---
abstract: 'We study central simple algebras with involution of the first kind that become hyperbolic over the function field of the conic associated to a given quaternion algebra $Q$. We classify these algebras in degree $4$ and give an example of such a division algebra with orthogonal involution of degree $8$ that does not contain ${(Q,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})}$, even though it contains $Q$ and is totally decomposable into a tensor product of quaternion algebras.'
address:
- |
    Université Paris 13 (LAGA)\
    CNRS (UMR 7539)\
    Université Paris 12 (IUFM)\
    93430 Villetaneuse, France
- |
    Département de Mathématiques\
    Université catholique de Louvain\
    Chemin du cyclotron, 2\
    B1348 Louvain-la-Neuve\
    Belgique
author:
- 'Anne Quéguiner-Mathieu'
- 'Jean-Pierre Tignol'
title: Algebras with involution that become hyperbolic over the function field of a conic
---

Given two central simple algebras with involution $(A,\sigma)$ and $(B,\tau)$ over a field $F$, we say that $(A,\sigma)$ contains $(B,\tau)$ if $A$ contains a $\sigma$-stable subalgebra isomorphic to $B$ over which the involution induced by $\sigma$ is conjugate to $\tau$. By the double centralizer theorem [@KMRT (1.5)], this is also equivalent to saying that $(A,\sigma)$ is isomorphic to a tensor product $(A,\sigma)\simeq(B,\tau)\otimes(C,\gamma)$ for some central simple algebra with involution $(C,\gamma)$ over $F$.

Let ${(Q,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})}$ be a quaternion division algebra over $F$, endowed with its canonical involution. We denote by $F_Q$ the function field of the associated conic, which is the Severi–Brauer variety of $Q$. Since ${\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}}$ is of symplectic type, it becomes hyperbolic over any field that splits $Q$, hence in particular over $F_Q$. From this, one may easily deduce that *any $(A,\sigma)$ that contains ${(Q,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})}$ becomes hyperbolic over $F_Q$.* The main theme of this paper is to investigate the reverse implication. In the case where $A$ is split and $\sigma$ is anisotropic, it is an easy consequence of the Cassels–Pfister subform theorem in the algebraic theory of quadratic forms that the converse holds, see Proposition \[split.prop\]. When $A$ is not split, the problem is much more delicate, and comparable to the characterization of quadratic forms that become *isotropic* over $F_Q$, which was studied by Hoffmann, Lewis, and Van Geel [@Hoff:thesis], [@HLVG], [@HVG]. Using an example from [@HVG] of a $7$-dimensional quadratic form over a suitable field $F$ that becomes isotropic over $F_Q$, we construct in §\[totdec.sec\] a division algebra $A$ of degree $8$ with an orthogonal involution $\sigma$ such that $(A,\sigma)$ becomes hyperbolic over $F_Q$ but does not contain ${(Q,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})}$, even though $A$ contains $Q$ and $(A,\sigma)$ decomposes into a tensor product of quaternion algebras with involution. This situation does not occur in lower degrees.

Algebras with involution that become split hyperbolic over $F_Q$ are considered in §\[split.sec\]. It is shown in Proposition \[split.cor\] that their anisotropic kernel contains ${(Q,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})}$ (if it is not trivial). This applies in particular to algebras of degree $2m$ with $m$ odd, see Corollary \[split2.cor\]. The case of algebras of degree $4$ is completely elucidated in §\[deg4.sec\], using Clifford algebras for orthogonal involutions and a relative cohomological invariant of degree $3$ due to Knus–Lam–Shapiro–Tignol for symplectic involutions. In the symplectic case, we classify the algebras of degree $4$ that become hyperbolic over $F_Q$ but do not contain ${(Q,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})}$, see Theorem \[deg4symp.thm\]. We show in §\[min.sec\] that our result is equivalent to the Hoffmann–Lewis–Van Geel classification of $5$-dimensional quadratic forms that become isotropic over $F_Q$ without containing a Pfister neighbour of the norm form of $Q$, see Corollary \[qf.cor\]. Orthogonal involutions on algebras of degree $8$ are considered in §§\[totdec.sec\] and \[deg8.sec\], using triality. In §\[totdec.sec\] we relate tensor products of quaternion algebras to quadratic forms of dimension $8$ with trivial discriminant. The algebras with involution that do not decompose into tensor products of quaternion algebras with involution and become hyperbolic over $F_Q$ are determined in §\[deg8.sec\], see Theorem \[deg8.thm\]. They are isotropic, and their anisotropic kernel contains ${(Q,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})}$. Finally, we use Laurent power series in §\[large.sec\] to construct algebras with involution of large degree that do not contain ${(Q,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})}$.

Notations and preliminary observations
======================================

We work over a base field $F$ of characteristic different from $2$, and only consider algebras with involution of the first kind. We refer the reader to [@KMRT] and to [@Lam] for background information on central simple algebras with involution and on quadratic forms. However, we depart from the notation in [@Lam] by using ${{\langle\!\langle{a_1,\dots, a_n}\rangle\!\rangle}}$ to denote the $n$-fold Pfister form $\otimes_{i=1}^n{{\left\langle{1,-a_i}\right\rangle}}$.

If $h\colon V\times V\to D$ is a regular hermitian or skew-hermitian form on a finite-dimensional vector space $V$ over a division algebra $D$, we denote by $\operatorname{ad}_h$ the involution on $\operatorname{End}_D(V)$ that is adjoint to $h$. In the particular case where $D$ is split and $h$ is the polar form of a quadratic form $q$, we also denote $\operatorname{ad}_q$ for $\operatorname{ad}_h$. Following Becher [@Becher], for any $n$-dimensional quadratic form $q$ over $F$, we denote by $\operatorname{Ad}_q$ the split orthogonal algebra with involution $(M_n(F),\operatorname{ad}_q)$.

Recall that a central simple $F$-algebra with involution $(A,\sigma)$ is hyperbolic if and only if it is isomorphic to $(\operatorname{End}_DV,\operatorname{ad}_h)$ for some hyperbolic hermitian or skew-hermitian form $h$ on a vector space $V$ over a division algebra $D$. In particular, $A$ admits a hyperbolic involution if and only if the index of $A$ divides $\frac12\deg(A)$. If so, then $A$ admits up to conjugation a unique orthogonal hyperbolic involution, and a unique symplectic one; see [@BST] or [@KMRT 6.B].

For any central simple $F$-algebra with involution $(A,\sigma)$ and any field extension $K/F$, we let $(A,\sigma)_K=(A\otimes_FK,\sigma\otimes{\operatorname{Id}})$. As pointed out in the introduction, we shall be mostly interested in the special case where $K=F_Q$ is the function field of the conic associated to a quaternion algebra $Q$, i.e. its Severi–Brauer variety. Throughout the paper, we fix the notation $$Q=(a,b)_F$$ with $a$, $b\in F^\times$, and we assume $Q$ is not split. We may then identify $F_Q$ with the quadratic extension $F(\sqrt{at^2+b})$ of $F(t)$, where $t$ is an indeterminate. For any central simple $F$-algebra $A$, we let $A(t)=A\otimes_FF(t)$ and $A[t]=A\otimes_FF[t]$.

Let $(A,\sigma)$ be a central simple $F$-algebra with involution. Assume $\sigma$ is anisotropic. The following conditions are equivalent:

1.  $(A,\sigma)_{F_Q}$ is hyperbolic;

2.  $A(t)$ contains an element $y$ satisfying $$\label{y.eq} 
        \sigma(y)=-y\qquad\text{and}\qquad y^2=at^2+b.$$

If conditions *(a)* and *(b)* hold, then $A[t]$ contains an element $y_0$ such that $$\label{y0.eq}
  \sigma(y_0)y_0=-(at^2+b).$$ Moreover, the following conditions are equivalent:

1.  $(A,\sigma)$ contains ${(Q,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})}$;

2.  $A[t]$ contains an element $y$ satisfying .

The equivalence of (a) and (b) readily follows from the description of anisotropic involutions that become hyperbolic over a quadratic extension in [@BST 3.3]. If $y\in A(t)$ satisfies , then the version of the Cassels–Pfister theorem for algebras with involution in [@T:96] yields an element $u\in A(t)$ such that $\sigma(u)u=1$ and $uy\in A[t]$. Then $y_0=uy$ satisfies .

If (a’) holds, then $A$ contains two skew-symmetric elements $i$, $j$ such that $i^2=a$, $j^2=b$, and $ji=-ij$. Then $y=it+j\in A[t]$ satisfies , so (b’) holds. Conversely, suppose (b’) holds and let $y\in A[t]$ satisfy . We then have $\sigma(y)y=-(at^2+b)$, hence the degree of $y$ is $1$ since $\sigma$ is anisotropic. Thus, $y=\lambda t+\mu$ for some $\lambda$, $\mu\in A$. It follows from that $\lambda$ and $\mu$ are skew-symmetric and satisfy $\lambda^2=a$, $\mu^2=b$, and $\mu\lambda=-\lambda\mu$, hence they generate a $\sigma$-stable subalgebra of $(A,\sigma)$ isomorphic to ${(Q,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})}$.

\[division.prop\] Let $(A,\sigma)$ be a central simple $F$-algebra with involution. Assume $A$ is division. If $(A,\sigma)_{F_Q}$ is hyperbolic, then $A$ contains $Q$ and $A[t]$ contains an element $y_1$ such that $y_1^2=at^2+b$.

Since $(A,\sigma)_{F_Q}$ is hyperbolic, the algebra $A_{F_Q}$ is not division, hence Merkurjev’s index reduction theorem [@Mer:92 Th. 1] shows that $A$ contains $Q$, hence also two elements $i$, $j$ such that $i^2=a$, $j^2=b$, and $ji=-ij$. Then $y_1=it+b$ satisfies $y_1^2=at^2+b$.

Thus, the algebra with involution $(A,\sigma)$ in Theorem \[totdec.thm\] below is such that $A(t)$ contains an element $y$ satisfying , and $A[t]$ contains elements $y_0$, $y_1$ satisfying $$\sigma(y_0)y_0=-(at^2+b),\qquad y_1^2=at^2+b,$$ but no element satisfying both equations.

