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\magnification = 1200
\centerline {\bf Pfister Involutions}
\bigskip
\centerline {E. Bayer--Fluckiger and R. Parimala}

\vskip 1cm
{\bf Introduction} 
\bigskip
Pfister forms play a prominent role in the algebraic theory of 
quadratic forms. On the other hand, involutions on central simple
algebras share many properties with quadratic forms. Hence it is
natural to look for an analog of the notion of Pfister form in 
the framework of algebras with involution. 
\bigskip
The aim of the present paper is to propose such a notion.
An $n$--fold Pfister involution 
(or Pfister involution, for short) will be
by definition a central simple algebra with an 
orthogonal involution which is 
a tensor product of $n$ quaternion algebras with involution.
We show that if $n = 4$, then after passing to any splitting field
of the algebra, the involution is induced by a Pfister form. 
For $n\le 3$, this
was already proved by D. Tao.
We also compute cohomological invariants of $2$-fold
Pfister involutions, and raise some
open questions.

\bigskip
\bigskip
{\bf 1. Definitions and notation}
\bigskip
Let $F$ be a field of characteristic $\not= 2$. Let $D$ be 
a central simple algebra over $F$. We say that an involution
$\sigma : D \to D$ is of the {\it first kind} if its restriction
to $F$ is the identity. After extension to a splitting field
of $D$, any involution of the first kind is induced by a symmetric
or by a skew-symmetric form. We say that the involution is of
the {\it orthogonal type} in the first case, and of the
{\it symplectic type} in the second case.
\bigskip
\noindent
{\bf Definition.}
Let $(D,\sigma)$ be a central simple algebra endowed with an involution
of the first kind. We say that $(D,\sigma)$ is an  {\it $n$--fold Pfister
involution} if $\sigma$ is of the
orthogonal type, and if there exist quaternion algebras $H_1$,\dots ,$H_n$,
and involutions $\sigma_i : H_i \to H_i$ such that 
$$(D,\sigma) \simeq (H_1,\sigma_1)\otimes... \otimes (H_n,\sigma_n).$$

\bigskip
Let $F_s$ be a separable closure of $F$, and 
set $H^n(F) = H^n({\rm Gal}(F_s/F), {\bf Z}/2{\bf Z})$.

\bigskip
\bigskip
{\bf 2. Statement of results and open questions}
\bigskip
Let $(D,\sigma)$ be a central simple algebra with an 
orthogonal involution
of the first kind. Let $K$ be a splitting field of $D$. Then
after tensoring with $K$, the involution is induced by a
quad\-ratic form $q$ defined over $K$. It is natural to ask
whether $(D,\sigma)$ is a Pfister involution if and only
if $q$ is similar to a Pfister form. 
This is proved by Tao [5] for algebras
of degree $\le 8$ (cf. [4], [2], p.150).
\bigskip
In the present paper, we show that for a $4$--fold Pfister involution
the associated quad\-ratic form is a $4$--fold Pfister form (see
\S 5, th. 2). It would
be interesting to know whether the converse also holds : this is
still an open question.
\bigskip
It is also natural to try to define $n$--dimensional cohomological
invariants of $n$--fold Pfister involutions. More precisely,
suppose that $(D,\sigma)$ is an $n$--fold Pfister involution. Let
$X_D$ be the Brauer--Severi variety of $D$ and $q_{F(X_D)}$ the
associated quadratic form. For $n\le 4$, we know that $q_{F(X_D)}$
is an $n$--fold Pfister form. Let 
$E_n (D) = Ker( H^n (F) \to H^n (F(X_D)).$
Does there exist an invariant $e_n(D,\sigma) \in H^n(F)/E_n (D)$ with
the property that $e_n(q_{F(X_D)})$ is equal to the
image of $e_n(D,\sigma)$ in $H^n(F(X_D))$ ? In \S 4, we define 
such an invariant for $2$--fold Pfister involutions.
\bigskip
\bigskip
{\bf 3. A  lemma}
\bigskip 
The following lemma will be used in \S 4 and \S 5 :
\bigskip \noindent
{\bf Lemma 1.} {\it Let $\tau$ be an orthogonal involution on $D$.
Let $\phi_{\tau} : D \otimes_FD \to {\rm End}_F(D)$ be the isomorphism
induced by $\phi_{\tau}(v,w)(x) = vx \tau(w)$. Let $u \in D^*$ be such
that $\tau(u) = u$. Then the involution $\tau \otimes {\rm Int}(u) o \tau$
transports under $\phi_{\tau}$ to $\sigma_q$, where $\sigma_q$ is
the adjoint involution on ${\rm End}_F(D)$ induced by the quadratic form
$q : D \to F$ given by $q(x) = {\rm Trd}(x u^{-1} \tau(x))$.}
\bigskip
\noindent
{\bf Proof.} See [2], II.1, page 133.



