Mixing
Constructing the first Chern class without using the classifying map for line bundles?
Clash Royale CLAN TAG#URR8PPP
up vote
5
down vote
favorite
I've gathered the gist of a particularly nice construction for Chern classes in a topological setting, but I can't quite figure out how to find the first class without making use of a classifying map. I've been told that this construction roughly goes through in AG without classifying maps (evidently first proposed by Grothendieck.)
Given a vector bundle $E to B$ with fiber $V$, we form the projectivization $mathbb P(E) to B$, which has a tautological sub-bundle $L$, where fibers for $(x,ell) in P$ where $x in B$ and $ell subset E_x$ is exactly $ell$.
A formal descripton is given in $10.1.5$ here.
Now, supposing that we have a description of $alpha in H^2(P,mathbb Z)$, and argue that powers of this element restrict to generators on $H^2(mathbb CP^n-1)$, and conclude with Leray Hirsch that $H^*(mathbb P(E))$ is a free module over $H^*(B)$. Expressing $c_1(L)^n$ as a linear combination of the first $n-1$ powers gives the chern classes for $E$.
Question 1: How can one define $c_1(L) in H^2(P,mathbb Z)$ without using the classifying map $B to mathbb CP^infty$ for line bundles?
Question 2: Can the following argument be made to work (of course by completing it?
Given the tautological bundle $L to mathbb P(E)$ one can use the association
$Vect^1(mathbb P(E)) to checkH^1(mathbb P(E))$ to obtain $alpha in H^1(P,mathbb C^times)$. Is there a way to map from $H^1(mathbb P(E),mathbb C^times) to H^2(mathbb P(E),mathbf Z)$ and use this to get the chern classes?
characteristic-classes
asked Jul 24 at 1:34
Andres Mejia
14.6k11343
 |Â
show 6 more comments
up vote
5
down vote
favorite
I've gathered the gist of a particularly nice construction for Chern classes in a topological setting, but I can't quite figure out how to find the first class without making use of a classifying map. I've been told that this construction roughly goes through in AG without classifying maps (evidently first proposed by Grothendieck.)
Given a vector bundle $E to B$ with fiber $V$, we form the projectivization $mathbb P(E) to B$, which has a tautological sub-bundle $L$, where fibers for $(x,ell) in P$ where $x in B$ and $ell subset E_x$ is exactly $ell$.
A formal descripton is given in $10.1.5$ here.
Now, supposing that we have a description of $alpha in H^2(P,mathbb Z)$, and argue that powers of this element restrict to generators on $H^2(mathbb CP^n-1)$, and conclude with Leray Hirsch that $H^*(mathbb P(E))$ is a free module over $H^*(B)$. Expressing $c_1(L)^n$ as a linear combination of the first $n-1$ powers gives the chern classes for $E$.
Question 1: How can one define $c_1(L) in H^2(P,mathbb Z)$ without using the classifying map $B to mathbb CP^infty$ for line bundles?
Question 2: Can the following argument be made to work (of course by completing it?
Given the tautological bundle $L to mathbb P(E)$ one can use the association
$Vect^1(mathbb P(E)) to checkH^1(mathbb P(E))$ to obtain $alpha in H^1(P,mathbb C^times)$. Is there a way to map from $H^1(mathbb P(E),mathbb C^times) to H^2(mathbb P(E),mathbf Z)$ and use this to get the chern classes?
characteristic-classes
asked Jul 24 at 1:34
Andres Mejia
14.6k11343
Sorry about the ramble below the line. I was just mostly mumbling to myself and I realized something might work (kind of.) I'll leave it in case it helps anyone else conceptually
– Andres Mejia
Jul 24 at 2:25
You could use the connecting morphism in the LES here: en.wikipedia.org/wiki/Exponential_sheaf_sequence (I guess you can repeat the same story on a general topological space, with $O_X$ replaced by the sheaf of continuous functions... but I don't know for sure.)
– Lorenzo
Jul 24 at 18:53
Note that since $O_X$ (sheaf of continuous functions) is soft when $X$ is a metric space, $H^1(O_X) = H^2(O_X) = 0$, so the chern class boundary map is an isomorphism, which exactly what we expect (the chern class determines the isomorphism class of the topological line bundle). (This sanity check suggests to me that the story with the LES would work out.) Would this answer your question? I can write up some details...
