ncg

bachelorthesis in physics
git clone git://popovic.xyz/ncg.git
Log | Files | Refs

6_electroncg.tex (21445B)


      1 \subsection{Noncommutative Geometry of Electrodynamics\label{sec:6}}
      2 In this chapter we go through a derivation Electrodynamics with
      3 the almost commutative manifold $M\times F_X$ and the abelian gauge group
      4 $U(1)$. The conclusion is an unified description of gravity and
      5 electrodynamics although in the classical level.
      6 
      7 The almost commutative Manifold $M\times F_X$ outlines a local gauge group
      8 $U(1)$. The inner fluctuations of the Dirac operator relate to $Y_\mu$ the
      9 gauge field of $U(1)$. According to the setup we ultimately arrive at two
     10 serious problems.
     11 
     12 First of all the operator $D_F$, in the Two-Point space $F_X$, must vanish
     13 such that a real structure can exists. However this implies that the electrons
     14 are massless.
     15 
     16 The second problem arises when looking at the Euclidean action for a free
     17 Dirac field
     18 \begin{align}
     19     S = - \int i \bar{\psi}(\gamma ^\mu\partial _\mu - m) \psi d^4x,
     20 \end{align}
     21 where $\psi,\ \bar{\psi}$ must be considered as two independent variables.
     22 This means that the fermionic action $S_f$ needs two independent Dirac spinors.
     23 Let us try and construct two independent Dirac spinors with our data, first
     24 take a look at the decomposition of the basis and of the total
     25 Hilbertspace $H = L^2(S) \otimes H_F$. For the orthonormal basis of $H_F$ we
     26 can write $\{e, \bar{e}\}$ , where $\{e\}$ is the orthonormal basis of
     27 $H_F^+$ and $\{\bar{e}\}$ the orthonormal basis of $H_F^-$. Accompanied with
     28 the real structure we arrive at the following relations
     29 \begin{align}
     30     J_F e &= \bar{e} \;\;\;\;\;\; J_F \bar{e} = e, \\
     31     \gamma_F e &= e  \;\;\;\;\;\;   \gamma_F \bar{e} = \bar{e}.
     32 \end{align}
     33 Along with the decomposition of $L^2(S) = L^2(S)^+ \oplus L^2(S)^-$ and  $\gamma = \gamma _M
     34 \otimes \gamma _F$ we can obtain the positive eigenspace
     35 \begin{align}
     36     H^+ = L^2(S)^+ \otimes H_F^+ \oplus L(S)^- \otimes H_F^-.
     37 \end{align}
     38 So, for an $\xi \in H^+$ we can write
     39 \begin{align}
     40     \xi = \psi _L \otimes e + \psi _R \otimes \bar{e},
     41 \end{align}
     42 where $\psi_L \in L^2(S)^+$ and $\psi _R \in L^2(S)^-$ are the two Wheyl
     43 spinors. We denote that $\xi$ is only determined by one Dirac spinor $\psi :=
     44 \psi_L + \psi _R$. Since \textbf{we require two independent spinors}, our
     45 conclusion is that the definition of the fermionic action gives too much
     46 restrictions to the Two-Point space $F_X$.
     47 \subsubsection{The Finite Space}
     48 To solve the two problems we simply enlarge (double) the Hilbertspace. This
     49 is visualized by introducing multiplicities in Krajewski Diagrams
     50 \cite{ncgwalter} which will also allow us to choose a nonzero Dirac operator
     51 that will connect the two vertices and preserve real structure making our
     52 particles massive and bringing anti-particles into the mix.
