\documentclass[ALCO, Unicode]{cedram}


\usepackage[utf8]{inputenc}
\usepackage[english]{babel}
\usepackage{amssymb}
\usepackage{amsthm}
\usepackage{enumitem}
\usepackage{amsmath}
\usepackage{setspace}
\usepackage{indentfirst}
\usepackage{mathtools}

\newcommand{\R}{\mathbb{R}}
\newcommand{\Z}{\mathbb{Z}}
\newcommand{\F}{\mathbb{F}}
\newcommand{\Q}{\mathbb{Q}}
\newcommand{\bC}{\mathbb{C}}
\newcommand{\PG}{\textnormal{PG}}
\newcommand{\Spec}{\textnormal{Spec}}
\newcommand{\Cay}{\textnormal{Cay}}
\newcommand{\Irr}{\textnormal{Irr}}
\newcommand{\tr}{\textnormal{tr}}
\newcommand{\wt}{\textnormal{wt}}
\newcommand{\Supp}{\textnormal{supp}}
\newcommand{\abs}[1]{|#1|}


%\theoremstyle{definition}
%\newtheorem*{definition}{Definition}
%\newtheorem{example}{Example}
%\newtheorem{remark}{Remark}
%\newtheorem{notation}{Notation}
\usepackage{tabstackengine}
%\setstacktabbedgap{1ex}
\savestack{\mathexpr}{$x=0,y=x\left[ \begin{array}{cc}u^{\alpha}+gt & ft^{\alpha^{-1}}\\ t & u\end{array}\right] ;u,t\in K$, $K$ a field}


\title{Twisted quadrics and $\alpha$-flocks}

\author{\firstname{Norman L.} \lastname{Johnson}}

\address{University of Iowa\\ 750 E. Foster Rd. \#306\\ Iowa
City\\ IA 52245}
\email{normjohnson0@icloud.com}


\keywords{twisted hyperbolic flocks, Klein quadric, j-planes, quasifibrations, T-copies, quaternion division rings}
  

\subjclass{51E20, 51E14}

\datereceived{2021-01-28}
\daterevised{2021-12-17}
\dateaccepted{2021-12-21}

\begin{document}
%\equalenv{author_thm}{cedram_thm}

\begin{abstract}
In this article, we provide a general study of what we call twisted quadrics
and consider flocks of the variant of $\alpha $-conics and $\alpha $%
-hyperbolic quadrics. We extend the notion of the Klein quadric to what we
call an $\alpha $-Klein quadric.  Blended kernel translation planes are
defined and analysed when considering $\alpha $-conical flocks and $\alpha $%
-twisted hyperbolic flocks.

The Thas--Walker constructions of conical flocks and flocks of hyperbolic
quadrics are extended to their $\alpha $-analogues.  Using the idea that
any derivable net can be embedded into a $3$-dimensional projective space
over a skewfield, allows us to formulate what might be called a projective
version of work previously given in an algebraic framework.  The theory of
deficiency one flocks is extended to both $\alpha $-conical flocks and $%
\alpha $-twisted hyperbolic flocks. $j$-planes are used to construct two
infinite classes of finite $\alpha $-hyperbolic flocks.
\end{abstract}
\maketitle

\section{Introduction}
This article is something of a culmination of results of flocks of
hyperbolic quadrics, flocks of quadratic cones and their connections to
translation planes admitting so-called regulus-inducing groups.  The study
of flocks of finite quadratic cones has generated the most interest.  The
theory of Thas and Bader--Lunardon \cite{Thas75,BaderLunar}
completely classifies finite flocks of hyperbolic quadrics.  
Then there are
the deficiency one flocks of quadratic cones and deficiency one flocks of
hyperbolic quadrics. 
These are connected by results of the author to
translation planes that admit certain Baer groups, (see \cite{BaerGroups}). 
By ingenious arguments, Thas and Payne \cite{paynethas} and, later by
other mathematicians, every such deficiency one partial flock of a finite quadratic cone may be extended uniquely to a flock of a quadratic
cone\textendash{}saying something very interesting about the associated Baer group
planes being derivable.  
However, the study of deficiency one hyperbolic
flocks is still alive, as there are a few\textendash{}just a few\textendash{}examples showing
that an extension is not always possible.

Jha and Johnson \cite{ConicalET} also analyze deficiency one conical and
hyperbolic flocks over an arbitrary field and show these are equivalent to a
translation plane admitting certain Baer groups.

There are appearances of flocks of quadratic cones, as in hyperbolic
fibrations with constant backhalf, which oddly enough are also connected to
translation planes admitting certain collineation groups (cyclic homology
groups of order $q+1$ if the field is $GF(q)$, and analogous groups when the
field $K$ is infinite, Johnson et al (\cite{Combin,handbook})).  As
certain $j$-planes (translation planes, also) admit such groups, there are
corresponding flocks of quadratic cones from $j$-planes, as well.  In this
way, in the infinite case, all $j$-planes over the field of real numbers
have been determined by the author \cite{Combin}.

In the finite case, all of these planes involve the analysis of translation
planes admitting regulus nets.  The only other derivable net (in the finite
case) is the twisted regulus net (twisted by an automorphism of $GF(q)$).

It might be noted that the twisted regulus net is also a regulus net over a
different scalar field, due to the embedding theory of the author \cite{johnsubplanecovered}.  In this general study, any derivable net may be
shown to be a ``pseudo-regulus'' net in some $3$-dimensional projective space
over a skewfield $F$. As an aside, the author recently completed a study of
all derivable nets over any skewfield (see \cite{CLASS}), where there is
quite a variety of interesting derivable nets other than the twisted regulus
net.  So, at least in the infinite case, the theory of flocks of different
colors and corresponding translation planes is far from complete. We shall
come back to the classification of derivable nets later in this article.

Concerning twisted regulus nets, which occur both in the finite and infinite
cases, when the associated field $K$ admits an automorphism $\alpha $ (they
all do, if $\alpha =1$), we can consider the twisted versions of conical
flocks and hyperbolic quadrics over arbitrary fields. 

The theory of $\alpha $-flocks of $\alpha$-cones has been completed by
Cherowitzo, Johnson and Vega \cite{BILLNORMOSCAR} and connects translation
planes admitting $\alpha $-regulus inducing elation groups with flocks of $%
\alpha $-cones (here the term is $\alpha $-flokki, coined by Kantor and
Pentilla \cite{kANTORpEN}, who pointed the way to this theory).

Furthermore, the author \cite{JohnHomolog} shows that the theory of
hyperbolic flocks and associated translation planes may be generalized by
considering twisted hyperbolic flocks and translation planes admitting
twisted regulus inducing homology groups.

It is known that there cannot be finite partial flocks of quadratic cones of
deficiency one, and the concept of deficiency one partial flocks of $\alpha $%
-cones has been analyzed in Cherowitzo, Johnson and Vega \cite{BILLNORMOSCAR}%
.  In that article, it was shown that all finite deficiency one $\alpha $%
-flocks ($\alpha $-flokki) may be extended to a flock, using an ingenious
argument of Sziklai \cite{Sick}.

There are, however, deficiency one partial hyperbolic flocks and there is a
corresponding equivalent theory of translation planes admitting certain Baer
homology groups.  In this article, we discuss the known examples of such
deficiency one hyperbolic flocks and further consider and complete the
analogous deficiency one twisted hyperbolic flock theory.  

It might be mentioned that for all of the previous work, the translation
planes were always of dimension $2$; that is, their spreads lie in $PG(3,K)$%
, for some field $K$.  For deficiency one twisted hyperbolic flocks, we
develop some theory on what we call blended kernels to completely describe
the translation planes associated, as they no longer are of dimension $2$.

In \cite{BILLNORMOSCAR}, the study of maximal partial $\alpha$-flokki was
considered and connected with maximal partial spreads that are called
``quasifibrations''.  These occur only in the infinite cases and are strange
and wonderful objects.  The quasifibrations studied previously admit
elation $\alpha $-regulus inducing groups.  Here we consider
quasifibrations that admit homology $\alpha $-regulus-inducing groups. \
Then there are the associated Baer groups but now acting on quasifibrations
that are not of dimension $2$ and have blended kernels. So, everything one
can say about translation planes and associated flocks and $\alpha $-flocks,
one can say about quasifibrations and maximal partial flocks called quasi $%
\alpha $-flocks.

We list the main result of the author, for convenience.

\begin{theorem}[Johnson {\cite{TWISTEDFLOCKS.tex}}]
Let $\Sigma $ be a translation plane with
spread in $PG(3,K)$, for $K$ an arbitrary field.  Let $\alpha $ denote an
automorphism of $K$, possibly trivial.  Assume that $\Sigma $ admits an
affine homology group one orbit of which, together with the axis and coaxis,
is a twisted regulus net.  Then all orbits are twisted regulus nets and the
spread may be coordinatized in the following form: Let $V_{4}$ be the
associated $4$-dimensional vector space over $K$. Letting $x$ and $y$ denote 
$2$-vectors, then the spread is:%
\begin{eqnarray*}
x &=&0,\text{ }y=0,\text{ }y=x\left[ 
\begin{array}{cc}
u^{\alpha } & 0 \\ 
0 & u%
\end{array}%
\right] ,\text{ } \\
and\text{ }y &=&x\left[ 
\begin{array}{cc}
F(t) & G(t) \\ 
1 & t%
\end{array}%
\right] \left[ 
\begin{array}{cc}
v^{\alpha } & 0 \\ 
0 & v%
\end{array}%
\right] ; \\
\forall u,t,v,uv &\not=&0,of\text{ }K,
\end{eqnarray*}%
and functions $F,G$ on $K$.  Furthermore, $F$ is bijective. The ``$\alpha $%
-twisted hyperbolic quadric'' has the form:%
\begin{equation*}
\{(x_{1},x_{2},x_{3},x_{4});\text{ such that }x_{1}x_{4}^{\alpha
}=x_{2}^{\alpha }x_{3}\}.
\end{equation*}%
Then there is a flock of the $\alpha $-twisted hyperbolic quadric with the
flock of planes of $PG(3,K)$, as follows:%
\begin{equation*}
\pi _{t}:-x_{1}G(t)^{\alpha }+x_{2}F(t)-x_{3}t^{\alpha }+x_{4}=0,and\text{ }%
\rho :x_{2}=x_{3},
\end{equation*}%
where the intersection with each plane is a non-degenerate $\alpha $-conic.
Conversely, a flock of a twisted hyperbolic flock corresponds to a
translation plane admitting an $\alpha $-regulus inducing homology group.
\end{theorem}

When $\alpha =1$, we have a flock of a hyperbolic quadric and an associated
translation plane admitting a regulus-inducing affine homology group. These
two geometries, the hyperbolic flocks and the translation planes are
equivalent.  There are exactly the following classes; the flocks are the
linear flock, where the associated planes of $PG(3,q)$ share a line, and the
Thas flocks, with a few exceptions. The Thas flocks correspond to the
regular nearfield planes and the exceptional flocks correspond to the
irregular nearfield planes and are due to a number of mathematicians from
various different points of view, Bader \cite{Bader}, Baker--Ebert \cite%
{BakerEbert}, Bonisoli \cite{BILLNORMOSCAR}, Johnson \cite{JohnHomolog}.
There are three irregular nearfield planes that correspond to hyperbolic
flocks of orders $11^{2},23^{2},59^{2}$ which Bader, Bonisoli and Johnson
found, independently for all three orders, and by Baker and Ebert for orders 
$11^{2},23^{2}$.  All of these mathematicians determined the
flocks/translation planes by using essentially different methods.  The main
point here is that the associated translation planes are all Bol planes;
which has been of considerable interest, and there is a complete
classification due to Thas and Bader-Lunardon \cite{BaderLunar,ThasMax}.  There are two (at least) possible formulations for this
classification, depending on when it is phrased in the associated
translation plane or in the hyperbolic flock.

\begin{theorem}[Bader--Lunardon {\cite{BaderLunar}}, Thas  {\cite{ThasMax}}]
Classification of
finite hyperbolic flocks/translation planes admitting regulus-inducing
homology groups in $PG(3,q)$.

