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\begin{document}
\author{Alexey A. Kryukov \\
%\\
%{\small Department of Mathematics,}\\
{\ University of Wisconsin}\\
{\small 5119 Helen C. White, 600 N. Park St., Madison, WI 53706}\\
%{\ University of Wisconsin}\\
%\\
\smallskip
{\small e-mail: aakrioukov@facstaff.wisc.edu}\\
\\
\smallskip
{\small Key words: conformal symmetry, spinors, complex space-time}\\
\smallskip
{\small PACS: 02.40.Ky, 02.40.Tt, 03.30.+p, 04.50.+h}\\
}
\title{Conformal Transformations of Space-Time as Vector Bundle Automorphisms}
\maketitle
\begin{abstract}
Conformal group of Minkowski space-time M is considered as a group of bundle automorphisms
of a vector bundle U over M. 4-component spin-vectors (4-spinors) are sections of a subbundle of the tangent bundle
over U. Isotropic 4-vectors are images of 4-spinors under projection. This leads to a
particularly clear interpretation of the spin properties of Nature.
\end{abstract}
\bigskip
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
%\newpage
\section{Introduction}
\setcounter{equation}{0}
According to Einstein, space-time continuum is a pseudo-Riemannian
manifold. Although it is a real manifold, complex numbers have proven to be
very useful in gravity and even more so in physics on space-time. This
naturally led to the idea of incorporating complex numbers into the
structure of space-time manifold.
The most far reaching and successful advance in this direction is the
twistor theory of Penrose \cite{Penrose,Ward-Wells}. In this theory, the
(conformally compactified and complexified) Minkowski space-time is
identified with the Grassmanian $G_{2,4}(C)$ of complex 2-planes in $C^{4}.$
The drawback of twistors is that they appear to be powerful in a rather
restrictive class of conformally flat and half-conformally flat metrics on
space-time. In addition, the idea that the usual manifold of points has to
be replaced with a manifold of complex 2-dimensional surfaces has not been
readily accepted by physicists. As a result, twistors are generally
considered a very useful tool for dealing with a significant number of
linear and non-linear differential equations relevant to physics. The
essence of the tool is the twistor transform \cite{Ward-Wells},
which permits interpretation of equations on space-time in terms of holomorphic
data on the associated twistor space.
The success of twistor theory contributes to the feeling that complex
structure is something immanent to space-time and should be considered as
more than just a tool for applications.
In this paper, the basic geometry of spinors and twistors is revisited
and a simple geometric interpretation of conformal symmetry is proposed.
The conformal group appears as a group of fibre preserving transformations of a
certain vector bundle $\pi :$ $U\rightarrow \bar{M}$ over the space-time manifold $%
\bar{M}$. The total space $U$ possesses a natural complex structure.
In this light the spin properties of nature also obtain a clear geometric interpretation.
Usually spin-vectors are considered as objects of algebraic nature. In fact,
they are elements of the space of spin representation of Lorentz group.
In Penrose's approach, spin-vectors are interpreted as isotropic flags \cite{Penrose1}.
This gives a semi-geometric representation of spin-vectors in Minkovski
space-time.
In the paper 4-component spin-vectors (4-spinors) appear as vectors on the total space $U$ of
the above mentioned vector bundle. In fact,
they are restricted to $\bar{M}$ sections of the tangent bundle $T(U)$ over $U$.
Isotropic 4-vectors are images of 4-spinors under projection $\pi$.
The entire approach has been inspired by the above mentioned strong belief
that the space-time manifold must be extended to a complex manifold. It was
the main method used by Einstein to interpret physics on space-time as a
physics of space-time. The fact that complex numbers are used extensively in
describing physics favors the idea of a complex space-time manifold.
On the other hand, any analytic (pseudo-) Riemannian space-time can be
isometrically embedded into a complex 4-manifold with a nice metric
structure, e.g., (indefinite-) K\"{a}hler, on it (see \cite{Alex1} and references there).
So the idea of extension of both space-time and Riemannian structure is easily fulfilled. The
problem is to find a correct extension. That means, in particular, to relate
the extension to conformal symmetry, which is the main underlying theme of
the twistor theory. It also means to find a natural geometric representation of 4-component
spin-vectors within the extension formalism.
Yet another major factor that led to the proposed interpretation was an
analysis of similarities and differences between conformal group action on
the spheres $%
S^{2}$ and $S^{4}$, and on (conformally compactified) Minkowski space-time $%
\bar{M}\simeq S^{1}\times S^{3}$. We start with the bundles $\pi :C_{\ast
}^{2}\rightarrow CP^{1}\simeq S^{2}$ and $\pi :C_{\ast }^{4}\rightarrow
HP^{1}\simeq S^{4}$ that provide a clear geometric framework for conformal
group action on $S^{2}$ and $S^{4}.$ Later these bundles are generalized to
a universal bundle construction applicable to $\bar{M}$ as well.
