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Classification of line bundles by second cohomology of a manifold
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In the book Loop spaces, Characteristic classes and geometric quantization by Brylinski I see following result when trying to motivate geometric description of $H^3(M,mathbbZ)$.
$H^2(M,mathbbZ)$ is the group of isomorphism classes of line bundles over $M$.
I guess they mean there is a natural isomorphism.
Can some one give a rough idea of what obvious second cohomology class we can think of given a line bundle over $M$ and what line bundle can we think of given an arbitrary second cohomology class.
Intuitive comments are also welcome.
I am familiar (not the proof details) with following result:
If $G$ is a group and $M$ is a $G$-module, then the $H^2(G, A)$ is in one-one correspondence with the set of
equivalence classes of extensions $E$ of $M$ by $G$, in which the action of $G$ on $M$ induced by conjugation in $E$ is the same as the action defined by the $G$- module $M$.
I am expecting some intuitive explanation that looks similar to this.
smooth-manifolds line-bundles derham-cohomology
asked Aug 14 at 11:27
Praphulla Koushik
4541220
 |Â
show 1 more comment
up vote
2
down vote
favorite
In the book Loop spaces, Characteristic classes and geometric quantization by Brylinski I see following result when trying to motivate geometric description of $H^3(M,mathbbZ)$.
$H^2(M,mathbbZ)$ is the group of isomorphism classes of line bundles over $M$.
I guess they mean there is a natural isomorphism.
Can some one give a rough idea of what obvious second cohomology class we can think of given a line bundle over $M$ and what line bundle can we think of given an arbitrary second cohomology class.
Intuitive comments are also welcome.
I am familiar (not the proof details) with following result:
If $G$ is a group and $M$ is a $G$-module, then the $H^2(G, A)$ is in one-one correspondence with the set of
equivalence classes of extensions $E$ of $M$ by $G$, in which the action of $G$ on $M$ induced by conjugation in $E$ is the same as the action defined by the $G$- module $M$.
I am expecting some intuitive explanation that looks similar to this.
smooth-manifolds line-bundles derham-cohomology
asked Aug 14 at 11:27
Praphulla Koushik
4541220
6
H^2(M,Z) classifies complex line bundles with respect to tensor product, by the first Chern class. One can find the isomorphism using the exponential sequence, found in any introductory text on complex algebraic geometry.
– Tobias Shin
Aug 14 at 11:38
@TobiasShin Yes, that is true. Given a complex line bundle, we can associate first(and only) chern class which gives an element of $H^2(M,mathbbZ)$.. Can you give some reference whose proof you like it better than others..
– Praphulla Koushik
Aug 14 at 11:47
@TobiasShin Can you make your comment as an answer.. I can upvote..
– Praphulla Koushik
Aug 14 at 14:14
@users who want to close. Reason some one mentioned for their close vote is "This question does not appear to be about research level mathematics within the scope defined in the help center." I can not argue that this is a research level question. I believe this is serious question. I am ok if you think otherwise but can you wait till Tobias responds for my comment and makes his comment as an answer.. Then you can close it..
– Praphulla Koushik
Aug 14 at 16:21
It's possible that this was all supposed to be obvious to me, and it's certainly all familiar in broad strokes, but I upvoted the question and all of the answers.
– Paul Siegel
Aug 17 at 19:07
 |Â
show 1 more comment
up vote
2
down vote
favorite
up vote
2
down vote
favorite
In the book Loop spaces, Characteristic classes and geometric quantization by Brylinski I see following result when trying to motivate geometric description of $H^3(M,mathbbZ)$.
$H^2(M,mathbbZ)$ is the group of isomorphism classes of line bundles over $M$.
I guess they mean there is a natural isomorphism.
Can some one give a rough idea of what obvious second cohomology class we can think of given a line bundle over $M$ and what line bundle can we think of given an arbitrary second cohomology class.
Intuitive comments are also welcome.
I am familiar (not the proof details) with following result:
If $G$ is a group and $M$ is a $G$-module, then the $H^2(G, A)$ is in one-one correspondence with the set of
equivalence classes of extensions $E$ of $M$ by $G$, in which the action of $G$ on $M$ induced by conjugation in $E$ is the same as the action defined by the $G$- module $M$.
I am expecting some intuitive explanation that looks similar to this.
smooth-manifolds line-bundles derham-cohomology
asked Aug 14 at 11:27
Praphulla Koushik
4541220
In the book Loop spaces, Characteristic classes and geometric quantization by Brylinski I see following result when trying to motivate geometric description of $H^3(M,mathbbZ)$.
$H^2(M,mathbbZ)$ is the group of isomorphism classes of line bundles over $M$.
I guess they mean there is a natural isomorphism.
Can some one give a rough idea of what obvious second cohomology class we can think of given a line bundle over $M$ and what line bundle can we think of given an arbitrary second cohomology class.
Intuitive comments are also welcome.
