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\renewcommand{\@oddfoot}{--- ~March, 2002~ ---\hfil
\raisebox{0.3ex}{\tiny --- ~D.~A.~Arbatsky~ ``QED. V. Vector
representation\dots''~ ---}\hfil --- ~\thepage~
---}
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\title{On quantization of electromagnetic field.\\
       V. Vector representation of little Lorentz group\\
       for light-like momentum.}
\author{D.~A.~Arbatsky\footnote{ http://daarb.narod.ru/ , http://wave.front.ru/ }}
\date{March, 2002}
\maketitle

\begin{abstract}
Using elementary geometric methods we prove the isomorphism of
the little Lorentz group for light-like momentum and the group of
motions of a Euclidian plane. In accordance with
Jordan-H\"older-Noether theorem we perform ``reduction'' of the
real and complex vector representations of this group. We also
prove indecomposability of these representations.
\end{abstract}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\myparagraph{Preliminary remarks.}
 There are quite many books devoted to the theory of the Lorentz
group~$\EuScript L$ and the theory of representations of this
group. Among them there are monographs specially devoted  to this
topic; for
example,~\cite{Naymark1958,GelMinShap1958,Fyodorov1979}. It is
mentioned in all them that the little Lorentz group~$\EuScript
L_k$ for a light-like momentum\footnote{The word ``momentum'' is
used because we keep in mind applications to the theory of
zero-mass fields; in particular electromagnetic. While we perform
purely mathematical investigation, we can suppose that we talk
about any isotropic vector~$k$: $k^2=0$.} $k$ is isomorphic to the
group of motions of a Euclidian plane~$E(2)$. The proof that is
usually provided is usually based on an analytic investigation of
the group $SL(2,\mathbb C)$ which is the universal covering group
of the Lorentz group.

Such an approach seems to be optimal if we want to study then
arbitrary (including two-valued) representations of the
group~$\EuScript L_k$. At the same time it turns out to be very
difficult to realize the geometric sense of the isomorphism of
$\EuScript L_k$ and $E(2)$\footnote{In particular,
Ryder~\cite{Ryder1987} noted about it that the physical meaning of
this result is unclear.}.


Let us specify here what is usually implied when we talk about
``geometric sense''. From the point of view of physical
applications the Lorentz group is most naturally defined by its
vector representation\footnote{Such a representation is sometimes
called ``fundamental''.}. And the little group $\EuScript L_k$ is
defined just as the stationary subgroup of the vector~$k$. The
group $E(2)$ in its turn is defined as the group of motions of a
Euclidean plane. Taking into account these definitions of the
groups $\EuScript L_k$ and $E(2)$ and using purely geometric
constructions we will find some two-dimensional plane, possessing
natural Euclidean structure, where the group $\EuScript L_k$ acts
as the group of motions.

Another question that we will study here (and which is the main
motivation for publishing of this paper) is the ``reduction'' of
vector representation of the group~$\EuScript L_k$, i.~e. we will
expose what irreducible representations it consists of. From the
point of view of the general representation theory, this case is
very particular and does not contain any specific difficulties.
Nevertheless, taking into account its great practical importance,
it seems to be useful to perform its investigation with using
constructions that are closest to our geometric intuition.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section*{Real vector representation}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\myparagraph{Notations and terminology.}
 {\it The Minkowski space\/} $M$ is a four-dimensional real linear
space~$\mathbb R^4$, where we have real symmetrical bilinear form
$g(\,\cdot\,,\,\cdot\,)$ with the signature $(+,-,-,-)$. This form
is called {\it scalar product}. For short, instead of writing
$g(a,b)$, we will write just~$a b$ or~$a\cdot b$.

Consider now the group of all linear transformations of the
Minkowski space~$M$ preserving the scalar product. The connected
component of this group containing neutral element is a group
also. This latter group we will call {\it the Lorentz group\/} and
we will denote it~$\EuScript L$.

In this paper we will consider only the vector representation of
the Lorentz group, i.~e. the representation in the Minkowski
space. This representation will be denoted as~$T$.

{\it The little group of a vector\/}~$k$ is the stationary
subgroup of the vector~$k$, i.~e. the subgroup of the Lorentz
group leaving the vector~$k$ unchanged. It will be denoted
as~$\EuScript L_k$.

A vector~$k$ is called {\it time-like}, if $k^2>0$; {\it
light-like\/} or {\it isotropic}, if $k^2=0$; and {\it
space-like}, if $k^2<0$.