– Lorenzo
Jul 24 at 19:06
@lorenzo I think so. As long as there is a construction involved (that is * chern class referenced in your second comment.) I’d love to see how the short exact sequence story goes.
– Andres Mejia
Jul 24 at 19:09
It's absolutely a construction - The Chern class map appears as a boundary map in the LES associated to a certain exact sequence of sheaves. One of these sheaves has the property that its $H^1$ is naturally isomorphic to the group of line bundles, and another one of these sheaves has the property that its $H^2$ is the usual $H^2$. The boundary map connecting them is the chern class. (Have you studied sheaves and sheaf cohomology?)
– Lorenzo
Jul 24 at 19:14
 |Â
show 6 more comments
up vote
5
down vote
favorite
up vote
5
down vote
favorite
I've gathered the gist of a particularly nice construction for Chern classes in a topological setting, but I can't quite figure out how to find the first class without making use of a classifying map. I've been told that this construction roughly goes through in AG without classifying maps (evidently first proposed by Grothendieck.)
Given a vector bundle $E to B$ with fiber $V$, we form the projectivization $mathbb P(E) to B$, which has a tautological sub-bundle $L$, where fibers for $(x,ell) in P$ where $x in B$ and $ell subset E_x$ is exactly $ell$.
A formal descripton is given in $10.1.5$ here.
Now, supposing that we have a description of $alpha in H^2(P,mathbb Z)$, and argue that powers of this element restrict to generators on $H^2(mathbb CP^n-1)$, and conclude with Leray Hirsch that $H^*(mathbb P(E))$ is a free module over $H^*(B)$. Expressing $c_1(L)^n$ as a linear combination of the first $n-1$ powers gives the chern classes for $E$.
Question 1: How can one define $c_1(L) in H^2(P,mathbb Z)$ without using the classifying map $B to mathbb CP^infty$ for line bundles?
Question 2: Can the following argument be made to work (of course by completing it?
Given the tautological bundle $L to mathbb P(E)$ one can use the association
$Vect^1(mathbb P(E)) to checkH^1(mathbb P(E))$ to obtain $alpha in H^1(P,mathbb C^times)$. Is there a way to map from $H^1(mathbb P(E),mathbb C^times) to H^2(mathbb P(E),mathbf Z)$ and use this to get the chern classes?
characteristic-classes
asked Jul 24 at 1:34
Andres Mejia
14.6k11343
I've gathered the gist of a particularly nice construction for Chern classes in a topological setting, but I can't quite figure out how to find the first class without making use of a classifying map. I've been told that this construction roughly goes through in AG without classifying maps (evidently first proposed by Grothendieck.)
Given a vector bundle $E to B$ with fiber $V$, we form the projectivization $mathbb P(E) to B$, which has a tautological sub-bundle $L$, where fibers for $(x,ell) in P$ where $x in B$ and $ell subset E_x$ is exactly $ell$.
A formal descripton is given in $10.1.5$ here.
Now, supposing that we have a description of $alpha in H^2(P,mathbb Z)$, and argue that powers of this element restrict to generators on $H^2(mathbb CP^n-1)$, and conclude with Leray Hirsch that $H^*(mathbb P(E))$ is a free module over $H^*(B)$. Expressing $c_1(L)^n$ as a linear combination of the first $n-1$ powers gives the chern classes for $E$.
Question 1: How can one define $c_1(L) in H^2(P,mathbb Z)$ without using the classifying map $B to mathbb CP^infty$ for line bundles?
Question 2: Can the following argument be made to work (of course by completing it?