Plane version:  The translation planes are nearfield planes; based upon the
regular nearfields of order $q^{2}$ and the three irregular nearfields of
orders $11^{2},23^{2}$ and $59^{2}$.

Flock version:  The hyperbolic flocks are exactly the Thas flocks and the
flocks of Bader, Baker--Ebert, Bonisoli, Johnson.
\end{theorem}

In this article, we consider partial flocks of twisted hyperbolic flocks of
deficiency one and show that these correspond to a certain type of
translation plane admitting a Baer group that we call an $\alpha $-Baer
homology group.  We also study partial flocks of twisted hyperbolic type
and of $\alpha $-flocks of deficiency one, and this time, there is an
associated translation plane admitting an $\alpha$-Baer elation group. 
The translation planes are equivalent to the partial twisted flocks of
deficiency one. However, the translation planes do not have spreads in $%
PG(3,K)$, but have what we call ``blended kernels''.

We also are interested in the Baer theory for $\alpha $-conical flocks, and
complete that theory as well. We list the main theorem for background. The
notation is changed to fit our definition of $\alpha $-regulus nets.

In this article, we define an $\alpha $-regulus net as follows: 
\begin{equation*}
x=0,y=x\left[ 
\begin{array}{cc}
u^{\alpha } & 0 \\ 
0 & u%
\end{array}%
\right] ;u\in K\text{, }\alpha \text{ an automorphism of }K.
\end{equation*}%
Where in Cherowitzo, Johnson, and Vega \cite{BILLNORMOSCAR}, this would have
been called an $\alpha ^{-1}$-regulus net.

\begin{theorem}[Cherowitzo--Johnson--Vega {\cite{BILLNORMOSCAR}}]
Let $K$ be any field and $%
\alpha $ an automorphism of $K$. Let $(x_{1},x_{2},x_{3},x_{4})$ denote
homogeneous coordinates of $PG(3,K)$.  Define $C_{\alpha }$ an $\alpha $%
-cone as $x_{1}^{\alpha }x_{2}=x_{3}^{\alpha +1}$, with vertex $(0,0,0,1)$.
A set of planes which partition the non-vertex points of $C_{\alpha }$ will
be called an $\alpha $-flokki (also an $\alpha $-conical flock). The plane
intersections are called $\alpha $-conics. Assume that $\pi $ is a
translation plane that admits an elation group $E^{\alpha }$, one component
of which together with the axis is an $\alpha $-regulus net has the
following form 
\begin{equation*}
x=0,y=x\left[ 
\begin{array}{cc}
u+g(t) & f(t) \\ 
t & u^{\alpha }%
\end{array}%
\right] ;t,u\in K\text{,}f,g\text{ functions of }K,
\end{equation*}%
when writing the $\alpha $-regulus in the form 
\begin{equation*}
x=0,y=x\left[ 
\begin{array}{cc}
u & 0 \\ 
0 & u^{\alpha }%
\end{array}%
\right] =\left[ 
\begin{array}{cc}
v^{\alpha ^{-1}} & 0 \\ 
0 & v%
\end{array}%
\right] ;t,u,v\in K.
\end{equation*}%
Then there is a corresponding $\alpha $-flokki with planes, corresponding
to 
\begin{equation*}
\rho _{t}:x_{1}t-x_{2}f(t)^{\alpha }+x_{3}g(t)^{\alpha }-x_{4};t\in K.
\end{equation*}%
Conversely, an $\alpha $-flokki may be written in the form $\left\{ \rho
_{t};t\in K\right\} $, for some functions $f(t)$ and $g(t)$, which
constructs a translation plane with the previous form. Hence, translation
planes with spreads in $PG(3,K)$ that admit $\alpha $-regulus inducing
elation groups are equivalent to flocks of an $\alpha $-quadratic cone.
\end{theorem}

We also give a Thas--Walker style construction-proof of the
Cherowitzo--Johnson--Vega result, which will allow a more complete way to deal
with deficiency one $\alpha $-conical flocks.

Concerning deficiency one partial flocks of quadratic cones and deficiency
one partial flocks of hyperbolic quadrics, we list for convenience the
following theorem of Jha and the author.  The approach by Jha and Johnson
using the Klein quadric will be generalized using what we shall call blended
kernel translation planes.  The two theorems corresponding to the elation
groups and homology groups shall be separated for clarity.

\begin{theorem}[Jha-Johnson {\cite{ConicalET}}]
Let $K$ be a field and let $\pi $ be a
translation plane with spread in $PG(3,K)$ that admits a full point-Baer
elation group. Then there is a corresponding partial conical flock of
deficiency one in $PG(3,K)$.  Conversely, any partial conical flock of
deficiency one constructs a translation plane with spread in $PG(3,K)$ that
admits a full point-Baer elation group. The partial conical flock of
deficiency one may be extended to a flock of a quadratic cone if and only if
the net containing the point-Baer axis is a regulus net sharing the axis.
\end{theorem}

\begin{theorem}[Jha--Johnson {\cite{ConicalET}}]
Let $K$ be a field and let $\Sigma $ be a
translation plane with spread in $PG(3,K)$ that admits a full point-Baer
homology group.  Then there is a corresponding partial hyperbolic flock of
deficiency one in $PG(3,K)$.  Conversely, any partial hyperbolic flock of
deficiency one constructs a translation plane with spread in $PG(3,K)$ that
admits a full point-Baer homology group. The partial hyperbolic flock may be
extended to a hyperbolic flock if and only if the net containing the
point-Baer axis and coaxis is a regulus net.
\end{theorem}

\begin{remark}
Note that in the above two theorems there is the term ``point-Baer'' in the
hypotheses.  A Baer subplane in an affine plane must have two properties: \
Every point of the plane must be on a line of the subplane (``point-Baer'')
and every line of the plane must be on a line of the subplane (``line-Baer'').
 These conditions are equivalent in the finite case but, in the infinite
case, Barlotti \cite{Barlotti} shows that they are not!  The author \cite%
{N.LNonBaer} constructs dual translation planes that admit a derivable net
which are not derivable!  So, when dealing with the infinite case, care
must be taken (the interested reader might also look at (25.3) of \cite%
{johnsubplanecovered} for more explanation). When we give our main extension
results, we will show that we may remove the point-Baer hypothesis in the
setting under consideration.
\end{remark}

We also point out that the $j=\frac{p^{s}-1}{2}-$planes of Johnson,
Pomareda, Wilke \cite{J-PLANES3} provide two infinite classes of finite $%
p^{s}$-twisted hyperbolic flocks.

\section{Blended kernels}

In the author's work \cite{johnsubplanecovered}, it is shown that every
derivable net may be embedded into a $3$-dimensional vector space over a
skewfield $K$, $PG(3,K)$, such that points, lines of the derivable net
correspond to lines and points not incident with a fixed line $N$ of $%
PG(3,K) $.  The Baer subplanes of the net correspond to planes that do not
contain $N$, and parallel classes of the derivable net correspond to the
planes of $PG(3,K)$ that contain $N$. It then follows that the collineation
group of the derivable net may be determined as $P\Gamma L(4,K)_{N}$.

The main result of this work shows that every derivable net is a classical
pseudo-regulus net, which is then a classical regulus net when $K$ is a
field.  Using certain natural subgroups of the derivable net, we may
realize an embedding in an associated $4$-dimensional vector space over $K$,
with left kernel mappings $(x_{1},x_{2},x_{3},x_{4})\rightarrow (\delta
x_{1},\delta x_{2},\delta x_{3},\delta x_{4})$, for $\delta \in K^{\ast }$.
 And thus it is possible to obtain the form for the derivable net, which
shall here be called the ``classical form'' is as follows:  Assume that we
are considering $V~$a left $K$-space.%
\begin{eqnarray*}
x &=&0,y=\delta x;\forall \delta \in K;\text{components are right spaces,} \\
P(a,b) &=&\{(ca,cb,da,db);\forall c,d\in K\};\text{ Baer subspaces are left
spaces.}
\end{eqnarray*}%
The derived net is 
\begin{eqnarray*}
x &=&0,y=x\delta ;\forall \delta \in K;\text{ components are left spaces}; \\
P(a,b) &=&\{(ac,bc,ad,bd);\forall c,d\in K\};\text{ Baer subspaces are right
spaces.}
\end{eqnarray*}

So, the form is ``classical'' due to the choice of the subgroup of $P\Gamma
L(4,K))_{N}$, that we use to represent the translation group back into the
constructed $4$-dimensional vector space. Suppose that $K$ has an
automorphism $\alpha $, and consider, for the moment, that $K$ is a field.
Then a classical regulus net $R$ has the form:%
\begin{eqnarray*}
R &:&x=0,y=x\left[ 
\begin{array}{cc}
u & 0 \\ 
0 & u%
\end{array}%
\right] \forall u\in K;\text{ the components and} \\
P(a,b) &=&\{(ac,bc,ad,bd);\forall c,d\in K\};Baer~subspaces\text{. }
\end{eqnarray*}%
Now it is known that there are also ``twisted regulus nets'', twisted by an
automorphism $\alpha $ of $K$, and the classical ones have the form:%
\begin{eqnarray*}
R^{\alpha } &:&x=0,y=x\left[ 
\begin{array}{cc}
u^{\alpha } & 0 \\ 
0 & u%
\end{array}%
\right] \forall u\in K;\text{ the components and} \\
P(a,b) &=&\{(ac^{\alpha },bc,ad^{\alpha },bd);\forall c,d\in K\};Baer\text{
components.}
\end{eqnarray*}%
Now note that the Baer subplanes are no longer subspaces over the original
kernel mappings. Since every derivable net is a classical regulus net, we
seem to have a contradiction.  The catch is that using the embedding
theory, a different ``kernel'' group could be used to turn the derivable net
into a classical regulus net.  In the twisted version, we use the new
kernel with mappings 
\begin{equation*}
(x_{1},x_{2},x_{3},x_{4})\rightarrow (\delta ^{\alpha }x_{1},\delta
x_{2},\delta ^{\alpha }x_{3},\delta x_{4});\text{ }for\text{ }\delta \in
K^{\ast }.
\end{equation*}%
 Call this group $K^{\ast \alpha }$. If we then define $x\cdot u=x\left[ 
\begin{array}{cc}
u^{\alpha } & 0 \\ 
0 & u%
\end{array}%
\right] ,$ where $x$ is a $2$-vector over $K$, we see that that the
components and the Baer subplanes are then both $K^{\alpha }$-subspaces, so
we have the classical derivable net but over a different but isomorphic
field $K^{\alpha }$.  Note that $R^{\alpha }$ has components that are $K$
and $K^{\alpha }$-subspaces, where the Baer subplanes are only $K^{\alpha }$%
-subspaces, when $\alpha \not=1$.

Now to switch ideas again, consider the twisted regulus net over $K$, and
form the derived net.  Now this net is also embeddable and becomes a
classical regulus net under a suitable field $L$. So, that we can appreciate
what the form must be, if we wish to transform the Baer ``components'' to
standard form components, we make a transformation of the Baer subspaces as
follows:

$P(a,b)=\{(ac^{\alpha },bc,ad^{\alpha },bd);\forall c,d\in K\};Baer$
components, and form the basis change transformation%
\begin{equation*}
\theta :(x_{1},x_{2},x_{3},x_{4})\rightarrow (x_{1},x_{3},x_{2},x_{4})\text{.%
}
\end{equation*}%
If $x^{\alpha ^{-1}}=(x_{1}^{\alpha ^{-1}},x_{2}^{\alpha ^{-1}})$, where $%
x=(x_{1},x_{2})$, we see that $P(a,b)$ under $\theta $ consists of the
following vectors: $(ac^{\alpha },ad^{\alpha },bc,bd),$ for $a\not=0,$ is
incident with $y=x^{\alpha ^{-1}}\left[ 
\begin{array}{cc}
u & 0 \\ 
0 & u%
\end{array}%
\right] ,$ for $x=(c^{\alpha },d^{\alpha })$ and $u=a^{-1}b$.  Note that $c$
and $d$ vary over $K$.  When $a=0$, we see we have the form of the derived
net as:%
\begin{equation*}
x=0,y=x^{\alpha ^{-1}}\left[ 
\begin{array}{cc}
u & 0 \\ 
0 & u%
\end{array}%
\right] \forall u\in K;
\end{equation*}%
But, now we are using a different kernel group:%
\begin{equation*}
^{\alpha }K^{\ast }:(x_{1},x_{2},x_{3},x_{4})\rightarrow (\delta ^{\alpha
}x_{1},\delta ^{\alpha }x_{2},\delta x_{3},\delta x_{4});\text{ }for\text{ }%
\delta \in K^{\ast }.
\end{equation*}%
Again, this group may be used in the embedded $3$-dimensional projective
space $PG(3,K)$, to create a different $4$-dimensional vector space over $%
^{\alpha }K$, again creating a classical regulus net with a different scalar
mapping and, hence, a different but isomorphic kernel.