Consider also the group of all transformations of a Euclidean
plane preserving distances. The connected component of this group
containing neutral element is a group also. We will call it {\it
the group of motions of a Euclidean plane\/} and denote it~$E(2)$.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\myparagraph{Reducing subspaces.}
 Following the Jordan-H\"older-Noether theorem let us immediately
guess a composition series of the representation~$T$, i.~e. such a
monotonic set of reducing subspaces of the representation~$T$ that
corresponding successive factor-representations are irreducible:
\begin{equation}
 \{0\}= M^0 \subset M^\parallel \subset M^\perp \subset M^4=M \ .
\label{admset}
\end{equation}
Here
 $M^0$ is the zero subspace in~$M$;
 $M^\parallel$ is the subspace containing all vectors parallel
     to~$k$;
 $M^\perp$ is the subspace containing all vectors orthogonal
     to~$k$;
 $M^4$ is another notation for the Minkowski space.

Each of these subspaces is obviously invariant with respect to the
action of the group~$\EuScript L_k$. The subrepresentations in the
subspaces $M^\parallel$ and $M^\perp$ we denote $T^\parallel$ and
$T^\perp$, correspondingly.

For the set~(\ref{admset}) we have the corresponding sequence of
the three successive factor-representations in the factor-spaces
 $M^\parallel/M^0=M^\parallel$,
 $M^\perp/M^\parallel=M^{\perp/\parallel}$ and
 $M^4/M^\perp=M^{4/\perp}$.
The first of these factor-representations is a subrepresentation.
We have already denoted it as $T^\parallel$. The second and the
third factor-representations we denote $T^{\perp/\parallel}$ and
$T^{4/\perp}$, correspondingly.

The irreducibility of the factor-representations $T^\parallel$ and
$T^{4/\perp}$ is undoubted because they are one-dimensional. As
regards the factor-representation~$T^{\perp/\parallel}$, its
irreducibility (real) will be proved in the
section~\ref{tperppar}.

In the sections~\ref{unt} and~\ref{tperp} we will prove two
lemmas that have as a consequence that besides the
subspaces~(\ref{admset}) the representation $T$ does not have any
other reducing subspaces. So, the composition
series~(\ref{admset}) is the only one. Therefore in this case the
statement of the Jordan-H\"older-Noether theorem that all
composition series are of the same length and corresponding
factor-representations of these series are equivalent turns out to
be trivial.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\myparagraph{Homomorphism $\EuScript L_k$ into $E(2)$.}
 \label{homom}
 Consider now in the Minkowski space a three-dimensional
hyperplane~$N$ defined by the equation:
\begin{equation}
 k\cdot x = const \neq 0  \ .
\label{plane}
\end{equation}
So far as the group~$\EuScript L_k$ leaves the vector~$k$
unchanged and preserves the scalar product, the hyperplane~$N$ is
invariant with respect to the action of the group~$\EuScript L_k$.

Let us introduce now an equivalence relation~$\sim$ for vectors
from~$N$:
\begin{equation}
 a\sim b \ \ \Longleftrightarrow \ \  a-b \,\in\, M^\parallel \ ,
\label{equiv}
\end{equation}
In other words, two vectors are equivalent if their ends belong to
a line parallel to~$k$. Because of~(\ref{admset}) and
(\ref{plane}) every such a line completely lies in the
hyperplane~$N$. Making factorization of~$N$ with respect to the
equivalence relation~(\ref{equiv}) we get some two-dimensional
plane\footnote{The plane $\widetilde N$ is not {\it imbedded\/}
into the Minkowski space.}~$\widetilde N$. We will denote points
from~$\widetilde N$ and lines corresponding to them on~$N$ by the
same symbols: $\widetilde a$, $\widetilde c$ etc.

The plane~$\widetilde N$ has natural Euclidean structure. Really,
let us take two arbitrary points $\widetilde a$ and $\widetilde c$
on~$\widetilde N$. Both of them are equivalence classes of points
on~$N$. Let us take a one arbitrary representative for each of the
two classes: $a$ and $c$, correspondingly. Let us calculate the
value
\begin{equation}
 \rho\,(a,c)=\sqrt{(a-c)^2} \ .
\label{distN}
\end{equation}
This value does not depend on the choice of the representatives
$a$ and $c$. Really, let us substitute, for example, $a$ with
$a+\alpha k$. Taking into account~(\ref{plane}), we have:
\[
 \rho\,(a+\alpha k,c) = \sqrt{(a-c+\alpha k)^2} =
 \sqrt{(a-c)^2 + 2\alpha(k a-k c) + \alpha^2k^2} =
 \sqrt{(a-c)^2} = \rho\,(a,c) \ .
\]

What is more, $(a-c)^2$ is always greater than zero and the
function $\rho\,(\,\cdot\,,\,\cdot\,)$ have all properties of
Euclidean metric, because the plane~$\widetilde N$ can be
identified with any of space-like two-dimensional planes lying
in~$N$.

Note now that the equivalence relation~(\ref{equiv}) and metric
$\rho\,(\,\cdot\,,\,\cdot\,)$ are invariant with respect to the
action of the little group~$\EuScript L_k$. So, we have
constructed a homomorphous mapping of the group~$\EuScript L_k$
into the group of motions~$E(2)$ of the Euclidean
plane~$\widetilde N$.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\myparagraph{Inverse mapping.} \label{inverse}
 Later on we will repeatedly use the following two statements that
are true in both the real and the complex cases.