\bigskip
\bigskip
{\bf 4. Invariants of 2-fold Pfister involutions}
\bigskip
Let $(H_1,\tau_1)$ and $(H_2,\tau_2)$ be two quaternion algebras
with involution, and set $(D,\tau) = (H_1,\tau_1) \otimes (H_2,\tau_2)$.
Suppose that $\tau$ is an orthogonal involution. By [1] we know that
${\rm disc}(\tau) = 1$ and that 
$(D,\tau) = (H_3,\tau_3) \otimes (H_4,\tau_4)$, where $H_3$, $H_4$ 
are quaternion algebras, and $\tau_3$, $\tau_4$ are their canonical
involutions. We have $C(D,\tau) = H_3 \times H_4$ and the factors
are unique up to switch. 
\bigskip
Let $X_D$ be the Severi--Brauer variety of $D$, and set
$$
E_2(D) = Ker( H^2(F) \to H^2(F(X_D)).
$$ Then $E_2(D)$ is the subgroup
of $H^2(F)$ generated by the class of $D$. Let us denote by
$[H_3]$ the class of $H_3$ in $H^2(F)/E_2(D)$. Then $[H_3]$ is
an invariant of $(D,\tau)$. 
\bigskip
If $D$ is split, then $H_3 \simeq H_4$, and
$(D,\tau) = (H_3,\tau_3) \otimes (H_4,\tau_4) \simeq
({\rm End}(H_3),\tau_q)$,
where $q : H_3 \to F$ is the quadratic form $q(x) = {\rm Trd}(x \tau_3(x)) 
= 2 {\rm Nrd}(x)$, cf. Lemma 1. Thus $q \simeq 2 N_{H_3}$, 
where $N_{H_3}$ 
is the norm form of the quaternion algebra $H_3$. This implies that
$e_2(q) \in H^2(F)$ is the class of $H_3$. Set
$e_2(D, \tau) = [H_3] \in H^2(F)/E(D)$. This is an invariant of the
algebra with involution $(D,\tau)$, and it coincides with $e_2(q)$ if
$D \simeq M_4(F)$ and $\tau = \tau_q$.

\bigskip
\noindent
{\bf Theorem 1.} {\it Let $(D,\tau) = (H_1,{\rm Int}(u) o \tau_1)
\otimes (H_2,{\rm Int}(v) o \tau_2)$ where the $H_i$ are quaternion
algebras, $\tau_i$ the canonical involution of $H_i$ and $u \in H_1$,
$v \in H_2$ satisfy $\tau_1(u) = -u$, $\tau_2(v) = -v$.
Then $e_2(D, \tau) = [H_1] + (-{\rm Nrd}(u)) \cup (-{\rm Nrd}(v))$
in $H^2(F)/E(D)$.}