For the rest of this section, we focus on the case where $\sigma$ is orthogonal. We then have a discriminant $\operatorname{disc}\sigma\in
F^{\times}/F^{\times2}$ and a Clifford algebra $C(A,\sigma)$, see [@KMRT §§7, 8]. Recall from [@KMRT (8.25)] that the center of $C(A,\sigma)$ is the quadratic étale $F$-algebra obtained by adjoining a square root of $\operatorname{disc}\sigma$. Therefore, if $\operatorname{disc}\sigma=1$ the algebra $C(A,\sigma)$ decomposes into a direct product of two components, which are central simple $F$-algebras, $$C(A,\sigma)\simeq C_+(A,\sigma)\times C_-(A,\sigma).$$ If $\deg A\equiv0\bmod4$, then the canonical involution $\underline{\sigma}$ on $C(A,\sigma)$ restricts to involutions $\sigma_+$ and $\sigma_-$ on $C_+(A,\sigma)$ and $C_-(A,\sigma)$. Moreover, the tensor product $C_+(A,\sigma)\otimes_FC_-(A,\sigma)$ is Brauer-equivalent to $A$, see [@KMRT (8.12), (9.14)].

\[orthog.prop\] Let $(A,\sigma)$ be a central simple $F$-algebra with orthogonal involution and $\deg A\equiv0\bmod4$. If $(A,\sigma)_{F_Q}$ is hyperbolic, then $\operatorname{disc}\sigma=1$ and at least one of $(C_+(A,\sigma),\sigma_+)$, $(C_-(A,\sigma),\sigma_-)$ is split and isotropic.

Since $\sigma_{F_Q}$ is hyperbolic, $\operatorname{disc}\sigma$ is a square in $F_Q$, hence also in $F$ since $F$ is quadratically closed in $F_Q$. By [@KMRT (8.31)], the Clifford algebra of any hyperbolic involution has a split component; therefore $C_+(A,\sigma)_{F_Q}$ or $C_-(A,\sigma)$ is split. The corresponding involution $\sigma_\pm$ is isotropic over $F_Q$ by the main theorem in [@G:clif].

Since the Brauer group kernel of the scalar extension map from $F$ to $F_Q$ is $\{0,[Q]\}$, the description of algebras $(A,\sigma)$ with $\sigma$ orthogonal and $\deg A\equiv0\bmod4$ such that $(A,\sigma)_{F_Q}$ is hyperbolic falls into two cases:

Case 1

:   $\bigl\{[C_+(A,\sigma)],[C_-(A,\sigma)]\bigr\}=
    \bigl\{0,[A]\bigr\}$;

Case 2

:   $\bigl\{[C_+(A,\sigma)],[C_-(A,\sigma)]\bigr\}=
    \bigl\{[Q],[A\otimes_FQ]\bigr\}$.

Following Garibaldi’s definition in [@G:16], case 1 is when $(A,\sigma)\in I^3$. Note that these two cases are not exclusive: they intersect if and only if $A$ is Brauer-equivalent to $Q$; then $(A,\sigma)$ becomes split hyperbolic over $F_Q$. This case is easily dealt with in §\[split.sec\]. If $\deg A=4$, the first case arises if and only if $(A,\sigma)$ is hyperbolic, and the second case if and only if $(A,\sigma)$ contains ${(Q,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})}$, see §\[deg4orth.sec\]. For algebras of degree $8$, case 1 is when $(A,\sigma)$ is totally decomposable, see §\[totdec.sec\]; case 2 is considered in §\[deg8.sec\].

Algebras with involution that become split hyperbolic over $F_Q$ {#split.sec}
================================================================

Since $F_Q$ is the function field of the Severi–Brauer variety of $Q$, the Brauer group kernel of the scalar extension from $F$ to $F_Q$ is $\{0,[Q]\}$. Therefore, if $(A,\sigma)$ is a central simple $F$-algebra with involution that becomes split hyperbolic over $F_Q$, then either $A$ is split or $A$ is Brauer-equivalent to $Q$. We consider each case separately.

\[split.prop\] Let $(A,\sigma)$ be a split central simple $F$-algebra with involution.

1.  If $\sigma$ is symplectic, then it is hyperbolic. In this case, $(A,\sigma)$ contains ${(Q,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})}$ if and only if $\deg A\equiv0\bmod4$.

2.  If $\sigma$ is orthogonal, then $(A,\sigma)\simeq\operatorname{Ad}_q$ for some quadratic form $q$. We have $(\operatorname{Ad}_q)_{F_Q}$ hyperbolic if and only if the anisotropic kernel of $q$ is a multiple of the norm form $n_Q$. Assuming this condition holds, $\operatorname{Ad}_q$ contains ${(Q,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})}$ if and only if the Witt index of $q$ is a multiple of $4$.

\(1) Since every alternating form on an $F$-vector space is hyperbolic, it follows that every split algebra with hyperbolic involution is hyperbolic. If $\rho$ is an orthogonal involution on $Q$, then ${(Q,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})}\otimes(Q,\rho)$ is a split algebra of degree $4$ with symplectic involution. If $\deg A=4m$, we have $$(A,\sigma)\simeq{(Q,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})}\otimes(Q,\rho)\otimes\operatorname{Ad}_q$$ for any quadratic form $q$ of dimension $m$ since all the hyperbolic involutions on $A$ are conjugate. Conversely, if $A$ contains $Q$, then the centralizer of $Q$ in $A$ is Brauer-equivalent to $Q$, hence of even degree. Therefore, $\deg A\equiv0\bmod4$.

\(2) By definition, $(\operatorname{Ad}_q)_{F_Q}$ is hyperbolic if and only if $q_{F_Q}$ is hyperbolic, hence the first statement follows from [@Lam X(4.11)]. Let $n_Q\otimes q_0$ be the anisotropic kernel of $q$. If the Witt index of $q$ is $4m$, then denoting by ${\mathbb{H}}$ the hyperbolic plane over $F$ we have $$q\simeq n_Q\otimes(q_0\perp m{\mathbb{H}}).$$ Since $\operatorname{Ad}_{n_Q}\simeq{(Q,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})}\otimes{(Q,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})}$ (see [@KMRT (11.1)]) it follows that $$\operatorname{Ad}_q\simeq{(Q,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})}\otimes{(Q,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})}\otimes\operatorname{Ad}_{q_0\perp m{\mathbb{H}}},$$ so $\operatorname{Ad}_q$ contains ${(Q,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})}$. Conversely, assume $\operatorname{Ad}_q$ contains ${(Q,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})}$ and let $(A_1,\sigma_1)$ be the centralizer of ${(Q,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})}$ in $(A,\sigma)$. Then $$\operatorname{Ad}_q\simeq{(Q,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})}\otimes(A_1,\sigma_1)$$ hence $\sigma_1$ is symplectic and $A_1$ is Brauer-equivalent to $Q$. Therefore, there is a hermitian $Q$-form $h_1$ such that $\sigma_1\simeq\operatorname{ad}_{h_1}$. The anisotropic kernel of $h_1$ has a diagonalization ${{\left\langle{a_1,\ldots,a_r}\right\rangle}}_Q$ with $a_1$, …, $a_r\in F^\times$; letting $m$ be the Witt index of $h_1$, we have $$h_1\simeq{{\left\langle{a_1,\ldots,a_r}\right\rangle}}_Q\perp m{\mathbb{H}}_Q.$$ Consider the following quadratic form over $F$: $$q_1={{\left\langle{a_1,\ldots,a_r}\right\rangle}}\perp m{\mathbb{H}}.$$ Then $(A_1,\sigma_1)\simeq{(Q,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})}\otimes\operatorname{Ad}_{q_1}$, hence $\operatorname{Ad}_q\simeq\operatorname{Ad}_{n_Q\otimes q_1}$, and therefore $q$ is isometric to $n_Q\otimes q_1$ up to a scalar factor. Now, $$n_Q\otimes q_1=n_Q\otimes{{\left\langle{a_1,\ldots,a_r}\right\rangle}}\perp m(n_Q\otimes{\mathbb{H}})
  = n_Q\otimes{{\left\langle{a_1,\ldots,a_r}\right\rangle}}\perp4m{\mathbb{H}},$$ and $n_Q\otimes{{\left\langle{a_1,\ldots, a_r}\right\rangle}}$ is anisotropic since ${{\left\langle{a_1,\ldots, a_r}\right\rangle}}_Q$ is anisotropic (see [@Scha Ch. 10, Th. 1.1]). Therefore, the Witt index of $q$ is $4m$.

For algebras that are Brauer-equivalent to $Q$, the result is as follows:

Let $(A,\sigma)$ be a central simple $F$-algebra with involution such that $A$ is Brauer-equivalent to $Q$.

1.  If $\sigma$ is symplectic, then $(A,\sigma)$ contains ${(Q,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})}$ and $(A,\sigma)_{F_Q}$ is hyperbolic.

2.  If $\sigma$ is orthogonal and $(A,\sigma)_{F_Q}$ is hyperbolic, then $(A,\sigma)$ is hyperbolic and contains ${(Q,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})}$.

\(1) It was already observed in the proof of Proposition \[split.prop\](2) that every central simple algebra with symplectic involution that is Brauer-equivalent to $Q$ contains ${(Q,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})}$. Therefore, every such algebra becomes hyperbolic over $F_Q$.

\(2) The first statement follows from the injectivity of the scalar extension map $W^-{(Q,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})}\to W(F_Q)$ proved independently in [@Dej:01] and [@PSS]. If $(A,\sigma)$ is hyperbolic, then $A=M_r(F)\otimes
  Q$ for some even integer $r$, and $\sigma$ is conjugate to $\tau\otimes{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}}$ for any symplectic (hyperbolic) involution $\tau$ on $M_r(F)$. In particular, $(A,\sigma)$ contains ${(Q,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})}$.

