# Fixed point groups of involutions of type
$\operatorname{O}(\operatorname{q},k)$ over a field of characteristic two

Mark Hunnell
Winston-Salem State University
<EMAIL_ADDRESS>

John Hutchens
Winston-Salem State University
<EMAIL_ADDRESS>
###### Abstract

For $\operatorname{O}(\operatorname{q},k)$, the orthogonal group over a field
$k$ of characteristic 2 with respect to a quadratic form $\operatorname{q}$,
we discuss the isomorphism classes of fixed points of involutions. When the
quadratic space is either totally singular or nonsingular, a full
classification of the isomorphism classes is given. We also give some
implications of these results for a general quadratic space over a field of
characteristic $2$.

Declarations of interest: none
This research did not receive any specific grant from funding agencies in the
public, commercial, or not-for-profit sectors.
Mathematics Subject Classification: 20G15
Keywords: orthogonal groups; quadratic forms; involutions; algebraic groups;
characteristic 2

##  1 Introduction

We continue the discussion from [18] on
$\operatorname{O}(\operatorname{q},k)$-conjugacy classes of involutions of
$\operatorname{O}(\operatorname{q},k)$, where $k$ is a field of characteristic
2, and turn our focus to the fixed point groups of such involutions. The goal
of this research is to describe generalized symmetric spaces of the form
$G(k)/H(k)$ where $G$ is an algebraic group defined over a field $k$, $H$ is
the fixed point group of some automorphism of order $2$ on $G$ and $G(k)$
(respectively $H(k)$) denotes the $k$-rational points of $G$ (resp. $H$). In
particular we want to extend Helminck’s study of $k$-involutions and symmetric
$k$-varieties [15] to include fields of characteristic $2$. This has been
studied for groups of type $\operatorname{G}_{2}$ and $\mathrm{A}_{n}$ in [22,
25] and over fields of characteristic not $2$ in [6, 3, 2, 4, 19, 20, 21]. We
also extend the results of Aschbacher and Seitz [1] who studied similar
structures for finite fields of characteristic $2$. Symmetric spaces were
first studied by Gantmacher in [12] in order to classify simple real Lie
groups. This was continued in [5] by Berger who provides a complete
classification of symmetric spaces for simple real Lie algebras.

We refer to [17] for notation and vocabulary concerning quadratic forms over
fields of characteristic $2$. Orthogonal and symplectic groups over fields of
characteristic $2$ have been studied extensively. In [14] Cheng Hao discusses
automorphisms of the orthogonal group over perfect fields of characteristic
$2$ when the quadratic form is nondefective. Pollak discusses orthogonal
groups over global fields of characteristic $2$ in the case the quadratic form
is nondefective in [24] and Connors writes about automorphism groups of
orthogonal groups over fields of characteristic $2$ in [7, 8, 9, 10] for a
nondegenerate quadratic form. In [27] Wiitala uses the Jordan canonical form
to study involutions of the orthogonal group over fields of characteristic 2
in the nondefective case. In [11] Jean Dieudonné discusses algebraic
homogeneous spaces over fields of characteristic $2$.

In the following we establish isomorphism classes for the fixed point groups
of involutions of orthogonal groups for quadratic spaces with a totally
singular quadratic form and a non-singular quadratic form. We also provide a
characterization of the fixed point group for a general quadratic form
defining an orthogonal group over a field of characteristic 2. When possible,
this is done without assuming the form is nondegenerate.

Over fields of characteristic not $2$ the $G(k)$-conjugacy classes of
automorphisms of order $2$ are in bijection with $G(k)$-conjugacy classes of
fixed point groups of involutions on $G$ as seen in [16]. However, in
characteristic $2$ there are examples of involutions that are not
$G(k)$-conjugate with isomorphic fixed point groups, see Theorem 5.23, as well
as Theorem 5.27 in [22].

In the Preliminaries section we recall definitions, notation, and useful
results concerning quadratic forms and bilinear forms over fields of
characteristic $2$ and associated algebraic groups. This section also includes
notation from the theory of symmetric $k$-varieties. Next we recall results
concerning involutions of $\operatorname{O}(\operatorname{q},k)$ when $k$ is a
field of characteristic $2$ in Section $3$. This is followed by describing
isomorphism classes of fixed point groups for different types of vector
spaces. In Section $4$ we consider the fixed point group for involutions of
$\operatorname{O}(\operatorname{q},k)$ where $\operatorname{q}$ is the
quadratic form of a totally singular vector space. Then in Section $5$ we
discuss the fixed point group of involutions where $\operatorname{q}$ is the
quadratic form of a nonsingular vector space. Finally, Section 6 has an
overview of the implications for a general vector space over a field of
characteristic $2$.

##  2 Preliminaries

The following definitions can be found in [17]. We also refer to [13] often,
which uses some of the vocabulary for orthogonal groups differently than [17].
When this happens, we default to the definitions of [17].

Let $k$ be a field of characteristic 2 and $W$ a vector space defined over
$k$. We call $\operatorname{q}:W\rightarrow k$ a _quadratic form_ if it
satisfies $\operatorname{q}(aw)=a^{2}\operatorname{q}(w)$ for all $a\in k$,
$w\in W$. There exists a corresponding symmetric bilinear form
$\mathrm{B}:W\times W\rightarrow k$ such that
$\operatorname{q}(w+w^{\prime})=\operatorname{q}(w)+\operatorname{q}(w^{\prime})+\mathrm{B}(w,w^{\prime})$
for all $w,w^{\prime}\in W$. Over fields of characteristic $2$ nonsingular
symmetric bilinear forms are also symplectic.

The pair $(W,\operatorname{q})$ is called a _quadratic space_. The subspace of
$W$

$\mathrm{rad}(W)=\\{x\in W\ |\ \mathrm{B}(x,w)=0,\text{ for all }w\in W\\}.$

is called the _radical_ of $W$.

Given a quadratic form, there exists a basis of $W$, consisting of
$x_{i},y_{i},g_{j}$, where $i\in\\{1,2,\ldots,r\\}$ and
$j\in\\{1,2,\ldots,s\\}$ and field elements $a_{i},b_{i},c_{j}\in k$ such that

$\operatorname{q}(w)=\displaystyle\sum_{i=1}^{r}(a_{i}e_{i}^{2}+e_{i}f_{i}+b_{i}f_{i}^{2})+\sum_{j=1}^{s}c_{j}z_{j}^{2}$

when $w=\sum_{i=1}^{r}(e_{i}x_{i}+f_{i}y_{i})+\sum_{j=1}^{s}z_{j}g_{j}$. We
denote this quadratic form by

$\operatorname{q}=[a_{1},b_{1}]\perp[a_{2},b_{2}]\perp\cdots\perp[a_{r},b_{r}]\perp\langle
c_{1},c_{2},\ldots,c_{s}\rangle$ (1)

and so
$\operatorname{rad}(W)=\operatorname{span}\\{g_{1},g_{2},\ldots,g_{s}\\}$. The
representation of $\operatorname{q}$ given in 1 is called the _quadratic
signature_ of $\operatorname{q}$. We say that such a quadratic form is of type
$(r,s)$. A nonzero vector $w\in W$ is an _isotropic vector_ if
$\operatorname{q}(w)=0$, $V$ is an _isotropic vector space_ if it contains
isotropic elements and _anisotropic_ otherwise. The space $W$ is called
_nonsingular_ if $s=0$, and _totally singular_ if $r=0$. Furthermore, $W$ is
called _nondefective_ if $s=0$ or $\operatorname{rad}(W)$ is anisotropic. A
_hyperbolic plane_ has a quadratic form isometric to the form $[0,0]$ and will
be denoted by $\mathbb{H}$. We will call $\operatorname{q}^{\prime}$ a
_subform_ of $\operatorname{q}$ if there exists a form $\mathrm{p}$ such that
$\operatorname{q}\cong\operatorname{q}^{\prime}\perp\mathrm{p}$. In this case,
$p$ is called a completion of $\operatorname{q}^{\prime}$ with respect to
$\operatorname{q}$.

Sometimes we will need to refer to a diagonal bilinear form

$\mathrm{B}(w,w^{\prime})=a_{1}w_{1}w^{\prime}_{1}+a_{2}w_{2}w^{\prime}_{2}+\cdots+a_{l}w_{l}w^{\prime}_{l},$

which we denote $\langle a_{1},a_{2},\cdots,a_{l}\rangle_{\mathrm{B}}$
following [17]. Throughout the paper we will use the calligraphy math font to
denote sets of vectors in $W$ such as
$\mathcal{U}=\\{u_{1},u_{2},\ldots,u_{l}\\}$ and
$\overline{\mathcal{U}}=\operatorname{span}\mathcal{U}$. The next result is
Proposition 2.4 from [17].

######  Proposition 2.1.

Let $\operatorname{q}$ be a quadratic form over $k$. Then

$\operatorname{q}\cong
m\times\mathbb{H}\perp\widetilde{\operatorname{q}_{r}}\perp\widetilde{\operatorname{q}_{s}}\perp
d\times\langle 0\rangle,$

with $\widetilde{q_{r}}$ nonsingular, $\widetilde{\operatorname{q}_{s}}$
totally singular and $\widetilde{\operatorname{q}_{r}}\perp\widetilde{q_{s}}$
anisotropic. The form
$\widetilde{\operatorname{q}_{r}}\perp\widetilde{\operatorname{q}_{s}}$ is
uniquely determined up to isometry. In particular $m$ and $d$ are uniquely
determined.

We call $m$ the _Witt index_ and $d$ the _defect_ of $\operatorname{q}$. If

$\operatorname{q}\cong
m\times\mathbb{H}\perp\widetilde{\operatorname{q}_{r}}\perp d\times\langle
0\rangle\perp\widetilde{\operatorname{q}_{s}},$

with respect to the basis

$\\{x_{1},y_{1},\ldots x_{m},y_{m},\ldots,x_{r},y_{r},g_{1},\ldots
g_{d},g_{d+1},\ldots,g_{s}\\},$

we will call

$\mathrm{def}(W)=\mathrm{span}\\{g_{1},\ldots g_{d}\\},$

the _defect_ of $W$. If $\mathcal{U}$ is a basis for a subspace
$\overline{\mathcal{U}}$ of $W$, we will refer to the restriction of
$\operatorname{q}$ to $\overline{\mathcal{U}}$ with respect to this basis by
$\operatorname{q}_{\mathcal{U}}$.

Let $G$ be an algebraic group, then an automorphism $\theta:G\rightarrow G$ is
an _involution_ if $\theta^{2}=\operatorname{id}$,
$\theta\neq\operatorname{id}$. The automorphism $\theta$ is a $k$-_involution_
if in addition $\theta(G(k))=G(k)$, where $G(k)$ denotes the $k$-rational
points of $G$. We define the fixed point group of $\theta$ in $G(k)$ by

$G(k)^{\theta}=\\{\gamma\in G(k)\ |\ \theta\gamma\theta^{-1}=\gamma\\},$

this is often denoted $H(k)$ or $H_{k}$ in the literature when there is no
ambiguity with respect to $\theta$. Let $\mathcal{I}_{\tau}$ denote
conjugation by $\tau$.

We often consider groups that leave a bilinear form or a quadratic form
invariant. If $\mathrm{B}$ is a bilinear form on a nonsingular vector space
$W$ we will denote the symplectic group of $(W,\operatorname{q})$ over a field
$k$ by

$\operatorname{Sp}(\mathrm{B},k)=\\{\varphi\in\operatorname{GL}(W)\ |\
\mathrm{B}(\varphi(w),\varphi(w^{\prime}))=\mathrm{B}(w,w^{\prime})\text{ for
}w,w^{\prime}\in W\\}.$

The classification of involutions for $\operatorname{Sp}(\mathrm{B},k)$ for a
field $k$ such that $\mathrm{char}(k)\neq 2$ has been studied in [3]. For any
quadratic space $W$ over a field $k$ we will denote the orthogonal group of
$(W,\operatorname{q})$ by

$\operatorname{O}(\operatorname{q},k)=\\{\varphi\in\operatorname{GL}(W)\ |\
\operatorname{q}(\varphi(w))=\operatorname{q}(w)\text{ for all }w\in W\\}.$

When $W$ is nonsingluar we have
$\operatorname{O}(\operatorname{q},k)\subset\operatorname{Sp}(\mathrm{B},k)$
if $\mathrm{B}$ is the bilinear form associated with $\operatorname{q}$.

We will need to make use of some facts about quadratic spaces stated in the
following lemmas. The first statement outlines some standard isometries for
quadratic forms over a field of characteristic $2$, and the second allows us
to express $\operatorname{q}$ using a different completion of the nonsingular
space. These results and more like them appear in [17].

######  Lemma 2.2.

Let $\operatorname{q}$ be a quadratic form on a vector space $W$ and suppose
$\alpha\in k$. Then the following are equivalent representations of $q$ on
$W$:

  1. 1.

$[a,a^{\prime}]=[a,a+a^{\prime}+1]=[a^{\prime},a]=[\alpha^{2}a,\alpha^{-2}a^{\prime}]$

  2. 2.

$[a,a^{\prime}]\perp[b,b^{\prime}]=[a+b,a^{\prime}]\perp[b,a^{\prime}+b^{\prime}]=[b,b^{\prime}]\perp[a,a^{\prime}]$

######  Lemma 2.3.

Let $a_{i},b_{i},c_{i}\in k$ for $1\leq i\leq n$. Suppose
$\\{a_{1},...,a_{n}\\}$ and $\\{b_{1},...,b_{n}\\}$ span the same vector space
over $k^{2}$ and
$\operatorname{q}=[a_{1},c_{1}]\perp\ldots\perp[a_{n},c_{n}]$. Then there
exist $c_{i}^{\prime}\in k$, $1\leq i\leq n$, such that
$\operatorname{q}=[b_{1},c_{1}^{\prime}]\perp\ldots\perp[b_{n},c_{n}^{\prime}]$.

For a map $\theta\in\operatorname{GL}(W)$ we define the _residual space_ of
$\theta$ to be the image of $\theta+\operatorname{id}$ and the _residue_ of
$\theta$ to be the dimension of the residual space.

##  3 Summary of involution results

Here we summarize the results from [18] that we need to talk about the
classification of the fixed points groups of each
$\operatorname{O}(\operatorname{q},k)$-conjugacy class of an involution.

If $W$ is a nonsingular vector space with respect to $\operatorname{q}$ we can
define a _transvection_ on $W$ induced by $w\in W$ and $a\in k$

$\tau_{w,a}(z)=z+a\mathrm{B}(w,z)w,$

and an _orthogonal transvection_ is of the form
$\tau_{w,\frac{1}{\operatorname{q}(w)}}$ for $w\in W$ with
$\operatorname{q}(w)\neq 0$. In this case we denote the transvection by
$\tau_{w}$.

If $\mathcal{U}$ is the set of mutually orthogonal vectors inducing a product
of orthogonal transvections $\tau$ we call $\mathcal{U}$ the _inducing set_
for

$\tau=\tau_{\mathcal{U}}=\tau_{u_{l}}\cdots\tau_{u_{2}}\tau_{u_{1}}.$

Let $\mathcal{U}=\\{u_{1},\ldots,u_{l}\\}$ and
$\mathcal{X}=\\{x_{1},\ldots,x_{l}\\}$ be sets of mutually orthogonal vectors
in a nonsingular subspace of $W$. In what follows we will factor involutions
into a product of involutions of a given type (orthogonal transvection, basic
null, basic radical), such a factorization is called _reduced_ if it uses the
minimum number of factors. We refer to the number of factors in a reduced
factorization as the _length_ of an involution.

######  Lemma 3.1.

Two orthogonal involutions given by reduced products of orthogonal
transvections are equal,
$\tau_{u_{l}}\cdots\tau_{u_{2}}\tau_{u_{1}}=\tau_{x_{l}}\cdots\tau_{x_{2}}\tau_{x_{1}}$,
if and only if $\overline{\mathcal{U}}=\overline{\mathcal{X}}$. and

$\left\langle\operatorname{q}(u_{1}),\operatorname{q}(u_{2}),\ldots,\operatorname{q}(u_{l})\right\rangle_{\mathrm{B}}\cong\left\langle\operatorname{q}(x_{1}),\operatorname{q}(x_{2}),\ldots,\operatorname{q}(x_{l})\right\rangle_{\mathrm{B}}.$

The classification of $\operatorname{O}(\operatorname{q},k)$-conjugacy classes
of involutions given by a product of transvections in a nonsingular space can
then be described.

