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\title{Birational geometry of quadrics}
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\author{Burt Totaro}
\date{  }

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\hyphenation{Knebusch Grass-mann-ian}

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\newtheorem{theorem}{Theorem}[section]
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{conjecture}[theorem]{Conjecture}

A central method in the theory of quadratic forms is the study of
function fields of projective quadrics.
In particular, it is important to ask
when there is a rational map from
one quadric over a field to another. This suggests the problem
of determining when two quadrics
are birational, which turns out to be much harder. The answer
is known for quadratic forms of dimension at most 7 (thus
for projective quadrics of dimension at most 5),
by Ahmad-Ohm and Roussey among others \cite{AO, Roussey}.

In particular, there is a conjectural characterization of which quadratic
forms are ruled (meaning that the associated quadric is birational
to $\P^1$ times some variety), Conjecture \ref{ruledness}.
The conjecture was known for quadratic forms
of dimension at most 9
\cite{Totaroaff}. One result of this paper is to prove
Conjecture \ref{ruledness} for odd-dimensional forms of dimension
at most 17, and also for forms of dimension 10 or 14
(Theorem \ref{ruled14}). We use
in particular a new structure theorem on 14-dimensional forms (Theorem
\ref{dim14}),
generalizing Izhboldin's theorem on 10-dimensional forms. Vishik gave
an example of a 16-dimensional form with first Witt index 2
which is not divisible by a binary form; such a form should
be ruled by the conjecture, but it fell outside previously known
classes of ruled forms. Nonetheless, we give a new construction
which shows that Vishik's form is ruled (Theorem \ref{vishikruled}).

In this paper we also solve the problem of birational classification
for a significant class of quadratic forms, Pfister neighbors
of dimension at most 16 (Corollary \ref{pfister}). (Among these forms,
only the case of ``special Pfister neighbors'' was known
before, which fails to include all Pfister neighbors in dimensions
9, 10, and 11.)
We also give several generalizations for a broader class
of quadratic forms: neighbors of multiples of a Pfister form
(Theorems \ref{main} and \ref{birat}), strengthening
Roussey's theorem in this direction.
We also give the first construction of a birational
map between two non-isomorphic half-neighbors in section
\ref{sectranspo}. This is genuinely
different from any previously constructed birational map
between quadrics.

Thanks to Alexander Vishik for allowing me to include his example
of a 16-dimensional form with splitting pattern $(2,2,2,2)$
which is not divisible by a binary
form (Lemma \ref{binary}).


\section{Known low-dimensional results}
\label{known}

Throughout we work over fields of characteristic not 2.

We say that two varieties over a field $k$ are birational if they
are birational over $k$. In particular, a quadric (meaning a smooth
projective quadric hypersurface)
over $k$ is rational (birational to projective space over $k$)
if and only if it is {\it isotropic}, meaning that it has a $k$-rational
point. The proof is standard
(stereographic projection). So the nontrivial problems of
birational geometry for quadrics are concerned only with anisotropic quadrics,
and we often tacitly assume that the quadrics we consider
are anisotropic. In particular, the whole subject would be
vacuous over an algebraically closed field.

A quadratic form $q$ (always assumed nondegenerate) on an
$n$-dimensional vector space over $k$ determines a projective
quadric $Q=\{q=0\}$ of dimension $n-2$ in $\P^{n-1}$. The function
field $k(q)$ means the field $k(Q)$ of rational functions on $Q$.
We say that two
quadratic forms $q_1$ and $q_2$ are {\it similar }if $q_1$
is isomorphic to $aq_2$ for some $a$ in $k^*$. Two forms
are similar if and only
if the corresponding projective quadrics are isomorphic.
The group of {\it similarity factors }$G(q)$ is
the group of elements $a$ in $k^*$ with $q$ isomorphic to $aq$.
Every form can be written as a diagonal form $\langle a_1,\ldots,a_n\rangle
=a_1x_1^2+\cdots +a_nx_n^2$ for some $a_i\in k^*$.
The {\it discriminant }$\dets q$
of a quadratic form $q=\langle a_1,\ldots,a_n\rangle$
is $(-1)^{n(n-1)/2} a_1\cdots a_n$ in $(k^*)/(k^*)^2=H^1(k,\F_2)$.

We recall the definition of the splitting pattern of a quadratic
form $q$. Let $k_0=k$, $q_0=q\an$ (the anisotropic part of $q$, well-defined
up to isomorphism), and then inductively define $k_{j}=k_{j-1}(q_{j-1})$
and $q_{j}=((q_{j-1})_{k_{j}})\an$. We stop when $q_h$ has dimension at most 1.
Then the {\it splitting pattern }${\bf i}(q)$ is the sequence
$(i_1,\ldots,i_h)$ where $i_j$ is the Witt index $i_W((q_{j-1})_{k_j})$.
The {\it splitting tower }of $q$ is the sequence of fields $k_0,\ldots,k_h$.
For the associated projective quadric $Q$, we write $i_j(Q)=i_j(q)$.

We say that two varieties over a field $k$, possibly
of different dimensions, are {\it stably birational }if their products
with some projective spaces over $k$ are birational. For
quadrics $X$ and $Y$, a simple argument shows that
stable birational equivalence is equivalent to the apparently
weaker condition that $X$ is isotropic over the function field
of $Y$ and vice versa, or equivalently that there are
rational maps from $X$ to $Y$ and from $Y$ to $X$ over $k$
\cite[Theorem X.4.25]{Lam}. We say that two quadratic forms over $k$
are birational if the associated projective quadrics
are birational.

By Karpenko and Merkurjev,
we know exactly when an anisotropic quadric $X$ is stably birational to
a variety of lower dimension; this happens if and only if the first
Witt index $i_1(X)$ is greater than 1 \cite[Theorem 3.1]{KM}.
(For the quadric $X$ associated to an anisotropic quadratic form $q$,
the first Witt index $i_1(X)$ is the maximal dimension of an isotropic
linear subspace for $q$ over the function field $k(X)$.)
Every quadric
$X$ is stably birational to any subquadric $Y$ of codimension at most
$i_1(X)-1$. This suggests the following conjecture, which
would imply in particular that an anisotropic quadric is ruled
(birational to $Y\times \P^1$ for some variety $Y$ over $k$) if and only
if its first Witt index is greater than 1.

\begin{conjecture}
\label{ruledness}
(Ruledness conjecture) Let $X$ be an anisotropic quadric
over a field. Then $X$ is birational to $Y\times \P^{i_1(X)-1}$ for some
subquadric $Y$ of codimension $i_1(X)-1$.
\end{conjecture}

Before this paper, Conjecture \ref{ruledness} was known
for quadratic forms of dimension at most 9 \cite[Lemma 3.3]{Totaroaff}.
A natural companion to Conjecture \ref{ruledness} is the
quadratic Zariski problem:

\begin{conjecture}
\label{qzp}
(Quadratic Zariski problem) If two quadrics of the same
dimension over a field are stably birational, then
they are birational.
\end{conjecture}

Conjecture \ref{qzp} is known for quadratic forms of dimension at most 7,
as we will discuss. A general reference on Conjecture \ref{qzp}
is \cite[Theorem 6.1]{Hoffmannhalf}.

Here are the known results about birational geometry of
low-dimensional quadrics in more detail; throughout we consider
only anisotropic quadrics over a field. Two anisotropic conics
are birational if and only they are isomorphic. Likewise,
two anisotropic quadric surfaces are birational if and only if
they are isomorphic, by Wadsworth
\cite[Theorem XII.2.2]{Lam}. This hides a special phenomenon, however:
a Pfister quadric surface is birational to $\P^1$ times
any conic contained in it, by Knebusch \cite[pp.~73--74]{Knebusch1}.
(For example: the quadric surface
$x_0^2-ax_1^2-bx_2^2+abx_3^2=0$ is birational to
$\P^1$ times the conic $x_0^2-ax_1^2-bx_2^2=0$, for any $a,b$
in $k^*$.)
A non-Pfister quadric surface $X$
(there are many equivalent conditions: $X$ has first Witt index 1,
or Picard number 1, or nontrivial discriminant)
is not stably birational to any curve. For the definitions of Pfister
forms and Pfister neighbors, see section \ref{neighbors}.

We state Hoffmann's bound on the first Witt index; several proofs
are given in the book
by Elman-Karpenko-Merkurjev \cite[Theorem 22.5, Corollary 26.6]{EKM}.
The bound is optimal, because equality holds for Pfister neighbors.

\begin{lemma}
\label{hoff}
An anisotropic quadratic form of dimension $2^r+s$, for $1\leq s\leq 2^r$,
has first Witt index at most $s$.
\end{lemma}

All 5-dimensional anisotropic forms (corresponding to projective
quadrics of dimension 3) have first Witt index 1
by Lemma \ref{hoff},
and so they are not stably birational
to any lower-dimensional variety. Two non-isomorphic
5-dimensional anisotropic forms
are stably birational if and only if they are
both neighbors of the same Pfister 3-form, by Hoffmann
\cite[main theorem]{Hoffmann5}. In that case, the two quadrics
are in fact birational, by Ahmad-Ohm \cite[Corollary 1.6,
Theorem 2.5]{AO}. Lam's textbook explains Ahmad-Ohm's
birational map between the quadrics $\langle 1,a,b,ab,c\rangle$
and $\langle 1,a,c,ac,b\rangle$ \cite[Theorem XII.2.15]{Lam}.
The map was generalized by Roussey \cite[Proposition 7.2.1]{Roussey},
and generalized further in this paper (Lemma \ref{transpo}).

The same thing happens in dimension 6: two 6-dimensional anisotropic
forms which are not isomorphic are stably birational
if and only if both are neighbors of the same
3-fold Pfister form. The proof for 6-dimensional forms with trivial
discriminant (Albert forms) uses Merkurjev's index reduction formula
\cite[Theorem 30.5]{EKM}, \cite[Theorem III.4.8, Corollary XII.2.13]{Lam}.
The proof for
nontrivial discriminant is by Laghribi \cite[Theorem 1.4(2)]{Laghribi6}.
Two 6-dimensional forms which are neighbors
of the same 3-fold Pfister form
are birational, again by Ahmad-Ohm \cite{AO}.
Also, a Pfister neighbor of dimension 6 is birational to $\P^1$
times (the quadric associated to) any 5-dimensional subform \cite{AO}.

