Timeline for Is SO(4) a subgroup of SU(3)?
Current License: CC BY-SA 4.0
9 events
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| Nov 18, 2021 at 19:15 | comment | added | user44143 | @mme, it’d be worth incorporating your new arguments into your answers (and explaining what $\mathfrak{g}$ is). | |
| Nov 18, 2021 at 17:33 | comment | added | mme | No, the same proof works: a simply connected compact surface without boundary has pi_2 = Z. The argument in my comment works in even more generality. | |
| Nov 18, 2021 at 17:14 | comment | added | Michael | I wonder now whether $SO_4$ has a double cover that embeds into $SU_3$. | |
| Nov 18, 2021 at 15:45 | history | edited | mme | CC BY-SA 4.0 |
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| Nov 18, 2021 at 15:44 | comment | added | mme | Of course; I made this explicit. | |
| Nov 17, 2021 at 3:31 | comment | added | Kapil | You should say that $H$ is a closed subgroup of $G$ otherwise one may have taken a winding line in a torus and the quotient is not a manifold. In your case, this is automatic since $H$ is compact. | |
| Nov 16, 2021 at 23:17 | history | edited | mme | CC BY-SA 4.0 |
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| Nov 16, 2021 at 23:16 | comment | added | mme | Here is an additional, more straightforward, argument. If $H \subset G$ is codimension 2, then $G/H$ is a surface. An element of $g$ acts trivially on $G/H$ if $gg'H = g'H$ for all $g'$, which amounts to saying that every conjugate of $g$ lies in $H$, which amounts to saying that $g$ lies in the intersection of all conjugates of $H$, which is a normal subgroup of $G$ (call it $K$). Now $G/K$ is a subgroup of the isometry group of a surface, hence at most 3 dimensional. When $\mathfrak g$ is simple $K$ must be 0-dimensional and $\dim G \le 3$. Conclude, because $\mathfrak{su}(3)$ is simple. | |
| Nov 16, 2021 at 14:01 | history | answered | mme | CC BY-SA 4.0 |