bf 1}$ gives rise to
the involution $A$ and does not act on $t$, although it does
permute the inflection points among themselves. The full symmetry
group of the pencil is a semi-direct product of the Heisenberg group
and $SL(2, {\bf F}_3)$. This is the affine group on a finite affine
plane. It is known as the Hessian group \cite{Jordan}, or as the
Clifford group.

There are many accounts of this material in the literature, from geometric
\cite{Grove}, undergraduate \cite{Gibson}, and modern \cite{Artebani} points
of view. It forms a recurrent theme in Klein's history of nineteenth
century mathematics \cite{Klein}. The fact that the inflection points form
a SIC was first noted by Lane Hughston \cite{Hughston}.

\section{The elliptic normal curve in prime dimensions}
Felix Klein and the people around him put considerable effort into the
description of elliptic curves embedded into projective spaces of dimension
higher than 2. They proceeded by means of explicit parametrisations of
the curve using Weierstrass' $\sigma$-function \cite{Bianchi, Hulek}. As far
as we are concerned now, we only need to know that the symmetries they
built into their curves is again the Heisenberg group supplemented with
the involution $x_a \leftrightarrow x_{-a}$ coming from the Clifford group.
An analysis of this group of symmetries leads directly to ``{\it une
configuration tr\`es-remarquable}'' originally discovered by Segre
\cite{Segre}. We will present it using some notational improvements
that were invented later \cite{Gross, ADF}.

Since $N = 2n-1$ is odd, the integer $n$ serves
as the multiplicative inverse of $2$ among the integers modulo $N$.
It is then convenient to write the Heisenberg group elements as

\begin{equation} D(i,j) = q^{nij}X^iZ^j \hspace{5mm} \Rightarrow
\hspace{5mm} D(i,j)D(k,l) = q^{n(jk-il)}D(i+k, j+l) = q^{jk-il}D(k,l)D(i,j) \ . \end{equation}

\noindent Let us also introduce explicit matrix representations of
the group generators:

\begin{equation} D(i,j) = q^{nij + bj}\delta_{a,b+i} \ ,
\hspace{10mm} A = \delta_{a+b,0} \ . \end{equation}

\noindent Note that the spectrum of the involution $A$ consists of
$n$ eigenvalues $1$ and $n-1$ eigenvalues $-1$. Hence $A$ splits the
vector space into the direct sum

\begin{equation} {\cal H}_N = {\cal H}_n^{(+)} \oplus {\cal H}_{n-1}^{(-)}
\ . \end{equation}

\noindent It is these subspaces that we should watch. In fact there
are altogether $N^2$ subspaces of dimension $n$ singled out in this
way, because there are $N^2$ involutions

\begin{equation} A_{ij} = D(i,j)AD(i,j)^\dagger
\ . \label{Aij} \end{equation}

\noindent
The eigenvectors of the various cyclic subgroups can be collected
into the $N+1$ MUB

\begin{equation} \triangle_{am}^{(k)} = \left\{\begin{array}{cll}
\delta_{am} & , & k = 0 \\ \ \\ \frac{1}{\sqrt{N}}q^{\frac{(a-m)^2}{2k}} &
, & 1 \leq k \leq N-1 \\ \ \\
\frac{1}{\sqrt{N}}q^{am} & , & k = \infty \end{array} \right. \
.\end{equation}

\noindent Here $k$ labels the basis, $m$ the vectors, and $a$ their
components. For $N = 3$ this coincides with form (\ref{MUB3}) given
earlier. Note that $N-1$ MUB have been written as circulant matrices,
which is a convenient thing to do.

