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{\bf \boldmath \Large

\vspace*{3cm}


{\color{Green}{\shadowbox{
   \begin{minipage}{14cm}
    \begin{center} 
     \vspace*{1cm}
      {\color{Cyan}{\Large Matrix Configurations for Spherical 4-branes \\
           and Non-Commutative Structures on $S^{4}$}} \\
        {\color{Cyan}{\sf hep-th/0402010}} \\
       {\color{Cyan}{\Large \it Ryuichi Nakayama and Yusuke Shimono}} \\
    \end{center}
   \end{minipage}
}}}


\vspace{1cm}
  \begin{center}
  {\Large  \it Journal Club at KEK presented by 
     {\color{Magenta}{Takehiro Azuma}}}\footnote{This slide is used
   for Takehiro Azuma's presentation at KEK. Therefore, it is not the
   authors but {\it the presenter Takehiro Azuma} that is responsible
   for any potential flaws in this slide.} \\ 

  {\Large May 21st 2004, 12:15 $\sim$}
  \end{center}

 \tableofcontents



\newpage
%########################### Page 2 ##########################
\section{Introduction}
  The studies of the fuzzy sphere are interesting from the following
  points of view: \\

 {\color{Green}{\ovalbox{Matrix-model side}}} \\

  The fuzzy spheres play an important role as the prototype of the 
  curved-space background. \\

  The IIB matrix model is the most promising candidate of the constructive 
  definition of the superstring theory. \\

  On the other hand, this has only a classical background, which makes the
  perturbation around the curved space impossible. \\
 
  In order to evade this difficulty, the matrix models with 
  {\color{Magenta}{the Chern-Simons term}} have been extensively 
  studied. \\
 {\normalsize{\color{Magenta}{\sf hep-th/0101102,0103192,0108002,
  0204256,0207115,0209057,0301055,0303120,0307007, \\
  0309264,0312241,0401038, 0403242,0405096}}} \\

 {\color{Green}{\ovalbox{String-theory side}}} \\

 The fuzzy spheres appear as the classical solutions in the
 presence of {\color{Magenta}{an external RR field}}
 ({\color{Magenta}{Myers effect}}): \\
 {\normalsize{\color{Magenta}{R.Myers, {\sf hep-th/9910053}}}} \\

\newpage
%####################### Page 3 ##############################
 The interests of the higher-dimensional fuzzy $S^{4}$ spheres:
  \begin{itemize}
   \item{Longitudinal 5-branes in the Matrix theory: \\
  {\normalsize{\color{Magenta}{J.Castelino, S.Lee and W.Taylor, 
  {\sf hep-th/9712105}}}} }
  \item{Four-dimensional noncommutative theory in the large-$N$ limit
      of the large-$N$ reduced model: \\
  {\normalsize{\color{Magenta}{Y.Kimura, {\sf hep-th/0204256}, \\
  T. Azuma, S. Bal, K.Nagao and J. Nishimura, {\sf hep-th/0405096}
  }}} 
  }
  \item{Application to the quantum Hall effect. \\
  {\normalsize{\color{Magenta}{K.Hasebe and Y. Kimura, 
  {\sf hep-th/0310274}}}} }
  \end{itemize}

\vspace*{1cm}
  The purpose of this paper:
  \begin{itemize}
   \item{The authors attempt to build the algebra of the fuzzy
    $S^{4}$ spheres, retaining the {\color{Blue}{$SO(3) \otimes SO(2)$
    symmetry}}.}
   \item{They build the associative star product for the 
    fuzzy $S^{4}$ spheres.}
  \end{itemize}

\newpage
%####################### Page 4-5 ###########################
\section{The algebra on the fuzzy $S^{4}$ sphere}
  To retain the {\color{Magenta}{$SO(5)$ symmetry of the fuzzy
  $S^{4}$ sphere}}, one has to introduce the extra degrees of
  freedom. \\

  The algebra is realized by the {\color{Magenta}{6-dimensional}} 
  homogeneous space {\color{Magenta}{$SO(5)/U(2)$}}. \\

  Namely, {\color{Blue}{the fuzzy $S^{2}$ sphere}} is attached
  on every point of the $S^{4}$ sphere.
  {\color{Blue}{
  \begin{eqnarray}
   SO(5)/U(2) &\sim& (SO(5)/SO(4)) \times (SO(4)/U(2)) \nonumber \\
   &\sim& \underbrace{(SO(5)/SO(4))}_{\textrm{$S^{4}$ sphere}}
  \times \underbrace{(SO(3)/U(1))}_{\textrm{$S^{2}$ sphere}}. \nonumber
  \end{eqnarray} }} 

  Such an algebra is constructed through the {\color{Magenta}{$n$-fold}}
 {\color{Blue}{symmetric
  tensor product}} of the 5-dimensional gamma matrices:
  {\color{Blue}{
  \begin{eqnarray}
   G_{A} &=& \underbrace{(\Gamma_{A} \otimes {\bf 1} \otimes \cdots 
   \otimes {\bf 1})_{\textrm{sym}}}_{\textrm{$n$-fold product}}
  + \cdots +
  ({\bf 1} \otimes \cdots \otimes {\bf 1} \otimes \Gamma_{A})_{\textrm{sym}},
   \nonumber
  \end{eqnarray} }}
  where $\Gamma_{A}$ satisfies the following Clifford algebra:
  \begin{eqnarray}
   \{ \Gamma_{A}, \Gamma_{B} \} = 2 \delta_{AB}, 
  \hspace{2mm} (A,B, \cdots = 1,2,3,4,5). \nonumber
  \end{eqnarray}
  More explicitly, they are constructed as
  {\color{Blue}{
  \begin{eqnarray}
   \Gamma_{a} = \sigma_{3} \otimes \sigma_{a}, \hspace{2mm}
   \Gamma_{4} = \sigma_{2} \otimes {\bf 1}_{2 \times 2}, \hspace{2mm}
   \Gamma_{5} = \sigma_{1} \otimes {\bf 1}_{2 \times 2}. \nonumber
  \end{eqnarray} }}
  \begin{itemize}
   \item{$G_{A}$ is realized by the {\color{Magenta}{$N = \frac{1}{6}
  (n+1)(n+2)(n+3)$}}-dimensional matrices: \\
  Unlike the $S^{2}$ fuzzy sphere, the $S^{4}$ fuzzy sphere is
  realized by a limited size of the matrices.}
   \item{$G_{A}$ {\color{Magenta}{does not close}} with respect to the
      commutator. \\

      {\color{Magenta}{$G_{A}$}} and the commutator 
      {\color{Magenta}{$G_{AB} = [G_{A}, G_{B}]$}} satisfy
      the following algebras:
     {\color{Blue}{\large
    \begin{eqnarray}
    &(1)& G^{2}_{A} = n(n+4) {\bf 1}_{N \times N}, \nonumber \\
    &(2)& G_{AB} G_{AB} = -16n(n+4) {\bf 1}_{N \times N},
     \nonumber \\
    &(3)& [G_{AB}, G_{C}] =
     4 ( - \delta_{AC} G_{B} + \delta_{BC} G_{A}),
     \nonumber \\
    &(4) & [G_{AB}, G_{CD}] =
     4 (\delta_{BC} G_{AD} + \delta_{AD} G_{BC}
      - \delta_{AC} G_{BD} - \delta_{BD} G_{AC}), 
      \nonumber \\
    &(5) & \epsilon_{ABCDE} G_{B} G_{C} G_{D} G_{E}
      = 8(n+2) G_{A}. \nonumber \\
     & & \Rightarrow G_{AB} = \frac{1}{n+2} \epsilon_{ABCDE}
     G_{C} G_{D} G_{E}. \nonumber     
    \end{eqnarray} }}
     }
   \end{itemize}

  Here, we break the $SO(5)$ rotational symmetry as 
  {\color{Magenta}{
   \begin{eqnarray}
     SO(5) \to SO(3) \otimes SO(2), \nonumber
   \end{eqnarray} }}
  and try to build the {\color{Blue}{$N = 2(2j+1)$}}-dimensional 
  representation. \\

  Namely, we take the following tensor product:
   \begin{eqnarray}
    \textrm{(Pauli matrices)} \otimes
    \textrm{(spin-$j$ rep. of $SO(3)$)}. \nonumber
   \end{eqnarray} 
  Then, the representation is constructed as \\