In dimension 7, Karpenko (following Izhboldin) determined
when two non-similar forms $\alpha$ and $\beta$
are stably birational \cite[Theorem 3.1, Corollary 3.4]{KarpenkoIzh}.
This happens exactly when the 8-dimensional form
$\alpha\perp\langle \dets (\alpha )\rangle$ is divisible
by a binary form and is similar to
$\beta\perp\langle \dets(\beta)\rangle$.
When that happens, Roussey proved that $\alpha$ and $\beta$
are birational \cite[Theorem 7.3.1]{Roussey}. Theorems
\ref{main} and \ref{birat} below generalize his result.

For quadratic forms of dimension 8 or more, results on birational
(as opposed to stably birational) equivalence
are much less complete. In particular,
two 8-dimensional forms $\sigma$ and $\tau$ which are ``half-neighbors''
($a\sigma\perp b\tau$ is a Pfister form for some nonzero $a$ and $b$)
are stably birational \cite[Corollary 2.6]{Hoffmannhalf}.
But it has not been known whether such forms
(if non-similar) are birational, even in a single example.
We give the first known construction of birational maps
between non-isomorphic half-neighbors in section \ref{sectranspo}.

All the quadrics in the previously known
constructions of birational maps between quadrics, by Ahmad-Ohm and Roussey,
are ``neighbors'' of multiples of Pfister forms, as we have seen
in low dimensions. We define these neighbors in the next section.


\section{Neighbors}
\label{neighbors}

In this section we define neighbors of multiples of Pfister forms
and the previously known results on their birational
classification.

The most important quadratic forms over any field are the Pfister forms,
whose definition we now recall.
First, for a nonzero element $a$ of a field $k$, the 1-fold
Pfister form $\l a \r$ means the 2-dimensional quadratic
form $\langle 1,-a\rangle$, that is, $x_1^2-ax_2^2$.
For nonzero elements $a_1,\ldots,a_n$ of $k$, the $n$-fold
{\it Pfister form }$\l a_1,\ldots, a_n\r$ means the tensor product
$\l a_1\r\otimes \cdots\otimes \l a_n\r$, a quadratic form
of dimension $2^n$.

A {\it neighbor }of a Pfister form $\rho$ means a form $\varphi$ which
is similar to a subform of $\rho$
of dimension greater than half the dimension of $\rho$. We can think
of Pfister neighbors as a class of the most special quadratic forms.
An anisotropic
quadratic form is a Pfister neighbor if and only if it is stably
birational to a Pfister form (meaning that the corresponding projective
quadrics are stably birational),
by Hoffmann \cite[Theorem 22.5, Corollary 26.6]{EKM}.
Moreover, a Pfister neighbor
$\varphi$ (as a quadratic form) uniquely determines the corresponding
Pfister form $\rho$ \cite[Remark 23.12]{EKM}.
On the other hand, it is not known whether two Pfister neighbors
which are stably birational and have the same dimension must
be birational. Until this paper, only the following case was known.
By definition, a {\it special Pfister neighbor }$\varphi$
is a quadratic form
that is similar to a form $P_0\perp -aP_1$ where $P_0$ is a
Pfister form, $P_1$ is a nonzero subform of $P_0$, and $a$ is
a nonzero element of $k$. Clearly $\varphi$ is a neighbor
of the Pfister form $P_0\otimes\l a \r$. The point of the notion
is that, by Ahmad-Ohm, any two special Pfister neighbors which
are stably birational (equivalently, which are neighbors of
the same Pfister form) and which have the same dimension
must be birational \cite[Theorem 1.6]{AO}. Their maps
are a special case of the transposition maps defined
in Lemma \ref{transpo}.

Ahmad-Ohm also showed that
every neighbor of codimension at most $n$ of an $n$-fold Pfister form
is special, whereas (over suitable fields) 
neighbors of codimension at least 5 in a 4-fold Pfister form
need not be special
\cite[2.5, 2.8]{AO}.
As a result, all quadratic forms which are
Pfister neighbors of dimension at most 8
are special (this explains several of the low-dimensional
results in section \ref{known}), whereas those of dimension 9 or 10 or 11
need not be special. Thus Pfister neighbors of dimension 9 or 10 or 11
were basic cases where the
quadratic Zariski problem remained open. But now Theorem \ref{main}
solves the ruledness and quadratic Zariski problems,
Conjectures \ref{ruledness} and \ref{qzp},
for all neighbors of an $r$-fold Pfister form
when $r$ is at most 4.

Now we consider a more general class of ``neighbors'' studied
by Ahmad-Ohm and Roussey.
First, define a {\it multiple }of an $r$-fold Pfister form $\rho$ with
$r$ at least 1
to be the tensor product of $\rho$ with some quadratic form;
thus we can write a multiple of $\rho$
as $\langle b_1,\ldots,b_n\rangle\otimes \rho
=b_1\rho \perp\cdots\perp b_n\rho$ for
some nonzero elements $b_i$
of the field $k$. Define a {\it neighbor }$\sigma$
of a multiple of a Pfister form
$\rho$ to be a nondegenerate
form which is similar to a subform of some multiple $\varphi\otimes
\rho$ of $\rho$ such that $\varphi$ has dimension at least 2
and $\dim(\varphi\otimes\rho)-\dim(\sigma)<\dim(\rho)$. The point of this
notion is that a neighbor $\sigma$ of a multiple $\varphi\otimes \rho$
of a
Pfister form $\rho$ is stably birational to $\varphi\otimes\rho$,
since the first Witt index of $\varphi\otimes\rho$ (if it
is anisotropic) is a multiple of
the dimension of $\rho$. (If $\varphi\otimes\rho$ is isotropic,
then again $\sigma$ is stably birational to $\varphi\otimes\rho$,
since both quadrics are rational: either $\rho$ is hyperbolic,
in which case $\varphi\otimes\rho$ is hyperbolic and so
$\sigma$ is isotropic, or $\rho$ is anisotropic, in which case
the Witt index of $\varphi\otimes\rho$ is a multiple of the
dimension of $\rho$, and hence $\sigma$ is isotropic.)
The standard proof of these results uses function
fields \cite[Corollary 23.7]{EKM}. There is also an elementary
proof by Wadsworth-Shapiro \cite[Theorem 2(c)]{WS}.

As in the case of Pfister neighbors, we can ask whether two neighbors
of the same multiple of a Pfister form which have the same dimension
must be birational. The previously known positive results were restricted
to a special case.
Define a {\it special neighbor }of a multiple of a Pfister form $\rho$
to be a neighbor of the product of $\rho$ with some $d$-dimensional
form that contains the product of $\rho$ with
some $(d-1)$-dimensional form. (Notice that $\rho$ is part of
the data in this definition. We are calling a ``special neighbor''
what Roussey called a ``very special neighbor'', for brevity
\cite[Definition 7.1.3]{Roussey}.)
Roussey's main result is the following.
He used this to complete the solution of the quadratic Zariski
problem for all quadratic forms of dimension 7, as discussed in section
\ref{known}.

\begin{theorem} (Roussey \cite[Theorem 7.3.1]{Roussey})
\label{atmost3}
Let $q_1$ and $q_2$ be 
special neighbors of the same multiple of an $r$-fold
Pfister form $\rho$ with $r\leq 3$. If $q_1$ and $q_2$
have the same dimension, then they are birational.
\end{theorem}

We have two generalizations of Theorem \ref{atmost3}. First,
we can remove the restriction to $r$ at most 3 (Theorem \ref{birat}).
Also, for $r$ at most 3, Theorem \ref{main} proves the same result
on birational classification
for arbitrary neighbors of a multiple of a Pfister form,
not just special neighbors.


\section{Construction of birational maps between neighbors}

We now solve the problem of birational classification
for all neighbors of an $r$-fold Pfister form
with $r$ at most 4. We also determine exactly which neighbors of
an $r$-fold Pfister form with $r$ at most 4
are ruled, thus proving the ruledness conjecture
(Conjecture \ref{ruledness})
in this case.
Only the case of special Pfister neighbors
had been known before, by Ahmad-Ohm in 1995 \cite{AO}.
Also, the main result works for arbitrary neighbors of multiples
of an $r$-fold Pfister form with $r$ at most 3. (Note that
any neighbor of a 4-fold Pfister form can be viewed as a neighbor
of a multiple of a 3-fold Pfister form.)

\begin{theorem}
\label{main}
Any two neighbors of a given multiple of a $r$-fold Pfister form
with $r$ at most 3 which have
the same dimension are birational.

