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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}) $.)

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  • $\begingroup$ Does $\mathbb R^*_+$ mean the multiplicative group of positive real numbers? Are you identifying it with the usual target of the Dieudonné determinant $\mathbb H^\times/[\mathbb H^\times, \mathbb H^\times]$ via the norm map? $\endgroup$
    – LSpice
    Feb 13, 2022 at 15:32
  • $\begingroup$ @LSpice: Yes and yes. This is already in Dieudonné's original paper. $\endgroup$
    – abx
    Feb 13, 2022 at 16:06

1 Answer 1

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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 $a=4$.

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    $\begingroup$ Is it clear that the determinant of $g$ as a real $4n$-dimensional matrix is the Dieudonné determinant? $\endgroup$
    – abx
    Feb 13, 2022 at 14:26
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    $\begingroup$ @abx I actually thought it was the definition. I'll check. (Well, it might be the square instead?) $\endgroup$
    – YCor
    Feb 13, 2022 at 14:28
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    $\begingroup$ I would guess the fourth power — think of the multiplication by a real number. Of course that would be sufficient, since these determinants live in $\mathbb{R}^*_{+}$. But how does one prove it? $\endgroup$
    – abx
    Feb 13, 2022 at 14:50
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    $\begingroup$ @abx if you know that the derived subalgebra of the real Lie algebra $\mathfrak{gl}_n(H)$ has codimension 1 and that $\mathrm{GL}_n(H)$ is Zariski-connected, then it should follow that the maps are multiplicatively proportional (on the set of real points). $\endgroup$
    – YCor
    Feb 13, 2022 at 15:15
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    $\begingroup$ Is there any way to prove this without looking at a set of generators? At least, that's the only proof I know for symplectic matrices—look at what happens on the torus and root groups, and then observe that they generate the entire symplectic group. I've always wondered about a more conceptual proof. $\endgroup$
    – LSpice
    Feb 13, 2022 at 15:34

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