I am familiar (not the proof details) with following result:
If $G$ is a group and $M$ is a $G$-module, then the $H^2(G, A)$ is in one-one correspondence with the set of
equivalence classes of extensions $E$ of $M$ by $G$, in which the action of $G$ on $M$ induced by conjugation in $E$ is the same as the action defined by the $G$- module $M$.
I am expecting some intuitive explanation that looks similar to this.
smooth-manifolds line-bundles derham-cohomology
asked Aug 14 at 11:27
Praphulla Koushik
4541220
edited Aug 14 at 13:49
asked Aug 14 at 11:27
Praphulla Koushik
4541220
asked Aug 14 at 11:27
Praphulla Koushik
4541220
asked Aug 14 at 11:27
Praphulla Koushik
4541220
4541220
6
H^2(M,Z) classifies complex line bundles with respect to tensor product, by the first Chern class. One can find the isomorphism using the exponential sequence, found in any introductory text on complex algebraic geometry.
– Tobias Shin
Aug 14 at 11:38
@TobiasShin Yes, that is true. Given a complex line bundle, we can associate first(and only) chern class which gives an element of $H^2(M,mathbbZ)$.. Can you give some reference whose proof you like it better than others..
– Praphulla Koushik
Aug 14 at 11:47
@TobiasShin Can you make your comment as an answer.. I can upvote..
– Praphulla Koushik
Aug 14 at 14:14
@users who want to close. Reason some one mentioned for their close vote is "This question does not appear to be about research level mathematics within the scope defined in the help center." I can not argue that this is a research level question. I believe this is serious question. I am ok if you think otherwise but can you wait till Tobias responds for my comment and makes his comment as an answer.. Then you can close it..
– Praphulla Koushik
Aug 14 at 16:21
It's possible that this was all supposed to be obvious to me, and it's certainly all familiar in broad strokes, but I upvoted the question and all of the answers.
– Paul Siegel
Aug 17 at 19:07
 |Â
show 1 more comment
6
H^2(M,Z) classifies complex line bundles with respect to tensor product, by the first Chern class. One can find the isomorphism using the exponential sequence, found in any introductory text on complex algebraic geometry.
– Tobias Shin
Aug 14 at 11:38
@TobiasShin Yes, that is true. Given a complex line bundle, we can associate first(and only) chern class which gives an element of $H^2(M,mathbbZ)$.. Can you give some reference whose proof you like it better than others..
– Praphulla Koushik
Aug 14 at 11:47
@TobiasShin Can you make your comment as an answer.. I can upvote..
– Praphulla Koushik
Aug 14 at 14:14
@users who want to close. Reason some one mentioned for their close vote is "This question does not appear to be about research level mathematics within the scope defined in the help center." I can not argue that this is a research level question. I believe this is serious question. I am ok if you think otherwise but can you wait till Tobias responds for my comment and makes his comment as an answer.. Then you can close it..
– Praphulla Koushik
Aug 14 at 16:21
It's possible that this was all supposed to be obvious to me, and it's certainly all familiar in broad strokes, but I upvoted the question and all of the answers.
– Paul Siegel
Aug 17 at 19:07
6
6
H^2(M,Z) classifies complex line bundles with respect to tensor product, by the first Chern class. One can find the isomorphism using the exponential sequence, found in any introductory text on complex algebraic geometry.
– Tobias Shin
Aug 14 at 11:38
H^2(M,Z) classifies complex line bundles with respect to tensor product, by the first Chern class. One can find the isomorphism using the exponential sequence, found in any introductory text on complex algebraic geometry.
– Tobias Shin
Aug 14 at 11:38
@TobiasShin Yes, that is true. Given a complex line bundle, we can associate first(and only) chern class which gives an element of $H^2(M,mathbbZ)$.. Can you give some reference whose proof you like it better than others..
– Praphulla Koushik
Aug 14 at 11:47
@TobiasShin Yes, that is true. Given a complex line bundle, we can associate first(and only) chern class which gives an element of $H^2(M,mathbbZ)$.. Can you give some reference whose proof you like it better than others..
– Praphulla Koushik
Aug 14 at 11:47
@TobiasShin Can you make your comment as an answer.. I can upvote..
– Praphulla Koushik
Aug 14 at 14:14
@TobiasShin Can you make your comment as an answer.. I can upvote..
– Praphulla Koushik
Aug 14 at 14:14
@users who want to close. Reason some one mentioned for their close vote is "This question does not appear to be about research level mathematics within the scope defined in the help center." I can not argue that this is a research level question. I believe this is serious question. I am ok if you think otherwise but can you wait till Tobias responds for my comment and makes his comment as an answer.. Then you can close it..
– Praphulla Koushik
Aug 14 at 16:21
@users who want to close. Reason some one mentioned for their close vote is "This question does not appear to be about research level mathematics within the scope defined in the help center." I can not argue that this is a research level question. I believe this is serious question. I am ok if you think otherwise but can you wait till Tobias responds for my comment and makes his comment as an answer.. Then you can close it..