%%%%%%%%%%%%%%%%%
\section{Geometry of spin}
\setcounter{equation}{0}
%%%%%%%%%
In the simplest case, a space-time manifold is a four-dimensional space $M$ of
vectors $\ X=$\ $(x^{0},...,x^{3})$. To measure distances and angles on $M$
an inner product of vectors is introduced: $(X,Y)=\eta _{ik}x^{i}y^{k}$. \
The norm $(X,X)=\left\| X\right\| _{M}^{2}$ must be independent of the
choice of coordinates in space-time. This leads to Minkowski tensor $\eta
_{ik}$ of signature $(+,-,-,-)$ and to the group of (restricted) Lorentz
transformations $SO(1,3)$ preserving $\left\| X\right\| _{M}^{2}.$ Space $M$
with the above metric is called Minkowski space.
In order to describe electrons, it is necessary to generalize vectors to
spin-vectors. For this, we identify $M$ with the space $H(2)$ of $2\times 2$
Hermitian matrices via the following isomorphism:
\begin{equation}
\label{2.1}
X=\ (x^{0},...,x^{3})\rightarrow p=\left(
\begin{array}{ll}
x^{0}+x^{3} & x^{1}+ix^{2} \\
x^{1}-ix^{2} & x^{0}-x^{3}
\end{array}
\right).
\end{equation}
Clearly, $\det p=\left\| X\right\| _{M}^{2}.$ \ \ If $\det p=0,$ i.e. $X$ is
a null vector, then \
\begin{equation}
\label{2.2}
\left(
\begin{array}{ll}
x^{0}+x^{3} & x^{1}+ix^{2} \\
x^{1}-ix^{2} & x^{0}-x^{3}
\end{array}
\right) =\left(
\begin{array}{l}
\xi \\
\eta
\end{array}
\right) \left(
\begin{array}{ll}
\bar{\xi} & \bar{\eta}
\end{array}
\right) =\left(
\begin{array}{l}
\xi \\
\eta
\end{array}
\right) \left(
\begin{array}{l}
\xi \\
\eta
\end{array}
\right) ^{+}
\end{equation}
for some vector $\left(
\begin{array}{l}
\xi \\
\eta
\end{array}
\right) \in C^{2}.$ Vectors $\left(
\begin{array}{l}
\xi \\
\eta
\end{array}
\right) $ are called spin-vectors or spinors. Because of the above relation,
spin-vectors are also called square-roots of vectors.
If $A\in $ $SL(2,C)$ acts on $C^{2}$\ and $\left(
\begin{array}{l}
\mu \\
\nu
\end{array}
\right) =A\left(
\begin{array}{l}
\xi \\
\eta
\end{array}
\right) $, then we have the induced action on $M:$ $p\rightarrow p^{\prime
}=ApA^{+}.$ This action preserves $\left\| X\right\| _{M}^{2}$ as $\det
p=\det ApA^{+}.$ It is therefore an action of the Lorentz group. If, on the
other hand, $L\in SO(1,3)$, then under the above isomorphism $X\rightarrow p$, $LX$
corresponds to $\ ApA^{+},$ where $A\in SL(2,C)$ is determined up to a sign.
Therefore $SL(2,C)$ is a 2-1 covering of the Lorentz group $SO(1,3)$.
It is known that the (restricted) Lorentz group is isomorphic to the
group $C(2)$\ of (proper) conformal transformations of the sphere $%
S^{2}.$ That is, it preserves angles between vectors tangent to the sphere.
The relation between $C(2)$ and $SL(2,C)$ has a nice interpretation in
terms of the stereographic projection. On Figure 1 $(x,y,z)$ are
Cartesian coordinates of a point $P$ on the unit sphere $S^{2}\subset R^{3}.$
Center $O$ of $S^{2}$ is at the origin of the coordinate system. Point $\xi $
on the complex plane $C$ through $O$ is the corresponding complex coordinate. We have
$\xi =\frac{x+iy}{1-z}$ and $SL(2,C)$, acting
via fractional linear transformations on the complex plane $C$, induces the
conformal group action on $S^{2}$.