######  Theorem 3.2.

Let $\tau_{u_{l}}\cdots\tau_{u_{2}}\tau_{u_{1}}$ and
$\tau_{x_{l}}\cdots\tau_{x_{2}}\tau_{x_{1}}$ be orthogonal involution written
as a product of transvections on $W$ such that
$\phi\in\operatorname{O}(\operatorname{q},k)$. Then

$\phi\tau_{u_{l}}\cdots\tau_{u_{2}}\tau_{u_{1}}\phi^{-1}=\tau_{x_{l}}\cdots\tau_{x_{2}}\tau_{x_{1}}$

if and only if

$\left\langle\operatorname{q}(u_{1}),\operatorname{q}(u_{2}),\ldots,\operatorname{q}(u_{l})\right\rangle_{\mathrm{B}}\cong\left\langle\operatorname{q}(x_{1}),\operatorname{q}(x_{2}),\ldots,\operatorname{q}(x_{l})\right\rangle_{\mathrm{B}},$

and $\dim(\overline{\mathcal{U}})=\dim(\overline{\mathcal{X}})$.

The last stipulation,
$\dim(\overline{\mathcal{U}})=\dim(\overline{\mathcal{X}})$, distinguishes
between elements $\tau$ whose reduced length is $\mathrm{res}(\tau)+1$, called
hyperbolic involutions, and those whose reduced length is
$\mathrm{res}(\tau)$, called diagonal.

There is one other type of nonsingular involution called a null involution.

######  Definition 3.3.

A plane $P=\operatorname{span}\\{x,y\\}$ is hyperbolic (or Artinian) if both
of the following are satisfied:

  1. 1.

$\operatorname{q}(x)=\operatorname{q}(y)=0$

  2. 2.

$B(x,y)\neq 0.$

If $x,y$ span a hyperbolic plane, we can rescale to assume $B(x,y)=1$.
Proposition 188.2 of [26] guarantees that every nonsingular nonzero isotropic
vector is contained in a hyperbolic plane.

######  Definition 3.4.

Let $\eta$ be an involution of $\operatorname{O}(\operatorname{q},k)$ where
$(W,\operatorname{q})$ is a quadratic space, and let $\mathbb{P}$ be the
orthogonal sum of two hyperbolic planes. Then an involution $\eta$ is called a
basic null involution in $\mathbb{P}$ if all of the following are satisfied:

  1. 1.

$\eta$ leaves $\mathbb{P}$ invariant

  2. 2.

$\eta$ fixes a 2-dimensional subspace of $\mathbb{P}$ where every vector has
norm zero

  3. 3.

$\eta|_{\mathbb{P}^{C}}=\operatorname{id}_{\mathbb{P}^{C}}$, where
$\mathbb{P}^{C}$ is the complement of $\mathbb{P}$ in $W$.

A _null involution_ is a product of basic null involutions. Note that if
$\eta$ is a null involution and $\tau$ is a transvection, then $\tau\eta$ is a
product of transvections [26], so we only need to consider the conjugacy of
null involutions themselves.

######  Proposition 3.5.

Two null involutions are $\operatorname{O}(\operatorname{q},k)$-conjugate if
and only if they have the same length.

Finally we consider involutions acting only on the radical of $W$.

######  Proposition 3.6.

$\operatorname{O}(\operatorname{q}_{\operatorname{rad}(W)},k)\cong\operatorname{GL}_{d}(k)\ltimes\operatorname{Mat}_{d,s-d}(k)$
where $d$ is the defect of $\operatorname{rad}(W)$.

###### Proof.

Let $d$ be the defect of $W$. We can choose a basis of the form

$\\{g_{1},\ldots,g_{d},g_{d+1},\ldots,g_{s}\\}$

where $\operatorname{q}(g_{i})=0$ for $1\leq i\leq d$

$\mathcal{G}_{\mathrm{is}}=\\{g_{1},\ldots,g_{d}\\},$

and the subspace spanned by

$\mathcal{G}_{\mathrm{an}}=\\{g_{d+1},\ldots,g_{s}\\}$

is anisotropic. We claim we can represent an isometry on $\mathrm{rad}(W)$ by

$\Psi=\begin{bmatrix}\psi&M\\\ 0&\operatorname{id}\end{bmatrix}.$

We must have $\Psi|_{\mathcal{G}_{\mathrm{an}}}=\operatorname{id}$ otherwise
there would exist two elements in $\mathcal{G}_{\mathrm{an}}$ with the same
norm, and since the space is totally singular the sum of the two vectors would
have norm zero. The restriction
$\Psi|_{\mathcal{G}_{\mathrm{is}}}=\psi\in\operatorname{GL}_{d}(k)$ since
$\psi$ must be invertible but has no other contraints due to the bilinear form
and the norm being identically zero for
$\overline{\mathcal{G}_{\mathrm{is}}}$. The isometric image of an element from
$\overline{\mathcal{G}_{\mathrm{is}}}$ cannot include any components from
$\overline{\mathcal{G}_{\mathrm{an}}}$ since that would imply that
$\overline{\mathcal{G}_{\mathrm{an}}}$ was not anisotropic. Lastly the $M$
block is free since we can add any isotropic vector from the the radical to
any vector in $W$ and leave the quadratic form (and bilinear form) invariant.

We can factor

$\Psi=\begin{bmatrix}\psi&M\\\
0&\operatorname{id}\end{bmatrix}=\begin{bmatrix}\psi&0\\\
0&\operatorname{id}\end{bmatrix}\begin{bmatrix}\operatorname{id}&M\\\
0&\operatorname{id}\end{bmatrix}$

where $\bigl{[}\begin{smallmatrix}\psi&M\\\
0&\operatorname{id}\end{smallmatrix}\bigr{]}\bigl{[}\begin{smallmatrix}\operatorname{id}&B\\\
0&\operatorname{id}\end{smallmatrix}\bigr{]}\bigl{[}\begin{smallmatrix}\psi&M\\\
0&\operatorname{id}\end{smallmatrix}\bigr{]}^{-1}=\bigl{[}\begin{smallmatrix}\operatorname{id}&\psi
B\\\ 0&\operatorname{id}\end{smallmatrix}\bigr{]}$. ∎

A _radical involution_ is an element
$\rho\in\operatorname{O}(\operatorname{q},k)$ or order $2$ defined on vectors
in the radical of $W$. Notice that a nontrivial orthogonal transformation on
$\operatorname{rad}(W)$ detects a defective vector in $W$, since if
$\rho(g)=g^{\prime}$ then
$\operatorname{q}(g+g^{\prime})=\operatorname{q}(g)+\operatorname{q}(g^{\prime})=0$.
A _basic radical involution_ is a radical involution
$\rho_{i}\in\operatorname{O}(\operatorname{q},k)$ such that
$\rho_{i}(g_{i})=g_{i}^{\prime}$, where $g_{i},g_{i}^{\prime}$ are linearly
independent vectors in $\operatorname{rad}(W)$ with
$\operatorname{q}(g_{i})=\operatorname{q}(g_{i}^{\prime})$, and $\rho_{i}$
acts trivially on the complement of $\overline{\\{g_{i},g_{i}^{\prime}\\}}$.
All radical involutions can be written as a finite product of basic radical
involutions.

######  Proposition 3.7.

Two basic radical involutions $\rho_{1},\rho_{2}$ are
$\operatorname{O}(\operatorname{q},k)$-conjugate if and only if $\rho_{1}$ and
$\rho_{2}$ act non-trivially on isometric vectors.

######  Corollary 3.8.

All radical involutions of length $m$ with quadratic signature isometric to

$\langle\operatorname{q}(g_{1}),\operatorname{q}(g_{2}),\ldots,\operatorname{q}(g_{m})\rangle,$

are $\operatorname{O}(\operatorname{q},k)$-conjugate.

##  4 Totally singular spaces

We are now ready to discuss the isomorphism classes of the fixed point groups
of our involutions of $\operatorname{O}(\operatorname{q},k)$, which can be
used to study the corresponding symmetric $k$-varieties. First, we discuss the
fixed point groups of involutions acting on a totally singular vector space.
This in turn leads to a discussion about fixed point groups of diagonal
involutions on nonsingular subspaces.

Let us consider a vector space $W$ such that $W=\operatorname{rad}(W)$. Recall
that in such a vector space any quadratic form will be of the form

$\operatorname{q}=\langle\operatorname{q}(g_{1}),\operatorname{q}(g_{2}),\ldots,\operatorname{q}(g_{s})\rangle,$

and by Proposition 2.1 we have an isometry

$\operatorname{q}\cong\langle
0,\ldots,0,\operatorname{q}(g_{1}),\operatorname{q}(g_{2}),\ldots,\operatorname{q}(g_{s-d})\rangle,$

where $d$ is the defect of $\operatorname{q}$.

By Proposition 3.6 we have
$\operatorname{O}(\operatorname{q}_{\operatorname{rad}(W)},k)\cong\operatorname{GL}_{d}(k)\ltimes\operatorname{Mat}_{d,s-d}(k)$.
For an element $\delta\in\operatorname{O}(\operatorname{q},k)$ to be fixed by
conjugation with the involution $\rho$ we need $\rho\delta\rho=\delta$.

######  Proposition 4.1.

If $\operatorname{q}$ is a quadratic form on $W=\mathrm{rad}(W)$ and
$\rho\in\operatorname{O}(\operatorname{q},k)$ is an involution on $W$ such
that $\rho(g_{i})=g_{i}^{\prime}$ and quadratic signature

$\langle\operatorname{q}(g_{1}),\operatorname{q}(g_{2}),\cdots,\operatorname{q}(g_{n})\rangle$

then
$\operatorname{O}(\operatorname{q},k)^{\mathcal{I}_{\rho}}\cong\operatorname{O}(\operatorname{q}_{\mathcal{G}\cup\mathcal{H}},k)\ltimes\operatorname{Mat}_{n,s-n}(k)$
where

$\displaystyle\mathcal{G}^{\prime}$
$\displaystyle=\\{g_{1}^{\prime},\ldots,g_{n}^{\prime}\\}$
$\displaystyle\mathcal{G}$ $\displaystyle=\\{g_{1},\ldots,g_{n}\\}$
$\displaystyle\mathcal{H}$ $\displaystyle=\\{h_{1},\ldots,h_{s-2n}\\}$

with $\rho(h_{i})=h_{i}$ and
$\overline{\mathcal{G}^{\prime}\cup\mathcal{G}\cup\mathcal{H}}=W$.

###### Proof.

Let
$\langle\operatorname{q}(g_{1}),\operatorname{q}(g_{2}),\ldots,\operatorname{q}(g_{n})\rangle$
be the signature of a radical involution $\rho$ such that
$\rho(g_{i})=g_{i}^{\prime}$ and $g_{i}^{\prime}\neq g_{i}$ for any $i$ or
$i^{\prime}$. Notice that $\rho(g_{i}+g_{i}^{\prime})=g_{i}+g_{i}^{\prime}$
and further that $\rho(g_{i})=g_{i}+(g_{i}+g_{i}^{\prime})$. Now we order a
basis for $W$,

$\mathcal{W}=\\{g_{1}+g_{1}^{\prime},g_{2}+g_{2}^{\prime},\ldots,g_{n}+g_{n}^{\prime},h_{1},h_{2},\ldots,h_{s-2n},g_{1},g_{2},\ldots,g_{n}\\}$

such that $\overline{\mathcal{W}}=W$ and $\rho(h_{i})=h_{i}$. With respect to
this basis we can represent $\rho$ as the block matrix

$\rho=\begin{bmatrix}\operatorname{id}&0&\operatorname{id}\\\
0&\operatorname{id}&0\\\ 0&0&\operatorname{id}\end{bmatrix},$

and so any
$\delta\in\operatorname{O}(\operatorname{q},k)^{\mathcal{I}_{\rho}}$

$\delta=\begin{bmatrix}\delta_{11}&\delta_{12}&\delta_{13}\\\
\delta_{21}&\delta_{22}&\delta_{23}\\\ \delta_{31}&\delta_{32}&\delta_{33}\\\
\end{bmatrix},$

such that $\delta\rho=\rho\delta$ is of the form

$\delta=\begin{bmatrix}\delta_{11}&\delta_{12}&\delta_{13}\\\
0&\delta_{22}&\delta_{23}\\\ 0&0&\delta_{11}\\\
\end{bmatrix}=\begin{bmatrix}\delta_{11}&0&0\\\ 0&\delta_{22}&\delta_{23}\\\
0&0&\delta_{11}\\\
\end{bmatrix}\begin{bmatrix}\operatorname{id}&\delta_{12}^{\prime}&\delta_{13}^{\prime}\\\
0&\operatorname{id}&0\\\ 0&0&\operatorname{id}\\\ \end{bmatrix}$

where $\delta_{ij}^{\prime}=\delta_{11}^{-1}\delta_{ij}$. Notice that
$\delta_{12}^{\prime}$ and $\delta_{13}^{\prime}$ can take any matrix value,
since $\overline{\\{g_{1}+g_{1}^{\prime},\ldots,g_{n}+g_{n}^{\prime}\\}}$ is
contained in the defect of $W$ and the span of the remaining basis vectors
would be

$\mathcal{H}\cup\mathcal{G}=\\{h_{1},\dots,h_{s-2n},g_{1},\ldots,g_{n}\\},$

Now the subgroup

$\mathsf{M}=\left\\{\begin{bmatrix}\delta_{11}&0&0\\\
0&\delta_{22}&\delta_{23}\\\ 0&0&\delta_{11}\\\
\end{bmatrix}\bigg{|}\bigl{[}\begin{smallmatrix}\delta_{22}&\delta_{23}\\\
0&\delta_{11}\end{smallmatrix}\bigr{]}\in\operatorname{O}(\operatorname{q}_{\mathcal{H}\cup\mathcal{G}},k,\overline{\mathcal{H}})\right\\},$

is such that
$\mathsf{M}\cong\operatorname{O}(\operatorname{q}_{\mathcal{H}\cup\mathcal{G}},k,\overline{\mathcal{H}})$
which is the subgroup of
$\operatorname{O}(\operatorname{q}_{\mathcal{H}\cup\mathcal{G}},k)$ that
leaves the subspace $\overline{\mathcal{H}}$ invariant. The subgroup

$\mathsf{M}^{\prime}=\left\\{\begin{bmatrix}\operatorname{id}&\delta_{12}^{\prime}&\delta_{13}^{\prime}\\\
0&\operatorname{id}&0\\\ 0&0&\operatorname{id}\\\
\end{bmatrix}\bigg{|}\delta_{12}^{\prime}\in\operatorname{Mat}_{n,d^{\prime}}(k),\delta_{13}^{\prime}\in\operatorname{Mat}_{n,s-n-d^{\prime}}(k)\right\\},$

is such that $\mathsf{M}^{\prime}\cong\operatorname{Mat}_{n,s-n}(k)$ and
$\mathsf{M}$ is a normal subgroup of
$\operatorname{O}(\operatorname{q},k)^{\mathcal{I}_{\rho}}$, so we have that
$\operatorname{O}(\operatorname{q},k)^{\mathcal{I}_{\rho}}\cong\operatorname{O}(\operatorname{q}_{\mathcal{H}\cup\mathcal{G}},k,\overline{\mathcal{H}})\ltimes\operatorname{Mat}_{n,s-n}(k)$.
∎

######  Remark 4.2.

Notice that if $W=\mathrm{rad}(W)$ is anisotropic then
$\operatorname{O}(\operatorname{q},k)=\\{id\\}$ so there are no involutions,
and on the other hand if $W$ is totally isotropic we have that
$\operatorname{O}(\operatorname{q},k)\cong\operatorname{GL}_{s}(k)$.

##  5 Nonsingular spaces

Now we look at fixed point groups of diagonal involutions on nonsingular
spaces.

######  Proposition 5.1.

Let $\tau=\tau_{u_{l}}\cdots\tau_{u_{2}}\tau_{u_{1}}$ be a reduced diagonal
orthogonal involution with inducing set $\mathcal{U}$, then
$\varphi\in\operatorname{O}(\operatorname{q},k)^{\mathcal{I}_{\tau}}$ if and
only if $\varphi(\overline{\mathcal{U}})=\overline{\mathcal{U}}$ .

###### Proof.

Let $\varphi\in\operatorname{O}(\operatorname{q},k)^{\mathcal{I}_{\tau}}$
notice that $\tau\varphi\tau^{-1}=\varphi$ if and only if
$\varphi\tau\varphi^{-1}=\tau$ and so

$\varphi\tau_{u_{l}}\cdots\tau_{u_{2}}\tau_{u_{1}}\varphi^{-1}=\tau_{\varphi(u_{l})}\cdots\tau_{\varphi(u_{2})}\tau_{\varphi(u_{1})}=\tau_{u_{l}}\cdots\tau_{u_{2}}\tau_{u_{1}}.$

By Lemma 3.1 $\varphi(\overline{\mathcal{U}})=\overline{\mathcal{U}}$.

