Why is $\operatorname{U}(n,\mathbb{H})\subset \operatorname{SL}(n,\mathbb{H}) $? This question is inspired by Maximal compact subgroup of $\mathrm{SL}(2,\mathbb{H})$. Consider the embedding $\operatorname{U}(n,\mathbb{H})\subset \operatorname{GL}(n,\mathbb{H})  $. Since $\operatorname{U}(n,\mathbb{H})\cong \operatorname{Sp}(n) $ is almost simple, we know that it is actually contained in $\operatorname{SL}(n,\mathbb{H})  $, the kernel of the Dieudonné determinant $\ \det: \operatorname{GL}(n,\mathbb{H})  \rightarrow \mathbb{R}^*_{+}\,$. Is there a direct way to prove   $\det A=1$ for $A\in \operatorname{U}(n,\mathbb{H})  $?
(Recall that $\ \det\, {}^{t}\overline{A}\neq \det A\ $ for $A$ general in $\operatorname{GL}(n,\mathbb{H}) $.)
 A: Yes, relying on the fact that elements of the symplectic group (over $\mathbf{C}$, and hence over $\mathbf{R}$) have determinant 1.
Indeed, an element $g$ of $\mathrm{U}(n,\mathbf{H})$ preserves the canonical Hermitian form $b$. Let $b'$ be any imaginary component of $b$. Then $b'$ is a $g$-invariant real-valued symplectic form on the $4n$-dimensional real space $\mathbf{H}^n$. So the determinant of $g$ as $4n$-dimensional matrix (which is the Dieudonné determinant) is $1$.
Edit (abx): actually the determinant $\det_{\mathbb{R}}(g)$ of $g$ as a real $4n$-dimensional matrix is $(\det_{D}( g))^4$, where $\det_D$ is the Dieudonné determinant — since $\det_D(g)\in\mathbb{R}^*_+$, this implies $\det_D(g)=1$ as required.
Indeed $\det_{\mathbb{R}}$ is a homomorphism from  $\operatorname{GL}(n,\mathbb{H}) $ to $\mathbb{R}^*$; such a homomorphism is trivial on the derived subgroup  $\operatorname{SL}(n,\mathbb{H}) $. Since $\det_D$ induces an isomorphism of
$\operatorname{GL}(n,\mathbb{H}) /\operatorname{SL}(n,\mathbb{H}) $ onto $\mathbb{R}^*_+$, we have $\det_{\mathbb{R}}=\varphi \circ \det_{D}$, where $\varphi :\mathbb{R}^*_+\rightarrow \mathbb{R}^*$ is a continuous homomorphism. Such a homomorphism is of the form $x\mapsto x^{a}$ for some $a\in\mathbb{R}$; thus $\det_{\mathbb{R}}=  (\det_{D})^{a}$. Taking $g= tI_n$ gives $a=4$.