  {\color{Red}{\shadowbox{\parbox{15cm}{
    \begin{eqnarray}
   & & {\hat X}^{a} = \frac{2}{3} \sigma_{3} \otimes T^{a}_{(j)}, \hspace{2mm}
   {\hat X}^{4} = \frac{1}{3} \sigma_{1} \otimes {\bf 1}_{2j+1}, \nonumber \\
   & & {\hat X}^{5} = \frac{1}{3} \sigma_{2} \otimes {\bf 1}_{2j+1}, 
    \textrm{ where} \nonumber \\
   & & a,b,c, \cdots = 1,2,3, \hspace{2mm} i,j,\cdots = 4,5, \nonumber \\
   & & A,B,C,\cdots = 1,2,3,4,5, \nonumber \\
   & & [T^{a}_{(j)}, T^{b}_{(j)}] = i \epsilon^{abc} T^{c}_{(j)}, \hspace{2mm} 
       (T^{a}_{(j)})^{2} = j(j+1). \nonumber
    \end{eqnarray} }}}}
 \newpage
%############################ Page 6 ############################
 This representation satisfies the following relations:
 {\color{Blue}{
  \begin{eqnarray}
  & &  [{\hat X}^{4}, {\hat X}^{5}] = \frac{3}{4j(j+1)} \epsilon_{45abc}
    {\hat X}^{a} {\hat X}^{b} {\hat X}^{c}, \nonumber \\
  & &  [{\hat X}^{A}, {\hat X}^{B}] = \epsilon_{ABCDE}
    {\hat X}^{C} {\hat X}^{D} {\hat X}^{E}, \textrm{ (otherwise)}. \nonumber
  \end{eqnarray} }}

  {\sf 
  In general, it is impossible to resurrect the $SO(5)$ rotational symmetry
  via the following rescaling:
  {\color{Blue}{
  \begin{eqnarray}
   {\hat X}^{a} = \frac{2\alpha}{3} \sigma_{3} \otimes T^{a}_{(j)}, 
    \hspace{2mm}
   {\hat X}^{i} = \frac{\beta}{3} \sigma_{i-3} \otimes {\bf 1}_{2j+1},
   \nonumber
  \end{eqnarray} }}
 which satisfies the following commutation relations:
  {\large 
  \begin{eqnarray}
   & & [{\hat X}^{a}, {\hat X}^{b}] 
  = \frac{\alpha}{\beta^{2}} \epsilon^{abcij} 
    ({\hat X}^{c} {\hat X}^{i} {\hat X}^{j}
   - {\hat X}^{i} {\hat X}^{c} {\hat X}^{j}
   + {\hat X}^{i} {\hat X}^{j} {\hat X}^{c}), \nonumber \\
   & & [{\hat X}^{a}, {\hat X}^{i}]
  = \frac{1}{\alpha} \epsilon^{aibcj} ({\hat X}^{b} {\hat X}^{c}
    {\hat X}^{j} - {\hat X}^{b} {\hat X}^{j} {\hat X}^{c}
     + {\hat X}^{j} {\hat X}^{b} {\hat X}^{c}), \nonumber \\
   & & [{\hat X}^{4}, {\hat X}^{5}] = \frac{3 \beta^{2}}{4j(j+1)
    \alpha^{3}} \epsilon^{45abc} {\hat X}^{a} {\hat X}^{b} {\hat X}^{c}.
   \nonumber
  \end{eqnarray}
  }
 The $SO(5)$ symmetry is resurrected only if {\color{Magenta}{
  $\frac{\alpha}{\beta^{2}} = \frac{1}{\alpha}
   = \frac{3 \beta^{2}}{4j(j+1)\alpha^{3}}$}}, which is realized
  {\color{Magenta}{only for $j=\frac{1}{2}$}}.
 } \\

 The matrix theory action for this fuzzy-4 sphere is
  {\color{Blue}{\large
  \begin{eqnarray}
  S &=& \int dt \textrm{tr} \left\{
  \frac{1}{2R_{M}} (D_{0} {\hat X}^{A})^{2}
  + \frac{1}{4} \left( [{\hat X}^{a}, {\hat X}^{b}] - \frac{\alpha}{\beta^{2}}
    \epsilon^{abCDE} {\hat X}^{C} {\hat X}^{D} {\hat X}^{E} \right)^{2}
   \right. \nonumber \\
  & &  \hspace{10mm} 
   + \frac{1}{2} \left( [{\hat X}^{a}, {\hat X}^{i}] - \frac{1}{\alpha}
    \epsilon^{aiCDE} {\hat X}^{C} {\hat X}^{D} {\hat X}^{E} \right)^{2}
  \nonumber \\
  & &   \hspace{10mm}
   + \left. \frac{1}{2} \left( [{\hat X}^{4}, {\hat X}^{5}] 
   - \frac{3\beta^{2}}{4j(j+1) \alpha^{3}}
    \epsilon^{45CDE} {\hat X}^{C} {\hat X}^{D} {\hat X}^{E} \right)^{2}
   \right\}. \nonumber
  \end{eqnarray} }}
}
\newpage
%######################## Page 7 #############################
{\bf \boldmath \Large
\section{Associative product of the fuzzy $S^{2}$ sphere}
 The construction of the associative product for the fuzzy $S^{2}$
 sphere plays a pivotal role in the following $S^{4}$ case. \\

 Here, we review the work for $S^{2}$ \\
 {\normalsize{\color{Magenta}{K. Hayasaka, R. Nakayama and Y. Takaya, 
  {\sf hep-th/0209240}}}} \\

  We start with the star product of the arbitrary representation:
  {\color{Blue}{\large
   \begin{eqnarray}
 & &  f(x) \star_{\lambda} g(x)
  = f(x) g(x) + \lambda J^{ab} (\partial_{a} f(x)) (\partial_{b} g(x)) 
 \nonumber \\
 &+& \sum^{\infty}_{n=2} \lambda^{n} \left( \sum^{n}_{m=2}
   \chi^{(n)}_{m,m} J^{a_{1} b_{1}} \cdots J^{a_{m} b_{m}}
    (\partial_{a_{1}} \cdots \partial_{a_{m}} f(x))
    (\partial_{b_{1}} \cdots \partial_{b_{m}} g(x)) \right). \nonumber
  \end{eqnarray} }}
 Here, $J^{ab}$ is defined as
  {\color{Red}{
  \begin{eqnarray}
   J^{ab}  = r^{2} \delta^{ab} - x^{a} x^{b}
    + i r \epsilon^{abc} x^{c}, \nonumber 
  \end{eqnarray} }}
 ($r=\sqrt{(x^{1})^{2} + (x^{2})^{2} + (x^{3})^{2}}$) and has 
  the following properties:
 {\large
  \begin{eqnarray}
  &(1)& x^{a} J^{ab} = x^{b} J^{ab} = 0, \nonumber \\
  &(2)& J^{ab} J^{ac} = J^{ba} J^{ca} = 0, \nonumber \\
  &(3)& J^{ab} (\partial_{a} J^{cb}) = J^{ca} (\partial_{a} J^{db}), 
   \nonumber \\
  &(4)& J^{ba} J^{dc} (\partial_{a} \partial_{c} J^{fe})
  = - J^{bf} J^{de} - J^{be} J^{df}, \nonumber \\
  &(5)& J^{a_{1} b_{1}} J^{a_{2} b_{2}} \cdots J^{a_{n} b_{n}}
    (\partial_{b_{1}} \cdots \partial_{b_{n}} J^{cd}) = 0, 
   \textrm{ (for $n \geq 3$)}. \nonumber
  \end{eqnarray}
  }
 
  From $(1)$ and $(2)$, {\color{Magenta}{$r$ is a constant for 
    $\star_{\lambda}$ product}}. \\
    This is a necessary condition for $\star_{\lambda}$ to be a 
    product on the $S^{2}$ sphere.
  {\color{Magenta}{
  \begin{eqnarray}
   f(r) \star_{\lambda} g(x^{a}) = g(x^{a})
     \star_{\lambda} f(r) = f(r) g(x^{a}). \nonumber
  \end{eqnarray} }}

\newpage
%##################### Page 8 ###################################
 We determine {\color{Magenta}{$\chi^{(n)}_{m,m}$}},
 to satisfy the associativity: \\