Now assume $\varphi\in\operatorname{O}(\operatorname{q},k)$ and
$\varphi(\overline{\mathcal{U}})=\overline{\mathcal{U}}$ then

$\langle\operatorname{q}(\varphi(u_{1})),\operatorname{q}(\varphi(u_{2})),\ldots,\operatorname{q}(\varphi(u_{l}))\rangle_{B}=\langle\operatorname{q}(u_{1}),\operatorname{q}(u_{2}),\ldots,\operatorname{q}(u_{l})\rangle_{B}.$

So by Theorem 3.2 we have $\varphi\tau\varphi^{-1}=\tau$. ∎

We begin with the case that the length of $\tau$ is maximized in the
nonsingular space. We denote by $\varphi_{\mathcal{U}}$ the matrix
representation of the resriction of $\varphi$ to the subspace
$\overline{\mathcal{U}}$ with respect to the basis $\mathcal{U}$. Recall that

$\operatorname{q}_{\mathcal{U}}\cong\langle\operatorname{q}(u_{1}),\operatorname{q}(u_{2}),\ldots,\operatorname{q}(u_{l})\rangle,$

since $\overline{\mathcal{U}}$ is a totally singular subspace of $W$. In this
case, Proposition 2.1 yields

$\operatorname{q}_{\mathcal{U}}\cong m^{\prime}\times\langle
0\rangle\perp\langle c_{m^{\prime}+1},\ldots,c_{l}\rangle.$

This corresponds to a basis of
$\mathcal{U}^{\prime}=\\{u_{1}^{\prime},u_{2}^{\prime},\ldots,u_{l}^{\prime}\\}$
such that $\operatorname{q}(u_{i}^{\prime})=c_{i}$ for $1\leq i\leq l$ and
$c_{i}=0$ for $1\leq i\leq m^{\prime}$. Since the vectors in
$\overline{\mathcal{U}}$ are not in $\operatorname{rad}(W)$ there exist non-
singular completions within $W$ that form a basis for a nonsingular subspace.
A nonsingular completion of $\operatorname{q}$ is of the form

$[0,\operatorname{q}(v_{1})]\perp[0,\operatorname{q}(v_{2})]\perp\ldots\perp[0,\operatorname{q}(v_{m^{\prime}})]\perp[c_{m^{\prime}+1},\operatorname{q}(v_{m^{\prime}+1})]\perp\ldots\perp[c_{l},\operatorname{q}(v_{l})],$

where $\mathrm{B}(u_{i}^{\prime},v_{i})=1$,
$\mathrm{B}(u_{i}^{\prime},v_{j})=0$ when $j\neq i$, and
$\mathcal{V}=\\{v_{1},\ldots,v_{l}\\}$ which spans a totally singular
subspace. Notice that for $1\leq i\leq m^{\prime}$ we can choose
$v_{i}^{\prime}=v_{i}+\operatorname{q}(v_{i})u_{i}^{\prime}$. We will refer to
the vectors of this completion as $\mathcal{V}^{\prime}$. Now

$\mathrm{B}(u_{i}^{\prime},v_{i}^{\prime})=\mathrm{B}(u_{i}^{\prime},v_{i})+\operatorname{q}(v_{i})\mathrm{B}(u_{i}^{\prime},u_{i}^{\prime})=\mathrm{B}(u_{i}^{\prime},v_{i})=1,$

and

$\operatorname{q}(v_{i}^{\prime})=\operatorname{q}(v_{i})+\operatorname{q}(v_{i})^{2}\operatorname{q}(u_{i}^{\prime})+\operatorname{q}(v_{i})B(v_{i},u_{i}^{\prime})=\operatorname{q}(v_{i})+\operatorname{q}(v_{i})=0.$

So there is a basis $\mathcal{W}$ of this nonsingular space of the form

$\mathcal{W}^{\prime}=\mathcal{U}^{\prime}\cup\mathcal{V}^{\prime}=\\{u_{1}^{\prime},v_{1}^{\prime},u_{2}^{\prime},v_{2}^{\prime},\ldots,u_{l}^{\prime},v_{l}^{\prime}\\},$

with the corresponding quadratic form

$\operatorname{q}_{\mathcal{W}}\cong
m^{\prime}\times[0,0]\perp[c_{m^{\prime}+1},d_{m^{\prime}+1}]\perp\ldots\perp[c_{l},d_{l}],$

where $\operatorname{q}(v_{i}^{\prime})=d_{i}$ for $1\leq i\leq l$.

Let us consider
$\varphi\in\operatorname{O}(\operatorname{q},k)^{\mathcal{I}_{\tau}}$. Since
$\varphi$ leaves $\overline{\mathcal{U}}$ invariant we have

$\varphi=\begin{bmatrix}\varphi_{\mathcal{U}}&\varphi_{\mathcal{U}}A\\\
0&\varphi_{\mathcal{V}}\end{bmatrix},$

where $\varphi_{\mathcal{V}}:\overline{\mathcal{V}}\to\overline{\mathcal{V}}$
and $A:\overline{\mathcal{V}}\to\overline{\mathcal{U}}$. Considering the
bilinear form we have

$\mathrm{B}(\varphi(u_{i}),\varphi(v_{j}))=\mathrm{B}(\varphi_{\mathcal{U}}(u_{i}),\varphi_{\mathcal{V}}(v_{j})+\varphi_{\mathcal{U}}(Av_{j}))=\mathrm{B}(\varphi_{\mathcal{U}}(u_{i}),\varphi_{\mathcal{V}}(v_{j}))=\mathrm{B}(u_{i},v_{j}).$

So, we have shown the following

######  Proposition 5.2.

The map $\varphi_{\mathcal{V}}$ is completely determined by
$\varphi_{\mathcal{U}}$.

We will now refer to $\varphi_{\mathcal{V}}=\varphi_{\mathcal{U}}^{*}$ and
using the matrix representation of $\varphi_{\mathcal{U}}$ we have
$\varphi_{\mathcal{U}}^{*}=(\varphi_{\mathcal{U}}^{-1})^{T}$. We can represent
$\varphi\in\operatorname{O}(\operatorname{q},k)^{\mathcal{I}_{\tau}}$ with the
block matrix

$\varphi=\begin{bmatrix}\varphi_{\mathcal{U}}&\varphi_{\mathcal{U}}A\\\
0&\varphi_{\mathcal{U}}^{*}\end{bmatrix}.$

We define the additive group

$\mathsf{A}(\operatorname{q}_{\mathcal{U}},k)=\\{A\in\text{Lin}(\overline{\mathcal{V}},\overline{\mathcal{U}})\
|\ \operatorname{q}(Av)=\mathrm{B}(v,Av)\\}.$

Notice that $\mathsf{A}(\operatorname{q}_{\mathcal{U}},k)$ embeds into
$\operatorname{O}(\operatorname{q},k)^{\mathcal{I}_{\tau}}$ as an additive
group since

$\operatorname{q}((A+C)v)=\operatorname{q}(Av)+\operatorname{q}(Cv)=\mathrm{B}(v,Av)+\mathrm{B}(v,Cv)=\mathrm{B}(v,(A+C)v),$

for $A,C\in\mathsf{A}(\operatorname{q}_{\mathcal{U}},k)$. We will write
$\varphi^{-*}$ for $(\varphi^{-1})^{*}$. The following results give us some
insight into the structure of the fixed point groups.

######  Lemma 5.3.

If $u\in\overline{\mathcal{U}}$ and $v\in\overline{\mathcal{V}}$, then
$\mathrm{B}(\varphi_{\mathcal{U}}(u),v)=\mathrm{B}(u,\varphi_{\mathcal{U}}^{-*}(v))$.

###### Proof.

$\displaystyle\mathrm{B}(u,\varphi_{\mathcal{U}}^{-*}(v))$
$\displaystyle=\mathrm{B}(\varphi(u),\varphi\varphi_{\mathcal{U}}^{-*}(v))$
$\displaystyle=\mathrm{B}(\varphi_{\mathcal{U}}(u),(\varphi_{\mathcal{U}}^{*}+\varphi_{\mathcal{U}}A)\varphi_{\mathcal{U}}^{-*}(v))$
$\displaystyle=\mathrm{B}(\varphi_{\mathcal{U}}(u),v+\varphi_{\mathcal{U}}A\varphi_{\mathcal{U}}^{-*}(v))$
$\displaystyle=\mathrm{B}(\varphi_{\mathcal{U}}(u),v),$

since $\varphi_{\mathcal{U}}A\varphi_{\mathcal{U}}^{-*}(v)\in\mathcal{U}$. ∎

In the next result we see that this gives us an isomorphism of the additive
group $\mathsf{A}(\operatorname{q}_{\mathcal{U}},k)$.

######  Lemma 5.4.

For
$\varphi_{\mathcal{U}}\in\operatorname{O}(\operatorname{q}_{\mathcal{U}},k)$
and $A\in\mathsf{A}(\operatorname{q}_{\mathcal{U}},k)$ we have that

$\Phi_{\varphi_{\mathcal{U}}}(A)=\varphi_{\mathcal{U}}A\varphi_{\mathcal{U}}^{-*},$

is an isomorphism of $\mathsf{A}(\operatorname{q}_{\mathcal{U}},k)$.

###### Proof.

The map $\Phi_{\varphi_{\mathcal{U}}}(A)$ is a homomorphism with respect to
addition in $\mathsf{A}(\operatorname{q}_{\mathcal{U}},k)$ and

$\displaystyle\operatorname{q}(\varphi_{\mathcal{U}}A\varphi_{\mathcal{U}}^{-*}(v))$
$\displaystyle=\operatorname{q}(A\varphi_{\mathcal{U}}^{-*}(v))$
$\displaystyle=B(\varphi_{\mathcal{U}}^{-*}(v),A\varphi_{\mathcal{U}}^{-*}(v))$
$\displaystyle=B(v,\varphi_{\mathcal{U}}A\varphi_{\mathcal{U}}^{-*}(v)),$

by Lemma 5.3 and the fact that
$\varphi_{\mathcal{U}}\in\operatorname{O}(\operatorname{q}_{\mathcal{U}},k)$,
we have the result. ∎

######  Proposition 5.5.

For
$\varphi=\bigl{[}\begin{smallmatrix}\varphi_{\mathcal{U}}&\varphi_{\mathcal{U}}A\\\
0&\varphi_{\mathcal{U}}^{*}\end{smallmatrix}\bigr{]}$ we have
$\varphi\in\operatorname{O}(\operatorname{q},k)^{\mathcal{I}_{\tau}}$ if and
only if $\varphi_{\mathcal{U}}\in\operatorname{O}(\operatorname{q},k)$ and
$\operatorname{q}(v+Av)=\operatorname{q}(\varphi_{\mathcal{U}}^{*}(v))$ for
$v\in\overline{\mathcal{V}}$ a nonsingular completion of
$\overline{\mathcal{U}}$ in $W$.

###### Proof.

This can be shown with the following computation. Let us compute

$\displaystyle\operatorname{q}(\varphi(v))$
$\displaystyle=\operatorname{q}(\varphi_{\mathcal{U}}^{*}(v)+\varphi_{\mathcal{U}}(Av))$
$\displaystyle\operatorname{q}(v)$
$\displaystyle=\operatorname{q}(\varphi_{\mathcal{U}}^{*}(v))+\operatorname{q}(\varphi_{\mathcal{U}}(Av))+\mathrm{B}(\varphi_{\mathcal{U}}^{*}(v),\varphi_{\mathcal{U}}(Av))$
$\displaystyle\operatorname{q}(v)$
$\displaystyle=\operatorname{q}(\varphi_{\mathcal{U}}^{*}(v))+\operatorname{q}(Av)+\mathrm{B}(v,Av)$
$\displaystyle\operatorname{q}(v)+\operatorname{q}(Av)+\mathrm{B}(v,Av)$
$\displaystyle=\operatorname{q}(\varphi_{\mathcal{U}}^{*}(v))$
$\displaystyle\operatorname{q}(v+Av)$
$\displaystyle=\operatorname{q}(\varphi_{\mathcal{U}}^{*}(v)).$

∎

######  Remark 5.6.

We note here that it is not necessarily the case that for every
$\varphi_{\mathcal{U}}\in\operatorname{O}(\operatorname{q}_{\mathcal{U}},k)$
that there exists an element of the form
$\bigl{[}\begin{smallmatrix}\varphi_{\mathcal{U}}&0\\\
0&\varphi_{\mathcal{U}}^{*}\end{smallmatrix}\bigr{]}\in\operatorname{O}(\operatorname{q},k)^{\mathcal{I}_{\tau}}$.
It is true, however, that for every
$\varphi_{\mathcal{U}}\in\operatorname{O}(\operatorname{q}_{\mathcal{U}},k)$
there exists $A_{\varphi}:\overline{\mathcal{V}}\to\overline{\mathcal{U}}$
such that
$\bigl{[}\begin{smallmatrix}\varphi_{\mathcal{U}}&\varphi_{\mathcal{U}}A_{\varphi}\\\
0&\varphi_{\mathcal{U}}^{*}\end{smallmatrix}\bigr{]}\in\operatorname{O}(\operatorname{q},k)^{\mathcal{I}_{\tau}}\subset\operatorname{O}(\operatorname{q},k)$,
which we state next.

######  Lemma 5.7.

If $\tau=\tau_{u_{l}}\cdots\tau_{u_{2}}\tau_{u_{1}}$ is a diagonal involution
on $W$ of length $l$, $\dim(W)=2l$, and
$\varphi_{\mathcal{U}}\in\operatorname{O}(\operatorname{q}_{\mathcal{U}},k)$,
then there exists
$\varphi=\bigl{[}\begin{smallmatrix}\varphi_{\mathcal{U}}&\varphi_{\mathcal{U}}A\\\
0&\varphi_{\mathcal{U}}^{*}\end{smallmatrix}\bigr{]}\in\operatorname{O}(\operatorname{q}_{\mathcal{U}},k)^{\mathcal{I}_{\tau}}$.

###### Proof.

We know that elements
$\varphi\in\operatorname{O}(\operatorname{q},k)^{\mathcal{I}_{\tau}}$ leave
$\overline{\mathcal{U}}$ invariant and must be of the form

$\varphi=\begin{bmatrix}\varphi_{\mathcal{U}}&A\\\
0&\varphi_{\mathcal{U}}^{*}\end{bmatrix}.$

We now consider $\varphi_{\mathcal{U^{\prime}}}$ where $\mathcal{U}^{\prime}$
is the basis of $\overline{\mathcal{U}}$ with respect to

$\operatorname{q}^{\prime}=m\times[0,0]\perp[\operatorname{q}(u^{\prime}_{m+1}),\operatorname{q}(v^{\prime}_{m+1})]\perp\ldots\perp[\operatorname{q}(u^{\prime}_{l}),\operatorname{q}(v^{\prime}_{l})]$

and
$[\operatorname{q}(u^{\prime}_{m+1}),\operatorname{q}(v^{\prime}_{m+1})]\perp\ldots\perp[\operatorname{q}(u^{\prime}_{l}),\operatorname{q}(v^{\prime}_{l})]$
is anisotropic, so that the Witt index of $\operatorname{q}$ is $m$. Then we
can write

$\varphi=\begin{bmatrix}\varphi_{\mathcal{U^{\prime}}}&A^{\prime}\\\
0&\varphi_{\mathcal{U^{\prime}}}^{*}\end{bmatrix},$

where

$\varphi_{\mathcal{U^{\prime}}}=\begin{bmatrix}P_{1}&P_{2}\\\
0&\operatorname{id}\end{bmatrix}\text{ which give us
}\varphi_{\mathcal{U^{\prime}}}^{*}=\begin{bmatrix}P_{1}^{*}&0\\\
P_{2}^{T}P_{1}^{*}&\operatorname{id}\end{bmatrix}.$

Notice that we can choose $P_{1}\in\operatorname{GL}_{m}(k)$ and
$P_{2}\in\operatorname{Mat}_{m,l-m}(k)$ since $P_{1}$ is acting on
$\overline{\\{u_{1}^{\prime},\ldots,u_{m}^{\prime}\\}}$ all of the vectors are
orthogonal to one another and all have norm zero. The map $P_{2}$ adds norm
zero vectors to the anisotropic vectors, and again all vectors in the space
$\mathcal{U}^{\prime}$ are mutually orthogonal, so there are no further
constraints.