     53 
     54 We start of with the same algebra $C^\infty(M, \mathbb{C}^2)$, corresponding
     55 to space $N= M\times X$. The Hilbertspace describes four particles, meaning
     56 it has four orthonormal basis elements. It describes \textbf{left handed
     57 electrons} and \textbf{right handed positrons}. This way we have
     58 $\{ \underbrace{e_R, e_L}_{\text{left-handed}}, \underbrace{\bar{e}_R,
     59 \bar{e}_L}_{\text{right-handed}}\}$ an orthonormal basis for $H_F =
     60 \mathbb{C}^4$. Accompanied with the real structure $J_F$ allowing us to
     61 interchange particles with antiparticles by the following equations
     62 \begin{align}
     63     &J_F e_R = \bar{e}_R, \\
     64     &J_F e_L = \bar{e_L}, \\
     65     \nonumber \\
     66     &\gamma _F e_R = -e_R,\\
     67     &\gamma_F e_L = e_L,
     68 \end{align}
     69 where $J_F$ and $\gamma_F$ have to following properties
     70 \begin{align}
     71     &J_F^2 = 1,\\
     72     & J_F \gamma_F  = - \gamma_F J_F.
     73 \end{align}
     74 By the means of $\gamma_F$ we have two options to decompose the total
     75 Hilbertspace $H$, firstly into
     76 \begin{align}
     77     H_F = \underbrace{H_F^+}_{\text{ONB } \{e_L, \bar{e}_L\}}
     78     \oplus \underbrace{H_F^-}_{\text{ONB } \{e_R, \bar{e}_R\}},
     79 \end{align}
     80 or alternatively into the eigenspace of particles and their
     81 antiparticles (electrons and positrons) which is preferred in literature and
     82 which will be used further out
     83 \begin{align}
     84     H_F = \underbrace{H_{e}}_{\text{ONB } \{e_L, e_R\}} \oplus
     85     \underbrace{H_{\bar{e}}}_{\text{ONB } \{\bar{e}_L, \bar{e}_R\}},
     86 \end{align}
     87 the shortening `ONB' means orthonormal basis.
     88 
     89 The action of $a \in A = \mathbb{C}^2$ on $H$ with respect to the ONB
     90 $\{e_L, e_R, \bar{e}_L, \bar{e}_R\}$ is represented by
     91 \begin{align}\label{eq:leftrightrepr}
     92     a =
     93     (a_1 , a_2 ) \mapsto
     94         \begin{pmatrix}
     95             a_1 &0 &0 &0\\
     96              0&a_1 &0 &0\\
     97             0 &0 &a_2 &0\\
     98             0 &0 &0 &a_2\\
     99         \end{pmatrix}
    100 \end{align}
    101 Do note that this action commutes wit the grading and that $[a, b^\circ] = 0$
    102 with $b:= J_F b^*J_F$ because both the left and the right action are given by
    103 diagonal matrices according to equation \eqref{eq:leftrightrepr}. Furthermore
    104 note that we are still left with $D_F = 0$ and the following spectral triple
    105 \begin{align}\label{eq:fedfail}
    106         \left( \mathbb{C}^2, \mathbb{C}^2, D_F=0; J_F =
    107         \begin{pmatrix}
    108             0 & C \\ C &0
    109         \end{pmatrix},
    110         \gamma _F =
    111         \begin{pmatrix}
    112             1 & 0 \\ 0 &-1
    113         \end{pmatrix}
    114         \right).
    115     \end{align}
    116 It can be represented in the following Krajewski diagram \cite{ncgwatler},
    117 with two nodes of multiplicity two, in figure \ref{fig:krajewski} bellow.
