Reconstructing Group from Covering Maps

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WWGD
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Hi, let p : E--->B be a covering map. Then we have a result that for every subgroup of ## \pi_1(B) ## we have an associated covering map. Now, going in sort-of the reverse direction, is there a way of figuring out what ## \pi_1(B) ## is, if we know a collection of covering maps for B; what information do we need for these to be able to reconstruct ## \pi_1 (B) ## up to isomorphism?
 
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mathwonk
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do you know what a "deck transformation" is?
 
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WWGD
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Yes, Wonk, these are the maps that factor through the covering map p, right, i.e., pof=p, right? Ah, I see, if the cover by X is a universal cover, then the group is equal to ## \pi_1(X) ##, right? What if X is not universal. It will still cover all other universal covers by its universal property, right?
 
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lavinia
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Hi, let p : E--->B be a covering map. Then we have a result that for every subgroup of ## \pi_1(B) ## we have an associated covering map. Now, going in sort-of the reverse direction, is there a way of figuring out what ## \pi_1(B) ## is, if we know a collection of covering maps for B; what information do we need for these to be able to reconstruct ## \pi_1 (B) ## up to isomorphism?
WWGD your question confuses me a bit so for clarity here is a question for you.

Suppose you have a space whose fundamental group is ZxZ. Suppose also that it has a deck transformation of order 2. Can you deduce the fundamental group of the quotient space? Is that the sort of thing you are asking?
 
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WWGD
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Hi lavinia, I assume the space whose fundamental group is ZxZ is the base of some covering. If so, is there any constraint for the top space, is the top space the universal covering space? I guess no, since the deck group would then be ZxZ. And if I looked it up correctly, if the covering is regular, then there is a homeomorphism ## E /( \frac {\mathbb Z}{2 \mathbb Z} ) \simeq B ##, where B is the base. But then we would have ## \pi_1( E/( \mathbb Z / 2\mathbb Z))=\mathbb Z \times \mathbb Z ## , which cannot be. I hope I understood your question correctly.
 
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  • #6
lavinia
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Hi lavinia, I assume the space whose fundamental group is ZxZ is the base of some covering. If so, is there any constraint for the top space, is the top space the universal covering space? I guess no, since the deck group would then be ZxZ. And if I looked it up correctly, if the covering is regular, then there is a homeomorphism ## E /( \frac {\mathbb Z}{2 \mathbb Z} ) \simeq B ##, where B is the base. But then we would have ## \pi_1( E/( \mathbb Z / 2\mathbb Z))=\mathbb Z \times \mathbb Z ## , which cannot be. I hope I understood your question correctly.
What I meant to say was that the space with fundamental group ZxZ covers another space and the covering is a 2 fold covering. Can you deduce the fundamental group of the base? I was asking if that was your question.
 
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WWGD
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What I meant to say was that the space with fundamental group ZxZ covers another space and the covering is a 2 fold covering. Can you deduce the fundamental group of the base? I was asking if that was your question.
Well, no, I don't think so; I was trying to construct a sort-of converse to the result that : given a cover p:E-->X, for every subgroup of ##G:=\pi_1(X)## , there is an associated covering map for every subgroup of G. Now, if we did not know the fundamental group of ##X## , but we knew all the covering maps p: E-->X, could we figure out ## G:= \pi_1(X) ## ? It is difficult to associate a subgroup for every cover, unless , I think, you can show that some are covers of others. And there is the additional issue : is it possible for different groups with non-trivial subgroups to share the exact same subgroups?

But your question is an interesting one; I am looking into it. And the idea itself of finding
possible spaces whose fundamental group is ZxZ is an interesting one too. One possibility is
that the space is the product of spaces whose fundamental group is Z, e.g., the circle (actually, I am thinking of the non-zero complexes too). I guess one choice would be the map ##z^2## from the circle to itself. But I don't know how to classify all the covers with those properties. It is a nice exercise. Let me think about it.
spaces does the circle cover? (Seems like a choice is ) If the deck transformation does not have order 2

EDIT: Sorry for that weird emoticon, I don't know how to get rid of it.
 
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  • #8
lavinia
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Well, no, I don't think so; I was trying to construct a sort-of converse to the result that : given a cover p:E-->X, for every subgroup of ##G:=\pi_1(X)## , there is an associated covering map for every subgroup of G. Now, if we did not know the fundamental group of ##X## , but we knew all the covering maps p: E-->X, could we figure out ## G:= \pi_1(X) ## ? It is difficult to associate a subgroup for every cover, unless , I think, you can show that some are covers of others. And there is the additional issue : is it possible for different groups with non-trivial subgroups to share the exact same subgroups?
This is why I was confused.
- If E is the universal cover then what group is the group of covering transformations of X?
-If E is not the universal cover, then even if you know the fundamental group of E and know the covering group, how do you figure out the the fundamental group of X? Assume even that E is a regular cover of X.
E.G. Can you construct a space whose fundamental group is ZxZ and has covering transformations of order 2 of two different X's?
 

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