\mystat{}
 {\it If a linear transformation changes some basis
$\{n^i\}_{i=1..4}$ so that scalar products of all elements of the
basis remain unchanged (including squares of the elements of the
basis), then this transformation preserves the scalar product.}

\mystat{}
 {\it If a linear transformation changes some basis
$\{n^i\}_{i=1..4}$ so that all values $(n^i-n^j)^2$ and $(n^i)^2$
remain unchanged, then this transformation preserves the scalar
product.}

The first statement follows from the fact that every vector can be
decomposed into a linear combination of the elements of basis. The
second statement follows from the first, if we take into account
the formula:
\[
 (n^i-n^j)^2=(n^i)^2+(n^j)^2-2n^in^j   \ .
\]

Let us fix now an arbitrary motion of the plane~$\widetilde N$. A
transformation from the little group~$\EuScript L_k$, that
transforms to this motion on~$\widetilde N$ under the homomorphism
constructed in the section~\ref{homom}, we will call {\it
desired}. Let us show that the desired transformation
from~$\EuScript L_k$ exists and it is unique. So, we will prove
that the homomorphism under consideration is really an
isomorphism.

For that consider a family of hypersurfaces defined by the
equations like
\begin{equation}
 x^2=const \ .
\label{hypsurf}
\end{equation}
The intersection of any of them with the hyperplane~$N$ is a
paraboloid. Every such a paraboloid is invariant with respect to
the desired transformation. And every line representing a point
from~$\widetilde N$ intersects such a paraboloid exactly in one
point, and for every point on the paraboloid there is exactly one
such a line that contains this point. Therefore we have defined
the action of the desired transformation on every paraboloid.

Let us take now on one of these paraboloids four points that do
not lie in any two-dimensional plane. The four vectors
$\{n^i\}_{i=1..4}$ that have their ends at these four points form
a basis in~$M$. The transformation on the paraboloid that we have
just defined preserves values $(n^i-n^j)^2$, because each of these
values is a square of the corresponding distance in~$\widetilde
N$. The values~$(n^i)^2$ are also preserved because they are
constant on the paraboloid. Let us linearly continue the
constructed transformation of the vectors from the basis to the
whole space~$M$. It follows from the second statement formulated
in the beginning of this section that we will get a transformation
preserving the scalar product in~$M$. And according to our
construction, this transformation preserves the
plane~(\ref{plane}), and therefore it leaves the vector~$k$
unchanged. So far as the initial motion of the plane~$\widetilde
N$ was supposed to belong to the connected group $E(2)$, the
constructed transformation in~$M$ also belongs to the connected
group~$\EuScript L_k$.

The constructed transformation is the only one that can pretend to
be the desired. On the other hand, the constructed transformation
from~$\EuScript L_k$ under the homomorphism from the
section~\ref{homom} transforms into some motion on~$\widetilde N$.
This motion, in accordance with the construction, acts on the four
points\footnote{The lines containing the ends of the vectors of
basis $\{n^i\}_{i=1..4}$}~$\widetilde n^i\in \widetilde N$ in
exactly the same way as the initial motion in~$\widetilde N$. So
far as a motion in a Euclidean plane is uniquely defined even by
its action on two not-coinciding points, the constructed
transformation from~$\EuScript L_k$ is the desired. So, we have
proved the

\mytheorem{} {\it The groups $\EuScript L_k$ and $E(2)$ are
isomorphic.}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\myparagraph{Factor-representation $T^{\perp/\parallel}$.}
\label{tperppar}
 Similarly to the constructed in the section~\ref{homom}
factor-plane, the factor-space $M^{\perp/\parallel}$ has natural
Euclidean structure\footnote{And what is more,
$M^{\perp/\parallel}$ is linear, and there is the corresponding
scalar product in it:
 $\widetilde a\cdot \widetilde c = a c$.}, this structure is
defined by the same formula~(\ref{distN}). So, the little
group~$\EuScript L_k$ acts in $M^{\perp/\parallel}$ as the group
of rotations of a Euclidean plane over a fixed point. Such a group
is denoted as $SO(2)$.