\bigskip
\noindent
{\bf Proof.} Suppose that $H_1 \otimes H_2$ is split. We may
assume that $H_1 = H_2$. Then we have
$$(D,\tau) = (H_1,{\rm Int}(u) o \tau_1)
\otimes (H_1,{\rm Int}(\theta) o {\rm Int}(u) o \tau_1),$$ where
$\theta = v u^{-1}$. Then $\tau = \tau_q$, where $q : H_1 \to F$ is the
quadratic form $$q(x) = {\rm Trd}(x {\theta}^{-1} {\rm Int}(u) o \tau_1(x)) =
{\rm Trd}(x u v^{-1} u \tau_1(x) u^{-1}),$$ cf. lemma 1. Let us define
$b_q : H_1 \times H_1 \to F$ by 
$b_q(x,y) = {\rm Trd}(x u v^{-1} u \tau_1(y) u^{-1})$.
Set $\lambda = {\rm Trd}(u v^{-1})$.
\medskip
If  $\lambda = 0$, then $q(1) = {\rm Trd}(uv^{-1}) = 0$.
Thus $q$ is isotropic and has discriminant $1$, hence it is
hyperbolic. As $\lambda = 0$, the elements $1, u, v, uv$ form
a quaternionic basis for $H_1$, so that $H_1 = (-{\rm N}(u)) \cup
(-{\rm N}(v))$ and $[H_1] + (-{\rm N}(u)) \cup (-{\rm N}(v)) = 0$.
This concludes the proof in case $\lambda = 0$.
\medskip
Let us suppose that $\lambda \not = 0$. Then $1$, $u$
and $v u^{-1} - {2 \over {\lambda}}$ are mutually orthogonal.
Indeed, $b_q(1, u) = -{\rm Trd}(u v^{-1} u^2 u^{-1}) =
-{{\rm Trd}(u v^{-1}  u^{-1}) \over {u^2}} = 0$, 
as ${\rm Trd}(v) = 0$.
We have $b_q(1, v u^{-1} - {2 \over {\lambda}}) = 
{\rm Trd}(u v^{-1} u (u^{-1}v - {2 \over {\lambda}}) u^{-1}) =
{\rm Trd}(1 - {2 \over {\lambda}} u v^{-1}) = 0$.
Further, $b_q(u, v u^{-1} - {2 \over {\lambda}}) =
{\rm Trd}(u^2 v^{-1} u ( u^{-1} v - {2 \over {\lambda}})u^{-1}) =
{\rm Trd}(-{\rm N}(u) u^{-1} + {2 \over {\lambda}}{\rm N}(u) v^{-1}) = 0$.
\medskip
We have $q(1) = \lambda$, $q(u) = {\rm Trd}(-u^2 v^{-1} u) = 
{\rm N}(u) \lambda$, and 
$$q( v u^{-1} - {2 \over {\lambda}} ) =
{\rm Trd}[(v u^{-1} - {2 \over {\lambda}}) 
u v^{-1} u (u^{-1} v  - {2 \over {\lambda}}) u^{-1})]$$
$$ =
{\rm Trd}[(u - {2 \over {\lambda}} u v^{-1} u) 
(u^{-1}v u^{-1} - {2 \over {\lambda}}u^{-1})].$$
Hence $q( v u^{-1} - {2 \over {\lambda}} ) = {\rm Trd}(v u^{-1} 
- {4 \over {\lambda}} + {4 \over {\lambda}^2} u v^{-1}) =
{\rm Trd}(v u^{-1}) - {4 \over {\lambda}} = 
{{\lambda} \over {{\rm N}(u v^{-1})}} - {4 \over {\lambda}}$.
\medskip
Since ${\rm disc}(q) = 1$, we have 

$$q \simeq \lambda < 1, {\rm N}(u), 
{1 \over {{\rm N}(u v^{-1})}} - {4 \over {\lambda}^2}, {\rm N}(u)
({1 \over {{\rm N}(u v^{-1})}} - {4 \over {\lambda}^2})>.$$

