Timeline for Which covers of Lie groups will I get
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9 events
| when toggle format | what | by | license | comment | |
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| Oct 1, 2010 at 7:01 | comment | added | S. Carnahan♦ | You should mentally replace "$H$" with "the image of $H$" in the above comment. | |
| Oct 1, 2010 at 4:27 | comment | added | S. Carnahan♦ | 1) I think this depends on whether or not you have fixed the covering maps for the topological spaces in question. Yes if so (quotient by intersection of deck groups), no if not (conjugacy problem). 2) Both $H$ and $G$ are connected subgroups of $H_1 \times H_2$, and their Lie algebras coincide as Lie subalgebras of the Lie algebra of $H_1 \times H_2$. | |
| Oct 1, 2010 at 3:59 | vote | accept | Ying Zhang | ||
| Oct 1, 2010 at 3:44 | comment | added | Ying Zhang | Sorry writing the other comment caused me some delay therefore I didn't see your comment earlier. Thank you very much for your clarification and pointing out my terrible notation errors! Now could you comment in general if two topological space have the same universal cover then do they necessarily have a unique minimal common cover? Also, do you mind giving more details why $G$ is the image of $H\subset H\times H\rightarrow H_1\times H_2$? | |
| Oct 1, 2010 at 3:38 | comment | added | Ying Zhang | Sorry for my confusing notation above. Now at least in the category of Lie groups, I think we are fine. The kernel of the map $H\rightarrow H_i$ is a discrete subrgoup in the center of $H$ which is our $A_i$ above. So we can make sense of $A_1\cap A_2$ as sungroups inside $H$, therefore there is a unique minimal common cover, and $G$ is isomorphic to that! | |
| Oct 1, 2010 at 3:35 | comment | added | S. Carnahan♦ | I think you want your groups to be subgroups of the center of H. Then the intersection makes sense. Minor points: you are writing your quotients backwards (should be $H/A$ instead of $A/H$), and should use \subset instead of \in for $H \subset H \times H$. | |
| Oct 1, 2010 at 3:13 | comment | added | Ying Zhang | Heuristically I want to say for any topological space having the same universal cover then I want to say there exist a "unique" minial common cover, is it true? If it were true I guess we are done, G isomorphic to that unique minimal comon cover, right? | |
| Oct 1, 2010 at 3:11 | comment | added | Ying Zhang | Ah! Based on what you told me, if we take $D$ to be another common cover of $H_1$ and $H_2$, then $H\in H\times H$ will factor through $D\in D\times D$ to $H_1\times H_2$ then, therefore if $H_1$ and $H_2$ has a unique "minimal" common cover $D$, $G$ would be isomorphic to $D$ then? Now say we have two groups $A_1$ and $A_2$ acting on $H$, and $A_1/H\cong H_1$ and $A_2/H\cong H_2$, then the minimal cover would be $A_1\cap A_2/H$? Note here I don't really know how to make sense of $A_1\cap A_2$. | |
| Oct 1, 2010 at 2:45 | history | answered | S. Carnahan♦ | CC BY-SA 2.5 |