\begin{figure}[ht]
\label{fig:1}
\centerline{\epsfig{figure=sphere1.eps}}
\centerline{\small{Figure 1}}
\end{figure}
To clarify the geometry of this correspondence, assume that the (unit) sphere $S^{2}$ is
embedded in the complex Euclidean space $C^{2}$ with coordinates $(\xi ,\eta ).$ The north pole of $%
S^{2} $ is taken to be at the origin $\xi =\eta =0$. The complex plane $C$\
is given by $\eta =1$ (see Figure 2). Let $A\in SL(2,C),$ $A=\left(
\begin{array}{ll}
a & b \\
c & d
\end{array}
\right) ,$ and \ $\left[
\begin{array}{l}
\mu \\
\nu
\end{array}
\right] $ denotes the complex line through $\left(
\begin{array}{l}
\mu \\
\nu
\end{array}
\right) $ in $C^{2}.$ Assume $\nu \neq 0.$ Then $\left[
\begin{array}{l}
\mu \\
\nu
\end{array}
\right] =\left[
\begin{array}{l}
\xi \\
1
\end{array}
\right] $ , where $\xi =\mu \nu ^{-1}$ and \
\begin{equation}
\label{2.3}
A\left[
\begin{array}{l}
\xi \\
1
\end{array}
\right] =\left(
\begin{array}{ll}
a & b \\
c & d
\end{array}
\right) \left[
\begin{array}{l}
\xi \\
1
\end{array}
\right] =\left[
\begin{array}{l}
(a\xi +b)(c\xi +d)^{-1} \\
\multicolumn{1}{c}{1}
\end{array}
\right].
\end{equation}
The case $\nu =0,\mu \neq 0$ is treated similarly. The resulting fractional
transformation $\xi \rightarrow (a\xi +b)(c\xi +d)^{-1}$ is a conformal
transformation of the compactified complex plane $\bar{C}$ which is
conformally equivalent to $S^{2}.$
With the above embedding of $S^{2}$ into $C^{2}$ the entire picture has
almost a mechanical interpretation. On Figure 2 $SL(2,C)$ acts on $%
C^{2}$ and therefore on the complex lines in $C^{2},$ i.e. on $CP^{1}.$ This
action gets passed through the ``poles'' (complex lines of $C^{2}$) to the
sphere $S^{2}.$ Location of the sphere is kept fixed, so $S^{2}$ can only
transform along its surface. The resulting transformation of $S^{2}$ is
conformal. In particular, for an action of the $SU(2)$ subgroup of $SL(2,C)$
the resulting transformation is an $SO(3)$ rotation yielding the isomorphism
$SO(3)\simeq SU(2)/Z_{2}.$
\begin{figure}[ht]
\label{fig:2}
\centerline{\epsfig{figure=sphere2.eps}}
\centerline{\small{Figure 2}}
\end{figure}
It is worth noticing that this mechanical interpretation can be put in
precise terms using the language of fibre bundles. Consider the $CP^{1}$%
-defining fibre bundle $\pi :C_{\ast }^{2}\rightarrow
CP^{1}\simeq S^{2},$ where $\ast $ in $C_{\ast }$ and $C_{\ast }^{2}$ means
``take away zero''. \ We would like to identify the space $CP^{1}$ \ of
complex lines in $C^{2}$ with the parametrizing unit sphere $S^{2}$ embedded in $%
C^{2}$ in the way described above (i.e. with the north pole of $S^{2}$ at
the origin of $C^{2}$). The group $SL(2,C)$ of fibre preserving \
transformations of the bundle then induces conformal transformations of
the base $S^{2}.$ The problem with this interpretation is that
$S^{2}$ can not be embedded in the total space $C_{\ast }^{2}$ in the required
way. For simplicity we then remove the fibre $l$ through $\left(
\begin{array}{l}
1 \\
0
\end{array}
\right) $ in $C_{\ast }^{2}$ and the corresponding point $L=\left[
\begin{array}{l}
1 \\
0
\end{array}
\right] $ \ of $CP^{1}.$ This gives the subbundle $\pi :C_{\ast }^{\prime 2}\rightarrow C$,
where $C_{\ast }^{\prime 2}=C_{\ast}^{2}-l$, and $C=CP^{1}-L$.
The base $C$ is identified with the affine complex plane $%
\eta =1$ (see Figure 2). Then the $SL(2,C)$ action on $C_{\ast }^{\prime
2} $ induces the action of conformal group $C(2)$ on $C=R^{2}$. The commutative diagram
below summarizes the situation:
\begin{figure}[ht]
\label{fig:3}
\centerline{\epsfig{figure=diagram1.eps}}
\end{figure}
Here $F\in SL(2,C)$ and $f\in C(2)$ is the induced action on $C$.
The sphere $S^{2}$ on Figure 2 is the conformal compactification of $C$. This means
simply that conformal transformations on $C$ (i.e. fractional linear
transformations obtained above) can be extended to act on the
compactification $\bar{C}$ $\simeq S^{2}$.