– Praphulla Koushik
Aug 14 at 16:21
It's possible that this was all supposed to be obvious to me, and it's certainly all familiar in broad strokes, but I upvoted the question and all of the answers.
– Paul Siegel
Aug 17 at 19:07
It's possible that this was all supposed to be obvious to me, and it's certainly all familiar in broad strokes, but I upvoted the question and all of the answers.
– Paul Siegel
Aug 17 at 19:07
 |Â
show 1 more comment
3 Answers
3
active
oldest
votes
up vote
9
down vote
accepted
I think $H^2(M;mathbbZ)$ cannot mean the de Rham cohomology group. The coefficients are wrong.
Anyway: $mathbbCP^infty$ is an amazing space. It is both a model for $K(mathbbZ,2)$ and a model of $BU(1)$. Homotopy classes into $K(mathbb Z,2)$ is in bijection with $H^2(M;mathbbZ)$ (By pulling back the fundamental class) and homotopy classes into $BU(1)$ classify (complex) line bundles (by pulling back the tautological bundle). This works well with natural group structures. This gives the required isomorphism.
The cohomology class in $H^2(M;mathbbZ)$ corresponding to the complex line bundle is the first Chern class.
I can recommend Milnor Stasheff "characteristic classes" for the classification of the line bundles. I can recommend Hatcher for the classification of cohomology in terms of homotopy classes into $K(mathbb Z,n)$.
answered Aug 14 at 11:39
Thomas Rot
2,96411733
After identifying deRham cohomology with singular cohomology...
– Praphulla Koushik
Aug 14 at 11:46
3
The point is that the coefficients are wrong. De Rham cohomology is isomorphic with the singular cohomology with $mathbbR$ coefficients. This does not recover the cohomology with coefficients in $mathbbZ$.
– Thomas Rot
Aug 14 at 11:47
1
You can also look at Chern-Weil theory if you want to stay in the de Rham world.
– Thomas Rot
Aug 14 at 11:49
:D :D I am very sorry.. I do not know what I was thinking when I said that... Yes, you are right.. isomorphism is from $H^2_dR(M)$ to $H^2(M,mathbbR)$..
– Praphulla Koushik
Aug 14 at 11:49
Can you explain your comment “you can look at Chern-Weil theory if you want to stay in deRham worldâ€Â...
– Praphulla Koushik
Aug 14 at 11:56
 |Â
show 7 more comments
up vote
6
down vote
For what it is worth, here is another approach, in the algebraic geometry style. By taking out the zero section, complex line bundles correspond bijectively to $mathbbC^*!$-principal bundles on $M$. Such bundles are classified by the cohomology group $H^1(M, mathcalO_M^*)$, where $mathcalO_M$ is the sheaf of $mathcalC^infty$ complex-valued functions on $M$. Now there is an exact sequence of sheaves
$$0rightarrow mathbbZlongrightarrow mathcalO_Mxrightarrow mathbfe mathcalO_M^*rightarrow 1 $$where $mathbfe(f):=exp(2pi if)$.
This gives rise to a cohomology exact sequence
$$H^1(M, mathcalO_M)longrightarrow H^1(M, mathcalO_M^*)xrightarrow partial H^2(M,mathbbZ)longrightarrow H^2(M, mathcalO_M),.$$But since $mathcalO_M$ is a fine sheaf, its higher cohomology vanishes, and $partial $ is an isomorphism.
answered Aug 14 at 14:27
abx
22k34277
I saw this approach some time back... Thanks for your answer +1.. Do you have any specific references in your thought where this is done in a better way than others?? It is done in Brylinkski's Loop spaces, Characteristic classes and Geometric quantization (I did not read that seriously though)...
– Praphulla Koushik
Aug 14 at 15:00
add a comment |Â
up vote
4
down vote
Although Milnor and Stasheff is an excellent suggestion, I was also led to wonder where one would find this result in more recent textbooks. Most ingredients are in May's "Concise introduction to algebraic topology". Specifically, May proves on page 177 that $H^2(X;mathbbZ)simeq [X,K(mathbbZ,2)]$. Here $K(mathbbZ,2)$ is officially defined as $B^2(mathbbZ)$, where $B$ is the simplicial classifying space functor. However, on page 121 May gives a sequence of exercises about Eilenberg-MacLane spaces. As a special case, we find that whenever $Z$ is a connected CW complex with $pi_2(Z)=mathbbZ$ and $pi_i(Z)=0$ for $ineq 2$, we have $Zsimeq K(mathbbZ,2)$. In particular, it is not hard to find fibrations whose long exact sequences prove that the spaces $Z=BU(1)$ and $Z=mathbbCP^infty$ have the required homotopy groups, so they are both homotopy equivalent to $K(mathbbZ,2)$. May also proves on page 197 that the set of isomorphism classes of line bundles on $X$ is naturally identified with $[X,BU(1)]$.
answered Aug 17 at 11:04
Neil Strickland
34.9k589181
1
One reference would be Husemoller's "Fibre bundles". In the 3rd edition it is theorem 3.4 in Chapter 17, p.250.