Let us define
$\mathcal{U^{\prime}}_{\mathbb{H}}=\\{u_{1}^{\prime},\ldots,u_{m}^{\prime}\\}$
and
$\mathcal{V^{\prime}}_{\mathbb{H}}=\\{v_{1}^{\prime},\ldots,v_{m}^{\prime}\\}$,
the defective subspaces of $\mathcal{U}$ and $\mathcal{V}$, where
$\overline{\mathcal{U}_{\mathbb{H}}^{\prime}\cup\mathcal{V}_{\mathbb{H}}^{\prime}}$
forms the hyperbolic subspace of $W=\overline{\mathcal{W}}$.

We need to establish that $A^{\prime}$ exists where

$A^{\prime}=\begin{bmatrix}A_{1}^{\prime}&A_{2}^{\prime}\\\
A_{3}^{\prime}&A_{4}^{\prime}\end{bmatrix}.$

We can choose $A_{2}^{\prime}=A_{3}^{\prime}=A_{4}^{\prime}=0$ and
$A_{1}^{\prime}:\overline{\mathcal{V^{\prime}}_{\mathbb{H}}}\to\overline{\mathcal{V^{\prime}}_{\mathbb{H}}}$
such that
$A_{1}^{\prime}v^{\prime}_{j}=\operatorname{q}(P_{2}^{T}P_{1}^{*}v^{\prime}_{j})P_{1}u^{\prime}_{j}$.
First notice that for $v_{j}^{\prime}$

$\displaystyle\operatorname{q}(\varphi(v_{j}^{\prime}))$
$\displaystyle=\operatorname{q}(A_{1}^{\prime}v_{j}^{\prime}+P_{1}^{*}v_{j}^{\prime}+P_{2}^{T}P_{1}^{*}v_{j}^{\prime})$
$\displaystyle=\operatorname{q}(A_{1}^{\prime}v_{j}^{\prime})+\operatorname{q}(P_{1}^{*}v_{j}^{\prime}+P_{2}^{T}P_{1}^{*}v_{j}^{\prime})+B(A_{1}^{\prime}v_{j}^{\prime},P_{1}^{*}v_{j}^{\prime}+P_{2}^{T}P_{1}^{*}v_{j}^{\prime})$
$\displaystyle=\operatorname{q}(P_{1}^{*}v_{j}^{\prime})+\operatorname{q}(P_{2}^{T}P_{1}^{*}v_{j}^{\prime})+B(A_{1}^{\prime}v_{j}^{\prime},P_{1}^{*}v_{j}^{\prime})$
$\displaystyle=\operatorname{q}(P_{2}^{T}P_{1}^{*}v_{j}^{\prime})+B(A_{1}^{\prime}v_{j}^{\prime},P_{1}^{*}v_{j}^{\prime})$
$\displaystyle=\operatorname{q}(P_{2}^{T}P_{1}^{*}v_{j}^{\prime})+B(\operatorname{q}(P_{2}^{T}P_{1}^{*}v^{\prime}_{j})P_{1}u^{\prime}_{j},P_{1}^{*}v_{j}^{\prime})$
$\displaystyle=0,$

since
$A_{1}^{\prime}v_{j}^{\prime},P_{1}^{*}v_{j}^{\prime}\in\mathcal{V}_{\mathbb{H}}$
and $B(u_{j}^{\prime},v_{j}^{\prime})=1$. We use Proposition 5.5 noting that
$\varphi=\left[\begin{smallmatrix}\varphi_{\mathcal{U}}&A\\\
0&\varphi_{\mathcal{U}}^{*}\end{smallmatrix}\right]=\left[\begin{smallmatrix}\varphi_{\mathcal{U}}&\varphi_{\mathcal{U}}\varphi_{\mathcal{U}}^{-1}A\\\
0&\varphi_{\mathcal{U}}^{*}\end{smallmatrix}\right]$. Now if
$v_{j}^{\prime}\in\mathcal{V^{\prime}}_{\mathbb{H}}$

$\displaystyle\operatorname{q}(v_{j}^{\prime}+\varphi_{\mathcal{U}}^{-1}(Av_{j}^{\prime}))$
$\displaystyle=\operatorname{q}(v_{j}^{\prime})+\operatorname{q}(\varphi_{\mathcal{U}}^{-1}(Av_{j}^{\prime}))+\mathrm{B}(v_{j}^{\prime},\varphi_{\mathcal{U}}^{-1}(Av_{j}^{\prime}))$
$\displaystyle=0+\operatorname{q}(\operatorname{q}(P_{2}P_{1}^{*}v_{j}^{\prime})u_{j}^{\prime})+\mathrm{B}(v_{j}^{\prime},\operatorname{q}(P_{2}P_{1}^{*}v_{j}^{\prime})u_{j}^{\prime})$
$\displaystyle=0+\operatorname{q}(P_{2}P_{1}^{*}v_{j}^{\prime})\mathrm{B}(v_{j}^{\prime},\operatorname{q}(P_{2}P_{1}^{*}v_{j}^{\prime})u_{j}^{\prime})$
$\displaystyle=\operatorname{q}(P_{2}P_{1}^{*}v_{j}^{\prime})$
$\displaystyle=\operatorname{q}(\varphi_{\mathcal{U}}^{*}(v_{j}^{\prime})).$

If $v_{j}$ is a basis vector not in
$\overline{\mathcal{V^{\prime}}_{\mathbb{H}}}$ then then $Av_{j}^{\prime}=0$
and Proposition 5.5 is trivially satisfied. ∎

We define $A_{\varphi}\in\operatorname{Mat}_{l,l}(k)$ as the diagonal matrix
with $\operatorname{q}(\varphi_{\mathcal{U}}^{*}(v_{j}^{\prime}))$ in the
$j$-th diagonal entry $1\leq j\leq m$ and $0$ elsewhere, i.e.
$A_{\varphi}v_{j}=\operatorname{q}(\varphi_{\mathcal{U}}^{*}(v_{j}^{\prime}))u_{j}$
for $1\leq j\leq m$ and $A_{\varphi}v_{j}^{\prime}=0$ for $m+1\leq j\leq l$.

######  Proposition 5.8.

$A_{\varphi\theta}=\theta_{\mathcal{U}}^{-1}A_{\varphi}\theta_{\mathcal{U}}^{*}+A_{\theta}$

###### Proof.

If we let

$\theta_{\mathcal{U}}=\begin{bmatrix}Q_{1}&Q_{2}\\\
0&\operatorname{id}\end{bmatrix},$

then we have

$\varphi=\begin{bmatrix}P_{1}&P_{2}&P_{1}A_{\varphi 1}&0\\\
0&\operatorname{id}&0&0\\\ 0&0&P_{1}^{*}&0\\\
0&0&P_{2}^{T}P_{1}^{*}&\operatorname{id}\end{bmatrix}\text{ and }\
\theta=\begin{bmatrix}Q_{1}&Q_{2}&Q_{1}A_{\theta 1}&0\\\
0&\operatorname{id}&0&0\\\ 0&0&Q_{1}^{*}&0\\\
0&0&Q_{2}^{T}Q_{1}^{*}&\operatorname{id}\end{bmatrix}.$

This gives us the product

$\varphi\theta=\begin{bmatrix}P_{1}Q_{1}&P_{1}Q_{2}+P_{2}&P_{1}A_{\varphi
1}Q_{1}+P_{1}Q_{1}A_{\theta 1}&0\\\ 0&\operatorname{id}&0&0\\\
0&0&P_{1}^{*}Q_{1}^{*}&0\\\
0&0&(P_{2}^{T}P_{1}^{*}+Q_{2}^{T})Q_{1}^{*}&\operatorname{id}\end{bmatrix}$

and computing
$(\theta_{\mathcal{U}}^{-1}A_{\varphi}\theta_{\mathcal{U}}^{*}+A_{\theta})v_{j}^{\prime}$
for $1\leq j\leq m$

$\displaystyle(\theta_{\mathcal{U}}^{-1}A_{\varphi}\theta_{\mathcal{U}}^{*}+A_{\theta})v_{j}^{\prime}$
$\displaystyle=P_{1}Q_{1}(Q_{1}^{-1}A_{\varphi}Q_{1}^{*}+A_{\theta})v_{j}^{\prime}$
$\displaystyle=Q_{1}^{-1}A_{\varphi}Q_{1}^{*}v_{j}^{\prime}+A_{\theta}v_{j}^{\prime}$
$\displaystyle=Q_{1}^{-1}\operatorname{q}(P_{2}^{T}P_{1}^{*}Q_{1}^{*}v_{j})Q_{1}u_{j}^{\prime}+\operatorname{q}(Q_{2}^{T}Q_{1}^{*}v_{j})u_{j}^{\prime}$
$\displaystyle=\operatorname{q}(P_{2}^{T}P_{1}^{*}Q_{1}^{*}v_{j})u_{j}^{\prime}+\operatorname{q}(Q_{2}^{T}Q_{1}^{*}v_{j})u_{j}^{\prime}$
$\displaystyle=\operatorname{q}((P_{2}^{T}P_{1}^{*}+Q_{2}^{T})Q^{*}v_{j})u_{j}^{\prime}$
$\displaystyle=A_{\varphi\theta}v_{j}^{\prime}.$

Notice that
$(\theta_{\mathcal{U}}^{-1}A_{\varphi}\theta_{\mathcal{U}}^{*}+A_{\theta})v_{j}^{\prime}=0$
for $m\leq j\leq l$. ∎

We make the identification

$\begin{bmatrix}\varphi_{\mathcal{U}}&\varphi_{\mathcal{U}}A_{\varphi}\\\
0&\varphi_{\mathcal{U}}^{*}\end{bmatrix}\mapsto(\varphi_{\mathcal{U}},0),$

which gives us the product

$(\varphi_{\mathcal{U}},A)(\theta_{\mathcal{U}},C)=(\varphi_{\mathcal{U}}\theta_{\mathcal{U}},\theta_{\mathcal{U}}^{-1}A\theta_{\mathcal{U}}^{*}+C)$
(2)

and

$(\varphi_{\mathcal{U}},A)^{-1}=(\varphi_{\mathcal{U}}^{-1},\varphi_{\mathcal{U}}A\varphi_{\mathcal{U}}^{-*}).$
(3)

######  Corollary 5.9.

$(\varphi_{\mathcal{U}},0)(\theta_{\mathcal{U}},0)=(\varphi_{\mathcal{U}}\theta_{\mathcal{U}},0)$

######  Lemma 5.10.

If $\tau$ is a diagonal involution on $W$ of length $l$ and $\dim(W)=2l$, then
$\varphi\in\operatorname{O}(\operatorname{q},k)^{\mathcal{I}_{\tau}}$ can be
written as $(\varphi_{\mathcal{U}},A)$ where
$\varphi_{\mathcal{U}}\in\operatorname{O}(\operatorname{q}_{\mathcal{U}},k)$
and $A\in\mathsf{A}(\operatorname{q}_{\mathcal{U}},k)$.

###### Proof.

By Proposition 5.2 we have determined that elements in
$\operatorname{O}(\operatorname{q},k)^{\mathcal{I}_{\tau}}$ are of the form

$\varphi=\begin{bmatrix}\varphi_{\mathcal{U}}&\varphi_{\mathcal{U}}M\\\
0&\varphi_{\mathcal{U}}^{*}\end{bmatrix}$

and we have made the identification

$\begin{bmatrix}\varphi_{\mathcal{U}}&\varphi_{\mathcal{U}}A_{\varphi}\\\
0&\varphi_{\mathcal{U}}^{*}\end{bmatrix}\mapsto(\varphi_{\mathcal{U}},0).$

This gives us by (2) that $(\varphi,A)=(\varphi,0)(\operatorname{id},A)$ such
that $A\in\mathsf{A}(\operatorname{q}_{\mathcal{U}},k)$ and we can set
$M=A_{\varphi}+A$. For $v\in\overline{\mathcal{V}}$, a nonsingular completion
of $\overline{\mathcal{U}}$, we check

$\displaystyle\operatorname{q}(v+(A_{\varphi}+A)v)$
$\displaystyle=\operatorname{q}(v)+\operatorname{q}(A_{\varphi}v+Av)+\mathrm{B}(v,A_{\varphi}v)+\mathrm{B}(v,Av)$
$\displaystyle=\operatorname{q}(v)+\operatorname{q}(A_{\varphi}v)+\operatorname{q}(Av)+\mathrm{B}(v,A_{\varphi}v)+\mathrm{B}(v,Av)$
$\displaystyle=\operatorname{q}(v)+\operatorname{q}(A_{\varphi}v)+\mathrm{B}(v,A_{\varphi}v)$
$\displaystyle=\operatorname{q}(v+A_{\varphi}v)$
$\displaystyle=\operatorname{q}(\varphi^{*}_{\mathcal{U}}(v))$

and so, by Proposition 5.5 we have the result. ∎

Finally we show that all elements of the fixed point group can be decomposed
as the semi-direct product of the two subgroups of
$\operatorname{O}(\operatorname{q},k)^{\mathcal{I}_{\tau}}$ described above.

######  Theorem 5.11.

When $\tau=\tau_{u_{l}}\cdots\tau_{u_{2}}\tau_{u_{1}}$ is a diagonal
involution of length $l$ on a nonsingular space of dimension $2l$

$\operatorname{O}(\operatorname{q},k)^{\mathcal{I}_{\tau}}\cong\operatorname{O}(\operatorname{q}_{\mathcal{U}},k)\ltimes\mathsf{A}(\operatorname{q}_{\mathcal{U}},k).$

###### Proof.

By Proposition 5.1 we know that $\varphi$ must leave $\overline{\mathcal{U}}$
invariant. By Proposition 5.2 we know that the action of $\varphi$ leaving
$\overline{\mathcal{V}}$ invariant is completely determined by
$\varphi_{\mathcal{U}}$. Lemma 5.7 tells us that for any
$\varphi_{\mathcal{U}}\in\operatorname{O}(\operatorname{q}_{\mathcal{U}},k)$
there exists
$(\varphi_{\mathcal{U}},0)\in\operatorname{O}(\operatorname{q},k)^{\mathcal{I}_{\tau}}$.
Now consider any
$(\varphi_{\mathcal{U}},C)\in\operatorname{O}(\operatorname{q},k)^{\mathcal{I}_{\tau}}$
then we can factor

$(\varphi_{\mathcal{U}},C)=(\varphi_{\mathcal{U}},0)(\operatorname{id},C)$

and we know
$(\varphi_{\mathcal{U}},0),(\varphi_{\mathcal{U}},C)\in\operatorname{O}(\operatorname{q},k)$
so we must have
$(\operatorname{id},C)\in\mathsf{A}(\operatorname{q}_{\mathcal{U}},k)$.

Now $\mathsf{A}(\operatorname{q}_{\mathcal{U}},k)$ is a normal subgroup of
$\operatorname{O}(\operatorname{q},k)^{\mathcal{I}_{\tau}}$, since

$\displaystyle(\theta_{\mathcal{U}},D)(\operatorname{id},C)(\theta_{\mathcal{U}},D)^{-1}$
$\displaystyle=(\theta_{\mathcal{U}},D)(\operatorname{id},C)(\theta_{\mathcal{U}}^{-1},\theta_{\mathcal{U}}D\theta_{\mathcal{U}}^{-*})$
$\displaystyle=(\theta_{\mathcal{U}},D+C)(\theta_{\mathcal{U}}^{-1},\theta_{\mathcal{U}}D\theta_{\mathcal{U}}^{-*})$
$\displaystyle=(\theta_{\mathcal{U}}\theta_{\mathcal{U}}^{-1},\theta_{\mathcal{U}}(D+C)\theta_{\mathcal{U}}^{-*}+\theta_{\mathcal{U}}D\theta_{\mathcal{U}}^{-*})$
$\displaystyle=(\operatorname{id},\theta_{\mathcal{U}}C\theta_{\mathcal{U}}^{-*}).$

∎

Here we look at the fixed point group of a diagonal involution on a
nonsingular vector space $W$ with $\dim(W)=2r$ with $2l<2r$.

######  Proposition 5.12.

An element
$\varphi\in\operatorname{O}(\operatorname{q},k)^{\mathcal{I}_{\tau}}$ where
$\tau=\tau_{\mathcal{U}}$ is a product of orthogonal transvections can be
written in the form

$\begin{bmatrix}\varphi_{\mathcal{U}}&\varphi_{\mathcal{U}}C&\varphi_{\mathcal{U}}A\\\
0&\varphi_{\mathcal{X}}&\varphi_{\mathcal{X}}D\\\
0&0&\varphi_{\mathcal{U}}^{*}\end{bmatrix},$

where
$\varphi_{\mathcal{U}}\in\operatorname{O}(\operatorname{q}_{\mathcal{U}},k)$,
$\varphi_{\mathcal{X}}\in\mathrm{Sp}(\mathrm{B}_{\mathcal{X}},k)$ with
$\varphi_{\mathcal{U}}^{*}$ is the unique linear map such that
$\mathrm{B}(\varphi_{\mathcal{U}}(u),\varphi_{\mathcal{U}}^{*}(v))=\mathrm{B}(u,v)$.

