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First Chern class for smooth line bundle
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For holomorphic line bundle we define its first Chern class by exponential sequence $$0to mathbb Z to mathcal O to mathcal O^* to 0 $$
and we can similarly define Chern class for smooth line bundle by the short exact sequence
$$0to mathbb Z to mathcal C^infty to (mathcal C^infty)^*to 0$$
Then there is a natural morphism from the first short exact sequence to the second one, so there is a natural map $H^2(mathbb Z)to H^2(mathbb Z)$. Is this map isomorphic? Similarly, is the map $H^1(mathcal O^*)to H^1((mathcal C^infty)^*)$ just the natural map of on equivalent classes of line bundles?
In fact I am almost sure this is true (because they looks natural), but I do not know how to show this formally?
algebraic-geometry complex-geometry line-bundles
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User X
507
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up vote
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For holomorphic line bundle we define its first Chern class by exponential sequence $$0to mathbb Z to mathcal O to mathcal O^* to 0 $$
and we can similarly define Chern class for smooth line bundle by the short exact sequence
$$0to mathbb Z to mathcal C^infty to (mathcal C^infty)^*to 0$$
Then there is a natural morphism from the first short exact sequence to the second one, so there is a natural map $H^2(mathbb Z)to H^2(mathbb Z)$. Is this map isomorphic? Similarly, is the map $H^1(mathcal O^*)to H^1((mathcal C^infty)^*)$ just the natural map of on equivalent classes of line bundles?
In fact I am almost sure this is true (because they looks natural), but I do not know how to show this formally?
algebraic-geometry complex-geometry line-bundles
asked yesterday
User X
507
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
For holomorphic line bundle we define its first Chern class by exponential sequence $$0to mathbb Z to mathcal O to mathcal O^* to 0 $$
and we can similarly define Chern class for smooth line bundle by the short exact sequence
$$0to mathbb Z to mathcal C^infty to (mathcal C^infty)^*to 0$$
Then there is a natural morphism from the first short exact sequence to the second one, so there is a natural map $H^2(mathbb Z)to H^2(mathbb Z)$. Is this map isomorphic? Similarly, is the map $H^1(mathcal O^*)to H^1((mathcal C^infty)^*)$ just the natural map of on equivalent classes of line bundles?
In fact I am almost sure this is true (because they looks natural), but I do not know how to show this formally?
algebraic-geometry complex-geometry line-bundles
asked yesterday
User X
507
For holomorphic line bundle we define its first Chern class by exponential sequence $$0to mathbb Z to mathcal O to mathcal O^* to 0 $$
and we can similarly define Chern class for smooth line bundle by the short exact sequence
$$0to mathbb Z to mathcal C^infty to (mathcal C^infty)^*to 0$$
Then there is a natural morphism from the first short exact sequence to the second one, so there is a natural map $H^2(mathbb Z)to H^2(mathbb Z)$. Is this map isomorphic? Similarly, is the map $H^1(mathcal O^*)to H^1((mathcal C^infty)^*)$ just the natural map of on equivalent classes of line bundles?
In fact I am almost sure this is true (because they looks natural), but I do not know how to show this formally?
algebraic-geometry complex-geometry line-bundles
asked yesterday
User X
507
asked yesterday
User X
507
asked yesterday
User X
507
asked yesterday
User X
507
507
add a comment |Â
add a comment |Â
1 Answer
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The map $Bbb Z to Bbb Z$ in your morphism of short exact sequences is just the identity, and so induces an isomorphism on $H^*(X;Bbb Z)$.
The second and third maps are inclusion, and think about what this says at the level of building up a vector bundle from transition functions: given a representative for an element of $H^1(X;mathcal O^*)$, we have charts $U_i subset X$ and holomorphic maps $U_i cap U_j to Bbb C^times$ that we think of as transition functions, the way to glue $U_i times Bbb C$ to $U_j times Bbb C$ over the overlap. The corresponding element of $H^1(X;(mathcal C^infty)^*)$ is the same maps $U_i cap U_j to Bbb C^times$, but we have forgotten that they are holomorphic (and just remembered that they are smooth). Thus the line bundle we construct is the same smooth manifold (it has the same charts and transition maps!), but we have forgotten the holomorphic structure it naturally inherits.
So the conclusion you get is that the map $c_1: H^1(X;mathcal O^*) = textHolLineBun to H^2(X;Bbb Z)$ factors through $H^1(X;(mathcal C^infty)^*) = textSmLineBun_Bbb C$.
(That is, the first Chern class is really an invariant of smooth complex line bundles, and to take the first Chern class of a holomorphic line bundle, just forget that it's holomorphic!)
answered 11 hours ago
Mike Miller
33.6k362127
Maybe stupid but why identity map must induce isomorphism on cohomology? I think this is not directly followed from definition.
– User X
2 hours ago
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
The map $Bbb Z to Bbb Z$ in your morphism of short exact sequences is just the identity, and so induces an isomorphism on $H^*(X;Bbb Z)$.
The second and third maps are inclusion, and think about what this says at the level of building up a vector bundle from transition functions: given a representative for an element of $H^1(X;mathcal O^*)$, we have charts $U_i subset X$ and holomorphic maps $U_i cap U_j to Bbb C^times$ that we think of as transition functions, the way to glue $U_i times Bbb C$ to $U_j times Bbb C$ over the overlap. The corresponding element of $H^1(X;(mathcal C^infty)^*)$ is the same maps $U_i cap U_j to Bbb C^times$, but we have forgotten that they are holomorphic (and just remembered that they are smooth). Thus the line bundle we construct is the same smooth manifold (it has the same charts and transition maps!), but we have forgotten the holomorphic structure it naturally inherits.