Moreover, let $q$ be any neighbor of a multiple $\varphi\otimes\rho$
of an $r$-fold Pfister form $\rho$ with $r$ at most 3. Then $q$ is birational
to the product of projective space of dimension
$\dim(q)-\dim(\varphi\otimes\rho)+\dim(\rho)$ with
a neighbor of $\varphi\otimes\rho$. In particular,
$q$ is ruled if it has codimension at most $\dim(\rho)-2$
in $\varphi\otimes\rho$.
\end{theorem}

\begin{corollary}
\label{pfister}
The ruledness and quadratic Zariski problems, Conjectures
\ref{ruledness} and \ref{qzp},
have a positive solution for neighbors of an $r$-fold
Pfister form with $r$ at most 4. More precisely,
if $\sigma$ is a neighbor of an $r$-fold Pfister form
with $r$ at most 4, and $\tau$ is any form
of the same dimension which is stably birational to $\sigma$,
then $\tau$ is birational to $\sigma$. Moreover,
if a quadratic form $\sigma$ is a Pfister neighbor of dimension $2^r+s$,
$1\leq s\leq 2^r$, $r\leq 3$, then $\sigma$ is birational to the
product of projective space of dimension $s-1$ with
a subquadric of $\sigma$; moreover $s$ here is equal
to the first Witt index $i_1(\sigma)$.
\end{corollary}

{\bf Proof of Corollary \ref{pfister}. }If $\sigma$
is a neighbor of a Pfister form $\rho$
and $\tau$ is any form stably birational
to $\sigma$, then $\tau$ is also a neighbor of $\rho$,
by Hoffmann \cite[Theorem 22.5, Corollary 26.6]{EKM}.
By Theorem \ref{main}, if $\sigma$ has the same dimension
as $\tau$, then they are birational. (Notice that, to apply
the theorem, we view $\rho=\l a_1,\ldots,a_r\r$ for $r\geq 1$
as a multiple of the Pfister form $\l a_1,\ldots,a_{r-1}\r$,
since the definition of
a ``neighbor of a multiple $\varphi\otimes \rho$'' requires
$\varphi$ to have dimension at least 2.) The ruledness statement
is immediate from Theorem \ref{main}. Finally,
for an anisotropic Pfister neighbor
$\sigma$ of dimension $2^r+s$, $1\leq s\leq 2^r$, it is immediate
from Hoffmann's bound on the first Witt index, Lemma \ref{hoff},
that $i_1(\sigma)$ is equal
to $s$. \qed

A special case of Corollary \ref{pfister} is that every Pfister
neighbor of codimension at most 7 satisfies the ruledness
and quadratic Zariski problems, Conjectures \ref{ruledness}
and \ref{qzp}. Indeed, these
conjectures hold for all neighbors of an $r$-fold Pfister
form with $r\leq 3$. A neighbor of codimension at most 7 of an $r$-fold
Pfister form with $r\geq 4$, on the other hand, can be viewed
as a neighbor of a multiple of a 3-fold Pfister form, and
so again Corollary \ref{pfister} proves the results we want.
Thus these conjectures hold for all Pfister neighbors
of dimension at most 16, as well as for those of dimensions from 25 to 32,
and so on. In very high dimensions, however, Corollary \ref{pfister}
gives less about Pfister neighbors than
Ahmad-Ohm's result that a neighbor of codimension at most $r$
in an $r$-fold Pfister form is special and therefore satisfies
the two conjectures \cite[Corollary 2.5]{AO}.

{\bf Proof of Theorem \ref{main}. }For $k$ finite, every anisotropic
quadratic form has dimension at most 2 and the theorem is trivially
true. So we can assume that $k$ is infinite.

Write the given multiple
of the Pfister form $\rho$ as $\langle a_1,\ldots,a_n\rangle\otimes\rho$,
where $n\geq 2$. 
Since $r$ is at most 3, the $r$-fold Pfister form $\rho$
admits a bilinear ``multiplication''
$xy$ on the vector space of $\rho$
such that $\rho(xy)=\rho(x)\rho(y)$ \cite[Theorem X.2.11]{Lam},
and for generic $x$ the map $y\mapsto xy$ is invertible.
Ahmad-Ohm introduced a rational map $F$
from the quadric $\langle a_1,\ldots,a_n\rangle\otimes\rho$
to the quadric $\langle a_1,\ldots,a_{n-1}\rangle\otimes\rho
\perp \langle a_n\rangle$, defined by
$$[x_1,\ldots,x_{n-1},y]\mapsto [yx_1,\ldots,yx_{n-1},\rho(y)].$$
\cite[after 1.3]{AO}. The map $F$
is dominant, and its general fibers are linear spaces.
Therefore the first quadric is birational to the product
of projective space of dimension $\dim(\rho)-1$ with the second quadric.
(The latter statement is true for $r$-fold Pfister forms for any $r$.
The advantage of $r\leq 3$ is that the general fibers are linear
subspaces in projective space, not just isomorphic to projective
spaces.)

Let $q$ be any neighbor of $\varphi\otimes\rho=\langle a_1,\ldots,a_n\rangle
\otimes\rho$. That is, after scaling, $q$ is a nondegenerate subform
of codimension less than $\dim(\rho)$ in $\varphi\otimes\rho$. The
projective quadric $Q$ corresponding to $q$ is the intersection of
the quadric corresponding to $\varphi\otimes\rho$ with a linear
subspace. Suppose that the rational map $F$ is defined at some point
of $Q$, and that the derivative
of $F$ restricted to $Q$ is surjective at that point.
Then the restriction of $F$
to $Q$ is dominant, and its general fibers are still linear
subspaces. That will imply that $q$ is birational to the product
of projective space of the appropriate dimension with the quadric
$\langle a_1,\ldots,a_{n-1}\rangle \otimes \rho\perp \langle
a_n\rangle$. Since the latter quadric does not depend on the form $q$,
that will complete the proof.

It remains to show that for any nondegenerate
subform $q$ of $\varphi\otimes\rho$ with codimension
less than $\dim(\rho)$,
$q$ can be embedded in $\varphi\otimes\rho$ in such a way that the
rational map $F$ is defined at some point of the corresponding
quadric $Q$ and that $F$ restricted
to $Q$ has surjective derivative at that point.

Consider the Grassmannian
of all linear subspaces of $\varphi\otimes\rho$ of dimension
equal to $\dim(q)$. An open dense subset of this Grassmannian
corresponds to subspaces whose intersection with the quadric
$\varphi\otimes\rho=0$ is smooth, meets the open set where
$F$ is defined, and such that $F$ restricted to this intersection
has surjective derivative at some point. (Here we use the assumption
that $\dim(q)$ is at least the dimension of the form
$\langle a_1,\ldots,a_{n-1}\rangle\otimes \rho\perp\langle a_n\rangle$.)
Then the desired conclusion follows from Lemma \ref{general}. \qed

\begin{lemma}
\label{general}
Let $\alpha$ be any nondegenerate subform of a nondegenerate
quadratic form $\beta$ over an infinite field $k$.
Let $U$ be a nonempty open subset
of the Grassmannian of linear subspaces of the vector
space $\beta$ of dimension $\dim(\alpha)$.
Then there is an isometry of $\beta$ which moves
$\alpha$ to a subspace that
corresponds to a point of $U$.
\end{lemma}

{\bf Proof. }By Witt's cancellation theorem, the orthogonal
group $O(\beta)$ over the algebraic closure of $k$
acts transitively on the open dense subset of the Grassmannian
corresponding to subspaces whose intersection with the quadric
is smooth (equivalently, subspaces on which the quadratic
form is nondegenerate). The group of $k$-points
$O(\beta)(k)$ is Zariski dense in the whole
orthogonal group, using that $k$ is infinite and the special
orthogonal group is connected \cite[Corollary 18.3]{Borel}. Therefore, there
is an element of $O(\beta)(k)$ which moves the given
subspace $\alpha\subset \beta$ into the given open set $U$. \qed


\section{The ruledness conjecture and the structure of
14-dimensional forms}
\label{ruledsect}

If a quadric is ruled, then Karpenko showed that its first Witt index
is greater than 1, and we conjecture the converse.
More precisely, Conjecture \ref{ruledness} predicts
that any quadric $X$ is birational to the product
of some subquadric with projective space of dimension $i_1(X)-1$.
Using the results of this paper, we prove this conjecture
in a larger range of dimensions than before, as follows.

\begin{theorem}
\label{ruled14}
The ruledness conjecture, Conjecture \ref{ruledness},
holds for odd-dimensional quadratic forms of dimension at most 17,
and also for even-dimensional forms of dimension at most 10 or equal
to 14.
\end{theorem}

{\bf Proof. }Throughout 
the proof, we consider only anisotropic forms.
In dimensions at most 9, the ruledness conjecture was known
\cite{Totaroaff}.

There is an appealing dichotomy for 10-dimensional forms:
if $i_1(q)>1$, then $q$ is either a Pfister neighbor or divisible by a binary
form, by Izhboldin \cite[proof of Conjecture 0.10]{IzhboldinAnn}.
We know that $i_1(q)=2$ by Hoffmann's bound for $i_1$, Lemma \ref{hoff}.
If $q$
is divisible by a binary form, then $q$ is ruled 
by Ahmad-Ohm (see Lemma \ref{ruled}). If $q$ is a Pfister neighbor,
then Corollary \ref{pfister} shows the new result that $q$ is ruled,
even though $q$ need not be a special Pfister neighbor.

The ruledness conjecture for 12-dimensional forms $q$ remains open.
In what follows we use Karpenko's determination of all possible 
values of the first Witt index for anisotropic forms of a given
dimension, although the cases we need were known earlier
\cite[Theorem 79.9]{EKM}. For a 12-dimensional form,
if the first Witt index is greater than 1, then
it is either 2 or 4. If $i_1(q)=4$, then
$q$ is a Pfister neighbor by Kahn \cite[Theorem 4]{Kahn}, simplified
by Izhboldin \cite[Corollary 1.6]{Izhboldinmax}.
So Conjecture \ref{ruledness} on ruledness
holds by Corollary \ref{pfister}. That is, the corresponding
projective quadric is birational to $\P^3$ times a 7-dimensional
quadric. If $i_1(q)=2$,
then the ruledness conjecture would follow from Vishik's
conjecture that such a form $q$ must be divisible by a binary
form \cite[p.~95]{Vishik}.

We now prove Conjecture \ref{ruledness} on ruledness for forms of odd dimension
at most 17. For forms $q$ of dimension 11 or 13, having first Witt
index greater than 1 implies that $q$ is a Pfister neighbor
by Kahn \cite[Theorem 4]{Kahn},
simplified by Izhboldin \cite[Corollary 1.6]{Izhboldinmax}.
Thus we know the ruledness conjecture
by Corollary \ref{pfister}. Next, a 15-dimensional form
with $i_1>1$ has $i_1=7$ or $i_1=3$. If $i_1=7$,
then $q$ is a Pfister neighbor by Knebusch
and the ruledness conjecture holds.
If $i_1=3$, then Vishik showed
that $q\perp \langle \dets(q)\rangle$
is a multiple of a 2-fold Pfister form \cite[p.~79]{Vishik}.
Therefore $q$ is a neighbor
of the type covered by Theorem \ref{main}, and so the ruledness
conjecture holds. Finally, for $q$ of dimension 17,
we have $i_1=1$ by Hoffmann's bound, Lemma \ref{hoff}, and so the ruledness
conjecture is vacuously true.