###### Proof.

The map $\varphi$ acts on $W$ using ordered basis

$\mathcal{W}=\\{u_{1},\ldots,u_{l},x_{1},\ldots,x_{r-l},y_{1},\ldots,y_{r-l},v_{1},\ldots,v_{l}\\}$

where $\mathcal{U}$ is the inducing set for $\tau$, $\mathcal{V}$ is a
nonsingular completion of $\mathcal{U}$ and $\tau(x)=x$ for all
$x\in\overline{\mathcal{X}}$ such that

$\mathcal{X}=\\{x_{1},\ldots,x_{r-l},y_{1},\ldots,y_{r-l}\\}.$

Notice that the blocks are zero under $\varphi_{\mathcal{U}}$ in the first
column since by Proposition 5.1 the subspace $\overline{\mathcal{U}}$ is left
invariant by any element in
$\operatorname{O}(\operatorname{q},k)^{\mathcal{I}_{\tau}}$ where

$\tau=\tau_{\mathcal{U}}=\tau_{u_{l}}\cdots\tau_{u_{2}}\tau_{u_{1}}.$

In the second column the third block is zero since $\overline{\mathcal{U}}$ is
left invariant by $\varphi$ and
$\overline{\mathcal{X}}\subset\overline{\mathcal{U}}^{\perp}$. ∎

We have established
$\mathrm{B}(\varphi_{\mathcal{U}}(u),\varphi_{\mathcal{U}}^{*}(v))=\mathrm{B}(u,v)$,
and we will see that $C$ and $D$ have a similar relationship.

######  Proposition 5.13.

Let $\mathcal{X}$ be a basis for
$\left(\overline{\mathcal{U}\cup\mathcal{V}}\right)^{\perp}$ such that
$\tau(x)=x$ for all $x\in\overline{\mathcal{X}}$. The map
$\varphi_{\mathcal{X}}:\overline{\mathcal{X}}\to\overline{\mathcal{X}}$ leaves
the bilinear form invariant.

###### Proof.

Let $x,x^{\prime}\in\overline{\mathcal{X}}$, then

$\displaystyle\mathrm{B}(x,x^{\prime})$
$\displaystyle=\mathrm{B}(\varphi(x),\varphi(x^{\prime}))$
$\displaystyle=\mathrm{B}(\varphi_{\mathcal{X}}(x)+\varphi_{\mathcal{U}}(Cx),\varphi_{\mathcal{X}}(x^{\prime})+\varphi_{\mathcal{U}}(Cx^{\prime}))$
$\displaystyle=\mathrm{B}(\varphi_{\mathcal{X}}(x),\varphi_{\mathcal{X}}(x^{\prime}))$

since $\overline{\mathcal{U}}\subset\overline{\mathcal{U}}^{\perp}$. ∎

The group $\operatorname{O}(\operatorname{q},k)^{\mathcal{I}_{\tau}}$ has a
subgroup isomorphic to $\operatorname{O}(\operatorname{q}_{\mathcal{U}},k)$
with elements of the form

$\begin{bmatrix}\varphi_{\mathcal{U}}&0&\varphi_{\mathcal{U}}A_{\varphi}\\\
0&\operatorname{id}&0\\\ 0&0&\varphi_{\mathcal{U}}^{*}\end{bmatrix}.$

Next, we include Theorem 14.1 from [13], which is used several times. The
following results lead to a description of the fixed point groups of
involutions on $\operatorname{O}(\operatorname{q},k)$.

######  Theorem 5.14.

Let $(W,\operatorname{q})$ be a quadratic space with maximal nonsingular
subspace $W_{1}$ where $2r=\mathrm{dim}(W_{1})$ and
$s=\mathrm{dim}(\mathrm{rad}(W))$ with $\operatorname{q}|_{\mathrm{rad}(W)}$
anisotropic. If $\sigma\in\operatorname{O}(W)$ then
$\sigma|_{\mathrm{rad}{(W)}}=\operatorname{id}_{\mathrm{rad}(W)}$,
$\sigma|_{W_{1}}=\sigma_{0}+\sigma_{1}$, where
$\sigma_{0}:W_{1}\to\mathrm{rad}(W)$ and
$\sigma_{1}\in\operatorname{Sp}(W_{1})$. If $\tau\in\mathrm{Sp}(W_{1})$ then
there exists $\sigma\in\operatorname{O}(W)$ with $\sigma_{1}=\tau$ if and only
if
$\operatorname{q}(\tau(w))+\operatorname{q}(w)\in\operatorname{q}(\mathrm{rad}(W))$
for all $w\in W_{1}$, moreover if $\sigma$ exists it is unique.

######  Proposition 5.15.

$\mathrm{B}(v,Cx)=\mathrm{B}(Dv,x)$.

###### Proof.

Consider

$\displaystyle 0$ $\displaystyle=\mathrm{B}(v,x)$
$\displaystyle=\mathrm{B}(\varphi(v),\varphi(x))$
$\displaystyle=\mathrm{B}(\varphi_{\mathcal{U}}(Av)+\varphi_{\mathcal{U}}^{*}(v)+\varphi_{\mathcal{X}}(Dv),\varphi_{\mathcal{U}}(Cx)+\varphi_{\mathcal{X}}(x))$
$\displaystyle=\mathrm{B}(\varphi_{\mathcal{X}}(Dv),\varphi_{\mathcal{X}}(x))+\mathrm{B}(\varphi_{\mathcal{U}}^{*}(v),\varphi_{\mathcal{U}}(Cx))$
$\displaystyle=\mathrm{B}(Dv,x)+\mathrm{B}(v,Cx).$

∎

Since $D$ is completely determined by $C$ we will denote $D$ by $C^{\dagger}$.
So, we can rewrite the result in Proposition 5.15

$\mathrm{B}(v,Cx)=\mathrm{B}(C^{\dagger}v,x)$ (4)

In fact we can choose a basis for $W$ such that
$\mathrm{B}_{\mathcal{U}\mathcal{V}}:\overline{\mathcal{U}}\times\overline{\mathcal{V}}\to
k$ is represented by $\operatorname{id}$ and
$\mathrm{B}_{\mathcal{X}}:\overline{\mathcal{X}}\times\overline{\mathcal{X}}\to
k$ is represented by a diagonal block matrix of the form

$J=\begin{bmatrix}0&\operatorname{id}\\\ \operatorname{id}&0\end{bmatrix},$

which gives us the equation $C=C^{\dagger T}J$ or in other words
$C^{\dagger}=JC^{T}$. From this we can see that
$(C+C^{\prime})^{\dagger}=C^{\dagger}+C^{\prime\dagger}$

We choose the basis

$\mathcal{W}=\\{u_{1},u_{2},\ldots,u_{l},x_{1},x_{2},\ldots,x_{r-l},y_{1},y_{2},\ldots,y_{r-l},v_{1},v_{2},\ldots,v_{l}\\}$

and write
$\varphi\in\operatorname{O}(\operatorname{q},k)^{\mathcal{I}_{\tau}}$ as

$\varphi=\begin{bmatrix}\varphi_{\mathcal{U}}&\varphi_{\mathcal{U}}C&\varphi_{\mathcal{U}}A\\\
0&\varphi_{\mathcal{X}}&\varphi_{\mathcal{X}}C^{\dagger}\\\
0&0&\varphi_{\mathcal{U}}^{*}\end{bmatrix}.$

The image of $x\in\overline{\mathcal{X}}$ under
$\varphi\in\operatorname{O}(\operatorname{q},k)^{\mathcal{I}_{\tau}}$ gives us
the following relations

$\displaystyle\operatorname{q}(x)$
$\displaystyle=\operatorname{q}(\varphi(x))$
$\displaystyle=\operatorname{q}(\varphi_{\mathcal{U}}(Cx)+\varphi_{\mathcal{X}}(x))$
$\displaystyle=\operatorname{q}(\varphi_{\mathcal{U}}(Cx))+\operatorname{q}(\varphi_{\mathcal{X}}(x))$
$\displaystyle=\operatorname{q}(Cx)+\operatorname{q}(\varphi_{\mathcal{X}}(x)).$

So further we see that

$\operatorname{q}(x+Cx)=\operatorname{q}(\varphi_{\mathcal{X}}(x)).$

From here we begin building an argument that for every pair
$(\varphi_{\mathcal{X}},C)$ such that
$\operatorname{q}(x+Cx)=\operatorname{q}(\varphi_{\mathcal{X}}(x))$, we have a
linear transformation $M$ such that the triple
$(\varphi_{\mathcal{U}}^{*},C^{\dagger},M)$ leaves $\operatorname{q}$
invariant on the image of $\overline{\mathcal{V}}$ while preserving
$\mathrm{B}$ on $\varphi(W)$.

######  Lemma 5.16.

Let $\operatorname{q}_{\mathcal{U}}$ be anisotropic and
$T_{x,\omega}\in\operatorname{GL}(W)$ be of the form

$T_{x,\omega}=\begin{bmatrix}\operatorname{id}_{\mathcal{U}}&C&M\\\
0&\tau_{x,\omega}&\tau_{x,\omega}C^{\dagger}\\\
0&0&\operatorname{id}_{\mathcal{V}}\end{bmatrix}$

with $\tau_{x,\omega}\in\operatorname{Sp}(\mathrm{B}_{\mathcal{X}},k)$ and
$\operatorname{q}(\tau_{x,\omega}(z)+Cz)=\operatorname{q}(z)$ for all
$z\in\mathcal{X}$. Then there exists an
$M:\overline{\mathcal{V}}\to\overline{\mathcal{U}}$ such that
$T_{x,\omega}\in\operatorname{O}(\operatorname{q},k)$.

###### Proof.

let $x\in\overline{\mathcal{X}}$. To show that we have such an $M$ for the
pair $(\tau_{x,\omega},C)$, consider

$\begin{bmatrix}\operatorname{id}&C&M\\\
0&\tau_{x,\omega}&\tau_{x,\omega}C^{\dagger}\\\
0&0&\operatorname{id}\end{bmatrix}$

We can assume that $x=x_{1}$ is a basis vector such that there exists
$y_{1}\in\overline{\mathcal{X}}$ with $\mathrm{B}(x_{1},y_{1})=1$.

Since $C:\overline{\mathcal{X}}\to\overline{\mathcal{U}}$ and
$\overline{\mathcal{U}}$ is a totally singular anisotropic subspace of $W$ we
know that the image of $\operatorname{q}_{\mathcal{U}}$ is
$k^{2}[\operatorname{q}(u_{1}),\operatorname{q}(u_{2}),\ldots,\operatorname{q}(u_{l})]$,
i.e. the $k^{2}$ vector space spanned by the norms of the elements of
$\mathcal{U}$.

Assume $\operatorname{q}(z)=\operatorname{q}(\tau_{x,\omega}(z)+Cz)$, then we
have
$\operatorname{q}(z)+\operatorname{q}(\tau_{x,\omega}(z))=\operatorname{q}(Cz)$
with $\operatorname{q}(Cz)\in
k^{2}[\operatorname{q}(u_{1}),\operatorname{q}(u_{2}),\ldots,\operatorname{q}(u_{l})]$.

We extend $\\{x_{1},y_{1}\\}$ to a basis $\\{x_{1},y_{1},\dots,x_{n},y_{n}\\}$
for $\mathcal{X}$ such that

$\operatorname{q}_{\mathcal{X}}=[\operatorname{q}(x_{1}),\operatorname{q}(y_{1})]\perp\cdots\perp[\operatorname{q}(x_{n}),\operatorname{q}(y_{n})].$

We can now be more explicit about the image of $C$. If
$\operatorname{q}(z)=\operatorname{q}(\tau_{x_{1},\omega}(z))$, then
$\operatorname{q}(Cz)=0$ and thus $Cz=0$ since $\overline{\mathcal{U}}$ is
anisotropic. Now suppose
$\operatorname{q}(z)\neq\operatorname{q}(\tau_{x_{1},\omega}(z))$ and let
$c_{jj^{\prime}}$ be the coefficient of $u_{j}$ of $Cx_{j^{\prime}}$ and
$d_{jj^{\prime}}$ be the coefficient of $u_{j}$ for $Cy_{j^{\prime}}$. Since
$\tau_{x_{1},\omega}(x_{j^{\prime}})=x_{j^{\prime}}$,
$\tau_{x_{1},\omega}(y_{j^{\prime}})=y_{j^{\prime}}$ for $j^{\prime}\neq 1$
and $\tau_{x_{1},\omega}(y_{1})=y_{1}+\omega x_{1}$ we have

$\displaystyle\displaystyle\sum_{j=1}^{l}c_{jj^{\prime}}^{2}\operatorname{q}(u_{j})$
$\displaystyle=0\text{ for all }j,j^{\prime}$ (5)
$\displaystyle\displaystyle\sum_{j=1}^{l}d_{jj^{\prime}}^{2}\operatorname{q}(u_{j})$
$\displaystyle=0\text{ for all }j\text{ and all }j^{\prime}\neq 1$ (6)
$\displaystyle\displaystyle\sum_{j=1}^{l}d_{j1}^{2}\operatorname{q}(u_{j})$
$\displaystyle=\operatorname{q}(y_{1})+\operatorname{q}(y_{1}+\omega x_{1})$
(7)

By (5) we have $c_{jj^{\prime}}=0$ for all $j$ and $j^{\prime}$ and by (6)
$d_{jj^{\prime}}=0$ for all $j^{\prime}\neq 1$. Solving for
$\operatorname{q}(x_{1})$ in (7) we have

$\operatorname{q}(x_{1})=\dfrac{1}{\omega}+\displaystyle\sum_{j=1}^{l}\left(\dfrac{d_{j1}}{\omega}\right)^{2}\operatorname{q}(u_{j}).$

Now we consider the image of $v_{i}$

$T_{x_{1},\omega}(v_{i})=v_{i}+\tau_{x_{1},\omega}(C^{\dagger}v_{i})+Mv_{i},$

and computing the norm of both sides of this equation we end up with

$\displaystyle\operatorname{q}(v_{i})$
$\displaystyle=\operatorname{q}(v_{i})+\operatorname{q}(Mv_{i})+\mathrm{B}(v_{i},Mv_{i})+\operatorname{q}(\tau_{x_{1},\omega}(C^{\dagger}v_{i}))$
$\displaystyle\operatorname{q}(Mv_{i})+\mathrm{B}(v_{i},Mv_{i})$
$\displaystyle=\operatorname{q}(\tau_{x_{1},\omega}(C^{\dagger}v_{i}))$
$\displaystyle\operatorname{q}(Mv_{i})+\mathrm{B}(v_{i},Mv_{i})$
$\displaystyle=\operatorname{q}(C^{\dagger}v_{i}+\omega\mathrm{B}(x_{1},C^{\dagger}v_{i})x_{1})$
$\displaystyle\operatorname{q}(Mv_{i})+\mathrm{B}(v_{i},Mv_{i})$
$\displaystyle=\operatorname{q}(C^{\dagger}v_{i})+\operatorname{q}(\omega\mathrm{B}(x_{1},C^{\dagger}v_{i})x_{1})+\mathrm{B}(C^{\dagger}v_{i},\omega\mathrm{B}(x_{1},C^{\dagger}v_{i})x_{1})$
$\displaystyle\operatorname{q}(Mv_{i})+\mathrm{B}(v_{i},Mv_{i})$
$\displaystyle=\operatorname{q}(C^{\dagger}v_{i})+\omega^{2}\mathrm{B}(x_{1},C^{\dagger}v_{i})^{2}\operatorname{q}(x_{1})+\omega\mathrm{B}(x_{1},C^{\dagger}v_{i})^{2}$

Recall $C^{\dagger}=JC^{T}$ and we have that

$C^{\dagger}v_{i}=\sum d_{ij^{\prime}}x_{j^{\prime}}+\sum
c_{ij^{\prime}}y_{j^{\prime}}=d_{i1}x_{1},$

since all other coefficients are zero. With the computation above we need
$Mv_{i}$ such that

$\displaystyle\operatorname{q}(Mv_{i})+\mathrm{B}(v_{i},Mv_{i})$
$\displaystyle=\operatorname{q}(C^{\dagger}v_{i})+\omega^{2}\mathrm{B}(x_{1},C^{\dagger}v_{i})^{2}\operatorname{q}(x_{1})+\omega\mathrm{B}(x_{1},C^{\dagger}v_{i})^{2}$
$\displaystyle=\operatorname{q}(d_{i1}x_{1})+\omega^{2}\mathrm{B}(x_{1},d_{i1}x_{1})^{2}\operatorname{q}(x_{1})+\omega\mathrm{B}(x_{1},d_{i1}x_{1})^{2}$
$\displaystyle=\operatorname{q}(d_{i1}x_{1})$
$\displaystyle=\dfrac{d_{i1}^{2}}{\omega}+\displaystyle\sum_{j=1}^{l}\left(\dfrac{d_{i1}d_{j1}}{\omega}\right)^{2}\operatorname{q}(u_{j})$