 {\color{Red}{\ovalbox{\parbox{15cm}{
  \begin{eqnarray}
    (f(x) \star_{\lambda} g(x)) \star_{\lambda} h(x) 
  = f(x) \star_{\lambda} (g(x) \star_{\lambda} h(x)). \nonumber
  \end{eqnarray} }}}}
  \begin{itemize}
   \item{Up to the first order: for both {\color{Magenta}{$(f(x) 
    \star_{\lambda} g(x)) \star_{\lambda} h(x)$}} and 
    {\color{Magenta}{$f(x) \star_{\lambda} (g(x) \star_{\lambda} 
     h(x))$}},
   {\color{Blue}{ 
   \begin{eqnarray}
    fgh + \lambda J^{ab} \{ (\partial_{a} f)(\partial_{b} g) h 
          + (\partial_{a} f) g (\partial_{b} h)
          + f (\partial_{a} g) (\partial_{b} h) \}. \nonumber
   \end{eqnarray} }}
     }
   \item{To the second order, the difference \\
    {\color{Magenta}{$(f(x) \star_{\lambda} g(x)) \star_{\lambda} h(x) 
  - f(x) \star_{\lambda} (g(x) \star_{\lambda} h(x))$}} is
  {\large
  \begin{eqnarray}
  & & \lambda^{2} [ \underbrace{(J^{ab} \partial_{a} J^{cd} - J^{ca} 
     \partial_{a} J^{db})}_{\textrm{vanishes due to $(3)$.}}
     (\partial_{c} f) (\partial_{d} g) (\partial_{b} h) \nonumber \\
  &+& \hspace{10mm} (1 - 2 \chi^{(2)}_{2,2}) J^{ac} J^{bd}
       (\partial_{a} \partial_{b} f \partial_{c} g \partial_{d} h
     -  \partial_{a} f \partial_{b} g \partial_{c} \partial_{d} h) ].
  \nonumber
  \end{eqnarray} }
   This gives the constraint {\color{Magenta}{$\chi^{(2)}_{2,2} = 
   \frac{1}{2}$}}.
   }
  \item{By the same token, we derive the relation
   {\color{Magenta}{\large
    \begin{eqnarray}
     \chi^{(n)}_{m,m} = \frac{1}{m!} \sum_{P} (m-1)^{P_{1}} (m-2)^{P_{2}} 
     \cdots 2^{P_{m-2}} 1^{P_{m-1}}. \nonumber
    \end{eqnarray} }}
   {\large $\{ P_{i} \}$ is the partition of $n-m$ into $m-1$ nonnegative
    integers.}
   {\large
   \begin{eqnarray}
    \chi^{(n)}_{3,3} = \frac{1}{3!} \left( 2^{0} 1^{n-3} + 2^{1} 1^{n-4}
    + \cdots + 2^{n-4} 1^{1} + 2^{n-3} 1^{0} \right) 
      = \frac{2^{n-2} - 1}{6}. \nonumber
   \end{eqnarray} }
   }
  \end{itemize}
 We thus obtain the following product: \\

  {\color{Red}{\shadowbox{\parbox{16cm}{
  \begin{eqnarray}
   f(x) \star_{\lambda} g(x) &=& 
    fg + \sum_{m=1}^{\infty} C_{m} (\lambda) J^{a_{1} b_{1}} \cdots 
  J^{a_{m} b_{m}} \nonumber \\
  & & \hspace{5mm} \times (\partial_{a_{1}} \cdots \partial_{a_{m}} f)
   (\partial_{b_{1}} \cdots \partial_{b_{m}} g), \textrm{ where } \nonumber \\
  C_{m} (\lambda) &=& \frac{\lambda^{m}}{m! (1-\lambda) (1-2 \lambda) \cdots 
   (1-(m-1)\lambda)}. \nonumber 
  \end{eqnarray} }}}}

\newpage
%##################### Page 9 #####################################
 For the {\color{Magenta}{unitary representation}}, this should 
 reproduce the ordinary $SU(2)$ algebra and its Casimir:
  {\color{Blue}{
   \begin{eqnarray}
    [x^{a}, x^{b}]_{\star} = 2i \lambda r \epsilon^{abc} x^{c}. \nonumber
   \end{eqnarray} }}
  Then the rescaling {\color{Magenta}{$y^{a} = \frac{x^{a}}{2 \lambda r}$}}
  gives
  {\color{Red}{
   \begin{eqnarray}
    [y^{a}, y^{b}]_{\star} = i \epsilon^{abc} y^{c}, \hspace{2mm}
    y^{a} \star_{\lambda} y^{a} = \frac{1+2 \lambda}{(2 \lambda)^{2}}
    = j(j+1). \nonumber
   \end{eqnarray} }}
  Then, $\lambda$ is determined as {\color{Red}{$\lambda = \frac{1}{2j},
  - \frac{1}{2j+2}$}}. \\

  {\color{Green}{\ovalbox{$\lambda = \lambda^{(A)}_{j} = \frac{1}{2j}$}}} \\

  In this case, the coefficient $C_{m}(\lambda)$ is given by
  {\color{Red}{
   \begin{eqnarray}
     C_{m}(\lambda^{(A)}_{j}) = \left\{ \begin{array}{ll} 
        \frac{(2j-m)!}{m! (2j)!} & \textrm{ (for $m \leq 2j$)}, \\
        \infty & \textrm{ (for $m > 2j$)}. \end{array} \right. \nonumber
   \end{eqnarray} }}
  This product is {\color{Red}{limited to the finite representation}}! \\
  We denote this product as {\color{Magenta}{$\star$}}. \\

  Especially for $j = \frac{1}{2}$, this gives
  {\color{Blue}{
  \begin{eqnarray}
    x^{a} \star x^{b} = r^{2} \delta_{ab} + i r \epsilon^{abc} x^{c}, 
     \hspace{2mm}
    1 \star x^{a} = x^{a} \star 1 = x^{a}. \nonumber  
  \end{eqnarray} }}
  This corresponds to the algebra of {\color{Magenta}{the Pauli matrices 
  $\sigma_{a}$}} and {\color{Magenta}{the unit matrices $1_{2 \times 2}$}}. 

\newpage
%################# Page 10 ####################################
   {\color{Green}{\ovalbox{$\lambda = \lambda^{(B)}_{j} =
     - \frac{1}{2j+2}$}}} \\

  In this case, the coefficient is given by
  {\color{Red}{
   \begin{eqnarray}
    C_{m} (\lambda^{(B)}_{j}) &=& \frac{(-1)^{m}}{m!} 
    \frac{(2j+1)!}{(2j+1+m)!},  \textrm{ (for any $m$)}. \nonumber
   \end{eqnarray} }}
  This has {\color{Red}{no divergence for any $m$}}. 
  Therefore, this product is applicable to {\color{Red}{an arbitrary 
  size of representation}}. \\
   We denote this product as {\color{Magenta}{$\bullet$}}. \\

  Especially for $j = \frac{1}{2}$, this gives
  {\color{Blue}{\large 
  \begin{eqnarray}
   x^{a} \bullet x^{b} &=& \frac{4}{3} (x^{a} x^{b} - \frac{1}{3} r^{2} 
   \delta^{ab} ) + \frac{1}{9} r^{2} \delta^{ab}
   - \frac{i}{3} r \epsilon^{abc} x^{c}, \nonumber \\
   x^{a} \bullet x^{b} \bullet x^{c} &=& 
   \frac{20}{9} x^{a} x^{b} x^{c} - \frac{i}{27} r^{3} \epsilon^{abc} + 
   \cdots. \nonumber
  \end{eqnarray} }}
 The algebra of the Pauli matrices {\color{Magenta}{is not realized}}. \\

 On the other hand, the integration on the sphere gives
  {\color{Blue}{\large
  \begin{eqnarray}
   \underbrace{\int d \Omega x^{a} \bullet x^{b} = \frac{4 \pi}{9} r^{2}
   \delta^{ab}}_{= \textrm{tr} \sigma^{a} \sigma^{b}}, 
   \hspace{2mm}
   \underbrace{\int d \Omega x^{a} \bullet x^{b} \bullet x^{c} 
  = - \frac{4 \pi i}{27} r^{3}
   \epsilon^{abc}}_{= \textrm{tr} \sigma^{a} \sigma^{b} \sigma^{c}}.
  \nonumber
  \end{eqnarray} }}
 The {\color{Red}{trace}} corresponds to {\color{Red}{the product of 
 the Pauli matrices}}.
}
\newpage
%##################### Page 11 ####################################
{\bf \boldmath \Large
\section{Noncommutative product on $(S^{4})_{j}$}
 We construct the star product on the {\color{Magenta}{fuzzy
 $S^{4}$ sphere}}. \\

 Here, we focus on the finite set of the function, and focus on
 the product {\color{Magenta}{$\star$ for $\lambda = \lambda^{(A)}_{j}$}}. \\

