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 