Setting

$Mv_{i}=\displaystyle\sum_{j=1}^{l}\dfrac{d_{i1}d_{j1}}{\omega}u_{j},$

we see that

$\displaystyle\operatorname{q}(Mv_{i})+\mathrm{B}(v_{i},Mv_{i})$
$\displaystyle=\operatorname{q}\left(\displaystyle\sum_{j=1}^{l}\dfrac{d_{i1}d_{j1}}{\omega}u_{j}\right)+\mathrm{B}\left(v_{i},\displaystyle\sum_{j=1}^{l}\dfrac{d_{i1}d_{j1}}{\omega}u_{j}\right)$
$\displaystyle=\dfrac{d_{i1}^{2}}{\omega}+\displaystyle\sum_{j=1}^{l}\left(\dfrac{d_{i1}d_{j1}}{\omega}\right)^{2}\operatorname{q}(u_{j}).$

Notice that this also preserves the bilinear form since

$\displaystyle\mathrm{B}(T_{x_{1},\omega}(v_{i}),T_{x_{1},\omega}(v_{i^{\prime}}))$
$\displaystyle=\mathrm{B}(v_{i}+\tau_{x_{1},\omega}(C^{\dagger}v_{i})+Mv_{i},v_{i^{\prime}}+\tau_{x_{1},\omega}(C^{\dagger}v_{i^{\prime}})+Mv_{i^{\prime}})$
$\displaystyle=\mathrm{B}\left(v_{i},\sum\frac{d_{i^{\prime}1}d_{j1}}{\omega}u_{j}\right)$
$\displaystyle+\mathrm{B}(\tau_{x_{1},\omega}(C^{\dagger}v_{i}),\tau_{x_{1},\omega}(C^{\dagger}v_{i^{\prime}}))+\mathrm{B}\left(\sum\frac{d_{i1}d_{j1}}{\omega}u_{j},v_{i^{\prime}}\right)$
$\displaystyle=\dfrac{d_{i^{\prime}1}d_{i1}}{\omega}\mathrm{B}(v_{i},u_{i})+\mathrm{B}(C^{\dagger}v_{i},C^{\dagger}v_{i^{\prime}})+\dfrac{d_{i1}d_{i^{\prime}1}}{\omega}\mathrm{B}(u_{i^{\prime}},v_{i^{\prime}})$
$\displaystyle=\dfrac{d_{i^{\prime}1}d_{i1}}{\omega}+\mathrm{B}(d_{i1}x_{1},d_{i^{\prime}1}x_{1})+\dfrac{d_{i1}d_{i^{\prime}1}}{\omega}$
$\displaystyle=0.$

∎

######  Proposition 5.17.

Let

$\varphi=\begin{bmatrix}\varphi_{\mathcal{U}}&\varphi_{\mathcal{U}}C&\varphi_{\mathcal{U}}A\\\
0&\varphi_{\mathcal{X}}&\varphi_{\mathcal{X}}C^{\dagger}\\\
0&0&\varphi_{\mathcal{U}}^{*}\end{bmatrix}$

such that $C:\overline{\mathcal{X}}\to\overline{\mathcal{U}}$ for
$\varphi_{\mathcal{X}}\in\operatorname{Sp}(\mathrm{B}_{\mathcal{X}},k)$ such
that $\operatorname{q}(x+Cx)=\operatorname{q}(\varphi_{\mathcal{X}})$ for all
$x\in\overline{\mathcal{X}}$ then there exists an
$M:\overline{\mathcal{V}}\to\overline{\mathcal{U}}$ such that

$\operatorname{q}(\varphi(v))=\operatorname{q}(\varphi_{\mathcal{U}}(Mv)+\varphi_{\mathcal{X}}(C^{\dagger}v)+\varphi_{\mathcal{U}}^{*}(v)).$

Moreover $\varphi\in\operatorname{O}(\operatorname{q},k)$.

###### Proof.

Let us order the basis as before and consider

$\begin{bmatrix}\varphi_{\mathcal{U}}&\varphi_{\mathcal{U}}C&\varphi_{\mathcal{U}}A\\\
0&\varphi_{\mathcal{X}}&\varphi_{\mathcal{X}}C^{\dagger}\\\
0&0&\varphi_{\mathcal{U}}^{*}\end{bmatrix}=\begin{bmatrix}\varphi_{\mathcal{U}}&0&\varphi_{\mathcal{U}}(A+M)\\\
0&\operatorname{id}&0\\\
0&0&\varphi_{\mathcal{U}}^{*}\end{bmatrix}\begin{bmatrix}\operatorname{id}&C&M\\\
0&\varphi_{\mathcal{X}}&\varphi_{\mathcal{X}}C^{\dagger}\\\
0&0&\operatorname{id}\end{bmatrix}$ (8)

First we choose $v_{j}$ with $1\leq j\leq m$ an isotropic basis vector
$\operatorname{q}(v_{j})=0$ in $\overline{\mathcal{V}}$. We choose
$Mv_{j}=(\operatorname{q}(\varphi_{\mathcal{X}}(C^{\dagger}v_{j}))+\operatorname{q}(\varphi_{\mathcal{U}}^{*}(v_{j})))u_{j}$

$\displaystyle\operatorname{q}(\varphi_{\mathcal{U}}(Mv_{j})+\varphi_{\mathcal{X}}(C^{\dagger}v_{j})+\varphi_{\mathcal{U}}^{*}(v_{j}))$
$\displaystyle=\operatorname{q}(\varphi_{\mathcal{U}}(Mv_{j}))+\operatorname{q}(\varphi_{\mathcal{X}}(C^{\dagger}v_{j}))$
$\displaystyle\ \hskip
28.45274pt+\operatorname{q}(\varphi_{\mathcal{U}}^{*}(v_{j}))+\mathrm{B}(\varphi_{\mathcal{U}}(Mv_{j}),\varphi_{\mathcal{U}}^{*}(v_{j})))$
$\displaystyle=\operatorname{q}(Mv_{j})+\operatorname{q}(\varphi_{\mathcal{X}}(C^{\dagger}v_{j}))$
$\displaystyle\ \hskip
56.9055pt+\operatorname{q}(\varphi_{\mathcal{U}}^{*}(v_{j}))+\mathrm{B}(Mv_{j},v_{j})=0$

Now let us choose $v_{j}$ such that $m<j\leq l$. Then
$\operatorname{q}(v_{j})\neq 0$ and $\varphi_{\mathcal{U}}^{*}(v_{j})=v_{j}$.
We can assume without loss of generality that $\overline{\mathcal{U}}$ is
anisotropic, since any isotropic vector from $\overline{\mathcal{U}}$ would
not contribute to the overall norm of $\varphi(v_{j})$ being both norm zero
and having a trivial bilinear form with all anisotropic basis vectors.

Since $\varphi_{\mathcal{X}}\in\operatorname{Sp}(\mathrm{B}_{\mathcal{X}},k)$
and $\mathcal{U}$ is anisotropic we have by Theoreom 2.1.9 in [23] that
$\varphi_{\mathcal{X}}$ can be written as a product of symplectic
transvections

$\varphi_{\mathcal{X}}=\tau_{x_{N},\omega_{N}}\cdots\tau_{x_{2},\omega_{2}}\tau_{x_{1},\omega_{1}}$

and by Theorem 5.14 $C$ is unique. So we can factor

$\begin{bmatrix}\operatorname{id}&C&M\\\
0&\varphi_{\mathcal{X}}&\varphi_{\mathcal{X}}C^{\dagger}\\\
0&0&\operatorname{id}\end{bmatrix}=\begin{bmatrix}\operatorname{id}&C_{N}&M_{N}\\\
0&\tau_{z_{N},\omega_{N}}&\tau_{z_{N},\omega_{N}}C_{N}^{\dagger}\\\
0&0&\operatorname{id}\end{bmatrix}\cdots\begin{bmatrix}\operatorname{id}&C_{1}&M_{1}\\\
0&\tau_{z_{1},\omega_{1}}&\tau_{z_{1},\omega_{1}}C_{1}^{\dagger}\\\
0&0&\operatorname{id}\end{bmatrix}$

where
$\varphi_{\mathcal{X}}=\tau_{z_{N},\omega_{N}}\cdots\tau_{z_{2},\omega_{2}}\tau_{z_{1},\omega_{1}}$.
By Lemma 5.16 we have that $M_{i}$ exists for $1\leq i\leq N$. Since each
$C_{i}$ is unique, the product on the right hand side gives us the unique $C$
and $M$ on the left hand side. The bilinear form is preserved by the product
since it is preserved by each $T_{x_{i},\omega_{i}}$.

By Lemma 5.7 we can choose $A+M$ such that the first factor on the right hand
side of (8) is an element of $\operatorname{O}(\operatorname{q},k)$. We have
constructed the second factor to be an element of
$\operatorname{O}(\operatorname{q},k)$, and so
$\varphi\in\operatorname{O}(\operatorname{q},k)$.

∎

In order to establish the isomorphism classes of the fixed point groups of
involutions on nonsingular vector spaces over fields of characteristic $2$ we
need a basis with the vectors organized by specific properties. Let

$\mathcal{U}=\\{u_{1},u_{2},\ldots,u_{l}\\}$

be the set of anisotropic mutually orthogonal vectors in $W$ inducing a
diagonal involution $\tau_{\mathcal{U}}$. We choose a new basis of
$\overline{\mathcal{U}}$ indicating the Witt index $m$ of
$\overline{\mathcal{U}}$. We call this basis

$\mathcal{U}^{\prime}=\\{u_{1}^{\prime},u_{2}^{\prime},\ldots,u_{m}^{\prime},u_{m+1}^{\prime},\ldots,u_{l}^{\prime}\\},$

so that $\overline{\mathcal{U}}=\overline{\mathcal{U}^{\prime}}$ and
$\mathrm{dim}_{k}(\overline{\mathcal{U}})=l$ and
$\operatorname{q}(u_{i}^{\prime})=0$ for $1\leq i\leq m$. We address the case
when $\mathrm{dim}_{k}(\overline{\mathcal{U}})=l-1$ at the end of this
section.

Now we can choose $\mathcal{V}^{\prime}$ to be the corresponding hyperbolic
pairs making up the nonsingular subspace
$\overline{\mathcal{U}\cup\mathcal{V}}=\overline{\mathcal{U}^{\prime}\cup\mathcal{V}^{\prime}}\subset
W$ such that $\operatorname{q}(v_{i}^{\prime})=0$ for $1\leq i\leq m$ with
$\dim(\overline{\mathcal{U}\cup\mathcal{V}})=2l$. We choose a new ordered
basis $\mathcal{W}^{\prime}$ for $W=\overline{\mathcal{W}}$

$\mathcal{W}=\\{u_{1},u_{2},\ldots,u_{l},v_{1},v_{2},\ldots,v_{l},x_{1},x_{2},\ldots,x_{r-l},y_{1},y_{2},\ldots,y_{r-l}\\}$

with
$\mathcal{W}^{\prime}=\mathcal{U}^{\prime}\cup\mathcal{X}\cup\mathcal{V}^{\prime}$
with the order below

$\mathcal{U}^{\prime}=\\{u_{1}^{\prime},u_{2}^{\prime},\ldots
u_{m}^{\prime},u_{m+1}^{\prime},\ldots,u_{l}^{\prime}\\},$
$\mathcal{X}=\\{x_{1},x_{2},\ldots,x_{r-l},y_{1},y_{2},\ldots,y_{r-l}\\}$

and

$\mathcal{V}^{\prime}=\\{v_{m+1}^{\prime},\ldots,v_{l}^{\prime},v_{1}^{\prime},v_{2}^{\prime},\ldots,v_{m}^{\prime}\\}.$

so that the isotropic vectors in $\mathcal{U}^{\prime}$ and
$\mathcal{V}^{\prime}$ are ordered first and last, respectively. We can
represent an element in
$\operatorname{O}(\operatorname{q},k)^{\mathcal{I}_{\tau_{\mathcal{U}}}}$ as
the matrix

$P_{\varphi}X_{\varphi}C_{\varphi}=\begin{bmatrix}\varphi_{\mathcal{U}}&\varphi_{\mathcal{U}}C&\varphi_{\mathcal{U}}M\\\
0&\varphi_{\mathcal{X}}&\varphi_{\mathcal{X}}C^{\dagger}\\\
0&0&\varphi_{\mathcal{U}}^{*}\end{bmatrix}$
$=\begin{bmatrix}P_{1}&P_{2}&P_{1}C_{1}+P_{2}C_{3}&P_{1}C_{2}+P_{2}C_{4}&P_{1}M_{1}+P_{2}M_{3}&P_{1}M_{2}+P_{2}M_{4}\\\
0&\operatorname{id}&C_{3}&C_{4}&M_{3}&M_{4}\\\
0&0&X_{1}&X_{2}&X_{1}C_{4}^{T}+X_{2}C_{3}^{T}&X_{1}C_{2}^{T}+X_{2}C_{1}^{T}\\\
0&0&X_{3}&X_{4}&X_{3}C_{4}^{T}+X_{4}C_{3}^{T}&X_{3}C_{2}^{T}+X_{4}C_{1}^{T}\\\
0&0&0&0&\operatorname{id}&P_{2}^{T}P_{1}^{*}\\\ 0&0&0&0&0&P_{1}^{*}\\\
\end{bmatrix}$

$\displaystyle P_{\varphi}$
$\displaystyle=\begin{bmatrix}P_{1}&P_{2}&0&0&0&A_{\varphi}\\\
0&\operatorname{id}&0&0&0&0\\\ 0&0&\operatorname{id}&0&0&0\\\
0&0&0&\operatorname{id}&0&0\\\ 0&0&0&0&\operatorname{id}&P_{2}^{T}P_{1}^{*}\\\
0&0&0&0&0&P_{1}^{*}\\\ \end{bmatrix}$ (9) $\displaystyle X_{\varphi}$
$\displaystyle=\begin{bmatrix}\operatorname{id}&0&0&0&0&0\\\
0&\operatorname{id}&C_{3}&C_{4}&M_{\varphi}&0\\\
0&0&X_{1}&X_{2}&X_{1}C_{4}^{T}+X_{2}C_{3}^{T}&0\\\
0&0&X_{3}&X_{4}&X_{3}C_{4}^{T}+X_{4}C_{3}^{T}&0\\\
0&0&0&0&\operatorname{id}&0\\\ 0&0&0&0&0&\operatorname{id}\\\ \end{bmatrix}$
(10) $\displaystyle C_{\varphi}$
$\displaystyle=\begin{bmatrix}\operatorname{id}&0&C_{1}&C_{2}&M_{1}&M_{2}\\\
0&\operatorname{id}&0&0&0&M_{4}\\\ 0&0&\operatorname{id}&0&0&C_{2}^{T}\\\
0&0&0&\operatorname{id}&0&C_{1}^{T}\\\ 0&0&0&0&\operatorname{id}&0\\\
0&0&0&0&0&\operatorname{id}\\\ \end{bmatrix}$ (11)

We use $\mathcal{U}_{is}$ to denote the isotropic vectors in $\mathcal{U}$ and
$\mathcal{U}_{an}$ to denote the anisotropic vectors in $\mathcal{U}$.

######  Proposition 5.18.

If $\varphi\in\operatorname{O}(\operatorname{q},k)^{\mathcal{I}_{\tau}}$ where
$\tau=\tau_{\mathcal{U}}$ then $\varphi=P_{\varphi}X_{\varphi}C_{\varphi}$
from (9),(10) and (11).