– Igor Belegradek
Aug 17 at 12:36
Thanks, I never saw that book carefully.. This gives enough reason to see that with care :) :) Thanks again... +1
– Praphulla Koushik
Aug 17 at 13:01
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
9
down vote
accepted
I think $H^2(M;mathbbZ)$ cannot mean the de Rham cohomology group. The coefficients are wrong.
Anyway: $mathbbCP^infty$ is an amazing space. It is both a model for $K(mathbbZ,2)$ and a model of $BU(1)$. Homotopy classes into $K(mathbb Z,2)$ is in bijection with $H^2(M;mathbbZ)$ (By pulling back the fundamental class) and homotopy classes into $BU(1)$ classify (complex) line bundles (by pulling back the tautological bundle). This works well with natural group structures. This gives the required isomorphism.
The cohomology class in $H^2(M;mathbbZ)$ corresponding to the complex line bundle is the first Chern class.
I can recommend Milnor Stasheff "characteristic classes" for the classification of the line bundles. I can recommend Hatcher for the classification of cohomology in terms of homotopy classes into $K(mathbb Z,n)$.
answered Aug 14 at 11:39
Thomas Rot
2,96411733
After identifying deRham cohomology with singular cohomology...
– Praphulla Koushik
Aug 14 at 11:46
3
The point is that the coefficients are wrong. De Rham cohomology is isomorphic with the singular cohomology with $mathbbR$ coefficients. This does not recover the cohomology with coefficients in $mathbbZ$.
– Thomas Rot
Aug 14 at 11:47
1
You can also look at Chern-Weil theory if you want to stay in the de Rham world.
– Thomas Rot
Aug 14 at 11:49
:D :D I am very sorry.. I do not know what I was thinking when I said that... Yes, you are right.. isomorphism is from $H^2_dR(M)$ to $H^2(M,mathbbR)$..
– Praphulla Koushik
Aug 14 at 11:49
Can you explain your comment “you can look at Chern-Weil theory if you want to stay in deRham worldâ€Â...
– Praphulla Koushik
Aug 14 at 11:56
 |Â
show 7 more comments
up vote
9
down vote
accepted
I think $H^2(M;mathbbZ)$ cannot mean the de Rham cohomology group. The coefficients are wrong.
Anyway: $mathbbCP^infty$ is an amazing space. It is both a model for $K(mathbbZ,2)$ and a model of $BU(1)$. Homotopy classes into $K(mathbb Z,2)$ is in bijection with $H^2(M;mathbbZ)$ (By pulling back the fundamental class) and homotopy classes into $BU(1)$ classify (complex) line bundles (by pulling back the tautological bundle). This works well with natural group structures. This gives the required isomorphism.
The cohomology class in $H^2(M;mathbbZ)$ corresponding to the complex line bundle is the first Chern class.
I can recommend Milnor Stasheff "characteristic classes" for the classification of the line bundles. I can recommend Hatcher for the classification of cohomology in terms of homotopy classes into $K(mathbb Z,n)$.
answered Aug 14 at 11:39
Thomas Rot
2,96411733
After identifying deRham cohomology with singular cohomology...
– Praphulla Koushik
Aug 14 at 11:46
3
The point is that the coefficients are wrong. De Rham cohomology is isomorphic with the singular cohomology with $mathbbR$ coefficients. This does not recover the cohomology with coefficients in $mathbbZ$.
– Thomas Rot
Aug 14 at 11:47
1
You can also look at Chern-Weil theory if you want to stay in the de Rham world.
– Thomas Rot
Aug 14 at 11:49
:D :D I am very sorry.. I do not know what I was thinking when I said that... Yes, you are right.. isomorphism is from $H^2_dR(M)$ to $H^2(M,mathbbR)$..
– Praphulla Koushik
Aug 14 at 11:49
Can you explain your comment “you can look at Chern-Weil theory if you want to stay in deRham worldâ€Â...
– Praphulla Koushik
Aug 14 at 11:56
 |Â
show 7 more comments
up vote
9
down vote
accepted
up vote
9
down vote
accepted
I think $H^2(M;mathbbZ)$ cannot mean the de Rham cohomology group. The coefficients are wrong.
Anyway: $mathbbCP^infty$ is an amazing space. It is both a model for $K(mathbbZ,2)$ and a model of $BU(1)$. Homotopy classes into $K(mathbb Z,2)$ is in bijection with $H^2(M;mathbbZ)$ (By pulling back the fundamental class) and homotopy classes into $BU(1)$ classify (complex) line bundles (by pulling back the tautological bundle). This works well with natural group structures. This gives the required isomorphism.