These ideas come into play when we have a translation plane with spread in $%
PG(3,K)$, with the standard kernel group $K^{\ast },$ but we have a twisted
regulus net, whose components are both $K$ and $K^{\alpha }$ subspaces. \
When we derive this twisted regulus net, the derived plane does not have
kernel $K$ and it does not have kernel $K^{\alpha }$, the kernel might be
said to be ``blended''.

\begin{definition}
Given two fields $K$ and $R$. If a translation plane $\pi $ is a union of
subspaces $P\cup L$, where all subspaces in $P$ are $2$-dimensional $K$%
-subspaces and all subspaces of $L$ are $2$-dimensional $R$-subspaces, we
shall say that $\pi $ is a translation plane with blended kernel $(K,R)$ of
dimension $2$.
\end{definition}

In the situation under consideration,  the kernel would be $Fix(\alpha )$,
and the blended kernel would be $(K,K^{\alpha })$.

\begin{definition}
Clearly the idea of blended kernel extends to a set of distinct fields $%
K_{i} $ for $i\in \lambda $ and $\cup P_{i}$ subspaces making up the spread
for $\pi $, where $P_{i}$ are $K_{i}$-subspaces.  Furthermore, the
dimensions need not always be equal.  We use the dimension $2$ only as this
is the situation in the current article.
\end{definition}

Translation planes with blended kernels have nice properties since the
collineation group must leave all $P_{i}$ invariant. When a translation
plane with blended kernel arises from net replacement, there are then a set
of interesting and mutually non-isomorphic translation planes obtained.

Consider the following translation planes:%
\begin{equation*}
x=0,y=x\left[ 
\begin{array}{cc}
u^{\alpha } & bt^{\alpha } \\ 
t & u%
\end{array}%
\right] ;u,t\in K\text{, }K\text{ a field.}
\end{equation*}%
$K$ could be finite or infinite.  In fact, the matrix set could also define
a quaternion division ring, a finite Hughes--Kleinfeld semifield plane or a
(generalized) Hughes--Kleinfeld semifield plane over an infinite field. We
note that this plane corresponds to an $\alpha $-conical flock and to an $%
\alpha $-twisted hyperbolic flock, (see Cherowitzo, Johnson, Vega \cite%
{BILLNORMOSCAR}, Johnson \cite{TWISTEDFLOCKS.tex}, \cite{CLASS}). Consider
the finite case, where the field $K$ is $GF(q)$. Then there are $q$ twisted
regulus nets corresponding to the $\alpha $-conical flock sharing $x=0$ and $%
q+1$ twisted regulus nets sharing $x=0$ and $y=0$.  Similarly, there are
the same/analogous twisted regulus nets when $K$ is an infinite field.

When one of the nets is derived, we obtain a blended translation plane with
respect to $(K,K^{\alpha })$ or with respect to $(K,^{\alpha }K)$.  Hence,
the full collineation group of the translation plane will leave the derived
net invariant.  Furthermore, it follows that given any two of these $2q+1$
possible translation planes or infinitely many possible translation planes
in the infinite field case, any isomorphism $\Gamma $ must map the derived
net of one plane to the derived net of the other plane, and then must act as
a collineation group of the original translation plane.  In these cases,
since the translation plane is always a semifield plane, we have that the $q$
possible blended translation planes obtained from deriving a twisted regulus
net sharing $x=0$ are all isomorphic, whereas, in the finite case, we would
obtain a set of $q+1$ mutually non-isomorphic blended translation planes. \
In the infinite case, the translation planes obtained from deriving a
twisted regulus net sharing $x=0$ and $y=0$ are isomorphic if and only if
the translation plane is a quaternion division ring plane.

Thus we have:

\begin{theorem}
The translation planes with blended kernel $(K,K^{\alpha })$ in the finite
case, where $K$ is $GF(q)$, produce exactly $q+2$ mutually non-isomorphic
blended translation planes.  When $K$ is infinite either there are
infinitely many mutually non-isomorphic blended translation planes or $%
\alpha $ has order $2$ and the associated translation plane is a quaternion
division ring plane, and there are exactly two non-isomorphic blended
translation planes.
\end{theorem}

\subsection{Lifting quasifibrations}

In this section, we give a short reminder of the concept of lifting a
translation plane of dimension $2$.  The term ``quasifibration'' might not be
well known, so we provide a more general definition for any finite dimension 
$n$.

\begin{definition}
Let $Q$ be any partial spread over a field $K$, where the associated vector
space is a $2k-$dimensional vector space over $K$ and there is a matrix
representation of $Q$ as a set of mutually disjoint $k$-dimensional $K$%
-subspaces, where $k$ is a positive integer (the components). If there is a
row among the matrix spread set of the form $[e_{1},e_{2},...,e_{k}]$, for
all $e_{i}\in K$, $Q$ shall be called a quasifibration of dimension $k$. \
Therefore, clearly $Q$ is a maximal partial spread.  If there is a vector
that is not incident with a component of $Q$, then $Q$ is said to be a
proper quasifibration. If the context is clear, we shall often just use the
term ``quasifibration'' to mean proper quasifibration.
\end{definition}

For this subsection, the reader is referred to Biliotti, Jha, Johnson \cite%
{BiliottiJhaJ}, and Johnson, Jha \cite{NEWLIFT} for any additional reference
material.  The main idea of lifting is that from any translation plane or
quasifibration with spread in $PG(3,K)$, where $K$ is a field that admits a
quadratic extension with non-trivial automorphism $\sigma $ of order $2$ of $%
F=K(\theta )$ fixing $K$ pointwise, then there is a translation plane
admitting a $\sigma $-twisted derivable net in $PG(3,K(\theta ))$ as follows:

If the quasifibration (there may not be a complete cover) is represented in
the form: 
\begin{equation*}
x=0,y=x\left[ 
\begin{array}{cc}
f(t,u) & g(t,u) \\ 
t & u%
\end{array}%
\right] ;t,u\in K
\end{equation*}%
then the ``lifted quasifibration has the following form 
\begin{equation*}
x=0,y=x\left[ 
\begin{array}{cc}
w^{\sigma } & (\theta f(t,u)+g(t,u))^{\sigma } \\ 
\theta t+u & w%
\end{array}%
\right] ;t,u\in K,\text{ }w\in F\text{,}
\end{equation*}%
and is a spread or a quasifibration equivalent to a $\sigma $-flock of a $%
\sigma $-cone (or a maximal partial structure).  Notice that we have a $%
\sigma $-twisted derivable net 
\begin{equation*}
x=0,y=x\left[ 
\begin{array}{cc}
w^{\sigma } & 0 \\ 
0 & w%
\end{array}%
\right] ;t,u\in K,\text{ }w\in F.
\end{equation*}%
Therefore, the derived plane is a translation plane/quasifibration with
blended kernel $(K,K^{\sigma })$.  For a translation plane that corresponds
to an $\alpha $-flock of an $\alpha $-cone, but is not a lifted plane, we
consider the following:

\subsection{The Kantor--Pentilla translation planes}

Kantor and Pentilla \cite{kANTORpEN} construct the following translation
planes:

\begin{equation*}
x=0,y=x\left[ 
\begin{array}{cc}
u^{2}+t^{5} & t^{14} \\ 
t & u%
\end{array}%
\right] ;t,u\in GF(2^{e}),\text{where }3\text{ does not divide }e.
\end{equation*}%
We note here that again we have a $2$-twisted derivable net.  The derived
translation plane has blended kernel $(K,K^{2})$, for $K=GF(2^{e})$.

\subsection{Blended quasifibrations}

There are examples of quasifibrations in $PG(3,K)$ that may be lifted to
quasifibrations or correspond to partial $\alpha $-conical flocks in
Cherowitzo, Johnson, Vega \cite{BILLNORMOSCAR}, and Biliotti, Jha, Johnson 
\cite{BiliottiJhaJ}.  All of these quasifibrations correspond to
quasifibrations that admit an $\alpha $-twisted regulus net and hence may be
derived.  All of these derived quasifibrations or translation planes are
with blended kernel $(K,K^{\alpha })$.

\subsection{Twisted quadrics}

\begin{itemize}
\item In preparation for the statement of the main results, we mention the
concept of ``twisted quadrics'' of $PG(3,K)$. Recall the idea of a twisted
regulus net:%
\begin{eqnarray*}
x &=&0,y=x\left[ 
\begin{array}{cc}
w^{\alpha } & 0 \\ 
0 & w%
\end{array}%
\right] ;t,u\in K,\text{ }w\in K\text{, }K\text{ a field,} \\
&&\text{ }\alpha \text{ an automorphism of }K. \\
P^{\alpha }(a,b) &=&\left\{ ((a,b)\left[ 
\begin{array}{cc}
u^{\alpha } & 0 \\ 
0 & u%
\end{array}%
\right] ,(a,b)\left[ 
\begin{array}{cc}
v^{\alpha } & 0 \\ 
0 & v%
\end{array}%
\right] );\forall u,v\in K\right\} ; \\
&&Baer\text{ components.}
\end{eqnarray*}%
We note that although the twisted regulus net is not a regulus over $K$, it
is a regulus over $K^{\alpha }$.

\item Consider a hyperbolic quadric $\Gamma $ over $K$ with homogeneous
coordinates $(x_{1},x_{2},x_{3},x_{4})$ such that 
\begin{equation*}
x_{1}x_{4}=x_{2}x_{3}.
\end{equation*}%
Now write $(x_{1},x_{2},x_{3},x_{4})$ over $K^{\alpha }$, so that the same
vector with the same vector basis but writing over $K^{\alpha }$, becomes $%
(x_{1}^{\alpha ^{-1}},x_{2},x_{3}^{\alpha ^{-1}},x_{4})$, that we have the
associated $\alpha $-twisted regulus as 
\begin{equation*}
x_{1}^{\alpha }x_{4}=x_{2}x_{3}^{\alpha }\text{. }
\end{equation*}%
(In the author's work, (see \cite{TWISTEDFLOCKS.tex}) on twisted hyperbolic
flocks, the form, 
\begin{equation*}
x_{1}x_{4}^{\alpha }=x_{2}^{\alpha }x_{3}
\end{equation*}%
is used, as this fits better with the algebraic form).

\item Consider a quadratic cone over $K$, with homogeneous coordinates

$(x_{1},x_{2},x_{3},x_{4})$ such that 
\begin{equation*}
x_{1}x_{4}=x_{2}^{2}.
\end{equation*}%
Note that there is a conic in the plane in $x_{2}=x_{3}$. Now consider $%
\alpha $-twisted regulus net over $K^{\alpha }$ as a regulus and then
consider the associated $\alpha $-quadric over $K^{\alpha }$ has the form 
\begin{equation*}
x_{1}^{\alpha }x_{4}=x_{2}^{\alpha +1}.
\end{equation*}
\end{itemize}

\section{Foundations and examples}

As mentioned, there are $\alpha $-conical flocks associated with translation
planes that admit certain elation groups and there are $\alpha $-hyperbolic
flocks associated with translation planes that admit certain homology
groups.  All of these planes may be derived to produce translation planes
with blended kernel $(K,K^{\alpha })$.  Now the elation and homology groups
become Baer collineation groups under derivation.

