\section{Introduction}

In \cite{PENROSE} Roger Penrose takes issue with the standard view of the graviton as a weak-field perturbation of a background spacetime.  He proposes the idea that each graviton should carry its measure of curvature, corresponding to a solution of the full nonlinear Einstein equations.  The issue of the graviton is of supreme importance, particularly when one wishes to construct a quantum theory of gravity.  The Penrose approach in \cite{PENROSE} presents the concept of left-handed and right-handed gravitons, which leads to a twistor theory naturally adapted to the description of complex spacetimes.  In this paper we will demonstrate a realization of Penrose's idea using a nonmetric description of complex GR which is different from twistor theory, and is derivable from the Ashtekar formulation.\par
\indent
In the complex Ashtekar theory of gravity the basic phase space variables are a self-dual $SO(3,C)$ connection and a densitized triad $\Omega_{Ash}=(A^a_i,\widetilde{\sigma}^i_a)$.  The action in 3+1 form is a canonical one form minus a linear combination of first class constraints smeared by auxilliary fields \cite{ASH1}, \cite{ASH2}, \cite{ASH3}

\begin{eqnarray}
\label{THEACTION}
I_{Ash}=\int{dt}\int_{\Sigma}d^3x\Bigl[\widetilde{\sigma}^i_a\dot{A}^a_i+A^a_0G_a-N^iH_i-i\underline{N}H\Bigr].
\end{eqnarray}

\noindent
The fields $N^i$, $A^a_0$ and $\underline{N}=N(\hbox{det}\widetilde{\sigma})^{-1/2}$ are respectively the shift vector, temporal component of a 4-D self-dual connection $A^a_{\mu}$, and the lapse density function.  The initial value constraints are the diffeomorphism constraint $H_i$, given by

\begin{eqnarray}
\label{THEACTION1}
H_i=\epsilon_{ijk}\widetilde{\sigma}^j_aB^k_a=0,
\end{eqnarray}

\noindent
the Gauss' law constraint $G_a$ which is given by

\begin{eqnarray}
\label{THEACTION2}
G_a=D_i\widetilde{\sigma}^i_a=0,
\end{eqnarray}

\noindent
and the Hamiltonian constraint $H$ by

\begin{eqnarray}
\label{THEACTION3}
H=\epsilon_{ijk}\epsilon^{abc}\widetilde{\sigma}^i_a\widetilde{\sigma}^j_b\Bigl(B^k_c+{\Lambda \over 3}\widetilde{\sigma}^k_c\Bigr)=0
\end{eqnarray}

\noindent
where $\Lambda$ is the cosmological constant.  There is a nontrivial solution to the system (\ref{THEACTION1}), (\ref{THEACTION2}) and (\ref{THEACTION3}) given by $\widetilde{\sigma}^i_a=-{3 \over \Lambda}B^i_a$ which enables one to construct a Hamilton--Jacobi functional $I_{CS}[A]$, namely the Chern--Simons functional of the spatial connection $A^a_i$.  The exponentiation of this functional yields the Kodama state, which was first discovered by Hideo Kodama \cite{KOD}

\begin{eqnarray}
\label{THEACTION4}
\boldsymbol{\psi}_{Kod}[A]=e^{-3(\hbar{G}\Lambda)^{-1}I_{CS}[A]}.
\end{eqnarray}

\noindent
Equation (\ref{THEACTION4}) exactly solves the classical constraints and also the quantum constraints of GR for a particular operator ordering \cite{POSLAMB}.  There are some objections to the use of $\boldsymbol{\psi}_{Kod}$ as a ground state for gravity, by analogy to the pathologies of the Chern--Simons functional when seen in the purely Yang--Mills context.  These pathologies include nonnormalizability and nonunitarity, as well as the lack of a reliable Hilbert space structure \cite{WITTEN1}.\par
\indent
We will address these objections by showing that there exists a well-defined theory of fluctuations about $\boldsymbol{\psi}_{Kod}$, seen as the ground state for some gravitational system.  In this paper we will show demonstrate that the fluctuations take on the interpretation of gravitons, and we will quantize these fluctuations and show that they admit a genuine Hilbert space.  This task has been carried out to some extent at the
linearized level in \cite{LINKO} in the Ashtekar variables.  In the present paper we will extend the demonstration to Lorentzian signature spacetimes, using a new set of phase space variables $\Omega_{Inst}=(\Psi_{ae},A^a_i)$ which will be defined later.  We will carry out the demonstration both at the linearized level and for the full nonlinear theory.  In this paper we will not address reality conditions, which is treated elsewhere.\par
\indent

\subsection{Organization of this paper}

The organization of this paper is as follows.  After transforming from the Ashtekar phase space $\Omega_{Ash}=(\widetilde{\sigma}^i_a,A^a_i)$ into the new phase space variables variables $\Omega_{Inst}=(\Psi_{ae},A^a_i)$ we expand the starting action including the initial value constraints about the action associated with the Kodama state $\boldsymbol{\psi}_{Kod}$.  Prior to embarking upon the full theory we first demonstrate the expected features of the graviton in the linearized approximation in Part I.  Section 3 performs a linearization about $\boldsymbol{\psi}_{Kod}$ on $\Omega_{Inst}$, producing the massless spin two polarizations in this limit.  We then perform a quantization
on $\Omega_{Inst}$, demonstrating the existence of a Hilbert space structure at the linearized level.  In Part Two we redo the previous exercises, now with respect to the full nonlinear theory.  First we put in place the requisite canonical structure for quantization, which entails the implementation of the kinematic constraints at the level of the starting action.  Then we introduce the auxilliary Hilbert space and use it as a basis for construction of wavefunctions annihilated by the Hamiltonian constraint of the full theory.  A time variable $T$ on configuration space emerges similarly to the case in the linearized theory, and the wavefunction evolves with respect to this time in the full theory.  The well-definedness of the quantization is linked to the convergence of solutions of the full Hamiltonian constraint with respect to time variable $T$, which we prove in this section.\par
\indent
To proceed from the Ashtekar phase space $\Omega_{Ash}$ into the new phase space $\Omega_{Inst}$, let us first make a substitution called the CDJ Ansatz

\begin{eqnarray}
\label{THEACTION41}
\widetilde{\sigma}^i_a=\Psi_{ae}B^i_e,
\end{eqnarray}

\noindent
where $\Psi_{ae}\indent{SO}(3,C)\otimes{SO}(3,C)$ is the CDJ matrix.\footnote{This Ansatz is attributable to Capovilla, Dell and Jacobson, was used in \cite{CAP} to write down a general solution to the Hamiltonian and the diffeomorphism constraints, which are algebraic in nature.  Equation (\ref{THEACTION4}) holds as long as $(\hbox{det}B)\neq{0}$ and $(\hbox{det}\Psi)\neq{0}$.  Since $\widetilde{\sigma}^i_a$ is dimensionless and $[A^a_i]=1$, then it
follows that $[\Psi_{ae}]=-2$.}  The action (\ref{THEACTION}) then becomes

\begin{eqnarray}
\label{THEACTION5}
I_{Inst}=\int{dt}\int_{\Sigma}d^3x\Bigl[\Psi_{ae}B^i_e\dot{A}^a_i+A^a_0G_a-N^iH_i-iNH\Bigr],
\end{eqnarray}

\noindent
where the corresponding constraints are given by

\begin{eqnarray}
\label{THEACTION6}
H_i=\epsilon_{ijk}B^j_aB^k_e\Psi_{ae}
\end{eqnarray}

\noindent
for the diffeomorphism constraint,

\begin{eqnarray}
\label{THEACTION7}
G_a=B^i_eD_i\Psi_{ae}+C_{be}\bigl(f_{abf}\delta_{ge}+f_{ebg}\delta_{af}\bigr)\Psi_{fg}\equiv\textbf{w}_e\{\Psi_{ae}\}
\end{eqnarray}

\noindent
for the Gauss' law constraint, and

\begin{eqnarray}
\label{THEACTION8}
H=(\hbox{det}B)\bigl({1 \over 2}Var\Psi+\Lambda\hbox{det}\Psi\bigr)=0
\end{eqnarray}

\noindent
for the Hamiltonian constraint where $Var\Psi=(\hbox{tr}\Psi)^2-\hbox{tr}\Psi^2$.  In (\ref{THEACTION7}) we have defined a magnectic helicity density matrix $C_{be}\equiv{A}^b_iB^i_e$ and we have made use of the definition of the covariant derivative of $\Psi_{ae}$, seen as a second-rank $SO(3,C)$ tensor.  In the language of the new phase space variables $\Omega_{Inst}=(\Psi_{ae},A^a_i)$, the Kodama state
corresponds to the solution $\Psi_{ae}=-{3 \over \Lambda}\delta_{ae}$.\par
\indent
The variation of the canonical one form corresponding to (\ref{THEACTION5}) is given by

\begin{eqnarray}
\label{THEACTION9}
\delta\boldsymbol{\theta}_{Inst}=\delta\Bigl(\int_{\Sigma}d^3x\Psi_{ae}B^i_e\delta{A}^a_i\Bigr)\nonumber\\
=\int_{\Sigma}d^3x\Bigl[B^i_e{\delta\Psi_{ae}}\wedge{\delta{A}^a_i}+\Psi_{ae}\epsilon^{ijk}{(D_j\delta{A}^e_k)}\wedge{\delta{A}^a_i}\Bigr],
\end{eqnarray}

\noindent
which owing to the second term is not a symplectic two form of canonical form.  This features poses an obstruction to the quantization of the theory, which we will show becomes eliminated when one restricts oneself to the linearized level.  In the nonlinear case a transformation from $\Omega_{Inst}$ into new densitized variables is required on the kinematic phase space $\Omega_{Kin}$.

\section{Expansion of the classical constraints relative to the pure Kodama state}

\noindent
We now return to the starting theory defined on $\Omega_{Inst}$ and expand the initial value constraints in fluctuations about the pure Kodama state $\boldsymbol{\psi}_{Kod}$.  We will use the Ansatz

\begin{equation}
\label{KODD}
\Psi_{ae}=-\bigl({3 \over \Lambda}\delta_{ae}+\epsilon_{ae}\bigr),
\end{equation}

\noindent
where $\epsilon_{ae}$ is the CDJ deviation matrix $\epsilon_{ae}$, which parametrizes deviations from $\boldsymbol{\psi}_{Kod}$.  Substitution of (\ref{KODD}) into (\ref{THEACTION5}) yields a canonical one form

\begin{eqnarray}
\label{ONEFOR2}
\boldsymbol{\theta}_{Inst}=-{3 \over \Lambda}\int{dt}\int_{\Sigma}d^3xB^i_a\dot{A}^a_i
-\int{dt}\int_{\Sigma}d^3x\epsilon_{ae}B^i_e\dot{A}^a_i.
\end{eqnarray}

\noindent
The first term of (\ref{ONEFOR2}) is the integral of a total derivative, which integrates to the Chern--Simons functional $I_{CS}$.\par
\indent
When one expands the constraints relative to $\boldsymbol{\psi}_{Kod}$, one finds that the $-{3 \over \Lambda}\delta_{ae}$ part of (\ref{KODD}) drops out for the constraints linear in $\Psi_{ae}$.  So for the Gauss' law constraint we have

\begin{eqnarray}
\label{KODD3}
B^i_eD_i\Psi_{ae}=-B^i_eD_i\bigl({3 \over \Lambda}\delta_{ae}+\epsilon_{ae}\bigr)=-B^i_eD_i\epsilon_{ae},
\end{eqnarray}

\noindent
since the homogeneous and isotropic part of the CDJ matrix is annihilated by the covariant derivative.  Likewise, for the diffeomorphism constraint the $-{3 \over \Lambda}\delta_{ae}$ part cancels out due to antisymmetry

\begin{eqnarray}
\label{KODD5}
H_i=-\epsilon_{ijk}B^j_aB^k_e\bigl({3 \over \Lambda}\delta_{ae}+\epsilon_{ae}\bigr)=\epsilon_{ijk}B^j_aB^k_e\epsilon_{ae}=0.
\end{eqnarray}

\noindent
For the Hamiltonian constraint, an imprint of $-{3 \over \Lambda}\delta_{ae}$ remains upon expansion due to the nonlinearity of the contraint.  This can be seen as the imprint of $\boldsymbol{\psi}_{Kod}$, which interacts with the fluctuations.  The Hamiltonian constraint uses the invariants of the CDJ matrix, namely the trace

\begin{eqnarray}
\label{THETRACE}
\hbox{tr}\Psi=-\bigl({9 \over \Lambda}+\hbox{tr}\epsilon\bigr)
\end{eqnarray}

\noindent
and the variance $Var\Psi$, given by

\begin{eqnarray}
\label{VAR}
Var\Psi=\epsilon_{abc}\epsilon_{efc}\bigl({3 \over \Lambda}\delta_{ae}+\epsilon_{ae}\bigr)\bigl({3 \over \Lambda}\delta_{bf}+\epsilon_{bf}\bigr)\nonumber\\
=\epsilon_{abc}\epsilon_{efc}\Bigl({9 \over {\Lambda^2}}\delta_{ae}\delta_{bf}
+{6 \over {\Lambda}}\delta_{ae}\epsilon_{bf}+\epsilon_{ae}\epsilon_{bf}\Bigr)
={{54} \over {\Lambda^2}}+{{12} \over {\Lambda}}\hbox{tr}\epsilon+Var\epsilon
\end{eqnarray}