 {\color{Green}{\ovalbox{$S^{2} \otimes S^{2}$ parameterization of $S^{4}$}}} \\

 {\color{Magenta}{$S^{4}$}} representation is expressed by
  the tensor product 
  {\large 
  \begin{eqnarray}
  & & (2 \times 2 \textrm{ matrices spanned by } {\bf 1}_{2 \times 2}, 
 \sigma_{a}) \nonumber \\  
 &\otimes& ((2j+1) \times (2j+1) \textrm{ matrices spanned by }
  {\bf 1}_{2j+1}, T^{a}_{(j)}). \nonumber
  \end{eqnarray}}
 We assign the $S^{2}$ coordinate for each representation as
  {\color{Magenta}{
  \begin{eqnarray}
   \begin{array}{ll}
   x^{1}, x^{2}, x^{3}: & \textrm{ (for the $2 \times 2$ rep. of $S^{2}$)}, \\
   y^{1}, y^{2}, y^{3}: & \textrm{ (for the $(2j+1) \times (2j+1)$
    rep. of $S^{2}$)}. \end{array} \nonumber
  \end{eqnarray}  }}
  We define the radii of the $S^{2}$ spheres as
  {\color{Blue}{
  \begin{eqnarray}
   r = \sqrt{(x^{a})^{2}}, \hspace{2mm}
   \rho = \sqrt{(y^{a})^{2}}. \nonumber
  \end{eqnarray} }}
  The correspondence with the matrices is
  {\color{Red}{
  \begin{eqnarray}
   \frac{\sigma^{a}}{2} \Leftrightarrow \frac{x^{a}}{2r}, \hspace{2mm}
   T^{a}_{(j)} \Leftrightarrow \frac{j y^{a}}{\rho}. \nonumber
  \end{eqnarray} }}
  We then find the following correspondence\\ (for {\color{Magenta}{
  $\alpha = \frac{3R}{2j}$, $\beta = 3R$}}):
  {\color{Red}{
  \begin{eqnarray}
    \begin{array}{ll} 
   {\hat X}^{a} = \frac{R}{j} \sigma_{3} \otimes T^{a}_{(j)}
     \Leftrightarrow X^{a} = \frac{R x^{3} y^{a}}{r \rho}, 
   & \textrm{ (for $a=1,2,3$),} \\
     {\hat X}^{i} = R \sigma_{i-3} \otimes {\bf 1}_{2j+1}
  \Leftrightarrow X^{i} = \frac{R x^{i-3}}{r},
   & \textrm{ (for $i=4,5$).} \end{array} \nonumber
  \end{eqnarray} }}
\newpage
%####################### Page 12 #################################
  This mapping {\color{Magenta}{$S^{2} \otimes S^{2} \to S^{4}$}}
  gives a {\color{Blue}{double cover}} of the $S^{4}$ sphere: \\

  The following two points on {\color{Magenta}{$S^{2} \otimes S^{2}$}}
  fall onto the same point on $S^{4}$:
  {\color{Blue}{
   \begin{eqnarray}
    &P:& \left( \frac{(x^{1}, x^{2}, x^{3})}{r}, \frac{(y^{1}, y^{2},
     y^{3})}{\rho} \right) \in S^{2}_{+} \hspace{2mm}
      (x^{3} \geq 0), \nonumber \\
    &P':& \left( \frac{(x^{1}, x^{2}, -x^{3})}{r}, \frac{(-y^{1},
       - y^{2}, -y^{3})}{\rho} \right) \in S^{2}_{-} \hspace{2mm}
      (x^{3} \leq 0). \nonumber
   \end{eqnarray} }}
  The overlapping point of $S^{2}$ (namely, the equator {\color{Blue}{
  $x^{3}=0$}}) constitutes $S^{1} \otimes S^{2}$. \\

  This {\color{Blue}{$S^{1} \otimes S^{2}$}} is mapped on the {\color{Magenta}{
  $S^{1}$ circle ${\cal C}$}}:
  \begin{eqnarray}
   {\cal C} &=& \{ (X^{1}, \cdots, X^{5}) | X^{1} = X^{2} = X^{3} = 0, 
    X^{4} + X^{5} = R^{2} \}. \nonumber
  \end{eqnarray}

  Therefore, the {\color{Blue}{inverse map $S^{4} \to S^{2}_{+} 
  \otimes S^{2}$}} is {\color{Blue}{multi-valued (indeterminate)}}
  on the circle:
  {\color{Magenta}{
  \begin{eqnarray}
   & & x^{1}(X) = \frac{rX^{4}}{R}, x^{2}(X) = \frac{r X^{5}}{R}, 
   x^{3}(X) = \frac{r D(X)}{R}, \nonumber \\
   & & y^{a} (X)= \frac{\rho X^{a}}{D(X)}, \textrm{ where } \nonumber \\
   & & D(X) = \sqrt{(X^{1})^{2} + (X^{2})^{2} + (X^{3})^{2}}. \nonumber 
  \end{eqnarray}  }}

  The mapping  {\color{Blue}{$S^{4} \to S^{2}_{-} \otimes S^{2}$}} is
  likewise obtained by the replacement {\color{Magenta}{$D(X) \Rightarrow
  - D(X)$}}. It is also {\color{Magenta}{indeterminate on ${\cal C}$}}.
\newpage
%##################### Page 13 #############################
  It is straightforward to build the noncommutative product on $(S^{4})_{j}$, 
  from the {\color{Magenta}{star product on fuzzy $S^{2}$ sphere}}: \\

  To this end, we combine the noncommutative product for each $S^{2}$:\\

  {\color{Red}{\shadowbox{\parbox{17cm}{
  \begin{eqnarray}
  & & F(X) \star G(X) \nonumber \\
   &=& F(X) G(X) +
   (r^{2} \delta^{ab} - x^{a} x^{b} + i r \epsilon^{abc} x^{c})
   \frac{\partial F(X)}{\partial x^{a}} \frac{\partial G(X)}{\partial x^{b}} 
   \nonumber \\
   & & + \lambda (\rho \delta^{ab} - y^{a} y^{b} + i \rho \epsilon^{abc}
    y^{c}) \frac{\partial F(X)}{\partial y^{a}} 
    \frac{\partial G(X)}{\partial y^{b}} \nonumber \\
   & & + \lambda (r^{2} \delta^{ab} - x^{a} x^{b} + i r \epsilon^{abc}
      x^{c}) \nonumber \\
   & & \times (\rho^{2} \delta^{de} - y^{d} y^{e} + i \rho \epsilon^{def}
      y^{f}) \frac{\partial^{2} F(X)}{\partial x^{a} \partial y^{d}}
      \frac{\partial^{2} G(X)}{\partial x^{b} \partial y^{e}} + \cdots 
    \nonumber
  \end{eqnarray} }}}} \\

  In this way, we can easily build the product for the $S^{4}$ sphere,
  by reducing the symmetry as \\
 {\color{Magenta}{$SO(5) \to SO(3) \otimes  SO(2)$}}.

\newpage
%###################### Page 14 ###############################
 {\color{Green}{\ovalbox{Functions on the noncommutative $(S^{4})_{j}$}}} \\

  The functions of the fuzzy $(S^{4})_{j}$ are given by the product of
  the functions on the two fuzzy $S^{2}$:
  \begin{itemize}
   \item{First $S^{2}$: spanned by {\color{Magenta}{$1, \frac{x^{a}}{r}$}}.}
   \item{Second $S^{2}$: spanned by
     {\color{Magenta}{$1, \frac{y^{a}}{\rho}, \cdots,
       \frac{y^{a_{1}} y^{a_{2}} \cdots y^{a_{2j}}}{\rho^{2j}}$}}.}
  \end{itemize}
  Then, the functions of the fuzzy $(S^{4})_{j}$ are given by 
  {\color{Red}{
  \begin{eqnarray}
   1, \frac{x^{a}}{r}, \frac{y^{a_{1}}}{\rho}, \frac{x^{a} y^{a_{1}}}{r \rho}, 
   \cdots,
   \frac{y^{a_{1}} \cdots y^{a_{2j}}}{\rho^{2j}},
   \frac{x^{a} y^{a_{1}} \cdots y^{a_{2j}}}{r \rho^{2j}}. \nonumber
  \end{eqnarray} }}

  The functions on $(S^{4})_{j}$ are expressed via $X^{A}$ as \\

  {\color{Red}{\ovalbox{\parbox{17cm}{
  \begin{eqnarray}
  & & 1, \hspace{2mm} \underbrace{\frac{X^{i}}{R}, \frac{D(X)}{R}}_{=
      \frac{x^{a}}{r}}, \cdots, \nonumber \\
  & & \underbrace{\frac{X^{a_{1}} \cdots X^{a_{2j}}}{D(X)^{n}}}_{=
      \frac{y^{a_{1}} \cdots y^{a_{2j}}}{\rho^{2j}} }, \hspace{2mm}
      \underbrace{\frac{X^{i} X^{a_{1}} \cdots X^{a_{2j}}}{R D(X)^{n}}}_{=
      \frac{x^{i-3} y^{a_{1}} \cdots y^{a_{2j}}}{r \rho^{2j}} }, \hspace{2mm}
      \underbrace{\frac{X^{3} X^{a_{1}} \cdots X^{a_{2j}}}{R D(X)^{n}}}_{=
      \frac{x^{3} y^{a_{1}} \cdots y^{a_{2j}}}{r \rho^{2j}} }. \nonumber
  \end{eqnarray} }}}} \\

  The number of the independent function is
  {\color{Magenta}{
  \begin{eqnarray}
   4 \times (2j+1)^{2} = N^{2}. \nonumber
  \end{eqnarray} }}
  Thus, the function of {\color{Blue}{$(S^{4})_{j}$}} corresponds to the
  {\color{Blue}{$2(2j+1) \times 2(2j+1)$ hermitian matrices}}. \\