It is possible to study Baer collineation groups in translation planes with
blended kernel $(K,K^{\alpha })$, without the assumption that the associated
net containing the Baer axis (Baer axis and coaxis) is derivable. It will
become apparent that such a translation plane will be missing exactly one
twisted regulus net (regulus net if $\alpha =1$) and then will correspond to
a partial flock of a $\alpha $-conical or $\alpha $-twisted hyperbolic flock
of deficiency one.  In fact, we show that any such deficiency one partial
flock corresponds to such a translation plane with blended kernel.

When $K$ is finite, the deficiency one conical flock may be extended to a
flock and this result has been proved by Payne and Thas \cite{paynethas}. \
Considering deficiency $1$ $\alpha $-conical flocks in the finite case also
may be extended to $\alpha $-conical flocks by result of Cherowitzo,
Johnson, and Vega \cite{BILLNORMOSCAR}.  We also extend the ideas of
deficiency one in this article.

\subsection{An unusual twisted hyperbolic flock}

We note that any deficiency $q-1$ partial spread in $PG(3,K)$, may be a
maximal partial spread but still be embeddable in an affine plane.  By a
result of Jungnickel \cite{JUNGNICKEL}, if the maximal partial spread is
embeddable then the affine plane is the unique extension and is a
translation plane.  The catch is that the translation plane will not have
dimension $2$. That is, the spread will not be in $PG(3,K)$, it will be in a
translation plane with blended kernel $(K,K^{\alpha })$, or a similar
variant.

Consider the following putative spread with blended kernel $%
(GF(4),GF(4)^{2}) $%
\begin{eqnarray*}
x &=&0,y=(x_{1}^{2},x_{1}^{2})\left[ 
\begin{array}{cc}
u & 0 \\ 
0 & u%
\end{array}%
\right] , \\
y &=&x\left[ 
\begin{array}{cc}
v^{2} & b+1 \\ 
1 & v%
\end{array}%
\right] \left[ 
\begin{array}{cc}
s & 0 \\ 
0 & s%
\end{array}%
\right] ;v,u,s\in GF(4)\text{, and }b+1\text{ and }b\text{ both }\not=0.
\end{eqnarray*}

Assume, for the moment, that this is a translation plane.  Recall that a
classical $K$-regulus net has the form 
\begin{equation*}
x=0,y=x\left[ 
\begin{array}{cc}
u & 0 \\ 
0 & u%
\end{array}%
\right] ;u\in K
\end{equation*}%
where all vectors are over a $4$-dimensional vector space $V_{4}$ with
standard scalar multiplication $K^{\ast }$.  Since $\left[ 
\begin{array}{cc}
v^{2} & b+1 \\ 
1 & v%
\end{array}%
\right] $ are non-singular matrices, a transformation by 
\begin{equation*}
\left[ 
\begin{array}{cc}
I & 0 \\ 
0 & \left[ 
\begin{array}{cc}
v^{2} & b+1 \\ 
1 & v%
\end{array}%
\right] ^{-1}%
\end{array}%
\right]
\end{equation*}%
will transform 
\begin{equation*}
x=0,y=x\left[ 
\begin{array}{cc}
v^{2} & b+1 \\ 
1 & v%
\end{array}%
\right] \left[ 
\begin{array}{cc}
s & 0 \\ 
0 & s%
\end{array}%
\right] ;\forall s\in GF(4)^{\ast }
\end{equation*}%
to the classical form.  Hence, we have a set of $4$ regulus nets sharing $%
x=0,y=0$.

Now while the remaining 
\begin{equation*}
x=0,y=(x_{1}^{2},x_{1}^{2})\left[ 
\begin{array}{cc}
u & 0 \\ 
0 & u%
\end{array}%
\right] ;u\in GF(4)
\end{equation*}%
admits the ``regulus-inducing'' group 
\begin{equation*}
\left[ 
\begin{array}{cc}
I_{2} & 0_{2} \\ 
0_{2} & \left[ 
\begin{array}{cc}
u & 0 \\ 
0 & u%
\end{array}%
\right]%
\end{array}%
\right] ;u\in GF(4)^{\ast }.
\end{equation*}%
of order $4-1$, this set of matrices is no longer invariant under the
mappings $GF(4)^{\ast }$; this is not a spread in $PG(3,GF(4))$, but is a
spread with a blended kernel.

Hence, this set is not a regulus.  It is the derived version of a $2$%
-twisted regulus net.

Now note that since we have $q=4$ regulus nets sharing $x=0$ and $y=0$,
there is a corresponding partial hyperbolic flock of deficiency one which
cannot be extended to a hyperbolic flock.  That is, if it could be so
extended, it would correspond to a translation plane with spread in $%
PG(3,16) $.  Then this would be the unique affine plane extension of the
corresponding partial spread in $PG(3,16)$ of deficiency $q-1=3$. Now to see
that we indeed have a spread, we derive the structure given.  We leave the
details to the reader to see that the derived structure has the following
form:%
\begin{equation*}
x=0,y=x\left[ 
\begin{array}{cc}
u^{2} & bt^{2} \\ 
t & u%
\end{array}%
\right] ;t,u\in GF(4).
\end{equation*}%
The regulus-inducing group now has the following form:%
\begin{equation*}
\left[ 
\begin{array}{cc}
\left[ 
\begin{array}{cc}
1 & 0 \\ 
0 & v%
\end{array}%
\right] & 0_{2} \\ 
0_{2} & \left[ 
\begin{array}{cc}
1 & 0 \\ 
0 & v%
\end{array}%
\right]%
\end{array}%
\right] ;v\in GF(4)^{\ast }\text{. }
\end{equation*}%
Here the group is what we call a full Baer homology group with axis $%
(x_{1},0,x_{3},0)$ and coaxis $(0,x_{2},0,x_{4}),$ where $x_{i}$ vary over $%
K $, $i=1,2,3,4$. We see that using the images of $y=x\left[ 
\begin{array}{cc}
s^{2} & b \\ 
1 & s%
\end{array}%
\right] $ the group will map these elements to $y=x\left[ 
\begin{array}{cc}
s^{2} & bv \\ 
v^{-1} & s%
\end{array}%
\right] ,\forall s,v\not=0$.  Letting $v^{-1}=t$, we see that $t^{2}=v$,
since $t\in GF(4)$.

So, we obtain a semifield translation plane of order $16$, and to check that
we have a spread we need only check that the non-zero matrices are
non-singular.  
The determinant is $s^{3}+bv^{3}$, which is either $1+b,1$ or 
$b.$ Taking $b$ and $b+1$ nonzero, we have proved that we have a spread.

\begin{itemize}
\item This spread is unusual in that it provides a $2$-flock of an $2-$cone,
a $2$-twisted hyperbolic flock and a partial hyperbolic flock of deficiency
one. Most unusual.
\end{itemize}

\subsection{The examples of deficiency one partial hyperbolic flocks}

In addition to the spread of the previous subsection, there are exactly five
other examples. 

There is a similar derivable translation plane of order $81$, due to Johnson
and Pomareda \cite{JohnPomer}, also of order $p^{4}$, for $p$ a prime, with
blended kernel also containing%
\begin{equation*}
x=0,y=(x_{1}^{p},x_{1}^{p})\left[ 
\begin{array}{cc}
u & 0 \\ 
0 & u%
\end{array}%
\right] ;u\in GF(p^{2}),\text{ }p\text{ a prime.}
\end{equation*}%
There is the associated derived spread has spread 
\begin{equation*}
x=0,y=x\left[ 
\begin{array}{cc}
u^{p} & bt^{-1} \\ 
t, & u%
\end{array}%
\right] ;u,t\in GF(p^{2}),\text{ }p=3,b\text{ an appropriate constant.}
\end{equation*}%
And the net listed above is the derived net of the twisted regulus net,
given when $t=0$ ($0^{-1}=0$). Consider the derivates of these two
translation planes.  This translation plane and the previous one, when
derived are derivable by a $p$-twisted regulus net and are also transitive
on the components other than $x=0$.  In Johnson and Cordero \cite{JohnCord}%
, there is a classification of all such translation planes.  Note that the
previous homology group of order $p^{2}-1$ now appears as a Baer group in
the plane with spread in $PG(3,p^{2})$.

These two planes of orders $16$ and $81$ of order $p^{2}$, are
extraordinarily exceptions, as seen in the following result.

\begin{theorem}[{Johnson-Cordero \cite{JohnCord}}]
Let $\pi $ be a translation plane in $%
PG(3,p^{2})$, for $p$ a prime that is derivable by a $p$-twisted regulus net
and is also transitive on the components distinct from $x=0$. If the plane
admits a Baer group of order $p^{2}-1$, then the plane is the translation
plane corresponding to either the Johnson partial hyperbolic flock of
deficiency one or  the Johnson-Pomareda partial hyperbolic flock of
deficiency one, of order $p^{4}$ for $p=2$ and $3$, respectively.
\end{theorem}

The four other partial hyperbolic flocks are of orders $25$ or $49$ and are
found by using a computer by Royle \cite{Royle}.

Since translation planes of order $p^{2}$, for $p$ a prime, do not admit
twisted regulus nets, the associated translation planes cannot be derivable.
 Hence, we have:

\begin{theorem}
The four Royle deficiency one partial hyperbolic flocks, two of order $5^{2}$
and two of order $7^{2}$, correspond to translation planes in $PG(3,GF(p))$,
for $p=5,7$ that admit Baer groups of order $p-1$.  There are $p$ orbits $%
(p^{2}+1-(p+1))/(p-1)$ such that the Baer axis and coaxis together with the $%
p$ orbits form a partial hyperbolic flock of deficiency one.
\end{theorem}

\begin{itemize}
\item All translation planes of orders $25$ and $49$ are known by computer
searches by Mathon and Royle \cite{MATHONROYLE} and Czerwinski and Oakden 
\cite{CzerOak}.

\item The two translation planes of order $25$ in question are identified as 
$A_{2}$ and $B_{5}$. These identifications might not be correct.  This is
due to the fact that $A_{2}$ is identified further in Czerwinski and Oakden 
\cite{CzerOak} as a Dickson nearfield plane.  But this plane corresponds
not to a partial hyperbolic flock but to an extended one, which would imply
that the Baer net is derivable. Hence, the partial hyperbolic flock is not
maximal or the plane was misidentified.

\item $B_{5}$ is identified as a Walker plane by Czerwinski and Oakden \cite%
{CzerOak}.  However, if it is meant that this is the Walker plane that
admits a group of order $5(5-1)$, this cannot correspond to a translation
plane that admits a Baer group (see the last section).  We must have a Baer
group of order $4$, which does not occur in the Walker plane.  Also,
another problem is that if $B_{5}$ is a Walker plane, the Walker plane is
derivable, by Jha--Johnson \cite{JJWalker}, and the translation plane in
question does not contain a regulus, according to the computer program. But,
even if it did, the partial hyperbolic flock would be extendable.

\item The translation planes order $49$ are identified as $S771$ and $S773$%
. Since the associated translation planes must admit a Baer group of order $%
7-1$, there is probably a Baer net of degree $7+1=8$ that is invariant (this
is just a speculation, at this point).  In both of these translation planes
there is an orbit of length $8$.  This would be very interesting, as it
would say that the components of the Baer net are permuted in one orbit.
\end{itemize}

\section{Quasifibrations as $T$-copies}

Previously, we mentioned quasifibrations of dimension $2$.  Therefore, we
have a quasifibration of the following form in $PG(3,K)$, for $K$ a field:%
\begin{equation*}
x=0,y=x\left[ 
\begin{array}{cc}
f(t,u) & g(t,u) \\ 
t & u%
\end{array}%
\right] ;\forall t,u\in K\text{, and }f\text{, and }g\text{ functions on }%
K\times K.
\end{equation*}%
This is a maximal partial spread and when $K$ is finite, it is clearly a
spread.  When this maximal partial spread is not a spread, it is a proper
quasifibration of dimension $2$.

At the time of the writing of this article, all of the known quasifibrations
were of dimension $2$ and all but one were constructed by what is called
``transcendental copying'', or $T$-copying, by which a spread over $K$ is
copied in $K(z)$, the rational function field over $K$.