Obviously, in a Euclidean plane there is no such a direction that
would be invariant under rotations. From this we get the

\mytheorem{}
 {\it The factor-representation $T^{\perp/\parallel}$ is irreducible
(in the real sense).}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\myparagraph{Indecomposability of representation $T$.} \label{unt}
 Let us suppose that the representation~$T$ is decomposable, i.~e.
the space~$M$ can be represented as a direct sum $M=M'\oplus M''$
of two non-trivial reducing spaces~$M'$ and $M''$. One of these
spaces, for example~$M'$, necessarily contains some vector that
does not belong to $M^\perp$. If we properly choose the constant
in the equation~(\ref{plane}), the end of this vector will lie in
the plane~$N$. It was shown in the section~\ref{inverse} that by
transformations from $\EuScript L_k$ this vector can be
transformed into any other vector, if the end of the latter lies
on the same paraboloid. In particular, it can be transformed into
four vectors $\{n^i\}_{i=1..4}$ forming a basis in~$M$. Therefore,
$M'$ must coincide with the whole~$M$. So, we have proved the

\mylemma{}
 {\it Every reducing subspace of the representation~$T$ either
coincides with the whole $M$ or lies in $M^\perp$.}

From this lemma we get the

\mytheorem{} {\it The representation~$T$ is indecomposable.}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\myparagraph{Indecomposability of representation $T^\perp$.}
\label{tperp}
 Consider a cylinder defined by equations:
\begin{equation}
 k\cdot x=0 \ , \qquad x^2 = const \neq 0 \ .
\label{cylinder}
\end{equation}
Let us prove that $\EuScript L_k$ acts transitively on it, i.~e.
every point on the cylinder can be transformed to any other. Let
us denote the vector that have the end in the initial point as
$e^1(0)$, and the vector that have the end in the final point as
$e^1(1)$. So far as the cylinder is connected, these two points
can be connected by a continuous curve $e^1(t)$, $t\in[0;1]$.

Let us introduce now a vector $e^2(t)$ which continuously depends
on~$t$. Let us require that the end of the vector $e^2(t)$ always
lies on the same cylinder of the type~(\ref{cylinder}).
Furthermore, we require that the scalar product $e^1(t)\cdot
e^2(t)=const$ does not depend on $t$. And what is more, we require
that the vectors $k$, $e^1(t)$ and $e^2(t)$ are linearly
independent when $t=0$ and, therefore, they are linearly
independent with any~$t$.

The vector $e^2(t)$ is not defined uniquely, of course. It always
can be substituted by~$e^2(t)+\alpha(t)\,k$, where~$\alpha(t)$ is
an arbitrary continuous function. We will not eliminate this
non-uniqueness in any way.

Let us use the three vectors $k$, $e^1(t)$ and $e^2(t)$ as a basis
in~$M^\perp$. Using the three-dimensional version of the first
statement from the section~\ref{inverse} we get a continuous
family of linear transformations in~$M^\perp$ which preserve the
scalar product in~$M^\perp$.

Let us prove now that these transformations can be completed as
transformations from~$\EuScript L_k$. For this let us take an
arbitrary vector~$n(0)$ that does not belong to the
subspace~$M^\perp$. The vectors $k$, $e^1(t)$, $e^2(t)$ and $n(0)$
form a basis in the Minkowski space~$M$. Let us show now that for
any~$t$ vector $n(t)$ can be chosen so that the following four
values do not depend on~$t$:
\begin{equation}
 k\cdot n(t) =const \ , \qquad e^1(t)\cdot n(t) =const \ ,
 \qquad
 e^2(t)\cdot n(t) =const \ , \qquad (n(t))^2 =const \ .
\label{fourth}
\end{equation}

The first three conditions with fixed~$t$ define in the Minkowski
space a line~$\widetilde n$, which is a point of the factor-plane
$\widetilde N$ constructed in the section~\ref{homom}. As we have
pointed out in the section~\ref{inverse}, on this line there is
exactly one point satisfying the fourth of the
equations~(\ref{fourth}). This is the point where the desired
vector~$n(t)$ has its end.

Using now the first statement from the section~\ref{inverse}, we
see that we have constructed a continuous family of
transformations from $\EuScript L_k$ transforming a given point of
the cylinder~(\ref{cylinder}) to any other given point. So, we
have proved the

\mylemma{}
 {\it Transformations of the little group $\EuScript L_k$ act on
the cylinder}~(\ref{cylinder}) {\it transitively.}

By the way, we have proved the following

\mystat{}
 {\it A linear transformation of the space $M^\perp$, preserving
the scalar product there, can be uniquely completed as a linear
transformation of~$M$, preserving the scalar product in~$M$.}

Let us suppose now that the representation~$T^\perp$ is
decomposable, i.~e. the space~$M^\perp$ decomposes into a direct
sum $M^\perp=M'\oplus M''$ of two reducing subspaces $M'$ and
$M''$. One of these two subspaces, for example~$M'$, necessarily
contain some vector that does not belong to~$M^\parallel$. The end
of this vector lies on the corresponding
cylinder~(\ref{cylinder}). But then, as we have just shown, the
whole cylinder must belong to $M'$. But the linear shell of the
cylinder~(\ref{cylinder}) coincides with~$M^\perp$ and, therefore,
$M'=M^\perp$. So, we have proved a

\mylemma{}
 {\it Every reducing subspace of the representation $T^\perp$
either coincides with~$M^\perp$ or belongs to $M^\parallel$.}

From this we have the

\mytheorem{}
 {\it The representation $T^\perp$ is indecomposable.}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section*{Complex vector representation}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\myparagraph{Complexification.} \label{compl}
 As we have shown in the paper~[IV], for investigation of
electromagnetic field it is useful to study also the complex
vector representation of the little group $\EuScript L_k$. In
order to get this representation we just should admit that
components of vectors under consideration can be complex numbers
and suppose that representation of the group $\EuScript L_k$ acts
there by the same formulas as in the real case. And the group
$\EuScript L_k$ does not change in any way.