The key observation is that the zeroth columns in the MUB all obey---we
suppress the index labelling components---

\begin{equation} A\triangle_0^{(k)} = \triangle_0^{(k)} \ . \end{equation}

\noindent Hence this set of $N+1$ vectors belongs to the $n$-dimensional
subspace ${\cal H}_n^{(+)}$ defined by the involution $A$. We can go on to
show that each of the $n$-dimensional eigenspaces defined by the
$N^2$ involutions $A_{ij}$ contain $N+1$ MUB vectors. Conversely,
each MUB vector belongs to $N$ subspaces. We have found the Segre configuration

\begin{equation} \left( N(N+1)_N, N^2_{N+1} \right) \end{equation}

\noindent containing $N^2 + N$ points and $N^2$ $(n-1)$-planes in
projective $(N-1)$-space, always assuming that $N$ is an odd prime.

The intersection properties of the Segre configuration are remarkable.
Two $n$-spaces in $2n-1$ dimensions intersect at least in a single ray.
With a total of $N^2$ such subspaces to play with we expect many vectors
to arise in this way. But let $\psi$ be such a vector. A minor calculation
shows that

\begin{equation} \psi  = A_{ij}A_{kl}\psi  =
q^{2(il-jk)}D(2i-2k, 2j-2k)\psi  \ . \end{equation}

\noindent Thus $\psi$ must be an eigenvector of some element in the Heisenberg
group, and hence the intersection of any two $n$-spaces is always one
of the $N(N+1)$ eigenvectors in the configuration. In the other direction
things are a little more complicated. Two vectors belonging to the same
basis are never members of same eigenspace, while two vectors of two
different MUB belong to a unique common eigenspace. Using
projective duality we obtain the dual configuration

\begin{equation} \left( N^2_{N+1}, N(N+1)_N \right) \end{equation}

\noindent consisting of $N^2$ $(n-1)$-spaces and $N^2 + N$ hyperplanes.
The intersection properties are precisely those of a finite affine plane
\cite{Dolgachev}.

These are the facts that so delighted Segre. A hundred years
later they delighted Wootters \cite{Wootters1}---although he phrased the
discussion directly in terms of the phase point
operators $A_{ij}$ rather than in terms of their eigenspaces.
A systematic study of prime power dimensions in Segre's spirit appears
not to have been made, although there are some results for $N = 9$ \cite{Horadam}.

But where is the SIC? It is hard to tell. When the dimension
$N = 2n-1 = 3$ we observe that $n-1 = 1$, so the dual Segre configuration
involves $N^2$ vectors, and these are precisely the SIC vectors (\ref{points}).
When $N \geq 5$ the Segre configuration contains not even a candidate
set of $N^2$ vectors. But at least, as a byproduct of the construction,
we find a set of $2n$ equiangular vectors in any $n$ dimensional Hilbert
space such that $2n-1$ is an odd prime. Explicitly they are

\begin{equation} \left[ \begin{array}{ccccc} \sqrt{2n-1} & 1 & 1 & \dots
& 1 \\ 0 & \sqrt{2} & \sqrt{2}q^{1\cdot 1^2} & \dots & \sqrt{2}q^{(2n-2)\cdot 1^2} \\
0 & \sqrt{2} & \sqrt{2}q^{1\cdot2^2} & \dots & \sqrt{2}q^{(2n-2)\cdot 2^2} \\
\vdots & \vdots & \vdots & & \vdots \\
0 & \sqrt{2} & \sqrt{2}q^{1\cdot (n-1)^2} & \dots & \sqrt{2}q^{(2n-2)(n-1)^2}
\end{array} \right] \ . \label{2n} \end{equation}

\noindent Such sets are of some interest in connection with pure state
quantum tomography \cite{Flammia}.

The elliptic curve itself has not been much in evidence in this section.
It is still there in the background though, and in any dimension it
contains $N^2$ distinguished $N$-torsion points. A study of the explicit
expression for the Heisenberg covariant elliptic curve shows that each
of its torsion points belong to one of the $N^2$ eigenspaces
${\cal H}_{n-1}^{(-)}$ \cite{Hulek}, and with the single exception of the
$N = 3$ example (\ref{points}) the known SICs never sit in such a subspace,
so the torsion points are not SICs. This is discouraging, but we will
find some consolation when we proceed to examine the $N = 4$ case.