The interesting part of $T$-copying is that the properties of the spread
from which it is copied are preserved. If there is a regulus-inducing group,
either an elation or homology group acting on a translation plane, and if
the $T$-copy exists, there will be a quasifibration with this same property.

\begin{definition}
A quasifibration of dimension $2$ and is not a spread, which admits $\alpha $%
-regulus inducing elation or homology group shall be called a quasi $\alpha $%
-flock, either of an $\alpha $-cone or of an $\alpha $-twisted quadric.  If
a matrix spread of $n$ dimensions over $K$ can be considered a partial
spread over $K(z)$, the rational function field over $K$, it is called a $T$%
-copy.  This also can work over infinite fields $K$, wherever the exponents
in the spread are meaningful in the rational function field extension.  If
there are automorphisms in a group $G$ within the entries of the spread
sets, then extend these automorphisms to $K(z)$, so that the automorphisms
leave $Fix(G(z))$ fixed pointwise.  In this setting, if this is also a $T$%
-copy, we use the term ``twisted $T$-copy''.
\end{definition}

The following are examples of $T$-copies as quasifibrations:%
\begin{eqnarray*}
x &=&0,y=x\left[ 
\begin{array}{cc}
u^{\sigma } & bt^{\rho } \\ 
t & u%
\end{array}%
\right] ;\forall t,u\in K=GF(q)(\theta )\text{, } \\
\sigma &\not=&1\text{, }\rho \text{ automorphisms of }GF(q),\text{ } \\
&&K\text{ a transcendental field extension of }GF(q), \\
&&b\text{ a non-square in }GF(q),q\text{ odd.}
\end{eqnarray*}%
This is an example of a quasi $\sigma $-flock of a $\sigma $-cone.

\begin{itemize}
\item If $\sigma =\rho $ then this maximal partial spread, is also a quasi $%
\sigma $-hyperbolic partial spread; 
\begin{equation*}
x=0,y=x\left[ 
\begin{array}{cc}
u^{\sigma } & bt^{\sigma } \\ 
t & u%
\end{array}%
\right] =(y=x\left[ 
\begin{array}{cc}
s^{\sigma } & b \\ 
1 & s%
\end{array}%
\right] \left[ 
\begin{array}{cc}
v^{\sigma } & 0 \\ 
0 & v%
\end{array}%
\right] ).
\end{equation*}%
Also, a twisted $T$-copy of this form becomes a spread by results of Johnson
and Jha \cite{RATIONALJJ}.

\item If we try the same analysis with the Johnson partial hyperbolic flock
of deficiency one, it is possible to show that we have a maximal partial
hyperbolic quasi flock of deficiency one.
\end{itemize}

\begin{eqnarray*}
x &=&0,y=x\left[ 
\begin{array}{cc}
u^{2} & bt^{-1} \\ 
t & u%
\end{array}%
\right] ;\forall t,u\in GF(4)(\theta )\text{, }b,b+1\text{ not }0\text{,} \\
&&\text{where }GF(4)(\theta )\text{ is a transcendental field extension of }%
GF(4).
\end{eqnarray*}%
Noting that $t^{-1}=t^{2}$, we see now we have a $T$-copy of the Johnson
partial hyperbolic flock of deficiency one.  Note that $u\rightarrow u^{2}$
is not bijective, thus we have a proper quasifibration.

Furthermore, if we take a twisted $T$-copy, we have a spread, and contains a
subplane that admits a partial hyperbolic flock of deficiency one.

Hence, we have the following two results Biliotti, Jha and Johnson \cite%
{BiliottiJhaJ}, for the first result, (slightly generalized).  More details
and the complete proof are in Johnson and Jha \cite{RATIONALJJ}.

\begin{enumerate}[label=(\arabic*)]
\item The $T$-copy of a Klein flock of a quadratic cone and flock of a
hyperbolic quadric is a proper quasifibration (not a spread).

\item The $T$-copy of the Johnson partial flock of deficiency one of a
hyperbolic quadric over $GF(4)$ is a proper quasifibration and a twisted $T$%
-copy is a spread that contains a subplane producing a deficiency one
hyperbolic flock.

\item Consider also%
\begin{eqnarray*}
x &=&0,y=x\left[ 
\begin{array}{cc}
u+nt^{3} & nt^{9}+n^{3}t \\ 
t & u%
\end{array}%
\right] ; \\
&&n\text{ non-square in }K\text{ a field of characteristic }3 \\
&&\text{(Biliotti--Jha--Johnson \cite[29.3.5]{BiliottiJhaJ}).}
\end{eqnarray*}%
These quasifibrations are called the ``generalized Ganley'' additive
quasifibrations.  When $K$ is infinite, there are fields that determine
proper quasifibrations.

\item The Cherowitzo--Johnson--Vega \cite[(2.6)]{BILLNORMOSCAR} 
quasifibrations: 
\begin{eqnarray*}
x &=&0,y=x\left[ 
\begin{array}{cc}
u^{\alpha } & -t^{3\alpha ^{-1}} \\ 
t & u%
\end{array}%
\right] ;\text{ }K\text{ an ordered field} \\
&&\text{ and }\alpha \text{ an automorphism of }K.
\end{eqnarray*}
Then this is a quasifibration providing a proper quasi $\alpha $-conical
flock, but is not a $T$-copy. 
\end{enumerate}

Recently, the author  considers Galois chains of
quasifibrations \cite{CHAINS}, where given any quasifibration of dimension $2$, over a
field $K$ and a Galois tower of quadratic extensions with base $K$, it is
possible to form an associated chain of quasifibrations.  The basic
question if the quasifibration is proper, that is, not a spread, is there
any way to determine if there is a quasifibration at link $k$ that becomes a
spread? There is a method developed for this question and it is shown that
there can never be a spread in a Galois chain of arbitrary length, with one
of the known proper quasifibrations as base.  

Therefore, although there are just a few proper quasifibrations to serve as
bases of Galois chains of quasifibrations, every link in the chain is a
proper quasifibration, as is each of the derived planes with blended kernel.
 In this way, we have infinitely many proper quasifibrations both of
dimension $2$ and infinitely many proper quasifibrations of dimension $4$.

\subsection{Baer theory of twisted flocks; Part I}

Our main results in this article extend the above result for partial $\alpha 
$-conical flocks and partial $\alpha $-twisted hyperbolic flocks of
deficiency one, over any field $K$ that admits an automorphism $\alpha $.

In Jha and Johnson \cite{ConicalET}, the analysis focused on understanding
the structure of the hyperbolic flocks and of flocks of quadratic cones. \
The analysis is that from a translation plane with spread in $PG(3,K)$
admitting a regulus-inducing elation group $E$ or a regulus-inducing
homology group $H$ actually involves partitioning the hyperbolic quadric or
quadratic cone by the invariant Baer subplanes of the associated group $H$
or $E$, respectively.

We shall give the pertinent theory providing the structure of the set of
invariant $2$-dimensional $K$-subspaces in the associated $4$-dimensional $K$%
-vector spaces.  But, also, we have noted that instead of a partition of $%
V_{4}$ over $K$ by regulus-inducing groups, we may also consider analogous
theory by twisted regulus-inducing groups, either elation or homology
groups.  We now consider a partition of the twisted quadrics; the twisted
conical flocks and the twisted hyperbolic flocks.  After the preliminary
material below, the extension of the theorem of Jha--Johnson \cite{ConicalET}
above may be directly extended by using $\alpha $-reguli instead of reguli.
 The idea is to work over $K^{a}$, when dealing with the Baer subplanes,
then all of the arguments extend directly.

Our main results for the hyperbolic situation are as follows.  However,
``translation plane'' may be replaced by ``quasifibration'' in the most general
version of the theorem.

\begin{theorem}
Let $\Sigma $ be a translation plane with blended kernel $(K,K^{\alpha })$
that admits a full Baer $\alpha $-homology group.  Then there is a
corresponding twisted hyperbolic flock of deficiency one in $PG(3,K)$.
Conversely, any partial twisted hyperbolic flock of deficiency one
constructs a translation plane of blended kernel $(K,K^{\alpha })$ that
admits a full Baer $\alpha $-homology group. The partial $\alpha $%
-hyperbolic flock may be extended to an $\alpha $-hyperbolic flock if and
only if the Baer net is a derived $\alpha $-regulus net.
\end{theorem}

The theorem for Baer $\alpha $-elation groups is then as follows:

\begin{theorem}
Let $\Sigma $ be a translation plane with blended kernel $(K,K^{\alpha })$
that admits a full Baer $\alpha $-elation group.  Then there is a
corresponding partial $\alpha $-flock of an $\alpha $-cone of deficiency one
in $PG(3,K)$. Conversely, any partial flock of deficiency one of an $\alpha $%
-cone constructs a translation plane of blended kernel $(K,K^{\alpha })$
that admits a full Baer $\alpha $-elation group. The partial $\alpha $-flock
may be extended to an $\alpha $-flock if and only if the Baer net is a
derived $\alpha $-regulus net.
\end{theorem}

The proofs will follow after some results on group orbits.  We shall state
the elation and homology Baer theorems separately, but will tend to combine
the proofs, as there are definite similarities.  First the Baer elation
group situation:

\begin{theorem}
Let $K$ be a field and let $\alpha $ be an automorphism of $K$. Let $V_{4}$
be a vector space over $K$ and let $x=0$ and $y=0$ be disjoint $2$%
-dimensional $K$ and $K^{\alpha }$ subspaces, where $x$ and $y$ are $2-$%
vectors. We note that $V_{4}$ may also be considered a $4$-dimensional
vector space over $K^{\alpha }$. Consider the following group in $GL(4,K)$: 
\begin{equation*}
E^{\alpha }=\left\{ \left[ 
\begin{array}{cc}
I_{2} & \left[ 
\begin{array}{cc}
u^{\alpha } & 0 \\ 
0 & u%
\end{array}%
\right] \\ 
0_{2} & I_{2}%
\end{array}%
\right] ;u\in K\right\} .
\end{equation*}%
We note that $E^{a}$ fixes $x=0$ pointwise.\  (1)(a) Then the elements of
the set of $E^{\alpha }$ invariant $2$-dimensional subspaces not equal to $%
x=0$ are each generated from a $1$-dimensional $K^{\alpha }$ subspace of $%
x=0 $ and another $1$-dimensional space $K^{\alpha }$. Two of these $2$%
-dimensional subspaces over $K^{\alpha }$ are disjoint if and only if they
are disjoint on $x=0$. (1)(b) Let $\left\{ x_{i};i\in \lambda \right\} $
denote the set of $1$-dimensional $K^{\alpha }$ subspaces of $x=0$ and for
each $x_{i}$, choose any other $1$-dimensional subspace $w_{i}$ not incident
with $x=0$ such that the generated $2$-dimensional subspace is $E^{\alpha }$%
-invariant. Then the set of all $2$-dimensional $K^{\alpha }$-subspaces of 
\begin{equation*}
\left\{ \left\langle x_{i},w_{i}\right\rangle ,i\in \lambda \right\}
\end{equation*}
is a partial spread net covering $x=0$. If we can form these subspaces
into a derivable net, then the derived net is an $\alpha $-regulus net over $%
K$.
\end{theorem}