The complexified Minkowski space will be denoted as~$M_{\mathbb
C}$. Generally, later on all spaces and representations will be
labeled by symbol of the corresponding field of scalars. If two
spaces are denoted the same but have different scalar label, then
we imply, that the complex space is the complexification of the
real. For example, the real Minkowski space will be denoted
as~$M_{\mathbb R}$. We will also imply that real spaces are
subsets of corresponding complex spaces. The representation of the
group $\EuScript L_k$ that acts in~$M_{\mathbb C}$ will be denoted
as $T_{\mathbb C}$.

An arbitrary vector $n\in M_{\mathbb C}$ can be represented as a
sum of its real and imaginary parts:
$n=n^\mathrm{Re}+in^\mathrm{Im}$. Here
$n^\mathrm{Re},n^\mathrm{Im}\in M_{\mathbb R}$. So far as
transformations of the group $\EuScript L_k$ are transformations
with real coefficients, we may suppose that a transformation of a
complex vector is a simultaneous transformations of its real and
imaginary parts.

All reducing subspaces and factor-spaces found in the real case
are obviously complexified. But the analogy between the
representations $T_{\mathbb R}$ and $T_{\mathbb C}$ is not full.
Reducibility and decomposability of complexified representations
must be investigated in addition. For proving of
indecomposability of complex representations the following lemma
will play an important role

\mylemma{}
 {\it Let us suppose that we have defined:
a vector~$n\in M_{\mathbb C}$; some two-dimensional
subspace~$P^2_{\mathbb R}$ in~$M_{\mathbb R}$, such that
 $n\in P^2_{\mathbb C}$; and some one-dimensional
subspace~$P^1_{\mathbb R}$ in~$P^2_{\mathbb R}$, such that
 $n\notin P^1_{\mathbb C}$. Then multiplying~$n$ by an appropriate
factor~$\theta\in\mathbb C$ we can make that
 $(\theta n)^\mathrm{Im}\in P^1_{\mathbb R}$ and
 $(\theta n)^\mathrm{Re}\notin P^1_{\mathbb R}$.}

In order to prove this lemma, let us multiply the initial
vector~$n$ by a number like $e^{i\varphi}$, where the
argument~$\varphi$ runs through the set of real numbers. When the
argument~$\varphi$ changes, the vectors $n^\mathrm{Re}(t)$ and
$n^\mathrm{Im}(t)$ will move on some ellipsis in the
plane~$P^2_{\mathbb R}$ (If $n^\mathrm{Re}$ and $n^\mathrm{Im}$
are parallel, the ellipsis degenerates into the segment). This
ellipsis intersects with any one-dimensional
subspace~$P^1_{\mathbb R}$ belonging to~$P^2_{\mathbb R}$. Now it
is obvious that with appropriate~$\varphi$ the imaginary part of
the vector will be in $P^1_{\mathbb R}$ and the real part will not
be there.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\myparagraph{Decomposability of
 $T^{\perp/\parallel}_{\mathbb C}$.} \label{tperpparcompl}
 As we have shown in the section~\ref{tperppar}, in the space
$M^{\perp/\parallel}_{\mathbb R}$ the little group $\EuScript L_k$
acts as the group $SO(2)$. So far as the group $SO(2)$ is Abelian,
as it follows from the Schur lemma, the complex representation
$T^{\perp/\parallel}_{\mathbb C}$ must be reducible. On the other
hand, so far as the group $SO(2)$ is compact, the representation
$T^{\perp/\parallel}_{\mathbb C}$ is equivalent to some unitary
representation and therefore it is fully reducible. The
corresponding subrepresentations we will denote as
$T^{(+1)/\parallel}_{\mathbb C}$ and
 $T^{(-1)/\parallel}_{\mathbb C}$.

