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How does $BU(1)$ “account for the first Chern class�
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Suppose $M$ is the connected sum of infinitely many $n$-dimensional complex projective space, i.e. $M:=#_ige 1mathbbCP^n$. Let $z$ be a holomorphic local coordinate for $M$ and let $V = TM$ be the bundle of complex tangent vectors having the form $afracpartialpartial z$ at each point, where $a$ is a complex number.
Then how do we show that $V$ is or is not a trivial vector bundle (i.e. the first Chern class does or does not vanish) via integration i.e. $int_Mc_1$?
I came across this statement in the wiki article for complex projective space: "There is a space $mathbbCP^infty$ which, in a sense, is the inductive limit of $mathbbCP^n$ as $ntoinfty$. It is $BU(1)$, the classifying space of $U(1)$, in the sense of homotopy theory, and so classifies complex line bundles; equivalently it accounts for the first Chern class."
What does it mean by it accounts for the first Chern class?
Thanks in advance!
algebraic-topology
asked yesterday
Multivariablecalculus
483312
 |Â
show 6 more comments
up vote
-1
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Suppose $M$ is the connected sum of infinitely many $n$-dimensional complex projective space, i.e. $M:=#_ige 1mathbbCP^n$. Let $z$ be a holomorphic local coordinate for $M$ and let $V = TM$ be the bundle of complex tangent vectors having the form $afracpartialpartial z$ at each point, where $a$ is a complex number.
Then how do we show that $V$ is or is not a trivial vector bundle (i.e. the first Chern class does or does not vanish) via integration i.e. $int_Mc_1$?
I came across this statement in the wiki article for complex projective space: "There is a space $mathbbCP^infty$ which, in a sense, is the inductive limit of $mathbbCP^n$ as $ntoinfty$. It is $BU(1)$, the classifying space of $U(1)$, in the sense of homotopy theory, and so classifies complex line bundles; equivalently it accounts for the first Chern class."
What does it mean by it accounts for the first Chern class?
Thanks in advance!
algebraic-topology
asked yesterday
Multivariablecalculus
483312
1
Why are you of the mind that the first Chern class of this bundle (if it even exists) does not vanish?
– Tyrone
yesterday
I was under the impression that the bundle is non-trivial. Maybe this assumption is unfounded? @Tyrone
– Multivariablecalculus
18 hours ago
3
I'm not even sure how you would like to define the tangent bundle in this context. The infinite sphere is tautologically weakly contractible and CW, so therefore contractible. Hence $H^2(sqcup_igeq 0S^infty)cong prod_igeq 0H^2(S^infty)cong prod_igeq 0 0=0$, so the first Chern class of any bundle will be $0$ since it belongs to the trivial group. Unless I am missing something in your details, any bundle over this space will be trivial for the reasons of (component-wise) contractibility.
– Tyrone
18 hours ago
Suppose we instead we consider $bigsqcup_ige 1mathbbCP^infty$. @Tyrone
– Multivariablecalculus
18 hours ago
Well we're still not working with a smooth manifold, so you might like to detail the definitions you wish to use a bit closer. In any case, I don't think you're going to end up with a sensible theory of integration so you might want to give some further details on what it is you really are looking for. Perhaps give some wider context to your question as well.
– Tyrone
18 hours ago
 |Â
show 6 more comments
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
Suppose $M$ is the connected sum of infinitely many $n$-dimensional complex projective space, i.e. $M:=#_ige 1mathbbCP^n$. Let $z$ be a holomorphic local coordinate for $M$ and let $V = TM$ be the bundle of complex tangent vectors having the form $afracpartialpartial z$ at each point, where $a$ is a complex number.
Then how do we show that $V$ is or is not a trivial vector bundle (i.e. the first Chern class does or does not vanish) via integration i.e. $int_Mc_1$?
I came across this statement in the wiki article for complex projective space: "There is a space $mathbbCP^infty$ which, in a sense, is the inductive limit of $mathbbCP^n$ as $ntoinfty$. It is $BU(1)$, the classifying space of $U(1)$, in the sense of homotopy theory, and so classifies complex line bundles; equivalently it accounts for the first Chern class."
What does it mean by it accounts for the first Chern class?
Thanks in advance!
algebraic-topology
asked yesterday
Multivariablecalculus
483312
Suppose $M$ is the connected sum of infinitely many $n$-dimensional complex projective space, i.e. $M:=#_ige 1mathbbCP^n$. Let $z$ be a holomorphic local coordinate for $M$ and let $V = TM$ be the bundle of complex tangent vectors having the form $afracpartialpartial z$ at each point, where $a$ is a complex number.
Then how do we show that $V$ is or is not a trivial vector bundle (i.e. the first Chern class does or does not vanish) via integration i.e. $int_Mc_1$?
I came across this statement in the wiki article for complex projective space: "There is a space $mathbbCP^infty$ which, in a sense, is the inductive limit of $mathbbCP^n$ as $ntoinfty$. It is $BU(1)$, the classifying space of $U(1)$, in the sense of homotopy theory, and so classifies complex line bundles; equivalently it accounts for the first Chern class."
What does it mean by it accounts for the first Chern class?
Thanks in advance!
algebraic-topology
asked yesterday
Multivariablecalculus
483312
edited 17 hours ago
asked yesterday
Multivariablecalculus
483312
asked yesterday
Multivariablecalculus
483312
asked yesterday
Multivariablecalculus
483312
483312
1
Why are you of the mind that the first Chern class of this bundle (if it even exists) does not vanish?