\medskip
We have to check that
$$(-{\rm N}(u)) \cup (-{\rm N}(v)) + [H_1] = (-{\rm N}(u)) \cup 
-({1 \over {{\rm N}(u v^{-1}) }} - {4 \over {\lambda}^2}).$$

Since $u$, $v$ are trace zero elements, there exists a trace zero
element $w$ such
that $1, u, w, uw$ is a quaternionic basis for $H_1$, and
$v = au + bw$, $[H_1] = (-{\rm N}(u)) \cup (- {\rm N}(w))$.
\medskip
Hence we have :
$$(-{\rm N}(u)) \cup (-{\rm N}(v)) + [H_1] = (-{\rm N}(u) \cup ({\rm N}(v)
{\rm N}(w)).$$

It suffices to show that ${\rm N}(v) {\rm N}(w) ({4 \over {\lambda}^2}
- {1 \over {{\rm N}(u v^{-1})}})$ is a value of $<1, {\rm N}(u)>$.
\medskip

We have ${\rm N}(v) {\rm N}(w) ({4 \over {\lambda}^2}
- {1 \over {{\rm N}(u v^{-1})}}) = (a^2 {\rm N}(u) + b^2 {\rm N}(w))
{\rm N}(w) ({4 \over {{\rm Trd} (u v^{-1})^2}} - {1 \over
{{\rm N}(u v^{-1}}})).$
\medskip
Note that $${\rm Trd}(u v^{-1}) = {{{\rm Trd}(v u^{-1}) 
\over {\rm N}(v u^{-1})}} = { {2a} \over {a^2 + b^2 {\rm N} (w u^{-1})}},$$
and 
$${\rm N}(uv^{-1}) = {1 \over {{\rm N}(v u^{-1})}} = 
{ {1} \over {a^2 + b^2 {\rm N} (w u^{-1})}}.$$

\medskip

We have 



$${\rm N}(v) {\rm N}(w) ({4 \over {\lambda}^2}
- {1 \over {{\rm N}(u v^{-1})}}) = (a^2 {\rm N}(u) + b^2 {\rm N}(w)) 
{\rm N}(w) (a^2 + b^2 {\rm N}(w u^{-1}))
[{{a^2 + b^2 {\rm N}(wu^{-1})} \over {a^2}} -1]$$
$$
= {{[a^2 {\rm N}(u) + b^2{\rm N}(w)]^2} \over 
{{\rm N}(u)^2 a^2}}{\rm N}(w)^2 b^2,
$$ 
which is a square, hence a norm. Therefore we have
$e_2(q) = (-{\rm N}(u)) \cup (-{\rm N}(v)) +[H_1]$, as claimed.

\bigskip
\bigskip
{\bf 5. $4$--fold Pfister involutions}
\bigskip

The aim of this section is to prove the following result :
\bigskip
\noindent
{\bf Theorem 2.} Let $(D,\sigma)$ be a $4$--fold Pfister
involution. Let $K$ be a splitting field of $D$, and let 
$q = q_K$ be the quadratic form induced by $\sigma$ over $K$.
Then $q$ is a Pfister form.

\bigskip
The proof will make use of the following lemmas :


\bigskip
\noindent
{\bf Lemma 2.} {\it Let $H_1$ and $H_2$ be two quaternion algebras,
and set $D = H_1 \otimes H_2$. Let $\tau$ be an orthogonal involution
with trivial discriminant
on $D$, and let $u \in D^*$ with $\tau (u) = u$. Then the quadratic
form $q : D \to F$ defined by $q(x) = {\rm Trd}(x u \tau(x))$ is in
$I^3(F)$.}
\bigskip
\noindent
{\bf Proof.} By lemma 1, the involution $\tau \otimes ({\rm Int}u o \tau)$
on $D \otimes D$ transports into $\sigma_q$ on ${\rm End}_F(D)$. Thus
${\rm disc}(q) = 1$ and $C_0(q) \simeq C(D \otimes D, \tau 
\otimes ({\rm Int u} o \tau))$ is split (see for instance [2],p. 150).
This implies that $C(q)$ is split and that $q$ is in $I^3(F)$.