\noindent
and the deteminant given by

\begin{eqnarray}
\label{DET}
-6\hbox{det}\Psi=\epsilon_{abc}\epsilon_{efg}\bigl({3 \over \Lambda}\delta_{ae}+\epsilon_{ae}\bigr)
\bigl({3 \over \Lambda}\delta_{bf}+\epsilon_{bf}\bigr)\bigl({3 \over \Lambda}\delta_{cg}+\epsilon_{cg}\bigr)\nonumber\\
=\epsilon_{abc}\epsilon_{efg}\Bigl({{54} \over {\Lambda^3}}\delta_{ae}\delta_{bf}\delta_{cg}
+{{27} \over {\Lambda^2}}\delta_{ae}\delta_{bf}\epsilon_{cg}
+{9 \over {\Lambda}}\delta_{ae}\epsilon_{bf}\epsilon_{cg}+\epsilon_{ae}\epsilon_{bf}\epsilon_{cg}\Bigr)\nonumber\\
={{162} \over {\Lambda^3}}+{{54} \over {\Lambda^2}}\hbox{tr}\epsilon+{9 \over {\Lambda}}Var\epsilon+6\hbox{det}\epsilon
\end{eqnarray}

\noindent
Combining (\ref{VAR}) and (\ref{DET}), then the Hamiltonian constraint is given by

\begin{eqnarray}
\label{HAMI}
\hbox{det}B\bigl(\Lambda\hbox{det}\Psi+{1 \over 2}Var\Psi\bigr)\nonumber\\
=-\hbox{det}B\Bigl({6 \over {\Lambda}}\hbox{tr}\epsilon+2Var\epsilon
+2\Lambda\hbox{det}\epsilon\Bigr)=0.
\end{eqnarray}

\noindent
\indent
At the classical level, the constraints can be written as a system of seven equations in nine unknowns

\begin{eqnarray}
\label{CONSTRAINTTT}
\epsilon_{ijk}B^j_aB^k_e\epsilon_{ae}=0;~~
\textbf{w}_e\{\epsilon_{ae}\}=0;~~
\hbox{tr}\epsilon+{\Lambda \over 3}Var\epsilon+{{\Lambda^2} \over 3}\hbox{det}\epsilon=0.
\end{eqnarray}

\noindent
The third equation of (\ref{CONSTRAINTTT}) has used $(\hbox{det}B)\neq{0}$, which is a required condition for the transformation (\ref{THEACTION4}) to be valid.  Therefore the analysis of this paper does not apply to flat spacetimes, where $B^i_a=0$.

\section{Part One: The linearized theory}

Having expanded $\Psi_{ae}$ as in (\ref{KODD}), we will now linearize the theory using the following expansion about a reference connection $\alpha^a_i$

\begin{eqnarray}
\label{NOWEXPAND}
A^a_i=\alpha^a_i+a^a_i,
\end{eqnarray}

\noindent
where $\vert{a}^a_i\vert<<\alpha^a_i$.  We must substitute (\ref{NOWEXPAND}) into (\ref{ONEFOR2}) and (\ref{CONSTRAINTTT}) and expand to linear order in $a^a_i$.  The Ashtekar $SO(3,C)$ magnetic
field $B^i_a=\epsilon^{ijk}\partial_jA^a_k+{1 \over }2\epsilon^{ijk}f_{abc}A^b_jA^c_k$ is given by

\begin{eqnarray}
\label{NOWEXPAND1}
B^i_a=\epsilon^{ijk}\partial_j(\alpha^a_k+a^a_k)+{1 \over 2}\epsilon^{ijk}f^{abc}(\alpha^b_j+a^b_j)(\alpha^c_k+a^c_k)\nonumber\\
=\beta^i_a[\alpha]+\epsilon^{ijk}(\partial_ja^a_k+f^{abc}\alpha^b_ja^c_k)+O(a^2).
\end{eqnarray}

\noindent
To make the physical content of the theory clear we will choose a reference connection $\alpha^a_i=\delta^a_i\alpha$, where $\alpha$ is a numerical constant.  Then we have

\begin{eqnarray}
\label{NOWEXPAND2}
B^i_a=\delta^i_a\alpha^2+\epsilon^{ijk}\partial_ja^a_k+\alpha(\delta^{ia}a^c_c-a^i_a)+\dots;~~C_{ae}=\delta_{ae}\alpha^3+\dots,
\end{eqnarray}

\noindent
where the dots signify higher order terms.  The canonical one form to linearized level, the second term of (\ref{ONEFOR2}), is given by

\begin{eqnarray}
\label{NOWEXPAND3}
\boldsymbol{\theta}_{Linear}=-{i \over G}\int_{\Sigma}d^3x\epsilon_{ae}B^i_e\dot{A}^a_i\nonumber\\
=-{i \over G}\int_{\Sigma}d^3x\epsilon_{ae}(\delta^i_e\alpha^2+\dots)(\dot{a}^a_i+\dots)
=-{i \over G}\alpha^2\int_{\Sigma}d^3x\epsilon_{ae}\dot{a}^a_e.
\end{eqnarray}

\noindent
To linearized order the theory exhibits a symplectic two form

\begin{eqnarray}
\label{NOWEXPAND31}
\boldsymbol{\Omega}_{Linear}=-{i \over G}\alpha^2\int_{\Sigma}d^3x{\delta\epsilon_{ae}}\wedge{\delta{a}_{ae}}\nonumber\\
=-{i \over G}\alpha^2\delta\Bigl(\int_{\Sigma}d^3x\epsilon_{ae}\delta{a}_{ae}\Bigr)=\delta\boldsymbol{\theta}_{Linear}.
\end{eqnarray}

\noindent
So at the unconstrained level one can read off the following elementary Poisson brackets from (\ref{NOWEXPAND31})

\begin{eqnarray}
\label{NOWEXPAND32}
\{a_{ae}(x,t),\epsilon_{bf}(y,t)\}=-i\Bigl({{\alpha^2} \over G}\Bigr)\delta_{ab}\delta_{ef}\delta^{(3)}(x,y).
\end{eqnarray}

\noindent
Since the constraints (\ref{CONSTRAINTTT}) are already of at least linear order in $\epsilon_{ae}$, then we need only expand them to zeroth order in $B^i_a$.  Hence the diffeomorphism constraint is given by

\begin{eqnarray}
\label{NOWEXPAND4}
H_i=\epsilon_{ijk}(\alpha^4\delta^j_a\delta^k_e)\epsilon_{ae}=\alpha^4\epsilon_{iae}\epsilon_{ae}=0,
\end{eqnarray}

\noindent
which implies that $\epsilon_{ae}=\epsilon_{ea}$ must be symmetric.  The Hamiltonian constraint to linearized order is given by

\begin{eqnarray}
\label{NOWEXPAND5}
\hbox{tr}\epsilon=0,
\end{eqnarray}

\noindent
which states that $\epsilon_{ae}$ is traceless to this order.  For the Gauss' law constraint we have

\begin{eqnarray}
\label{NOWEXPAND6}
G_a=\alpha^2\delta^i_e\partial_i\epsilon_{ae}+\alpha^3\delta_{be}\bigl(f_{abf}\delta_{ge}+f_{ebg}\delta_{af}\bigr)\epsilon_{fg}\nonumber\\
=\alpha^2\partial_e\epsilon_{ae}+\alpha^3f_{agf}\epsilon_{fg}=0.
\end{eqnarray}

\noindent
The second term on the right hand side of (\ref{NOWEXPAND6}) vanishes since $\epsilon_{ae}$ is symmetric from (\ref{NOWEXPAND4}), and the Gauss' law constraint reduces to

\begin{eqnarray}
\label{NOWEXPAND7}
\partial_e\epsilon_{ae}=0,
\end{eqnarray}

\noindent
which states that $\epsilon_{ae}$ is transverse.  Since upon implementation of the linearized constraints $\epsilon_{ae}$ is symmetric, traceless and transverse then it corresponds to a spin two field.

\subsection{Massless spin two polarizations}

We will now make contact with the conventional formalism, as is best seen in momentum space, using a plane waveform for $\epsilon_{ae}$.  From (\ref{NOWEXPAND4}) and (\ref{NOWEXPAND5}) the most general form for $\epsilon_{ae}$ is given by the parametrization of its diagonal and off-diagonal parts, $\varphi_f$ and $\Psi_f$ respectively\footnote{We have omitted the time dependence, since the initial value constraints are solved with respect to a given spatial hypersurface $\Sigma_t$ for each time $t$.}

\begin{displaymath}
\epsilon_{ae}=
\left(\begin{array}{ccc}
\varphi_1 & \Psi_3 & \Psi_2\\
\Psi_3 & \varphi_2 & \Psi_1\\
\Psi_2 & \Psi_1 & \varphi_3\\
\end{array}\right)
e^{\vec{k}\cdot\vec{r}},
\end{displaymath}

\noindent
subject to the tracelessness condition $\hbox{tr}\epsilon=\varphi_1+\varphi_2+\varphi_3=0$, where $\vec{k}=(k_1,k_2,k_3)$ is the wave vector of the gravitational wave.  The linearized Gauss' law constraint (\ref{NOWEXPAND7}) is given by

\begin{displaymath}
k_e\epsilon_{ae}=
\left(\begin{array}{ccc}
\varphi_1 & \Psi_3 & \Psi_2\\
\Psi_3 & \varphi_2 & \Psi_1\\
\Psi_2 & \Psi_1 & \varphi_3\\
\end{array}\right)
\left(\begin{array}{c}
k_1\\
k_2\\
k_3\\
\end{array}\right)
=
\left(\begin{array}{c}
0\\
0\\
0\\
\end{array}\right)
,
\end{displaymath}

\noindent
which can be rewritten as

\begin{displaymath}
\left(\begin{array}{ccc}
0 & k_3 & k_2\\
k_3 & 0 & k_1\\
k_2 & k_1 & 0\\
\end{array}\right)
\left(\begin{array}{c}
\Psi_1\\
\Psi_2\\
\Psi_3\\
\end{array}\right)
=-
\left(\begin{array}{ccc}
k_1 & 0 & 0\\
0 & k_2 & 0\\
0 & 0 & k_3\\
\end{array}\right)
\left(\begin{array}{c}
\varphi_1\\
\varphi_2\\
\varphi_3\\
\end{array}\right)
.
\end{displaymath}

\noindent
To make the physical content more apparent in terms of gravitation modes, let us use a wave vector of the form $\vec{k}=(k_1,0,0)$, which corresponds to a wave travelling
in the $\textbf{x}$ direction of a Cartesian coordinate system.  For $k_2=k_3=0$ this is given in matrix form by

\begin{displaymath}
\left(\begin{array}{ccc}
0 & 0 & 0\\
0 & 0 & k_1\\
0 & k_1 & 0\\
\end{array}\right)
\left(\begin{array}{c}
\Psi_1\\
\Psi_2\\
\Psi_3\\
\end{array}\right)
=-
\left(\begin{array}{ccc}
k_1 & 0 & 0\\
0 & 0 & 0\\
0 & 0 & 0\\
\end{array}\right)
\left(\begin{array}{c}
\varphi_1\\
\varphi_2\\
\varphi_3\\
\end{array}\right)
,
\end{displaymath}

\noindent
This yields the equations

\begin{eqnarray}
\label{WAVE30}
0=\varphi_1k_1;~~k_3\Psi_3=0;~~k_1\Psi_2=0,
\end{eqnarray}

\noindent
from which we have that $\varphi_1=\Psi_2=\Psi_3=0$.  But since $\epsilon_{ae}$ is traceless with $\varphi_1=0$, then $\varphi_3=-\varphi_2$.  The deviation matrix is then of the form

\begin{displaymath}
\epsilon_{ae}=\varphi
\left(\begin{array}{ccc}
0 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & -1\\
\end{array}\right)
e^{\vec{k}\cdot\vec{r}}+\Psi
\left(\begin{array}{ccc}
0 & 0 & 0\\
0 & 0 & 1\\
0 & 1 & 0\\
\end{array}\right)
e^{\vec{k}\cdot\vec{r}}.
\end{displaymath}

\noindent
We have obtained the two polarizations of a massless spin two field in $SO(3,C)$ language.  A similar result can be obtained for waves travelling in the  $\textbf{y}$ and the $\textbf{z}$ directions.\par
\indent
For the general case where the wave vector $\vec{k}$ is not aligned with the coordinate directions the, Gauss' law constraint can be written as

\begin{eqnarray}
\label{WAVE28}
\Psi_f=\hat{J}^g_f\varphi_g.
\end{eqnarray}

\noindent
This expresses the off-diagonal elements $\Psi_f$ as the image of the diagonal elements $\varphi_f$ with respect to a propagator $\hat{J}^g_f$.  When $\hat{J}^g_f$ exists, then equation (\ref{WAVE28}) in matrix form is given by

\begin{displaymath}
\left(\begin{array}{c}
\Psi_1\\
\Psi_2\\
\Psi_3\\
\end{array}\right)
=-
\left(\begin{array}{ccc}
0 & k_3 & k_2\\
k_3 & 0 & k_1\\
k_2 & k_1 & 0\\
\end{array}\right)^{-1}
\left(\begin{array}{ccc}
k_1 & 0 & 0\\
0 & k_2 & 0\\
0 & 0 & k_3\\
\end{array}\right)
\left(\begin{array}{c}
\varphi_1\\
\varphi_2\\
\varphi_3\\
\end{array}\right)
,
\end{displaymath}

\noindent
which hinges upon the ability to invert the off-diagonal matrix of wave vector components.  This is given by

\begin{displaymath}
\left(\begin{array}{c}
\Psi_1\\
\Psi_2\\
\Psi_3\\
\end{array}\right)
=(2k_1k_2k_3)^{-1}
\left(\begin{array}{ccc}
-k_1^3 & k_1k_2^2 & k_3^2k_1\\
k_1^2k_2 & -k_2^3 & k_2k_3^2\\
k_3^2k_1 & k_2^2k_3 & -k_3^3\\
\end{array}\right)
\left(\begin{array}{c}
\varphi_1\\
\varphi_2\\
\varphi_3\\
\end{array}\right)
,
\end{displaymath}

\noindent
whence one sees that we must have $k_1k_2k_3\neq{0}$.  Note that $\hat{J}^g_f$ does not exist for the previous case of propagation along the coordinate directions.\par
\indent
This solution for $\epsilon_{ae}$ contains two degrees of freedom per point and can be written completely in terms of the traceless diagonal elements $\varphi_g$ via the relation

\begin{eqnarray}
\label{WAVE29}
\epsilon_{ae}=\bigl((e^g)_{ae}+(E^f)_{ae}\hat{J}^g_f\bigr)\varphi_g\equiv(\hat{T}^g)_{ae}\varphi_g.
\end{eqnarray}

\noindent
In equation (\ref{WAVE29}), $\hat{T}^g_{ae}$ is an operator which implements an embedding map from the two dimensional space $(\varphi_1,\varphi_2)$ into the six dimensional space $\epsilon_{ae}$, taking the kinematics of the Gauss' law constraint into account.