Finally, Conjecture \ref{ruledness} on ruledness for dimension 14
follows from Theorem \ref{dim14} below, in view of Theorem
\ref{main}. \qed (Theorem \ref{ruled14}).

The following result is a satisfying extension of Izhboldin's theorem 
on 10-dimensional forms to dimension 14.
The proof uses Vishik's results and methods,
but they have to be combined in an elaborate way.

\begin{theorem}
\label{dim14}
Let $q$ be a 14-dimensional anisotropic quadratic form with
first Witt index greater than 1. Then $q$ is either

(1) a Pfister neighbor, and $i_1(q)=6$,

(2) divisible by a binary form, and $i_1(q)=2$, or

(3) a subform of a 16-dimensional multiple of a 2-fold Pfister
form, and $i_1(q)=2$.
\end{theorem}

Vishik has shown that Theorem
\ref{dim14} does not extend to dimension 16: there is a 16-dimensional
form with first Witt index 2 which is not divisible by a binary form
and hence, for dimension reasons, is not a neighbor of any multiple
of a Pfister form (Theorem \ref{binary}).
Nonetheless, we will prove the ruledness conjecture
for Vishik's example in Theorem \ref{vishikruled}.

Although cases (2) and (3) in Theorem \ref{dim14} overlap,
neither can be omitted. In particular, over a rational
function field $k=k_0(a,b,c,d,e,f)$, the 14-dimensional
form $\l a,b\r \otimes \langle c,d,e,f\rangle
-\langle c,d\rangle$ satisfies (3). But it is not divisible
by a binary form, because it has Witt index 3 over the
extension field $k(\sqrt{-cd})$, with anisotropic part
the 8-dimensional form $\l a,b\r\otimes \langle e,f\rangle$.
(A multiple of a binary form is either hyperbolic or
has even Witt index, by the results mentioned in section \ref{neighbors}.)

{\bf Proof of Theorem \ref{dim14}. }For an anisotropic form
of dimension 14, Karpenko's theorem gives that the first
Witt index is 1, 2, or 6 \cite[Theorem 79.9]{EKM}.
If $i_1(q)=6$, then $q$ is a Pfister neighbor by Knebusch
\cite[Corollary 8.2]{Knebusch2} and we are done.
So we can assume that $i_1(q)=2$.

Next, we will show that $q$ must have splitting pattern
either $(2,2,2,1)$ or $(2,1,1,2,1)$. We use Chow groups
as in \cite{EKM} to prove this,
although the argument could also be formulated in terms of
motivic decompositions as in \cite{Vishik};
the equivalence between the two languages is described
in \cite[Theorem 94.1]{EKM}. Let $Q$ be the projective
quadric of dimension 12
associated to $q$. For any projective quadric, write $D=\dim Q$ (in the current
example, $D=12$) and $d=\lfloor D/2\rfloor$. 
A basis for the Chow groups modulo 2
of $Q^2$ over the algebraic closure $\kbar$ is given by the products
of powers $h^i$ and classes $l_i$ for $0\leq i\leq d$,
where $h$ denotes the hyperplane class
and $l_i$ denotes a linear space of dimension $i$ on $X_{\kbar}$.
A basic discrete invariant of a quadratic form
is the image of the group of algebraic cycles on $Q^2$ (defined over $k$)
in the Chow groups modulo 2 of $Q^2_{\kbar}$, viewed as a vector
space over $\F_2$. Here the cycles
$h^i\times h^j$ are always defined over $k$, and so we simply remove
them from any cycle we consider.
The remaining basis elements of the middle dimension and above
for $Q^2_{\kbar}$ are pictured below. 
\begin{figure}[h] \label{bigtriangles}
\hspace{2cm}
\begin{picture}(300,80)
\put(28.5,10){\circle{5}}
\put(48.5,10){\circle{5}}
\put(68.5,10){\circle{5}}
\put(88.5,10){\circle{5}}
\put(108.5,10){\circle{5}}
\put(128.5,10){\circle{5}}
\put(148.5,10){\circle{5}}
\put(168.5,10){\circle{5}}
\put(188.5,10){\circle{5}}
\put(208.5,10){\circle{5}}
\put(228.5,10){\circle{5}}
\put(248.5,10){\circle{5}}
\put(268.5,10){\circle{5}}
\put(288.5,10){\circle{5}}
\put(38.5,20){\circle{5}}
\put(58.5,20){\circle{5}}
\put(78.5,20){\circle{5}}
\put(98.5,20){\circle{5}}
\put(118.5,20){\circle{5}}
\put(138.5,20){\circle{5}}
\put(178.5,20){\circle{5}}
\put(198.5,20){\circle{5}}
\put(218.5,20){\circle{5}}
\put(238.5,20){\circle{5}}
\put(258.5,20){\circle{5}}
\put(278.5,20){\circle{5}}
\put(48.5,30){\circle{5}}
\put(68.5,30){\circle{5}}
\put(88.5,30){\circle{5}}
\put(108.5,30){\circle{5}}
\put(128.5,30){\circle{5}}
\put(188.5,30){\circle{5}}
\put(208.5,30){\circle{5}}
\put(228.5,30){\circle{5}}
\put(248.5,30){\circle{5}}
\put(268.5,30){\circle{5}}
\put(58.5,40){\circle{5}}
\put(78.5,40){\circle{5}}
\put(98.5,40){\circle{5}}
\put(118.5,40){\circle{5}}
\put(198.5,40){\circle{5}}
\put(218.5,40){\circle{5}}
\put(238.5,40){\circle{5}}
\put(258.5,40){\circle{5}}
\put(68.5,50){\circle{5}}
\put(88.5,50){\circle{5}}
\put(108.5,50){\circle{5}}
\put(208.5,50){\circle{5}}
\put(228.5,50){\circle{5}}
\put(248.5,50){\circle{5}}
\put(78.5,60){\circle{5}}
\put(98.5,60){\circle{5}}
\put(218.5,60){\circle{5}}
\put(238.5,60){\circle{5}}
\put(88.5,70){\circle{5}}
\put(228.5,70){\circle{5}}
\end{picture}
% \caption[]{$T_{2,r,n-r}$ Dynkin diagram}
\end{figure}
The rows represent the cycles
of a given dimension, with the bottom representing middle-dimensional
cycles. The $i$th row from the bottom starts with the classes
$h^0\times l_i,\ldots,h^{d-i}\times l_d$, and then come the classes
$l_d\times h^{d-i},\ldots,l_i\times h^0$. For $D$ even, the middle-dimensional
basis element $l_d\times l_d$ does not occur in the decomposition
of any cycle defined over $k$ for $Q$ not hyperbolic \cite[Lemma 73.2]{EKM},
and so is not pictured.

By \cite[Lemma 73.13]{EKM}, all cycles
of the middle dimension or above on $Q^2$, viewed over $\kbar$,
can only involve basis elements
which lie in a sequence of triangles determined by the splitting pattern.
For example, if $q$ had splitting pattern $(2,1,3,1)$,
these ``shell
triangles'' would be:
\begin{figure}[ht] \label{shelltriangles}
\hspace{2cm}
\begin{picture}(300,40)
\put(28.5,10){\circle{5}}
\put(48.5,10){\circle{5}}
\put(68.5,10){\circle{5}}
\put(88.5,10){\circle{5}}
\put(108.5,10){\circle{5}}
\put(128.5,10){\circle{5}}
\put(148.5,10){\circle{5}}
\put(168.5,10){\circle{5}}
\put(188.5,10){\circle{5}}
\put(208.5,10){\circle{5}}
\put(228.5,10){\circle{5}}
\put(248.5,10){\circle{5}}
\put(268.5,10){\circle{5}}
\put(288.5,10){\circle{5}}
\put(38.5,20){\circle{5}}
\put(98.5,20){\circle{5}}
\put(118.5,20){\circle{5}}
\put(198.5,20){\circle{5}}
\put(218.5,20){\circle{5}}
\put(278.5,20){\circle{5}}
\put(108.5,30){\circle{5}}
\put(208.5,30){\circle{5}}
\end{picture}
% \caption[]{$T_{2,r,n-r}$ Dynkin diagram}
\end{figure}

By \cite[Lemma 73.20]{EKM}, since $i_1(Q)=2$, there is a 13-dimensional
cycle $\alpha$
on $Q^2$ whose class over $\kbar$ contains $h^0\times l_1$,
the top of the first shell triangle. By
\cite[Corollary 80.14]{EKM} (on ``outer excellent connections''),
it follows that $\alpha$ also contains $l_5\times h^4$, shown
as a black dot in the picture.
\begin{figure}[ht] \label{outer}
\hspace{2cm}
\begin{picture}(300,30)
\put(28.5,10){\circle{5}}
\put(48.5,10){\circle{5}}
\put(68.5,10){\circle{5}}
\put(88.5,10){\circle{5}}
\put(108.5,10){\circle{5}}
\put(128.5,10){\circle{5}}
\put(148.5,10){\circle{5}}
\put(168.5,10){\circle{5}}
\put(188.5,10){\circle{5}}
\put(208.5,10){\circle{5}}
\put(228.5,10){\circle{5}}
\put(248.5,10){\circle{5}}
\put(268.5,10){\circle{5}}
\put(288.5,10){\circle{5}}
\put(38.5,20){\circle*{5}}
\put(58.5,20){\circle{5}}
\put(78.5,20){\circle{5}}
\put(98.5,20){\circle{5}}
\put(118.5,20){\circle{5}}
\put(138.5,20){\circle{5}}
\put(178.5,20){\circle{5}}
\put(198.5,20){\circle*{5}}
\put(218.5,20){\circle{5}}
\put(238.5,20){\circle{5}}
\put(258.5,20){\circle{5}}
\put(278.5,20){\circle{5}}
\end{picture}
% \caption[]{$T_{2,r,n-r}$ Dynkin diagram}
\end{figure}
(This theorem is
the deepest ingredient of the proof, in that it uses
the Steenrod operations on Chow groups modulo 2.) By
the above statement on shell triangles, it follows
that $i_h(q)$ is not 2 and that if $i_h(q)$ is 1, then $i_{h-1}(q)$
is at least 2. Here $h$ denotes the height of $q$, meaning the length
of the splitting pattern $(i_1,\ldots,i_h)$.

