This question is inspired by [Maximal compact subgroup of $\mathrm{SL}(2,\mathbb{H})$](https://mathoverflow.net/questions/195060/maximal-compact-subgroup-of-mathrmsl2-mathbbh). 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}) $.)



Why is $\operatorname{U}(n,\mathbb{H})\subset \operatorname{SL}(n,\mathbb{H}) $?
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 $a=4$.