\subsection{Quantization and Hilbert space of the linearized theory}

\noindent
We will now perform a quantization by promoting Poisson brackets (\ref{NOWEXPAND32}) to equal-time commutators

\begin{eqnarray}
\label{NOWEXPAND32}
\bigl[\hat{a}_{ae}(x,t),\hat{\epsilon}_{bf}(y,t)\bigr]=\alpha^{-2}(\hbar{G})\delta_{ab}\delta_{ef}\delta^{(3)}(x,y).
\end{eqnarray}

\noindent
Since the variables are complex, then to ensure square integrability of the wavefunctions of the auxilliary Hilbert space we will use a Gaussian measure

\begin{eqnarray}
\label{NOWEXPAND321}
D\mu=\prod_{x,a,e}\delta{a}_{ae}(x)\hbox{exp}\Bigl[-\mu\int_{\sigma}d^3x\overline{a}_{ae}(x)a_{ae}(x)\Bigr]
\end{eqnarray}

\noindent
for normalization, where $\mu$ is a numerical constant of mass dimension $[\mu]=1$.  For the auxilliary Hilbert space we will use holomorphic plane waves $\boldsymbol{\psi}$ which are eigenstates of the momentum operators, given by

\begin{eqnarray}
\label{NOWEXPAND322}
\boldsymbol{\psi}_{\lambda}[a]=\hbox{exp}\Bigl[-\alpha^4\mu^{-1}(\hbar{G})^{-2}\int_{\Sigma}d^3x\lambda^{*}_{ae}(x)\lambda_{ae}(x)\Bigr]\hbox{exp}\Bigl[\alpha^2(\hbar{G})^{-1}\int_{\Sigma}d^3x\lambda_{ae}(x)a_{ae}(x,t)\Bigr],
\end{eqnarray}

\noindent
where the pre-factor is a normalization factor and $\lambda_{ae}$ labels the state.  The action of the operators on (\ref{NOWEXPAND321}) are given by

\begin{eqnarray}
\label{NOWEXPAND323}
\hat{a}_{ae}(x,t)\boldsymbol{\psi}_{\lambda}=a_{ae}(x,t)\boldsymbol{\psi}_{\lambda};~~
\hat{\epsilon}_{ae}(x,t)\boldsymbol{\psi}=\alpha^{-2}(\hbar{G}){\delta \over {\delta{a}_{ae}(x,t)}}\boldsymbol{\psi}_{\lambda}.
\end{eqnarray}

\noindent
There are two possibilities for quantization of gravitons, depending on whether the Gauss' law propagator $\hat{J}^f_g$ exists or not.  In the latter case one may parametrize the momentum space degrees of freedom by

\begin{displaymath}
\epsilon_{ae}=\pi_1
\left(\begin{array}{ccc}
0 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & -1\\
\end{array}\right)
+\pi_2
\left(\begin{array}{ccc}
0 & 0 & 0\\
0 & 0 & 1\\
0 & 1 & 0\\
\end{array}\right)
.
\end{displaymath}

\noindent
Since the constraints (\ref{CONSTRAINTTT}) do not constrain the configuration space then we are free to choose $a_{ae}$, each choice tantamount to the choice of a gauge.  Let us make choose the connection

\begin{displaymath}
a_{ae}=a_1
\left(\begin{array}{ccc}
0 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & -1\\
\end{array}\right)
+a_2
\left(\begin{array}{ccc}
0 & 0 & 0\\
0 & 0 & 1\\
0 & 1 & 0\\
\end{array}\right)
.
\end{displaymath}

\noindent
Then the commutation relations (\ref{NOWEXPAND2}) reduce to

\begin{eqnarray}
\label{NOWEXPAND324}
\bigl[\hat{a}_1(x,t),\hat{\pi}_1(y,t)\bigr]=\bigl[\hat{a}_2(x,t),\hat{\pi}_2(y,t)\bigr]=\alpha^{-2}(\hbar{G})\delta^{(3)}(x,y).
\end{eqnarray}

\noindent
But since the phase space must have six dimensions per point at the level prior to implementation of the Hamiltonian constraint, we need a third variable $a$ with conjugate momentum $\pi=\hbox{tr}\epsilon$ satisfying the relation

\begin{eqnarray}
\label{NOWEXPAND38}
\bigl[\hat{a}(x,t),\hat{\pi}(y,t)\bigr]=\alpha^{-2}(\hbar{G})\delta^{(3)}(x,y).
\end{eqnarray}

\noindent
This enables us to directly implement the Hamiltonian constraint in the more general case where $\hat{J}_f^g$ exists, the Hamiltonian constraint can be implemented at the quantum level completely in terms of the diagonal
elements $\varphi_f$.  The quantum Hamiltonian constraint to linearized order is then given by

\begin{eqnarray}
\label{NOWEXPAND39}
\hat{H}\boldsymbol{\psi}=\alpha^{-2}(\hbar{G}){\delta \over {\delta{a}(x,t)}}\boldsymbol{\psi}=0,
\end{eqnarray}

\noindent
which states that $\boldsymbol{\psi}$ is independent of $a$.  If we interpret $a$ as a time variable on configuration space, then this means that $\boldsymbol{\psi}$ is independent of time.  Then the most general
solution is $\boldsymbol{\psi}[a_1,a_2]$, which depends on the two physical degrees of freedom which are orthogonal to the time direction.  In this case the wavefunctionals solving the constraints
are given by $\boldsymbol{\psi}[a_1,a_2]\indent{L}^2(a_1,a_2;D\mu)$, the set of square integrable functions of $a_1$ and $a_2$ in the measure (\ref{NOWEXPAND321}).\par
\indent

\section{Part Two: the full nonlinear theory}

Having demonstrated the existence of a well-defined Hilbert space structure for complex gravity at the linearized level on the phase space $\Omega_{Inst}$, we will now demonstrate the same for the full, nonlinear theory.\footnote{The intent is to show that the physical degrees of freedom of the full theory are preserved under linearization.}  First, we must show that the full nonlinearized version of the constraints (\ref{CONSTRAINTTT}) admit a solution at the classical level.  The diffeomorphism constraint is

\begin{eqnarray}
\label{NONLIN}
H_i=\epsilon_{ijk}B^j_kB^a_e\Psi_{ae}=(\hbox{det}B)(B^{-1})^f_i\epsilon_{fae}\Psi_{ae}=0.
\end{eqnarray}

\noindent
Since $(\hbox{det}B)\neq{0}$ by assumption, then equation (\ref{NONLIN}) states that $\Psi_{ae}=\Psi_{ea}$ is symmetric.  This is the case independently of linearization and holds for all
connections $A^a_i$ with $(\hbox{det}B)\neq{0}$.  According to \cite{WEYL}, a complex symmetric 3 by 3 matrix can be diagonalized when there exist three linearly independent eigenvectors.  In the case of $\epsilon_{ae}$ this is the case when $(\hbox{det}\epsilon)\neq{0}$, which enables us to write the following polar decomposition

\begin{eqnarray}
\label{NONLIN1}
\epsilon_{ae}=(e^{\theta\cdot{T}})_{af}\lambda_f(e^{-\theta\cdot{T}})_{fe}.
\end{eqnarray}

\noindent
In equation (\ref{NONLIN1}) $\vec{\theta}=(\theta^1,\theta^2,\theta^3)$ are three complex rotation parameters, which implement a transformation of the eigenvalues $\lambda_f=(\lambda_1,\lambda_2,\lambda_3)$ into a new Lorentz frame.\par
\indent
On account of the cyclic property of the trace, the Hamiltonian constraint can now be written completely in terms of the eigenvalues

\begin{eqnarray}
\label{NONLIN2}
H=\hbox{tr}\epsilon+{\Lambda \over 3}Var\epsilon+{{\Lambda^2} \over 3}\hbox{det}\epsilon=\lambda_1+\lambda_2+\lambda_3\nonumber\\
+{{2\Lambda} \over 3}\bigl(\lambda_1\lambda_2+\lambda_2\lambda_3+\lambda_3\lambda_1\bigr)+{{\Lambda^2} \over 3}\lambda_1\lambda_2\lambda_3=0.
\end{eqnarray}

\noindent
Just as we applied the quantization procedure to $\hbox{tr}\epsilon=0$ in the linerized theory, we will as well apply the quantization procedure to (\ref{NONLIN2}), which brings into question the role of the Gauss' law constraint.  Using the parametrization (\ref{NONLIN1}) the full nonlinear Gauss' law constraint can be written as

\begin{eqnarray}
\label{NONLIN3}
G_a=\textbf{w}_e\{\lambda_f(e^{-\theta\cdot{T}})_{fa}(e^{-\theta\cdot{T}})_{fe}\}=0.
\end{eqnarray}

\noindent
Note that (\ref{NONLIN2}) is independent of $\vec{\theta}$ and depends only on $\lambda_f$, which we will regard as the physical degrees of freedom.  From this perspective, we will regard (\ref{NONLIN3}) as a condition for determining the angles $\vec{\theta}$.  The set of connections $A^a_i$ defines an equivalence class of angles $\vec{\theta}\equiv\vec{\theta}[\vec{\lambda};A]$ labelled by each
triple of eigenvalues $\lambda_f$ satisfying (\ref{NONLIN2}).\par
\indent
All that remains then is to show that the canonical structure upon the indentification (\ref{NONLIN1}) reduces accordingly to a canonical structure on the reduced phase space under $H_i$ and $G_a$.  This can be seen from the relation

\begin{eqnarray}
\label{NONLIN4}
\epsilon_{ae}B^i_e\dot{A}^a_i=\lambda_f((e^{-\theta\cdot{T}})_{fe}B^i_e)((e^{-\theta\cdot{T}})_{fa}\dot{A}^a_i,
\end{eqnarray}

\noindent
whence the $SO(3,C)$ matrices rotate the internal indices of $B^i_a$ and $\dot{A}^a_i$.  Since the velocity $\dot{A}^a_i$ lives in the tangent space to configuration space, then it transforms the same way as $B^i_a$ under $SO(3,C)$ gauge transformations, namely inhomogeneously.  We can then make the identifications

\begin{eqnarray}
\label{NONLIN5}
\dot{a}^a_i=(e^{-\theta\cdot{T}})_{fa}\dot{A}^a_i;~~b^i_a=(e^{-\theta\cdot{T}})_{fe}B^i_a,
\end{eqnarray}

\noindent
and regard the new connection $a^a_i$ as specially adapted to an `intrinsic' $SO(3,C)$ frame corresponding to the eigenvalues $\lambda_f$.  So we can redefine a new theory, starting at the level after implementation of the diffeomorphism and Gauss' law constraints, with canonical one form

\begin{eqnarray}
\label{NONLIN6}
\boldsymbol{\theta}_{Kin}=\int_{\Sigma}d^3x\lambda_fb^i_f\delta{a}^f_i
\end{eqnarray}

\noindent
where the angles $\vec{\theta}$ are ignorable.  However, (\ref{NONLIN6}) is presently not in a form suitable for quantization.  This is because its functional variation

\begin{eqnarray}
\label{THEACTION9}
\delta\boldsymbol{\theta}_{Kin}=
=\int_{\Sigma}d^3x\Bigl[b^i_f{\delta\lambda_f}\wedge{\delta{a}^f_i}+\lambda_f\epsilon^{ijk}{(D_j\delta{a}^f_k)}\wedge{\delta{a}^f_i}\Bigr],
\end{eqnarray}

\noindent
where $D_i\equiv(D^{ae})_i=\delta^{ae}\partial_i+f^{abe}a^b_i$ is the covariant derivative with respect to the connection $a^f_i$, in direct analogy to (\ref{THEACTION9}) is not of symplectic form owing to the presence of the second term.  Additionally, there is a mismatch in degrees of freedom between the momentum space and the configurations space.  To have a cotangent bundle structure, we need three configuration space degrees of freedom corresponding to the eigenvalues $\lambda_f$.  Since the constraints do not place any restriction on the connection $A^a_i$, then provided $\hbox{det}\neq{0}$ we are free the choose any connection we wish.  For the purposes of this paper we will
limit ourself to a diagonal connection $A^a_i=\delta^a_iA^a_a$ with no summation over $a$.  We will see that this choice eliminates the second term of (\ref{THEACTION9}), while providing the requisite canonical structure necessary for quantization of the full, nonlinear theory.