    118     \begin{figure}[H] \centering
    119     \begin{tikzpicture}[
    120         dot/.style = {draw, circle, inner sep=0.06cm},
    121         bigdot/.style = {draw, circle, inner sep=0.09cm},
    122         no/.style = {},
    123         ]
    124         \node[no](a) at (0,0) [label=left:$\textbf{1}^\circ$] {};
    125         \node[no](b) at (0, -1) [label=left:$\textbf{1}^\circ$] {};
    126         \node[no](c) at (0.5, 0.5) [label=above:$\textbf{1}$] {};
    127         \node[no](d) at (1.5, 0.5) [label=above:$\textbf{1}$] {};
    128         \node[dot](d0) at (1.5,0) [] {};
    129         \node[dot](d0) at (0.5,-1) [] {};
    130         \node[bigdot](d0) at (1.5,0) [] {};
    131         \node[bigdot](d0) at (0.5,-1) [] {};
    132         \end{tikzpicture}
    133         \caption{Krajewski diagram of the spectral triple from equation
    134         \ref{eq:fedfail}
    135         \label{fig:krajewski}
    136     }
    137     \end{figure}
    138 \subsubsection{A noncommutative Finite Dirac Operator}
    139 To extend our spectral triple with a non-zero Operator, we need to take a
    140 closer look at the Krajewski diagram in figure \ref{fig:krajewski} above. Notice that edges only exist
    141 between multiple vertices, meaning we can construct a Dirac operator mapping
    142 between the two vertices. The operator can be represented by the following matrix
    143 \begin{align}\label{eq:feddirac}
    144     D_F =
    145     \begin{pmatrix}
    146     0 & d & 0 & 0 \\
    147     \bar{d} & 0 & 0 & 0 \\
    148     0 & 0 & 0 & \bar{d} \\
    149     0 & 0 & d & 0
    150     \end{pmatrix}
    151 \end{align}
    152 We can now define the finite space $F_{ED}$.
    153 \begin{align}
    154     F_{ED} := (\mathbb{C}^2, \mathbb{C}^4, D_F; J_F, \gamma_F)
    155 \end{align}
    156 where $J_F$ and $\gamma_F$ are as in equation \eqref{eq:fedfail} and $D_F$
    157 from equation \eqref{eq:feddirac}.
    158 
    159 \subsubsection{Almost commutative Manifold of Electrodynamics}
    160 The almost commutative manifold $M\times F_{ED}$ has KO-dimension 2, and is
    161 represented by the following spectral triple
    162 \begin{align}\label{eq:almost commutative manifold}
    163     M\times F_{ED} := \big(C^\infty(M,\mathbb{C}^2),\ L^2(S)\otimes
    164     \mathbb{C}^4,\
    165     D_M\otimes 1 +\gamma _M \otimes D_F;\; J_M\otimes J_F,\ \gamma_M\otimes
    166     \gamma _F\big).
    167 \end{align}
    168 The algebra didn't change, thus we can decompose it like before
    169 \begin{align}
    170     C^\infty(M, \mathbb{C}^2) = C^\infty (M) \oplus C^\infty (M).
    171 \end{align}
    172 As for the Hilbertspace, we can decomposition it in the following way
    173 \begin{align}
    174     H = (L^2(S) \otimes H_e ) \oplus (L^2(S) \otimes H_{\bar{e}}).
    175 \end{align}
    176 Note that the one component of the algebra is acting on $L^2(S) \otimes H_e$,
    177 and the other one acting on $L^2(S) \otimes H_{\bar{e}}$. In other words the components of
    178 the decomposition of both the algebra and the Hilbertspace match by the action of
    179 the algebra.
    180 
    181 The derivation of the gauge theory is the same for $F_{ED}$ as for the
    182 Two-Point space $F_X$. We have $\mathfrak{B}(F) \simeq U(1)$ and for an
    183 arbitrary gauge field $B_\mu = A_\mu - J_F A_\mu J_F^{-1}$ we can write
    184 \begin{align} \label{field}
    185     B_\mu =
    186     \begin{pmatrix}
    187         Y_\mu & 0 & 0 & 0 \\
    188         0 & Y_\mu& 0 & 0 \\
    189         0 & 0 & Y_\mu& 0 \\
    190         0 & 0 & 0 & Y_\mu
    191     \end{pmatrix} \;\;\;\;\;\ \text{for} \;\;\ Y_\mu (x) \in \mathbb{R}.