Given the tautological bundle $L to mathbb P(E)$ one can use the association
$Vect^1(mathbb P(E)) to checkH^1(mathbb P(E))$ to obtain $alpha in H^1(P,mathbb C^times)$. Is there a way to map from $H^1(mathbb P(E),mathbb C^times) to H^2(mathbb P(E),mathbf Z)$ and use this to get the chern classes?
characteristic-classes
asked Jul 24 at 1:34
Andres Mejia
14.6k11343
edited Jul 25 at 1:03
asked Jul 24 at 1:34
Andres Mejia
14.6k11343
asked Jul 24 at 1:34
Andres Mejia
14.6k11343
asked Jul 24 at 1:34
Andres Mejia
14.6k11343
14.6k11343
Sorry about the ramble below the line. I was just mostly mumbling to myself and I realized something might work (kind of.) I'll leave it in case it helps anyone else conceptually
– Andres Mejia
Jul 24 at 2:25
You could use the connecting morphism in the LES here: en.wikipedia.org/wiki/Exponential_sheaf_sequence (I guess you can repeat the same story on a general topological space, with $O_X$ replaced by the sheaf of continuous functions... but I don't know for sure.)
– Lorenzo
Jul 24 at 18:53
Note that since $O_X$ (sheaf of continuous functions) is soft when $X$ is a metric space, $H^1(O_X) = H^2(O_X) = 0$, so the chern class boundary map is an isomorphism, which exactly what we expect (the chern class determines the isomorphism class of the topological line bundle). (This sanity check suggests to me that the story with the LES would work out.) Would this answer your question? I can write up some details...
– Lorenzo
Jul 24 at 19:06
@lorenzo I think so. As long as there is a construction involved (that is * chern class referenced in your second comment.) I’d love to see how the short exact sequence story goes.
– Andres Mejia
Jul 24 at 19:09
It's absolutely a construction - The Chern class map appears as a boundary map in the LES associated to a certain exact sequence of sheaves. One of these sheaves has the property that its $H^1$ is naturally isomorphic to the group of line bundles, and another one of these sheaves has the property that its $H^2$ is the usual $H^2$. The boundary map connecting them is the chern class. (Have you studied sheaves and sheaf cohomology?)
– Lorenzo
Jul 24 at 19:14
 |Â
show 6 more comments
Sorry about the ramble below the line. I was just mostly mumbling to myself and I realized something might work (kind of.) I'll leave it in case it helps anyone else conceptually
– Andres Mejia
Jul 24 at 2:25
You could use the connecting morphism in the LES here: en.wikipedia.org/wiki/Exponential_sheaf_sequence (I guess you can repeat the same story on a general topological space, with $O_X$ replaced by the sheaf of continuous functions... but I don't know for sure.)
– Lorenzo
Jul 24 at 18:53
Note that since $O_X$ (sheaf of continuous functions) is soft when $X$ is a metric space, $H^1(O_X) = H^2(O_X) = 0$, so the chern class boundary map is an isomorphism, which exactly what we expect (the chern class determines the isomorphism class of the topological line bundle). (This sanity check suggests to me that the story with the LES would work out.) Would this answer your question? I can write up some details...
– Lorenzo
Jul 24 at 19:06
@lorenzo I think so. As long as there is a construction involved (that is * chern class referenced in your second comment.) I’d love to see how the short exact sequence story goes.
– Andres Mejia
Jul 24 at 19:09
It's absolutely a construction - The Chern class map appears as a boundary map in the LES associated to a certain exact sequence of sheaves. One of these sheaves has the property that its $H^1$ is naturally isomorphic to the group of line bundles, and another one of these sheaves has the property that its $H^2$ is the usual $H^2$. The boundary map connecting them is the chern class. (Have you studied sheaves and sheaf cohomology?)
– Lorenzo
Jul 24 at 19:14
Sorry about the ramble below the line. I was just mostly mumbling to myself and I realized something might work (kind of.) I'll leave it in case it helps anyone else conceptually
– Andres Mejia
Jul 24 at 2:25
Sorry about the ramble below the line. I was just mostly mumbling to myself and I realized something might work (kind of.) I'll leave it in case it helps anyone else conceptually
– Andres Mejia
Jul 24 at 2:25
You could use the connecting morphism in the LES here: en.wikipedia.org/wiki/Exponential_sheaf_sequence (I guess you can repeat the same story on a general topological space, with $O_X$ replaced by the sheaf of continuous functions... but I don't know for sure.)
