Here is a question I get from sitting in my Lie algebra class: Fix a Lie algebra $\mathfrak{h}$, we know there is a unique simply connected Lie group $H$ which serves as the universal cover of other connected Lie groups with the same Lie algebra. Now assuming $H$ is a $n_1$ sheeted cover of $H_1$, and a $n_2$ sheeted cover of $H_2$, (we are not restricting the Deck groups yet). we know $\phi: \mathfrak{h_1}\cong \mathfrak{h_2} (\cong \mathfrak{h})$ but we only cares about the isomorphism between the first two factors. Now as we travel along all $a\in \mathfrak{h_1}$, $(a, \phi (a))$ will form a Lie subalgebra in $\mathfrak{h_1}\bigoplus\mathfrak{h_2}$, denoted by $\mathfrak{h'}$. By the existence theorem of Lie subgroups we have a uniqueLie subgroup $G\subset H_1\times H_2$ corresponding to $\mathfrak{h'}$. Note $\mathfrak{h_1},\mathfrak{h_2}, \mathfrak{h'}$ are isomorphic Lie subalgebras in $\mathfrak{h_1}\bigoplus\mathfrak{h_2}$, however since they sort of "sit in different directions" in the ambient Lie algebra when we form the exponential map we are producing non isomorphic Lie subgroups. Anyhow, Project down from $G$ to $H_1$ and $H_2$ induce an iso on the Lie algebra therefore we know $G$ actually covers $H_i$. That's the set up of the question.

Now if I know $n_1$ and $n_2$ are coprime to each other, then automatically $G$ has to be the universal cover $H$. However, if say the maximal common divisor of $n_1$ and $n_2$ is 2, and that there are two non isomorphic Lie groups that could cover both $H_1$ and $H_2$, namely H and the one doubly covered by H donoted $\tilde{H}$, (here I need some compatible condition on the Deck group, but let assume $\tilde{H}$ covers $H_i$). So my question is which one is isomorphic to $G$ when I draw the graph? I am worried a little bit that the there may not a canonical way of choosing this isomorphism $\phi$ to make this question make sense, (or is there)? My guess is that if there is a "canonical choice", then the $G$ we get from the graph should be the smaller $\tilde{H}$, while my colleage thinks that by choosing different $\phi$, you can get both. When $H_1$ and $H_2$ have a lot of non isomorphic common covers, then by choosing different $\phi$ you can produce all of them as subgroups of $H_1\times H_2$? (fixing $H_1$ and $H_2$.) I really hope someone ccan shed lights on this, thanks very much!


If we fix universal covering maps $H \to H_1$ and $H \to H_2$, then $G$ is uniquely defined as the image of the diagonally embedded $H \subset H \times H$ under the covering map to $H_1 \times H_2$. If you transform one of the covering maps from $H$ using the deck group, you will get a group isomorphic to $G$.

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  • $\begingroup$ 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 $A_1\cap A_2$. $\endgroup$ – Ying Zhang Oct 1 '10 at 3:11
  • $\begingroup$ 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? $\endgroup$ – Ying Zhang Oct 1 '10 at 3:13
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    $\begingroup$ 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 $H \subset H \times H$. $\endgroup$ – S. Carnahan Oct 1 '10 at 3:35
  • $\begingroup$ 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! $\endgroup$ – Ying Zhang Oct 1 '10 at 3:38
  • $\begingroup$ 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 $H\subset H\times H\rightarrow H_1\times H_2$? $\endgroup$ – Ying Zhang Oct 1 '10 at 3:44

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