    192 \end{align}
    193 There is one single $U(1)$ gauge field $Y_\mu$, carrying the action of the
    194 gauge group
    195 \begin{align}
    196    \text{$\mathfrak{B}$}(M\times F_{ED}) \simeq C^\infty(M, U(1))
    197 \end{align}
    198 
    199 The space $N = M\times X$ consists of two copies of $M$.
    200 If $D_F = 0$ we have infinite distance between the two copies, yet now we have
    201 adjusted the spectral triple to have a nonzero Dirac operator. The new
    202 Dirac operator still has a commuting relation with the algebra $[D_F, a] = 0$
    203 $\forall a \in A$, and we should note that the distance between the two
    204 copies of $M$ is still infinite. This is purely an mathematically abstract
    205 observation and doesn't affect physical results.
    206 
    207 \subsubsection{Spectral Action}
    208 In this chapter we bring all our results together to establish an
    209 Action functional to describe a physical system. It turns out that
    210 the Lagrangian of the almost commutative manifold $M\times F_{ED}$
    211 corresponds to the Lagrangian of Electrodynamics on a curved
    212 background manifold (+ gravitational Lagrangian), consisting of the spectral
    213 action $S_b$ (bosonic) and of the fermionic action $S_f$.
    214 
    215 The simplest spectral action of a spectral triple $(A, H, D)$ is given by the
    216 trace of a function of $D$. We also consider inner fluctuations of the Dirac
    217 operator
    218 \begin{align}
    219     D_\omega = D + \omega + \varepsilon' J\omega J^{-1},
    220 \end{align}
    221 where $\omega = \omega ^* \in \Omega_D^1(A)$.
    222 \begin{mydefinition}
    223     Let $f:\mathbb{R} \rightarrow \mathbb{R}$ be a suitable function
    224     \textbf{positive and even}. The spectral action is then
    225     \begin{align}
    226         S_b [\omega] := \text{Tr}\big(f(\frac{D_\omega}{\Lambda})\big)
    227     \end{align}
    228     where $\Lambda$ is a real cutoff parameter. The minimal condition on $f$
    229     is that $f(\frac{D_\omega}{\Lambda})$ is  a trace class operator. A trace
    230     class operator is a compact operator with a well defined finite trace
    231     independent of the basis. The subscript $b$ in $S_b$ stands for bosonic,
    232     because in physical applications $\omega$ will describe bosonic fields.
    233 
    234     In addition to the bosonic action $S_b$, we can define a topological spectral
    235     action $S_{top}$. Leaning on the grading $\gamma$ the topological spectral action is
    236     \begin{align}
    237         S_{\text{top}}[\omega] := \text{Tr}(\gamma\
    238         f(\frac{D_\omega}{\Lambda})).
    239     \end{align}
    240 \end{mydefinition}
    241 \begin{mydefinition}\label{def:fermionic action}
    242     The fermionic action is defined by
    243     \begin{align}
    244         S_f[\omega, \psi] = (J\tilde{\psi}, D_\omega \tilde{\psi})
    245     \end{align}
    246     with $\tilde{\psi} \in H_{cl}^+ := \{\tilde{\psi}: \psi \in H^+\}$, where
    247     $H_{cl}^+$ is a set of Grassmann variables in $H$ in the $+1$-eigenspace
    248     of the grading $\gamma$.
    249 \end{mydefinition}
    250 
    251 %---------------------- APPENDIX ?????????????--------------------
    252 Grassmann variables are a set of Basis vectors of a vector space, they
    253 form a unital algebra over a vector field $V$, where the generators are
    254 anti commuting, that is for Grassmann variables $\theta _i, \theta _j$  we have
    255 \begin{align}
    256     &\theta _i \theta _j = -\theta _j \theta _i, \\
    257     &\theta _i x = x\theta _j \;\;\;\; x\in V, \\
    258     &(\theta_i)^2 = 0 \;\;\; (\theta _i \theta _i = -\theta _i \theta _i).