– Lorenzo
Jul 24 at 18:53
You could use the connecting morphism in the LES here: en.wikipedia.org/wiki/Exponential_sheaf_sequence (I guess you can repeat the same story on a general topological space, with $O_X$ replaced by the sheaf of continuous functions... but I don't know for sure.)
– Lorenzo
Jul 24 at 18:53
Note that since $O_X$ (sheaf of continuous functions) is soft when $X$ is a metric space, $H^1(O_X) = H^2(O_X) = 0$, so the chern class boundary map is an isomorphism, which exactly what we expect (the chern class determines the isomorphism class of the topological line bundle). (This sanity check suggests to me that the story with the LES would work out.) Would this answer your question? I can write up some details...
– Lorenzo
Jul 24 at 19:06
Note that since $O_X$ (sheaf of continuous functions) is soft when $X$ is a metric space, $H^1(O_X) = H^2(O_X) = 0$, so the chern class boundary map is an isomorphism, which exactly what we expect (the chern class determines the isomorphism class of the topological line bundle). (This sanity check suggests to me that the story with the LES would work out.) Would this answer your question? I can write up some details...
– Lorenzo
Jul 24 at 19:06
@lorenzo I think so. As long as there is a construction involved (that is * chern class referenced in your second comment.) I’d love to see how the short exact sequence story goes.
– Andres Mejia
Jul 24 at 19:09
@lorenzo I think so. As long as there is a construction involved (that is * chern class referenced in your second comment.) I’d love to see how the short exact sequence story goes.
– Andres Mejia
Jul 24 at 19:09
It's absolutely a construction - The Chern class map appears as a boundary map in the LES associated to a certain exact sequence of sheaves. One of these sheaves has the property that its $H^1$ is naturally isomorphic to the group of line bundles, and another one of these sheaves has the property that its $H^2$ is the usual $H^2$. The boundary map connecting them is the chern class. (Have you studied sheaves and sheaf cohomology?)
– Lorenzo
Jul 24 at 19:14
It's absolutely a construction - The Chern class map appears as a boundary map in the LES associated to a certain exact sequence of sheaves. One of these sheaves has the property that its $H^1$ is naturally isomorphic to the group of line bundles, and another one of these sheaves has the property that its $H^2$ is the usual $H^2$. The boundary map connecting them is the chern class. (Have you studied sheaves and sheaf cohomology?)
– Lorenzo
Jul 24 at 19:14
 |Â
show 6 more comments
1 Answer
1
active
oldest
votes
up vote
1
down vote
The above argument can indeed be modified (and it appears that Grothendieck did this in his original paper.) This was also already mentioned by Lorenzo in the comments, but it took me a few days to (maybe) understand what was going on.
There is a short exact sequence of sheaves
$$A(X,mathbb Z) to A(X,mathbb C) to A(X,mathbb C^times) $$
that gives rise to a long exact sequence in sheaf cohomology where the connecting homomorphism $delta:checkH^1(X,mathbb C^times) to checkH^2(X,mathbb Z)$, provides the isomorphism we needed.
answered Jul 27 at 18:16
Andres Mejia
14.6k11343
I'm not going to accept this though, in case there are other nice ideas!
– Andres Mejia
Jul 28 at 19:26
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
The above argument can indeed be modified (and it appears that Grothendieck did this in his original paper.) This was also already mentioned by Lorenzo in the comments, but it took me a few days to (maybe) understand what was going on.
There is a short exact sequence of sheaves
$$A(X,mathbb Z) to A(X,mathbb C) to A(X,mathbb C^times) $$
that gives rise to a long exact sequence in sheaf cohomology where the connecting homomorphism $delta:checkH^1(X,mathbb C^times) to checkH^2(X,mathbb Z)$, provides the isomorphism we needed.
answered Jul 27 at 18:16
Andres Mejia
14.6k11343
I'm not going to accept this though, in case there are other nice ideas!
– Andres Mejia
Jul 28 at 19:26
add a comment |Â
up vote
1
down vote
The above argument can indeed be modified (and it appears that Grothendieck did this in his original paper.) This was also already mentioned by Lorenzo in the comments, but it took me a few days to (maybe) understand what was going on.