\subsection{Canonical structure}

\noindent
We will now put in place the canonical structure required to quantize the fluctuations about the Kodama state.  After rotation of all variables into the intrinsic $SO(3,C)$ frame, one sees that a choice of diagonal configuration space variables admits a canonical structure.  The canonical one form is given by\footnote{We have identified the diagonal elements of $\epsilon_{ae}$ with its eigenvalues, which means that the variables have been adapted
to a $SO(3,C)$ frame solving the Gauss' law constraint.}

\begin{eqnarray}
\label{JACOBI}
\boldsymbol{\theta}_{Kin}={i \over G}\int_{\Sigma}d^3x\epsilon_{ae}B^i_e\delta{A}^a_i\nonumber\\
={i \over G}\int_{\Sigma}d^3x\biggl(\epsilon_{11}A^2_2A^3_3\delta{A}^1_1+\epsilon_{22}A^3_3A^1_1\delta{A}^2_2+\epsilon_{33}A^1_1A^2_2\delta{A}^3_3\biggr).
\end{eqnarray}

\noindent
Now define densitized momentum variables $\widetilde{\epsilon}_{ae}=\epsilon_{ae}(\hbox{det}A)$, where $(\hbox{det}A)\neq{0}$.  Hence we have

\begin{eqnarray}
\label{JACOBI1}
\widetilde{\epsilon}_{11}=\epsilon_1(A^1_1A^2_2A^3_3);~~\widetilde{\epsilon}_{22}=\epsilon_2(A^1_1A^2_2A^3_3);~~\widetilde{\epsilon}_{33}=\epsilon_3(A^1_1A^2_2A^3_3).
\end{eqnarray}

\noindent
In the densitized variables (\ref{JACOBI1}), then (\ref{JACOBI}) is given by

\begin{eqnarray}
\label{JACOBI2}
\boldsymbol{\theta}_{Kin}={i \over G}\int_{\Sigma}d^3x
\biggl(\widetilde{\epsilon}_{11}\Bigl({{\delta{A}^1_1} \over {A^1_1}}\Bigr)+
\widetilde{\epsilon}_{22}\Bigl({{\delta{A}^2_2} \over {A^2_2}}\Bigr)+
\widetilde{\epsilon}_{33}\Bigl({{\delta{A}^3_3} \over {A^3_3}}\Bigr)\biggr).
\end{eqnarray}

\noindent
Next, rewrite (\ref{JACOBI2}) in the form

\begin{eqnarray}
\label{JACOBI3}
\boldsymbol{\theta}_{Kin}={i \over G}\int_{\Sigma}d^3x
\biggl((\widetilde{\epsilon}_{11}-\widetilde{\epsilon}_{33}){{\delta{A}^1_1} \over {A^1_1}}+
(\widetilde{\epsilon}_{22}-\widetilde{\epsilon}_{33}){{\delta{A}^2_2} \over {A^2_2}}\nonumber\\
+\widetilde{\epsilon}_{33}\Bigl({{\delta{A}^1_1} \over {A^1_1}}+{{\delta{A}^2_2} \over {A^2_2}}+{{\delta{A}^3_3} \over {A^3_3}}\Bigr)\biggr)
\end{eqnarray}

\noindent
and make the following change of variables

\begin{eqnarray}
\label{JACOBI4}
\Pi_1=\Bigl({\Lambda \over {3a_0^3}}\Bigr)(\widetilde{\epsilon}_{11}-\widetilde{\epsilon}_{33});~~
\Pi_2=\Bigl({\Lambda \over {3a_0^3}}\Bigr)(\widetilde{\epsilon}_{22}-\widetilde{\epsilon}_{33});~~\Pi=\Pi_1=\Bigl({\Lambda \over {3a_0^3}}\Bigr)\widetilde{\epsilon}_{33}
\end{eqnarray}

\noindent
where $a_0$ is a numerical constant of mass dimension $[a_0]=1$.  Then for the configuration space make the definition

\begin{eqnarray}
\label{JACOBI5}
{{\delta{A}^1_1} \over {A^1_1}}=\delta{X};~~{{\delta{A}^2_2} \over {A^2_2}}=\delta{Y};~~{{\delta{A}^1_1} \over {A^1_1}}+{{\delta{A}^2_2} \over {A^2_2}}+{{\delta{A}^3_3} \over {A^3_3}}=\delta{T}.
\end{eqnarray}

\noindent
Equation (\ref{JACOBI5}) provides global coordinates $(X,Y,T)$ on the kinematic onfiguration space $\Gamma_{Kin}$, given by

\begin{eqnarray}
\label{JACOBI6}
X=\hbox{ln}\Bigl({{A^1_1} \over {a_0}}\Bigr);~~Y=\hbox{ln}\Bigl({{A^2_2} \over {a_0}}\Bigr);~~T=\hbox{ln}\Bigl({{A^1_1A^2_2A^3_3} \over {a_0^3}}\Bigr).
\end{eqnarray}

\noindent
The canonical one form corresponding to (\ref{JACOBI7}) is given by

\begin{eqnarray}
\label{ONEFORM7}
\boldsymbol{\theta}_{Kin}=\Bigl({{3ia_0^3} \over {G\Lambda}}\Bigr)\int_{\Sigma}d^3x\int{dt}\bigl(\Pi\dot{T}+\Pi_1\dot{X}+\Pi_2\dot{Y}\bigr).
\end{eqnarray}

\noindent
With the variables as defined, (\ref{ONEFORM7}) yields a symplectic two form

\begin{eqnarray}
\label{JACOBI7}
\boldsymbol{\Omega}_{Kin}
=\Bigl({{3ia_0^3} \over {G\Lambda}}\Bigr)\int_{\Sigma}d^3x\Bigl({\delta\Pi_1}\wedge{\delta{X}}+{\delta\Pi_2}\wedge{\delta{Y}}+{\delta\Pi}\wedge{\delta{T}}\Bigr)=\delta\boldsymbol{\theta}_{Kin}.
\end{eqnarray}

\noindent
The mass dimensions of the dynamical variables are

\begin{eqnarray}
\label{JACOBI8}
[\Pi_1]=[\Pi_2]=[\Pi]=[X]=[Y]=[T]=0,
\end{eqnarray}

\noindent
and we have the Poisson brackets

\begin{eqnarray}
\label{BRACK}
\{\hat{T}(x,t),\hat{\Pi}(y,t)\}=\{\hat{X}(x,t),\hat{\Pi}_1(y,t)\}=\{\hat{Y}(x,t),\hat{\Pi}_2(y,t)\}=\Bigl({{G\Lambda} \over {3ia_0^3}}\Bigr)\delta^{(3)}(x,y)
\end{eqnarray}

\noindent
The advantage of the choice of dimensionless variables is that the Hamiltonian constraint, the third equation of (\ref{CONSTRAINTTT}), can be written as a dimensionless equation.  First rewrite it in terms of the densitized variables

\begin{eqnarray}
\label{CANNOW}
(\hbox{det}A)^{-1}\hbox{tr}\widetilde{\epsilon}+{\Lambda \over 3}(\hbox{det}A)^{-2}Var\widetilde{\epsilon}+{{\Lambda^2} \over 3}(\hbox{det}A)^{-3}\hbox{det}\widetilde{\epsilon}=0.
\end{eqnarray}

\noindent
Multiplication of (\ref{CANNOW}) by ${\Lambda \over 3}(\hbox{det}A)^3a_0^{-9}$ and using (\ref{JACOBI4}) enables the Hamiltonian constraint to be written as

\begin{eqnarray}
\label{CANNOW1}
H=e^{2T}\bigl(3\Pi+\Pi_1+\Pi_2\bigr)+e^T\bigl(3\Pi^2+2(\Pi_1+\Pi_2)\Pi\nonumber\\
+\Pi_1\Pi_2\bigr)+3\Pi(\Pi+\Pi_1)(\Pi+\Pi_2)=0.
\end{eqnarray}

\noindent
Let us define the following operators

\begin{eqnarray}
\label{OPERATE}
Q^{(1)}=\Pi+{1 \over 3}(\Pi_1+\Pi_2);\nonumber\\
Q^{(2)}=\Pi^2+{2 \over 3}(\Pi_1+\Pi_2)\Pi+{1 \over 3}\Pi_1\Pi_2;\nonumber\\
O=\Pi(\Pi+\Pi_1)(\Pi+\Pi_2).
\end{eqnarray}

\noindent
Then upon dividing by a factor of $3$ the Hamiltonian constraint can be written as

\begin{eqnarray}
\label{OPERATE1}
H=e^TQ^{(1)}+e^{2T}Q^{(2)}+O=0.
\end{eqnarray}

\subsection{Quantization and auxilliary Hilbert space}

To pass over into the quantum theory we promote Poisson brackets (\ref{BRACK}) to commutators.  Upon quantization, the variables $\Pi$, $\Pi_1$ and $\Pi_2$ become
promoted to operators $\hat{\Pi}$, $\hat{\Pi}_1$, and $\hat{\Pi}_2$ and $T$, $X$ and $Y$ to operators $\hat{T}$, $\hat{X}$ and $\hat{Y}$ satisfying the nontrivial equal time commutation relations

\begin{eqnarray}
\label{QUANTIZATION}
\bigl[\hat{T}(x,t),\hat{\Pi}(y,t)\bigr]=\bigl[\hat{X}(x,t),\hat{\Pi}_1(y,t)\bigr]=\bigl[\hat{Y}(x,t),\hat{\Pi}_2(y,t)\bigr]=\mu\delta^{(3)}(x,y)
\end{eqnarray}

\noindent
where we have defined the constant

\begin{eqnarray}
\label{HAVEDEFINED}
\mu=\Bigl({{\hbar{G}\Lambda} \over {3a_0^3}}\Bigr).
\end{eqnarray}

\noindent
In the functional Schr\"odinger representation, holomorphic in $X$, $Y$ and $T$, the operators act respectively by multiplication

\begin{eqnarray}
\label{QUANTUM1}
\hat{T}(x,t)\boldsymbol{\psi}=T(x,t)\boldsymbol{\psi};~~
\hat{X}(x,t)\boldsymbol{\psi}=X(x,t)\boldsymbol{\psi};~~
\hat{Y}(x,t)\boldsymbol{\psi}=Y(x,t)\boldsymbol{\psi}
\end{eqnarray}

\noindent
and by functional differentiation

\begin{eqnarray}
\label{QUANTUM2}
\hat{\Pi}(x,t)\boldsymbol{\psi}=\mu{\delta \over {\delta{T}(x,t)}}\boldsymbol{\psi};~~
\hat{\Pi}_1(x,t)\boldsymbol{\psi}=\mu{\delta \over {\delta{X}(x,t)}}\boldsymbol{\psi};~~
\hat{\Pi}_2(x,t)\boldsymbol{\psi}=\mu{\delta \over {\delta{Y}(x,t)}}\boldsymbol{\psi}.
\end{eqnarray}

\noindent
The wavefunction $\boldsymbol{\psi}$ is determined from the following resolution of the identity

\begin{eqnarray}
\label{QUANTUM3}
I=\prod_x\int{\delta\mu}\bigl\vert{X},Y,T\bigr>\bigl<X,Y,T\bigr\vert,
\end{eqnarray}

\noindent
whence the state diagonal in the configuration variables is given by

\begin{eqnarray}
\label{QUANTUM4}
\boldsymbol{\psi}(X,Y,T)=\bigl<X,Y,T\bigl\vert\boldsymbol{\psi}\bigr>.
\end{eqnarray}

\noindent
Since the variables are complex, we choose a measure Gaussian in $X$ and $Y$ to ensure normalizable wavefunctions.  This is given by\footnote{Note, we do not include a measure in $T$, because we will interpret $T$ as a time variable and one does not normalize a wavefunction in time.}

\begin{eqnarray}
\label{NOMRAL}
D\mu(X,Y)=\prod_x\delta{X}\delta\overline{X}\delta{dY}\delta\overline{Y}\hbox{exp}\Bigl[-\nu^{-1}\int_{\Sigma}d^3x(\vert{X}\vert^2+\vert{Y}\vert^2)\Bigr]
\end{eqnarray}

\noindent
where $\nu$ is a numerical constant of mass dimension $[\nu]=-3$.  For the auxilliary Hilbert space we will use eigenstates of the momentum operators

\begin{eqnarray}
\label{EIGENSTATE}
\boldsymbol{\psi}=\bigl<X,Y,T\bigl\vert\alpha,\beta,\lambda\bigr>
=e^{\mu^{-1}(\alpha\cdot{X}+\beta\cdot{Y}+\lambda\cdot{T})},
\end{eqnarray}

\noindent
where the dot signifies an integration over 3-space $\Sigma$ as in

\begin{eqnarray}
\label{EIGENSTATE1}
U\cdot{V}=\int_{\Sigma}d^3xU(x)V(x).
\end{eqnarray}