\bigskip
\noindent
{\bf Lemma 3.} {\it Let $H_1$ and $H_2$ be two quaternion algebras,
and set $D = H_1 \otimes H_2$. Let $\tau_i$ be the canonical involution
on $H_i$, and set $\tau = \tau_1 \otimes \tau_2$. For any $u \in D^*$ such
that ${\rm Nrd}(u) \in F^{*2}$, we define the quadratic form
$q_u : D \to F$  by $q_u(x) = {\rm Trd}(x u \tau(x))$. Then $q_u$ is
a Pfister form.}

\bigskip
\noindent
{\bf Proof.} It suffices to show that if $q_u$ is isotropic then
it is split. Let us suppose $q_u$ isotropic. By a general position
argument, there exists $y \in D^*$ such that $q_u(y) = 0$. The
quadratic forms $q_u$ and $q_{yu \tau(y)}$ are isometric under the
map $D \to D$, $x \mapsto x y^{-1}$. Hence replacing $u$ by $y u \tau(y)$, 
we may assume that $u$ has the additional properties
${\rm Trd}(u) = 0$, $\tau (u) = u$.
\medskip
For $x \in H_1$, we have $\tau(x) = x^{-1} {\rm Nrd}(x)$ so that 
$q_u(x) = {\rm Trd}(x u x^{-1} {\rm Nrd}(x)) = 
{\rm Nrd}(x) {\rm Trd} (x u x^{-1}) = 0.$ Thus $H_1$ is totally
isotropic for $q_u$ : indeed, $H_1$ is totally isotropic for 
$q_z$ for any $z\in D^*$ with ${\rm Trd}(z) = 0$, $\tau (z) = z$.
\medskip
The involution ${\rm Int}(u) o \tau$ on $D$ has trivial discriminant,
hence $D = H_1' \otimes H_2'$ and ${\rm Int}(u) o \tau = \tau_1' \otimes
\tau_2'$, where $\tau_i'$ denotes the canonical involution on $H_i'$.
Let $j \in H_2'$ with ${\rm Trd}(j) = 0$ and let $(H_1')^0$ be the
set of trace zero elements in $H_1'$. Then $W = (H_1')^0 j$ is a
3--dimensional subspace of $D$ elementwise fixed by ${\rm Int}(u) o \tau$ 
such that for every $y \in W$, we have $y^2 \in F$. Set
$T = \{ y \in D, \ {\rm Trd}(u \tau y) = 0 \}$. Then ${\rm dim}(T) = 15$
and ${\rm dim}(T \cap W) \ge 2$. Let $V_j \subset T \cap W$ be
a 2--dimensional subspace. Then for $y \in V_j$, we have 
${\rm Trd}(u \tau(y) = 0$, $y^2 \in F$ and ${\rm Int}(u) o \tau(y) = y$.
This implies that $u \tau(y) = yu$, hence $\tau (u \tau(y) = yu = u \tau(y)$,
i.e. $u \tau (y)$ is symmetric under $\tau$. Since for $y \in V_j$
we have $y^2 \in F$, we have
${\rm Trd}(yu \tau (y) = {\rm Trd} (y^2 u) = y^2 {\rm Trd}(u) = 0$.
Thus $V_j$ is totally isotropic for $q_u$.
\medskip
Further for $y \in V_j$, the element $u \tau(y)$ being symmetric 
for $\tau$ with trace zero, $H_1$ is totally isotropic for
$q_{u \tau(y)}$. 
Hence for $x, z \in H_1$, we have
${\rm Trd}(xu \tau(y) \tau(z)) = 0$. In particular,
setting $z = 1$, we get, for $x \in H_1$, 
${\rm Trd}(x u \tau(y)) = 0$. Hence $H_1$ is orthogonal to $V_j$ with
respect to $q_u$.
\medskip
Let $i, j, k$ be a basis of trace zero elements in
$H_2'$ with $i = jk$, $jk = -kj$.
Suppose that $V_j \subset H_1$. Then we have $V_j \subset H_1^0$.
If $V_k \subset H_1$, then we have $V_j + V_k \subset H^0_1$, and
$V_j \cap V_k = 0$. Hence we get ${\rm dim} (V_j + V_k) = 4$,
contradicting ${\rm dim}(H_1^0) = 3$. 
\medskip
Thus there exists $y \in V_j$ or $V_k$ such that
$y \not \in H_1$, $y$ totally isotropic for $q_u$ and 
${\rm Trd}(x u \tau(y)) = 0$ for all $x \in H_1$.
Thus $H_1 \oplus Fy$ is a 5--dimensional totally isotropic
subspace for $q_u$. Therefore the
anisotropic rank of $q_u$ is at most 6 and by Lemma 2, we
have $q_u \in I^3(F)$. Hence $q_u$ is hyperbolic.