The cohomology class in $H^2(M;mathbbZ)$ corresponding to the complex line bundle is the first Chern class.
I can recommend Milnor Stasheff "characteristic classes" for the classification of the line bundles. I can recommend Hatcher for the classification of cohomology in terms of homotopy classes into $K(mathbb Z,n)$.
answered Aug 14 at 11:39
Thomas Rot
2,96411733
I think $H^2(M;mathbbZ)$ cannot mean the de Rham cohomology group. The coefficients are wrong.
Anyway: $mathbbCP^infty$ is an amazing space. It is both a model for $K(mathbbZ,2)$ and a model of $BU(1)$. Homotopy classes into $K(mathbb Z,2)$ is in bijection with $H^2(M;mathbbZ)$ (By pulling back the fundamental class) and homotopy classes into $BU(1)$ classify (complex) line bundles (by pulling back the tautological bundle). This works well with natural group structures. This gives the required isomorphism.
The cohomology class in $H^2(M;mathbbZ)$ corresponding to the complex line bundle is the first Chern class.
I can recommend Milnor Stasheff "characteristic classes" for the classification of the line bundles. I can recommend Hatcher for the classification of cohomology in terms of homotopy classes into $K(mathbb Z,n)$.
answered Aug 14 at 11:39
Thomas Rot
2,96411733
edited Aug 14 at 11:48
answered Aug 14 at 11:39
Thomas Rot
2,96411733
answered Aug 14 at 11:39
Thomas Rot
2,96411733
answered Aug 14 at 11:39
Thomas Rot
2,96411733
2,96411733
After identifying deRham cohomology with singular cohomology...
– Praphulla Koushik
Aug 14 at 11:46
3
The point is that the coefficients are wrong. De Rham cohomology is isomorphic with the singular cohomology with $mathbbR$ coefficients. This does not recover the cohomology with coefficients in $mathbbZ$.
– Thomas Rot
Aug 14 at 11:47
1
You can also look at Chern-Weil theory if you want to stay in the de Rham world.
– Thomas Rot
Aug 14 at 11:49
:D :D I am very sorry.. I do not know what I was thinking when I said that... Yes, you are right.. isomorphism is from $H^2_dR(M)$ to $H^2(M,mathbbR)$..
– Praphulla Koushik
Aug 14 at 11:49
Can you explain your comment “you can look at Chern-Weil theory if you want to stay in deRham worldâ€Â...
– Praphulla Koushik
Aug 14 at 11:56
 |Â
show 7 more comments
After identifying deRham cohomology with singular cohomology...
– Praphulla Koushik
Aug 14 at 11:46
3
The point is that the coefficients are wrong. De Rham cohomology is isomorphic with the singular cohomology with $mathbbR$ coefficients. This does not recover the cohomology with coefficients in $mathbbZ$.
– Thomas Rot
Aug 14 at 11:47
1
You can also look at Chern-Weil theory if you want to stay in the de Rham world.
– Thomas Rot
Aug 14 at 11:49
:D :D I am very sorry.. I do not know what I was thinking when I said that... Yes, you are right.. isomorphism is from $H^2_dR(M)$ to $H^2(M,mathbbR)$..
– Praphulla Koushik
Aug 14 at 11:49
Can you explain your comment “you can look at Chern-Weil theory if you want to stay in deRham worldâ€Â...
– Praphulla Koushik
Aug 14 at 11:56
After identifying deRham cohomology with singular cohomology...
– Praphulla Koushik
Aug 14 at 11:46
After identifying deRham cohomology with singular cohomology...
– Praphulla Koushik
Aug 14 at 11:46
3
3
The point is that the coefficients are wrong. De Rham cohomology is isomorphic with the singular cohomology with $mathbbR$ coefficients. This does not recover the cohomology with coefficients in $mathbbZ$.
– Thomas Rot
Aug 14 at 11:47
The point is that the coefficients are wrong. De Rham cohomology is isomorphic with the singular cohomology with $mathbbR$ coefficients. This does not recover the cohomology with coefficients in $mathbbZ$.
– Thomas Rot
Aug 14 at 11:47
1
1
You can also look at Chern-Weil theory if you want to stay in the de Rham world.
– Thomas Rot
Aug 14 at 11:49
You can also look at Chern-Weil theory if you want to stay in the de Rham world.
– Thomas Rot
Aug 14 at 11:49
:D :D I am very sorry.. I do not know what I was thinking when I said that... Yes, you are right.. isomorphism is from $H^2_dR(M)$ to $H^2(M,mathbbR)$..
– Praphulla Koushik
Aug 14 at 11:49
:D :D I am very sorry.. I do not know what I was thinking when I said that... Yes, you are right.. isomorphism is from $H^2_dR(M)$ to $H^2(M,mathbbR)$..
– Praphulla Koushik
Aug 14 at 11:49
Can you explain your comment “you can look at Chern-Weil theory if you want to stay in deRham worldâ€Â...