In order to make this result more concrete, let us introduce in
the real space $M^{\perp/\parallel}_{\mathbb R}$ an orthonormal
basis~$\{\tilde e^i\}_{i=1,2}$. Then the matrices of the
representation $T^{\perp/\parallel}_{\mathbb R}$ will take the
form:
\[
 \left(\begin{array}{cc}
          \cos\varphi & -\sin\varphi\\
          \sin\varphi & \cos\varphi
 \end{array}\right)
\]

If now in the complex plane
 $M^{\perp/\parallel}_{\mathbb C}$ we use complex ``spiral''
basis~$\{\tilde e^{(\lambda)}\}_{\lambda=\pm1}$:
\[
 \tilde e^{(\pm1)} =
   \frac{\tilde e^1 \pm i\,\tilde e^2}{\sqrt{\,2\,}} \ ,
\]
the matrices of the representation
 $T^{\perp/\parallel}_{\mathbb C}$ become diagonal:
\[
\left(\begin{array}{cc}
          e^{-i\varphi} & 0             \\
          0             & e^{+i\varphi}
\end{array}\right)
\]
So, we have proved the

\mytheorem{}
 {\it The representation $T^{\perp/\parallel}_{\mathbb C}$ has exactly
two reducing subspaces:
 $M^{(+1)/\parallel}_{\mathbb C}$ and
 $M^{(-1)/\parallel}_{\mathbb C}$.
The sum of the corresponding subrepresentations is the
representation $T^{\perp/\parallel}_{\mathbb C}$:
\[
 T^{\perp/\parallel}_{\mathbb C} =
      T^{(+1)/\parallel}_{\mathbb C} \oplus
      T^{(-1)/\parallel}_{\mathbb C} \ .
\]}

For the electromagnetic field (see~[IV]) this theorem means that
there exist plane waves with helicity $+1$ and~$-1$~--- the
well-known fact.

Let us notice also that:
\begin{itemize}
\item[a.]
So far as in the Minkowski space we have not initially introduced
any orientation, there is no orientation on the plane
$M^{\perp/\parallel}_{\mathbb R}$. Therefore we can not say which
space is denoted $M^{(+1)/\parallel}_{\mathbb C}$, and which
$M^{(-1)/\parallel}_{\mathbb C}$. In fact, by fixation of these
notations we introduce orientation.
\item[b.]
If we extend the Lorentz group to the {\it full\/} Lorentz group
by including there space reflections, then the little group also
expands. The group $SO(2)$ acting in the plane
$M^{\perp/\parallel}_{\mathbb R}$ expands to $O(2)$. Under
reflections the complex spaces
 $M^{(+1)/\parallel}_{\mathbb C}$ and
 $M^{(-1)/\parallel}_{\mathbb C}$
will be transformed to each other and the complex representation
in the space $M^{\perp/\parallel}_{\mathbb C}$ will become
irreducible.
\end{itemize}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\myparagraph{Subrepresentations $T^{(+1)}_{\mathbb C}$ and
$T^{(-1)}_{\mathbb C}$.} \label{tpm}
 In accordance with the decomposability of the
factor-rep\-re\-sen\-ta\-tion $T^{\perp/\parallel}_{\mathbb C}$
shown in the section~\ref{tperpparcompl}, in the complex Minkowski
space $M_{\mathbb C}$ we can find two two-dimensional reducing
subspaces of the representation $T_{\mathbb C}$. These two
subspaces we will denote $M^{(+1)}_{\mathbb C}$ and
$M^{(-1)}_{\mathbb C}$. The corresponding one-dimensional
factor-spaces
 $M^{(+1)}_{\mathbb C}/M^\parallel_{\mathbb C} =
  M^{(+1)/\parallel}_{\mathbb C}$ and
 $M^{(-1)}_{\mathbb C}/M^\parallel_{\mathbb C} =
  M^{(-1)/\parallel}_{\mathbb C}$ were introduced in the
section~\ref{tperpparcompl}.

Now it is clear that in the complex case the series~(\ref{admset})
can be thickened in two ways:
\begin{equation}
 \{0\}= M^0_{\mathbb C} \subset
 M^\parallel_{\mathbb C} \subset
 M^{(+1)}_{\mathbb C} \subset
 M^\perp_{\mathbb C} \subset
 M^4_{\mathbb C} = M_{\mathbb C} \ ,
\label{admsetplus}
\end{equation}
\begin{equation}
 \{0\}= M^0_{\mathbb C} \subset
 M^\parallel_{\mathbb C} \subset
 M^{(-1)}_{\mathbb C} \subset
 M^\perp_{\mathbb C} \subset
 M^4_{\mathbb C} = M_{\mathbb C} \ ,
\label{admsetminus}
\end{equation}

It will be shown in the section~\ref{tperpcompl}, that the complex
vector representation does not have other reducing subspaces. So,
besides~(\ref{admsetplus}) and~(\ref{admsetminus}), the
representation $T_{\mathbb C}$ does not have other composition
series.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\myparagraph{Indecomposability of representation
 $T_{\mathbb C}$.} \label{tcompl}
 Let us suppose that the representation $T_{\mathbb C}$ is
decomposable, i.~e. the space $M_{\mathbb C}$ can be represented
as a sum
\begin{equation}
 M_{\mathbb C}=M'_{\mathbb C}\oplus M''_{\mathbb C}
\label{decompt}
\end{equation}
of two reducing subspaces $M'_{\mathbb C}$ and $M''_{\mathbb C}$.
Then one of these subspaces, for example~$M'_{\mathbb C}$,
necessarily contains some vector~$n$ that does not belong
to~$M^\perp_{\mathbb C}$. According to the lemma from the
section~\ref{compl}, without loss of generality we can assume that
$n^\mathrm{Im}\in M^\perp_{\mathbb R}$ and $n^\mathrm{Re}\notin
M^\perp_{\mathbb R}$.