Focusing on the anisotropic case in the propositions above, we have:

\[split.cor\] Let $(A,\sigma)$ be a central simple $F$-algebra with involution. Assume $\sigma$ is anisotropic and $(A,\sigma)_{F_Q}$ is split hyperbolic. Then either

-   $A$ is split, $\sigma$ is orthogonal, and there is a quadratic form $q$ such that $$(A,\sigma)\simeq\operatorname{Ad}_{n_Q\otimes
        q}\simeq{(Q,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})}\otimes{(Q,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})}\otimes\operatorname{Ad}_q;$$ or

-   $A$ is Brauer-equivalent to $Q$, $\sigma$ is symplectic, and there is a quadratic form $q$ such that $$(A,\sigma)\simeq{(Q,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})}\otimes\operatorname{Ad}_q.$$

In both cases, $(A,\sigma)$ contains ${(Q,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})}$.

Using the results in this section, we may describe the algebras with involution of degree $2\bmod4$ that become hyperbolic over $F_Q$:

\[split2.cor\] Let $(A,\sigma)$ be a central simple $F$-algebra with involution. If $\deg A\equiv2\bmod4$ and $(A,\sigma)_{F_Q}$ is hyperbolic, then $A_{F_Q}$ is split. If moreover $\sigma$ is anisotropic, then it is symplectic and there is an odd-dimensional quadratic form $q$ over $F$ such that $(A,\sigma)\simeq{(Q,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})}\otimes\operatorname{Ad}_q$.

Since $\deg A\equiv2\bmod4$, we have $A\simeq M_r(H)$ for some odd integer $r$ and some quaternion algebra $H$ over $F$. If $H$ is division, $A$ does not admit a hyperbolic involution. Therefore, the hypothesis that $(A,\sigma)_{F_Q}$ is hyperbolic implies that $H_{F_Q}$ is split, hence $(A,\sigma)_{F_Q}$ is split hyperbolic. The last statement then readily follows from Corollary \[split.cor\].

In view of this corollary, we only consider central simple algebras of degree divisible by $4$ in the following sections.

Algebras of degree 4 {#deg4.sec}
====================

Throughout this section, $A$ is a central simple algebra of degree $4$ over an arbitrary field $F$ of characteristic different from $2$. Involutions on $A$ are classified by cohomological invariants (see [@KMRT Ch. 4]), which we use to give an explicit description of the involutions on $A$ that become hyperbolic over $F_Q$.

The orthogonal case {#deg4orth.sec}
-------------------

This case is easy to handle using Clifford algebras.

\[deg4orth.prop\] Let $\sigma$ be an orthogonal involution on $A$.

1.  $(A,\sigma)$ is hyperbolic if and only if $\operatorname{disc}\sigma=1$ and one of the components of $C(A,\sigma)$ is split;

2.  $(A,\sigma)$ contains ${(Q,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})}$ if and only if $\operatorname{disc}\sigma=1$ and one of the components of $C(A,\sigma)$ is isomorphic to $Q$;

3.  $A$ contains $Q$ if and only if $A\otimes_FQ$ is Brauer-equivalent to a quaternion algebra.

\(1) This readily follows from [@BST 2.5].

\(2) If $(A,\sigma)$ contains ${(Q,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})}$, then there is a quaternion $F$-algebra $Q'$ with canonical involution ${\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}}$ such that $$(A,\sigma)\simeq{(Q,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})}\otimes(Q',{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}}).$$ By [@KMRT (15.12)], this relation holds if and only if $C(A,\sigma)\simeq Q\times Q'$. This proves (2).

\(3) If $A$ contains (a copy of) $Q$, then the centralizer of $Q$ in $A$ is a quaternion $F$-algebra $Q'$, and we have $$A\simeq Q\otimes_FQ'.$$ This relation holds if and only if $A\otimes_FQ$ is Brauer-equivalent to $Q'$, hence (3) follows.

Let $\sigma$ be an orthogonal involution on $A$. Then $(A,\sigma)_{F_Q}$ is hyperbolic if and only if $(A,\sigma)$ is hyperbolic or contains ${(Q,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})}$.

As noticed in the introduction, any $(A,\sigma)$ containing ${(Q,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})}$ is hyperbolic over $F_Q$, so we only have to prove the converse. Assume $(A,\sigma)_{F_Q}$ hyperbolic. Since $F$ is quadratically closed in $F_Q$, the discriminant of $\sigma$ is trivial, so $C(A,\sigma)$ is a direct product of two quaternion algebras. At least one of these quaternion algebras splits over $F_Q$ by Proposition \[deg4orth.prop\](1), hence that component must be either split or isomorphic to $Q$. It follows that $(A,\sigma)$ is either hyperbolic or contains ${(Q,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})}$, by Proposition \[deg4orth.prop\].

Note that a central simple algebra of degree $4$ with hyperbolic involution does not necessarily contain ${(Q,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})}$, even if it contains $Q$: if $Q'$ is a quaternion $F$-algebra such that $Q\otimes_FQ'$ has index $2$, then $(A,\sigma)=(M_2(F),{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})\otimes(Q',{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})$ is hyperbolic over $F$ and satisfies $C(A,\sigma)\simeq M_2(F)\times Q'$ by [@KMRT (15.12)]. Therefore, Proposition \[deg4orth.prop\](2) shows that $(A,\sigma)$ does not contain ${(Q,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})}$, even though Proposition \[deg4orth.prop\](3) shows that $A$ contains $Q$. This situation does not occur in the symplectic case, in view of Proposition \[deg4symp.prop\] below.

The symplectic case
-------------------

Symplectic involutions on $A$ are classified up to conjugation by a relative invariant $\Delta$ with values in the Galois cohomology group $H^3(F,\mu_2)$, see[^1] [@KMRT (16.9)]. We denote by $[A]$ the Brauer class of $A$, viewed as an element of $H^2(F,\mu_2)$, and for $\lambda\in F^\times$ we denote by $(\lambda)$ the square class of $\lambda$, viewed as an element in $H^1(F,\mu_2)$. Using the invariant $\Delta$, we show:

\[deg4symp.prop\] Let $\sigma$ be a symplectic involution on $A$. If $A$ contains $Q$ and $(A,\sigma)_{F_Q}$ is hyperbolic, then $(A,\sigma)$ contains ${(Q,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})}$.

Since $A$ contains $Q$, it decomposes as $A=Q\otimes Q'$ for some quaternion algebra $Q'=(a',b')_F$. Let $\tau'$ be an orthogonal involution on $Q'$ of discriminant $a'$. The involution $\tau={\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}}\otimes\tau'$ on $A=Q\otimes Q'$ is of symplectic type, and clearly hyperbolic over $F_Q$ since $(A,\tau)$ contains ${(Q,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})}$. Similarly, for every invertible $y\in\operatorname{Sym}(Q',\tau')$, the involution $$\tau_y=\operatorname{Int}(1\otimes y)\circ\tau={\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}}\otimes(\operatorname{Int}(y)\circ\tau)$$ is such that $(A,\tau_y)$ contains ${(Q,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})}$. We prove below that $\sigma$ is conjugate to $\tau_y$ for a suitable $y$. By [@KMRT (16.18)], the relative discriminant $\Delta_\tau(\tau')$ is given by $$\Delta_\tau(\tau')=(\operatorname{Nrp}_\tau(1\otimes y))\cup [A]\in H^3(F,\mu_2),$$ where $\operatorname{Nrp}_\tau$ is the Pfaffian norm, as defined in [@KMRT (2.9)]. Since this relative discriminant classifies symplectic involutions on $A$ up to conjugation, it suffices to show:

\[deg4symp.lem\] There exists an invertible element $y\in \operatorname{Sym}(Q',\tau')$ such that $$\Delta_\tau(\sigma)=(\operatorname{Nrp}_\tau(1\otimes y))\cup [A].$$

The involution $\sigma$ can be written as $\sigma=\operatorname{Int}(x)\circ \tau$ for some $x\in\operatorname{Sym}(A,\sigma)$, hence we already have $$\label{disc.eq} 
  \Delta_\tau(\sigma)=(\operatorname{Nrp}_\tau(x))\cup [A],$$ and we want to prove we can substitute for $x$ some element $1\otimes y$ with $y\in\operatorname{Sym}(Q',\tau')$. Since both $\sigma$ and $\tau$ become hyperbolic over $F_Q$, the relative discriminant $\Delta_\tau(\sigma)$ is killed by $F_Q$, hence $(\operatorname{Nrp}_\tau(x))\cup [Q']=(\lambda)\cup[Q]$ for some $\lambda\in
  F^\times$. By the common slot lemma [@A Lemma 1.7], we may even assume $$(\operatorname{Nrp}_\tau(x))\cup [Q']=(\lambda)\cup[Q']=(\lambda)\cup[Q],$$ from which we deduce $$\label{la.eq}
  (\lambda)\cup([Q]+[Q'])=0,\quad\text{and}\quad(\lambda\operatorname{Nrp}_\tau(x))\cup
  [Q']=0.$$ Hence the relative discriminant of $\sigma$ is $$\Delta_\tau(\sigma)=(\operatorname{Nrp}_\tau(x))\cup([Q]+[Q'])=(\lambda
  \operatorname{Nrp}_\tau(x))\cup([Q]+[Q'])=(\lambda \operatorname{Nrp}_\tau(x))\cup[Q].$$ Moreover, the quadratic space $(\operatorname{Sym}(A,\sigma),\operatorname{Nrp}_\tau)$ is an Albert quadratic space for the biquaternion algebra $A$ by [@KMRT (16.8)]. Hence, its Clifford invariant is $e_2(\operatorname{Nrp}_\tau)=[A]=[Q]+[Q']$. We deduce from (\[la.eq\]) that the quadratic form ${{\langle\!\langle{\lambda}\rangle\!\rangle}}\otimes\operatorname{Nrp}_\tau$ has trivial Arason invariant $e_3$, and hence is hyperbolic by the Arason–Pfister Hauptsatz. So there exists an element $x'\in\operatorname{Sym}(A,\sigma)$ such that $\lambda\operatorname{Nrp}_\tau(x)=\operatorname{Nrp}_\tau(x')$, and we have proven $$\label{Q'.eq} 
  \Delta_\tau(\sigma)=(\operatorname{Nrp}_\tau(x'))\cup [Q]
  \quad\text{and}\quad(\operatorname{Nrp}_\tau(x'))\cup [Q']=0$$ for some $x'\in\operatorname{Sym}(A,\sigma)$.