We will use systematically what is known about low-dimensional forms.
In particular, the possible splitting patterns
in dimension 10 are $(2,2,1)$, $(2,1,2)$, $(2,1,1,1)$, $(1,2,2)$,
$(1,1,2,1)$, $(1,1,1,2)$, and $(1,1,1,1,1)$ \cite{Vishik}.
The splitting pattern
of $q$ must be 2 followed by one of these sequences, and the previous
paragraph leaves only two possibilities, $(2,2,2,1)$ or $(2,1,1,2,1)$.

An anisotropic form $q$ of dimension congruent to 2 modulo 4
is divisible by a binary form if and only if it has splitting
pattern $(\text{even},\ldots,\text{even},1)$, by Knebusch's results
\cite[Theorem 7.1, Theorem 7.5]{Vishik}. More precisely, in that
case, $q$ is divisible by $\l \dets q\r$. Thus, if our form
$q$ has splitting pattern $(2,2,2,1)$, then it is divisible
by a binary form, as we want.

It remains to show that if $q$ has splitting pattern
$(2,1,1,2,1)$, then $q$ is a codimension-2 subform of a multiple
of a 2-fold Pfister form, $\l a,b\r \otimes \langle c,d,e,f\rangle$.
To begin, we claim that over $K:=k(\sqrt{\dets q})$, $q$ has trivial
Clifford invariant in $H^2(K,\F_2)$ \cite[section 14]{EKM}
as well as trivial discriminant. (Merkurjev's theorem
says that these properties are equivalent to $q_K$ being in the ideal
$I^3$ of the Witt ring \cite[section 16]{EKM}, but we do not need that here.)
To see that $q_K$ has trivial Clifford invariant, we use Knebusch's
theorem that the splitting pattern of $q$ over any field
extension is a ``specialization''
of that of $q$, meaning that it is obtained
by possibly adding some sets of adjacent
entries and possibly removing
some initial entries \cite[Theorem 7.5]{Vishik}.
Over the quadratic extension $K$, the last entry of the splitting pattern
of $q_K$ must be a power of 2 at least 2, but we see that it cannot
be 2, and so it must be at least 4. This means that $q_K$
has trivial discriminant and Clifford invariant \cite[Proposition 25.14]{EKM}.

It follows that the Clifford invariant of $q$ in $H^2(k,\F_2)$
is killed by passage to $k(\sqrt{\dets q})$, and so we can write
$c(q)=\{ \dets q,b\} $ for some $b$ in $k^*$ \cite[Theorem 99.13]{EKM}.
As a result,
the 16-dimensional form $\varphi=
q\perp\langle b,-b\,\dets q\rangle$ has trivial discriminant
and Clifford invariant.

Now let $p$ be a 15-dimensional form that contains $q$ and is contained
in $\varphi$. Since $\varphi$ has trivial discriminant and
Clifford invariant, the highest Witt index
$i_h(p)$ must be 3, and so the splitting pattern of $p$ is either
$(3,1,3)$, $(1,1,1,1,3)$ or (if $p$ is isotropic)
$(1,1,1,3)$, by Vishik \cite[Table 1, p.~84]{Vishik}. We will show
that it must be $(3,1,3)$. Note that this does not follow just from
$p$ being of codimension 1 in a form with trivial discriminant
and Clifford invariant; we need to use what
we know about the 14-dimensional form $q$ that $p$ contains.

As a preliminary, we continue our analysis of the Chow groups of $Q^2$,
using that $q$ has splitting
pattern $(2,1,1,2,1)$; this could all be said equivalently
in terms of motivic decompositions of $Q$.
We have already shown, using $i_1(Q)=2$,
that there is a 7-dimensional cycle
$\alpha$
on $Q^2$ whose class over $\kbar$ contains $h^0\times l_1$,
the top of the first shell triangle,
as well as $l_5\times h^4$, the top of the fourth shell triangle 
from the right.
By \cite[Corollary 73.23]{EKM}, the components of any cycle $y$ on $Q^2$
occur in pairs: if a basis element at position $i$ from the left
in the $j$th shell triangle occurs in $y$, then so does 
the basis element at position $i$ from the left in the $j$th
shell triangle from the right. So $\alpha$ contains the four basis elements
pictured.
\begin{figure}[ht] \label{four}
\hspace{2cm}
\begin{picture}(300,30)
\put(28.5,10){\circle{5}}
\put(48.5,10){\circle{5}}
\put(68.5,10){\circle{5}}
\put(88.5,10){\circle{5}}
\put(108.5,10){\circle{5}}
\put(128.5,10){\circle{5}}
\put(148.5,10){\circle{5}}
\put(168.5,10){\circle{5}}
\put(188.5,10){\circle{5}}
\put(208.5,10){\circle{5}}
\put(228.5,10){\circle{5}}
\put(248.5,10){\circle{5}}
\put(268.5,10){\circle{5}}
\put(288.5,10){\circle{5}}
\put(38.5,20){\circle*{5}}
\put(118.5,20){\circle*{5}}
\put(198.5,20){\circle*{5}}
\put(278.5,20){\circle*{5}}
\end{picture}
% \caption[]{$T_{2,r,n-r}$ Dynkin diagram}
\end{figure}
In view of the lemma on shell triangles, the class of $\alpha$
cannot involve any other basis elements, and so it is the sum
of the four basis elements shown, $h^0\times l_1$,
$h^4\times l_5$, $l_5\times h^4$, and $l_1\times h^0$.

Now suppose that the 15-dimensional form $p$ containing $q$
has splitting pattern $(1,1,1,1,3)$ or $(1,1,1,3)$. Let $P$
be the quadric associated to $p$. Because
$q$ has first Witt index 2, we see that over every extension field
of $k$, the conditions $i_W(p)>1$ and $i_W(q)>0$ on the Witt
indices are equivalent. Therefore, by \cite[Theorem 4.15]{Vishik},
there is a middle-dimensional cycle $\beta$ on $P^2$
associated to $\alpha$, consisting (in $CH_*(P^2_{\overline{k}})/2$)
of the sum of the basis elements $h^1\times l_1$, $h^5\times
l_5$, $l_5\times h^5$, and $l_1\times h^1$.
\begin{figure}[ht] \label{fourp}
\hspace{2cm}
\begin{picture}(300,10)
\put(28.5,5){\circle{5}}
\put(48.5,5){\circle*{5}}
\put(68.5,5){\circle{5}}
\put(88.5,5){\circle{5}}
\put(108.5,5){\circle{5}}
\put(128.5,5){\circle*{5}}
\put(148.5,5){\circle{5}}
\put(168.5,5){\circle{5}}
\put(188.5,5){\circle*{5}}
\put(208.5,5){\circle{5}}
\put(228.5,5){\circle{5}}
\put(248.5,5){\circle{5}}
\put(268.5,5){\circle*{5}}
\put(288.5,5){\circle{5}}
\end{picture}
% \caption[]{$T_{2,r,n-r}$ Dynkin diagram}
\end{figure}

Recall the definition of the splitting tower $k_0,\ldots,k_h$
of $p$ from section \ref{known}.
Let $E$ be the field $k_2$ if $p$ has splitting pattern
$(1,1,1,1,3)$ or $k_1$ if $p$ is isotropic with splitting pattern
$(1,1,1,3)$. Then $r:=(p_E)\an$ has dimension 11 and splitting pattern
$(1,1,3 )$; let $R$ be the corresponding projective
quadric. The cycle $\beta$ on $P_E$ determines a middle-dimensional cycle
$\gamma$ on $R$ which consists (in $CH_*(R^2_{\overline{E}})/2$)
of the sum of the basis elements $h^3\times l_3$ and $l_3\times h^3$
\cite[Remark 72.5]{EKM}.
\begin{figure}[ht] \label{two}
\hspace{2cm}
\begin{picture}(300,10)
\put(68.5,5){\circle{5}}
\put(88.5,5){\circle{5}}
\put(108.5,5){\circle{5}}
\put(128.5,5){\circle*{5}}
\put(148.5,5){\circle{5}}
\put(168.5,5){\circle{5}}
\put(188.5,5){\circle*{5}}
\put(208.5,5){\circle{5}}
\put(228.5,5){\circle{5}}
\put(248.5,5){\circle{5}}
\end{picture}
% \caption[]{$T_{2,r,n-r}$ Dynkin diagram}
\end{figure}
(This comes from a correspondence which chops off the left two
and right two dots, in the decomposition of a given cycle on $P^2$.)
But this contradicts what is known about the discrete invariants
of an 11-dimensional form $r$ with splitting pattern $(1,1,3)$.
Namely, Vishik showed that the class of 
every middle-dimensional cycle on $R^2$
is a linear combination of the two sums
of basis elements shown  \cite[Lemma 7.10]{Vishik}.
\begin{figure}[h] \label{twocycles}
\hspace{2cm}
\begin{picture}(300,20)
\put(68.5,0){\circle{5}}
\put(88.5,0){\circle*{5}}
\put(108.5,0){\circle{5}}
\put(128.5,0){\circle*{5}}
\put(148.5,0){\circle{5}}
\put(168.5,0){\circle{5}}
\put(188.5,0){\circle*{5}}
\put(208.5,0){\circle{5}}
\put(228.5,0){\circle*{5}}
\put(248.5,0){\circle{5}}
\put(68.5,20){\circle*{5}}
\put(88.5,20){\circle{5}}
\put(108.5,20){\circle*{5}}
\put(128.5,20){\circle{5}}
\put(148.5,20){\circle*{5}}
\put(168.5,20){\circle*{5}}
\put(188.5,20){\circle{5}}
\put(208.5,20){\circle*{5}}
\put(228.5,20){\circle{5}}
\put(248.5,20){\circle*{5}}
\end{picture}
% \caption[]{$T_{2,r,n-r}$ Dynkin diagram}
\end{figure}
Again, one can refer
to \cite[Theorem 94.1]{EKM} and \cite[Proposition 3.5]{Vishik}
for the translation between Chow groups and motives.

