Infinite cyclic cover corresponding to non-zero cohomology class $alpha in H^1(x,mathbb Z)$
2
$begingroup$
I want to understand the following sentence:
Let X a compact (complex) manifold which has a non-zero cohomology class $alpha in H^1(X,mathbb Z)$. Let $pi: bar Xto X$ be the corresponding infinite cyclic covering.
What does this mean? It seems that an infinite cyclic covering is a cover with fiber $mathbb Z$.
But why does such a covering exist, and how is it related to the cohomology class?
geometry differential-geometry algebraic-topology complex-geometry covering-spaces
asked Jan 1 at 12:45
klirkklirk
2,624530
$endgroup$
|
show 4 more comments
2
$begingroup$
I want to understand the following sentence:
Let X a compact (complex) manifold which has a non-zero cohomology class $alpha in H^1(X,mathbb Z)$. Let $pi: bar Xto X$ be the corresponding infinite cyclic covering.
What does this mean? It seems that an infinite cyclic covering is a cover with fiber $mathbb Z$.
But why does such a covering exist, and how is it related to the cohomology class?
geometry differential-geometry algebraic-topology complex-geometry covering-spaces
asked Jan 1 at 12:45
klirkklirk
2,624530
$endgroup$
$begingroup$
$H^1(X;G) = text{Hom}(pi_1, G)$.
$endgroup$
– Mike Miller
Jan 1 at 14:03
$begingroup$
Your hypothetical Stiefel-Whitney class would live in $ H^1(X,mathbb Z/2)$, not in $ H^1(X,mathbb Z)$.
$endgroup$
– Georges Elencwajg
Jan 1 at 14:44
$begingroup$
@GeorgesElencwajg One more reason why this seems to be the wrong approach. I guess I can delete that part.
$endgroup$
– klirk
Jan 1 at 15:41
$begingroup$
@MikeMiller: Any ideas how to proceed from here? The only thing I could think of was that $alpha: pi_1(X) to mathbb Z = pi_1(K(mathbb Z,1))$ and so $alpha$ is induced by a map from $Xto K(mathbb Z,1)=S^1$. But this is not the map I look for.
$endgroup$
– klirk
Jan 1 at 15:45
1
$begingroup$
@klirk It's standard covering space theory that homomorphisms $pi_1(X) to Bbb Z$ classify infinite cyclic covers. Do you know covering space theory? If not, look at Hatcher chapter 1.3.
$endgroup$
– Balarka Sen
Jan 1 at 16:48
|
show 4 more comments
2
2
2
1
$begingroup$
I want to understand the following sentence:
Let X a compact (complex) manifold which has a non-zero cohomology class $alpha in H^1(X,mathbb Z)$. Let $pi: bar Xto X$ be the corresponding infinite cyclic covering.
What does this mean? It seems that an infinite cyclic covering is a cover with fiber $mathbb Z$.
But why does such a covering exist, and how is it related to the cohomology class?
geometry differential-geometry algebraic-topology complex-geometry covering-spaces
asked Jan 1 at 12:45
klirkklirk
2,624530
$endgroup$
I want to understand the following sentence:
Let X a compact (complex) manifold which has a non-zero cohomology class $alpha in H^1(X,mathbb Z)$. Let $pi: bar Xto X$ be the corresponding infinite cyclic covering.
What does this mean? It seems that an infinite cyclic covering is a cover with fiber $mathbb Z$.
But why does such a covering exist, and how is it related to the cohomology class?
geometry differential-geometry algebraic-topology complex-geometry covering-spaces
geometry differential-geometry algebraic-topology complex-geometry covering-spaces
asked Jan 1 at 12:45
klirkklirk
2,624530
asked Jan 1 at 12:45
klirkklirk
2,624530
edited Jan 1 at 15:41
klirk
asked Jan 1 at 12:45
klirkklirk
2,624530
asked Jan 1 at 12:45
klirkklirk
2,624530
asked Jan 1 at 12:45
klirkklirk
2,624530
2,624530
$begingroup$
$H^1(X;G) = text{Hom}(pi_1, G)$.
$endgroup$
– Mike Miller
Jan 1 at 14:03
$begingroup$
Your hypothetical Stiefel-Whitney class would live in $ H^1(X,mathbb Z/2)$, not in $ H^1(X,mathbb Z)$.
$endgroup$
– Georges Elencwajg
Jan 1 at 14:44
$begingroup$
@GeorgesElencwajg One more reason why this seems to be the wrong approach. I guess I can delete that part.
$endgroup$
– klirk
Jan 1 at 15:41
$begingroup$
@MikeMiller: Any ideas how to proceed from here? The only thing I could think of was that $alpha: pi_1(X) to mathbb Z = pi_1(K(mathbb Z,1))$ and so $alpha$ is induced by a map from $Xto K(mathbb Z,1)=S^1$. But this is not the map I look for.
$endgroup$
– klirk
Jan 1 at 15:45
1
$begingroup$
@klirk It's standard covering space theory that homomorphisms $pi_1(X) to Bbb Z$ classify infinite cyclic covers. Do you know covering space theory? If not, look at Hatcher chapter 1.3.