For the Baer homology situation, we have:

\begin{theorem}
Let $K$ be a field and let $\alpha $ be an automorphism of $K$. Let $V_{4}$
be a vector space over $K$ and let $x=0$ and $y=0$ be disjoint $2$%
-dimensional $K$ and $K^{\alpha }$ subspaces, where $x$ and $y$ are $2-$%
vectors. We note that $V_{4}$ may also be considered a $4$-dimensional
vector space over $K^{\alpha }$. Consider the following group in $GL(4,K)$,%
\begin{equation*}
H^{\alpha }=\left\{ \left[ 
\begin{array}{cc}
I_{2} & 0_{2} \\ 
0_{2} & \left[ 
\begin{array}{cc}
u^{\alpha } & 0 \\ 
0 & u%
\end{array}%
\right]%
\end{array}%
\right] ;u\in K^{\ast }\right\} .
\end{equation*}%
Let $x=0$ and $y=0$ be as assumed.  (2)(a) Then all $H^{\alpha }$-invariant 
$2$-dimensional subspaces are generated by a $1$-dimensional $K^{\alpha }-$%
space from $x=0$ and a $1$-dimensional $K^{\alpha }-$space of $y=0$. (2)
(b) Let $\left\{ x_{i};i\in \lambda \right\} $ denote the set of $1$%
-dimensional $K^{\alpha }$ subspaces of $x=0$ and let $\left\{ y_{i};i\in
\lambda \right\} $ be the set of all $1$-dimensional $K^{\alpha }$-subspaces
of $y=0$. Let $\Gamma $ be any bijective mapping from $\left\{ x_{i};i\in
\lambda \right\} $ onto $\left\{ y_{i};i\in \lambda \right\} $.  Then the
set of $2$-dimensional $K^{\alpha }$-subspaces 
\begin{equation*}
\left\{ (x_{i},\Gamma (x_{i}));i\in \lambda \right\}
\end{equation*}
defines a partial spread net.  If these subspaces can be formed into a
derivable net then the derived net is an $\alpha $-regulus net whose
components are $K$-subspaces and $K^{\alpha }$-subspaces, and whose $1$%
-dimensional $K^{\alpha }$-subspaces on $x=0$ and $y=0$ are completely
covered.
\end{theorem}

\begin{proof}
(1)(a). $E^{\alpha }=\left\{ \left[ 
\begin{array}{cc}
I_{2} & \left[ 
\begin{array}{cc}
u^{\alpha } & 0 \\ 
0 & u%
\end{array}%
\right] \\ 
0_{2} & I_{2}%
\end{array}%
\right] ;u\in K\right\} $.  Consider any vector $(x_{1},x_{2},x_{3},x_{4})$
and consider the $E^{\alpha }$ orbit of this vector.  This is 
\begin{equation*}
\left\langle (x_{1},x_{2},x_{1}u^{\alpha }+x_{3},x_{2}u+x_{4});u\in
K\right\rangle .
\end{equation*}%
This subtracting $(x_{1},x_{2},x_{3},x_{4})$ from a general term, we see
that we have the vectors $(0,0,x_{1}u^{\alpha },x_{2}u)$.  Recall that $%
V_{4}$ is a $K^{\alpha }$-vector space as well, and we have the scalar
multiplication $\cdot $ given by $(x_{1},x_{2},x_{3},x_{4})\cdot
u=(x_{1}u^{\alpha },x_{2}u,x_{3}u^{\alpha },x_{4}u),$ we see that we must
have $(0,0,x_{1},x_{2})$ and the $1$-dimensional $K^{\alpha }$ space within
this subspace, in any two dimensional subspace that is invariant $E^{\alpha
} $.  This subspace is $2$-dimensional over $K$ if and only if $%
x_{1}=x_{2}=0$, for $\alpha \not=1$.  We consider $\Omega ^{\alpha }$ as
the set of ``points'' that are either $E^{\alpha }$ or $H^{\alpha }$ $2$%
-dimensional $K^{\alpha }$ orbits:  The set of points has the following
form 
\begin{equation*}
\left\{ (x_{1}^{\alpha }x_{4},x_{1}x_{3}^{\alpha },x_{2}^{\alpha
}x_{4},x_{2}x_{3}^{\alpha });x_{i}\in L\text{.}\right\}
\end{equation*}%
Abstractly, this forms the basis for the $3$-dimensional projective
intersection with the $K^{\alpha }$-Klein quadric, which will be discussed
in the subsequent material.  We note the following: The $E^{\alpha }$ and $%
H^{\alpha }$-orbits define  ``points'' and furthermore, if a vector $%
(x_{1},x_{2},x_{3},x_{4})\rightarrow $ $(x_{1}^{\alpha
}x_{4},x_{1}x_{3}^{\alpha },x_{2}^{\alpha }x_{4},x_{2}x_{3}^{\alpha })$ then
every $K$-space generated by this vector $(x_{1}\beta ,x_{2}\beta
,x_{3}\beta ,x_{4}\beta )~$maps to the same ``point''/modulo $K$ $\rightarrow
(x_{1}^{a}x_{4}\beta ^{\alpha +1},x_{1}x_{3}^{\alpha }\beta ^{\alpha
+1},x_{2}^{\alpha }x_{4}\beta ^{\alpha +1},x_{2}x_{3}^{\alpha }\beta
^{\alpha +1})\equiv $ $(x_{1}^{\alpha }x_{4},x_{1}x_{3}^{\alpha
},x_{2}^{\alpha }x_{4},x_{2}x_{3}^{\alpha })$.  So, in the associated
flocks, the plane intersections could be visualized as $K$-vectors whose
corresponding mapped $K^{\alpha }$-Klein quadric objects that arise from
either $E^{\alpha }$ or $H^{\alpha }$-$2$-dimensional or $K^{\alpha }$%
-subspace/orbits.  In this manner, we may consider plane intersections as
either covering $\Omega ^{\alpha }$ in the $H^{\alpha }$-case or by covering
an associated $\alpha $-flock of an $\alpha $-quadratic cone. Now assume
that two of the $E^{\alpha }$-subspaces are not disjoint%
\begin{equation*}
\left\langle (x_{1},x_{2},x_{3},x_{4})E^{\alpha }\right\rangle \cap
\left\langle (y_{1},y_{2},y_{3},y_{4})E^{\alpha }\right\rangle .
\end{equation*}%
Assume that%
\begin{equation*}
(x_{1},x_{2},x_{1}u^{\alpha }+x_{3},x_{2}u+x_{4})\cdot \delta
=(y_{1},y_{2},y_{1}v^{\alpha }+y_{3},y_{2}v+y_{4}),
\end{equation*}%
for $u,v,\delta \in K$. Then $\left\langle (x_{1},x_{2})\right\rangle =$\ $%
\left\langle (y_{1},y_{2})\right\rangle $, and hence these subspaces share a 
$1$-dimensional subspace on $x=0.$ It now follows directly that the two
subspaces are equal.  This proves (1)(a). Now it is clear that $\left\{
\left\langle x_{i},w_{i}\right\rangle ,i\in \lambda \right\} $ is a set of $%
E^{\alpha }$-invariant $K^{\alpha }$-subspaces that are mutually disjoint
and completely cover $x=0$.  Let $w_{i}=(x_{1,i},x_{2,i},x_{3,i},x_{4,i})$,
where the $2$-vector over $K^{\alpha }$ on $x=0$ is $x_{i}=(x_{1,i},x_{2,i})$%
. The rest of the theorem will now follow directly once we have worked out
the form of the derived nets. (2)(a) Now consider the group 
\begin{equation*}
H^{\alpha }=\left\{ \left[ 
\begin{array}{cc}
I_{2} & 0_{2} \\ 
0_{2} & \left[ 
\begin{array}{cc}
u^{\alpha } & 0 \\ 
0 & u%
\end{array}%
\right]%
\end{array}%
\right] ;u\in K^{\ast }\right\} .
\end{equation*}%
Consider any vector $(x_{1},x_{2},x_{3},x_{4})$ not in either $x=0$ or $y=0$
and consider the $H^{\alpha }$-orbit, $\left\langle
(x_{1},x_{2},x_{3}u^{\alpha },x_{4}u);u\in K^{\ast }\right\rangle $.  By
subtracting the generating vector, and then realizing that we have a $%
K^{\alpha }$-subspace, it follows that $(0,0,x_{3},x_{4})$ is a non-zero
vector in the generated $K^{\alpha }$-subspace. Now subtracting the original
vector from this preceding vector, we also have $(x_{1},x_{2},0,0)$ a
non-zero vector in the generated $K^{\alpha }$-space.  This says that the
only $H^{\alpha }$-invariant 2-dimensional subspaces are generated from two
non-zero vectors on each of the two invariant $K$-subspaces $x=0$ and $y=0$.
 Moreover, it now follows that the only way that two of these invariant $2$%
-dimensional $K^{\alpha }$-subspaces can non-trivially intersect is if they
intersect on either $x=0$ or $y=0$. Therefore, $\left\{ (x_{i},\Gamma
(x_{i});i\in \lambda \right\} $ defines a partial spread net of $K^{\alpha }$%
-invariant $2$-dimensional subspaces that cover $x=0$ and $y=0$.  This
completes the proofs of both (2)(a) and (b).
\end{proof}

\section{The $K^{\protect\alpha }$-Klein quadric}

The reader will notice that we are essentially using ideas of Thas and
Walker, Thas \cite{ThasMax}, coupled with insights about derivable nets, for
these extension notions.

Consider a $6$-dimensional vector space $V_{6}$ over $K$ in the standard
manner and over $K^{\alpha }$ with special scalar multiplication $\cdot $ as
follows:%
\begin{equation*}
(x_{1},x_{2},x_{3},x_{4},x_{5},x_{6})\cdot \delta =(x_{1}\delta ^{\alpha
},x_{2}\delta ,x_{3}\delta ^{\alpha },x_{4}\delta ,x_{5}\delta ^{\alpha
},x_{6}\delta )\text{,}
\end{equation*}%
and written over the associated standard basis, we obtain vectors as 
\begin{equation*}
(x_{1}^{\alpha ^{-1}},x_{2},x_{3}^{\alpha ^{-1}},x_{4},x_{5}^{\alpha
^{-1}},x_{6}),
\end{equation*}%
then the standard hyperbolic form $\Omega _{5}$ becomes 
\begin{equation*}
x_{1}^{\alpha }x_{6}+x_{2}x_{5}^{\alpha }+x_{3}^{\alpha }x_{4}=0.
\end{equation*}%
Now embed $PG(2,K)$ as $x_{3}=x_{4}=0$. We note that the group $E^{\alpha }$
now acts on $PG(3,K).$

We consider the associated $\alpha $-hyperbolic form, which can be given by $%
x_{1}^{\alpha }x_{6}=x_{2}x_{5}^{\alpha }$, without loss of generality. \
Now consider the set of points within $x_{2}=x_{5}$, to obtain $%
x_{1}^{\alpha }x_{6}=x_{2}^{\alpha +1}$, an $\alpha $-conic.  We realize
that $E^{\alpha }$ leaves the associated homogeneous points on the $\alpha $%
-regulus of Baer subspaces pointwise fixed (as ``points''). Identifying the $%
\alpha $-regulus projectively as the $\alpha $-twisted hyperbolic quadric,
we see that we may consider the $\alpha $-conical flock as a set of points
within the $\alpha $-twisted quadric in $PG(3,K)$ as follows:  Consider a
point of $PG(3,K)$ say $v_{0}=(0,0,0,0,0,1)$ as the vertex of the cone, then
a set of planes of $PG(3,K)$ that partition the points of the lines from $%
v_{0}$ to the $\alpha $-conic $x_{1}^{\alpha }x_{6}=x_{2}^{\alpha +1}$, then
the planes correspond to the $K$-components of a translation plane with
spread in $PG(3,K)$ that admit an $\alpha $-regulus-inducing elation group $%
E^{\alpha }$.  The converse that from a translation plane with spread in $%
PG(3,K)$ with group $E^{\alpha }$ produces an $\alpha $-flock of the $\alpha 
$-quadratic cone is now immediate.

Hence, this provides an alternative construction of the main theorem of
Cherowitzo--Johnson--Vega \cite{BILLNORMOSCAR}.  We shall use this approach
when considering the Baer groups and the deficiency one $\alpha $-conical
flocks.

\begin{itemize}
\item This shows that $\alpha $-flocks and the set of Baer components of the
translation plane have a bijective correspondence.

\item The results of Johnson \cite{TWISTEDFLOCKS.tex} show that we may
identity the $K^{\alpha }$-subspaces of the invariant $H^{\alpha }$%
-2-dimensional subspaces with the elements $\Omega _{3}^{\alpha },$ the
twisted hyperbolic quadric in $PG(3,K)$ and the Baer components of the
associated $\alpha $-regulus nets associated with the translation plane
admitting $H^{\alpha }$.