    259 \end{align}
    260 %---------------------- APPENDIX ?????????????--------------------
    261 \begin{myproposition}
    262     The spectral action of the almost commutative manifold $M$ with $\dim(M)
    263     =4$ with a fluctuated Dirac operator is
    264     \begin{align}
    265         \text{Tr}(f\frac{D_\omega}{\Lambda}) \sim \int_M \mathcal{L}(g_{\mu\nu},
    266          B_\mu, \Phi) \sqrt{g}\ d^4x + O(\Lambda^{-1}),
    267     \end{align}
    268     where
    269     \begin{align}
    270         \mathcal{L}(g_{\mu\nu}, B_\mu, \Phi) =
    271         N\mathcal{L}_M(g_{\mu\nu})
    272         \mathcal{L}_B(B_\mu)+
    273         \mathcal{L}_\phi(g_{\mu\nu}, B_\mu, \Phi).
    274     \end{align}
    275     The Lagrangian $\mathcal{L}_M$ is of the spectral triple $(C^\infty(M) ,
    276     L^2(S), D_M)$, represented by the following term
    277     \begin{align}\label{lagr}
    278         \mathcal{L}_M(g_{\mu\nu}) := \frac{f_4 \Lambda ^4}{2\pi^2} -
    279         \frac{f_2 \Lambda^2}{24\pi ^2}s - \frac{f(0)}{320\pi^2} C_{\mu\nu
    280         \varrho \sigma}C^{\mu\nu \varrho \sigma},
    281     \end{align}
    282     where $C^{\mu\nu \varrho \sigma}$ is the Weyl tensor defined in terms of the Riemannian
    283     curvature tensor $R_{\mu\nu \varrho \sigma}$ and the Ricci tensor
    284     $R_{\nu\sigma} = g^{\mu\varrho} R_{\mu\nu \varrho\sigma}$ such that
    285     \begin{align}
    286         C^{\mu\nu\varrho\sigma}C_{\mu\nu\varrho\sigma}=
    287         R_{\mu\nu\varrho\sigma}R^{\mu\nu\varrho\sigma} -
    288         2R_{\nu\sigma}R^{\nu\sigma} + \frac{1}{2}s^2.
    289     \end{align}
    290     The kinetic term of the gauge field is described by the Lagrangian
    291     $\mathcal{L}_B$, which takes the following shape
    292     \begin{align}
    293         \mathcal{L}_B(B_\mu) := \frac{f(0)}{24\pi^2}
    294         \text{Tr}(F_{\mu\nu}F^{\mu\nu}).
    295     \end{align}
    296     Lastly $\mathcal{L}_\phi$ is the scalar-field Lagrangian with a boundary
    297     term, given by
    298     \begin{align}
    299         \mathcal{L}_\phi(g_{\mu\nu}, B_\mu, \Phi) :=
    300         &-\frac{2f_2\Lambda^2}{4\pi^2}\text{Tr}(\Phi^2) + \frac{f(0)}{8\pi^2}
    301         \text{Tr}(\Phi^4) + \frac{f(0)}{24\pi^2}
    302         \Delta(\text{Tr}(\Phi^2))\nonumber\\
    303         &+ \frac{f(0)}{48\pi^2}s\text{Tr}(\Phi^2)
    304         \frac{f(0)}{8\pi^2}\text{Tr}((D_\mu \Phi)(D^\mu \Phi)).