###### Proof.

The matrix $P_{\varphi}$ is isomorphic to an element in
$\operatorname{O}(\operatorname{q}_{\mathcal{U}},k)$ with $A_{\varphi}$
unique. The matrix $X_{\varphi}$ is isomorphic to an element in
$\operatorname{O}(\operatorname{q}_{\mathcal{U}_{\mathrm{an}}\cup X},k)\cong
H\subset\operatorname{Sp}(\mathrm{B}_{\mathcal{X}},k)$, where $[C_{3}|C_{4}]$
is the unique linear map from Theorem 14.1 in [13]. The maps
$C_{1},C_{2}:\overline{X}\to\overline{\mathcal{U}}_{is}$ have isotropic images
in
$\mathrm{rad}(\overline{\mathcal{U}\cup\mathcal{X}})=\overline{\mathcal{U}}$
and so have zero norm and zero bilinear form on
$\overline{\mathcal{U}\cup\mathcal{X}}$. Since
$P_{\varphi},X_{\varphi}\in\operatorname{O}(\operatorname{q},k)$ and
$\varphi\in\operatorname{O}(\operatorname{q},k)^{\mathcal{I}_{\tau}}$ then
$C_{\varphi}\in\operatorname{O}(\operatorname{q},k)$ and
$P_{\varphi}X_{\varphi}C_{\varphi}=\varphi$ which leaves
$\overline{\mathcal{U}}$ invariant. ∎

We show that each of the matrices $P_{\varphi},X_{\varphi}$ and $C_{\varphi}$
corresponds to a subgroup of
$\operatorname{O}(\operatorname{q},k)^{\mathcal{I}_{\tau}}$, and we denote
these subgroups by $\mathsf{P},\mathsf{X}$ and $\mathsf{C}$ respectively. Next
we see what further restrictions are necessary for $\mathsf{P},\mathsf{C}$ and
$\mathsf{X}$ to be subgroups of $\operatorname{O}(\operatorname{q},k)$.

For the following characterizations of these subgroups we assume that the
basis for $\overline{\mathcal{U}\cup\mathcal{V}}$ has the following order

$\\{u_{1}^{\prime},u_{2}^{\prime},\ldots
u_{m}^{\prime},u_{m+1}^{\prime},\ldots,u_{l}^{\prime},v_{m+1}^{\prime},\ldots,v_{l}^{\prime},v_{1}^{\prime},v_{2}^{\prime},\ldots,v_{m}^{\prime}\\}$
(12)

where $m$ is the Witt index of $\overline{\mathcal{U}\cup\mathcal{V}}$ and
$\operatorname{q}(u_{i}^{\prime})=\operatorname{q}(v_{i}^{\prime})=0$ for
$1\leq i\leq m$.

######  Proposition 5.19.

An element
$C_{\varphi}\in\mathsf{C}\subset\operatorname{O}(\operatorname{q},k)$ has
$C_{1},C_{2}\in\operatorname{Mat}_{m,s-l}(k)$ and
$M_{1}\in\operatorname{Mat}_{m,l-m}(k)$, $M_{4}=M_{1}^{T}$ and
$M_{2}\in\operatorname{Mat}_{m,m}(k)$ such that

$M_{2}+M_{2}^{T}=C_{1}C_{2}^{T}+C_{2}C_{1}^{T}.$

###### Proof.

Using the ordered basis (12) we see that the matrix representation for
$\mathrm{B}_{\mathcal{U}\cup\mathcal{V}}$ is of the form

$[\mathrm{B}_{\mathcal{U}\cup\mathcal{V}}]=\begin{bmatrix}0&\operatorname{id}_{l-m}\\\
\operatorname{id}_{m}&0\end{bmatrix}.$

Now notice that
$M_{1}:\overline{\mathcal{V}_{\mathrm{is}}}\to\overline{\mathcal{U}_{\mathrm{an}}}$
and
$M_{4}:\overline{\mathcal{V}_{\mathrm{an}}}\to\overline{\mathcal{U}_{\mathrm{is}}}$.
We will compute $\mathrm{B}(Mv_{i},v_{j})$ for different values of $i$ and
$j$. First, we notice

$\displaystyle\mathrm{B}(Mv_{i},v_{j})$
$\displaystyle=(Mv_{i})^{T}[\mathrm{B}_{\mathcal{U}\cup\mathcal{V}}]v_{j}$
$\displaystyle=v_{i}^{T}\begin{bmatrix}M_{1}&M_{2}\\\
0&M_{4}\end{bmatrix}^{T}\begin{bmatrix}0&\operatorname{id}_{l-m}\\\
\operatorname{id}_{m}&0\end{bmatrix}v_{j}$
$\displaystyle=v_{i}^{T}\begin{bmatrix}M_{1}^{T}&0\\\
M_{2}^{T}&M_{4}^{T}\end{bmatrix}\begin{bmatrix}0&\operatorname{id}_{l-m}\\\
\operatorname{id}_{m}&0\end{bmatrix}v_{j}$
$\displaystyle=v_{i}^{T}\begin{bmatrix}0&\operatorname{id}_{l-m}\\\
\operatorname{id}_{m}&0\end{bmatrix}\begin{bmatrix}M_{4}^{T}&M_{2}^{T}\\\
0&M_{1}^{T}\end{bmatrix}v_{j}$

so, we have the following relations

$\displaystyle\mathrm{B}(Mv_{i},v_{j})$
$\displaystyle=\mathrm{B}(M_{1}v_{i},v_{j})$
$\displaystyle=\mathrm{B}(v_{i},M_{2}^{T}v_{j}+M_{1}^{T}v_{j})$
$\displaystyle=\mathrm{B}(v_{i},M_{1}^{T}v_{j})$

or

$\displaystyle\mathrm{B}(v_{i},Mv_{j})$
$\displaystyle=\mathrm{B}(M_{2}v_{j}+M_{4}v_{j},v_{i})$
$\displaystyle=\mathrm{B}(M_{4}v_{j},v_{i})$
$\displaystyle=\mathrm{B}(v_{j},M_{4}^{T}v_{i})$

when $1\leq j<m+1\leq i\leq l$ and

$\displaystyle\mathrm{B}(Mv_{i},v_{j})$
$\displaystyle=\mathrm{B}(M_{2}v_{i}+M_{4}v_{i},v_{j})$
$\displaystyle=\mathrm{B}(M_{2}v_{i},v_{j})$
$\displaystyle=\mathrm{B}(v_{i},M_{2}^{T}v_{j}+M_{1}^{T}v_{j})$
$\displaystyle=\mathrm{B}(v_{i},M_{2}^{T}v_{j})$

when $1\leq i,j\leq m$.

Now we compute $\mathrm{B}(C_{\varphi}v_{i},C_{\varphi}v_{j})$ for $1\leq
i,j\leq m$

$\displaystyle 0$
$\displaystyle=\mathrm{B}(C_{\varphi}v_{i},C_{\varphi}v_{j})$
$\displaystyle=\mathrm{B}(v_{i}+(C_{1}^{T}+C_{2}^{T}+M_{4}+M_{2})v_{i},v_{j}+(C_{1}^{T}+C_{2}^{T}+M_{4}+M_{2})v_{j})$
$\displaystyle=\mathrm{B}(v_{i},M_{2}v_{j})+\mathrm{B}(C_{1}^{T}v_{i},C_{2}^{T}v_{j})+\mathrm{B}(C_{2}^{T}v_{i},C_{1}^{T}v_{j})+\mathrm{B}(M_{2}v_{i},v_{j})$

and so, we have

$\mathrm{B}(v_{i},(M_{2}+M_{2}^{T})v_{j})=\mathrm{B}(v_{i},(C_{1}C_{2}^{T}+C_{2}C_{1}^{T})v_{j})$

and since $\mathrm{B}$ is nondegenerate on this nonsingular space we have
$M_{2}+M_{2}^{T}=C_{1}C_{2}^{T}+C_{2}C_{1}^{T}$.

On the other hand, if $1\leq i\leq m<j\leq l$ we have

$\displaystyle 0$
$\displaystyle=\mathrm{B}(C_{\varphi}v_{i},C_{\varphi}v_{j})$
$\displaystyle=\mathrm{B}(v_{i}+(C_{1}^{T}+C_{2}^{T}+M_{4}+M_{2})v_{i},v_{j}+M_{1}v_{j})$
$\displaystyle=\mathrm{B}(v_{i},M_{1}v_{j})+B(M_{4}v_{i},v_{j})$
$\displaystyle\mathrm{B}(v_{i},M_{1}v_{j})$
$\displaystyle=\mathrm{B}(v_{i},M_{4}^{T}v_{j})$

and so, we have $M_{1}=M_{4}^{T}$. Further notice that the images of
$C_{1},C_{2}$ and $M_{1}$ are in the totally isotropic subspace of
$\overline{\mathcal{U}}$ so have zero norm and zero bilinear form values on
$\overline{\mathcal{U}\cup\mathcal{X}}$.

There is a further constraint on the map $M_{2}$. To see this we consider the
norm of the image under $C_{\varphi}$ on $v_{i}$ with $1\leq i\leq m$

$\displaystyle\operatorname{q}(C_{\varphi}v_{i})$
$\displaystyle=\operatorname{q}(v_{i}+(C_{1}^{T}+C_{2}^{T}+M_{4}+M_{2})v_{i})$
$\displaystyle\operatorname{q}(v_{i})$
$\displaystyle=\operatorname{q}(v_{i})+\operatorname{q}((C_{1}^{T}+C_{2}^{T})v_{i})+\operatorname{q}(M_{4}v_{i}+M_{2}v_{i})+\mathrm{B}(v_{i},M_{4}v_{i}+M_{2}v_{i})$
$\displaystyle\mathrm{B}(v_{i},M_{2}v_{i})$
$\displaystyle=\operatorname{q}((C_{1}^{T}+C_{2}^{T})v_{i})+\operatorname{q}(M_{1}^{T}v_{i})$

this computation shows us that the diagonal of $M_{2}$ consists of entries of
the form
$\operatorname{q}((C_{1}^{T}+C_{2}^{T})v_{i})+\operatorname{q}(M_{1}^{T}v_{i}))$.
∎

######  Lemma 5.20.

The subgroup $\mathsf{P}$ is isomorphic to
$\operatorname{O}(\operatorname{q}_{\mathcal{U}},k)$.

###### Proof.

This the same group from Lemma 5.7 and Proposition 5.8. ∎

######  Lemma 5.21.

The subgroup $\mathsf{X}$ is isomorphic to
$\operatorname{O}(\operatorname{q}_{\mathcal{U}_{\mathrm{an}}^{\prime}\cup\mathcal{X}},k)$

###### Proof.

This is the group described within [13] by Theorem 14.1 and the related
discussion. Theorem 14.1 also appears in this document as Theorem 5.14. ∎

######  Lemma 5.22.

$\mathsf{X}\mathsf{C}=\mathsf{C}\mathsf{X}\cong\mathsf{X}\ltimes\mathsf{C}$.

###### Proof.

Notice that $\mathsf{XC}$ is closed under multiplication. We show that
$\mathsf{XC}=\mathsf{CX}$ by verifying that $\mathsf{C}$ is a normal subgroup
of $\mathsf{XC}$. Consider $X_{\varphi}\in\mathsf{X}$ from (10) and compute
the inverse

$X_{\varphi}^{-1}=\begin{bmatrix}\operatorname{id}&0&0&0&0&0\\\
0&\operatorname{id}&C_{3}X_{4}^{T}+C_{4}X_{3}^{T}&C_{3}X_{2}^{T}+C_{4}X_{1}^{T}&C_{3}C_{4}^{T}+C_{4}C_{3}^{T}+M_{\varphi}&0\\\
0&0&X_{4}^{T}&X_{2}^{T}&C_{4}^{T}&0\\\ 0&0&X_{3}^{T}&X_{1}^{T}&C_{3}^{T}&0\\\
0&0&0&0&\operatorname{id}&0\\\ 0&0&0&0&0&\operatorname{id}\\\ \end{bmatrix}$

We define $\varphi_{\mathcal{X}}$ as the symplectic matrix
$\bigl{[}\begin{smallmatrix}X_{1}&X_{2}\\\
X_{3}&X_{4}\end{smallmatrix}\bigr{]}$ with the standard symplectic basis for
$\overline{\mathcal{X}}$. Now
$X_{\varphi}C_{\varphi}X_{\varphi}^{-1}\in\mathsf{C}$. ∎

We define the group $\mathsf{C}$ described in Proposition 5.19 to be
$\mathsf{A}(\operatorname{q}_{\mathcal{U}},k)$.

######  Theorem 5.23.

The involution $\tau=\tau_{\mathcal{U}}$ written as a product of transvections
induced by vectors $\mathcal{U}$ that are mutually orthogonal has a fixed
point group
$\operatorname{O}(\operatorname{q},k)^{\mathcal{I}_{\tau}}\cong\operatorname{O}(\operatorname{q}_{\mathcal{U}},k)\ltimes(\operatorname{O}(\operatorname{q}_{\mathcal{U}_{\mathrm{an}}\cup\mathcal{X}},k)\ltimes\mathsf{A}(\operatorname{q}_{\mathcal{U}},k)$.

###### Proof.

We choose a basis for $\overline{\mathcal{U}}$ such that the basis vectors
corresponding to the maximal defect in $\overline{\mathcal{U}}$ are listed
first and the corresponding defect in $\overline{\mathcal{V}}$, the
nonsingular completion of $\overline{\mathcal{U}}$ in $W$, has its basis
vectors listed last. Then by Proposition 5.12
$\varphi\in\operatorname{O}(\operatorname{q},k)^{\mathcal{I}_{\tau}}$ has the
following form

$\varphi=\begin{bmatrix}\varphi_{\mathcal{U}}&\varphi_{\mathcal{U}}C&\varphi_{\mathcal{U}}M\\\
0&\varphi_{\mathcal{X}}&\varphi_{\mathcal{X}}C^{\dagger}\\\
0&0&\varphi_{\mathcal{U}}^{*}\end{bmatrix}$

$=\begin{bmatrix}P_{1}&P_{2}&P_{1}C_{1}+P_{2}C_{3}&P_{1}C_{2}+P_{2}C_{4}&P_{1}M_{1}+P_{2}M_{3}&P_{1}(M_{2}+A_{\varphi})+P_{2}M_{4}\\\
0&\operatorname{id}&C_{3}&C_{4}&M_{3}&M_{4}\\\
0&0&X_{1}&X_{2}&X_{1}C_{4}^{T}+X_{2}C_{3}^{T}&X_{1}C_{2}^{T}+X_{2}C_{1}^{T}\\\
0&0&X_{3}&X_{4}&X_{3}C_{4}^{T}+X_{4}C_{3}^{T}&X_{3}C_{2}^{T}+X_{4}C_{1}^{T}\\\
0&0&0&0&\operatorname{id}&P_{2}^{T}P_{1}^{*}\\\ 0&0&0&0&0&P_{1}^{*}\\\
\end{bmatrix}$

Notice that this factors as $\varphi=P_{\varphi}X_{\varphi}C_{\varphi}$ from
(9), (10) and (11) and since $\varphi\in\operatorname{O}(\operatorname{q},k)$
and we have shown that
$P_{\varphi},X_{\varphi}\in\operatorname{O}(\operatorname{q},k)$ so then
$C_{\varphi}\in\operatorname{O}(\operatorname{q},k)$. By Lemma 5.22 we have
shown the semi-direct product relationship between $\mathsf{X}$ and
$\mathsf{C}$ and in a similar way one can verify $\mathsf{XC}$ is a normal
subgroup of $\mathsf{P}(\mathsf{XC})$ so we have

$\operatorname{O}(\operatorname{q},k)^{\mathcal{I}_{\tau}}\cong\operatorname{O}(\operatorname{q}_{\mathcal{U}},k)\ltimes(\operatorname{O}(\operatorname{q}_{\mathcal{U}_{\mathrm{an}}\cup\mathcal{X}},k)\ltimes\mathsf{A}_{\mathcal{U}}(\operatorname{q},k)).$

∎

The only change when the involution is hyperbolic is that the dimension of
$\overline{\mathcal{U}\cup\mathcal{V}}$ is $2l-2$ instead of $2l$. We can see
from this fact that there are involutions that are not
$\operatorname{O}(\operatorname{q},k)$-conjugate that have the same fixed
point group. This provides other examples of two distinct $G(k)$-conjugacy
classes with the same fixed point group over a field of characteristics $2$.

######  Remark 5.24.

In the case that the inducing set $\mathcal{U}$ for the involution
$\tau=\tau_{\mathcal{U}}$ spans an anisotropic vector space
$\overline{\mathcal{U}}$, we have
$\operatorname{O}(\operatorname{q}_{\mathcal{U}},k)=\\{\operatorname{id}\\}$
and the matrices $C$ and $M$ are unique up to a given $\varphi_{\mathcal{X}}$.
So we have
$\operatorname{O}(\operatorname{q},k)^{\mathcal{I}_{\tau}}\cong\operatorname{O}(\operatorname{q}_{\mathcal{U}\cup\mathcal{X}},k)$.

If $\overline{\mathcal{U}}$ is totally isotropic then
$\operatorname{O}(\operatorname{q}_{\mathcal{U}},k)\cong\operatorname{GL}(\overline{\mathcal{U}})$
and
$\operatorname{O}(\operatorname{q}_{\mathcal{U}_{\mathrm{an}}\cup\mathcal{X}},k)\cong\operatorname{O}(\operatorname{q}_{\mathcal{X}},k)$.

