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Let $U=\mathrm{Spin}(2n)$, which is a simply connected compact simple Lie group, and let $\mathfrak{u}_0=\mathfrak{so}(2n)$, the Lie algebra of $U$. If $\mathfrak{g}_0$ is a noncompact dual of $\mathfrak{u}_0$ in the complexified Lie algebra $\mathfrak{g}=\mathfrak{so}(2n,\mathbb{C})$, namely $\mathfrak{g}_0=\mathfrak{u}_0^\theta+\sqrt{-1}\mathfrak{u}_0^{-\theta}$ for some involutive automorphism $\theta$ of $\mathfrak{u}_0$, where $\mathfrak{u}_0^{\pm\theta}=\{X\in\mathfrak{u}_0\mid\theta(X)=\pm X\}$. Then there exists a noncompact closed subgroup $G$ of $G_\mathbb{C}=\mathrm{Spin}(2n,\mathbb{C})$ with the Lie algebra $\mathfrak{g}_0$. For example, if $\mathfrak{g}_0=\mathfrak{so}(m,2n-m)$, then $G=\mathrm{Spin}(m,2n-m)$.

QUESTION

What is $G$ when $\mathfrak{g}_0=\mathfrak{so}^*(2n)$?

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  • $\begingroup$ What do you mean by "Then there exists a noncompact closed subgroup $G$ of $G_{\Bbb C}={\rm Spin}(2n,{\Bbb C})$" ? Do you mean a closed subgroup with Lie algebra $\mathfrak g_0$? $\endgroup$
    – Mikhail Borovoi
    Commented Apr 9, 2018 at 17:47
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    $\begingroup$ If this is what you mean, then the answer is $G={\rm Spin}^*(2n)$, the universal cover of the group $G={\rm SO}^*(2n)$. The latter is the group of quaternionic $n\times n$ -matrices with determinant 1, preserving a nondegenerate skew-hermitian form, for example, the form with matrix ${\rm diag }(i,i,\dots,i)$. $\endgroup$
    – Mikhail Borovoi
    Commented Apr 9, 2018 at 17:58
  • $\begingroup$ @MikhailBorovoi Thank you for your answer, professor Borovoi. Yes, that is what I mean. Now let $G=\mathrm{Spin}^*(2n)$. It is known that $\mathrm{Spin}(6)\cong\mathrm{SU}(4)$. Thus, is it true that $\mathrm{Spin}^*(6)\cong\mathrm{SU}(3,1)$? $\endgroup$
    – Hebe
    Commented Apr 10, 2018 at 4:01
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    $\begingroup$ Yes, this is true, see Helgason's book, X.6.4, case (vii). $\endgroup$
    – Mikhail Borovoi
    Commented Apr 10, 2018 at 4:17

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