There is a short exact sequence of sheaves
$$A(X,mathbb Z) to A(X,mathbb C) to A(X,mathbb C^times) $$
that gives rise to a long exact sequence in sheaf cohomology where the connecting homomorphism $delta:checkH^1(X,mathbb C^times) to checkH^2(X,mathbb Z)$, provides the isomorphism we needed.
answered Jul 27 at 18:16
Andres Mejia
14.6k11343
I'm not going to accept this though, in case there are other nice ideas!
– Andres Mejia
Jul 28 at 19:26
add a comment |Â
up vote
1
down vote
up vote
1
down vote
The above argument can indeed be modified (and it appears that Grothendieck did this in his original paper.) This was also already mentioned by Lorenzo in the comments, but it took me a few days to (maybe) understand what was going on.
There is a short exact sequence of sheaves
$$A(X,mathbb Z) to A(X,mathbb C) to A(X,mathbb C^times) $$
that gives rise to a long exact sequence in sheaf cohomology where the connecting homomorphism $delta:checkH^1(X,mathbb C^times) to checkH^2(X,mathbb Z)$, provides the isomorphism we needed.
answered Jul 27 at 18:16
Andres Mejia
14.6k11343
The above argument can indeed be modified (and it appears that Grothendieck did this in his original paper.) This was also already mentioned by Lorenzo in the comments, but it took me a few days to (maybe) understand what was going on.
There is a short exact sequence of sheaves
$$A(X,mathbb Z) to A(X,mathbb C) to A(X,mathbb C^times) $$
that gives rise to a long exact sequence in sheaf cohomology where the connecting homomorphism $delta:checkH^1(X,mathbb C^times) to checkH^2(X,mathbb Z)$, provides the isomorphism we needed.
answered Jul 27 at 18:16
Andres Mejia
14.6k11343
answered Jul 27 at 18:16
Andres Mejia
14.6k11343
answered Jul 27 at 18:16
Andres Mejia
14.6k11343
answered Jul 27 at 18:16
Andres Mejia
14.6k11343
14.6k11343
I'm not going to accept this though, in case there are other nice ideas!
– Andres Mejia
Jul 28 at 19:26
add a comment |Â
I'm not going to accept this though, in case there are other nice ideas!
– Andres Mejia
Jul 28 at 19:26
I'm not going to accept this though, in case there are other nice ideas!
– Andres Mejia
Jul 28 at 19:26
I'm not going to accept this though, in case there are other nice ideas!
– Andres Mejia
Jul 28 at 19:26
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2860923%2fconstructing-the-first-chern-class-without-using-the-classifying-map-for-line-bu%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sorry about the ramble below the line. I was just mostly mumbling to myself and I realized something might work (kind of.) I'll leave it in case it helps anyone else conceptually
– Andres Mejia
Jul 24 at 2:25
You could use the connecting morphism in the LES here: en.wikipedia.org/wiki/Exponential_sheaf_sequence (I guess you can repeat the same story on a general topological space, with $O_X$ replaced by the sheaf of continuous functions... but I don't know for sure.)
– Lorenzo
Jul 24 at 18:53
Note that since $O_X$ (sheaf of continuous functions) is soft when $X$ is a metric space, $H^1(O_X) = H^2(O_X) = 0$, so the chern class boundary map is an isomorphism, which exactly what we expect (the chern class determines the isomorphism class of the topological line bundle). (This sanity check suggests to me that the story with the LES would work out.) Would this answer your question? I can write up some details...
– Lorenzo
Jul 24 at 19:06
@lorenzo I think so. As long as there is a construction involved (that is * chern class referenced in your second comment.) I’d love to see how the short exact sequence story goes.
– Andres Mejia
Jul 24 at 19:09
It's absolutely a construction - The Chern class map appears as a boundary map in the LES associated to a certain exact sequence of sheaves. One of these sheaves has the property that its $H^1$ is naturally isomorphic to the group of line bundles, and another one of these sheaves has the property that its $H^2$ is the usual $H^2$. The boundary map connecting them is the chern class. (Have you studied sheaves and sheaf cohomology?)
– Lorenzo
Jul 24 at 19:14