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 