\noindent
So we have that

\begin{eqnarray}
\label{EIGENSTATE2}
\hat{\Pi}_1\bigl\vert\alpha,\beta,\lambda\bigr>=\alpha\bigl\vert\alpha,\beta,\lambda\bigr>;~~
\hat{\Pi}_2\bigl\vert\alpha,\beta,\lambda\bigr>=\beta\bigl\vert\alpha,\beta,\lambda\bigr>;~~
\hat{\Pi}\bigl\vert\alpha,\beta,\lambda\bigr>=\lambda\bigl\vert\alpha,\beta,\lambda\bigr>.
\end{eqnarray}

\noindent
The overlap between two unnormalized states is given, using (\ref{NOMRAL}) for a measure, by

\begin{eqnarray}
\label{OVERLAP}
\bigl\vert\bigl<{\alpha,\beta}\bigl\vert{\alpha^{\prime},\beta^{\prime}}\bigr>\bigr\vert^2
=e^{-\nu\mu^{-2}\vert\alpha-\alpha^{\prime}\vert^2}e^{-\nu\mu^{-2}\vert\beta-\beta^{\prime}\vert^2}e^{\mu^{-1}(\lambda\cdot{T}+\lambda^{*}\cdot\overline{T})}.
\end{eqnarray}

\noindent
There is always a nontrivial overlap between the states, which is a consequence of Gaussian measure needed for the holomorphic representation.  Hence, the states as defined by (\ref{EIGENSTATE}) form an overcomplete set.\par
\indent
Note that the operators (\ref{OPERATE}) have the following action on the auxilliary states

\begin{eqnarray}
\label{EIGENSTATE3}
\hat{Q}^{(1)}\vert\alpha,\beta,\lambda\bigr>=\bigl(\lambda+{1 \over 3}(\alpha+\beta)\bigr)\bigl\vert\alpha,\beta,\lambda\bigr>;\nonumber\\
\hat{Q}^{(2)}\vert\alpha,\beta,\lambda\bigr>=(\lambda+\gamma^{-}(\alpha,\beta))(\lambda+\gamma^{+}(\alpha,\beta))\bigl\vert\alpha,\beta,\lambda\bigr>;\nonumber\\
\hat{O}\vert\alpha,\beta,\lambda\bigr>=\lambda(\lambda+\alpha)(\lambda+\beta)\bigl\vert\alpha,\beta,\lambda\bigr>,
\end{eqnarray}

\noindent
where $\gamma^{\pm}(\alpha,\beta)$ are the roots of

\begin{eqnarray}
\label{EIGENSTATE31}
\lambda^2+{2 \over 3}(\alpha+\beta)\lambda+{1 \over 3}\alpha\beta=0.
\end{eqnarray}

\noindent
For the quantum Hamiltonian constraint we will choose an operator ordering with the momenta to the left of the coordinates, as in

\begin{eqnarray}
\label{EIGENSTATE4}
\hat{H}=\hat{Q}^{(1)}e^T+\hat{Q}^{(2)}e^{2T}+\hat{O}.
\end{eqnarray}

\subsection{Construction of the states}

\noindent
The quantum Hamiltonian constraint can be written as

\begin{eqnarray}
\label{THEQUANTUM}
\hat{O}\bigl\vert\boldsymbol{\psi}\bigr>=-\bigl(\hat{Q}^{(1)}e^T+\hat{Q}^{(2)}e^{2T}\bigr)\bigl\vert\boldsymbol{\psi}\bigr>.
\end{eqnarray}

\noindent
Let there be states $\boldsymbol{\psi}_0\indent{Ker}\{\hat{O}\}$.  Then acting on (\ref{THEQUANTUM}) with $\hat{O}^{-1}$, assumed to be invertible, we obtain

\begin{eqnarray}
\label{THEQUANTUM1}
\bigl\vert\boldsymbol{\psi}\bigr>=\bigl\vert\boldsymbol{\psi}_0\bigr>-\bigl(\hat{O}^{-1}\hat{Q}^{(1)}e^T+\hat{O}^{-1}\hat{Q}^{(2)}e^{2T}\bigr)\bigl\vert\boldsymbol{\psi}\bigr>.
\end{eqnarray}

\noindent
Equation (\ref{THEQUANTUM1}) can be rearranged into the form

\begin{eqnarray}
\label{THEQUANTUM2}
\bigl(1+\hat{q}_1+\hat{q}_2\bigr)\bigl\vert\boldsymbol{\psi}\bigr>=\bigl\vert\boldsymbol{\psi}_0\bigr>,
\end{eqnarray}

\noindent
where we have defined

\begin{eqnarray}
\label{THEQUANTUM3}
\hat{q}_1\equiv\hat{O}^{-1}\hat{Q}^{(1)};~~\hat{q}_2\equiv\hat{O}^{-1}\hat{Q}^{(2)}.
\end{eqnarray}

\noindent
Then (\ref{THEQUANTUM2}) can be rearranged into the form

\begin{eqnarray}
\label{THEQUANTUM4}
\bigl\vert\boldsymbol{\psi}\bigr>=\Bigl({1 \over {1+\hat{q}_1+\hat{q}_2}}\Bigr)\bigl\vert\boldsymbol{\psi}_0\bigr>.
\end{eqnarray}

\noindent
For labelling purposes let $\bigl\vert\lambda\bigr>\indent{Ker}\{\hat{O}\}$.  The action of the individual operators is are given by

\begin{eqnarray}
\label{LEGION9}
\hat{q}_1\bigl\vert\lambda\bigr>=\hat{O}^{-1}\hat{Q}^{(1)}e^T\bigl\vert\lambda\bigr>
=\hat{O}^{-1}\hat{Q}^{(1)}\bigl\vert\lambda+\mu^{\prime}\bigr>
=E^{(1)}_{\lambda+\mu^{\prime}}(\alpha,\beta)\bigl\vert\lambda+\mu^{\prime}\bigr>,
\end{eqnarray}

\noindent
and for $\hat{q}_2$ by

\begin{eqnarray}
\label{LEGION10}
\hat{q}_2\bigl\vert\lambda\bigr>=\hat{O}^{-1}\hat{Q}^{(2)}e^{2T}\bigl\vert\lambda\bigr>
=\hat{O}^{-1}\hat{Q}^{(2)}\bigl\vert\lambda+2\mu^{\prime}\bigr>
=E^{(2)}_{\lambda+2\mu^{\prime}}(\alpha,\beta)\bigl\vert\lambda+2\mu^{\prime}\bigr>,
\end{eqnarray}

\noindent
where

\begin{eqnarray}
\label{WHERE}
E^{(1)}_{\lambda+k\mu^{\prime}}(\alpha,\beta)={{\lambda+k\mu^{\prime}+{1 \over 3}(\alpha+\beta)} \over {(\lambda+k\mu^{\prime})(\lambda+k\mu^{\prime}+\alpha)(\lambda+k\mu^{\prime}+\beta)}};\nonumber\\
E^{(2)}_{\lambda+k\mu^{\prime}}(\alpha,\beta)={{(\lambda+k\mu^{\prime}+\gamma^{-}(\alpha,\beta))
(\lambda+k\mu^{\prime}+\gamma^{+}(\alpha,\beta))} \over {(\lambda+k\mu^{\prime})(\lambda+k\mu^{\prime}+\alpha)(\lambda+k\mu^{\prime}+\beta)}}.
\end{eqnarray}

\noindent
Then the full solution to (\ref{THEQUANTUM4}) can be written as

\begin{eqnarray}
\label{LEGION12}
\bigl\vert\boldsymbol{\psi}\bigr>=\Bigl(\sum_{n=0}^{\infty}(-1)^n(\hat{q}_1+\hat{q}_2)^n\Bigr)\bigl\vert\lambda\bigr>
=\sum_{n=1}^{\infty}(-1)^n\hat{q}_{k_n}\hat{q}_{k_{n-1}}\dots\hat{q}_{k_2}\hat{q}_{k_1}\bigl\vert\lambda\bigr>.
\end{eqnarray}

\noindent
At each order $n$ there are $2^n$ terms in the expansion, and the summation must be made over all permutations of indices with the operator ordering preserved.  The indices $k_n$ take on the value of $1$ or $2$.  To illustrate for the first two terms, starting with the first term we have

\begin{eqnarray}
\label{LEGION13}
\hat{q}_{k_1}\bigl\vert\lambda\bigr>=E^{(k_1)}_{\lambda+k_1\mu^{\prime}}\bigl\vert\lambda+k_1\mu^{\prime}\bigr>,
\end{eqnarray}

\noindent
where we have suppressed the $(\alpha,\beta)$ labels to avoid cluttering up the notation.  For the second term we have

\begin{eqnarray}
\label{LEGION14}
\hat{q}_{k_2}\hat{q}_{k_1}\bigl\vert\lambda\bigr>=E^{(k_1)}_{\lambda+k_1\mu^{\prime}}E^{(k_2)}_{\lambda+(k_1+k_2)\mu^{\prime}}
\bigl\vert\lambda+(k_1+k_2)\mu^{\prime}\bigr>.
\end{eqnarray}

\noindent
The $N^{th}$ term is given by

\begin{eqnarray}
\label{LEGION15}
\prod_{n=1}^N\hat{q}_{k_n}\bigl\vert\lambda\bigr>=\prod_{n=1}^NE^{(k_n)}_{\lambda+(k_1+k_2+\dots{k}_n)\mu^{\prime}}\bigl\vert\lambda+(k_1+k_2+\dots{k}_n)\mu^{\prime}\bigr>
\end{eqnarray}

\noindent
Our main concern is the convergence of the full series (\ref{LEGION12}), which we will show using norm inequalities.  For large $k$, equation (\ref{WHERE}) implies that

\begin{eqnarray}
\label{LEGION16}
\vert{E}^{(k_1)}_{\lambda+k\mu^{\prime}}\vert\leq{1 \over {k\mu^{\prime}}}.
\end{eqnarray}

\noindent
So for large $N$, the labels $\alpha$ and $\beta$ become unimportant and each term in the product in (\ref{LEGION15}) satisfies the following bound

\begin{eqnarray}
\label{LEGION17}
\bigl\vert{E}^{k_n}_{\lambda+(k_1+k_2+\dots{k}_N)\mu^{\prime}}\bigr\vert\leq{1 \over {N\mu^{\prime}}}.
\end{eqnarray}

\noindent
But there are $2^N$ terms, corresponding to the different permutations of the indices $k_1k_2\dots{k}_N$.  So the full series is bounded by

\begin{eqnarray}
\label{LEGION18}
\sum_N\Bigl(\prod_{n=1}^N\hat{q}_{k_n}\Bigr)\bigl\vert\lambda\bigr>\leq\sum_N{{2^N} \over {N!}}\Bigl({{e^{2\vert{T}\vert}} \over {\mu^{\prime}}}\Bigr)^N
=\hbox{exp}\Bigl({{2e^{2\vert{T}\vert}} \over {\mu^{\prime}}}\Bigr),
\end{eqnarray}

\noindent
which is a convergent function.  The solution to the Hamiltonian constraint converges, since each term of (\ref{LEGION18}) vanishes as $N\rightarrow\infty$, $\forall{T}$.

\section{Conclusion}

\noindent
This paper provides a first step toward the realization of Penrose's idea of the nonlinear graviton.  We have demonstrated the existence of gravitons, both at the linearized and at the non linearized level, using a new description of nonmetric complex general relativity defined on the phase space $\Omega_{Inst}=(\widetilde{\sigma}^i_a,A^a_i)$.  The conventional method in the linearization of gravity is to expand the spacetime metric in fluctuations about a fixed background, typically a flat Minkowski spacetime.  In this paper we have chosen for a background spacetimes whose semiclassical orbits arise from the Kodama state $\boldsymbol{\psi}_{Kod}$.  We have expanded the theory
on $\Omega_{Inst}$ in fluctuations about $\boldsymbol{\psi}_{Kod}$, constructing the Hilbert space of states annihilated by the constraints both at the linearized and at the nonlinearized level.  In each case there were two physical degrees of freedom per point, implying the preservation of these degrees of freedom under linearization.  Additionally, it is hoped that the results of this paper have provided an addressal of the issues surrounding the Kodama
state $\boldsymbol{\psi}_{Kod}$ raised in \cite{WITTEN1} and in \cite{NORMKOD}.  Clearly, the Kodama state provides a natural background for quantizable flucutations when one uses the phase space variables on $\Omega_{Inst}$.  It is clear also that when restricted to the diagonal connection $A^a_i=\delta^a_iA^a_a$ used in this paper, that the integrand of the Kodama state is proportional to $(\hbox{det}A)=A^1_1A^2_2A^3_3$, which plays the role of a time variable on configuration space.  The addressal of the issue of the normalizabilty of $\boldsymbol{\psi}_{Kod}$ then is simply that one does not normalize a wavefunction in time.  However, one should normalize wavefunctions with respect to the physical degrees of freedom orthogonal to the time direction.  And we have done so using a Gaussian measure for normalization, which ensures square integrability of the wavefunctions on the gravitational Hilbert space.  The well-definedness of these wavefunctions with respect to their time dependence arises due to the convergence of the infinite series constituting the solution to the quantum Hamiltonian constraint.  The notion of the Chern--Simons functional as a time variable on configuration space has been proposed in \cite{SOO} and \cite{SOO1}.


\section{Introduction}

Through the understanding of D-branes and string duality (see, for example, the text book by Polchinski~\cite{Pol}), a more fundamental underling theory called M-theory has been expected and each string theory is realized as various limits in M-theory. Although this M-theory is expected, we have only poor understanding. Its low energy effective theory is given by the eleven dimensional supergravity and it would be a theory of membrane compared from a string theory which is a theory of string. The strong string coupling limit of IIA string theory opens up the eleventh space and is described by M-theory. It is conjectured that M-theory in an infinite momentum frame is described by BFSS matrix model~\cite{Banks:1996vh}.