\section{The SIC in 4 dimensions}
In an $N = 4$ dimensional Hilbert space there is a parting of the ways,
in the sense that the MUB and the SIC are defined using two different
versions of the Heisenberg group. The elliptic curve stays with $H(4)$.
Using an argument concerning line bundles and employing ingredients such
as the Riemann-Roch theorem, it can be shown that an elliptic
normal curve in projective 3-space (not confined to any projective plane)
is the non-singular intersection of two quadratic polynomials. If we insist
that it is transformed into itself by the Heisenberg group in its clock
and shift representation (\ref{group}), it follows \cite{Hulek} that
these quadratic polynomials are

\begin{equation} Q_0 = x_0^2 + x_2^2 + 2ax_1x_3 \ , \hspace{8mm} Q_1 =
x_1^2 + x_3^2 + 2ax_0x_2 \ . \end{equation}

\noindent The extra symmetry under the involution $A$, defined in (\ref{A}),
again appears
automatically. We can diagonalise these quadratic forms by means of a unitary
transformation of our Hilbert space. In the new coordinates we have

\begin{equation} Q_0 = z_0^2 + iz_1^2 + a(iz_2^2 + z_3^2) \ , \hspace{8mm}
Q_1 = iz_2^2 - z_3^2 + a(z_0^2 - iz_1^2) \ . \end{equation}

\noindent Note that $Q_0 = Q_1 = 0$ implies

\begin{equation} z_0^4 +z_1^4 + z_2^4 + z_3^4 = 0 \ . \end{equation}

\noindent Hence the elliptic curve lies on a quartic surface.

The new basis that we have introduced has a natural interpretation in
terms of the involution $A$. First of all, by acting on $A$ with the
Heisenberg group as in eq. (\ref{Aij}) we obtain only four involutions
altogether, rather than $N^2$ as in the odd prime case. Their spectra
are $(1,1,1,-1)$, and in the new basis they are all represented by
diagonal matrices. Hence each basis vector is inverted by one involution,
and left invariant by the others. In projective 3-space they correspond
to four reference points, and one can show that the 16 tangents of the 16 torsion
points on the curve divide into 4 sets of 4 each coming together at one of
the 4 reference points \cite{Hulek}. Each such set is an orbit under the
subgroup of elements of order 2.

In our preferred basis the generators of the Heisenberg group appear in the form

\begin{equation} Z = e^{\frac{i\pi}{4}} \left( \begin{array}{rrrr} 0 & 1 & 0 & 0 \\
-i & 0 & 0 & 0 \\
0 & 0 & 0 & -i \\ 0 & 0 & -1 & 0 \end{array}\right) \ , \hspace{9mm}
X = e^{\frac{i\pi}{4}} \left( \begin{array}{rrrr}
0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ -i & 0 & 0 & 0 \\ 0 & i & 0 & 0 \end{array}
\right) \ . \end{equation}

\noindent Finding a set of 16 SIC-vectors covariant under the Heisenberg group
is now a matter of simple guesswork. One answer, ignoring overall phases and
normalisation, is

\begin{equation} \left[
\begin{array}{rrrrrrrrrrrrrrrr} x & x & x & x & i & i & - i & - i & i &  i
& - i & - i & i & i & - i & - i \\ 1 & 1 & - 1 & - 1 & x & x & x & x & i & -i
& i & - i & 1 & - 1 & 1 & - 1 \\ 1 & -1 & 1 & -1 & 1 & - 1 & 1 & -1 & x &
x & x & x & - i & i & i & - i \\
1 & - 1 & - 1 & 1 & -i & i & i & - i & - 1 & 1 & 1 & - 1 & x & x & x & x
\end{array} \right] \ , \label{SIC4} \end{equation}