Let us introduce now some real vector $e^1$. In the case if the
vector $n^\mathrm{Im}$ does not belong to
 $M^\parallel_{\mathbb R}$, $e^1$ is just another notation
for $n^\mathrm{Im}$. But if $n^\mathrm{Im}$ belongs to
$M^\parallel_{\mathbb R}$, vector $e^1$ is chosen as any vector
such that $e^1\in M^\perp_{\mathbb R}$ and
 $e^1\notin M^\parallel_{\mathbb R}$.

Consider now the subgroup of the little group $\EuScript L_k$ that
leaves the vector $e^1$ unchanged. According to the definition of
the vector $e^1$, this subgroup in any case leaves unchanged the
vector $n^\mathrm{Im}$. From the results of the
section~\ref{tperp} we get that the elements of this group can be
defined by some vector $e^2$, which always has its end lying on
some cylinder~(\ref{cylinder}), the condition $e^1\cdot e^2=const$
is satisfied, and the vectors $k$, $e^1$ and $e^2$ are linearly
independent. The end of the vector $e^2$ runs through some line
parallel to~$k$: $e^2(t)=e^2(0)+t k$. The equations~(\ref{fourth})
for the vector $n^\mathrm{Re}(t)$ take the form:
\[
 k\cdot n^\mathrm{Re}(t) =const        \ , \qquad
 e^1\cdot n^\mathrm{Re}(t) =const      \ , \qquad
 e^2(t)\cdot n^\mathrm{Re}(t) =const   \ , \qquad
 (n^\mathrm{Re}(t))^2 =const           \ .
\]
The third equation can be written with more details:
\begin{equation}
 (e^2(0)+t k)\cdot n^\mathrm{Re}(t) =const \ .
\label{thirdeq}
\end{equation}
Obviously, if we suppose that
 $(n^\mathrm{Re}(t)-n^\mathrm{Re}(0))\in M^\parallel$, then
the equation~(\ref{thirdeq}) for $t\neq0$ can not be satisfied.
Therefore, the subspace~$M'_{\mathbb C}$ must contain some real
vector belonging to
 $M^\perp_{\mathbb R}$ and not belonging to
 $M^\parallel_{\mathbb R}$. But then, according to the second
lemma in the section~\ref{tperp}, we
 have~$M^\perp_{\mathbb R}\subset M'_{\mathbb C}$ and, therefore,
$M^\perp_{\mathbb C}\subset M'_{\mathbb C}$.

Let us come back again to the complex vector
 $n\in M'_{\mathbb C}$,
for which
 $n^\mathrm{Re}\notin M^\perp_{\mathbb R}$ and
 $n^\mathrm{Im}\in M^\perp_{\mathbb R}$.
So far as we have already proved that
 $M^\perp_{\mathbb C}\subset M'_{\mathbb C}$,
we get that in $M'_{\mathbb C}$ there is also such a vector which
has zero imaginary part and which has the real part equal to the
real part of vector~$n$. In other words, we have proved that
$M'_{\mathbb C}$ contains a real vector that does not belong to
$M^\perp_{\mathbb R}$. But then, according to the lemma of the
section~\ref{unt}, $M_{\mathbb R}\subset M'_{\mathbb C}$, and,
therefore, spaces $M'_{\mathbb C}$ and $M_{\mathbb C}$ coincide.

So, we have proved the

\mylemma{}
 {\it Every reducing subspace of the representation~$T_{\mathbb C}$
either coincides with the whole space $M_{\mathbb C}$ or belongs
to $M^\perp_{\mathbb C}$.}

From this we get the

\mytheorem{}
 {\it The representation~$T_{\mathbb C}$ is indecomposable.}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\myparagraph{Indecomposability of representations
$T^\perp_{\mathbb C}$, $T^{(+1)}_{\mathbb C}$ and
$T^{(-1)}_{\mathbb C}$.} \label{tperpcompl}
 Let us prove the following lemma:

\mylemma{}
 {\it Every subrepresentation of representation $T^\perp_{\mathbb C}$
contains the subrepresentation $T^\parallel_{\mathbb C}$.}

In order to prove this lemma, let us denote the subrepresentation
under consideration as $T'_{\mathbb C}$. The reducing subspace
where it acts we denote $M'_{\mathbb C}$.

Consider some non-zero vector~$n$ from the subspace
 $M'_{\mathbb C}$. If it lies in~$M^\parallel_{\mathbb C}$,
then the statement of the lemma is satisfied. Suppose now that
this vector does not belong to~$M^\parallel_{\mathbb C}$.