$endgroup$
– Balarka Sen
Jan 1 at 16:48
|
show 4 more comments
$begingroup$
$H^1(X;G) = text{Hom}(pi_1, G)$.
$endgroup$
– Mike Miller
Jan 1 at 14:03
$begingroup$
Your hypothetical Stiefel-Whitney class would live in $ H^1(X,mathbb Z/2)$, not in $ H^1(X,mathbb Z)$.
$endgroup$
– Georges Elencwajg
Jan 1 at 14:44
$begingroup$
@GeorgesElencwajg One more reason why this seems to be the wrong approach. I guess I can delete that part.
$endgroup$
– klirk
Jan 1 at 15:41
$begingroup$
@MikeMiller: Any ideas how to proceed from here? The only thing I could think of was that $alpha: pi_1(X) to mathbb Z = pi_1(K(mathbb Z,1))$ and so $alpha$ is induced by a map from $Xto K(mathbb Z,1)=S^1$. But this is not the map I look for.
$endgroup$
– klirk
Jan 1 at 15:45
1
$begingroup$
@klirk It's standard covering space theory that homomorphisms $pi_1(X) to Bbb Z$ classify infinite cyclic covers. Do you know covering space theory? If not, look at Hatcher chapter 1.3.
$endgroup$
– Balarka Sen
Jan 1 at 16:48
$begingroup$
$H^1(X;G) = text{Hom}(pi_1, G)$.
$endgroup$
– Mike Miller
Jan 1 at 14:03
$begingroup$
$H^1(X;G) = text{Hom}(pi_1, G)$.
$endgroup$
– Mike Miller
Jan 1 at 14:03
$begingroup$
Your hypothetical Stiefel-Whitney class would live in $ H^1(X,mathbb Z/2)$, not in $ H^1(X,mathbb Z)$.
$endgroup$
– Georges Elencwajg
Jan 1 at 14:44
$begingroup$
Your hypothetical Stiefel-Whitney class would live in $ H^1(X,mathbb Z/2)$, not in $ H^1(X,mathbb Z)$.
$endgroup$
– Georges Elencwajg
Jan 1 at 14:44
$begingroup$
@GeorgesElencwajg One more reason why this seems to be the wrong approach. I guess I can delete that part.
$endgroup$
– klirk
Jan 1 at 15:41
$begingroup$
@GeorgesElencwajg One more reason why this seems to be the wrong approach. I guess I can delete that part.
$endgroup$
– klirk
Jan 1 at 15:41
$begingroup$
@MikeMiller: Any ideas how to proceed from here? The only thing I could think of was that $alpha: pi_1(X) to mathbb Z = pi_1(K(mathbb Z,1))$ and so $alpha$ is induced by a map from $Xto K(mathbb Z,1)=S^1$. But this is not the map I look for.
$endgroup$
– klirk
Jan 1 at 15:45
$begingroup$
@MikeMiller: Any ideas how to proceed from here? The only thing I could think of was that $alpha: pi_1(X) to mathbb Z = pi_1(K(mathbb Z,1))$ and so $alpha$ is induced by a map from $Xto K(mathbb Z,1)=S^1$. But this is not the map I look for.
$endgroup$
– klirk
Jan 1 at 15:45
1
1
$begingroup$
@klirk It's standard covering space theory that homomorphisms $pi_1(X) to Bbb Z$ classify infinite cyclic covers. Do you know covering space theory? If not, look at Hatcher chapter 1.3.
$endgroup$
– Balarka Sen
Jan 1 at 16:48
$begingroup$
@klirk It's standard covering space theory that homomorphisms $pi_1(X) to Bbb Z$ classify infinite cyclic covers. Do you know covering space theory? If not, look at Hatcher chapter 1.3.
$endgroup$
– Balarka Sen
Jan 1 at 16:48
|
show 4 more comments
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$begingroup$
$H^1(X;G) = text{Hom}(pi_1, G)$.
$endgroup$
– Mike Miller
Jan 1 at 14:03
$begingroup$
Your hypothetical Stiefel-Whitney class would live in $ H^1(X,mathbb Z/2)$, not in $ H^1(X,mathbb Z)$.
$endgroup$
– Georges Elencwajg
Jan 1 at 14:44
$begingroup$
@GeorgesElencwajg One more reason why this seems to be the wrong approach. I guess I can delete that part.
$endgroup$
– klirk
Jan 1 at 15:41
$begingroup$
@MikeMiller: Any ideas how to proceed from here? The only thing I could think of was that $alpha: pi_1(X) to mathbb Z = pi_1(K(mathbb Z,1))$ and so $alpha$ is induced by a map from $Xto K(mathbb Z,1)=S^1$. But this is not the map I look for.
$endgroup$
– klirk
Jan 1 at 15:45
1
$begingroup$
@klirk It's standard covering space theory that homomorphisms $pi_1(X) to Bbb Z$ classify infinite cyclic covers. Do you know covering space theory? If not, look at Hatcher chapter 1.3.
$endgroup$
– Balarka Sen
Jan 1 at 16:48