\noindent where

\begin{equation} x = \sqrt{2 + \sqrt{5}} \ . \end{equation}

\noindent All scalar products have the same modulus because

\begin{equation} (x^2-1)^2 = |x+1 + i(x-1)|^2 \ . \end{equation}

\noindent Thanks to our change of basis, this is significantly more memorable
than the standard solutions \cite{Zauner, Renes} (and it was in fact arrived
at, without considering the Heisenberg group at all, by Belovs \cite{Belovs}).
The whole set is organised
into 4 groups, where each group sits at a standard distance from the 4 basis
vectors that are naturally singled out by the elliptic curve. The normalised vectors
obey eq. (\ref{MUS}) for a Minimum Uncertainty State, even though our basis
is unusual.

The otherwise mysterious invariance of the SIC vectors under some element
of the Clifford group of order 3 is now easy to see. We focus
on the group of vectors

\begin{equation} \left[ \begin{array}{rrrr} x & x & x & x \\ 1 & 1 & -1 & -1 \\
1 & -1 & 1 & -1 \\ 1 & -1 & -1 & 1 \end{array} \right] \ . \end{equation}

\noindent They form an orbit under the subgroup of elements of order 2. When
we project them to the subspace orthogonal to the first basis vector we have
4 equiangular vectors in a 3 dimensional subspace. Each projected vector
will be invariant under a rotation of order 3 belonging to the symmetry
group of this tetrahedron. An example leaving the first vector invariant is

\begin{equation} R = \left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\
0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{array} \right) \ . \end{equation}

\noindent It is straightforward to check that the rotation $R$ belongs to
the Clifford group, and is indeed identical to one of "Zauner's unitaries"
\cite{Zauner}.

Each of the four involutions $A$ admit a "square root" belonging to the
Clifford group, such as

\begin{equation} F = \left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\
0 & 1 & 0 & 0 \\ 0 & 0 & 0 & i \end{array} \right) \hspace{5mm}
\Rightarrow \hspace{5mm} F^2 = A = \left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\ 0 & 0 & 0 & -1 \end{array} \right)\ . \end{equation}

\noindent Acting with these unitaries on the SIC (\ref{SIC4}) will give a
set of altogether 16 different SICs, collectively forming an orbit under the
Clifford group \cite{Zauner, Marcus}.

Note that the 16 SIC points in projective space do not actually sit on
the elliptic curve. In this sense the step from $N = 3$ to $N = 4$ is
non-trivial. In an arbitrary even dimension $N = 2n$
the involution $A$, see (\ref{A}), has a spectrum consisting of $n+1$
eigenvalues $1$ and $n-1$ eigenvalues $-1$. When $N = 4$ this singles out
a unique ray but in higher dimensions this is not so, so generalising
to arbitrary even dimension will not be easy.

\section{Minimum Uncertainty States in four dimensions}
Eddington and his surface have not yet appeared. The group on whose twofold
cover his Fundamental Theory hinged was not the Heisenberg group over
the ring of integers modulo 4, but a different Heisenberg group of
the form $H(2)\otimes H(2)$ \cite{Eddington}. This group can be
represented by real matrices, and is in fact the group which gives rise
to the complete set of MUB in 4 dimensions. What can we do with it?

There does not exist a SIC which is covariant under Eddington's group.
In fact the group $H(2)^{\otimes k}$ admits such an orbit only if
$k = 1$ or $k = 3$ \cite{Godsil}.
As a substitute we can look for an orbit of 16 Minimum Uncertainty
States with
respect to the maximal set of MUB. Such an orbit does exist, and is
given by the 16 vectors

\begin{equation} \left[ \begin{array}{cccccccccccccccc} x & x & x & x &
\alpha & \alpha & - \alpha & - \alpha & \alpha & \alpha & - \alpha &
- \alpha & \alpha & \alpha & - \alpha & - \alpha \\
\alpha & \alpha & - \alpha & - \alpha & x & x & x & x &
\alpha & - \alpha & \alpha & - \alpha & \alpha &
- \alpha & \alpha & - \alpha \\
\alpha & - \alpha & \alpha & - \alpha & \alpha & - \alpha & \alpha
& - \alpha & x & x & x & x & \alpha & -\alpha & - \alpha & \alpha \\
\alpha &  -\alpha & - \alpha & \alpha & \alpha & - \alpha & - \alpha
& \alpha & \alpha & - \alpha & - \alpha & \alpha & x & x & x & x
\end{array} \right] \ , \label{Edd} \end{equation}