Let us now pick a pure quaternion $i'\in Q'$ such that $i'^2=a'$ and $\tau'=\operatorname{Int}(i')\circ{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}}$. Denoting by $Q^0$ the vector space of pure quaternions in $Q$, we have $$\operatorname{Sym}(A,\tau)=(1\otimes\operatorname{Sym}(Q',\tau'))\oplus(Q^0\otimes i').$$ Hence $x'$ can be written as $x'=x_0+1\otimes\xi_1+\xi_2\otimes i'$ for some $x_0\in F$, some pure quaternion $\xi_1\in
  Q'^0\cap\operatorname{Sym}(Q',\tau')$ and some $\xi_2\in Q^0$. The pure quaternion $\xi_1$ being $\tau'$-symmetric, it anticommutes with $i'$, and changing of quaternionic basis if necessary, we may assume that $\xi_1^2=b'x_1^2$ for some $x_1\in F$. Similarly, we may assume that $\xi_2^2=ax_2^2$ for some $x_2\in F$. Note that we allow $x_i=0$ for $i=0$, $1$, and $2$ since some term might be zero in the decomposition of $x'$. We then have $$\label{x'.eq} 
  \operatorname{Nrp}_\tau(x')=x_0^2-b'x_1^2-aa'x_2^2,$$ and the following lemma finishes the proof:

\[fin.lem\] There exists $z_0$, $z_1$, $y_0$, $y_1$ and $y_2$ such that $\operatorname{Nrp}_\tau(x')(z_0^2-az_1^2)=y_0^2-b'y_1^2+a'b'y_2^2\not=0$.

Indeed, if we let $y=y_0+y_1j'+y_2i'j'$, where $j'$ is a pure quaternion in $Q'$ such that $j'^2=b'$ and $i'j'=-j'i'$, we have $\operatorname{Nrp}_\tau(1\otimes y)=Nrd_{Q'}(y)=y_0^2-b'y_1^2+a'b'y_2^2$ (see [@KMRT (2.11)]). Hence, $$\begin{aligned}
\Delta_\tau(\sigma)& =(\operatorname{Nrp}_\tau(x'))\cup
[(a,b)_F]=(\operatorname{Nrp}_\tau(x')(z_0^2-az_1^2))\cup
[(a,b)_F]\\
& =(\operatorname{Nrd}_{Q'}(y))\cup[A]=(\operatorname{Nrp}_\tau(1\otimes y))\cup[A],
\quad\text{ with }y\in\operatorname{Sym}(Q',\tau').\end{aligned}$$

If ${{\left\langle{1,-b',a'b'}\right\rangle}}$ is isotropic, then it is universal, hence represents $\operatorname{Nrp}_\tau(x')$ and we are done. Otherwise, denote $\mu=\operatorname{Nrp}_\tau(x')$; by (\[x’.eq\]), we have $(\mu)\cup[(aa',b')_F]=0$, and combining with , we get $(\mu)\cup[(a,b')]=0$. In terms of quadratic forms, this means that the $3$-fold Pfister form $$\label{pform.eq}
  {{\langle\!\langle{\mu,a,b'}\rangle\!\rangle}}={{\left\langle{1,-\mu,-a,\mu b'}\right\rangle}}\perp{{\left\langle{\mu
  a,ab',-\mu ab'}\right\rangle}}\perp{{\left\langle{-b'}\right\rangle}}$$ is hyperbolic. On the other hand, by , the quadratic form ${{\left\langle{1,-b',-aa',-\mu}\right\rangle}}$ is isotropic. Multiplying by $(-ab'\mu)$, we get that $-a'b'\mu$ is represented by the form ${{\left\langle{\mu a, ab',-\mu ab'}\right\rangle}}$. Hence, in view of , the quadratic form ${{\left\langle{1,-\mu,-a,\mu b',-a'b'\mu}\right\rangle}}$, which is a $5$-dimensional subform of ${{\langle\!\langle{\mu,a,b'}\rangle\!\rangle}}$, is necessarily isotropic. Since the quadratic forms ${{\left\langle{1,-a}\right\rangle}}$ and ${{\left\langle{1,-b',a'b'}\right\rangle}}$ are anisotropic, this completes the proof.

Using Proposition \[deg4symp.prop\], we may characterize the symplectic involutions on $A$ that become hyperbolic over $F_Q$:

\[deg4symp.thm\] Let $\sigma$ be a symplectic involution on $A$. If $(A,\sigma)_{F_Q}$ is hyperbolic, then either

1.  $(A,\sigma)\simeq{(Q,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})}\otimes(Q',\rho)$ for some quaternion $F$-algebra with orthogonal involution $(Q',\rho)$, or

2.  $(A,\sigma)\simeq\operatorname{Ad}_{{{\langle\!\langle{\lambda}\rangle\!\rangle}}}\otimes(Q',{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})$ for some quaternion $F$-algebra $Q'$ and some $\lambda\in F^\times$ with the following properties: $Q\otimes_FQ'$ is a division algebra and the norm forms $n_Q$, $n_{Q'}$ satisfy ${{\langle\!\langle{\lambda}\rangle\!\rangle}}\cdot n_Q\simeq{{\langle\!\langle{\lambda}\rangle\!\rangle}}\cdot n_{Q'}$.

Conversely, if $(A,\sigma)$ is of either type above, then $(A,\sigma)_{F_Q}$ is hyperbolic.

Note that (a) and (b) are mutually exclusive since in the first case $A\otimes_FQ$ has Schur index $1$ or $2$, whereas it has index $4$ in case (b).

Suppose $(A,\sigma)$ is not as in case (a) and $(A,\sigma)_{F_Q}$ is hyperbolic. Then Proposition \[deg4symp.prop\] shows that $A$ does not contain $Q$. In particular, $A$ is not division, by Proposition \[division.prop\], so $A\simeq M_2(Q')$ for some quaternion $F$-algebra $Q'$, and $Q\otimes_FQ'$ is a division algebra by Proposition \[deg4orth.prop\](3). As observed in the proof of Proposition \[split.prop\](2), $(A,\sigma)$ contains $(Q',{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})$, hence $$(A,\sigma)\simeq\operatorname{Ad}_{{{\langle\!\langle{\lambda}\rangle\!\rangle}}}\otimes(Q',{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})
  \qquad\text{for some $\lambda\in F^\times$}.$$ The algebra $A$ carries a hyperbolic symplectic involution $\tau$, and we have by [@KMRT (16.21)] $$\Delta_\tau(\sigma)=(\lambda)\cup [Q'].$$ Since $(A,\sigma)_{F_Q}$ is hyperbolic, this invariant vanishes over $F_Q$. Hence, by [@Lam X(4.11)] and Arason’s common slot lemma we may assume as in the proof of Lemma \[deg4symp.lem\] that $(\lambda)\cup
  [Q']=(\lambda)\cup [Q]$. Therefore, ${{\langle\!\langle{\lambda}\rangle\!\rangle}}\cdot
  n_{Q'}\simeq{{\langle\!\langle{\lambda}\rangle\!\rangle}}\cdot n_Q$, which shows that $(A,\sigma)$ is as in case (b).

It is clear that $(A,\sigma)_{F_Q}$ is hyperbolic in case (a). In case (b) the algebra $A$ carries a hyperbolic symplectic involution $\tau$ and $\Delta_\tau(\sigma)=(\lambda)\cup[Q']$ vanishes over $F_Q$. Therefore, $(A,\sigma)_{F_Q}$ is hyperbolic.

$F_Q$-minimal quadratic forms of dimension 5 {#min.sec}
============================================

A quadratic form $\varphi$ over $F$ is called *$F_Q$-minimal* if $\varphi_{F_Q}$ is isotropic and $\psi_{F_Q}$ is anisotropic for every proper subform $\psi\subset\varphi$. In this section, we show that Theorem \[deg4symp.thm\] can be used to recover (and is in fact equivalent to) the description of $F_Q$-minimal forms of dimension $5$ due to Hoffmann, Lewis, and Van Geel [@HLVG Prop. 4.1].

A general procedure to construct central simple algebras with involution that become hyperbolic over $F_Q$ uses Clifford algebras. Recall from [@Lam V(1.9), V(2.4)] that for any quadratic form $\varphi$ of odd dimension $2m+1$ over $F$ the even Clifford algebra $C_0(\varphi)$ is central simple over $F$ of degree $2^m$. It carries a canonical involution $\tau_0$, which is the restriction of the involution on the full Clifford algebra that leaves invariant every vector in the underlying vector space of $\varphi$. The involution $\tau_0$ is orthogonal if $m\equiv0$ or $3\bmod4$ and symplectic otherwise, see [@KMRT (8.4)].

\[qf.prop\] Let $\varphi$ be a quadratic form of odd dimension over $F$.

1.  If $\varphi_{F_Q}$ is isotropic, then $(C_0(\varphi),\tau_0)_{F_Q}$ is hyperbolic. The converse holds if $\dim\varphi=5$.

2.  If $\varphi$ contains a subform similar to ${{\left\langle{1,-a,-b}\right\rangle}}$, then $(C_0(\varphi),\tau_0)$ contains ${(Q,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})}$. The converse holds if $\dim\varphi=5$.

\(1) The first statement readily follows from [@KMRT (8.5)] and the second from [@KMRT (15.21)].

\(2) Suppose the underlying vector space of $\varphi$ contains orthogonal vectors $e_0$, $e_1$, $e_2$ satisfying for some $\lambda\in F^\times$ $$\varphi(e_0)=\lambda,\qquad \varphi(e_1)=-\lambda a,\qquad
  \varphi(e_2)=-\lambda b.$$ Then the products $e_0e_1$ and $e_0e_2$ generate a $\tau_0$-stable subalgebra of $C_0(\varphi)$ isomorphic to ${(Q,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})}$.