    305     \end{align}
    306 \end{myproposition}
    307 \begin{proof}
    308      The dimension of the manifold $M$ is $\dim(M) = \text{Tr}(id) =4$. For
    309      an $x \in M$, we have an asymptotic expansion of the term
    310      $\text{Tr}(f(\frac{D_\omega}{\Lambda}))$ as $\Lambda$ goes to infinity,
    311      which can be written as
    312      \begin{align}
    313          \text{Tr}(f(\frac{D_\omega}{\Lambda})) \simeq& \ 2f_4 \Lambda ^4
    314          a_0(D_\omega ^2)+ 2f_2\Lambda^2 a_2(D_\omega^2)\nonumber \\&+ f(0) a_4(D_\omega^4)
    315          +O(\Lambda^{-1}).\label{eq:trheatkernel}
    316      \end{align}
    317      We have to note here that the heat kernel coefficients are zero for uneven $k$,
    318      and they are dependent on the fluctuated Dirac operator
    319      $D_\omega$. We can rewrite the heat kernel coefficients in terms of $D_M$,
    320      for the first two terms $a_0$ and $a_2$ we use $N:=
    321      \text{Tr}(\mathbbm{1}_{H_F})$ and one obtains
    322      \begin{align}
    323          a_0(D_\omega^2) &= Na_0(D_M^2),\\
    324          a_2(D_\omega^2) &= Na_2(D_M^2) - \frac{1}{4\pi^2}\int_M
    325          \text{Tr}(\Phi^2)\sqrt{g}d^4x.
    326      \end{align}
    327      For $a_4$ we extend in terms of coefficients of $F$ from equation
    328      \eqref{eq: a_4}
    329      \begin{align}
    330          &\frac{1}{360}\text{Tr}(60RE)= -\frac{1}{6}S(NR + 4
    331          \text{Tr}(\Phi^2))\\
    332         \nonumber\\
    333          &E^2 = \frac{1}{16}R^2\otimes 1 + 1\otimes \Phi^4 - \frac{1}{4}
    334          \gamma^\mu\gamma^\nu \gamma^\varrho\gamma^\sigma
    335          F_{\mu\nu}F^{\mu\nu}+\nonumber\\
    336          &\;\;\;\;\;\;\;+\gamma^\mu\gamma^\nu\otimes(D_\mu\Phi)(D_\nu
    337          \Phi)+\frac{1}{2}s\otimes \Phi^2 + \ \text{traceless terms},\\
    338          \nonumber\\
    339          &\frac{1}{360}\text{Tr}(180E^2) = \frac{1}{8}R^2N + 2\text{Tr}(\Phi^4)
    340          + \text{Tr}(F_{\mu\nu}F^{\mu\nu}) +\nonumber\\
    341          &\;\;\;\;\;\;\;+2\text{Tr}((D_\mu\Phi)(D^\mu\Phi))
    342          + s\text{Tr}(\Phi^2)\\
    343          \nonumber\\
    344          &\frac{1}{360}\text{Tr}(-60\Delta E)=
    345          \frac{1}{6}\Delta(NR+4\text{Tr}(\Phi^2)).
    346      \end{align}
    347      The cross terms of the trace in $\Omega_{\mu\nu}^E\Omega^{E\mu\nu}$
    348      vanishes because of the antisymmetric property of the Riemannian
    349      curvature tensor, reading
    350      \begin{align}
    351          \Omega_{\mu\nu}^E\Omega^{E\mu\nu} = \Omega_{\mu\nu}^S\Omega^{S\mu\nu}
    352          \otimes 1 - 1\otimes F_{\mu\nu}F^{\mu\nu} + 2i\Omega_{\mu\nu}^S
    353          \otimes F^{\mu\nu}.
    354      \end{align}
    355      The trace  of the cross term $\Omega^{S}_{\mu\nu}$ vanishes because
    356      \begin{align}
    357          \text{Tr}(\Omega^{S}_{\mu\nu}) = \frac{1}{4}
    358          R_{\mu\nu\varrho\sigma}\text{Tr}(\gamma^\mu\gamma^\nu) = \frac{1}{4}
    359          R_{\mu\nu\varrho\sigma}g^{\mu\nu} =0,
    360      \end{align}
    361      then the trace of the whole term is given by
    362      \begin{align}
    363          \frac{1}{360}\text{Tr}(30\Omega^E_{\mu\nu}\Omega^{E\mu\nu}) =
    364          \frac{N}{24}R_{\mu\nu\varrho\sigma}R^{\mu\nu\varrho\sigma}
    365          -\frac{1}{3}\text{Tr}(F_{\mu\nu}F^{\mu\nu}).