\item This also shows that twisted hyperbolic flocks and the set of Baer
components of the translation plane have a bijective correspondence.
\end{itemize}

\subsection{Baer theory; Part II}

We now complete the proof of our Baer theorems. We begin with the
translation plane admitting a Baer elation group. The reader should note
that the group%
\begin{equation*}
E^{\alpha }=\left\{ \left[ 
\begin{array}{cc}
I_{2} & \left[ 
\begin{array}{cc}
u^{\alpha } & 0 \\ 
0 & u%
\end{array}%
\right] \\ 
0_{2} & I_{2}%
\end{array}%
\right] ;u\in K\right\} ,
\end{equation*}%
has this form when considering the axis as $x=0$. It might be expected that
any associated translation plane admitting $E^{\alpha }$ has $x=0$ as a
component.  However, this may not be the case, and will not be the case
when $E^{\alpha }$ is considered a Baer group and, in this case, $x=0$
becomes a Baer subplane.

If the translation plane admits $E^{\alpha }$ as a Baer group then the
components that lie on the Baer axis (i.e. $x=0$) are $E^{\alpha }$
subspaces, so we have the situation of a translation plane with blended
kernel $(K,K^{\alpha })$.  Now, we still have the orbits of $K$-components
as before, we just have one less $\alpha $-regulus derivable net than in the
case when $E^{\alpha }$ is an elation group.  For clarity, we now speak of
a Baer elation group.  Since the other $\alpha $-regulus nets still
correspond to sets of $E^{\alpha }$ orbits of $K^{\alpha }$-subspaces, we
clearly have a deficiency one $\alpha $-conical flock.

Now suppose that we have a deficiency one $\alpha $-conical flock.  Then on
each line of the $\alpha $-cone, we are missing exactly one point.  This
point corresponds to an $E^{\alpha }$-orbit and since no two of these are on
the same line of the $\alpha $-cone, then the points are mutually disjoint.
 Let $\Lambda $ denote the set of corresponding $2$-dimensional $K^{\alpha
} $-subspaces.  Let $\Omega $ denote the partial spread consisting of the $%
\alpha $-regulus nets corresponding to the deficiency one $\alpha $-conical
flock.  Since now $\Lambda \cup \left\{ \Omega -(x=0)\right\} $, covers all
of the $E^{\alpha }$-orbits in the associated vector space over $K^{\alpha }$%
, we have a spread, but now with blended kernel $(K,K^{\alpha })$, admitting 
$E^{\alpha }$ as a Baer elation group.

Now the question of when the deficiency one $\alpha $-conical flock
situation occurs depends on whether the set $\Lambda $ is a derivable net or
not.  This is valid, as this is the only method that can produce a set of $%
K $-components to return to a translation plane with kernel $K$ admitting $%
E^{\alpha }$ as an elation group. Note that it is possible that the net in
question is derivable without the associated Baer subplanes being $K$%
-subspaces, but it is only when this occurs that the deficiency one partial $%
\alpha $-conical flock may be extended to a flock.

This completes the proof of the deficiency one $\alpha $-conical flock
theorem. Now assume that we have a translation plane with blended kernel $%
(K,K^{\alpha })$ admitting 
\begin{equation*}
H^{\alpha }=\left\{ \left[ 
\begin{array}{cc}
I_{2} & 0_{2} \\ 
0_{2} & \left[ 
\begin{array}{cc}
u^{\alpha } & 0 \\ 
0 & u%
\end{array}%
\right]%
\end{array}%
\right] ;u\in K^{\ast }\right\}
\end{equation*}%
as a Baer homology group, where now $x=0$ and $y=0$ are $K$-Baer subplanes.
In this setting, we are missing one $\alpha $-regulus net of $H^{\alpha }$%
-invariant subplanes and hence the Baer subplanes of the other $\alpha $%
-regulus nets produce a deficiency one twisted hyperbolic flock.

Now assume that we have a deficiency one twisted hyperbolic flock.  From
the analysis of the author in \cite{TWISTEDFLOCKS.tex}, it is clear that we
may construct a partial spread $\Pi $ which consists of a set of $K$-$2$%
-dimensional subspaces that admit $H^{\alpha }$ and is a set of $\alpha $%
-regulus nets sharing $x=0$ and $y=0$.  Recall that our $\alpha $%
-hyperbolic quadric is a hyperbolic quadric with respect to $K^{\alpha }$
(since the $\alpha $-regulus components and Baer components are all $%
K^{\alpha }$-subspaces).  Hence, there are two sets of ruling lines with
respect to $K^{\alpha }$.  It follows that on each $K^{\alpha }$-ruling
line of each of the two sets of ruling lines, there is exactly one missing
point. These points correspond to a set of mutually disjoint $K^{\alpha }$%
-subspaces that are disjoint $H^{\alpha }$-orbits and thus completely cover $%
x=0$ and $y=0$.  Let $W$ denote the set of these $K^{\alpha }$-subspaces in
the associated vector space.  Consider $W\cup \Pi $.  This partial spread
is a spread, since it covers all of the $H^{\alpha }$-invariant $2$%
-dimensional $K^{\alpha }$-subspaces. This translation plane has blended
kernel $(K,K^{\alpha })$ and admits $H^{\alpha }$ as a Baer homology group.

For the question of when the deficiency one partial twisted hyperbolic flock
may be extended, it depends on whether $W$ is a derivable net with Baer
components being $K$-subspaces, similar to the argument for the Baer elation
situation. This completes the proofs of the Baer theory for $\alpha $%
-twisted flocks--except for that question about ``point Baer'' that is in the
hypotheses of the corresponding Baer theory for flocks of quadratic cones
and for hyperbolic quadrics. So, we have shown that with an $\alpha $-flock
of an $\alpha $-quadratic cone, there is an associated translation plane
admitting an $\alpha $-regulus-inducing elation group $E^{\alpha }$ and,
conversely, such a translation plane constructs the $\alpha $-conical flock.
 And, we have shown that flocks of an $\alpha $-twisted hyperbolic quadric
and translation planes admitting $\alpha $-regulus-inducing groups are
equivalent.  The real problem seems to come in the deficiency one theory.

Suppose that we have a translation plane of either type.  Choose any of the 
$\alpha $-regulus nets and suppose that the sublanes involved are ``point
Baer'' but not ``line Baer''.  Now derive the net, what is obtained?  This
cannot be a translation plane any longer, just a Sperner space with blended
kernel $(K,K^{\alpha })$ (which could be $K$, if $\alpha =1$).  But, this
structure still admits a full ``point Baer'' elation or full ``point Baer''
homology group. Therefore, we obtain a deficiency one $\alpha $-flock,
quadratic or hyperbolic, now with a point missing on each of the relative
lines of the cone or of the $\alpha $-rulings. Our proof still shows that we
can recover a spread-a ``translation plane'' --not a Sperner space. So, all of
the point Baer subplanes must be Baer subplanes, since all of the $\alpha $%
-regulus nets can be derived to affine planes. This completes the proofs of
the theorems on Baer theory.

\section{Quasi-flocks}

Quasifibrations have been mentioned previously. Much of the work on flocks
and associated translation planes has been done algebraically.  We have a
certain form of a translation plane and using this form, we create an
associated flock, using that algebraic representation.  More generally, we
may use the forms defining partial spreads to create what appear to be
flocks but are actually quasifibration/maximal partial flocks.  If we
define a ``quasi-flock'' as a partial flock that has the basic form of a flock
but does not satisfy the covering criterion, we have necessarily a maximal
partial flock.  Proper quasi-flocks like proper quasifibrations are
infinite.

In any case, all our results may be phrased more generally in terms of
quasifibrations. Without listing all of various theorems, here is an omnibus
theorem:

\begin{theorem}
Quasifibrations with $\alpha $-regulus inducing groups are equivalent to
quasi $\alpha $-flocks.
\end{theorem}

\section{The Baer forms}

Here we indicate what the translation planes with blended kernel $%
(K,K^{\alpha })$ look like that admit Baer groups.

\begin{itemize}
\item Baer elation translation planes would have the following form:%
\begin{equation*}
x=0,y=(x_{1}^{\alpha ^{-1}},x_{2}^{\alpha ^{-1}})\left[ 
\begin{array}{cc}
v & z(v) \\ 
0 & v%
\end{array}%
\right] ;\forall v\in K\text{,}
\end{equation*}%
\begin{equation*}
y=x\left[ 
\begin{array}{cc}
1 & -u^{\alpha } \\ 
0 & 1%
\end{array}%
\right] \left[ 
\begin{array}{cc}
F(t) & G(t) \\ 
t & 0%
\end{array}%
\right] \left[ 
\begin{array}{cc}
1 & u \\ 
0 & 1%
\end{array}%
\right]
\end{equation*}%
for all $u,t\not=0$ $\in K$. $z$ a function on $K$ so that $z(0)=0$, $%
z(1)=1, $ and functions $F$ and $G$ on $K$.

\item A partial $\alpha $-conical flock may be extended if and only if $%
z(v)=0$ for all $v\in K$.

\item Baer Homology translation planes would have the following form:%
\begin{equation*}
x=0,y=(x_{1}^{\alpha ^{-1}},x_{2}^{\alpha ^{-1}})\left[ 
\begin{array}{cc}
n(v) & 0 \\ 
0 & v%
\end{array}%
\right] ;\forall v\in K\text{, }
\end{equation*}%
$n$ a function on $K$ so that $n(0)=0$, $n(1)=1$ and 
\begin{equation*}
y=x\text{ }\left[ 
\begin{array}{cc}
1 & 0 \\ 
0 & u^{-\alpha }%
\end{array}%
\right] \left[ 
\begin{array}{cc}
g(t) & f(t) \\ 
1 & t%
\end{array}%
\right] \left[ 
\begin{array}{cc}
1 & 0 \\ 
0 & u%
\end{array}%
\right] ;\forall u\not=0,t\not=0\text{ in }K,
\end{equation*}%
for functions $g$ and $f$ on $K$.

\item The deficiency one partial twisted hyperbolic flock may be extended if
and only if $n(v)=v$ $\forall v\in K$.
\end{itemize}

\begin{proof}
The elation group changes form when considering $x=0$ as $(0,x_{2},0,x_{4})$
in the formulation of the axis of the elation group.  The components
incident with the Baer axis are $K^{\alpha }$-subspaces that are left fixed
by the Baer elation group.  We leave, as an exercise, to check out the
remaining parts of the form. Similarly, the homology group changes form when
considering $x=0$ as $(0,x_{2},0,x_{4})$ and $y=0$ as $(x_{1},0,x_{3},0)$. \
Again the components incident with the Baer axis and coaxis are $K^{\alpha }$%
-subspaces that are left fixed by the Baer homology group.  Here is another
exercise to check out the remaining parts of the form.  Note that there are
two distinct uses of the notation of $x=0$ and $y=0$; one use is when
considering the group axis/coaxis and the other use is when considering the
form of the translation plane, when $x=0$ and $y=0$ are Baer subplanes.
\end{proof}

\section{The algebraic and $\protect\alpha $-Klein methods; $\protect\alpha $
and $\protect\alpha ^{-1}$-flocks}

When the ideas of flocks of quadratic cone and flocks of hyperbolic quadrics
were introduced, there were a variety of new studies, in the infinite case,
and also later with what we are calling $\alpha $-flocks of $\alpha $%
-quadratic cones and $\alpha $-twisted hyperbolic flocks. Many of these were
algebraic in nature.  When this was done, there was essentially no
connection between $\alpha $-regulus-inducing groups, elation and homology,
that really becomes the essence of understanding the $\alpha $-Klein
methods.  Moreover, there is no uniformity with the notation of the $\alpha 
$-cones of $\alpha $-twisted quadrics.

\begin{itemize}
\item In Cherowitzo, Johnson, and Vega \cite{BILLNORMOSCAR}, it was pointed
out that whenever an $\alpha $-conical flock is constructed, there is also
an $\alpha ^{-1}$-conical flock which may be constructed and is isomorphic
to the original.  How these two examples may be understood using the $%
\alpha $-Klein method is by a translation of vectors $%
(x_{1},x_{2},x_{3},x_{4})\rightarrow (x_{4},x_{3},x_{2},x_{1})$, which
changes the $\alpha $-conic that is used to the associated $\alpha ^{-1}$%
-conic.  To see this, just note that 
\begin{equation*}
\left\{ x=0,y=x\left[ 
\begin{array}{cc}
u^{\alpha } & 0 \\ 
0 & u%
\end{array}%
\right] ;u\in K\right\} \rightarrow \left\{ x=0,y=x\left[ 
\begin{array}{cc}
u & 0 \\ 
0 & u^{\alpha ^{-1}}%
\end{array}%
\right] ;u\in K\right\}.
\end{equation*}%
This mapping works for the $\alpha $-twisted hyperbolic flocks.  There is
also a corresponding change of functions defining the $\alpha $ or $\alpha
^{-1}$-flocks.  This is the mapping $(x_{1},x_{2},x_{3},x_{4})\rightarrow
(x_{2},x_{1},x_{4},x_{3})$, which does the same thing, changing the $\alpha $%
-twisted regulus net to the $\alpha ^{-1}$-twisted regulus net.