The quantization of a membrane worldvolume theory is very challenging
and one of difficulty is the nonlocality associated with the deformation
of membrane without changing its volume (see, for example, a review by Taylor~\cite{tay}). In string theory, the open
string and closed string duality appears in many situations and has
provided many powerful techniques. One important idea behind BFSS matrix
model is also based on the open-closed string duality and the worldvolume theory of
multiple D0-brane, (which is governed by open string fluctuations) describes the target space dynamics, i.e. the gravity in the target space (which is governed by closed string fluctuations). Therefore another
direction to approach to M-theory is studying the effective action for
multiple Membrane.

Recently Bagger and Lambert (BL) constructed a new three dimensional N=8
superconformal theory using a three algebra~\cite{Bagger:2006sk} (see also~\cite{Gustavsson:2007vu} by Gustavsson). Since BL theory
satisfies all the properties which multiple membrane should have, it is
expected to describe multiple membranes. For BL theory with $SO(4)$ gauge symmetry, the moduli space~\cite{Bagger:2007vi, Lambert:2008et, Distler:2008mk} is discussed and the theory is conjectured to describe a two membrane system in an orbifold space~\cite{Lambert:2008et, Distler:2008mk}. Soon after the work
by Bagger-Lambert, Aharony, Bergman, Jafferis and Maldacena (ABJM) generalize their idea and constructed three
dimensional N=6 superconformal theories which contain BL theory as a special case~\cite{Aharony:2008ug}. ABJM also show the membrane configuration in
the eleven dimensional orbifold space time ($R^{1,2}\times (R^8/Z_k)$ and $k$ is the
level of Chern-Simons coupling) for their N=6 theory
with $U(N)\times U(N)$ gauge symmetry. Since the matter fields are charged under $U(1)$ in $U(N)=U(1)\times SU(N)$ and then $U(1)$ is not decoupled from $SU(N)$ in ABJM theory, BL theory with $SO(4)=SU(2)\times SU(2)$ may not
describe a multiple membrane system. However $U(1)$ gauge coupling is IR
free and the BL theory and ABJM theory with $U(2)\times U(2)$ gauge symmetry
may be connected by a renormalization flow. The target space superalgebra is studied in BL theory with the central element which suggests the target space is an eleven dimensional space~\cite{Passerini:2008qt}. It is also discussed that BL theory with the Nambu-Poisson algebra turns out be an action of single M5-brane~\cite{Ho:2008nn}. Therefore we may still
expect that BL theory describes multiple membranes. If so,
it worths studying a possibility that multiple membrane dynamics can describe
a target space dynamics, as parallel to that the D-brane dynamics describes the target space dynamics.

One important consequence of open-closed string duality is probing the
target space from D-brane scattering using D-brane effective theory,
i.e. Super Yang-Mills theory (SYM)~\cite{Douglas:1996yp} where the one loop effective potential reproduces the Newton potential in the target space. We then expect a similar correspondence in M-theory, and in
this note we study the one loop effective potential around a  relatively
moving membrane background in BL theory and see if the potential is understood as the Newton potential in the target space. Since the target space is discussed to be an orbifold $R^{1,2}\times (R^8/Z_k)$, we study a small $k$ case in order to probe the whole spacetime otherwise the one
spacial direction is effectively compactified in a large $k$
case ($Z_k$ is a subgroup of a $U(1)$ and we can always define one spacial direction generated by this $U(1)$ for any value of $k$. We call this direction the compactified direction.).
However the coupling constant is proportional to $1/k$, the theory
is strongly coupled for a small $k$ and the perturbation will not be a
good expansion. Despite of that we still expect the one loop effective
potential qualitatively gives a correct answer, since we expect that an
one loop open membrane amplitude can be reinterpreted as a tree closed
membrane amplitude, and also we treat a small deviation from BPS state. This situation is similar to BFSS matrix model. One should take a large N limit (N is the size of matrix) to recover the eleven dimensional Lorentz invariance, and the matrix model should give a controllable description at a shorter distance than the Plank
length~\cite{Douglas:1996yp} at which we may expect the spacetime no longer looks like a eleven (or ten) dimensional classical spacetime. Despite of these, even for a finite N (N is the size of matrix), the one loop effective potential reproduces the Newton potential.

With this expectation in mind, we study the membrane scattering and compute the one loop effective potential in BL theory. The membrane scattering in ABJM theory is mentioned in~\cite{Berenstein:2008dc} and that in Lorentzian BL theory is discussed in~\cite{Verlinde:2008di}. From our calculations, we find that the potential is understood as the Newton potential and the total dimension of target space, which is read from the exponent of the power law behavior, is ten. The potential does not show a different behavior depending on the value of $k$. Thus in a large $k$ limit, the potential consistently becomes the one computed from D2-brane SYM action. This result suggests that the open membrane, described as a perturbation from the background, always wraps the compactified direction even $k$ is finite and small, and the BL theory can probe only
remaining ten dimensions.

Using the three dimensional SYM action for multiple
D2-branes, the membrane scattering has been discussed.
Polchinski and Pouliot discussed the
membrane scattering with momentum transfer along the eleventh
direction (M-momentum transfer) corresponds to an instanton process~\cite{Polchinski:1997pz}. We will see that the same happens in BL theory.

In the next section, we review the moduli space of BL theory and
introduce the general form of small velocity which corresponds to motion
of membranes. In Sec.3, we compute the one loop effective potential
around backgrounds with several velocities and discuss what BL theory
can probe about the target space. In Sec.4, we summarize and conclude. \\\\

During the preparation of present paper, we received the paper~\cite{Baek:2008ws}. The authors calculate 1-loop effective potential of ABJM theory and see an agreement with the Newton potential on $AdS_4 \times S^7/{\bf Z}_k$.

\vspace{10mm}

\section{Moduli space and position of Membranes}

In this note, we treat BL theory with $SO(4)$ gauge symmetry.
The moduli space of this theory has been studied in~\cite{Lambert:2008et, Distler:2008mk} at which the scalar potential vanishes. The Lagrangian is
\begin{align}
 {\cal L} &= -\frac{1}{2} D_\mu X^{A,I} D^\mu X_{A,I}
 +\frac{i}{2}\overline{\Psi}^A \Gamma^\mu D_\mu \Psi_A
 +\frac{ig}{4}f_{ABCD}\overline{\Psi}^B \Gamma^{IJ} X^{C,I}X^{D,J} \Psi^A
 \nonumber\\
 &\;\;\;
 -\frac{g^2}{12}\left[ f_{ABCD}X^{A,I}X^{B,J}X^{C,K}
 \right] \left[ f_{EFG}^{~~~~~D}X^{E,I}X^{F,J}X^{G,K}
 \right]
 \nonumber\\
 &\;\;\;
 +
  \frac{g}{2}\epsilon^{\mu\nu\lambda}\left[
  f_{ABCD}A_\mu^{AB}\partial_\nu A_{\lambda}^{CD}
  +\frac{2g}{3}f_{AEF}^{~~~~~G}f_{BCDG}A_\mu^{AB}A_\nu^{CD}A_\lambda^{EF}
 \right] ,
 \\
 D_\mu X^{A,I} &= \partial_\mu X^{A,I} + g \tilde{A}_{\mu B}^{A}X^{B,I}
 ,
 \hspace{3ex}
 \tilde{A}_{\mu B}^{A} \equiv f^A_{~~BCD}A_\mu^{CD}X^{B,I}
 ,
\end{align}
where $f_{ABCD}$ is the structure constant for the three algebra and
$f_{ABCD}=\epsilon_{ABCD}$, ($A=1,\cdots, 4$ etc), for ${\cal A}_4$ algebra which is equivalent with $SO(4)$ gauge symmetry. The indices $I,J,K (=1,\cdots, 8)$ are those of $SO(8)$ global symmetry and the spacetime signature is $(-,+,+)$. This Lagrangian has $N=8$ superconformal symmetry and supersymmetry requires the coupling constants are same and the value of coupling constant $g=2\pi/k$ is quantized ($k\in Z$), because of Chern-Simons term.

After a suitable gauge transformation, the vacuum configuration (with the gauge fields and fermions are zero) is
\begin{align}
  \langle X^{A,I} \rangle &=\left(\begin{array}{c}
  0\\ 0\\ r_1^I \\ r_2^I
  \end{array}\right)
  ,
  \hspace{3ex}
  \begin{array}{l}
  A=1,\cdots,4,
  \\
  I = 1,\cdots, 8,
  \end{array}
\end{align}
where $r_1^{I}$ and $r_2^{I}$ are real values, and the index $A=1,\cdots, 4$ is the index for the three algebra.
There are two sets of eight values $r_1^I$ and $r_2^I$, and then $r_1^I$ and $r_2^I$ are related with the position of two membranes in the eight dimensional transverse directions in the target space.
The moduli space should be divided by the gauge symmetry. The discrete symmetry ${\cal O}(2,Z)\in SO(4)$ act on two vectors like:
\begin{align}
  \left(\begin{array}{cc}
  -1&0\\
  0 &1
  \end{array}\right)
  &\hspace{3ex}
  :\hspace{3ex}
  r_1^I \rightarrow - r_1^I ,
  \hspace{3ex}
   r_2^I \rightarrow r_2^I ,
  \label{d1}
  \\
  \left(\begin{array}{cc}
  1&0\\
  0 &-1
  \end{array}\right)
  &\hspace{3ex}
  :\hspace{3ex}
  r_1^I \rightarrow r_1^I , \hspace{3ex}
  r_2^I \rightarrow -r_2^I ,
  \\
  \left(\begin{array}{cc}
  ~0&1~\\
  ~1&0~
  \end{array}\right)
  &\hspace{3ex}
  :\hspace{3ex}
  r_1^I \rightarrow r_2^I , \hspace{3ex}
  r_2^I \rightarrow r_1^I ,
  \label{d3}
\end{align}
and the moduli space becomes $( (R^8/Z_2)\times (R^8/Z_2) )/Z_2$. The moduli space should be further divided by the continuous gauge symmetry. Since the gauge fields have the Chern-Simons coupling, the continuous symmetry which keeps the Chern-Simons term invariant and $\tilde{A}_\mu^{A,B}=0$ is $Z_k\in U(1)$:
\begin{align}
  z^{I} \rightarrow e^{i\theta}z^I ,
  \hspace{3ex}
  \theta = \frac{\pi n}{k} ,
  \hspace{3ex}
  n\in Z ,
  \label{s1}
\end{align}
where $z^I = r_1^I + i r_2^I$. Then the moduli space is $( R^8\times R^8 )/D_{2k}$ where $D_{2k}$ is a dihedral group and for $k=1$ it is just $( R^8\times R^8 )/ ( Z_2\times Z_2 )$ and the target space is expected to $R^{1,2}\times (R^8/Z_2)$.
This $Z_k$ is a subgroup of $U(1)$, and this $U(1)$ generate one spacial direction and we call this direction the compactified direction even for a finite $k$. In the large $k$ limit, this direction is identified and the BL theory reduces to the weakly coupled IIA theory~\cite{Distler:2008mk}.

Using $SO(8)$ global symmetry, the form of $\langle X^{A,I} \rangle$ can be written
\begin{align}
  \langle X^{A,I} \rangle &=\left(\begin{array}{ccccc}
  0 &\cdots &0 & 0 & 0\\
  0 &\cdots &0 & 0 & 0\\
  0 &\cdots &0 & b_0 & 0\\
  0 &\cdots &0 & 0 & a_0
  \end{array}\right)
  ,
\end{align}
\begin{figure}[t]
\begin{center}
\includegraphics[height=7cm]{bl1.eps}
\put(-26,104){$X^7$}
\put(-151,186){$X^8$}
\put(-60,92){$(b_0,0)$}
\put(-141,164){$(0,a_0)$}
\vspace{-5ex}
\caption{The positions of two membranes in $X^{I}$ coordinate, ($b_0,0$) and ($0,a_0$). The ellipse is the compactified direction generated by the $U(1)$ subgroup. The aria of the ellipse (the shaded region) is $\pi a_0b_0$.
}
\label{f1}
\end{center}
\end{figure}
and the position of membranes in $X^I$ coordinates and the compactified direction are plotted in Fig.~\ref{f1}.

When $a_0\neq 0$ and $b_0=0$, one can integrate out massive gauge fields and obtain $SU(2)$ (plus free $U(1)$) SYM theory, i.e. D2-brane action, at the leading order in $1/a_0$~\cite{Mukhi:2008ux}. If $b_0$ then turns on, $SU(2)$ gauge symmetry is broken down to $U(1)$ and the masses of massive gauge bosons are given $ga_0b_0$. Therefore $ga_0b_0=g_{YM}L$ where $g_{YM}$ is the gauge coupling of SYM and $L$ is the distance between two branes. Since only the product $g_{YM}L$ appears in the Lagrangian, there is an ambiguity for $g_{YM}$ (and $L$). We know there is a symmetry under the exchange of $a_0$ and $b_0$, and in $g\rightarrow 0, (k\rightarrow \infty)$ limit the theory reduces to the D2-brane system, and thus we choose $g_{YM}=g$ and $L=a_0b_0$ in this note.