This contradiction shows that the 15-dimensional form $p$
containing our form $q$ must have splitting pattern $(3,1,3)$.
But Vishik showed that for a 15-dimensional form $p$ with
splitting pattern $(3,1,3)$,
the 16-dimensional form $p\perp \langle
\dets p\rangle$ is a multiple of a 2-fold Pfister form,
$\l a,b\r\otimes \langle c,d,e,f\rangle$ \cite[p.~81]{Vishik}.
Thus the 14-dimensional form $q$ is a codimension-2 subform
of a multiple of a 2-fold Pfister form. \qed






\section{Transposition maps}
\label{sectranspo}

In this section, we construct a new family of birational
maps between two quadrics. For any Pfister form
$\rho$ over a field $k$, any form $\varphi$,
any two subforms $\sigma$ and $\tau$
of $\rho$ on which the quadratic form is not zero,
and any $b,c$ in $k^*$, we define a birational
equivalence
$$\varphi\otimes\rho\perp b\sigma\perp c\tau
\simb \varphi\otimes\rho\perp c\sigma\perp b\tau.$$
We call these maps ``transpositions.''
They generalize all the previously known birational
maps between different quadrics: Ahmad-Ohm's
map between two special Pfister neighbors, and Roussey's
map between two special neighbors of a multiple of
a Pfister form, which can be viewed as the case where
the subform $\sigma$ is all of $\rho$. The maps here
do not include the other maps introduced in Corollary
\ref{pfister}, birational maps between arbitrary neighbors
rather than just special neighbors.

The new transposition map is not restricted to neighbors
of multiples of Pfister forms. In particular, it gives
the first birational equivalence between two half-neighbors
which are not similar. By definition, two forms $\varphi$
and $\psi$ of dimension $2^n$ are called
half-neighbors (of each other)
if $d\varphi\perp e\psi$ is a Pfister form
for some $d,e$ in $k^*$. It is elementary that two half-neighbors
are stably birational \cite[Corollary 2.6]{Hoffmannhalf},
and this is a prominent case where
stable birational equivalence is not known to imply
birational equivalence. In particular, Hoffmann showed
that over a rational function field $k=k_0(a,b,x_1,x_2)$
or $k=k_0((a))((b))((x_1))((x_2))$, the following 8-dimensional forms
are half-neighbors but not similar. Take $\pi=\l -a,-x_1,-x_2\r$,
$\tau=\langle -1,-a,bx_1,abx_2\rangle$, $\varphi=(\tau\perp
\pi)\an$, and $\psi=(\tau\perp -b\pi)\an$. That is,
\begin{align*}
\varphi &=\langle x_1,x_2,ax_1,ax_2,x_1x_2,ax_1x_2,bx_1,abx_2\rangle \\
 &= x_1\l -a,-x_1x_2\r\perp x_1x_2\l -a\r \perp bx_1\l -ax_1x_2\r
\end{align*}
and
\begin{align*}
-bx_1\psi &= \langle x_1,x_2,ax_1,ax_2,bx_1,abx_1,x_1x_2,a
\rangle \\
 &= x_1\l -a,-x_1x_2\r\perp bx_1\l -a\r \perp x_1x_2\l -ax_1x_2\r.
\end{align*}
So the new transposition map shows that $\varphi$ 
and $\psi$ are birational. It remains an open question whether
all pairs of two 8-dimensional half-neighbors are birational, however.
Dimension 8 is the lowest where the quadratic
Zariski problem, Conjecture \ref{qzp}, remains open.

\begin{lemma}
\label{transpo}
For any Pfister form
$\rho$ over a field $k$, any form $\varphi$,
any two subforms $\sigma$ and $\tau$
of $\rho$ on which the quadratic form is not zero,
and any $b,c$ in $k^*$, there is a birational
equivalence
$$\varphi\otimes\rho\perp b\sigma\perp c\tau
\simb \varphi\otimes\rho\perp c\sigma\perp b\tau.$$
\end{lemma}

{\bf Proof. }A Pfister form $\rho$ is strongly multiplicative: there is
a ``multiplication''
$xy$ on the vector space of $\rho$ which is a rational function of $x$ and
(for generic $x$)
invertible linear in $y$ such that $\rho(xy)=\rho(x)\rho(y)$
\cite[Theorem X.2.11]{Lam}.
We can also assume that $(ax)(y)=a(xy)$ for $a\in k$ (use the given
product for $x$ in an affine hyperplane, and scale the product elsewhere
to make this property hold).

Write $\varphi=\langle a_1,\ldots,a_{n}\rangle$, $n\geq 0$.
In terms of a multiplication as above, we define the birational map
by
$$[r_1,\ldots,r_{n},s,t]\mapsto [t(sr_1),\ldots,t(sr_{n}),\rho(t)s,
\rho(s)t].$$
Here $r_1,\ldots,r_{n}$ belong to the vector space $\rho$,
while $s$ and $t$ are in the linear subspaces $\sigma$
and $\tau$, respectively.

To show that this map is birational, we would like to define the inverse
by
$$[u_1,\ldots,u_{n},v,w]\mapsto [v^{-1}(w^{-1}u_1),\ldots,
v^{-1}(w^{-1}u_{n}),\rho(v)^{-1}v,\rho(w)^{-1}w].$$
Note that $v^{-1}$ in the formula does not mean an element
of the vector space $\rho$, but merely the inverse of the
linear map $x\mapsto vx$.
Here $\rho(v)$ and $\rho(w)$ are nonzero for generic $v$ and $w$
in the subspaces $\sigma$ and $\tau$ by assumption. But for this
formula to make sense, we also need to know that some point $v$
in $\sigma$ lies in the open set of points such that
the linear map $x\mapsto vx$ is defined and invertible,
and likewise for $\tau$. Since
the quadratic form on $\sigma$ (say) is nonzero, it has a subspace
(say 1-dimensional) on which the quadratic form is nondegenerate.
Applying Lemma \ref{general} to that subspace gives
a new embedding of the quadratic form $\sigma$ into $\rho$ with the desired
property. \qed


\section{Birational maps between special neighbors}

In this section, Theorem \ref{birat} determines the birational type
of {\it special }neighbors of multiples of Pfister forms.
When the Pfister form is $r$-fold with $r$ at most 3,
Theorem \ref{main} is a more general result, covering
arbitrary neighbors. One advantage of special neighbors
even for $r$ at most 3 is that
all the birational maps we need are composites of certain
very explicit ``transposition maps'' between two quadrics
defined by Roussey, where one can describe exactly what is being
blown up and blown down.

First, we recall Ahmad-Ohm's observation \cite[after 1.3]{AO}
on the ruledness of
special neighbors in a
multiple of a Pfister form. The proof is modeled on Knebusch's
proof for the case of special Pfister neighbors
\cite[pp.~73--74]{Knebusch1}. The general proof is written out
in \cite[Lemma 3.2]{Totaroaff}, for example.

\begin{lemma}
\label{ruled}
Let $\rho$ be a Pfister form, $\varphi$ any quadratic
form of dimension at least 1, $\eta$ a nonzero subform of $\rho$,
$a$ an element of $k^*$. Then the quadric associated to
$\varphi\otimes\rho\perp a\eta$ is birational to projective
space of dimension $\dim(\eta)-1$ times the quadric associated
to $\varphi\otimes\rho\perp \langle a\rangle$.
\end{lemma}

The key birational maps we use to prove Theorem \ref{birat}
are the transposition maps of Lemma \ref{transpo}, which
were defined in the case at hand by Roussey. He also proved
a criterion for when two special neighbors are similar,
which we state for the reader's convenience although it
is not needed for what follows.

\begin{lemma}
\label{roussey}
(Transposition lemma)
Let $\rho$ be a Pfister form, $\varphi$
any quadratic form, and $c$ and $d$ nonzero elements of $k$.
Then the special neighbors $\varphi\otimes \rho\perp c\rho
\perp \langle d\rangle$ and $\varphi\otimes\rho \perp
d\rho\perp \langle c\rangle$ are birational.

Moreover, for $\rho$ of dimension at least 2, these two forms
are similar if and only if $cd$ belongs to the group
of similarity factors of $\varphi\otimes\rho$.
\end{lemma}

To prove Theorem \ref{atmost3}, Roussey uses the birational maps
of Lemma \ref{roussey} together with another class of maps that
are only defined when $\rho$ is an $r$-fold Pfister form with
$r\leq 3$. It turns out, however, that just by repeated use of the maps in
Lemma \ref{roussey}, we can prove the following statement,
with no restriction on $r$.

\begin{theorem}
\label{birat}
Let $q_1$ and $q_2$ be 
special neighbors of the same multiple of a
Pfister form $\rho$. If $q_1$ and $q_2$
have the same dimension, then they are birational.
\end{theorem}

{\bf Proof. }By Lemma \ref{ruled}, it suffices to consider
quadratic forms of the form $\sigma\otimes\rho\perp \langle a \rangle$.
That is, if $(\sigma\oplus \langle a\rangle)\otimes \rho$
is isomorphic to $(\tau\oplus \langle b\rangle)\otimes \rho$,
with $\sigma$ of dimension at least 1, then we have to show
that $\sigma\otimes\rho\perp \langle a\rangle$ and
$\tau\otimes\rho\perp \langle b\rangle$ are birational.
For $\sigma $ of dimension 1 (so the forms we consider
are special Pfister neighbors), this was proved by Ahmad-Ohm
\cite[Theorem 1.6]{AO}. We remark that all the birational maps
they used are composites of the maps provided by the transposition
lemma, Lemma \ref{roussey} (which they proved for $\sigma$
of dimension 1).
We can therefore assume that $\sigma$
has dimension at least 2.