    366      \end{align}
    367      Finally plugging the results into the coefficient $a_4$ and simplifying
    368      one gets
    369      \begin{align}
    370          a_4(x, D_\omega^4) &= Na_4(x, D_M^2) + \frac{1}{4\pi^2}\bigg(\frac{1}{12} s
    371          \text{Tr}(\Phi^2) + \frac{1}{2}\text{Tr}(\Phi^4) \nonumber \\
    372          &+ \frac{1}{4}
    373          \text{Tr}((D_\mu\Phi)(D^\mu \Phi)) + \frac{1}{6}
    374          \Delta\text{Tr}(\Phi^2) + \frac{1}{6}
    375          \text{Tr}(F_{\mu\nu}F^{\mu\nu})\bigg).
    376      \end{align}
    377      The only thing left is to substitute the heat kernel coefficients into the
    378      heat kernel expansion in equation \eqref{eq:trheatkernel}.
    379 \end{proof}
    380 
    381 \subsubsection{Fermionic Action}
    382 We remind ourselves the definition of the fermionic action in definition
    383 \ref{def:fermionic action} and the manifold we are dealing with in equation
    384 \eqref{eq:almost commutative manifold}. The Hilbertspace $H_F$ is separated
    385 into the particle-antiparticle states with ONB $\{e_R, e_L, \bar{e}_R,
    386 \bar{e}_L\}$. The orthonormal basis of $H_F^+$ is $\{e_L, \bar{e}_R\}$ and
    387 consequently for $H_F^-$, $\{e_R, \bar{e}_L\}$. The decomposition of a spinor
    388 $\psi \in L^2(S)$ in each of the eigenspaces $H_F^\pm$ is $\psi = \psi_R+
    389 \psi_L$. Meaning for an arbitrary $\psi \in H^+$ we can write
    390 \begin{align}
    391     \psi = \chi_R \otimes e_R + \chi_L \otimes e_L + \psi_L \otimes
    392     \bar{e}_R+
    393     \psi_R \otimes \bar{e}_L,
    394 \end{align}
    395 where $\chi_L, \psi_L \in L^2(S)^+$ and $\chi_R, \psi_R \in L^2(S)^-$.
    396 
    397 Since the fermionic action yields too much restriction on $F_{ED}$ (modified
    398 Two-Point space $F_X$) one redefines it by taking into account the fluctuated Dirac
    399 operator
    400 \begin{align}
    401     D_\omega = D_M \otimes i + \gamma^\mu \otimes B_\mu + \gamma_M \otimes
    402     D_F.
    403 \end{align}
    404 The Fermionic Action is
    405 \begin{align}
    406 S_F = (J\tilde{\xi}, D_\omega\tilde{\xi})
    407 \end{align}
    408 for a $\xi \in H^+$. Then the straight forward calculation gives \begin{align}
    409     \frac{1}{2}(J\tilde{\xi}, D_\omega\tilde{\xi})
    410         &=\frac{1}{2}(J\tilde{\xi}, (D_M \otimes
    411         i)\tilde{\xi})\label{eq:fermionic1}\\
    412         &+\frac{1}{2}(J\tilde{\xi}, (\gamma^\mu \otimes B_\mu)
    413         \tilde{\xi})\label{eq:fermionic2}\\
    414         &+\frac{1}{2}(J\tilde{\xi}, (\gamma_M\otimes
    415         D_F)\tilde{\xi})\label{eq:fermionic3},
    416 \end{align}
    417 (note that we add the constant $\frac{1}{2}$ to the action).