\noindent where

\begin{equation} x = \sqrt{2 + \sqrt{5}} \ , \hspace{8mm} \alpha = e^{ia} \ ,
\hspace{8mm} \cos{a} = \frac{\sqrt{5}-1}{2\sqrt{2 + \sqrt{5}}} \ . \end{equation}

\noindent I omit the lengthy proof that these 16 vectors really are
Minimum Uncertainty States \cite{Asa}. Although this is not a SIC,
in a way it comes close to being one. Like the SICs (\ref{points}) and
(\ref{SIC4}), it can be arrived at using the following procedure: Introduce a
vector $(x, e^{i\mu_1}, \dots , e^{i\mu_{N-1}})^{\rm T}$, and adjust the value
of $x$ so that the normalised vector solves eq. (\ref{MUS}) for a Minimum
Uncertainty State. Next introduce a complex Hadamard matrix, that is to
say a unitary matrix all of whose matrix elements have the same modulus.
Such matrices exist in any dimension, although their classification problem
is unsolved if the dimension exceeds 5 \cite{Tadej}. By multiplying with
an overall factor $\sqrt{N}$, and then multiplying the columns with phase
factors, we can ensure that all matrix elements in the first row equal
1. Replace these elements with $x$. Next multiply the rows
with phase factors until one of the columns equals the vector we introduced.
The result is a set of $N$ vectors with all mutual scalar products taking
the value that characterises a SIC. Next permute the entries of the original
vector cyclically, and afterwards try to adjust the phases $\mu_a$ so that the
resulting $N$ vectors are again equiangular with the mutual scalar products
characterising a SIC. Extending the new vectors using an Hadamard matrix
in the same way as before then gives $N$ equiangular vectors each of which
belongs to a separate group of $N$ equiangular vectors. Before we can say
that we have constructed a SIC we must check that all scalar products
between pairs of vectors not belonging to the same group take the SIC values.
The vectors (\ref{Edd}) fail to form a SIC only because the last step fails.

Finally we come back to Eddington's lecture room. In the treatise that
he read \cite{Hudson}
it is explained that an orbit of $H(2)\otimes H(2)$ gives a realisation of
the Kummer configuration $16_6$, consisting of 16 points and 16 planes in
projective 3-space, such that each point belongs to 6 planes and each plane
contains 6 points. The above set of Minimum Uncertainty States realises this
configuration. As an example, the 6 vectors

\begin{equation} \left[
\begin{array}{cccccc} - \alpha & - \alpha & - \alpha & - \alpha & - \alpha
& - \alpha \\
x & x & \alpha & - \alpha
& \alpha & - \alpha \\ \alpha & -\alpha & x & x & - \alpha & \alpha \\
-\alpha & \alpha & - \alpha & \alpha & x & x
\end{array} \right] \end{equation}

\noindent are orthogonal to the row vector

\begin{equation} ( \begin{array}{ccccc} x & \alpha & \alpha &
\alpha \end{array} ) \ , \end{equation}

\noindent or in other words the corresponding 6 points belong to the
corresponding plane. This is a purely group theoretical property and does
not require the vectors to be Minimum Uncertainty States. Still, Eddington's
story suggests that our 16 special vectors may have some use, somewhere.

\section*{Acknowledgments}
I thank Subhash Chaturvedi for telling me about the Segre configuration, at a
point in time when neither of us knew about Segre. Both of us give our best
wishes to Tony!

\section*{References}

\medskip