We can similarly consider the $HP^{1}$-defining bundle
$\pi :H_{\ast }^{2}\rightarrow HP^{1}\simeq S^{4}$ with
quaternion lines as fibres. Thus, let $(p,q)$ be coordinates in $H_{\ast
}^{2}.$ By removing the quaternion line through $\left(
\begin{array}{l}
1 \\
0
\end{array}
\right) $ in $H_{\ast }^{2}$ and the corresponding point $L$ below it in $%
HP^{1},$ we get the subbundle $\pi :H_{\ast }^{\prime2}\rightarrow H$.
Here $HP^{1}-L=H$ is identified with the affine quaternion
space $q=1$ in $H_{\ast}^{2}$. The group $SL(2,H)$ of fibre preserving transformations of the
bundle induces the action of conformal group $C(4)$ on $H=R^{4}$. This gives the following
diagram:
\begin{figure}[ht]
\label{fig:4}
\centerline{\epsfig{figure=diagram2.eps}}
\end{figure}
On the diagram $F\in SL(2,H)$ and $f\in C(4)$ is the induced action.
The sphere $S^{4}$ itself is the conformal compactification of the Euclidean
space $R^{4}$. Notice also that $HP^{1}\simeq S^{4}$ can be considered as a totally real
submanifold of the Grassmanian $G_{2,4}(C)$.
In case of a Lorentzian metric we can also embed the conformally
compactified Minkowski space $\bar{M}\simeq S^{1}\times S^{3}$ into the
Grassmanian $G_{2,4}(C).$ For this we use the isomorphism $M\rightarrow
H(2)$ as in (\ref{2.1}). Namely, we identify each point $p$ with the corresponding complex
plane $\left[
\begin{array}{l}
p \\
I
\end{array}
\right] $ through the columns of the $4\times 2$ matrix $\left(
\begin{array}{l}
p \\
I
\end{array}
\right) $\vspace{1pt}.
Let $SU(2,2)$ be the group of $4\times 4$ matrices acting on $C^{4}$ and
preserving the form
$\Phi (Z)=2Im(z_{1}\bar{z}_{3}+z_{2}\bar{z}_{4})$.
The action of $SU(2,2)$ on $C^{4}$ induces the action
of conformal group $C(1,3)$ on $M$. If $g\in SU(2,2)$, then
\begin{equation}
\label{2.4}
g\left[
\begin{array}{l}
p \\
I
\end{array}
\right] =\left(
\begin{array}{ll}
A & B \\
C & D
\end{array}
\right) \left[
\begin{array}{l}
p \\
I
\end{array}
\right] =\left[
\begin{array}{c}
(Ap+B)(Cp+D)^{-1} \\
I
\end{array}
\right].
\end{equation}
\vspace{1pt}
In particular, when $C=B=0,$ and $A\in SL(2,C)$, $p\rightarrow ApA^{+},$
i.e. we recover the group of Lorentz transformations.
The action of $C(1,3)$ can be extended to the compactified Minkowski space $%
\bar{M}\simeq S^{1}\times S^{3}$, and the corresponding embedding $\bar{M}%
\hookrightarrow G_{2,4}(C)$ is totally real \cite{Ward-Wells}.
Is there any fibre bundle structure in this case similar to the ones
we have in two and four dimensional Euclidean cases? The bundle structure $%
C_{\ast }^{4}\rightarrow \bar{M}$ would not be possible here as the 2-$C$ subspaces of $C^{4}$
parametrized by $\bar{M}$ intersect nontrivially.
To get a bundle structure consider a disjoint union of the $C^{2}$
subspaces of $C^{4}$ representing elements of $G_{2,4}(C)$ and the resulting universal
bundle $C^{2}\rightarrow \widetilde{U}\rightarrow G_{2,4}(C)$. The embedding $\varphi :\bar{%
M}\rightarrow G_{2,4}(C)$ produces the pullback bundle $L\rightarrow
U\rightarrow \bar{M}$ . Here a typical fibre $L$ is a ``real'' $C^{2}$ plane
(i.e. a plane $\left[
\begin{array}{l}
p \\
I
\end{array}
\right] $ or $\left[
\begin{array}{l}
I \\
p
\end{array}
\right] $ for some $p\in H(2)$). The total space $U$ is simply a ``portion''
of the universal bundle $\pi :\widetilde{U}\rightarrow G_{2,4}(C)$ above the
submanifold $\varphi (\bar{M})$ of $G_{2,4}(C)$.