For the rest of the proof, suppose $\dim\varphi=5$ and $(C_0(\varphi),\tau_0)$ contains ${(Q,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})}$. The centralizer of ${(Q,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})}$ is a quaternion algebra with orthogonal involution $(Q',\rho)$ such that $$(C_0(\varphi),\tau_0)={(Q,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})}\otimes(Q',\rho).$$ Let $V\subset\operatorname{Sym}(C_0(\varphi),\tau_0)$ be the vector space of $\tau_0$-symmetric elements of trace $0$. The map $x\mapsto x^2$ defines a quadratic form $s\colon V\to F$ that is similar to $\varphi$ by the equivalence $\mathsf{B}_2\equiv\mathsf{C}_2$, see [@KMRT (15.16)]. Let $y\in Q'$ be a $\rho$-skew-symmetric unit and let $Q^0\subset Q$ be the vector space of pure quaternions. The restriction of $s$ to the subspace $Q^0\otimes y\subset V$ is similar to ${{\left\langle{1,-a,-b}\right\rangle}}$, hence the proof is complete.

\[qf.cor\] A $5$-dimensional quadratic form $\varphi$ over $F$ is $F_Q$-minimal if and only if the following conditions hold:

1.  $\varphi$ is similar to a Pfister neighbour of the $3$-fold Pfister form ${{\langle\!\langle{a,b,\lambda}\rangle\!\rangle}}$ for some $\lambda\in F^\times$, and

2.  $C_0(\varphi)\simeq M_2(Q')$ for some quaternion $F$-algebra $Q'$ such that $Q\otimes_FQ'$ is a division algebra.

Proposition \[qf.prop\] shows that $\varphi$ is $F_Q$-minimal if and only if $(C_0(\varphi),\tau_0)$ is hyperbolic but $(C_0(\varphi),\tau_0)$ does not contain ${(Q,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})}$. By Theorem \[deg4symp.thm\], this condition is equivalent to $$\label{eq:qf}
    (C_0(\varphi),\tau_0)\simeq\operatorname{Ad}_{{{\langle\!\langle{\lambda}\rangle\!\rangle}}}\otimes(Q',{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})$$ for some quaternion $F$-algebra $Q'=(a',b')_F$ and some $\lambda\in
  F^\times$ with $Q\otimes Q'$ a division algebra and ${{\langle\!\langle{a,b,\lambda}\rangle\!\rangle}}\simeq{{\langle\!\langle{a',b',\lambda}\rangle\!\rangle}}$. It follows from the isomorphism  that $\varphi$ is similar to a Pfister neighbour of ${{\langle\!\langle{a',b',\lambda}\rangle\!\rangle}}$, by [@KMRT p. 271]. Thus, (a) and (b) hold if $\varphi$ is $F_Q$-minimal. Conversely, if $C_0(\varphi)\simeq M_2(Q')$ for some quaternion $F$-algebra $Q'=(a',b')_F$, then as observed in the proof of Proposition \[split.prop\](2) we have $$(C_0(\varphi),\tau_0)\simeq\operatorname{Ad}_{{{\langle\!\langle{\mu}\rangle\!\rangle}}}\otimes(Q',{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})
  \qquad\text{for some $\mu\in F^\times$}.$$ It then follows from [@KMRT p. 271] that $\varphi$ is similar to a Pfister neighbour of ${{\langle\!\langle{a',b',\mu}\rangle\!\rangle}}$. If (a) holds, then ${{\langle\!\langle{a,b,\lambda}\rangle\!\rangle}}\simeq{{\langle\!\langle{a',b',\mu}\rangle\!\rangle}}$ and by the common slot lemma we may assume $\lambda=\mu$. Thus, $\varphi$ is $F_Q$-minimal if (a) and (b) hold.

Totally decomposable orthogonal involutions of degree 8 {#totdec.sec}
=======================================================

In this section, $A$ denotes a central simple $F$-algebra of degree $8$ and $\sigma$ is an orthogonal involution on $A$. The algebra with involution $(A,\sigma)$ is called *totally decomposable* if there are $\sigma$-stable quaternion subalgebras $Q_1$, $Q_2$, $Q_3$ in $A$ such that $A=Q_1\otimes Q_2\otimes
Q_3$. Denoting by $\sigma_i$ the restriction of $\sigma$ to $Q_i$ for $i=1$, $2$, $3$, we then have $$(A,\sigma)=(Q_1,\sigma_1)\otimes(Q_2,\sigma_2)\otimes(Q_3,\sigma_3).$$ The totally decomposable algebras of degree $8$ are characterized by the property that $\operatorname{disc}\sigma=1$ and one of the components of the Clifford algebra $C(A,\sigma)$ is split, see [@KMRT (42.11)]. Proposition \[totdec.prop\] below shows how to use this criterion to relate totally decomposable algebras to quadratic forms.

Recall that for any quadratic form $\varphi$ of dimension $8$ with $\operatorname{disc}\varphi=1$ the even Clifford algebra decomposes into a direct product of central simple $F$-algebras of degree $8$, $$C_0(\varphi)\simeq C_+(\varphi)\times C_-(\varphi).$$ The canonical involution $\tau_0$ on $C_0(\varphi)$ restricts to orthogonal involutions $\tau_+$, $\tau_-$ on $C_+(\varphi)$ and $C_-(\varphi)$, and we have $$(C_+(\varphi),\tau_+)\simeq(C_-(\varphi),\tau_-).$$ It is easily checked that $(C_+(\varphi),\tau_+)$ is totally decomposable.

\[totdec.prop\] For every central simple algebra of degree $8$ with totally decomposable orthogonal involution $(A,\sigma)$, there is a quadratic form $\varphi$ with $\dim\varphi=8$ and $\operatorname{disc}\sigma=1$ such that $$(A,\sigma)\simeq(C_+(\varphi),\tau_+).$$ The form $\varphi$ is uniquely determined by $(A,\sigma)$ up to similarity. Moreover,

-   the algebra $A$ is split if and only if $\varphi$ is a multiple of a $3$-fold Pfister form; in that case $(A,\sigma)\simeq\operatorname{Ad}_\varphi$;

-   the algebra $(A,\sigma)$ contains ${(Q,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})}$ if and only if $\varphi$ contains a subform similar to ${{\left\langle{1,-a,-b}\right\rangle}}$.

Furthermore, the following conditions are equivalent:

1.  $(A,\sigma)$ is isotropic;

2.  $(A,\sigma)$ is hyperbolic;

3.  $\varphi$ is isotropic.

Since $(A,\sigma)$ is totally decomposable, it follows from [@KMRT (42.11)] that one of the components $C_+(A,\sigma)$ of the Clifford algebra is split. The canonical involution $\sigma_+$ on $C_+(A,\sigma)$ is orthogonal, hence there is a quadratic form $\varphi$ of dimension $8$ such that $$(C_+(A,\sigma),\sigma_+)\simeq\operatorname{Ad}_\varphi.$$ By triality (see [@KMRT (42.3)]) we have $$(C_0(\varphi),\tau_0)\simeq(A,\sigma)\times(A,\sigma).$$ Therefore, $\operatorname{disc}\varphi=1$ and $(A,\sigma)\simeq
  (C_+(\varphi),\tau_+)$. Conversely, triality also shows that if $(A,\sigma)\simeq(C_+(\varphi),\tau_+)$, then the canonical involution $\underline{\sigma}$ on $C(A,\sigma)$ satisfies $$(C(A,\sigma),\underline{\sigma})\simeq\operatorname{Ad}_\varphi\times(A,\sigma),$$ hence the form $\varphi$ is uniquely determined up to similarity.

The Clifford algebra of $\varphi$ splits if and only if $\varphi$ is a multiple of a $3$-fold Pfister form, by [@Scha Ch. 2, Th. 14.4] and [@Lam X(5.6)]. When that condition holds we have $(A,\sigma)\simeq\operatorname{Ad}_\varphi$ by [@KMRT (35.1)].

If $\varphi$ contains a multiple of ${{\left\langle{1,-a,-b}\right\rangle}}$, then the same argument as in the proof of Proposition \[qf.prop\](2) shows that $(C_0(\varphi),\tau_0)$ contains ${(Q,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})}$. Projecting on each component, it follows that $(C_+(\varphi),\tau_+)$ contains ${(Q,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})}$. Conversely, if $(A,\sigma)$ contains ${(Q,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})}$, then we have $$\label{eq:2}
    (A,\sigma)={(Q,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})}\otimes(A_1,\sigma_1)$$ for some central simple algebra with symplectic involution $(A_1,\sigma_1)$ of degree $4$. By [@KMRT (15.19)] there is a $5$-dimensional quadratic form $\psi$ such that $\operatorname{disc}\psi=1$ and $(A_1,\sigma_1)\simeq(C_0(\psi),\tau_0)$. Letting $\tau'_0$ be the canonical involution on $C_0({{\left\langle{ab,-a,-b}\right\rangle}})$, we may rewrite as $$\label{eq:1}
    (A,\sigma)\simeq(C_0({{\left\langle{ab,-a,-b}\right\rangle}}),\tau'_0)\otimes
    (C_0(\psi),\tau_0).$$ In view of the canonical embedding $$C_0({{\left\langle{ab,-a,-b}\right\rangle}})\otimes_FC_0(\psi)\hookrightarrow
  C_0({{\left\langle{ab,-a,-b}\right\rangle}}\perp\psi),$$ which is compatible with the canonical involutions, yields $$(A,\sigma)\simeq(C_+({{\left\langle{ab,-a,-b}\right\rangle}}\perp\psi),\tau_+).$$ Uniqueness of $\varphi$ shows that $\varphi$ is similar to ${{\left\langle{ab,-a,-b}\right\rangle}}\perp\psi$, hence it contains a subform similar to ${{\left\langle{1,-a,-b}\right\rangle}}$.