\newpage
%################### Page 15 ##################################
  {\sf  The spherical harmonics of fuzzy $S^{4}$ is given by  
  {\color{Magenta}{$Y_{l_{1} l_{2} l_{3} m}$}}.\\

  The degree of freedom is (up to order $l$)
  \begin{eqnarray}
   \sum_{l_{1}=0}^{l} \underbrace{\sum_{l_{2}=0}^{l_{1}} \sum_{l_{3}=0}^{l_{2}}
   \sum_{m=-l_{3}}^{l_{3}} 1}_{\textrm{\footnotesize 
   degree of freedom for the $l_{1}$-order
   polynomials}} = \frac{(l+1)(l+2)^{2}(l+3)}{12}. \nonumber
  \end{eqnarray}
  This is not a square number. This makes it difficult to establish the
  isomorphism 
  \begin{eqnarray}
   \textrm{(matrices)} \Leftrightarrow \textrm{(polynomial on $S^{4}$)}.
  \nonumber
  \end{eqnarray}
  This construction avoids this problem, since they introduce a 
  {\color{Magenta}{non-polynomial function}}. \\

  These functions are not well-defined {\color{Magenta}{
  on the equator ${\cal C}$}}, since 
  these functions are indeterminate on  ${\cal C}$. \\

 However, these singularities are not serious in the 
 {\color{Magenta}{noncommutative field theory}}. \\

 The derivative is given by the commutator
 {\color{Magenta}{$\nabla_{A} F(X) = \frac{i}{R} [X^{A}, F(X)]_{\star}$}}. \\

 We define the star product on the equator ${\cal C}$ {\color{Magenta}{
 by the limit outside ${\cal C}$}}, such that the singularities
 are canceled on ${\cal C}$. \\

 In this sense, the products and derivatives are ``well-defined''. \\

 This is evident from the {\color{Magenta}{matrix $\Leftrightarrow$ function}}
 correspondence, because the matrix configuration {\color{Magenta}{does not
 have any singularity}}. 
}
}
\newpage
%################### page 16 #################################
{\bf \boldmath \Large
\section{Noncommutative product on $(S^{4})_{j=\frac{1}{2}}$}
  In this section, we scrutinize {\color{Magenta}{the simplest case
  $j = \frac{1}{2}$}} in more detail. \\

  In principle, we can obtain the explicit form of the star product
  by replacing {\color{Magenta}{$(x^{a},y^{a})$}} with
  {\color{Magenta}{$X^{A}$}}. \\

  On the other hand, it is easier to determine the product 
  {\color{Magenta}{for $j=\frac{1}{2}$}}, such that it reproduces the 
  multiplication rule. \\

  For {\color{Magenta}{$j=\frac{1}{2}$}}, we have the following functions:
  {\color{Blue}{
  \begin{eqnarray}
  & &  1, \hspace{2mm} \underbrace{\frac{X^{i}}{R} = \frac{x^{i-3}}{r}, 
  \hspace{2mm} \frac{D(X)}{R} = \frac{x^{3}}{r}}_{=\frac{x^{a}}{r}}, 
  \hspace{2mm} \underbrace{\frac{X^{a}}{D} = \frac{y^{a}}{\rho}
    }_{=\frac{y^{a}}{\rho}}, \nonumber \\
  & & \underbrace{\frac{X^{a}}{R} = \frac{x^{3} y^{a}}{r \rho}, \hspace{2mm}
    \frac{X^{i}X^{a}}{RD(X)} = \frac{x^{i-3} y^{a}}{r \rho}}_{\frac{x^{b}
     y^{a}}{r \rho}}. \nonumber
  \end{eqnarray} }}

  The star products of these functions can be built from the $S^{2}$
  algebra (and likewise for $\frac{y^{a}}{\rho}$):
   \begin{eqnarray}
    1 \star 1 = 1, \hspace{2mm}
    1 \star \frac{x^{a}}{r} = \frac{x^{a}}{r} \star 1 = \frac{x^{a}}{r},
    \hspace{2mm}
    \frac{x^{a}}{r} \star \frac{x^{b}}{r} = \delta^{ab} + i \epsilon^{abc}
    \frac{x^{c}}{r}. \nonumber 
   \end{eqnarray}

  This gives the following multiplication laws: 
\newpage
%################### Page 17 ##############################
 \vspace*{-2cm}
 {\normalsize 
  \begin{eqnarray}
%%%(1)
 \frac{X^a}{R} \star \frac{X^b}{R}
  & = &
  \delta_{ab}+i\epsilon_{abc}\frac{X^c}{\Sq}, \qquad 
 %%%(2)
 \frac{X^i}{R} \star \frac{X^j}{R}
  = 
 \delta_{ij}+i\epsilon_{ij}\frac{\Sq}{R}, \label{mt1} \nonumber \\
 %%%(3)
 \frac{X^a}{R} \star \frac{X^i}{R}
 & = &
i\epsilon_{ij}\frac{X^j X^a}{R \Sq}, \qquad 
 %%%(4)
 \frac{X^i}{R} \star \frac{X^a}{R}
  = 
 -i\epsilon_{ij}\frac{X^j X^a}{R \Sq}, \nonumber \\
 %%%(5)
\frac{X^a}{\Sq} \star \frac{\Sq}{R}
& = &
\frac{X^a}{R}, \qquad \qquad \qquad 
%%%(6)
\frac{\Sq}{R} \star \frac{X^a}{\Sq}
 = 
\frac{X^a}{R}, \nonumber \\
%%%(7)
\frac{X^i}{R} \star \frac{\Sq}{R} 
& = &
-i \epsilon_{ij} \frac{X^j}{R}, \qquad \qquad 
%%%(8)
\frac{\Sq}{R} \star \frac{X^i}{R}
 = 
i \epsilon_{ij} \frac{X^j}{R}, \nonumber \\
%%%(9)
\frac{X^i}{R} \star \frac{X^j X^a}{R \Sq}
& = &
\delta_{ij}\frac{X^a}{\Sq}+i\epsilon_{ij}\frac{X^a}{R}, \nonumber \\
%%%(10)
\frac{X^i X^a}{R \Sq} \star \frac{X^j}{R}
& = &
\delta_{ij}\frac{X^a}{\Sq}+i\epsilon_{ij}\frac{X^a}{R}, \nonumber \\
%%%(11)
 \frac{X^a}{\Sq} \star \frac{X^b}{\Sq}
  & = &
  \delta_{ab}+i\epsilon_{abc}\frac{X^c}{\Sq}, \nonumber \\
 %%%(12)
\frac{X^i X^a}{R \Sq} \star \frac{X^b}{\Sq}
& = &
\delta_{ab}\frac{X^i}{R}+i\epsilon_{abc}\frac{X^i X^c}{R \Sq}, \nonumber \\
%%%(13)
\frac{X^a}{\Sq} \star \frac{X^i X^b}{R \Sq}
& = &
\delta_{ab}\frac{X^i}{R}+i\epsilon_{abc}\frac{X^i X^c}{R \Sq}, \nonumber \\
%%%(14)
\frac{X^a}{R} \star \frac{X^b}{\Sq}
& = &
\delta_{ab}\frac{\Sq}{R}+i\epsilon_{abc}\frac{X^c}{R}, \nonumber \\
%%%(15)
\frac{X^a}{\Sq} \star \frac{X^b}{R}
& = &
\delta_{ab}\frac{\Sq}{R}+i\epsilon_{abc}\frac{X^c}{R}, \nonumber \\
%%%(16)
\frac{X^i X^a}{R \Sq} \star \frac{\Sq}{R}
& = &
-i\epsilon_{ij}\frac{X^j X^a}{R \Sq}, \nonumber \\
%%%(17)
\frac{\Sq}{R} \star \frac{X^i X^a}{R \Sq}
& = & 
i \epsilon_{ij}\frac{X^j X^a}{R \Sq}, \nonumber \\
%%%(18)
\frac{X^a}{R} \star \frac{\Sq}{R}
& = &
\frac{X^a}{\Sq}, \qquad 
%%%(19)
\frac{\Sq}{R} \star \frac{X^a}{R}
 = 
\frac{X^a}{\Sq}, \nonumber \\
%%%(20)
\frac{X^a}{R} \star \frac{X^i X^b}{R \Sq}
& = &
i\epsilon_{ij}\delta_{ab}\frac{X^j}{R}
-\epsilon_{ij}\epsilon_{abc}\frac{X^j X^c}{R \Sq}, \nonumber \\
%%%(21)
\frac{X^i X^a}{R \Sq} \star \frac{X^b}{R}
& = &
-i\epsilon_{ij}\delta_{ab}\frac{X^j}{R}
+\epsilon_{ij}\epsilon_{abc}\frac{X^j X^c}{R \Sq}, \nonumber \\
%%%(23)
\frac{X^i X^a}{R \Sq} \star  \frac{X^j X^b}{R \Sq}
& = &
\delta_{ij}\delta_{ab}+i\delta_{ij}\epsilon_{abc}\frac{X^c}{\Sq} \nonumber \\
&& \qquad  +i\epsilon_{ij}\delta_{ab}\frac{\Sq}{R}
-\epsilon_{ij}\epsilon_{abc}\frac{X^c}{R}, \nonumber \\
%%%(22)
\frac{\Sq}{R} \star \frac{\Sq}{R}
& = & 1. \nonumber
\end{eqnarray} }
\newpage
%################ Page 18 ###################################
  The product of $j = \frac{1}{2}$ is {\color{Magenta}{at most of the second
  order of the derivative}}. \\