By the above, the action of
$SU(2,2)$ on $C_{\ast }^{4}$ and therefore on $U$
induces the conformal
group action on $M$ and on the conformal compactification $\bar{M}\simeq
S^{1}\times S^{3}$ of $M$. Therefore, given an ($SU(2,2)$-generated) automorphism
$F$ of $U$ and the induced map $f\in C(1,3)$ we have the following commutative diagram:
\begin{figure}[ht]
\label{fig:5}
\centerline{\epsfig{figure=diagram3.eps}}
\end{figure}
Notice that unlike the earlier diagrams, the base manifold here
is compact (i.e. it is conformally compactified). We could similarly
``compactify'' the previous diagrams by embedding the bundles
$\pi :C_{\ast }^{2}\rightarrow S^{2}$ and
$\pi :H_{\ast }^{4}\rightarrow S^{4}$ into
the universal bundles on $CP^{1}$ and $HP^{1}$ respectively.
\section{Bundle algebra}
\setcounter{equation}{0}
Based on the previous section we accept the following
hypothesis: The (compactified) Minkowski space-time manifold is the base of a
fibre bundle $L\rightarrow U\rightarrow \bar{M}$ . Here $\bar{M}$ is
embedded in the Grassmanian $G_{2,4}(C)$ as a totally real 4-dimensional manifold (see
Section 2).
The bundle itself is the subbundle of the universal bundle over $G_{2,N}(C)$.
It is the pullback bundle induced by the embedding $\varphi :\bar{M}\rightarrow G_{2,4}(C)$.
The typical fibre $L$ is a real complex two-dimensional plane, i.e. a plane $\left[
\begin{array}{l}
p \\
I
\end{array}
\right] $ or $\left[
\begin{array}{l}
I \\
p
\end{array}
\right] $, where $p\in H(2).$ For simplicity, the non compact version $M$ of
Minkowski space will be often used. It is then identified with the manifold
of real planes $\left[
\begin{array}{l}
p \\
I
\end{array}
\right] ,$ where $p\in H(2).$ The compactification step will be always
assumed.
Let us find first of all the largest subgroup $G$\ of the group of
linear transformations $GL(4,C)$\ of $C^{4}$ which induce fibre preserving
transformations of the bundle $\pi : U\rightarrow \bar{M}$. \ We have
\begin{equation}
\label{3.1}
g\left[
\begin{array}{l}
p \\
I
\end{array}
\right] =\left(
\begin{array}{ll}
A & B \\
C & D
\end{array}
\right) \left[
\begin{array}{l}
p \\
I
\end{array}
\right] =\left[
\begin{array}{c}
(Ap+B)(Cp+D)^{-1} \\
I
\end{array}
\right].
\end{equation}
Transformation $g$ will be fibre preserving iff
\begin{equation}
\label{3.2}
\lbrack (Ap+B)(Cp+D)^{-1}\rbrack ^{+}=(Ap+B)(Cp+D)^{-1}.
\end{equation}
In other words,
\begin{equation}
\label{3.3}
(pA^{+}+B^{+})(Cp+D)=(pC^{+}+D^{+})(Ap+B),
\end{equation}
or,
\begin{equation}
\label{3.3b}
pA^{+}Cp-pC^{+}Ap+pA^{+}D+B^{+}Cp-D^{+}Ap-pC^{+}B+B^{+}D-D^{+}B=0.
\end{equation}
This yields the following three equations:
\begin{equation}
\label{3.4}
B^{+}D=D^{+}B,
\end{equation}
\begin{equation}
\label{3.4b}
A^{+}C=C^{+}A,
\end{equation}
and
\begin{equation}
\label{3.4c}
p(A^{+}D-C^{+}B)=(D^{+}A-B^{+}C)p.
\end{equation}
The last equation must be true for every $p$. In particular, $%
A^{+}D-C^{+}B=D^{+}A-B^{+}C$, i.e. $A^{+}D-C^{+}B$ is hermitian. Moreover,
since the space $Mat(2\times 2,C)$ of complex $2\times 2$ matrices has an
Hermitian basis, (\ref{3.4c}) yields
\begin{equation}
\label{3.5}
A^{+}D-C^{+}B=D^{+}A-B^{+}C=rI,
\end{equation}
where $r$ must be a real number not equal to zero. It is easy to check that
the set $G$ of such matrices $\left(
\begin{array}{ll}
A & B \\
C & D
\end{array}
\right) $ forms a group. If we impose an extra condition $\det g=1$ in
addition to (\ref{3.4})-(\ref{3.4c}), we obtain the group $SU(2,2)$ discussed
earlier. As always, it is possible to satisfy $\det g=1$ without changing $%
(Ap+B)(Cp+D)^{-1}$ by multiplying $A,B,C,$ and $D$ by an appropriate real
number. Therefore, the linear group of fibre preserving transformations of the bundle
$\pi : U\rightarrow \bar{M}$ can be identified with the group $SU(2,2)$.