– Praphulla Koushik
Aug 14 at 11:56
Can you explain your comment “you can look at Chern-Weil theory if you want to stay in deRham worldâ€Â...
– Praphulla Koushik
Aug 14 at 11:56
 |Â
show 7 more comments
up vote
6
down vote
For what it is worth, here is another approach, in the algebraic geometry style. By taking out the zero section, complex line bundles correspond bijectively to $mathbbC^*!$-principal bundles on $M$. Such bundles are classified by the cohomology group $H^1(M, mathcalO_M^*)$, where $mathcalO_M$ is the sheaf of $mathcalC^infty$ complex-valued functions on $M$. Now there is an exact sequence of sheaves
$$0rightarrow mathbbZlongrightarrow mathcalO_Mxrightarrow mathbfe mathcalO_M^*rightarrow 1 $$where $mathbfe(f):=exp(2pi if)$.
This gives rise to a cohomology exact sequence
$$H^1(M, mathcalO_M)longrightarrow H^1(M, mathcalO_M^*)xrightarrow partial H^2(M,mathbbZ)longrightarrow H^2(M, mathcalO_M),.$$But since $mathcalO_M$ is a fine sheaf, its higher cohomology vanishes, and $partial $ is an isomorphism.
answered Aug 14 at 14:27
abx
22k34277
I saw this approach some time back... Thanks for your answer +1.. Do you have any specific references in your thought where this is done in a better way than others?? It is done in Brylinkski's Loop spaces, Characteristic classes and Geometric quantization (I did not read that seriously though)...
– Praphulla Koushik
Aug 14 at 15:00
add a comment |Â
up vote
6
down vote
For what it is worth, here is another approach, in the algebraic geometry style. By taking out the zero section, complex line bundles correspond bijectively to $mathbbC^*!$-principal bundles on $M$. Such bundles are classified by the cohomology group $H^1(M, mathcalO_M^*)$, where $mathcalO_M$ is the sheaf of $mathcalC^infty$ complex-valued functions on $M$. Now there is an exact sequence of sheaves
$$0rightarrow mathbbZlongrightarrow mathcalO_Mxrightarrow mathbfe mathcalO_M^*rightarrow 1 $$where $mathbfe(f):=exp(2pi if)$.
This gives rise to a cohomology exact sequence
$$H^1(M, mathcalO_M)longrightarrow H^1(M, mathcalO_M^*)xrightarrow partial H^2(M,mathbbZ)longrightarrow H^2(M, mathcalO_M),.$$But since $mathcalO_M$ is a fine sheaf, its higher cohomology vanishes, and $partial $ is an isomorphism.
answered Aug 14 at 14:27
abx
22k34277
I saw this approach some time back... Thanks for your answer +1.. Do you have any specific references in your thought where this is done in a better way than others?? It is done in Brylinkski's Loop spaces, Characteristic classes and Geometric quantization (I did not read that seriously though)...
– Praphulla Koushik
Aug 14 at 15:00
add a comment |Â
up vote
6
down vote
up vote
6
down vote
For what it is worth, here is another approach, in the algebraic geometry style. By taking out the zero section, complex line bundles correspond bijectively to $mathbbC^*!$-principal bundles on $M$. Such bundles are classified by the cohomology group $H^1(M, mathcalO_M^*)$, where $mathcalO_M$ is the sheaf of $mathcalC^infty$ complex-valued functions on $M$. Now there is an exact sequence of sheaves
$$0rightarrow mathbbZlongrightarrow mathcalO_Mxrightarrow mathbfe mathcalO_M^*rightarrow 1 $$where $mathbfe(f):=exp(2pi if)$.
This gives rise to a cohomology exact sequence
$$H^1(M, mathcalO_M)longrightarrow H^1(M, mathcalO_M^*)xrightarrow partial H^2(M,mathbbZ)longrightarrow H^2(M, mathcalO_M),.$$But since $mathcalO_M$ is a fine sheaf, its higher cohomology vanishes, and $partial $ is an isomorphism.
answered Aug 14 at 14:27
abx
22k34277
For what it is worth, here is another approach, in the algebraic geometry style. By taking out the zero section, complex line bundles correspond bijectively to $mathbbC^*!$-principal bundles on $M$. Such bundles are classified by the cohomology group $H^1(M, mathcalO_M^*)$, where $mathcalO_M$ is the sheaf of $mathcalC^infty$ complex-valued functions on $M$. Now there is an exact sequence of sheaves
$$0rightarrow mathbbZlongrightarrow mathcalO_Mxrightarrow mathbfe mathcalO_M^*rightarrow 1 $$where $mathbfe(f):=exp(2pi if)$.