  Then, the star product is constrained as
{\normalsize
 \begin{eqnarray}
F(X) \star G(X)
%%%%%%%%%%
= && FG  \nonumber \\
%%%%%%%%%%
& + & 
%%%
L^{(1)}_{a,b}
\frac{\partial F}{\partial X^a}
\frac{\partial G}{\partial X^b}
%%%
+
%%%
L^{(2)}_{i,j}
\frac{\partial F}{\partial X^i}
\frac{\partial G}{\partial X^j}
%%%
\nonumber \\
%%%
& + &
%%%
L^{(3)}_{a,i}
\frac{\partial F}{\partial X^a}
\frac{\partial G}{\partial X^i}
%%%
+
%%%
L^{(4)}_{i,a}
\frac{\partial F}{\partial X^i}
\frac{\partial G}{\partial X^a}
%%%
\nonumber \\
%%%%%%%%%%%%%%%%%%%%
& + &
%%%
L^{(5)}_{ab,c}
\frac{\partial^2 F}{\partial X^a \partial X^b}
\frac{\partial G}{\partial X^c}
%%% 
+
%%%
L^{(6)}_{a,bc}
\frac{\partial F}{\partial X^a}
\frac{\partial^2 G}{\partial X^b \partial X^c}
%%%%%%%%%%
\nonumber \\
%%%%%%%%%%
& + &
%%%
L^{(7)}_{ab,i}
\frac{\partial^2 F}{\partial X^a \partial X^b}
\frac{\partial G}{\partial X^i} 
%%%
+
%%%
L^{(8)}_{a,bi}
\frac{\partial F}{\partial X^a}
\frac{\partial^2 G}{\partial X^b \partial X^i}
%%%%%%%%%%
\nonumber \\
%%%%%%%%%%
& + &
%%%
L^{(9)}_{ai,b}
\frac{\partial^2 F}{\partial X^a \partial X^i}
\frac{\partial G}{\partial X^b}
%%%
+
%%%
L^{(10)}_{i,ab}
\frac{\partial F}{\partial X^i}
\frac{\partial^2 G}{\partial X^a \partial X^b}
%%%%%%%%%%
\nonumber \\
%%%%%%%%%%
& + &
%%%
L^{(11)}_{ai,j}
\frac{\partial^2 F}{\partial X^a \partial X^i}
\frac{\partial G}{\partial X^j} 
%%%
+
%%%
L^{(12)}_{a,ij}
\frac{\partial F}{\partial X^a }
\frac{\partial^2 G}{\partial X^i \partial X^j}
%%%%%%%%%%
\nonumber \\
%%%%%%%%%%
&+&
%%%
L^{(13)}_{i,aj}
\frac{\partial F}{\partial X^i}
\frac{\partial G}{\partial X^a \partial X^j}
%%%
+
%%%
L^{(14)}_{ij,a}
\frac{\partial^2 F}{\partial X^i \partial X^j}
\frac{\partial G}{\partial X^a}
%%%%%%%%%%
\nonumber \\
%%%%%%%%%%
&+&
%%%
L^{(15)}_{ij,k}
\frac{\partial^2 F}{\partial X^i \partial X^j}
\frac{\partial G}{\partial X^k} 
%%%
+
%%%
L^{(16)}_{i,jk}
\frac{\partial F}{\partial X^i}
\frac{\partial^2 G}{\partial X^j \partial X^k}
%%%%%%%%%%
\nonumber \\
%%%%%%%%%%%%%%%%%%%%
&+&
%%%
L^{(17)}_{ab,cd}
\frac{\partial^2 F}{\partial X^a \partial X^b}
\frac{\partial^2 G}{\partial X^c \partial X^d}
%%%
+
%%%
L^{(18)}_{ij,kl}
\frac{\partial^2 F}{\partial X^i \partial X^j}
\frac{\partial^2 G}{\partial X^k \partial X^l}
%%%%%%%%%%
\nonumber \\
%%%%%%%%%%
& + &
%%%
L^{(19)}_{ab,ci}
\frac{\partial^2 F}{\partial X^a \partial X^b}
\frac{\partial^2 G}{\partial X^c \partial X^i}
%%%
+
%%%
L^{(20)}_{ai,bc}
\frac{\partial^2 F}{\partial X^a \partial X^i}
\frac{\partial^2 G}{\partial X^b \partial X^c}
%%%%%%%%%%
\nonumber \\
%%%%%%%%%%
& + &
%%%
L^{(21)}_{ab,ij}
\frac{\partial^2 F}{\partial X^a \partial X^b}
\frac{\partial^2 G}{\partial X^i \partial X^j}
%%%
+
%%%
L^{(22)}_{ij,ab}
\frac{\partial^2 F}{\partial X^i \partial X^j}
\frac{\partial^2 G}{\partial X^a \partial X^b}
%%%%%%%%%%
\nonumber  \\
%%%%%%%%%%
&+&
%%%
L^{(23)}_{ai,bj}
\frac{\partial^2 F}{\partial X^a \partial X^i}
\frac{\partial^2 G}{\partial X^b \partial X^j}
%%%%%%%%%%
\nonumber \\
%%%%%%%%%%
& + &
%%%
L^{(24)}_{ai,jk}
\frac{\partial^2 F}{\partial X^a \partial X^i}
\frac{\partial^2 G}{\partial X^j \partial X^k}
%%%
+
%%%
L^{(25)}_{ij,ak}
\frac{\partial^2 F}{\partial X^i \partial X^j}
\frac{\partial^2 G}{\partial X^a \partial X^k}. \nonumber
%%%%%%%%%%
\end{eqnarray}} 
 The coefficients {\color{Magenta}{$L^{(1)}_{a,b} \sim L^{(25)}_{ij,ak}$}}
 are determined such that
 {\color{Red}{
  \begin{itemize}
   \item{The product should agree with the multiplication laws 
   of each function.}
   \item{$R = \sqrt{(X^{A})^{2}}$ should be constant; namely
  \begin{eqnarray}
    f(R) \star G(X) = G(X) \star f(R) = f(R) G(X). \nonumber
  \end{eqnarray} }
  \end{itemize} }}
  One of such solutions is given by
\newpage
%######################## Page 19 ############################
{\normalsize
\begin{eqnarray}
L^{(1)}_{a,b}
&=&
R^2 \delta_{ab} -X^a X^b +i R^2 \epsilon_{abc}\frac{X^c}{\Sq},
\label{L1}
\nonumber \\
%%%%%
L^{(2)}_{i,j}
&=&
R^2 \delta_{ij} -X^i X^j +i R^2 \epsilon_{ij}\frac{\Sq}{R},
\label{L2} \nonumber \\ 
%%%%%
L^{(3)}_{a,i}
&=&
-X^a X^i +i R^2 \epsilon_{ij}\frac{X^j X^a}{R\Sq},
\label{L3} \nonumber \\ 
%%%%%
L^{(4)}_{i,a}
&=&
-X^a X^i -i R^2 \epsilon_{ij}\frac{X^j X^a}{R\Sq},
\label{L4} \nonumber \\ 
%%%%%
L^{(5)}_{ab,c}
&=&
\frac{R^2-\Sq^2}{2}(X^a\delta_{bc}+X^b\delta_{ac})
-\frac{R^2-\Sq^2}{\Sq^2}X^a X^b X^c \nonumber\nonumber \\&&
+i\frac{R^2-\Sq^2}{2}(X^b \epsilon_{acd}+X^a\epsilon_{bcd})\frac{X^d}{\Sq},
\nonumber \\ 
%%%%%
L^{(6)}_{a,bc}
&=&
\frac{R^2-\Sq^2}{2}(X^c\delta_{ab}+X^b\delta_{ac})
-\frac{R^2-\Sq^2}{\Sq^2}X^a X^b X^c \nonumber \nonumber \\&&
+i\frac{R^2-\Sq^2}{2}(X^c \epsilon_{abd}+X^b\epsilon_{acd})\frac{X^d}{\Sq},
 \nonumber \\ 
%%%%%
L^{(7)}_{ab,i}
&=&
-\frac{1}{2}X^a X^b X^i +i \frac{R}{\Sq}\epsilon_{ij}X^j X^a X^b,
\nonumber \\ 
%%%%%
L^{(8)}_{a,bi}
&=&
-\Sq^2 X^i\delta_{ab} +X^a X^b X^i -i \Sq \epsilon_{abc}X^c X^i
+i R\Sq \delta_{ab}\epsilon_{ij}X^j \nonumber \nonumber \\&&
-i\frac{R}{\Sq}\epsilon_{ij}X^j X^a X^b
-R \epsilon_{ij}\epsilon_{abc}X^j X^c,
\nonumber \\\ 
%%%%%
L^{(9)}_{ai,b}
&=&
-\Sq^2 X^i\delta_{ab} +X^a X^b X^i -i \Sq \epsilon_{abc}X^c X^i
-i R\Sq \delta_{ab}\epsilon_{ij}X^j  \nonumber \nonumber \\&&
+i\frac{R}{\Sq}\epsilon_{ij}X^j X^a X^b
+R \epsilon_{ij}\epsilon_{abc}X^j X^c,
\nonumber \\ 
%%%%%
L^{(10)}_{i,ab}
&=&
-\frac{1}{2}X^a X^b X^i - i \frac{R}{\Sq}\epsilon_{ij}X^j X^a X^b,
 \nonumber \\ 
%%%%%
L^{(11)}_{ai,j}
&=&
\frac{\Sq^2}{2}X^a \delta_{ij}+i R^2 X^a \epsilon_{ij}\frac{\Sq}{R},
 \nonumber \\ 
%%%%%
L^{(12)}_{a,ij}
&=&
-i\frac{R}{2}X^a (\epsilon_{ik}X^j+\epsilon_{jk}X^i)\frac{X^k}{\Sq},
 \nonumber \\ 
%%%%%
L^{(13)}_{i,aj}
&=&
\frac{\Sq^2}{2}X^a \delta_{ij}+i R^2 X^a \epsilon_{ij}\frac{\Sq}{R},
 \nonumber \\ 
%%%%%
L^{(14)}_{ij,a}
&=&
i\frac{R}{2}X^a (\epsilon_{ik}X^j+\epsilon_{jk}X^i)\frac{X^k}{\Sq},
 \nonumber \\ 
%%%%%
L^{(15)}_{ij,k}
&=&
-i \frac{R\Sq}{2}(\delta_{ik}\epsilon_{jl}+\delta_{jk}\epsilon_{il})X^l
+\frac{\Sq^2}{4}(\epsilon_{ik}\epsilon_{jl}+\epsilon_{il}\epsilon_{jk})X^l,
\nonumber \\ 
%%%%%
L^{(16)}_{i,jk}
&=&
i \frac{R\Sq}{2}(\delta_{ij}\epsilon_{kl}+\delta_{ik}\epsilon_{jl})X^l 
-\frac{\Sq^2}{4}(\epsilon_{ij}\epsilon_{kl}+\epsilon_{ik}\epsilon_{jl})X^l.
 \nonumber
\end{eqnarray}
\newpage
%####################### Page 20 ######################