The group $SU(2,2)$ is usually defined as the group of transformations
preserving Hermitian form $\Phi $ of signature $(+,+,-,-)$ in $C^{4}.$ In
appropriate coordinates this form is given by $\Phi =\left(
\begin{array}{ll}
0 & iI \\
-iI & 0
\end{array}
\right) $, so that $\Phi (Z)=2Im(z_{1}\bar{z}_{3}+z_{2}\bar{z}_{4})$ for any vector $Z=(z_{1},z_{2},z_{3},z_{4})\in C^{4}.$
This form defines a quadric $Q_{7}$ (of dimension 7) given by $\Phi (Z)=0$.
In canonical coordinates equation of the quadric reduces to $\left|
z_{1}\right| ^{2}+$ $\left| z_{2}\right| ^{2}-\left| z_{3}\right|
^{2}-\left| z_{4}\right| ^{2}=0.$ We claim that $Q_{7}$ is the set of points
in $C^{4}$ swept by the real planes in $C^{4}$, i.e. by the fibres of
$\pi : U\rightarrow \bar{M}$.
In fact, if $p=\left(
\begin{array}{ll}
x & z \\
\bar{z} & y
\end{array}
\right) ,$ where $x,y\in R$ and $z\in C,$\ then solving the system of
equations
\begin{equation}
\label{3.6}
\alpha \left(
\begin{array}{l}
x \\
\bar{z} \\
1 \\
0
\end{array}
\right) +\beta \left(
\begin{array}{l}
z \\
y \\
0 \\
1
\end{array}
\right) =\left(
\begin{array}{l}
a \\
b \\
c \\
d
\end{array}
\right) \in C^{4},
\end{equation}
we have $\alpha =c$, $\beta =d$, and
\begin{equation}
\label{3.7}
\left\{
\begin{array}{c}
cx+dz=a \\
c\bar{z}+dy=b
\end{array}
\right\}.
\end{equation}
Determinant of the system is zero and solvability requires
$Im(a\bar{c}+b\bar{d})=0.$ Replacing $(a,b,c,d)$ with
$(z_{1},z_{2},z_{3},z_{4})$, we have $Im(z_{1}\bar{z}_{3}+z_{2}\bar{z}_{4})=0$ i.e.
$\Phi (Z)=0$. This condition
is also sufficient, if we include the planes $\left[
\begin{array}{l}
I \\
p
\end{array}
\right] $ . As a result, points of the quadric $Q_{7}$ and only these points
are reachable by the real planes $\left[
\begin{array}{l}
p \\
I
\end{array}
\right] $ and $\left[
\begin{array}{l}
I \\
p
\end{array}
\right] $ in $C^{4}.$
Real planes $\left[
\begin{array}{l}
p \\
I
\end{array}
\right] $ \ (as well as the planes $\left[
\begin{array}{l}
I \\
p
\end{array}
\right] $) intersect nontrivially in $C^{4}$, so that we have no uniqueness property for
the system above. That is why the bundle $\pi : U\rightarrow \bar{M}$ can not be
embedded in the space $C^{4}$ where the planes live. Let us impose an extra condition $xy-\left| z\right|
^{2}=0$ on $p$. Then the corresponding system has a unique solution for any
point $Z=(z_{1},z_{2},z_{3},z_{4})\in Q_{7}$ for which $z_{1}\bar{z}_{3}+z_{2}\bar{z}_{4}\neq 0$.
In fact, solving the system (\ref{3.7}) in this case we have:
\begin{equation}
\label{3.10}
x=\frac{\left| z_{1}\right| ^{2}}{z_{1}\bar{z}_{3}+z_{2}\bar{z}_{4}},\
y=\frac{\left| z_{2}\right| ^{2}}{z_{1}\bar{z}_{3}+z_{2}\bar{z}_{4}},\
z=\frac{z_{1}\bar{z}_{2}}{z_{1}\bar{z}_{3}+z_{2}\bar{z}_{4}}.
\end{equation}
Notice that the solvability condition
$Im(z_{1}\bar{z}_{3}+z_{2}\bar{z}_{4})=0$ ensures that
$z_{1}\bar{z}_{3}+z_{2}\bar{z}_{4}$ is real.