This gives rise to a cohomology exact sequence
$$H^1(M, mathcalO_M)longrightarrow H^1(M, mathcalO_M^*)xrightarrow partial H^2(M,mathbbZ)longrightarrow H^2(M, mathcalO_M),.$$But since $mathcalO_M$ is a fine sheaf, its higher cohomology vanishes, and $partial $ is an isomorphism.
answered Aug 14 at 14:27
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answered Aug 14 at 14:27
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answered Aug 14 at 14:27
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answered Aug 14 at 14:27
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I saw this approach some time back... Thanks for your answer +1.. Do you have any specific references in your thought where this is done in a better way than others?? It is done in Brylinkski's Loop spaces, Characteristic classes and Geometric quantization (I did not read that seriously though)...
– Praphulla Koushik
Aug 14 at 15:00
add a comment |Â
I saw this approach some time back... Thanks for your answer +1.. Do you have any specific references in your thought where this is done in a better way than others?? It is done in Brylinkski's Loop spaces, Characteristic classes and Geometric quantization (I did not read that seriously though)...
– Praphulla Koushik
Aug 14 at 15:00
I saw this approach some time back... Thanks for your answer +1.. Do you have any specific references in your thought where this is done in a better way than others?? It is done in Brylinkski's Loop spaces, Characteristic classes and Geometric quantization (I did not read that seriously though)...
– Praphulla Koushik
Aug 14 at 15:00
I saw this approach some time back... Thanks for your answer +1.. Do you have any specific references in your thought where this is done in a better way than others?? It is done in Brylinkski's Loop spaces, Characteristic classes and Geometric quantization (I did not read that seriously though)...
– Praphulla Koushik
Aug 14 at 15:00
add a comment |Â
up vote
4
down vote
Although Milnor and Stasheff is an excellent suggestion, I was also led to wonder where one would find this result in more recent textbooks. Most ingredients are in May's "Concise introduction to algebraic topology". Specifically, May proves on page 177 that $H^2(X;mathbbZ)simeq [X,K(mathbbZ,2)]$. Here $K(mathbbZ,2)$ is officially defined as $B^2(mathbbZ)$, where $B$ is the simplicial classifying space functor. However, on page 121 May gives a sequence of exercises about Eilenberg-MacLane spaces. As a special case, we find that whenever $Z$ is a connected CW complex with $pi_2(Z)=mathbbZ$ and $pi_i(Z)=0$ for $ineq 2$, we have $Zsimeq K(mathbbZ,2)$. In particular, it is not hard to find fibrations whose long exact sequences prove that the spaces $Z=BU(1)$ and $Z=mathbbCP^infty$ have the required homotopy groups, so they are both homotopy equivalent to $K(mathbbZ,2)$. May also proves on page 197 that the set of isomorphism classes of line bundles on $X$ is naturally identified with $[X,BU(1)]$.
answered Aug 17 at 11:04
Neil Strickland
34.9k589181
1
One reference would be Husemoller's "Fibre bundles". In the 3rd edition it is theorem 3.4 in Chapter 17, p.250.
– Igor Belegradek
Aug 17 at 12:36
Thanks, I never saw that book carefully.. This gives enough reason to see that with care :) :) Thanks again... +1
– Praphulla Koushik
Aug 17 at 13:01
add a comment |Â
up vote
4
down vote
Although Milnor and Stasheff is an excellent suggestion, I was also led to wonder where one would find this result in more recent textbooks. Most ingredients are in May's "Concise introduction to algebraic topology". Specifically, May proves on page 177 that $H^2(X;mathbbZ)simeq [X,K(mathbbZ,2)]$. Here $K(mathbbZ,2)$ is officially defined as $B^2(mathbbZ)$, where $B$ is the simplicial classifying space functor. However, on page 121 May gives a sequence of exercises about Eilenberg-MacLane spaces. As a special case, we find that whenever $Z$ is a connected CW complex with $pi_2(Z)=mathbbZ$ and $pi_i(Z)=0$ for $ineq 2$, we have $Zsimeq K(mathbbZ,2)$. In particular, it is not hard to find fibrations whose long exact sequences prove that the spaces $Z=BU(1)$ and $Z=mathbbCP^infty$ have the required homotopy groups, so they are both homotopy equivalent to $K(mathbbZ,2)$. May also proves on page 197 that the set of isomorphism classes of line bundles on $X$ is naturally identified with $[X,BU(1)]$.
answered Aug 17 at 11:04
Neil Strickland
34.9k589181
1
One reference would be Husemoller's "Fibre bundles". In the 3rd edition it is theorem 3.4 in Chapter 17, p.250.