Since we would like to discuss the scattering of membranes, we introduce the small time dependence into $X^{A,I}$. Solving the equations of motion for $\tilde{A}_\mu^{AB}$ and $X^{A,I}$ under $\tilde{A}_\mu^{A,B}=0$, we obtain
\begin{align}
  \langle X^{A,I} \rangle &=\left(\begin{array}{ccccc}
  0 &\cdots &0 & 0 & 0\\
  0 &\cdots &0 & 0 & 0\\
  v_1t &\cdots &v_6t & b_0+v_7t & v_8t\\
  u_1t &\cdots &u_6t & u_7t & a_0+u_8t
  \end{array}\right)
  ,
\end{align}
and except that $v_8=(b_0u_7)/a_0$, all the $v$ and $u$ are free. We note that the constraint $v_8=(b_0u_7)/a_0$ comes from the equation of motion for $\tilde{A}^{3,4}_\mu$ which is the gauge field corresponds to the continuous symmetry~\eqref{s1} and means that the momentum along the compactified direction is set to be zero. This may be the similar situation to that one light-cone direction is compactified and the momentum along that direction is set to be constant in BFSS matrix model.

\section{Membranes scattering and gravitational potential}

In the previous section, we review the moduli space and the general form of velocity which satisfies the equation of motion. In this section we study the one loop effective potential around the background with non-zero velocities. In string theory, D-brane scattering has been discussed using SYM theory and the one loop effective potential reproduces the Newton potential in the target space. Thus we expect we can probe the target space from the one loop potential which we will compute in this section.

From the relation which comes from the gauge field $\tilde{A}_\mu^{3,4}$, one spacial direction is special and there is no momentum transfer along the direction. Although we expect that the target space is eleven dimensions, this observation implies we can only probe ten dimensions, not eleven dimensions. On the other hand, the action has $SO(8)$ global symmetry and (supersymmetric) conformal symmetry and we may expect we can probe eleven dimension according to the discussion by~\cite{Banks:1997uu}. Thus we compute the gravitational potential by applying the idea of computing the gravitational potential from SYM theory, to clarify which observation is correct.\\

Before going to the calculation,
we notice that the regularization in Chern-Simons theory is not simple. A dimensional regularization naively breaks the gauge invariance due to the difficulty of analytic continuation of $\epsilon_{\mu\nu\rho}$. Another regularization is adding Yang-Mills term and a careful study on the regularization methods has been done in~\cite{Chen:1992ee}. The one loop corrections in BL theory have been discussed  with these regularization procedure~\cite{Gustavsson:2008bf}. In our calculation of one loop graphs, a dimensional regularization can be adapted.

\subsection{For $v_7\neq 0$ and $u_8\neq 0$}

We first study the case where only $v_7$ and $u_8$ are non-zero. In
order to study the one loop effective potential, we just have to keep
quadratic terms in the Lagrangian around the background. Then the relevant terms in the Lagrangian becomes $ {\cal L} ={\cal L}_1+{\cal L}_2 +{\cal L}_f$,
\begin{align}
 {\cal L}_{1}&=
 \sum_{\alpha=1,2} g \epsilon^{\mu\nu\rho} \tilde{A}^{\alpha}_\mu \partial_\nu \tilde{B}^{\alpha}_\rho
 -\frac{1}{2} [ \partial_\mu X^{2,7} + g b \tilde{A}^1_{\mu} ]^2
 -(\partial_t b) g \tilde{A}^1_{t} X^{2,7}
 -\frac{1}{2} [ \partial_\mu X^{1,8} + g a \tilde{B}^1_{\mu} ]^2
 -( \partial_t a) g \tilde{B}^1_{t} X^{1,8}
\nonumber\\
 &\;\;\;
 -\frac{1}{2} [ \partial_\mu X^{1,7} - g b \tilde{A}^2_{\mu} ]^2
 +(\partial_t b) g \tilde{A}^2_{t} X^{1,7}
 -\frac{1}{2} [ \partial_\mu X^{2,8} + g a \tilde{B}^2_{\mu} ]^2
 -( \partial_t a) g \tilde{B}^2_{t} X^{2,8}
 ,
 \nonumber
 \\
 {\cal L}_2&=
 g \epsilon^{\mu\nu\rho} \tilde{A}^3_\mu \partial_\nu \tilde{B}^3_\rho
 -\frac{1}{2} [ \partial_\mu X^{3,7} ]^2
 -\frac{1}{2} [ \partial_\mu X^{3,8} + g a \tilde{B}^3_{\mu} ]^2
 -( \partial_t a) g \tilde{B}^3_{t} X^{3,8}
 -\frac{1}{2} [ \partial_\mu X^{4,7} - g b \tilde{B}^3_{\mu} ]^2
 \nonumber
 \\
 &\;\;\;
 +(\partial_t b) g \tilde{B}^3_{t} X^{4,7}
 -\frac{1}{2} [ \partial_\mu X^{4,8} ]^2
 ,
\nonumber
\\
 {\cal L}_f &=
 \frac{1}{2}
 X^{A,I} ( \Box -g^2a^2b^2 ) X^{A,I}
 +\frac{1}{2}
 X^{A',I} \Box X^{A',I}
 +\overline{\Psi}^{A''} \Gamma^\mu \partial_\mu \Psi_{A''}
 +\frac{i}{2}gab [ \overline{\Psi}^2 \Gamma^{78} \Psi^1 - \overline{\Psi}^1 \Gamma^{78} \Psi^2 ]
  ,
 \nonumber
 \\
 &\;\;\;
  (A=1,2,
  \hspace{2ex}
  A'=3,4,
    \hspace{2ex}
  A''=1,\cdots,4,
    \hspace{2ex}
  I=1,\cdots,6
)
  \nonumber
\end{align}
where $a\equiv a_0+u_8t$ and $b\equiv b_0+v_7t$, and we have used the following notation
\begin{align}
 \tilde{A}_\mu^\alpha \equiv \frac{1}{2}\epsilon^{\alpha\beta\gamma}\tilde{A}_{\mu \beta\gamma}
 ,\hspace{3ex}
 \tilde{B}_\mu^\alpha \equiv \tilde{A}_\mu^{\alpha 4}
 ,
 \hspace{3ex}
 ( \alpha = 1,\cdots,3, \mbox{ etc} )
 .
\end{align}
\begin{figure}[t]
\begin{center}
\includegraphics[height=7cm]{bl2.eps}
\vspace{-7ex}
\caption{The arrows denote the direction of velocity.
}
\label{f2}
\end{center}
\end{figure}
\noindent
In this case the direction of velocity is normal to the compactified direction (Fig.~\ref{f2}).

We integrate out $\tilde{A}_\mu^3$ which gives that $\tilde{B}_\mu^3$ is written by a derivative of scalar field, i.e. $\tilde{B}_\mu^3=\partial_\mu B$.
Substitute this expression into ${\cal L}_2$, we obtain
\begin{align}
 {\cal L}_2&=
 -\frac{1}{2} [ \partial_\mu X^{3,7} ]^2
 -\frac{1}{2} [ \partial_\mu X^{3,8} + g (\partial_\mu aB) ]^2
 -\frac{1}{2} [ \partial_\mu X^{4,7} - g (\partial_\mu bB) ]^2
 -\frac{1}{2} [ \partial_\mu X^{4,8} ]^2
 ,
\end{align}
after by using a partial integral. Thus we have four massless scalar fields and the contribution to the one loop effective action from this part becomes
\begin{align}
 V_2^{\rm 1\; loop}(a_0,b_0;u_8,v_7) = \int \! d^3x \; 4\times \frac{1}{2} \ln \det \Box
 .
\end{align}
We can also easily compute the contribution from ${\cal L}_f$ which are twelve massive scalars with the mass $gab$, sixteen massless fermion and eight massive fermions with the mass$^2$ $g^2a^2b^2\pm g\partial_t (ab)$. Then we obtain
\begin{align}
 V_f^{\rm 1 \; loop}(a_0,b_0;u_8,v_7) &=  \int \! d^3x \;
 12\times \frac{1}{2} \ln \det ( \Box - g^2a^2b^2 )
 +12\times \frac{1}{2} \ln \det \Box
 -16\times \frac{1}{2} \ln \det \Box
 \nonumber \\
 &\;\;\;\; \;\;\;\;
 -8\times \frac{1}{2} \Big[ \ln \det ( \Box - g^2a^2b^2 +g(\partial_t ab) )
 + \ln \det ( \Box - g^2a^2b^2 -g(\partial_t ab) ) \Big]
 .
\end{align}
Now we study ${\cal L}_{1}$. We similarly integrate out $\tilde{B}_\mu^{a}$ using the equation of motion and we obtain,
\begin{align}
 {\cal L}_{1}&= -\frac{1}{4a^2} [ \partial_{\mu}\tilde{A}_{\nu}^{1} - \partial_{\nu}\tilde{A}^{1}_{\mu} ]^2 - \frac{1}{2}[\partial_{\mu} X^{2,7}+gb\tilde{A}_{t}^{1}]^2
 +(\partial_t b)g\tilde{A}^{1}_{t}X^{2,7}
 \nonumber \\
 &\;\;\;
 -\frac{1}{4a^2} [ \partial_{\mu}\tilde{A}_{\nu}^{2} - \partial_{\nu}\tilde{A}_{\mu}^{2} ]^2-\frac{1}{2}[\partial_{\mu} X^{1,7}-gb\tilde{A}_{2}]^2
 -(\partial_t b)g\tilde{A}_{t}X^{1,7}
 .
\end{align}
The Lagrangian ${\cal L}_1$ is exactly same as the quadratic part of two D2-brane action with the time dependent gauge coupling $a$.  Thus we immediately see that if $u_8=0$, the one loop effective potential is exactly same as that of two D2-brane scattering with the gauge coupling $a_0$ and the distance
between two D2 brane in $X^7$ direction is $gb$. Then in this case, we have
\begin{align}
 V_{1}^{\rm 1 \; loop}(a_0,b_0;u_8=0,v_7) &= \int \! d^3x \;
 2\times \frac{1}{2} \Big[
 \ln \det ( \Box - g^2a^2b^2 +(2g\partial_t ab) )
 \nonumber \\
 &\;\;\;\; \;\;\;\;
 +\ln \det ( \Box - g^2a^2b^2 -(2g\partial_t ab) )
 \Big]
 ,
\end{align}
and in total the one loop effective potential by expanding $v_7t \ll b_0$ is
\begin{align}
 V^{\rm 1 \; loop}(a_0,b_0;u_8=0,v_7) &= \int \! \frac{d^3p}{(2\pi)^3} \;
 \frac{ 2 (ga_0\partial_t b)^4 }{ ( p^2+g^2a_0^2b^2 )^4}
 +{\cal O}((v_7t)^6).
\end{align}
We notice that the terms with the second order in $v_7$ cancel out.
It gives the following potential at the leading order
\begin{align}
 V^{\rm 1 \; loop}(a_0,b_0;u_8=0,v_7) &=
 c_{YM} \frac{ (a_0\partial_t b)^4 }{ ga_0^5b_0^5 }
 = c_{YM} \frac{ (\partial_t L)^4 }{ g_{YM}L^5 },
\end{align}
where $c_{YM}$ is the numerical coefficient computed from D2-brane scattering using SYM theory, and $g_{YM}=g$ and $L=a_0b$ from the matching with D2-brane action in the $g\rightarrow 0$ limit.
Since there is a discrete symmetry which exchange $a$ and $b$, the one loop effective potential of the case $v_7=0$ is same as that of two D2-brane scattering with the gauge coupling $b_0$ and the distance $a$. Then in this case we have a same form
\begin{align}
 V^{\rm 1 \; loop}(a_0,b_0;u_8,v_7=0) &= \int \! \frac{d^3p}{(2\pi)^3} \;
 \frac{ (g\partial_t ab_0)^4 }{ ( p^2+g^2a^2b_0^2 )^4}
 \sim c_{YM}\frac{ (\partial_t ab_0)^4 }{ ga_0^5b_0^5 }
 = c_{YM}\frac{ (\partial_t L)^4 }{ g_{YM}L^5}
 ,
\end{align}
where $g_{YM}=g$ and $L=ab_0$.