The equivalence of (a), (b), (c) is clear if $A$ is split, since then $(A,\sigma)\simeq\operatorname{Ad}_\varphi$ and $\varphi$ is a $3$-fold Pfister form, hence it is isotropic if and only if it is hyperbolic. For the rest of the proof, we may thus assume $A$ is not split. If $(A,\sigma)$ is hyperbolic, then it follows from [@G:clif] that the split component of $(C(A,\sigma),\underline{\sigma})$ is isotropic, hence (b)$\Rightarrow$(c). Conversely, if (c) holds, then [@KMRT (8.5)] shows that $(C_0(\varphi),\tau_0)$ is hyperbolic, hence $(C_+(\varphi),\tau_+)$ also is hyperbolic, proving (c)$\Rightarrow$(b). The equivalence of (a) and (b) readily follows from [@BFPQM Prop. 2.10].

To give an example of a division $F$-algebra of degree $8$ with a totally decomposable orthogonal involution that is hyperbolic over $F_Q$ but does not contain ${(Q,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})}$, we use an example of $F_Q$-minimal quadratic form of dimension $7$ due to Hoffmann and Van Geel [@HVG]. For the rest of this section, we fix the following notation: $F_1=F_0(t,u)$ is the function field in two independent indeterminates over an arbitrary field $F_0$ of characteristic different from $2$, and $F=F_1((a))((b))$ is the iterated Laurent series field in two indeterminates $a$, $b$. In accordance with our running notation, $Q$ denotes the quaternion algebra $(a,b)_F$. Let $$\varphi_0={{\left\langle{1+t,u}\right\rangle}}\perp{{\left\langle{-a}\right\rangle}}{{\left\langle{1,u}\right\rangle}}
\perp{{\left\langle{-b}\right\rangle}}{{\left\langle{1,t+u}\right\rangle}}\perp{{\left\langle{ab}\right\rangle}}{{\left\langle{t}\right\rangle}}$$ (see [@HVG p. 43]) and $$\varphi=\varphi_0\perp{{\left\langle{ab}\right\rangle}}{{\left\langle{t(1+t)(t+u)}\right\rangle}},$$ so $\dim\varphi=8$ and $\operatorname{disc}\varphi=1$. Let also $$(A,\sigma)=(C_+(\varphi),\tau_+)\quad(=(C_0(\varphi_0),\tau_0)),$$ a central simple $F$-algebra of degree $8$ with a totally decomposable involution.

\[totdec.thm\] The algebra with involution $(A,\sigma)$ does not contain ${(Q,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})}$, yet $(A,\sigma)_{F_Q}$ is hyperbolic. Moreover, $A$ is a division algebra.

Since ${{\left\langle{1,t}\right\rangle}}\simeq{{\left\langle{1+t,t(1+t)}\right\rangle}}$ and ${{\left\langle{t,u}\right\rangle}}\simeq{{\left\langle{t+u,tu(t+u)}\right\rangle}}$, we have $$\label{eq:3}
    \varphi_0\perp {{\left\langle{t(1+t),-at,-btu(t+u), ab, abu}\right\rangle}} \simeq
    {{\left\langle{1,t,u}\right\rangle}}{{\langle\!\langle{a,b}\rangle\!\rangle}}.$$ Since the right side is hyperbolic over $F_Q$, it follows that $(\varphi_0)_{F_Q}$ is isotropic, and therefore $(A,\sigma)_{F_Q}$ is hyperbolic by Proposition \[qf.prop\](1) or \[totdec.prop\].

To show $(A,\sigma)$ does not contain ${(Q,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})}$, we prove $\varphi$ does not contain any subform similar to ${{\left\langle{1,-a,-b}\right\rangle}}$. As in [@HVG p. 43], we decompose $\varphi$ as $$\label{decqf} 
  \varphi=\alpha\perp{{\left\langle{-a}\right\rangle}}\beta\perp{{\left\langle{-b}\right\rangle}}\gamma
  \perp{{\left\langle{ab}\right\rangle}}\delta,$$ where $\alpha={{\left\langle{1+t,u}\right\rangle}}$, $\beta={{\left\langle{1,u}\right\rangle}}$, $\gamma={{\left\langle{1,t+u}\right\rangle}}$, and $\delta={{\left\langle{t,t(1+t)(t+u)}\right\rangle}}$. By Springer’s theorem, the isometry classes of $\alpha$, $\beta$, $\gamma$, and $\delta$ over $F_1$ are uniquely determined by $\varphi$. If three of those quadratic forms represent a common value $\lambda$, then $\varphi$ contains a $3$-dimensional subform of ${{\left\langle{\lambda}\right\rangle}}{{\langle\!\langle{a,b}\rangle\!\rangle}}$, hence a subform similar to ${{\left\langle{1,-a,-b}\right\rangle}}$. We claim that the converse also holds. Indeed, assume first that $\varphi$ contains ${{\left\langle{\lambda}\right\rangle}}{{\left\langle{1,-a,-b}\right\rangle}}$ for some $\lambda\in
  F_1^\times$. Writing $\varphi={{\left\langle{\lambda}\right\rangle}}{{\left\langle{1,-a,-b}\right\rangle}}\perp\varphi'$ as in , we get, by uniqueness of the forms $\alpha$, $\beta$ and $\gamma$ up to isometry, that all three represent $\lambda$. Consider now the general situation, where $\lambda\in
  F^\times$ need not be in $F_1$; modifying it by a square if necessary, we may write it as $\lambda_0$, $-a\lambda_0$, $-b\lambda_0$ or $ab\lambda_0$ for some $\lambda_0\in
  F_1^\times$. The same argument as above then proves the claim.

Thus, to prove that $\varphi$ does not contain any subform similar to ${{\left\langle{1,-a,-b}\right\rangle}}$, we have to show that no three of the quadratic forms $\alpha$, $\beta$, $\gamma$, and $\delta$ have any common value over $F_1$. This can be checked after some scalar extension. For instance, by [@HVG Lemma(4.4)(iii)], the only common value of $\alpha$ and $\beta$ over $F_0(t)((u))$ is the square class of $u$. On the other hand, applying [@Lam VI(1.3)], one may check that $\gamma$ and $\delta$ are both isomorphic over $F_0((t))((u))$ to ${{\left\langle{1,t}\right\rangle}}$, which does not represent $u$. On the other hand, over $F_0(t)((t+u))$ the form $\gamma$ only represents the square classes of $1$ and $t+u$, whereas $\delta$ only represents the square classes of $t$ and $t(1+t)(t+u)$. Therefore, $\gamma$ and $\delta$ have no common value in $F_1$.

To complete the proof, we show $A$ is a division algebra. Taking the Clifford invariant of each side of and applying [@Lam V(3.13)], we obtain the following equality in the Brauer group of $F$: $$[A]+[C_0({{\left\langle{t(1+t), -at, -btu(t+u), ab, abu}\right\rangle}})]=[Q].$$ The even Clifford algebra of the $5$-dimensional form is easily computed: $$C_0({{\left\langle{t(1+t), -at, -btu(t+u), ab, abu}\right\rangle}})\simeq (-u,bt)_F\otimes
  (ab(t+u),-au(1+t))_F,$$ hence $$\begin{aligned}
    A & \simeq (a,b)_F\otimes (-u,bt)_F\otimes (ab(t+u),-au(1+t))_F\\
      & \simeq (-u,t)_F\otimes (a(t+u),u(1+t)(t+u))_F \otimes (b,1+t)_F.
  \end{aligned}$$ Since $b$ is a uniformizing parameter for the $b$-adic valuation on $F$, it follows from [@Pierce §19.6, Prop.] that the right side is a division algebra if (and only if) the algebra $$B=(-u,t)_{F_1((a))}\otimes (a(t+u),u(1+t)(t+u))_{F_1((a))}\otimes
  F_1\bigl(\sqrt{1+t}\bigr)((a))$$ is division. (Alternatively, one may view $A$ as a ring of twisted Laurent series over $B$ in an indeterminate whose square is $b$.) Now, $a(t+u)$ is a uniformizing parameter for the $a$-adic valuation on $F_1\bigl(\sqrt{1+t}\bigr)((a))$, hence the same argument shows that $B$ is a division algebra if (and only if) the algebra $$C=(-u,t)_{F_1}\otimes F_1\bigl(\sqrt{1+t}, \sqrt{u(1+t)(t+u)}\bigr)$$ is division. Since $1+t$ and $u(t+u)$ are squares in $F_0(u)((t))$, we may embed $C$ in the quaternion algebra $(-u,t)_{F_0(u)((t))}$, which is clearly division. Therefore, $A$ is a division algebra.

Non-totally decomposable orthogonal involutions of degree 8 {#deg8.sec}
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In this section, we consider the case of central simple algebras with orthogonal involution $(A,\sigma)$ of degree $8$ that are not totally decomposable. These algebras do not contain any quaternion algebra with canonical involution $(H,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})$, since the centralizer of $H$ would be an algebra of degree $4$ with symplectic involution, hence decomposable by [@KMRT (16.16)]; the algebra $(A,\sigma)$ would then be totally decomposable.