So the conclusion you get is that the map $c_1: H^1(X;mathcal O^*) = textHolLineBun to H^2(X;Bbb Z)$ factors through $H^1(X;(mathcal C^infty)^*) = textSmLineBun_Bbb C$.
(That is, the first Chern class is really an invariant of smooth complex line bundles, and to take the first Chern class of a holomorphic line bundle, just forget that it's holomorphic!)
answered 11 hours ago
Mike Miller
33.6k362127
Maybe stupid but why identity map must induce isomorphism on cohomology? I think this is not directly followed from definition.
– User X
2 hours ago
add a comment |Â
up vote
0
down vote
The map $Bbb Z to Bbb Z$ in your morphism of short exact sequences is just the identity, and so induces an isomorphism on $H^*(X;Bbb Z)$.
The second and third maps are inclusion, and think about what this says at the level of building up a vector bundle from transition functions: given a representative for an element of $H^1(X;mathcal O^*)$, we have charts $U_i subset X$ and holomorphic maps $U_i cap U_j to Bbb C^times$ that we think of as transition functions, the way to glue $U_i times Bbb C$ to $U_j times Bbb C$ over the overlap. The corresponding element of $H^1(X;(mathcal C^infty)^*)$ is the same maps $U_i cap U_j to Bbb C^times$, but we have forgotten that they are holomorphic (and just remembered that they are smooth). Thus the line bundle we construct is the same smooth manifold (it has the same charts and transition maps!), but we have forgotten the holomorphic structure it naturally inherits.
So the conclusion you get is that the map $c_1: H^1(X;mathcal O^*) = textHolLineBun to H^2(X;Bbb Z)$ factors through $H^1(X;(mathcal C^infty)^*) = textSmLineBun_Bbb C$.
(That is, the first Chern class is really an invariant of smooth complex line bundles, and to take the first Chern class of a holomorphic line bundle, just forget that it's holomorphic!)
answered 11 hours ago
Mike Miller
33.6k362127
Maybe stupid but why identity map must induce isomorphism on cohomology? I think this is not directly followed from definition.
– User X
2 hours ago
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The map $Bbb Z to Bbb Z$ in your morphism of short exact sequences is just the identity, and so induces an isomorphism on $H^*(X;Bbb Z)$.
The second and third maps are inclusion, and think about what this says at the level of building up a vector bundle from transition functions: given a representative for an element of $H^1(X;mathcal O^*)$, we have charts $U_i subset X$ and holomorphic maps $U_i cap U_j to Bbb C^times$ that we think of as transition functions, the way to glue $U_i times Bbb C$ to $U_j times Bbb C$ over the overlap. The corresponding element of $H^1(X;(mathcal C^infty)^*)$ is the same maps $U_i cap U_j to Bbb C^times$, but we have forgotten that they are holomorphic (and just remembered that they are smooth). Thus the line bundle we construct is the same smooth manifold (it has the same charts and transition maps!), but we have forgotten the holomorphic structure it naturally inherits.
So the conclusion you get is that the map $c_1: H^1(X;mathcal O^*) = textHolLineBun to H^2(X;Bbb Z)$ factors through $H^1(X;(mathcal C^infty)^*) = textSmLineBun_Bbb C$.
(That is, the first Chern class is really an invariant of smooth complex line bundles, and to take the first Chern class of a holomorphic line bundle, just forget that it's holomorphic!)
answered 11 hours ago
Mike Miller
33.6k362127
The map $Bbb Z to Bbb Z$ in your morphism of short exact sequences is just the identity, and so induces an isomorphism on $H^*(X;Bbb Z)$.
The second and third maps are inclusion, and think about what this says at the level of building up a vector bundle from transition functions: given a representative for an element of $H^1(X;mathcal O^*)$, we have charts $U_i subset X$ and holomorphic maps $U_i cap U_j to Bbb C^times$ that we think of as transition functions, the way to glue $U_i times Bbb C$ to $U_j times Bbb C$ over the overlap. The corresponding element of $H^1(X;(mathcal C^infty)^*)$ is the same maps $U_i cap U_j to Bbb C^times$, but we have forgotten that they are holomorphic (and just remembered that they are smooth). Thus the line bundle we construct is the same smooth manifold (it has the same charts and transition maps!), but we have forgotten the holomorphic structure it naturally inherits.
So the conclusion you get is that the map $c_1: H^1(X;mathcal O^*) = textHolLineBun to H^2(X;Bbb Z)$ factors through $H^1(X;(mathcal C^infty)^*) = textSmLineBun_Bbb C$.
(That is, the first Chern class is really an invariant of smooth complex line bundles, and to take the first Chern class of a holomorphic line bundle, just forget that it's holomorphic!)
answered 11 hours ago
Mike Miller
33.6k362127
answered 11 hours ago
Mike Miller
33.6k362127
answered 11 hours ago
Mike Miller
33.6k362127
answered 11 hours ago
Mike Miller
33.6k362127
33.6k362127
Maybe stupid but why identity map must induce isomorphism on cohomology? I think this is not directly followed from definition.
– User X
2 hours ago
add a comment |Â
Maybe stupid but why identity map must induce isomorphism on cohomology? I think this is not directly followed from definition.
– User X
2 hours ago
Maybe stupid but why identity map must induce isomorphism on cohomology? I think this is not directly followed from definition.
– User X
2 hours ago
Maybe stupid but why identity map must induce isomorphism on cohomology? I think this is not directly followed from definition.
– User X
2 hours ago
add a comment |Â
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