%%%%%
\begin{eqnarray}
L^{(17)}_{ab,cd}
&=&
\frac{\Sq^2}{4}(R^2-\Sq^2)(\delta_{ac}\delta_{bd}+\delta_{ad}\delta_{bc})
+\frac{R^2-\Sq^2}{2\Sq^2}X^a X^b X^c X^d \nonumber \nonumber \\&&
+i\frac{R^2-\Sq^2}{4}(X^a X^c \epsilon_{bde}+X^a X^d\epsilon_{bce}
+X^b X^d \epsilon_{ace}+X^b X^c\epsilon_{ade})\frac{X^e}{\Sq} 
\nonumber \nonumber \\&&
-\frac{R^2-\Sq^2}{4}(\epsilon_{ace}\epsilon_{bdf}
+\epsilon_{ade}\epsilon_{bcf})X^e X^f,
 \nonumber \\ 
%%%%%
L^{(18)}_{ij,kl}
&=&
-\frac{\Sq^4}{4}(\delta_{ik}\delta_{jl}+\delta_{il}\delta_{jk})
+\frac{\Sq^4}{4}(\epsilon_{ik}\epsilon_{jl}+\epsilon_{il}\epsilon_{jk}),
 \nonumber \\ 
%%%%%
L^{(19)}_{ab,ci}
&=&
-\frac{\Sq^2}{2}X^i(X^b\delta_{ac}+X^a\delta_{bc})
-i\frac{\Sq}{2}X^i(X^b\epsilon_{acd}+X^a\epsilon_{bcd})X^d\nonumber \\&&
+i\frac{R\Sq}{2}(X^b\delta_{ac}+X^a\delta_{bc})\epsilon_{ij}X^j
-\frac{R}{2}(X^b\epsilon_{acd}+X^a\epsilon_{bcd})\epsilon_{ij}X^j X^d,
\nonumber \nonumber \\ 
%%%%%
L^{(20)}_{ai,bc}
&=&
-\frac{\Sq^2}{2}X^i(X^c\delta_{ab}+X^b\delta_{ac})
-i\frac{\Sq}{2}X^i(X^c\epsilon_{abd}+X^b\epsilon_{acd})X^d \nonumber \nonumber \\&&
-i\frac{R\Sq}{2}(X^c\delta_{ab}+X^b\delta_{ac})\epsilon_{ij}X^j
+\frac{R}{2}(X^c\epsilon_{abd}+X^b\epsilon_{acd})\epsilon_{ij}X^j X^d,
 \nonumber \\ 
%%%%%
L^{(21)}_{ab,ij}
&=&
-\frac{1}{2}\epsilon_{ik}\epsilon_{jl}X^k X^l X^a X^b,
\nonumber \\ 
%%%%%
L^{(22)}_{ij,ab}
&=&
-\frac{1}{2}\epsilon_{ik}\epsilon_{jl}X^k X^l X^a X^b,
 \nonumber \\ 
%%%%%
L^{(23)}_{ai,bj}
&=&
R^2\Sq^2\delta_{ij}\delta_{ab}-\Sq^2 X^i X^j\delta_{ab}
-\frac{R^2-\Sq^2}{2}X^a X^b \delta_{ij} \nonumber \nonumber \\&&
+i R^2\Sq\delta_{ij}\epsilon_{abc}X^c
-i \Sq X^i X^j \epsilon_{abc}X^c \nonumber \nonumber \\&&
+iR^2\Sq^2\delta_{ab}\epsilon_{ij}\frac{\Sq}{R}
-R\Sq^2\varepsilon_{ij}\epsilon_{abc}X^c,
 \nonumber \\ 
%%%%%
L^{(24)}_{ai,jk}
&=&
\frac{\Sq^2}{4}X^a(X^k\delta_{ij}+X^j\delta_{ik}),
\nonumber \\ 
%%%%%
L^{(25)}_{ij,ak}
&=&
\frac{\Sq^2}{4}X^a(X^j\delta_{ik}+X^i\delta_{jk}). \nonumber
%%%%%
\end{eqnarray} }
\newpage
%##################### Page 21 ##############################
 The {\color{Magenta}{covariant derivatives}} are defined (and given) as
{\normalsize
\begin{eqnarray}
\nabla_a G(X) &=& \frac{i}{R}[X_{a}, G(X)]_{\star} \nonumber \\
  &=& \frac{i}{R} \left( (L^{(1)}_{a,b} - L^{(1)}_{b,a})
     \frac{\partial G}{\partial X^{b}}
     + (L^{(3)}_{a,i} - L^{(4)}_{i,a}) \frac{\partial G}{\partial X^{i}}
     + (L^{(6)}_{a,bc} - L^{(5)}_{bc,a}) \frac{\partial^{2} G}{\partial X^{b}
       \partial X^{c}} \nonumber  \right. \\
  & & \left.   + (L^{(8)}_{a,bi} - L^{(9)}_{bi,a}) \frac{\partial^{2} G}{
        \partial X^{b} \partial X^{i}}
     + (L^{(12)}_{a,ij} - L^{(11)}_{ij,a}) \frac{\partial^{2} G}{
        \partial X^{i} \partial X^{j}} \right) \nonumber \\
&\equiv& -2\frac{R}{\Sq}\epsilon_{abc}X^c\frac{\partial G}{\partial X^b}
-2\frac{1}{\Sq}\epsilon_{ij}X^j X^a\frac{\partial G}{\partial X^i} \nonumber \\
&&-\frac{R^2-\Sq^2}{R\Sq}(\epsilon_{acd}X^b+\epsilon_{bcd}X^a)X^d
\frac{\partial^2 G}{\partial X^b \partial X^c} \nonumber \\ &&
+\frac{2}{R}\Sq\epsilon_{abc}X^c X^i
\frac{\partial^2 G}{\partial X^b \partial X^i} 
-2i\epsilon_{ij}\epsilon_{abc}X^j X^c
\frac{\partial^2 G}{\partial X^b \partial X^i} \nonumber \\&&
+\frac{1}{\Sq}(\epsilon_{ik}X^j+\epsilon_{jk}X^i)X^k X^a
\frac{\partial^2 G}{\partial X^i \partial X^j}, \nonumber
\end{eqnarray}

