Timeline for Noncompact dual of $\mathrm{Spin}(2n)$ corresponding to $\mathfrak{so}^*(2n)$
Current License: CC BY-SA 3.0
6 events
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| Apr 10, 2018 at 5:25 | history | edited | Hebe | CC BY-SA 3.0 |
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| Apr 10, 2018 at 4:17 | comment | added | Mikhail Borovoi | Yes, this is true, see Helgason's book, X.6.4, case (vii). | |
| Apr 10, 2018 at 4:01 | comment | added | Hebe | @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 $\mathrm{Spin}^*(6)\cong\mathrm{SU}(3,1)$? | |
| Apr 9, 2018 at 17:58 | comment | added | Mikhail Borovoi | 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 ${\rm diag }(i,i,\dots,i)$. | |
| Apr 9, 2018 at 17:47 | comment | added | Mikhail Borovoi | 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 $\mathfrak g_0$? | |
| Apr 9, 2018 at 14:50 | history | asked | Hebe | CC BY-SA 3.0 |