    418 For the term in \eqref{eq:fermionic1} we calculate
    419 \begin{align}
    420     \frac{1}{2}(J\tilde{\xi}, (D_M\otimes 1)\tilde{\xi}) &=
    421     \frac{1}{2}(J_M\tilde{\chi}_R,D_M\tilde{\psi}_L)+\nonumber
    422     \frac{1}{2}(J_M\tilde{\chi}_L,D_M\tilde{\psi}_R)+
    423     \\&+\frac{1}{2}(J_M\tilde{\psi}_L,D_M\tilde{\psi}_R)+\nonumber
    424     \frac{1}{2}(J_M\tilde{\chi}_R,D_M\tilde{\chi}_L)\\
    425     &= (J_M\tilde{\chi},D_M\tilde{\chi}).
    426 \end{align}
    427 For the term in \eqref{eq:fermionic2} we have
    428 \begin{align}
    429     \frac{1}{2}(J\tilde{\xi}, (\gamma^\mu \otimes B_\mu)\tilde{\xi})&=
    430     -\frac{1}{2}(J_M\tilde{\chi}_R, \gamma^\mu Y_\mu\tilde{\psi}_R)
    431     -\frac{1}{2}(J_M\tilde{\chi}_L, \gamma^\mu Y_\mu\tilde{\psi}_R)+\nonumber\\
    432     &+\frac{1}{2}(J_M\tilde{\psi}_L, \gamma^\mu Y_\mu\tilde{\chi}_R)+
    433     \frac{1}{2}(J_M\tilde{\psi}_R, \gamma^\mu Y_\mu\tilde{\chi}_L)=\nonumber\\
    434     &= -(J_M\tilde{\chi}, \gamma^\mu Y_\mu\tilde{\psi}).
    435 \end{align}
    436 And for \eqref{eq:fermionic3} we can write
    437 \begin{align}
    438     \frac{1}{2}(J\tilde{\xi}, (\gamma_M\otimes D_F)\tilde{\xi})&=
    439     +\frac{1}{2}(J_M\tilde{\chi}_R, d\gamma_M\tilde{\chi}_R)
    440     +\frac{1}{2}(J_M\tilde{\chi}_L, \bar{d}\gamma_M\tilde{\chi}_L)+\nonumber\\
    441     &+\frac{1}{2}(J_M\tilde{\chi}_L, \bar{d}\gamma_M\tilde{\chi}_L)
    442     +\frac{1}{2}(J_M\tilde{\chi}_R, d\gamma_M\tilde{\chi}_R)=\nonumber\\
    443     &= i(J_M\tilde{\chi}, m\tilde{\psi}).
    444 \end{align}
    445 A small problem arises, we obtain a complex mass parameter $d$, but we can
    446 write $d:=im$ for $m\in \mathbb{R}$, which stands for the real mass.
    447 
    448 Finally the fermionic action of $M\times F_{ED}$ takes the form
    449     \begin{align}
    450         S_f = -i\big(J_M\tilde{\chi}, \gamma(\nabla^S_\mu - i\Gamma_\mu)
    451         \tilde{\Psi}\big) + \big(S_M\tilde{\chi}_L, \bar{d}\tilde{\psi}_L\big) -
    452         \big(J_M\tilde{\chi}_R, d \tilde{\psi}_R\big).
    453     \end{align}
    454 Ultimately we arrive at the full Lagrangian of the almost commutative
    455 manifold $M\times F_{ED}$, which is the sum of the purely gravitational
    456 Lagrangian
    457 \begin{align}\label{eq:final1}
    458         \mathcal{L}_{grav}(g_{\mu\nu})=4\mathcal{L}_M(g_{\mu\nu})+
    459         \mathcal{L}_\phi (g_{\mu\nu}),
    460     \end{align}
    461 and the Lagrangian of electrodynamics
    462 \begin{align}\label{eq:final2}
    463         \mathcal{L}_{ED} = -i\bigg\langle
    464         J_M\tilde{\chi},\big(\gamma^\mu(\nabla^S_\mu - iY_\mu) -m\big)\tilde{\psi})
    465         \bigg\rangle
    466         +\frac{f(0)}{6\pi^2} Y_{\mu\nu}Y^{\mu\nu}.
    467     \end{align}
    468