In the solution above we assumed that
$z_{1}\bar{z}_{3}+z_{2}\bar{z}_{4}\neq 0$. Let $S$ be the subset of $Q_{7}$, where this
condition is violated. Then any point $Z\in Q_{7}\backslash S$ has a unique real plane
$\left[
\begin{array}{l}
p \\
I
\end{array}
\right]$
passing through $Z$ and such that $\det p=0$.
Condition $\det p=xy-\left| z\right| ^{2}=0$ defines a light cone $Q_{3}$ at
the origin of $M$. The above result shows that there is a well defined
projection from $Q_{7}\backslash S$ into $Q_{3}$ along real planes in $%
C^{4}$.
Real planes defined by $\det p=0$ represent a special family of real
planes. By considering light cones at every point of $M$ and imposing
the corresponding conditions on real planes we can obtain at each point of
$M$ a light cone $Q_{3}^{\prime }$ and the corresponding quadric
$Q_{7}^{\prime }$ above it with a well defined projection from $Q_{7}^{\prime}\backslash S$
into $Q_{3}^{\prime}$.
In fact, the group $SU(2,2)$ acts transitively on real
planes, i.e. on $\bar{M}$. So if we start with the real planes satisfying $%
\det p=0$ and apply $SU(2,2)$ transformations, we are going to get all
possible real planes. Moreover, translations form a subgroup in $SU(2,2)$
which acts transitively on $M.$ By applying a translation $\left(
\begin{array}{ll}
I & B \\
0 & I
\end{array}
\right) $ to the plane $\left[
\begin{array}{l}
p \\
I
\end{array}
\right] $ we obtain a new plane $\left[
\begin{array}{c}
p+B \\
I
\end{array}
\right]$. Translation of the cone $\det p=0$ gives a new cone $%
Q_{3}^{\prime }$ defined by $\det (p-B)=0$. An easy check shows that the planes above it can
intersect only along the set $S$ and that they sweep out the entire quadric $%
Q_{7}$.
More generally, consider $G_{2,4}(C)$ as a quadric $Q$ in $CP^{5}$ obtained via Pl\"{u}cker
embedding. Let $p\in \bar{M}\subset Q$ and let $\overline{T}_{p}Q$ be the tangent space to $Q$
at $p$. Then $\overline{T}_{p}Q\cap Q\cap M$ can be identified with the light cone $Q_{3}(p)$
at $p$ (see \cite{Ward-Wells}). Cosequently, $T_{p}U\cap U$ can be identified with the quadric
$Q_{7}(p)$ of the real $C^{2}$ planes above $Q_{3}(p)$, i.e. with $\pi ^{-1}(Q_{3}(p))$. Notice
also that this identification defines a complex structure on $U$.
Consider now the light cone
$Q_{3}(0)\in T_{0}\bar{M}$ at the origin of $\bar{M}$. Here $T_{0}\bar{M}$ is the tangent
space to $\bar{M}$ at $p=0$. Let $Q_{7}(0)=\pi ^{-1}(Q_{3}(0))\in T_{0}U$.
As $SU(2,2)$ acts on $C^4$ it generates the action on the fibres of $U$. On the quadric $Q_{7}(0)$
this action consists
of translations $Q_{7}(0)\rightarrow Q_{7}(p)$ and of $SU(2,2)$-transformations
preserving $Q_{7}(p)$. The induced action on $\bar{M}$ is conformal. On the light cone
$Q_{3}(0)$ it consists of translations
$Q_{3}(0)\rightarrow Q_{3}(p)$ and of the point $p$ preserving conformal transformations.
On a tangent space $T_{p}U$ at a given point $p\in \bar{M}$ we have an action of the group
$G_{p}$ of
$SU(2,2)$-transformations preserving $Q_{7}(p)$ and the induced action of the point $p$
preserving conformal transformations on $T_{p}\bar{M}$.
\section{What is spin?}
\setcounter{equation}{0}
The bundle structure $\pi :U\rightarrow \bar{M}$ together with the group of automorphisms on it
leads to a particularly clear interpretation of spin. A 4-spinor $Z$ at a
point $p\in \bar{M}$, for example, is a vector of $T_{p}U\simeq C^{4}.$ It
is an element of the space of fundamental representation of the group
$G_{p}$ of $SU(2,2)$-transformations preserving $Q_{7}(p)$.
Historically, such objects go back to Dirac's 4-component spinors \cite{Dirac}.
If $Z$ is isotropic, projection $\pi(Z)$ of $Z$ is defined and belongs to
$T_{p}(\bar{M})$ (identified with $M$).