We start with a couple of lemmas of independent interest related to triality. Let $Q_1$, $Q_2$, $Q_3$ be quaternion $F$-algebras such that $Q_1\otimes_FQ_2\otimes_FQ_3$ is split. By a well-known result due to Albert and to Pfister, this condition implies that $Q_1$, $Q_2$, and $Q_3$ have a common maximal subfield, see [@KMRT (16.30)]. Therefore, we may write $$Q_1=(c,d_1)_F,\qquad Q_2=(c,d_2)_F,\qquad Q_3=(c,d_3)_F$$ for some $c$, $d_1$, $d_2$, $d_3\in F^\times$ such that the quaternion algebra $(c,d_1d_2d_3)_F$ is split. For $\alpha=1$, $2$, $3$, let $\rho_\alpha$ be the orthogonal involution on $Q_\alpha$ with $\operatorname{disc}\rho_\alpha=c$. The involution $\rho_\alpha$ is uniquely determined up to conjugation by [@KMRT (7.4)].

\[quat.lem\] For $\{\alpha,\beta,\gamma\}=\{1,2,3\}$ we have $$(Q_\alpha,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})\otimes(Q_\beta,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})\simeq
  \operatorname{Ad}_{{{\langle\!\langle{d_\alpha}\rangle\!\rangle}}}\otimes(Q_\gamma,\rho_\gamma) \simeq
  \operatorname{Ad}_{{{\langle\!\langle{d_\beta}\rangle\!\rangle}}}\otimes(Q_\gamma,\rho_\gamma).$$

By Tao’s computation of the Clifford algebra of a decomposable involution [@Tao:94] we have $$\begin{aligned}
    C\bigl((Q_\alpha,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})\otimes(Q_\beta,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})\bigr)& \simeq
    Q_\alpha\times Q_\beta\simeq (c,d_\alpha)_F\times (c,d_\beta)_F,\\
    C\bigl(\operatorname{Ad}_{{{\langle\!\langle{d_\alpha}\rangle\!\rangle}}}\otimes(Q_\gamma,\rho_\gamma)\bigr)&
    \simeq (c,d_\alpha)_F\times(c,d_\alpha d_\gamma)_F\simeq
    (c,d_\alpha)_F\times(c,d_\beta)_F,\\
    C\bigl(\operatorname{Ad}_{{{\langle\!\langle{d_\beta}\rangle\!\rangle}}}\otimes(Q_\gamma,\rho_\gamma)\bigr)&
    \simeq (c,d_\beta)_F\times(c,d_\beta d_\gamma)_F\simeq
    (c,d_\beta)_F\times(c,d_\alpha)_F.
  \end{aligned}$$ Since central simple algebras with orthogonal involutions of degree 4 are classified by their Clifford algebra (see [@KMRT (15.7)]), the lemma follows.

Now, for $\alpha=1$, $2$, $3$, let $(A_\alpha,\sigma_\alpha)$ be a central simple $F$-algebra with orthogonal involution of degree 8 such that for $\beta$, $\gamma$ with $\{\alpha,\beta,\gamma\}=\{1,2,3\}$, $$(A_\alpha,\sigma_\alpha)\simeq\operatorname{Ad}_{{{\left\langle{1,-1,1,-d_\beta}\right\rangle}}}\otimes
(Q_\alpha,\rho_\alpha) \simeq \operatorname{Ad}_{{{\left\langle{1,-1,1,-d_\gamma}\right\rangle}}}\otimes
(Q_\alpha,\rho_\alpha).$$ Thus, $(A_\alpha,\sigma_\alpha)$ is Witt-equivalent to $(Q_\beta,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})\otimes (Q_\gamma,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})$ by Lemma \[quat.lem\].

\[tria.lem\] The triple $\bigl((A_1,\sigma_1), (A_2,\sigma_2), (A_3,
  \sigma_3)\bigr)$ is trialitarian, in the sense that for $\{\alpha,\beta,\gamma\}=\{1,2,3\}$ we have $$C(A_\alpha,\sigma_\alpha)\simeq (A_\beta,\sigma_\beta) \otimes
  (A_\gamma,\sigma_\gamma).$$ (See [@KMRT p. 548].)

By triality, it suffices to prove the isomorphism for $\alpha=1$, $\beta=2$, and $\gamma=3$. By definition, $(A_1,\sigma_1)$ is an orthogonal sum of the algebra $M_2(Q_1)$ with a hyperbolic involution and of $\operatorname{Ad}_{{{\langle\!\langle{d_2}\rangle\!\rangle}}}\otimes(Q_1,\rho_1)$, so by Lemma \[quat.lem\] $$(A_1,\sigma_1)\simeq \bigl((M_2(F),{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})\otimes (Q_1,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})\bigr)
  \boxplus\bigl((Q_2,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})\otimes(Q_3,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})\bigr).$$ By [@KMRT (15.12)] we have $$C\bigl((M_2(F),{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})\otimes (Q_1,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})\bigr)\simeq (M_2(F),{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})
  \times (Q_1,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})$$ and $$C\bigl((Q_2,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})\otimes(Q_3,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})\bigr)\simeq (Q_2,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})\times
  (Q_3,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}}).$$ Arguing as in Garibaldi’s “Orthogonal Sum Lemma” [@G:clif Lemma 3.2], we get $$(C(A_1,\sigma_1),\underline{\sigma_1})\simeq
  (C_+(A_1,\sigma_1),\sigma_+) \times (C_-(A_1,\sigma_1),\sigma_-)$$ with $$(C_+(A_1,\sigma_1),\sigma_+)\simeq
  \bigl((M_2(F),{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})\otimes(Q_2,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})\bigr) \boxplus
  \bigl((Q_1,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})\otimes(Q_3,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})\bigr)$$ and $$(C_-(A_1,\sigma_1),\sigma_-)\simeq
  \bigl((M_2(F),{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})\otimes(Q_3,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})\bigr) \boxplus
  \bigl((Q_1,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})\otimes(Q_2,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})\bigr).$$ Thus, $(C_+(A_1,\sigma_1),\sigma_+)$ is Witt-equivalent to $(Q_1,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})\otimes(Q_3,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})$, hence it is isomorphic to $(A_2,\sigma_2)$. Likewise, $(C_-(A_1,\sigma_1),\sigma_-)$ is isomorphic to $(A_3,\sigma_3)$.

\[deg8.thm\] Let $(A,\sigma)$ be a central simple $F$-algebra with orthogonal involution of degree $8$. Assume $(A,\sigma)$ is not totally decomposable. Then $(A,\sigma)_{F_Q}$ is hyperbolic if and only if there is a quaternion $F$-algebra $Q'$ with the following properties:

-   $\operatorname{ind}(Q\otimes_FQ')\leq2$, and

-   $(A,\sigma)$ is Witt-equivalent to $(Q,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})\otimes(Q',{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})$.

When these equivalent properties hold, we can find $c$, $d$, $d'\in
  F^\times$ such that $$Q\simeq(c,d)_F,\qquad Q'\simeq(c,d')_F,$$ and $$(A,\sigma)\simeq\operatorname{Ad}_{{{\left\langle{1,-1,1,-d}\right\rangle}}}\otimes(Q'',\rho'')$$ where $Q''=(c,dd')_F$ and $\rho''$ is an orthogonal involution on $Q''$ with $\operatorname{disc}\rho''=c$.

Clearly, $(A,\sigma)_{F_Q}$ is hyperbolic if $(A,\sigma)$ is Witt-equivalent to an algebra containing ${(Q,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})}$. Conversely, suppose $(A,\sigma)_{F_Q}$ is hyperbolic. Since $(A,\sigma)$ is not totally decomposable, it is not hyperbolic. If $A$ is split, Proposition \[split.prop\] shows that the anisotropic kernel of $(A,\sigma)$ is $\operatorname{Ad}_{n_Q}\simeq{(Q,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})}\otimes{(Q,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})}$, hence $$(A,\sigma)\simeq\operatorname{Ad}_{{{\left\langle{1,-1,1,-a}\right\rangle}}}\otimes\operatorname{Ad}_{{{\langle\!\langle{b}\rangle\!\rangle}}}.$$ For the rest of the proof, we may thus assume $A$ is not split. By Proposition \[orthog.prop\], one of the components of the Clifford algebra, $C_+(A,\sigma)$ say, is split by $F_Q$. However, $C_+(A,\sigma)$ is not split since $(A,\sigma)$ is not totally decomposable, hence $C_+(A,\sigma)$ is Brauer-equivalent to $Q$. As was observed in Proposition \[orthog.prop\], $(C_+(A,\sigma),\sigma_+)_{F_Q}$ is isotropic. If it is hyperbolic, then $(A,\sigma)$ is hyperbolic by the main theorem of [@G:clif], a contradiction. Therefore, the anisotropic kernel of $(C_+(A,\sigma),\sigma_+)$ has degree $4$. It has discriminant $1$ by triality, hence $(C_+(A,\sigma),\sigma_+)$ is Witt-equivalent to a product $(Q',{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})\otimes(Q'',{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})$ for some quaternion $F$-algebras $Q'$, $Q''$ such that $Q'\otimes Q''$ is Brauer-equivalent to $Q$. We may therefore find $c$, $d$, $d'$, $d''\in F^\times$ such that $$Q\simeq(c,d)_F,\qquad Q'\simeq(c,d')_F,\qquad Q''\simeq(c,d'')_F.$$ Letting $\rho$ (resp. $\rho'$, resp. $\rho''$) be an orthogonal involution on $Q$ (resp. $Q'$, resp. $Q''$) with discriminant $c$, we have by Lemma \[quat.lem\] $$(C_+(A,\sigma),\sigma_+)\simeq \operatorname{Ad}_{{{\left\langle{1,-1,1,-d'}\right\rangle}}}\otimes
  (Q,\rho)\simeq \operatorname{Ad}_{{{\left\langle{1,-1,1,-d''}\right\rangle}}}\otimes(Q,\rho).$$ By Lemma \[tria.lem\], it follows that $(A,\sigma)$ is isomorphic to $$\operatorname{Ad}_{{{\left\langle{1,-1,1,-d}\right\rangle}}}\otimes(Q',\rho') \qquad\text{or}\qquad
  \operatorname{Ad}_{{{\left\langle{1,-1,1,-d}\right\rangle}}}\otimes(Q'',\rho''),$$ hence it is Witt-equivalent to $${(Q,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})}\otimes(Q',{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}}) \qquad\text{or}\qquad
  {(Q,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})}\otimes(Q'',{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}}).$$ Interchanging $Q'$ and $Q''$ if necessary, we thus obtain the stated description of $(A,\sigma)$.