\item In the author's work on $\alpha $-twisted hyperbolic quadrics (see 
\cite{TWISTEDFLOCKS.tex}), the process of translating components of $\alpha $%
-regulus-inducing group $H^{\alpha }$ to the associated $\alpha $-twisted
hyperbolic quadrics does not use the understanding of the ``points'' being the
invariant $2$-dimensional subspaces over $K^{\alpha }$ (not equal to the two
components $x=0$ and $y=0$, the axis and coaxis of $H^{\alpha }$). In fact,
in the algebraic method, it may be seen that the projective connection is
accomplished with the invariant $2$-dimensional subspaces over $K^{\alpha
^{-1}}$.  This does not cause any difficulty as, similar to the elation
case, given an $\alpha $-twisted hyperbolic flock, there is always an
isomorphic $\alpha ^{-1}$-twisted hyperbolic flock.
\end{itemize}

\section{Quaternion division ring planes are flock planes}

To appreciate how the subject of division rings comes into the discussion,
we recall the main theorem of the classification of subplane covered nets,
Johnson \cite{johnsubplanecovered}, which we have previously discussed. \
Here we take a more in-depth view, in order to bring in quaternion division
rings.

We mention only the classification of derivable nets:

\begin{theorem}[Johnson {\cite{johnsubplanecovered}}]
Let $D$ be a derivable net.  Then
there is an embedding of $D$ into a $3$-dimensional projective space $\Sigma 
$ over a skewfield $K$, as follows:  Let $N$ be a fixed line of $\Sigma $,
then the embedding maps the points, lines, Baer subplanes, parallel classes
of $D$ into $PG(3,K)$ as lines, points, planes that do not contain $N$ and
planes that do contains $N$ of $\Sigma $, respectively.  The full
collineation group of $D$ is $P\Gamma L(4,K)_{N}$.  Using the collineation
group, then there is a contraction method that shows that $D$ may be
represented in a $4$-dimensional vector space over $K$ as a pseudo-regulus
net.  If $K$ is a field then the derivable net is a regulus-net.
\end{theorem}

The author's article on the classification of derivable nets, \cite{CLASS}%
, tries to understand the nature of other possible derivable nets that can
sit in the same vector space as the embedded and contracted pseudo-regulus
net.

This is where quaternion division rings come in.  There are four classes of
such derivable nets that can sit in the same $4$-dimensional vector space as
does the contracted pseudo-regulus;  we look at the classification of a
derivable net as a comparison to an existing derivable net.  There is
exactly one type of derivable net that is a pseudo-regulus net but sits in a
embedded/contracted $4$-dimensional vector space over a field; the
comparison derivable net is a regulus net.  This type is the set of
quaternion division rings realized as derivable nets.

It is not really necessary to have the non-commutative algebra definition of
a quaternion division ring for this discussion, but we will require the
matrix definition of a quaternion division ring spread, Johnson \cite{CLASS}:

\begin{definition}
The matrix construction of a quaternion division ring plane in any dimension 
$2$ matrix spread set is as follows: 
\begin{eqnarray*}
x &=&0,y=x\left[ 
\begin{array}{cc}
u^{\sigma } & bt^{\sigma } \\ 
t & u%
\end{array}%
\right] ;t,u\in F(\theta )\text{, a Galois quadratic extension of a field }F%
\text{,} \\
\text{ }b &\in &F\text{, }\sigma \text{ the induced automorphism.}
\end{eqnarray*}%
This example, in the finite case, has been seen earlier in the article. \
But in the previous finite version, or quasifibration version, required $b$
to be a non-square in $F(\theta )$, where the skewfield version requires $b$
to be a non-square in $F$.
\end{definition}

But, the interesting fact about the quaternion division ring planes is that
they are equivalent to $\sigma $-flocks of a $\sigma $-cone, and equivalent
to also $\sigma $-hyperbolic flocks.  The flocks, in both cases, are
linear, although not the same linear structure, as they are two completely
different linear flocks, \cite{TWISTEDFLOCKS.tex}.

\section{Lifting skewfield planes}

The quaternion division ring planes almost always have what are called
``central extensions'' of skewfields $S$, which are quadratic in this case. 
Considering the associated translation planes, the central extensions are
(analogous) to dimension $2$ translation planes, as they can be represented
over the $3$-dimensional projective space with respect to the original
skewfield $S$.  Whenever a skewfield $S$, quaternion or not, has a
quadratic central extension (meaning that the generating quadratic
polynomial is irreducible over the center of $S$), the corresponding
translation plane can be lifted, just as in the commutative dimension $2$
situation, Hiramine, Matsumoto, Oyama \cite{HIRAMINE}.  Lifted spreads of
dimension two provide a wealth of examples of $\sigma $-flocks of $\sigma $%
-quadratic cones.  

By lifting non-commutative skewfield planes that admit central quadratic
extensions, the constructed spreads are semifield spreads in non-commutative 
$3$-dimensional projective spaces, Johnson, Jha \cite{NEWLIFT}.  This is
interesting also as the quaternion division ring spreads, as dimension $2$
spreads over fields, can also be lifted by the standard procedure. Thus, we
have two mutually non-isomorphic spreads, one in $PG(3,K)$, for $K$ a field,
and another in $PG(3,L)$, for $L$ a non-commutative skewfield, constructed
from the same spread, by essentially the same method.

\section{Examples of twisted hyperbolic flocks}

In \cite{MONOMIALJ}, the author shows how to use the Kantor--Knuth flocks of
the quadratic cone in $PG(3,p^{r})$, $p$ odd, to construct the $%
j=(p^{s}-1)/2 $-planes of Johnson, Pomareda, and Wilke \cite{J-PLANES3},
which provide several infinite classes of $p^{s}\,$-twisted hyperbolic
quadrics.  In a $j$-plane, from Johnson, Pomareda and Wilke \cite{J-PLANES3}%
, there is always a cyclic homology group of the following form: 
\begin{equation*}
H^{\alpha }=\left\{ \left[ 
\begin{array}{cc}
\left[ 
\begin{array}{cc}
u^{2j+1} & 0 \\ 
0 & u%
\end{array}%
\right] & 0_{2} \\ 
0_{2} & I_{2}%
\end{array}%
\right] ;u\in GF(p^{r}=q)^{\ast }\right\} ,\text{ }
\end{equation*}%
so when $j=(p^{s}-1)/2$, $2j+1=p^{s},$ which provides the necessary $\alpha
=p^{s}-$twisted hyperbolic flocks.  To obtain the form of the group used in
this article, and to see the form that the twisted hyperbolic flocks take, a
basis change is required, basically by taking the inverses of all of the
matrices, which would make this into a ``right'' $\alpha $-twisted
regulus-inducing group instead of a ``left'' $\alpha $-twisted
regulus-inducing group (we need to invert $x=0$ and $y=0$)$.$  

\begin{itemize}
\item There is also a related $j=(p^{s}-1)/2+(q-1)/2-$plane that also
provides an infinite class of $p^{s}-$twisted hyperbolic flocks, obtained by
a derivation replacement of the set of regulus nets of ``odd'' determinant
type.
\end{itemize}

%\subsection{$x=0,y=x\left[ \begin{array}{cc}u^{\alpha}+gt & ft^{\alpha^{-1}}\\ t & u\end{array}\right] ;u,t\in K$, $K$ $a$ $Field.$}
%\subsection{$\ensuremath{x=0,y=x
%\begin{array}{cc}u^{\alpha}+gt & ft^{\alpha^{-1}}\end{array}}$}


\subsection{\mathexpr}

%\subsection{Suppose $X=\protect\begin{bmatrix}
%1 & 2 \protect\\
%3 & 4
%\protect\end{bmatrix}$}

%\subsection{$\protect\begin{array}{c}1\protect\end{array}$}
In this subsection, we look at translation planes that provide both $\alpha $%
-flocks of $\alpha $-conics and $\alpha $-twisted hyperbolic flocks.  That
this form above exactly describes the translation planes that accomplish
this is shown in Johnson \cite{TWISTEDFLOCKS.tex}, and are the
Hughes--Kleinfeld semifield planes and their infinite generalizations.  

\begin{itemize}
\item However, in that work the Hughes--Kleinfield planes and their $\alpha
^{-1}$-flocks and $\alpha $-twisted hyperbolic flocks were all labeled
linear.  However, this would only be valid in the $\alpha ^{-1}\,$-flock
case, when $\alpha ^{2}=1$ and $g=0$, so that correction should be noted.

\item We have noticed this form previously when $\alpha ^{2}=1$, and $\alpha
\not=1$ and $g=0$.  In this setting, we have both an $\alpha $-flock of an $%
\alpha$-conic and an $\alpha $-twisted hyperbolic flock.  These flocks are
linear and occur also for the quaternion division rings.

\item In this more general setting, by noting that 
\begin{equation*}
x=0,y=x\left[ 
\begin{array}{cc}
u^{\alpha }+gt & ft^{\alpha ^{-1}} \\ 
t & u%
\end{array}%
\right] =\left[ 
\begin{array}{cc}
v+gt & ft^{\alpha ^{-1}} \\ 
t & v\alpha ^{-1}%
\end{array}%
\right]
\end{equation*}%
then put in the expression for $\alpha ^{-1}$-flokki ($\alpha ^{-1}$-flock
of an $\alpha ^{-1}$-cone, as there is a notation change here) to obtain:%
\begin{equation*}
\rho _{t}:x_{1}t-x_{2}f(t)^{\alpha ^{-1}}+x_{3}g(t)^{\alpha
^{-1}}-x_{4};t\in K.
\end{equation*}%
and for $f(t)=ft^{\alpha ^{-1}},$ and $g(t)=gt$, we have:%
\begin{equation*}
\rho _{t}:x_{1}t-x_{2}f^{\alpha ^{-1}}t^{\alpha ^{-2}}+x_{3}g^{\alpha
^{-1}}t^{\alpha ^{-1}}-x_{4};t\in K.
\end{equation*}
\end{itemize}

Then, we note that this same spread is 
\begin{equation*}
=\left\{ 
\begin{array}{c}
x=0,y=x\left[ 
\begin{array}{cc}
u^{\alpha } & 0 \\ 
0 & u%
\end{array}%
\right] ;\text{ } \\ 
y=x\left[ 
\begin{array}{cc}
t^{\alpha }+g=F(t) & f=G(t) \\ 
1 & t%
\end{array}%
\right] \left[ 
\begin{array}{cc}
v^{\alpha } & 0 \\ 
0 & v%
\end{array}%
\right] ;u,t,v\not=0\text{ in }GF(q)%
\end{array}%
\right\} ,
\end{equation*}%
we see that we obtain a $\alpha$-twisted linear hyperbolic flock, 
\begin{equation*}
\pi _{t}:-x_{1}G(t)^{\alpha }+x_{2}F(t)-x_{3}t^{\alpha }+x_{4}=0,and\text{ }%
\rho :x_{2}=x_{3},
\end{equation*}%
which is given by:%
\begin{equation*}
\pi _{t}:-x_{1}f^{\alpha }+x_{2}(t^{\alpha }+g)-x_{3}t^{\alpha }+x_{4}=0,and%
\text{ }\rho :x_{2}=x_{3}.
\end{equation*}

\longthanks{The author would like to express his thanks to the referee for many helpful
suggestions as to the style and clarity of this article.
}


\bibliographystyle{mersenne-plain}
\bibliography{ALCO_Johnson_560}


\end{document}