– Tyrone
yesterday
I was under the impression that the bundle is non-trivial. Maybe this assumption is unfounded? @Tyrone
– Multivariablecalculus
18 hours ago
3
I'm not even sure how you would like to define the tangent bundle in this context. The infinite sphere is tautologically weakly contractible and CW, so therefore contractible. Hence $H^2(sqcup_igeq 0S^infty)cong prod_igeq 0H^2(S^infty)cong prod_igeq 0 0=0$, so the first Chern class of any bundle will be $0$ since it belongs to the trivial group. Unless I am missing something in your details, any bundle over this space will be trivial for the reasons of (component-wise) contractibility.
– Tyrone
18 hours ago
Suppose we instead we consider $bigsqcup_ige 1mathbbCP^infty$. @Tyrone
– Multivariablecalculus
18 hours ago
Well we're still not working with a smooth manifold, so you might like to detail the definitions you wish to use a bit closer. In any case, I don't think you're going to end up with a sensible theory of integration so you might want to give some further details on what it is you really are looking for. Perhaps give some wider context to your question as well.
– Tyrone
18 hours ago
 |Â
show 6 more comments
1
Why are you of the mind that the first Chern class of this bundle (if it even exists) does not vanish?
– Tyrone
yesterday
I was under the impression that the bundle is non-trivial. Maybe this assumption is unfounded? @Tyrone
– Multivariablecalculus
18 hours ago
3
I'm not even sure how you would like to define the tangent bundle in this context. The infinite sphere is tautologically weakly contractible and CW, so therefore contractible. Hence $H^2(sqcup_igeq 0S^infty)cong prod_igeq 0H^2(S^infty)cong prod_igeq 0 0=0$, so the first Chern class of any bundle will be $0$ since it belongs to the trivial group. Unless I am missing something in your details, any bundle over this space will be trivial for the reasons of (component-wise) contractibility.
– Tyrone
18 hours ago
Suppose we instead we consider $bigsqcup_ige 1mathbbCP^infty$. @Tyrone
– Multivariablecalculus
18 hours ago
Well we're still not working with a smooth manifold, so you might like to detail the definitions you wish to use a bit closer. In any case, I don't think you're going to end up with a sensible theory of integration so you might want to give some further details on what it is you really are looking for. Perhaps give some wider context to your question as well.
– Tyrone
18 hours ago
1
1
Why are you of the mind that the first Chern class of this bundle (if it even exists) does not vanish?
– Tyrone
yesterday
Why are you of the mind that the first Chern class of this bundle (if it even exists) does not vanish?
– Tyrone
yesterday
I was under the impression that the bundle is non-trivial. Maybe this assumption is unfounded? @Tyrone
– Multivariablecalculus
18 hours ago
I was under the impression that the bundle is non-trivial. Maybe this assumption is unfounded? @Tyrone
– Multivariablecalculus
18 hours ago
3
3
I'm not even sure how you would like to define the tangent bundle in this context. The infinite sphere is tautologically weakly contractible and CW, so therefore contractible. Hence $H^2(sqcup_igeq 0S^infty)cong prod_igeq 0H^2(S^infty)cong prod_igeq 0 0=0$, so the first Chern class of any bundle will be $0$ since it belongs to the trivial group. Unless I am missing something in your details, any bundle over this space will be trivial for the reasons of (component-wise) contractibility.
– Tyrone
18 hours ago
I'm not even sure how you would like to define the tangent bundle in this context. The infinite sphere is tautologically weakly contractible and CW, so therefore contractible. Hence $H^2(sqcup_igeq 0S^infty)cong prod_igeq 0H^2(S^infty)cong prod_igeq 0 0=0$, so the first Chern class of any bundle will be $0$ since it belongs to the trivial group. Unless I am missing something in your details, any bundle over this space will be trivial for the reasons of (component-wise) contractibility.
– Tyrone
18 hours ago
Suppose we instead we consider $bigsqcup_ige 1mathbbCP^infty$. @Tyrone
– Multivariablecalculus
18 hours ago
Suppose we instead we consider $bigsqcup_ige 1mathbbCP^infty$. @Tyrone
– Multivariablecalculus
18 hours ago
Well we're still not working with a smooth manifold, so you might like to detail the definitions you wish to use a bit closer. In any case, I don't think you're going to end up with a sensible theory of integration so you might want to give some further details on what it is you really are looking for. Perhaps give some wider context to your question as well.
– Tyrone
18 hours ago
Well we're still not working with a smooth manifold, so you might like to detail the definitions you wish to use a bit closer. In any case, I don't think you're going to end up with a sensible theory of integration so you might want to give some further details on what it is you really are looking for. Perhaps give some wider context to your question as well.
– Tyrone
18 hours ago
 |Â
show 6 more comments
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1
Why are you of the mind that the first Chern class of this bundle (if it even exists) does not vanish?
– Tyrone
yesterday
I was under the impression that the bundle is non-trivial. Maybe this assumption is unfounded? @Tyrone
– Multivariablecalculus
18 hours ago
3
I'm not even sure how you would like to define the tangent bundle in this context. The infinite sphere is tautologically weakly contractible and CW, so therefore contractible. Hence $H^2(sqcup_igeq 0S^infty)cong prod_igeq 0H^2(S^infty)cong prod_igeq 0 0=0$, so the first Chern class of any bundle will be $0$ since it belongs to the trivial group. Unless I am missing something in your details, any bundle over this space will be trivial for the reasons of (component-wise) contractibility.
– Tyrone
18 hours ago
Suppose we instead we consider $bigsqcup_ige 1mathbbCP^infty$. @Tyrone
– Multivariablecalculus
18 hours ago
Well we're still not working with a smooth manifold, so you might like to detail the definitions you wish to use a bit closer. In any case, I don't think you're going to end up with a sensible theory of integration so you might want to give some further details on what it is you really are looking for. Perhaps give some wider context to your question as well.
– Tyrone
18 hours ago