##  6 General quadratic spaces

We now remove the restriction that $W$ is nonsingular. Let $W_{1}$ denote a
nonsingular subspace of $W$ of maximal dimension, then
$W=W_{1}\perp\operatorname{rad}(W)$. A triple $(\rho,Y,\tau)$ will denote an
element in $\operatorname{O}(\operatorname{q},k)$ where $\tau$ is a product of
symplectic transvections,
$\operatorname{q}(\tau(w)+\rho(Yw))=\operatorname{q}(w)$ and $\rho$ is a
product of basic radical involutions.

######  Proposition 6.1.

An involution in $\operatorname{O}(\operatorname{q},k)$ is a map of the form

$(\rho,Y,\tau)=\begin{bmatrix}\rho&\rho Y\\\ 0&\tau\end{bmatrix},$

where
$\tau=\tau_{\mathcal{U}}\in\operatorname{Sp}(B_{\mathcal{U}\cup\mathcal{V}},k)$,
$\overline{\mathcal{U}\cup\mathcal{V}}\subset W_{1}$,
$Y:W_{1}\to\mathrm{rad}(W)$,
$\rho\in\operatorname{O}(\operatorname{q}_{\mathrm{rad}(W)},k)$ and
$\operatorname{q}(\tau(w)+\rho(Yw))=\operatorname{q}(w)$ with
$\tau^{2}=\operatorname{id}$, $\rho^{2}=id$ and $Y=\rho Y\tau$.

###### Proof.

The map $(\rho,Y,\tau)\in\operatorname{O}(\operatorname{q},k)$ must leave
$\mathrm{rad}(W)$ invariant. If you square $(\rho,Y,\tau)$ and set it equal to
the identity we need $\rho^{2}=\operatorname{id}_{\mathrm{rad}(W)}$ and
$\tau^{2}=\operatorname{id}$. In [23] O’Meara shows us that $\tau$ can be
written as a product of symplectic transvections induced by mutually
orthogonal vectors in $W_{1}$. We call this set of vectors
$\mathcal{U}=\\{u_{1},\ldots,u_{l}\\}$ and consider a non-singular completion
of this totally singular subspace of $W_{1}$ completed by the vectors
$\mathcal{V}=\\{v_{1},\ldots,v_{l}\\}$ such that $\mathrm{B}(u_{i},v_{j})=1$
if $i=j$ and is zero otherwise. So
$\tau\in\operatorname{Sp}(\mathrm{B}_{\mathcal{U}\cup\mathcal{V}},k)$. We have
$\rho\in\operatorname{O}(\operatorname{q}_{\mathrm{rad}(W)},k)$ since the
radical is left invariant by elements in
$\operatorname{O}(\operatorname{q},k)$. The linear transformation
$Y:W_{1}\to\mathrm{rad}(W)$ can exist since adding on radical vectors leaves
the bilinear form invariant. The quadratic form is left invariant as long as
$\operatorname{q}(\tau(w)+\rho(Yw))=\operatorname{q}(w)$. ∎

######  Corollary 6.2.

If $w\in W$ is such that $\tau(w)=w$ then $\rho(Yw)=Yw$.

######  Proposition 6.3.

For an involution $(\rho,Y,\tau)\in\operatorname{O}(\operatorname{q},k)$ of
the form

$(\rho,Y,\tau)=\begin{bmatrix}\rho&\rho Y\\\ 0&\tau\end{bmatrix},$

where
$\tau=\tau_{\mathcal{U}}\in\operatorname{Sp}(\mathrm{B}_{\mathcal{U}\cup\mathcal{V}},k)$,
$\overline{\mathcal{U}\cup\mathcal{V}}\subset W_{1}$,
$Y:W_{1}\to\mathrm{rad}(W)$,
$\rho\in\operatorname{O}(\operatorname{q}_{\mathrm{rad}(W)},k)$ and
$\operatorname{q}(\tau(w)+\rho(Yw))=\operatorname{q}(w)$ we can choose a basis
$\mathcal{U}^{\prime}$ such that with respect to the new basis $(\rho,Y,\tau)$
is of the form

$\begin{bmatrix}\rho&Y_{0}\\\ 0&\tau_{Y}\end{bmatrix}$

with $\tau_{Y}\in\operatorname{O}(\operatorname{q},k)$ as a product of
orthogonal transvections, $Y_{0}v_{i}=0$, and $\operatorname{q}(Y_{0}w)=0$.

###### Proof.

Let
$\tau=\tau_{a_{l},u_{l}}\cdots\tau_{a_{2},u_{2}}\tau_{a_{1},u_{1}}\in\operatorname{Sp}(\mathrm{B}_{\mathcal{U}\cup\mathcal{V}},k)$.
So that $l$ is the length of the transvection we will choose $a_{i}\neq 0$ for
$1\leq i\leq l$. Let $\mathcal{X}$ be a basis for
$(\overline{\mathcal{U}\cup\mathcal{V}})^{\perp}\cap W_{1}$ and notice that if
$w\in\overline{\mathcal{U}\cup\mathcal{X}}$ that
$(\rho,Y,\tau)(w)=\tau(w)+\rho(Yw)=w+\rho(Yw)$ and that
$\operatorname{q}(Yw)=0$ since $\mathrm{B}(w,\rho(Yw))=0$.

Next for $v_{i}\in\mathcal{V}$

$\displaystyle\operatorname{q}((\rho,Y,\tau)(v_{i}))$
$\displaystyle=\operatorname{q}(\tau(v_{i})+\rho(Yv_{i}))$
$\displaystyle\operatorname{q}(v_{i})$
$\displaystyle=\operatorname{q}\left(v_{i}+a_{i}u_{i}+\rho(Yv_{i})\right)$
$\displaystyle\operatorname{q}(v_{i})$
$\displaystyle=\operatorname{q}(v_{i})+a_{i}^{2}\operatorname{q}(u_{i})+a_{i}\mathrm{B}(v_{i},u_{i})+\operatorname{q}(Yv_{i})$
$\displaystyle\operatorname{q}(Yv_{i})$
$\displaystyle=a_{i}^{2}\operatorname{q}(u_{i})+a_{i}.$

Now we see that

$\displaystyle(\rho,Y,\tau)(v_{i})$ $\displaystyle=\tau(v_{i})+\rho(Yv_{i})$
$\displaystyle=v_{i}+a_{i}u_{i}+\rho(Yv_{i})$
$\displaystyle=v_{i}+a_{i}\mathrm{B}\left(v_{i},u_{i}+\frac{1}{a_{i}}\rho(Yv_{i})\right)\left(u_{i}+\frac{1}{a_{i}}\rho(Yv_{i})\right)$
$\displaystyle=\tau_{a_{i},u_{i}+\frac{1}{a_{i}}\rho(Yv_{i})}(v_{i})$
$\displaystyle=\tau_{u_{i}+\frac{1}{a_{i}}\rho(Yv_{i})}(v_{i})$

since

$\displaystyle\operatorname{q}\left(u_{i}+\frac{1}{a_{i}}\rho(Yv_{i})\right)$
$\displaystyle=\operatorname{q}(u_{i})+\frac{1}{a_{i}^{2}}\operatorname{q}(Yv_{i})$
$\displaystyle=\operatorname{q}(u_{i})+\frac{1}{a_{i}^{2}}(a_{i}^{2}\operatorname{q}(u_{i})+a_{i})$
$\displaystyle=\frac{1}{a_{i}}.$

Now we define

$\mathcal{U}_{Y}=\left\\{u_{1}+\frac{1}{a_{1}}\rho(Yv_{1}),u_{2}+\frac{1}{a_{2}}\rho(Yv_{2}),\ldots,u_{l}+\frac{1}{a_{l}}\rho(Yv_{l})\right\\}$

and consider the maximal nonsingular subspace of $W$,
$W_{1}^{\prime}=\overline{\mathcal{U}_{Y}\cup\mathcal{X}\cup\mathcal{V}}$.
Further we define

$\tau_{Y}=\tau_{u_{l}+\frac{1}{a_{l}}\rho(Yv_{l})}\cdots\tau_{u_{2}+\frac{1}{a_{2}}\rho(Yv_{2})}\tau_{u_{1}+\frac{1}{a_{1}}\rho(Yv_{1})}$

and $Y_{0}w=Yw$ for $w\in\overline{\mathcal{U}\cup\mathcal{X}}$ and $Y_{0}v=0$
for $v\in\overline{\mathcal{V}}$. Using this basis $(\rho,Y,\tau)$ is of the
form

$\begin{bmatrix}\rho&Y_{0}\\\ 0&\tau_{Y}\end{bmatrix}$

and we have $\tau_{Y}\in\operatorname{O}(\operatorname{q},k)$ is a product of
orthogonal transvections, $Y_{0}v_{i}=0$, and $\operatorname{q}(Y_{0}w)=0$ for
all $w\in W_{1}^{\prime}$. ∎

######  Proposition 6.4.

The involution $\left[\begin{smallmatrix}\rho&\rho Y\\\
0&\tau\end{smallmatrix}\right]$ fixes pointwise elements of the form
$u_{i}+\frac{1}{a_{i}}\rho(Yv_{i})$. Moreover
$Y_{0}\left(u_{i}+\frac{1}{a_{i}}\rho(Yv_{i})\right)=0$.

###### Proof.

Let $\tau=\tau_{\mathcal{U}}$ and $u_{i}\in\mathcal{U}$ with
$v_{i}\in\mathcal{V}$ such that $\mathrm{B}(u_{i},v_{i})=1$. Since
$(\rho,Y,\tau)$ is an involution we have that $\rho^{2}=\operatorname{id}$,
$\tau^{2}=\operatorname{id}$ and $Y=\rho Y\tau$. First, we note

$Yv_{i}=\rho(Y\tau(v_{i}))=\rho(Y(v_{i}+a_{i}u_{i})).$

We compute

$\displaystyle(\rho,Y,\tau)\left(u_{i}+\frac{1}{a_{i}}\rho(Yv_{i})\right)$
$\displaystyle=\tau(u_{i})+\rho(Yu_{i})+\rho\left(\frac{1}{a_{i}}\rho(Yv_{i})\right)$
$\displaystyle=u_{i}+\rho(Yu_{i})+\frac{1}{a_{i}}Yv_{i}$
$\displaystyle=u_{i}+\rho(Yu_{i})+\frac{1}{a_{i}}\rho(Y(v_{i}+a_{i}u_{i}))$
$\displaystyle=u_{i}+\frac{1}{a_{i}}\rho(Yv_{i}).$

Since

$(\rho,Y,\tau)\left(u_{i}+\frac{1}{a_{i}}\rho(Yv_{i})\right)=\tau_{Y}\left(u_{i}+\frac{1}{a_{i}}\rho(Yv_{i})\right)$

we have $Y_{0}\left(u_{i}+\frac{1}{a_{i}}\rho(Yv_{i})\right)=0$. ∎

We conclude with a statement about the conjugacy classes and the fixed point
groups for involutions of $\operatorname{O}(\operatorname{q},k)$ in the
general case for $\operatorname{q}$ along with remarks on special cases for a
general quadratic space.

######  Proposition 6.5.

Let $(\rho_{1},Y_{1},\tau_{1})\in\operatorname{O}(\operatorname{q},k)$ be an
involution such that
$\tau_{1}\in\operatorname{O}(\operatorname{q}_{W_{1}^{\prime}},k)$ and
$\rho_{1}\in\operatorname{O}(\operatorname{q}_{\mathrm{rad}(W)},k)$, then
$(\rho_{1},Y_{1},\tau_{1})$ is
$\operatorname{O}(\operatorname{q},k)$-conjugate to
$(\rho_{2},Y_{2},\tau_{2})\in\operatorname{O}(\operatorname{q},k)$ if and only
if

  1. 1.

$\rho_{1}$ is conjugate to $\rho_{2}$ by
$\psi\in\operatorname{O}(\operatorname{q}_{\mathrm{rad}(W)},k)$

  2. 2.

$\tau_{1}$ is conjguate to $\tau_{2}$ by
$\mu\in\operatorname{Sp}(\mathrm{B}_{W_{1}^{\prime}},k)$

  3. 3.

there exists $Z:W_{1}^{\prime}\to\mathrm{rad}(W)$ such that
$\operatorname{q}(w)=\operatorname{q}(\mu(w))+\operatorname{q}(Zw)$ for all
$w\in W_{1}^{\prime}$ and

$\psi(\rho_{1}Z+Z\tau_{1}+Y_{1})\mu^{-1}=Y_{2}.$

###### Proof.

The list of properties in the proposition is equivalent to
$\mathcal{I}_{(\psi,Z,\mu)}((\rho_{1},Y_{1},\tau_{1}))=(\rho_{2},Y_{2},\tau_{2})$.
To see that we compute

$\begin{bmatrix}\psi&\psi Z\\\
0&\mu\end{bmatrix}\begin{bmatrix}\rho_{1}&\rho_{1}Y_{1}\\\
0&\tau_{1}\end{bmatrix}\begin{bmatrix}\psi^{-1}&Z\mu^{-1}\\\
0&\mu^{-1}\end{bmatrix}=\begin{bmatrix}\psi\rho_{1}\psi^{-1}&\psi(\rho_{1}Z+Z\tau_{1}+Y_{1})\mu^{-1}\\\
0&\mu\tau_{1}\mu^{-1}\end{bmatrix}.$

∎

######  Proposition 6.6.

Let $(\rho,Y,\tau)\in\operatorname{O}(\operatorname{q},k)$ be an involution
such that $\tau\in\operatorname{O}(\operatorname{q}_{W_{1}^{\prime}},k)$ and
$\rho\in\operatorname{O}(\operatorname{q}_{\mathrm{rad}(W)},k)$, then
$(\psi,D,\mu)\in\operatorname{O}(\operatorname{q},k)$ is fixed pointwise by
$\mathcal{I}_{(\rho,Y,\tau)}$ if and only if

  1. 1.

$\psi$ is fixed by conjugation by $\rho$ in
$\operatorname{O}(\operatorname{q}_{\mathrm{rad}(W)},k)$

  2. 2.

$\mu$ is fixed by conjugation by
$\tau\in\operatorname{O}(\operatorname{q}_{W_{1}^{\prime}},k)$

  3. 3.

there exists $Z:W_{1}^{\prime}\to\mathrm{rad}(W)$ such that
$\operatorname{q}(w)=\operatorname{q}(\mu(w))+\operatorname{q}(Zw)$ for all
$w\in W_{1}^{\prime}$ and

$Y+\psi^{-1}Y\mu=Z+\rho Z\tau$

###### Proof.

The list of properties above is equivalent to an element of the form
$(\psi,Z,\mu)\in\operatorname{O}(\operatorname{q},k)$ being fixed by
$\mathcal{I}_{(\rho,Y,\tau)}$. We can compute

$\begin{bmatrix}\rho&\rho Y\\\ 0&\tau\end{bmatrix}\begin{bmatrix}\psi&\psi
Z\\\ 0&\mu\end{bmatrix}\begin{bmatrix}\rho&\rho Y\\\
0&\tau\end{bmatrix}=\begin{bmatrix}\rho\psi\rho&\rho\psi Y\tau+(\rho\psi
Z+\rho Y\mu)\tau\\\ 0&\tau\mu\tau\end{bmatrix}.$

So we have $\rho\psi\rho=\psi$ and $\tau\mu\tau=\mu$ right away. If we
consider

$\displaystyle\psi Z$ $\displaystyle=\rho\psi Y\tau+(\rho\psi Z+\rho
Y\mu)\tau$ $\displaystyle=\rho\psi Y\tau+\rho\psi Z\tau+\rho Y\mu\tau$
$\displaystyle=\psi\rho Y\tau+\psi\rho Z\tau+\rho Y\tau\mu$
$\displaystyle=\psi Y+\psi\rho Z\tau+Y\mu$ $\displaystyle Z+\rho Z\tau$
$\displaystyle=Y+\psi^{-1}Y\mu$

we have the result, recalling that in order for $(\rho,Y,\tau)$ to be order
$2$ we need $Y=\rho Y\tau$. ∎

######  Remark 6.7.

When $\mathrm{rad}(W)$ is anisotropic
$\operatorname{O}(\operatorname{q}_{\mathrm{rad}(W)},k)=\\{\operatorname{id}\\}$
and $Z$ is unique to a given $\mu$.

When $\mathrm{rad}(W)$ is totally isotropic we have
$\operatorname{O}(\operatorname{q}_{\mathrm{rad}(W)},k)\cong\operatorname{GL}_{s}(k)$,
$Z\in\operatorname{Mat}_{s,2r}(k)$ and
$\mu\in\operatorname{O}(\operatorname{q}_{W_{1}^{\prime}},k)$.

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