Examples of arbitrarily large degree {#large.sec}
====================================

Let $(A,\sigma)$ be a central simple $F$-algebra with involution of orthogonal or symplectic type. Consider the (iterated) Laurent series fields $F_1=F((x))$, $F_2=F((x))((y))$, and the quaternion $F_2$-algebra $H=(x,y)_{F_2}$. Let $\rho$ be any involution of orthogonal or symplectic type on $H$, and let $$(A_1,\sigma_1)=(A,\sigma)\otimes_F\operatorname{Ad}_{{{\langle\!\langle{x}\rangle\!\rangle}}},\qquad
(A_2,\sigma_2)=(A,\sigma)\otimes_F (H,\rho).$$ If $(A,\sigma)_{F_Q}$ is hyperbolic, then $(A_1,\sigma_1)_{F_Q}$ and $(A_2,\sigma_2)_{F_Q}$ also are hyperbolic, since they contain a hyperbolic factor.

\[large.thm\] Assume $(A,\sigma)$ is anisotropic. Then $(A_1,\sigma_1)$ and $(A_2,\sigma_2)$ are anisotropic. Moreover, the following conditions are equivalent:

1.  $(A,\sigma)$ contains ${(Q,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})}$;

2.  $(A_1,\sigma_1)$ contains ${(Q,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})}$;

3.  $(A_2,\sigma_2)$ contains ${(Q,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})}$.

Let $\xi_1=\bigl(
  \begin{smallmatrix}
    1&0\\0&-1
  \end{smallmatrix}\bigr)$, $\eta_1=\bigl(
  \begin{smallmatrix}
    0&x\\1&0
  \end{smallmatrix}\bigr)\in M_2(F_1)$, so $$\operatorname{ad}_{{{\langle\!\langle{x}\rangle\!\rangle}}}(\xi_1)=\xi_1 \qquad\text{and}\qquad
  \operatorname{ad}_{{{\langle\!\langle{x}\rangle\!\rangle}}}(\eta_1)=-\eta_1.$$ Let also $(a_i)_{i\in I}$ be an $F$-basis of $A$, so $(a_i\otimes1,
  a_i\otimes\xi_1, a_i\otimes\eta_1, a_i\otimes\xi_1\eta_1)_{i\in I}$ is an $F_1$-basis of $A_1$. We extend the $x$-adic valuation $v_1$ on $F_1$ to a map $$g_1\colon
  A_1\to\bigl({\textstyle\frac12}\mathbb{Z}\bigr)\cup\{\infty\}$$ defined by $$g_1\bigl(\sum_{i\in I} a_i\otimes(\alpha_i+\beta_i\xi_1
  +\gamma_i\eta_1+\delta_i\xi_1\eta_1)\bigr) =\min_{i\in I}\bigl(
  v_1(\alpha_i), v_1(\beta_i), v_1(\gamma_i)+{\textstyle\frac12},
  v_1(\delta_i)+{\textstyle\frac12}\bigr)$$ for $\alpha_i$, $\beta_i$, $\gamma_i$, $\delta_i\in F_1$. It is readily verified that the map $g_1$ satisfies the following conditions for $s$, $t\in A_1$ and $\alpha\in F_1$:

-   $g_1(1)=0$ and $g_1(s)=\infty$ if and only if $s=0$;

-   $g_1(s+t)\geq\min\bigl(g_1(s), g_1(t)\bigr)$ and $g_1(s\alpha)=g_1(s)+v_1(\alpha)$;

-   $g_1(st)\geq g_1(s)+g_1(t)$.

(It suffices to prove the last inequality for $s$, $t$ in the above $F_1$-base of $A_1$, see [@TW1 Lemma 1.2].) The map $g_1$ defines a filtration of $A_1$, and the associated graded ring $\operatorname{gr}(A_1)$ is $$\operatorname{gr}(A_1)\simeq A\otimes_F M_2(F[x,x^{-1}])$$ with the grading defined by $$\begin{aligned}
    \operatorname{gr}(A_1)_\lambda & = A\otimes
    \begin{pmatrix}
    x^\lambda&0\\0&x^\lambda
    \end{pmatrix}
    &&\text{for $\lambda\in\mathbb{Z}$},\\
    \operatorname{gr}(A_1)_\lambda & = A\otimes
    \begin{pmatrix}
    0&x^{\lambda+\frac12}\\ x^{\lambda-\frac12}&0
    \end{pmatrix}
    &&\text{for $\lambda\in\bigl({\textstyle\frac12}\mathbb{Z}\bigr)
      \setminus\mathbb{Z}$.} 
  \end{aligned}$$ Therefore, $\operatorname{gr}(A_1)$ is a graded simple algebra, and $g_1$ is a $v_1$-gauge in the sense of [@TW1]. The involution $\sigma_1$ preserves $g_1$. On $\operatorname{gr}(A_1)_0$, the induced involution $\widetilde{\sigma_1}$ is $\sigma\otimes\operatorname{Id}$, hence it is anisotropic. Therefore, $\sigma_1$ is anisotropic by [@TW2 Cor. 2.3], $g_1$ is the unique $v_1$-gauge that is preserved by $\sigma_1$ by [@TW2 Th. 2.2], and we have $$\label{eq:4}
    g_1(\sigma_1(s)s)=2g_1(s)\qquad\text{for all $s\in A_1$}.$$ Now, suppose $(A_1,\sigma_1)$ contains ${(Q,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})}$; it then contains elements $i$, $j$ such that $$\label{eq:5}
    i^2=a,\quad j^2=b,\quad ji=-ij,\quad\sigma_1(i)=-i,\quad
    \sigma_1(j)=-j.$$ Then by we have $g_1(i)=\frac12g_1(-a)=0$ and, similarly, $g_1(j)=0$. The images $\widetilde{i}$, $\widetilde{j}$ of $i$, $j$ in $\operatorname{gr}_1(A_1)_0$ satisfy conditions similar to . Since $\operatorname{gr}(A_1)_0\simeq A\times A$ we may consider a projection $\operatorname{gr}(A)_0\to A$, which is a homomorphism of algebras with involution $\pi\colon(\operatorname{gr}(A_1)_0,\widetilde{\sigma_1})\to
  (A,\sigma)$. The images $\pi(\widetilde{i})$, $\pi(\widetilde{j})$ generate a copy of ${(Q,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})}$ in $(A,\sigma)$. Thus, $(A,\sigma)$ contains ${(Q,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})}$ if $(A_1,\sigma_1)$ contains ${(Q,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})}$. The converse is clear.

The argument for $(A_2,\sigma_2)$ follows the same lines. Let $\xi_2$, $\eta_2\in H$ be such that $$\xi_2^2=x,\qquad \eta_2^2=y,\qquad\eta_2\xi_2=-\xi_2\eta_2.$$ Note that if $\rho$ is orthogonal its discriminant is represented by the quadratic form ${{\left\langle{x,y,-xy}\right\rangle}}$, hence it is the square class of $x$, $y$, or $-xy$. Therefore, we may assume $\rho=\operatorname{Int}(\xi_2)\circ{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}}$, $\operatorname{Int}(\eta_2)\circ{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}}$, or $\operatorname{Int}(\xi_2\eta_2)\circ{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}}$. In each case (and also if $\rho$ is symplectic) we have $\rho(\xi_2)=\pm\xi_2$ and $\rho(\eta_2)=\pm\eta_2$.

Let $v_2\colon F_2\to \mathbb{Z}^2\cup\{\infty\}$ be the $(x,y)$-adic valuation such that $v_2(x^\lambda
  y^\mu)=(\lambda,\mu)$ for $\lambda$, $\mu\in \mathbb{Z}$, where $\mathbb{Z}^2$ is endowed with the right-to-left lexicographic ordering. Considering again an $F$-basis $(a_i)_{i\in I}$ of $A$, we extend $v_2$ to a map $$g_2\colon
  A_2\to\bigl({\textstyle\frac12}\mathbb{Z}\bigr)^2\cup\{\infty\}$$ defined by $$\begin{gathered}
  g_2\bigl(\sum_{i\in I} a_i\otimes(\alpha_i+\beta_i\xi_2
  +\gamma_i\eta_2+\delta_i\xi_2\eta_2)\bigr) =\\ \min_{i\in
    I}\bigl(v_2(\alpha_i), v_2(\beta_i)+({\textstyle\frac12},0),
  v_2(\gamma_i)+(0,{\textstyle\frac12}),
  v_2(\delta_i)+({\textstyle\frac12,\frac12})\bigr)
  \end{gathered}$$ for $\alpha_i$, $\beta_i$, $\gamma_i$, $\delta_i\in F_2$. The map $g_2$ is a $v_2$-gauge on $A_2$ with associated graded ring $$\operatorname{gr}(A_2)=A\otimes(x,y)_{F[x,x^{-1},y,y^{-1}]}.$$ The involution $\sigma_2$ preserves $g_2$ and the induced involution $\widetilde{\sigma_2}$ on $\operatorname{gr}(A_2)_0=A$ is $\sigma$. Therefore, the same arguments as for $(A_1,\sigma_1)$ show that $(A_2,\sigma_2)$ is anisotropic, and that $(A_2,\sigma_2)$ contains ${(Q,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})}$ if and only if $(A,\sigma)$ contains ${(Q,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})}$.

Theorem \[large.thm\] applies in particular to the division algebra with orthogonal involution $(A,\sigma)$ of Theorem \[totdec.thm\], and yields central simple algebras with anisotropic involution $(A_1,\sigma_1)$ and $(A_2,\sigma_2)$ of degree $16$ that do not contain ${(Q,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})}$, even though they are hyperbolic over $F_Q$. The involution $\sigma_1$ is orthogonal and $\operatorname{ind}A_1=8$, while the involution $\sigma_2$ may be of orthogonal or symplectic type and $A_2$ is division. Of course, these constructions can be iterated to obtain examples of algebras with anisotropic involution of arbitrarily large degree that become hyperbolic over $F_Q$ and do not contain ${(Q,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})}$. Such examples can also be derived from the central simple algebras of degree $4$ with symplectic involution in case (b) of Theorem \[deg4symp.thm\], although no division algebra can be obtained in this way since the algebras in case (b) of Theorem \[deg4symp.thm\] have index $2$.

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[^1]: The discussion in [@KMRT] is in terms of the $3$-fold Pfister form $j$ whose Arason invariant is $\Delta$.