Wadsworth and Shapiro proved the following chain lemma,
characterizing
when two multiples of a Pfister form are isomorphic,
by a quick elementary argument \cite[Theorem 1]{WS}.

\begin{lemma}
\label{chain}
Let $\rho$ be a Pfister form over a field $k$.
Two multiples $\langle a_1,\ldots,a_m\rangle \otimes \rho$
and $\langle b_1,\ldots,b_m\rangle \otimes \rho$ of $\rho$
are isomorphic if and only if the sequence $a_1,\ldots,a_m$ of
elements of $k^*$ can be transformed into the sequence
$b_1,\ldots,b_m$ by the following moves.

(1) For $1\leq i<j\leq m$, we can change
$a_i,a_j$ to any other $b_i,b_j$ in $k^*$
such that $\langle a_i,a_j\rangle\cong \langle b_i,b_j\rangle$.

(2) For $1\leq i\leq m$, we can change $a_i$ to $ca_i$
for any $c\in G(\rho)$.
\end{lemma}

Let us say that $\langle a_1,\ldots,a_m;a_{m+1}\rangle
\sim \langle b_1,\ldots,b_m;b_{m+1}\rangle$,
for nonzero elements $a_i$ and $b_i$ of the field $k$,
if the forms $\langle a_1,\ldots,a_m\rangle\otimes \rho\perp \langle
a_{m+1}\rangle$ and $\langle b_1,\ldots,b_m\rangle \otimes\rho
\perp\langle b_{m+1}\rangle$ are birational. Thus, we need to show
that for $m\geq 2$, if $\langle a_1,\ldots,a_{m+1}\rangle\otimes \rho$
is isomorphic to $\langle b_1,\ldots,b_{m+1}\rangle\otimes \rho$,
then $\langle a_1,\ldots,a_m;a_{m+1}\rangle \sim
\langle b_1,\ldots,b_m;b_{m+1}\rangle$.

By Lemma \ref{chain}, it suffices to prove this birational
equivalence when we
perform any of the moves (1) or (2) on a sequence
$(a_1,\ldots,a_m,a_{m+1})$. If we multiply $a_i$ by an element of
$G(\rho)$ for $i\leq m$, then we do not change the form
$\langle a_1,\ldots,a_m\rangle\otimes\rho\perp \langle a_{m+1}\rangle$,
and so we certainly have the desired birational equivalence. If we
multiply $a_{m+1}$ by an element of $G(\rho)$, then the form
$\langle a_1,\ldots,a_m\rangle\otimes\rho\perp \langle a_{m+1}\rangle$
changes by a scalar, and so again we
have a birational equivalence. This covers move (2).
Performing move (1) on two of the numbers $a_1,\ldots,a_m$ does not change
the form we are considering. Finally, by permuting $a_1,\ldots,a_m$
if necessary, it remains to show: if $\langle c_1,c_2
\rangle$ is isomorphic to $\langle d_1,d_2\rangle$, then
$$\langle a_1,\ldots,a_{m-1},c_1;c_2\rangle
\sim \langle a_1,\ldots,a_{m-1},d_1;d_2\rangle.$$

We prove that as follows, using $m\geq 2$.
\begin{align*}
\langle a_1,\ldots,a_{m-1},c_1;c_2\rangle &\sim \langle a_1,
\ldots,a_{m-2},c_1,a_{m-1};c_2\rangle \\
&\sim \langle a_1,\ldots,a_{m-2},c_1,c_2;a_{m-1}\rangle \text{ by
transposition lemma}\\
&\sim \langle a_1,\ldots,a_{m-2},d_1,d_2;a_{m-1}\rangle\\
&\sim \langle a_1,\ldots,a_{m-2},d_1,a_{m-1};d_2\rangle \text{ by
transposition lemma}\\
&\sim \langle a_1,\ldots,a_{m-1},d_1;d_2\rangle .
\end{align*}
That is, the birational map we want is a composite of two of the
birational maps between quadrics provided by the transposition lemma,
Lemma \ref{roussey}.
Theorem \ref{birat} is proved.
\qed


\section{Ruledness of Vishik's example}
\label{newruled}

Vishik recently gave an example of an anisotropic quadratic form
$q$ of dimension 16
which has first Witt index greater than 1 but is not divisible
by a binary form. (Since the dimension is a power of 2, it follows
that $q$ is not a neighbor of any
multiple of a Pfister form.) Thus Conjecture \ref{ruledness}
on ruledness predicts
that $q$ is ruled, but all previous constructions of anisotropic quadrics
which are ruled have only applied to neighbors of multiples of Pfister
forms, making this a natural test case. In this section, we show
that this form is indeed ruled, in fact by lines in the quadric.
The formula we use to prove this
is related to the transposition map of Lemma
\ref{transpo}, but is notably more complex.

One natural context for Vishik's example is the problem of characterizing
the quadratic forms which are divisible by binary forms. 
If $q$ is divisible by a binary form, then the higher Witt indices
$i_1,\ldots,i_h$ are all even except possibly the last, by the
results mentioned in section \ref{neighbors}. For
quadratic forms $q$ of dimension congruent to 2 modulo 4, it follows
from Knebusch's results
that the converse holds: if $i_1,\ldots,i_{h-1}$ are all even
(in this case, $i_h$ will equal 1), then $q$ is divisible
by the binary form $\langle \langle \dets q \rangle \rangle$
\cite[Theorem 7.1, Theorem 7.5]{Vishik}. For forms of dimension 4 or 8,
if the higher Witt indices are all even, then again $q$ is divisible
by a binary form, and the same is conjectured for $q$ of dimension 12
\cite[p.~95]{Vishik}. A naive generalization of that conjecture
is destroyed by the following example.

\begin{lemma}
\label{binary}
(Vishik) The 16-dimensional form 
$$q=\l -d,-e,-f \r\perp b\langle 1,d,e,f\rangle \perp
c\langle 1,d,e,f\rangle $$
over the rational function field $k=k_0(b,c,d,e,f)$ is anisotropic,
with splitting pattern $(2,2,2,2)$, but is not divisible by a binary form.
\end{lemma}

{\bf Proof. }We begin
by computing the splitting pattern of $q$.
If we extend the base field by adjoining the square root
of $c(bdef)$, then $q$ becomes the anisotropic Pfister form
$\l -d,-e,-f,-b\r$, and so $q$ is anisotropic of dimension 16 over $k$.
Moreover, we can write down other field extensions over which
$q$ has anisotropic part of dimension 12, 8, or 4:
adjoin the square root of $c(-bd)$,
the square root of $c(-b)$, or the square roots
of $-b$ and $c(-d)$.

Next, we observe that for every field extension $E$ of $k$
such that $q_E$ has anisotropic part of dimension less than 16,
$q_E$ becomes divisible by the binary form $\l c(bdef)\r$. Indeed,
over $E(\sqrt{c(bdef)}$, $q$ becomes the Pfister form
$\l -d, -e,-f,-b\r$; but $q$ is isotropic over $E$ and hence
over $E(\sqrt{c(bdef)}$, so it is hyperbolic over the latter field.
That means that $q$ is divisible by $\l c(bdef)\r$ over $E$, as we want.

As a result, all the higher Witt indices of $q$ are even; given
the anisotropic parts of $q$ we have observed, it follows that
$q$ has splitting pattern $(2,2,2,2)$. (Equivalently, the possible
dimensions of the anisotropic part of $q$ over all field extensions of
$k$ are $16,12,8,4,0$.)

Finally, we show that $q$ is not divisible by a binary form over $k$.
It suffices to show this over the field extension $E=k_0
(d,e,f,b)((t))$ where $t:=bcdef$.
We can write $q=\l -d,-e,-f,-b\r-bdef\l t\r\otimes \langle
1,d,e,f\rangle$ in the Witt group. If $q$ is divisible by
a binary form $\l u\r$, then $q$ becomes hyperbolic over $E(\sqrt{u})$,
from which the properties of Pfister forms imply that
$\l -d,-e,-f,-b\r$ and (consequently) $-bdef\l t \r\otimes \langle
1,d,e,f\rangle$ become hyperbolic over $E(\sqrt{u})$. But modulo squares,
every $u$ in $E^*$ is either of the form $c$ or $ct$ for some
$c$ in $k_0(d,e,f,b)^*$. And the passage from $E$ to $E(\sqrt{ct})$
cannot kill $\l -d,-e,-f,-b\r$, while the passage from $E$
to $E(\sqrt{c})$ cannot kill $-bdef\l t \r\otimes \langle
1,d,e,f\rangle$. Thus $q$ is not divisible by a binary form.
\qed

Since Vishik's 16-dimensional form has first Witt index 2,
Conjecture \ref{ruledness} on ruledness
predicts that it is ruled. But since it is not a neighbor
of a multiple of a Pfister form,
no earlier criterion for ruledness applies. We
now prove that this form is in fact ruled. The construction applies
to a broader class of forms, including some in arbitrarily large
dimensions. In fact, we find that these quadrics are ruled by lines.
That has the consequence that any two codimension-1
subquadrics of such a quadric are birational to each other, as in the proof
of Theorem \ref{main}.

\begin{theorem}
\label{vishikruled}
Let $\rho$ be an $r$-fold Pfister form with $r$ at most 3,
$\varphi$ any form,
$\sigma$ a nondegenerate subform of dimension at least 2
in $\rho$, and $b$ and $c$ nonzero elements of $k$.
Then the form $\varphi\otimes\rho\perp b\sigma\perp c\sigma$
is ruled by lines.
\end{theorem}

{\bf Proof. }We will only write out the proof in the case
$\varphi=\langle 1\rangle$. For an arbitrary form
$\varphi=\langle a_1,\ldots,a_n\rangle$,
we just replace the element $r\in\rho$
in the formulas below by a sequence
$r_1,\ldots,r_n\in\rho$.