– Igor Belegradek
Aug 17 at 12:36
Thanks, I never saw that book carefully.. This gives enough reason to see that with care :) :) Thanks again... +1
– Praphulla Koushik
Aug 17 at 13:01
add a comment |Â
up vote
4
down vote
up vote
4
down vote
Although Milnor and Stasheff is an excellent suggestion, I was also led to wonder where one would find this result in more recent textbooks. Most ingredients are in May's "Concise introduction to algebraic topology". Specifically, May proves on page 177 that $H^2(X;mathbbZ)simeq [X,K(mathbbZ,2)]$. Here $K(mathbbZ,2)$ is officially defined as $B^2(mathbbZ)$, where $B$ is the simplicial classifying space functor. However, on page 121 May gives a sequence of exercises about Eilenberg-MacLane spaces. As a special case, we find that whenever $Z$ is a connected CW complex with $pi_2(Z)=mathbbZ$ and $pi_i(Z)=0$ for $ineq 2$, we have $Zsimeq K(mathbbZ,2)$. In particular, it is not hard to find fibrations whose long exact sequences prove that the spaces $Z=BU(1)$ and $Z=mathbbCP^infty$ have the required homotopy groups, so they are both homotopy equivalent to $K(mathbbZ,2)$. May also proves on page 197 that the set of isomorphism classes of line bundles on $X$ is naturally identified with $[X,BU(1)]$.
answered Aug 17 at 11:04
Neil Strickland
34.9k589181
Although Milnor and Stasheff is an excellent suggestion, I was also led to wonder where one would find this result in more recent textbooks. Most ingredients are in May's "Concise introduction to algebraic topology". Specifically, May proves on page 177 that $H^2(X;mathbbZ)simeq [X,K(mathbbZ,2)]$. Here $K(mathbbZ,2)$ is officially defined as $B^2(mathbbZ)$, where $B$ is the simplicial classifying space functor. However, on page 121 May gives a sequence of exercises about Eilenberg-MacLane spaces. As a special case, we find that whenever $Z$ is a connected CW complex with $pi_2(Z)=mathbbZ$ and $pi_i(Z)=0$ for $ineq 2$, we have $Zsimeq K(mathbbZ,2)$. In particular, it is not hard to find fibrations whose long exact sequences prove that the spaces $Z=BU(1)$ and $Z=mathbbCP^infty$ have the required homotopy groups, so they are both homotopy equivalent to $K(mathbbZ,2)$. May also proves on page 197 that the set of isomorphism classes of line bundles on $X$ is naturally identified with $[X,BU(1)]$.
answered Aug 17 at 11:04
Neil Strickland
34.9k589181
answered Aug 17 at 11:04
Neil Strickland
34.9k589181
answered Aug 17 at 11:04
Neil Strickland
34.9k589181
answered Aug 17 at 11:04
Neil Strickland
34.9k589181
34.9k589181
1
One reference would be Husemoller's "Fibre bundles". In the 3rd edition it is theorem 3.4 in Chapter 17, p.250.
– Igor Belegradek
Aug 17 at 12:36
Thanks, I never saw that book carefully.. This gives enough reason to see that with care :) :) Thanks again... +1
– Praphulla Koushik
Aug 17 at 13:01
add a comment |Â
1
One reference would be Husemoller's "Fibre bundles". In the 3rd edition it is theorem 3.4 in Chapter 17, p.250.
– Igor Belegradek
Aug 17 at 12:36
Thanks, I never saw that book carefully.. This gives enough reason to see that with care :) :) Thanks again... +1
– Praphulla Koushik
Aug 17 at 13:01
1
1
One reference would be Husemoller's "Fibre bundles". In the 3rd edition it is theorem 3.4 in Chapter 17, p.250.
– Igor Belegradek
Aug 17 at 12:36
One reference would be Husemoller's "Fibre bundles". In the 3rd edition it is theorem 3.4 in Chapter 17, p.250.
– Igor Belegradek
Aug 17 at 12:36
Thanks, I never saw that book carefully.. This gives enough reason to see that with care :) :) Thanks again... +1
– Praphulla Koushik
Aug 17 at 13:01
Thanks, I never saw that book carefully.. This gives enough reason to see that with care :) :) Thanks again... +1
– Praphulla Koushik
Aug 17 at 13:01
add a comment |Â
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6
H^2(M,Z) classifies complex line bundles with respect to tensor product, by the first Chern class. One can find the isomorphism using the exponential sequence, found in any introductory text on complex algebraic geometry.
– Tobias Shin
Aug 14 at 11:38
@TobiasShin Yes, that is true. Given a complex line bundle, we can associate first(and only) chern class which gives an element of $H^2(M,mathbbZ)$.. Can you give some reference whose proof you like it better than others..
– Praphulla Koushik
Aug 14 at 11:47
@TobiasShin Can you make your comment as an answer.. I can upvote..
– Praphulla Koushik
Aug 14 at 14:14
@users who want to close. Reason some one mentioned for their close vote is "This question does not appear to be about research level mathematics within the scope defined in the help center." I can not argue that this is a research level question. I believe this is serious question. I am ok if you think otherwise but can you wait till Tobias responds for my comment and makes his comment as an answer.. Then you can close it..
– Praphulla Koushik
Aug 14 at 16:21
It's possible that this was all supposed to be obvious to me, and it's certainly all familiar in broad strokes, but I upvoted the question and all of the answers.
– Paul Siegel
Aug 17 at 19:07