\bigskip
\noindent
{\bf Proof of theorem 2.} Since $(D,\sigma)$ is Pfister, we may
write $(D,\sigma)$ as 
$$
(D_1,\sigma_1) \otimes (D_2,\sigma_2) \otimes (D_3,\sigma_3)
\otimes (D_4,\sigma_4),
$$
where the $D_i$'s are quaternion algebras and $\sigma_i$'s are
orthogonal involutions. Using [1], we may write 
$(D_1,\sigma_1) \otimes (D_2,\sigma_2)$ as
$(H_1,\tau_1) \otimes (H_2,\tau_2)$,
where the $H_i$'s are quaternion algebras and the $\tau_i$'s their
canonical involutions. To prove the theorem, we may assume that
$D$ is split. 
\medskip
Using [1] again, we may write
$(D_3,\sigma_3) \otimes (D_4,\sigma_4) = (H_1 \otimes H_2, {\rm Int}(u ^{-1})
o(\tau_1 \otimes \tau_2)$, with $u \in H_1 \otimes H_2$ satisfying
${\rm Nrd}(u) \in F^{*2}$. By lemma 1, the form $q$ associated
to $\sigma$ is given by 
$q(x) = {\rm Trd}(x u \tau_1 \otimes \tau_2 (x))$. 
By lemma 3, the form $q$ is indeed
a $4$--fold Pfister form.



 
\bigskip
\bigskip
{\bf Bibliography}
\bigskip \noindent
[1] M. Knus, R. Parimala, R. Sridharan, Pfaffians, central simple algebras
and similitudes, {\it Math. Z.} {\bf 206} (1991), 589--604.
\medskip
\noindent
[2] M. Knus, A. Merkurjev, M. Rost, J.-P. Tignol, {\it The book
of involutions}, AMS Coll. Pub., Vol. 44, Providence, 1998.
\medskip \noindent
[3] W. Scharlau, {\it Quadratic and hermitian forms}, Grundlehren
der Math. Wiss., Vol. 270, Springer--Verlag, Heidelbert, 1985.
\medskip \noindent
[4] D. Tao, The generalised even Clifford algebra,
{\it J. Algebra} {\bf 172} (1995), 184--204.
\medskip \noindent
[5] D. Tao, Pfister--form--like behavior of algebras with involution,
unpublished manuscript.

\bigskip
\bigskip


E. Bayer Fluckiger

D\'epartement de Math\'ematiques

Ecole Polytechnique F\'ed\'erale de Lausanne

1015 Lausanne 

Switzerland

eva.bayer@epfl.ch

\bigskip

R. Parimala

Tata Institute of Fundamental Research

Homi Bhambha Road

Mumbai 400005

India

parimala@math.tifr.res.in




\bye 

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