For both $v_7$ and $u_8$ are non-zero, the calculation is involved and we introduce a proper gauge fixing term and compute the one loop effective potential. In order that the computation becomes simple, first we rescale $\tilde{A}_{\mu}^{\alpha=1,2}= aA_{\mu}^{\alpha=1,2}$ to have canonical kinetic terms
 $\frac{1}{2}A^{\alpha}[\Box-(gab)^2]A^{\alpha}$.
Next, we introduce the following gauge fixing
\begin{align}
 {\cal L}_{gf} &=
-\frac{1}{2}\left[\partial^{\mu}A^{1}_{\mu}+gabX^{2,7}-\frac{ \partial^{\mu}a}{a} A^{1}_{\mu}\right]^2
-\frac{1}{2}\left[\partial^{\mu}A^{2}_{\mu}-gabX^{1,7}-\frac{\partial^{\mu}a   }{a^2}A^{2}_{\mu}\right]^2.
 \end{align}
 The ghost Lagrangian may be suggested from ${\cal L}_{gf}$ as
\begin{eqnarray}
{\cal L}_{gh} = \sum_{\alpha=1,2}
\tilde{\bar{c}}^{\alpha}a\left[\partial^{\mu} \frac{1}{a}\partial_{\mu} -g^2ab^2-\frac{\partial^{\mu} a}{a^2}\partial_{\mu} \right]\tilde{c}^{\alpha}
,
\label{ghlag}
\end{eqnarray}
which follows from the gauge symmetry of ${\cal L}_{1}$:
\begin{eqnarray}
&&\delta A^{1}_{\mu}=\frac{1}{a}\partial_{\mu}\Lambda^1, \quad \delta X^{2,7}=-gb\Lambda^1.
\nonumber \\
&&\delta A^{2}_{\mu}=\frac{1}{a}\partial_{\mu}\Lambda^2, \quad \delta X^{2,8}=gb\Lambda^2.
\end{eqnarray}
However, note that these ghost may allow a background dependent field rescaling
$\tilde{c}\rightarrow f(a,b)\tilde{c}$ and $\tilde{\bar{c}}\rightarrow f(a,b)^{-1}\tilde{\bar{c}}$ with some function $f(a,b)$. The normalization is fixed such that the ghost Lagrangian has the supersymmetry after adding superpartners appropriately. In stead of fixing the normalization from supersymmetry, we can determine the correct normalization from the requirement that the total Lagrangian has the discrete symmetry under the exchange~\eqref{d1}-\eqref{d3}. It
is simply achieved by the ghost redefinition $\tilde{c}=ac$ and $\tilde{\bar{c}}=a^{-1}\bar{c}$ in (\ref{ghlag}), then we claim that correct ghost Lagrangian is
\begin{eqnarray}
{\cal L}_{gh}=\sum_{\alpha=1,2}\bar{c}^{\alpha}\left[\Box-g^2a^2b^2-2\frac{\partial^{\mu}a\partial_{\mu}a}{a^2}\right]c^{\alpha}.
\label{aght}
\end{eqnarray}
Then ${\cal L}_1+{\cal L}_{gf}+{\cal L}_{gh}$ becomes
\begin{eqnarray}
&&{\cal L}_1+{\cal L}_{gf}+{\cal L}_{gh} =
\sum_{\alpha=1,2}\frac{1}{2}X^{\alpha,7}(\Box-g^2a^2b^2)X^{\alpha,7}  +2g\partial^{\mu}(ab)\tilde{A}^{1}_{\mu}X^{2,7}
-2g\partial^{\mu}(ab)\tilde{A}^{2}_{\mu}X^{1,7}
\nonumber \\
&&\hspace{6ex}
 +\sum_{\alpha=1,2}\left[
\frac{1}{2}\tilde{A}^{\alpha\;\mu}(\Box-g^2a^2b^2)\tilde{A}^{\alpha}_{\mu}+\frac{1}{2}\frac{\partial^{\mu}a\partial^{\nu}a}{a^2}\tilde{A}^{\alpha}_{\mu}\tilde{A}^{\alpha}_{\nu}-\frac{\partial^{\mu}a\partial_{\mu}a}{a^2}\tilde{A}^{\alpha\;\mu}\tilde{A}^{\alpha}_{\mu}\right]
\nonumber \\
&&\hspace{6ex}
 +\sum_{\alpha=1,2}\bar{c}^{\alpha}\left[\Box-g^2a^2b^2-2\frac{\partial^{\mu}a\partial_{\mu}a}{a^2}\right]c^{\alpha}
 .
\end{eqnarray}
We compute the one loop effective action as  a perturbation with  $v=\partial_t(ab)$ and $\partial_t a$. The terms proportional to $(\partial_t a)^2$ and $(\partial_t a)^4$ cancel out between the gauge fields and ghosts, and because of this the ghost action~\eqref{aght} is consistent with the discrete symmetry~\eqref{d1}-\eqref{d3}. We can see the second order in terms of velocity $v=\partial_t(ab)$ cancels out as expected from supersymmetry.
This is because the boson loop contribution from ${\cal L}_{1}+{\cal L}_{gf}+{\cal L}_{gh}+{\cal L}_f$ is
\begin{align}
\int \! \frac{d^3p}{(2\pi)^3} \; \frac{4g^2[\partial(ab)]^2}{(p^2+g^2a^2b^2)^2}
\end{align}
and it is canceled by the fermion loop contribution from ${\cal L}_f$.
We can also easily see that the third order of $v$ vanishes and the potential starts from the fourth order in $v$,
\begin{align}
 V^{\rm 1\;loop}(a_0,b_0;u_8,v_7) &\sim c_{YM} \frac{[(ub_0+a_0v)]^4}{ga_0^5b_0^5}
 = c_{YM}\frac{(\partial_t L)^4}{g_{YM}L^5}
 .
\end{align}
In summary, we obtain that the form of one loop effective potential is given by $(\partial_t L)^4/L^5$  and the exponent $5$ for $L$ is consistent with the gravitational potential in ten dimensional space.

\vspace{2ex}

From the above result that there are no $1/a^6$ or $1/b^6$ terms in the potential, when $a_0\neq 0$ and $b_0=0$ the potential vanishes at the one loop,
\begin{align}
 V^{\rm 1\;loop}(a_0,b_0=0;u_8,v_7=0) &= 0
 . \label{a-pote}
\end{align}
Since Bagger-Lambert theory is a superconformal theory, the canonical dimension of $a$ is half and the possible form for the potential has the following form
\begin{align}
 V_{\rm eff}(a_0,b_0=0;u_8,v_7=0) \propto \frac{\partial_t a}{a^6}
 ,
\end{align}
If the coefficient is not zero, we may claim that $a$ is the distance between two membranes and the target space is a eleven dimensional space from a similar argument on a scale invariant $SO(8)$ symmetric theory~\cite{Banks:1997uu}. But, as (\ref{a-pote}), the coefficient is zero in Bagger-Lambert theory.

\vspace{1ex}

These results suggest that the membrane fluctuations  connecting two
membranes always wrap the compactified direction generated by $\tilde{A}_\mu^{3,4}$  even $k$ is finite, and therefore the one loop effective potential only probes ten dimensions. Since in the large $a_0$ limit the action at the leading terms in $1/a_0$ is same as the action for D2-branes, this result is natural.

\subsection{For $v_8\neq 0$ and $u_7\neq 0$}

We study the case $v_8\neq 0$ and $u_7=a_0v_8/b_0\neq 0$. In the previous case,  the membranes are pulled normal to the compactified direction. On the other in the case $v_8\neq 0$ and $u_7\neq 0$, the direction of velocity is tangent to the compactified direction (Fig.~\ref{f3}), (but notice that the momentum along the compactified direction is always zero).
\begin{figure}[t]
\begin{center}
\includegraphics[height=7cm]{bl3.eps}
\vspace{-7ex}
\caption{The arrows denote the direction of velocity.
}
\label{f3}
\end{center}
\end{figure}
\noindent
We may expect the result is different from the previous case.

The relevant term of Lagrangian after the redefinition
$\tilde{A}_\mu^\alpha = a_0A_\mu^\alpha$ and $\tilde{B}_\mu^\alpha = b_0 B_\mu^\alpha$ becomes $ {\cal L} = {\cal L}_1 + {\cal L}_2 + {\cal L}_f$,
\begin{align}
 {\cal L}_1 &=
  \sum_{\alpha=1,2} gL \epsilon^{\mu\nu\rho} A^{\alpha}_\mu \partial_\nu B^{\alpha}_\rho
 -\frac{1}{2} [ \partial_\mu X^{1,7} - gL {A}^2_{\mu} + gVt{B}_\mu^1]^2
 -\frac{1}{2} [ \partial_\mu X^{2,7} + gL {A}^1_{\mu} + gVt{B}_\mu^2 ]^2
 \nonumber \\
 &\;\;\;
 -\frac{1}{2} [ \partial_\mu X^{1,8} + gL {B}^1_{\mu}  - gVt{A}_\mu^2]^2
 -\frac{1}{2} [ \partial_\mu X^{2,8} + gL {B}^2_{\mu}  + gVt{A}_\mu^1]^2
 \nonumber \\
 &\;\;\;
 - V g ( {B}_t^1 X^{1,7} + {B}_t^2 X^{2,7})
 - V g ( -{A}_t^2 X^{1,8} + {A}_t^1 X^{2,8})
 ,
 \label{L1}
 \\
 {\cal L}_2 &=
 gL \epsilon^{\mu\nu\rho} {A}^3_\mu \partial_\nu {B}^3_\rho
 - V g {B}_t^3 X^{3,7} + V g {B}_t^3 X^{4,8}
 -\frac{1}{2} [ \partial_\mu X^{3,7} + gVt{B}_\mu^3]^2
 -\frac{1}{2} [ \partial_\mu X^{4,8} - gVt{B}_\mu^3]^2
 \nonumber \\
 &\;\;\;
 -\frac{1}{2} [ \partial_\mu X^{4,7} - gL {B}^3_{\mu} ]^2
 -\frac{1}{2} [ \partial_\mu X^{3,8} + gL {B}^3_{\mu} ]^2
 ,
 \label{L3}
 \\
 {\cal L}_f &=
 \frac{1}{2} X^{A,I} \Big[ \Box -g^2 (L - \frac{V^2}{L}t^2 )^2 \Big] X^{A,I}
 +\frac{1}{2}  X^{A',I} \Box X^{A,I}
 +\frac{i}{2} \overline{\Psi}^{A''} \Gamma^\mu \partial_\mu \Psi_{A''}
 \nonumber
 \\
 &\;\;\;
 +\frac{i}{2}g(L-\frac{V^2}{L}t^2) ( \overline{\Psi}^2 \Gamma^{78} \Psi^1 - \overline{\Psi}^1 \Gamma^{78} \Psi^2 )
 ,
 \nonumber
 \\
 &\;\;\;
 (A=1,2, \hspace{2ex}
 A'=3,4, \hspace{2ex}
 A''=1,\cdots,4 \hspace{2ex}
 I=1,\cdots,6 ),
 \label{Lf}
\end{align}
where $L=a_0b_0$ and $V=a_0v_8=b_0u_7$. The background always appears in the combination $L$ and $V$ and this Lagrangian can not be understood as SYM with time dependent gauge coupling and/or time dependent Higgs fields after integrating out $B_\mu^\alpha$ fields.
In this case the 1-loop effective potential becomes (We discuss on the calculation of the potential in appendix~A.)
\begin{eqnarray}
 V^{\rm 1\;loop}(L,V)&=&\frac{1}{g\pi}\int d^3x\;\left[\frac{V^4}{4L^5}-\frac{g^2V^4}{2L^3}t^2\right].
\end{eqnarray}
In $u_7\rightarrow 0$ with fixed $V$ limit it becomes D2-brane like potential.
Again it is suggested that 2-branes feel large ten dimension through this potential. \\

\vspace{2ex}

Before closing this section, we give two comments.

(1) Since the physical mass scale is $ab-u_7v_8t^2$, we expect that if $ab-u_7v_8t^2=a_0b_0$ is kept fixed the effective
potential is zero. However $ab-u_7v_8t^2=a_0b_0$ implies all the velocities $u_{7,8}$ and $v_{7,8}$ are zero.

(2) We look again at the equations of motion for the gauge fields $\tilde{B}_\mu^\alpha$ and $\tilde{A}_\mu^\alpha$ (with fermions are zero) which are
\begin{align}
  0 & =
   - X^{4,I} D_\mu X^{\alpha,I}
  + X^{\alpha,I} D_\mu X^{4,I}
  +\frac{1}{2} \epsilon^{\mu\nu\rho} ( \tilde{F}_{\nu\rho}^\alpha -g\epsilon^{\alpha\beta\gamma}\tilde{B}_\nu^\beta\tilde{B}_\rho^\gamma   ) ,
  \label{be}
  \\
  0 & =
   \epsilon_{\alpha\beta\gamma} X^{\beta,I} D_\mu X^{\gamma,I}
  +\epsilon^{\mu\nu\rho} ( \partial_\nu \tilde{B}_\rho^\alpha -g\epsilon^{\alpha\beta\gamma}\tilde{A}_\nu^\beta\tilde{B}_\rho^\gamma   ) ,
\end{align}
where $\tilde{F}^{\alpha}_{\mu\nu}$ is $SU(2)$ gauge field strength constructed by $\tilde{A}_\mu^\alpha$.
Then the momentum along the compactified direction is non zero
($-X^{4,I}D_\mu X^{3,I}+X^{3,I}D_\mu X^{4,I}\neq 0$)
 when $\tilde{A}_\mu^\alpha$ has a magnetic monopole configuration (with $\tilde{B}_\mu^\alpha =0$). (The monopole instanton configuration in ABJM theory is discussed in~\cite{Hosomichi:2008ip}.) This is consistent with the membrane scattering from three dimensional SYM with M-momentum discussed by Polchinski-Pouliot~\cite{Polchinski:1997pz}. Therefore we expect that higher loop contributions do not change the form of leading potential, $\propto (\partial_t L)^4/L^5$ and the eleventh direction cannot be probed perturbatically. The eleventh direction can be probed through a non-perturbative process.

\section{Conclusion and Discussions}

In this note, we studied membrane scattering from Bagger-Lambert theory and read out the dimensions of the target space from the one loop effective potential. We understand the membranes propagating between two membranes always wrap on the one spacial direction which becomes the compactified direction when the level of Chern-Simons coupling $k$ becomes infinite. This special direction cannot be probed and the membrane can only probe ten dimensions in perturbation, though the Bagger-Lambert theory has $SO(8)$ and scale symmetries. As similar to the membrane scattering from SYM theory, the eleventh direction can be probed through non perturbative effects.

\bigskip\bigskip

\subsection*{Acknowledgements}

The authors would like to thank all the members of string group in Taiwan and especially Pei-Ming Ho for useful discussion. This work is supported by National Center for Theoretical Sciences, Taiwan,
(No. NSC 97-2119-M-002-001, NSC97-2119-M-007-001).

\bigskip\bigskip