We can assume (after scaling $\sigma$) that $\sigma$
represents 1. Write $\sigma$ as the orthogonal sum $\sigma
=\langle 1\rangle \perp \tau$. Take a standard bilinear multiplication
on $\rho$, corresponding to the multiplication on the complex
numbers, quaternions, or octonions, so that $\rho(xy)=\rho(x)\rho(y)$.
A reference for these algebra structures
is \cite[Proposition 33.9, Proposition 33.18]{KMRT}.
We can assume that $\sigma$
contains the identity element 1 of the algebra as the basis for its subform
$\langle 1\rangle$. 
For $x$ in $\rho$, let $x_0$ be the
``real part'' of $x$, the coefficient of 1 in the decomposition
of $\rho$ as the direct sum of $k 1$ and its orthogonal complement, and define
the standard involution on $\rho$ by $\overline{x}=2x_0\cdot 1-x$. We
automatically have $x\overline{x}=\rho(x)1$, $\overline{xy}=
\overline{y}\; \overline{x}$, and $x(\overline{x}y)=(yx)\overline{x}=\rho(x)y$ 
\cite[Proposition 33.9]{KMRT}.

Then we define a rational map from the given quadric
$\rho\perp b\sigma\perp c\sigma$ to the codimension-1
subquadric $\rho\perp b\sigma\perp c\tau$ by
$$[r,s,t]\mapsto [[t(t_0s-s_0t)]r, s_0\rho(t)s-t_0\rho(s)t,
-\rho(t)(t_0s-s_0t)].$$
Here it is clear that the second and third coordinates are in the
linear subspace $\sigma$, and the third coordinate has
``real part'' equal to 0 so that it lies in the smaller subspace $\tau$.

We have to check that this map sends the quadric $\rho(r)+b\rho(s)+c\rho(t)=0$
into the quadric given by the same formula in $\rho\perp b\sigma
\perp c\tau$. Write $\langle , \rangle $ for the bilinear form
which corresponds to the quadratic form $\rho$, in the sense
that $\rho(x)=\langle x,x\rangle$. 
Since
$\rho(x+y)=\rho(x)+2\langle x,y\rangle +\rho(y)$,  we find that
$2\langle x,y\rangle =x\overline{y}+y\overline{x}=2(x\overline{y})_0$
and hence $\langle x,y\rangle =(x\overline{y})_0$.
We can now begin to compute
our quadratic form on the output of the above map. The tricky point
involves the middle term of the output:
\begin{align*}
\rho(s')&=\rho(s_0\rho(t)s-t_0\rho(s)t)\\
&= s_0^2\rho(s)\rho(t)^2-2s_0t_0\rho(s)\rho(t)\langle s,t\rangle
+t_0^2\rho(s)^2\rho(t)\\
&= \rho(s)\rho(t)\rho(t_0s-s_0t).
\end{align*}
Given that, plus multiplicativity of the form $\rho$,
we get a simple formula for the quadratic form
applied to the output of the above map:
\begin{align*}
\rho(r')+b\rho(s')+c\rho(t')
&=\rho(t)\rho(t_0s-s_0t)[\rho(r)+b\rho(s)+c\rho(t)]
\end{align*}
which is 0 if $\rho(r)+b\rho(s)+c\rho(t)=0$. Thus we have defined
a rational map $f$ from our quadric to a codimension-1 subquadric.

The inclusion from
the subquadric $\rho\perp b\sigma \perp c\tau$ into the whole
quadric is a section of the above rational map. (Indeed, if $t_0=0$,
then $t^2=-t\overline{t}=-\rho(t)1$, which implies that
the map as written is $s_0\rho(t)$ times the identity map.)
We can write down another section of this rational map
using the transposition map from Lemma \ref{transpo}. Explicitly,
the restriction of our map to the subquadric
$\rho\perp b\tau \perp c\sigma$ defined by $s_0=0$ is exactly
($-t_0$ times) the transposition map
$$[r,s,t]\mapsto [-(ts)r,\rho(s)t,\rho(t)s],$$
which is a birational map to the base quadric $\rho\perp b\sigma
\perp c\tau$. Therefore, the inverse of this map is a
section of our rational map, and it is easily checked to be given
by the formula
$$[u,v,w]\mapsto [(w\overline{v})u,\rho(v)w,\rho(w)v].$$

Next, we show that the line spanned by
these two sections maps into our quadric $q=\rho\perp b\sigma
\perp c\sigma$. In other words, for
general $[u,v,w]$ in the base quadric $\rho\perp b\sigma
\perp c\tau$, we know that $[u,v,w]$ and $[(w\overline{v})u,
\rho(v)w,\rho(w)v]$ lie on $q=0$, and we want to show that
the whole line spanned by these two points lies on $q=0$. That follows
if we can show that the inner product of these two points is 0.
But we can write that inner product as
\begin{align*}
& \langle u,(w\overline{v})u\rangle+b\langle v,\rho(v)w\rangle
+c\langle w,\rho(w)v\rangle .
\end{align*}
Here 
\begin{align*}
\langle u,(w\overline{v})u\rangle &= [((w\overline{v})u)\overline{u}]_0\\
&= [\rho(u)(w\overline{v})]_0\\
&= \rho(u)\langle v,w\rangle .
\end{align*}
Therefore, the above inner product is
\begin{align*}
&= (\rho(u)+b\rho(v)+c\rho(w))\langle v,w\rangle\\
&= 0,
\end{align*}
as we want.

Now define a map $X\dashrightarrow Y\times \P^1$
by $[r,s,t]\mapsto (f(r,s,t),[s_0,t_0])$, where
$X$ is the quadric $\rho\perp b\sigma \perp c\sigma$,
$Y$ is the subquadric $\rho\perp b\sigma\perp c\tau$,
and $f:X\dashrightarrow Y$ is the above rational map.
The above two sections of $f$ are the fibers
over $[0,1]$ and $[1,0]$ in $\P^1$.
We define a map back from $Y\times \P^1$ to $X$ as a suitable linear
combination of those two sections:
\begin{align*}
[u,v,w],[d,e]&\mapsto d\rho(w)(u,v,w)+e((w\overline{v})u,\rho(v)w,
\rho(w)v)\\
&= [(d\rho(w)+e(w\overline{v}))u,d\rho(w)v+e\rho(v)w,\rho(w)(dw+ev)].
\end{align*}
We will show that these two rational maps are inverses of each other,
which will
imply that $X$ is ruled by lines, since the map from $Y\times \P^1$ to $X$
clearly sends the $\P^1$ fibers to lines. It suffices
to show that the composition from $X$ to $Y\times \P^1$ and back
is the identity.

This composition is given by
\begin{align*}
[r,s,t]\mapsto [[t(t_0s-s_0t)]r, s_0\rho(t)s-t_0\rho(s)t,
-\rho(t)(t_0s-s_0t)],[s_0,t_0]\\
\mapsto [(s_0\rho(t)^2\rho(t_0s-s_0t)
-t_0\rho(t)(t_0s-s_0t)(s_0\rho(t)\overline{s}-t_0\rho(s)\overline{t}))
[(t(t_0s-s_0t))r], \\
s_0\rho(t)^2\rho(t_0s-s_0t)(s_0\rho(t)s-t_0\rho(s)t)
-t_0\rho(s_0\rho(t)s-t_0\rho(s)t)\rho(t)(t_0s-s_0t), \\
\rho(t)^2\rho(t_0s-s_0t)(-s_0\rho(t)(t_0s-s_0t)+t_0(s_0\rho(t)s-t_0\rho(s)t))].
\end{align*}

We will show that this composition is equal to
$\rho(t)^2\rho(t_0s-s_0t)(s_0^2\rho(t)-t_0^2\rho(s))$ times the
identity map $[r,s,t]$. This is immediate for the third term
(the coefficients of $s$ cancel out).
It holds for the second term, using the identity checked earlier
that $\rho(s_0\rho(t)s-t_0\rho(s)t)= \rho(s)\rho(t)\rho(t_0s-s_0t)$.
To check it for the first term, in view of the identity
$\overline{x}(xr)=\rho(x)r$, it will suffice to show
that 
$$s_0\rho(t)^2\rho(t_0s-s_0t)
-t_0\rho(t)(t_0s-s_0t)(s_0\rho(t)\overline{s}-t_0\rho(s)\overline{t})$$
is equal to $\rho(t)(s_0^2\rho(t)-t_0^2\rho(s))\overline{t(t_0s-s_0t)}$.
This is indeed true. Namely,
since $t_0s-s_0t$ has zero ``real part'', we have
$\overline{t(t_0s-s_0t)}=-(t_0s-s_0t)\overline{t}$, and so
the right side of the proposed equality is 
$$ = -\rho(t)(s_0^2\rho(t)-t_0^2\rho(s))(t_0s\overline{t}-s_0\rho(t)).$$
The left side of the proposed equality, on the other hand, is
\begin{align*}
&=s_0\rho(t)^2(t_0^2\rho(s)-2s_0t_0\langle s,t\rangle +s_0^2\rho(t))
-t_0\rho(t)(2s_0t_0\rho(s)\rho(t)-s_0^2\rho(t)t\overline{s}
-t_0^2\rho(s)s\overline{t})\\
&= s_0\rho(t)^2(t_0^2\rho(s)+s_0^2\rho(t))-2s_0t_0^2\rho(s)\rho(t)^2
-t_0\rho(t)(s_0^2\rho(t)-t_0^2\rho(s))s\overline{t}\\
&=-\rho(t)(s_0^2\rho(t)-t_0^2\rho(s))(t_0s\overline{t}-s_0\rho(t)).
\end{align*}
where the second equation uses that $t\overline{s}
=2\langle s,t\rangle-s\overline{t}$. \qed











 





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\end{document}
