Timeline for A Krull-Schmidt Theorem for Lie groups?
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Mar 7, 2023 at 0:06 | comment | added | LSpice | Even if there is no centre, a semisimple group need not be the direct product of its normal, almost simple subgroups. Consider, for example, the quotient $G$ of $\operatorname{SU}_2 \times \operatorname{SU}_2$ by the diagonally embedded central $\mu_2$. The relevant subgroups of this are both $\operatorname{SU}_2$, but the quotient of $G$ by one of the $\operatorname{SU}_2$s is the adjoint group $\operatorname{SU}_2/\mu_2$, not $\operatorname{SU}_2$. | |
| Feb 5, 2015 at 13:53 | comment | added | Henrik Winther | @MikhailBorovoi I see the problem now. $U(n)$ is only covered by $SU(n)\times S^1$, not equal to. If I find the time I will edit the answer to account for this situation. Thank you. | |
| Feb 3, 2015 at 9:54 | comment | added | Mikhail Borovoi | Excuse me, I meant that $U(n)$ is not a direct product of a semisimple group and a torus. | |
| Feb 2, 2015 at 8:38 | comment | added | Henrik Winther | @MikhailBorovoi I would call $SU(n)$ semi-simple. It does not need to decompose further. There is no torus in this case. | |
| Feb 1, 2015 at 21:15 | comment | added | Mikhail Borovoi | It is true that any connected compact Lie group decomposes uniquely into a product of a semisimple group and a central torus. However, this product does not have to be direct. For example, $SU(n)$ is not a direct product of a semisimple group and a torus. | |
| Jan 28, 2015 at 21:07 | history | edited | Henrik Winther | CC BY-SA 3.0 |
added 12 characters in body
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| Jan 28, 2015 at 20:53 | history | answered | Henrik Winther | CC BY-SA 3.0 |