If the vectors $k$, $n^\mathrm{Re}$ and $n^\mathrm{Im}$ are
linearly dependent then, according to the lemma from the
section~\ref{compl}, we can without loss of generality think that
 $n^\mathrm{Im}\in M^\parallel_{\mathbb R}$ and
 $n^\mathrm{Re}\notin M^\parallel_{\mathbb R}$.
According to the first lemma from the section~\ref{tperp}, using
transformations of the little group $\EuScript L_k$ we can change
the vector
 $n^\mathrm{Re}$ so that we add to it an arbitrary vector from
 $M^\parallel_{\mathbb R}$. At the same time the vector
 $n^\mathrm{Im}$ remains unchanged. Therefore
 $M^\parallel_{\mathbb R}\subset M'_{\mathbb C}$ and,
therefore, $M^\parallel_{\mathbb C}\subset M'_{\mathbb C}$.

But if the vectors $k$, $n^\mathrm{Re}$ and $n^\mathrm{Im}$ are
linearly dependent, then, according to the statement of the
section~\ref{tperp}, choosing a proper transformation from the
little group $\EuScript L_k$ we can add to the vector
$n^\mathrm{Re}$ a vector parallel to~$k$ so that the vector
$n^\mathrm{Im}$ remains unchanged. Therefore, again
 $M^\parallel_{\mathbb R}\subset M'_{\mathbb C}$ and
 $M^\parallel_{\mathbb C}\subset M'_{\mathbb C}$.
So, the lemma is proved.

From this lemma we get the two theorems:

\mytheorem{}
 {\it The representation $T^\perp_{\mathbb C}$ is
indecomposable.}

\mytheorem{}
 {\it The representations $T^{(+1)}_{\mathbb C}$ and
 $T^{(-1)}_{\mathbb C}$ are indecomposable.}

Furthermore, from the lemma and the theorem of the
section~\ref{tperppar} we get the

\mytheorem{}
 {\it Besides $M^\parallel_{\mathbb C}$, $M^{(+1)}_{\mathbb C}$,
$M^{(-1)}_{\mathbb C}$ (and $M^\perp_{\mathbb C}$) the
representation $T^\perp_{\mathbb C}$ does not have any other
reducing subspaces.}

So, the representation $T_{\mathbb C}$ does not have any other
composition series, besides~(\ref{admsetplus})
and~(\ref{admsetminus}).

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\section*{}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\myparagraph{Matrices of representations $T_{\mathbb R}$ and
$T_{\mathbb C}$.}
 Many of the obtained results can be represented by patterns for
matrices of the representations $T_{\mathbb R}$ and
 $T_{\mathbb C}$. If in the spaces $M_{\mathbb R}$ and
$M_{\mathbb C}$ we choose appropriate bases, then these templates
become block-triangular:
\[
 \left(\begin{array}{cccc}      \cline{1-1}
 \multicolumn{1}{|c|}{1} & \ \cdot & \cdot & \cdot \\ \cline{1-3}
  0  & \multicolumn{2}{|c|}{} & \cdot \\
  0  & \multicolumn{2}{|c|}{\smash{\raisebox{1.4ex}{$SO(2)$}}}
     & \cdot \\ \cline{2-4}
  0  &   \ 0   &   0   & \multicolumn{1}{|c|}{1}   \\ \cline{4-4}
 \end{array}\right)
 \qquad
 \left(\begin{array}{cccc}      \cline{1-1}
 \multicolumn{1}{|c|}{1}  & \cdot & \cdot & \cdot \\ \cline{1-2}
  0  & \multicolumn{1}{|c|}{e^{-i\varphi}} & 0 & \cdot \\ \cline{2-3}
  0  &   0   & \multicolumn{1}{|c|}{e^{+i\varphi}} & \cdot \\ \cline{3-4}
  0  &   0   &   0   & \multicolumn{1}{|c|}{1}   \\ \cline{4-4}
 \end{array}\right)
\]
And it is known that on the place of every point we have non-zero
element, in general case.

Studying in what subspaces and factor-spaces different blocks of
these matrices act I leave for the reader.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{thebibliography}{99}

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M.~A.~Naymark {\it\rql Lineynye predstavleniya gruppy
Lorentza\rqr}, M.: GIFML (1958).

\bibitem{GelMinShap1958}
I.~M.~Gel'fand, R.~A.~Minlos, Z.~Ya.~Shapiro {\it\rql
Predstavleniya gruppy vrascheniy i gruppy Lorentza\rqr}, M.: GIFML
(1958).

\bibitem{Fyodorov1979}
F.~I.~Fyodorov {\it\rql Gruppa Lorentza\rqr}, M.: Nauka (1979).

\bibitem{Ryder1987}
L.~Ryder {\it\rql Kvantovaya teoriya polya\rqr}, M.: Mir (1987).
[L.~H.~Ryder {\it ``Quantum field theory''}, Cambridge: Cambridge
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\end{thebibliography}

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\end{document}