\begin{eqnarray}
\nabla_i G(X) &=& \frac{i}{R} [X_{i}, G(X)]_{\star} \nonumber \\
  &=& \frac{i}{R} \left( (L^{(2)}_{i,j} - L^{(2)}_{j,i}) \frac{
    \partial G}{\partial X^{j}}
   + (L^{(4)}_{i,a} - L^{(3)}_{a,i}) \frac{\partial G}{\partial X^{a}}
   + (L^{(10)}_{i,ab} - L^{(7)}_{ab,i}) \frac{\partial^{2} G}{\partial X^{a}
      \partial X^{b}} \right. \nonumber \\
  & & \left. + (L^{(13)}_{i,aj} - L^{(11)}_{aj,i})
     \frac{\partial^{2} G}{\partial X^{a} \partial X^{j}}
     + (L^{(16)}_{i,jk} - L^{(15)}_{jk,i})
     \frac{\partial^{2} G}{\partial X^{j} \partial X^{k}} \right) \nonumber \\
&\equiv& -2\Sq \epsilon_{ij}
\frac{\partial G}{\partial X^j}
-2\frac{1}{\Sq}\epsilon_{ij}X^j X^a
\frac{\partial G}{\partial X^a} \nonumber \\
&&-2\frac{1}{\Sq}\epsilon_{ij}X^j X^a X^b
\frac{\partial^2 G}{\partial X^a \partial X^b}
-2\Sq X^a \epsilon_{ij}
\frac{\partial^2 G}{\partial X^a \partial X^j} \nonumber \\
&&-i\frac{\Sq^2}{2R}(\epsilon_{ij}\epsilon_{kl}+\epsilon_{ik}\epsilon_{jl})X^l
\frac{\partial^2 G}{\partial X^j \partial X^k}. \nonumber
\end{eqnarray}}

 The field strength of the {\color{Magenta}{gauge field $A_{A}(X)$}}
 is (for {\color{Blue}{$j=\frac{1}{2}$, $\alpha = \beta = 3R$}}):
  {\color{Blue}{
  \begin{eqnarray}
   F_{AB}(X) &=& \frac{i}{R^{2}} \left( [{\cal X}^{A}, {\cal X}^{B}]_{\star}
     - \frac{1}{3R} \epsilon_{ABCDE} {\cal X}_{C} \star {\cal X}_{D}
     \star {\cal X}_{E} \right), \nonumber
  \end{eqnarray} }}
 where {\color{Magenta}{${\cal X}^{A} = X^{A} + R A^{A}$}}. 


}
\newpage
%######################### Page 22 ################################
{\bf \boldmath \Large
\section{$B$-field background}
 In this section, we build the star product on the $S^{4}$ sphere,
 in relation to the $B$-fields.

 We introduce the following $S^{2} \times S^{2}$ parameterization of 
 the $S^{4}$ sphere:
 {\color{Blue}{\large 
 \begin{eqnarray}
  & &  X^{1} = R \cos \theta_{1} \sin \theta_{2} \cos \varphi_{2}, \hspace{2mm}
   X^{2} = R \cos \theta_{1} \sin \theta_{2} \sin \varphi_{2}, \nonumber \\
  & & X^{3} = R \cos \theta_{1} \cos \theta_{2}, \hspace{2mm}
   X^{4} = R \sin \theta_{1} \cos \varphi_{1}, \hspace{2mm}
   X^{5} = R \sin \theta_{1} \sin \varphi_{1}. \nonumber
  \end{eqnarray} }}
 This gives a double cover of the $S^{4}$ sphere. In $S^{4}$, the following 
 two points are identified:
  \begin{eqnarray}
   P &=& (\theta_{1}, \varphi_{1}, \theta_{2}, \varphi_{2}), \nonumber \\
   P' &=& (\pi - \theta_{1}, \varphi_{1}, \pi - \theta_{2}, \varphi_{2} + \pi).
  \nonumber
  \end{eqnarray}

 The $B$-field background which has the maximal symmetry of the two spheres
 $S^{2} \times S^{2}$ is
  {\color{Blue}{
  \begin{eqnarray}
   B = \frac{n_{1}}{2} \sin \theta_{1} d \theta_{1} d \varphi_{1}
     + \frac{n_{2}}{2} \sin \theta_{2} d \theta_{2} d \varphi_{2}. \nonumber
  \end{eqnarray} }}
  This is rewritten in terms of $X^{A}$ as
  {\color{Blue}{
 \begin{eqnarray}
  B &=& \frac{n_{2}}{4D(X)^{3}} \epsilon_{abc} X^{a} d X^{b} \wedge d X^{c}
   +  \frac{n_{1}}{4RD(X)} \epsilon_{ij} dX^{i} \wedge dX^{j} \nonumber \\
    &=& \frac{1}{2} B_{AB} dX^{A} \wedge dX^{B}. \nonumber
 \end{eqnarray} }}
  $B_{AB}$ has the {\color{Blue}{tangential condition 
 $B_{AB} X^{A} = 0$}}. 
  {\color{Blue}{
 \begin{eqnarray}
  & & B_{ab} = \frac{n_{2}}{2D(X)^{3}} \epsilon_{abc} X^{C}, \hspace{2mm}
  B_{ij} = \frac{n_{1}}{2RD(X)} \epsilon_{ij}, \nonumber \\
  & & B_{ai} = - B_{ia} = \frac{n_{1}}{2RD(X)^{3}} \epsilon_{ij} 
  X^{j} X^{a}. \nonumber
 \end{eqnarray} }}
 They are {\color{Blue}{singular at the equator ${\cal C}$}}, in which
 {\color{Magenta}{$D(X) = 0$}}. 
\newpage
%##################### Page 23 ####################################
 The inverse matrix {\color{Magenta}{$\alpha^{AB}$}}, such that \\
 {\color{Magenta}{$\alpha^{AB} B_{BC} = \delta_{AC} - X^{A} X^{C}/R^{2}$}},
 is
 {\color{Blue}{
 \begin{eqnarray}
  & & \alpha^{ab} = - \frac{2D(X)}{n_{2}} \epsilon_{abc} X^{c}, \hspace{2mm}
  \alpha^{ij} = - \frac{2D(X)^{3}}{n_{1}R} \epsilon_{ij}, \nonumber \\
  & & \alpha^{ai} = - \alpha^{ia} = - \frac{2D(X)}{n_{1}R} \epsilon_{ij}
  X^{j} X^{a}. \nonumber
 \end{eqnarray} }}
 This defines the Poisson bracket:
 {\color{Red}{
  \begin{eqnarray}
   \{ F(X), G(X) \}_{PB} = \frac{1}{2} \alpha^{AB}
  \partial_{A} F(X) \partial_{B} G(X). \nonumber
  \end{eqnarray} }}
 Via $\alpha^{AB}$, we define the noncommutative product as
 {\color{Red}{
 \begin{eqnarray}
&& F(X) \star' G(X) \nonumber \\
& =& F(X)  G(X) + i \ \alpha^{AB} \ 
\partial_A F(X) \ \partial_B G(X) \nonumber \\&&
- \frac{1}{2} \alpha^{AB} \ \alpha^{CD}
\ \partial_A \partial_C F(X) \ \partial_B \partial_D G(X)  \nonumber \\
&& - \frac{1}{3} \ \alpha^{AB} \ \left(\partial_B \ \alpha^{CD} 
\right) \ \left\{ \partial_A \partial_C F(X) \ \partial_D G(X) \right. 
  \nonumber \\ 
 && \left. - 
\partial_C F(X) \ \partial_A \partial_D G(X) \ \right\}  +\cdots. \nonumber
\end{eqnarray} }}


\newpage
%####################### Page 24 ########################33
\section{Conclusion}
 In this paper, the authors built the noncommutative product for
 {\color{Magenta}{the fuzzy $S^{4}$ sphere}}. \\

 To this end, they broke the symmetry {\color{Magenta}{$SO(5)$}} to
 {\color{Magenta}{$SO(3) \times SO(2)$}}. \\

 This has enabled us to apply the algebra of {\color{Magenta}{the fuzzy
 $S^{2}$ sphere}}. 

}
 \end{document}