Transformation of $Z$ under the action of $G_{p}$ on $T_{p}(U)$ is seen on $T_{p}(\bar{M})$
as a point $p$ preserving conformal transformation. In particular, for the subgroup $L$ of
$G_{p}$ given by the matrices $\left(
\begin{array}{ll}
A & 0 \\
0 & D
\end{array}
\right) $, $\ $with $A\in SL(2,C),$ $X=\pi(Z)$ transforms as a
vector under the action of the group of Lorentz.
Transformation properties of 4-spinors under rotations now have a
simple geometric interpretation. In a nutshell they are due to the fact that
a real plane $C^{2}$ and the flipped upside down plane have the same image under the
bundle projection. In particular, given a point $p\in M$ consider a rotation
of a vector $V$\ of $T_{p}(M)$\ through the angle of $2\pi$. To be
specific, consider the matrix $A=\left(
\begin{array}{ll}
e^{i\varphi } & 0 \\
0 & e^{-i\varphi }
\end{array}
\right) \in SU(2)$ and the corresponding matrix $g=\left(
\begin{array}{ll}
A & 0 \\
0 & A
\end{array}
\right) \in G_{p}$. Under the action of $g$ any vector $Z$ in the real plane
defined by $V$ will transform as
\begin{equation}
\label{4.3}
Z=\left(
\begin{array}{l}
z_{1} \\
z_{2} \\
z_{3} \\
z_{4}
\end{array}
\right) \longrightarrow \left(%
\begin{array}{l}
e^{i\varphi }z_{1} \\
e^{-i\varphi }z_{2} \\
e^{i\varphi }z_{3} \\
e^{-i\varphi }z_{4}
\end{array}
\right).
\end{equation}According to (\ref{3.10}) the components of vector
$V=\pi(Z)$ will transform as
\begin{equation}
\label{4.2}
x^{\prime }=x,\
y^{\prime }=y,\
z^{\prime }=e^{2i\varphi }z.
\end{equation}
As $\varphi $ changes from $0$ to $\pi $, $gZ$ changes from $Z$ to $-Z$.
Instead, $\tilde{g}V=\pi(gZ)$ changes from $V$ to $V^{\prime }=V$
describing a rotation through $2\pi$ in
the $z$-plane. This is so because the set of vectors $\{-Z\}$ generates the same real plane as
the set ${Z}$.
The matrix form of the relation between vectors and 4-spinors is clarified
as well: a single point $p$\ of $\bar{M}$ is the image under $\pi $\
of the complex 2-dimensional plane -- the fibre $\pi ^{-1}(p)$. Planes are
described by pairs of vectors, i.e. by matrices. So the nature of projection
$\pi $ leads to the correspondence{\small \ }\textit{points of} $%
M\longleftrightarrow $\textit{hermitian 2x2 matrices.}
\section{Concluding remarks}
\setcounter{equation}{0}
The main hypothesis advocated here is that the
space-time manifold is the base of the fibre bundle $\pi :U\rightarrow \bar{M}$ described above.
Unlike the twistor theory, points of space-time in this approach are not secondary objects.
Complex structure appears on
the total space $U$ of points rather than on Grassmanian $G_{2,4}(C)$. In this respect it is
more similar to the bundle $CP^{3}\rightarrow HP^{1}$ used in obtaining the instanton solutions
of Yang-Mills equations \cite{Atiyah}.
The space-time manifold itself does not need to be conformally flat or half-coformally flat (see
below).
When accepted, this hypothesis leads to a very natural interpretation
of conformal symmetry. Conformal group action is induced on $\bar{M}$ from the action of a
group of fibre-preserving transformations on $U$. The latter group is generated by the action
of $SU(2,2)$ on $C^4$.
With this interpretation 4-component spin-vectors turn out to be vectors tangent to $U$. Whenever they can
be projected on $T_p\bar{M}$ by $\pi$, they yield vectors which are transformed by a conformal
transformation.
There are several remaining questions:
More general fibre preserving transformations of $U$ are possible than those generated by the
$SU(2,2)$ action on $C^4$. It is important to investigate the induced action of these
transformations on $\bar{M}$.
The bundle $\pi :U\rightarrow \bar{M}$ can be defined even if $M$ has no conformal symmetry
at all. Consequences of this, in particular, a possibility to include the full gravity
in this context must be studied.
By the above $U$ has a complex structure. It is useful to know when this complex structure
is integrable giving $U$ as a complex 4-dimensional manifold.
What are the consequences of this interpretation of conformal symmetry in physics?
Some of these questions are addressed in a forthcoming paper \cite{Alex2}.
\bigskip
{\small \ }\textit{Acknowledgements.} I would like to express my deep gratitude to Malcolm Forster
for his constant support during preparation of this paper. This work was supported in part
by the University of Wisconsin summer research grant.
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\end{document}