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Universal Bundle — Understand the basic definition
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I read two versions of discussions on universal bundles. I could not really see how the two definitions are really the same.
From Wiki. The universal bundle in the theory of fiber bundles with structure group a given topological group G, is a specific bundle over a classifying space BG, such that every bundle with the given structure group G over M is a pullback by means of a continuous map M $to$ BG.
From ncatlab For G a topological group there is a notion of G-principal bundles P→X over any topological space X. Under continuous maps f:X→Y there is a notion of pullback of principal bundles f∗:GBund(Y)→GBund(X).
A universal G-principal bundle is a G-principal bundle, which is usually written EG→BG, such that for every CW-complex X the map
[X,BG]→GBund(X)/∼
from homotopy classes of continuous functions X→BG given by [f]↦f∗EG, is an isomorphism.
In this case one calls BG a classifying space for G-principal bundles.
The universal principal bundle is characterized, up to equivalence, by its total space EG being contractible.
Can one explain how the twos are the same? How do we intuitively define Universal Bundle?
algebraic-topology differential-topology fiber-bundles principal-bundles
edited Jul 26 at 13:25
Armando j18eos
2,35711125
asked Jul 26 at 2:04
annie heart
549616
add a comment |Â
up vote
0
down vote
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I read two versions of discussions on universal bundles. I could not really see how the two definitions are really the same.
From Wiki. The universal bundle in the theory of fiber bundles with structure group a given topological group G, is a specific bundle over a classifying space BG, such that every bundle with the given structure group G over M is a pullback by means of a continuous map M $to$ BG.
From ncatlab For G a topological group there is a notion of G-principal bundles P→X over any topological space X. Under continuous maps f:X→Y there is a notion of pullback of principal bundles f∗:GBund(Y)→GBund(X).
A universal G-principal bundle is a G-principal bundle, which is usually written EG→BG, such that for every CW-complex X the map
[X,BG]→GBund(X)/∼
from homotopy classes of continuous functions X→BG given by [f]↦f∗EG, is an isomorphism.
In this case one calls BG a classifying space for G-principal bundles.
The universal principal bundle is characterized, up to equivalence, by its total space EG being contractible.
Can one explain how the twos are the same? How do we intuitively define Universal Bundle?
algebraic-topology differential-topology fiber-bundles principal-bundles
edited Jul 26 at 13:25
Armando j18eos
2,35711125
asked Jul 26 at 2:04
annie heart
549616
The latter is just an unwinding of the first, though it's not clear a priori that such an object exists or is unique modulo equivalence. What do you want from an 'intuitive' definition that the first one you cite doesn't have?
– anomaly
Jul 26 at 2:14
Thanks, "The latter is just an unwinding of the first," -> this 2nd sounds totally a different statement - that has some but little relation to the 1st
– annie heart
Jul 26 at 2:53
I think the equivalency follows from there always being an appropriate lifting when the total space is contractible. The universal property has to do with that...
– Chris Custer
Jul 26 at 4:46
"Fiber $G$-bundles" and "principal $G$-bundles" are different objects, where $G$ is a topological group; so, fixed $G$, the universal fiber $G$-bundle is not the universal principal $G$-bundle.
– Armando j18eos
Jul 26 at 13:25
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I read two versions of discussions on universal bundles. I could not really see how the two definitions are really the same.
From Wiki. The universal bundle in the theory of fiber bundles with structure group a given topological group G, is a specific bundle over a classifying space BG, such that every bundle with the given structure group G over M is a pullback by means of a continuous map M $to$ BG.
From ncatlab For G a topological group there is a notion of G-principal bundles P→X over any topological space X. Under continuous maps f:X→Y there is a notion of pullback of principal bundles f∗:GBund(Y)→GBund(X).
A universal G-principal bundle is a G-principal bundle, which is usually written EG→BG, such that for every CW-complex X the map
[X,BG]→GBund(X)/∼
from homotopy classes of continuous functions X→BG given by [f]↦f∗EG, is an isomorphism.
In this case one calls BG a classifying space for G-principal bundles.
The universal principal bundle is characterized, up to equivalence, by its total space EG being contractible.
Can one explain how the twos are the same? How do we intuitively define Universal Bundle?
algebraic-topology differential-topology fiber-bundles principal-bundles
edited Jul 26 at 13:25
Armando j18eos
2,35711125
asked Jul 26 at 2:04
annie heart
549616
I read two versions of discussions on universal bundles. I could not really see how the two definitions are really the same.
From Wiki. The universal bundle in the theory of fiber bundles with structure group a given topological group G, is a specific bundle over a classifying space BG, such that every bundle with the given structure group G over M is a pullback by means of a continuous map M $to$ BG.
From ncatlab For G a topological group there is a notion of G-principal bundles P→X over any topological space X. Under continuous maps f:X→Y there is a notion of pullback of principal bundles f∗:GBund(Y)→GBund(X).
A universal G-principal bundle is a G-principal bundle, which is usually written EG→BG, such that for every CW-complex X the map
[X,BG]→GBund(X)/∼
from homotopy classes of continuous functions X→BG given by [f]↦f∗EG, is an isomorphism.
In this case one calls BG a classifying space for G-principal bundles.
The universal principal bundle is characterized, up to equivalence, by its total space EG being contractible.
Can one explain how the twos are the same? How do we intuitively define Universal Bundle?
algebraic-topology differential-topology fiber-bundles principal-bundles
edited Jul 26 at 13:25
Armando j18eos
2,35711125
asked Jul 26 at 2:04
annie heart
549616
edited Jul 26 at 13:25
Armando j18eos
2,35711125
edited Jul 26 at 13:25
Armando j18eos
2,35711125
edited Jul 26 at 13:25
Armando j18eos
2,35711125
2,35711125
asked Jul 26 at 2:04
annie heart
549616
asked Jul 26 at 2:04
annie heart
549616
asked Jul 26 at 2:04
annie heart
549616
549616
The latter is just an unwinding of the first, though it's not clear a priori that such an object exists or is unique modulo equivalence. What do you want from an 'intuitive' definition that the first one you cite doesn't have?
– anomaly
Jul 26 at 2:14
Thanks, "The latter is just an unwinding of the first," -> this 2nd sounds totally a different statement - that has some but little relation to the 1st
– annie heart
Jul 26 at 2:53
I think the equivalency follows from there always being an appropriate lifting when the total space is contractible. The universal property has to do with that...
– Chris Custer
Jul 26 at 4:46
"Fiber $G$-bundles" and "principal $G$-bundles" are different objects, where $G$ is a topological group; so, fixed $G$, the universal fiber $G$-bundle is not the universal principal $G$-bundle.
– Armando j18eos
Jul 26 at 13:25
add a comment |Â
The latter is just an unwinding of the first, though it's not clear a priori that such an object exists or is unique modulo equivalence. What do you want from an 'intuitive' definition that the first one you cite doesn't have?
– anomaly
Jul 26 at 2:14
Thanks, "The latter is just an unwinding of the first," -> this 2nd sounds totally a different statement - that has some but little relation to the 1st
– annie heart
Jul 26 at 2:53
I think the equivalency follows from there always being an appropriate lifting when the total space is contractible. The universal property has to do with that...
– Chris Custer
Jul 26 at 4:46
"Fiber $G$-bundles" and "principal $G$-bundles" are different objects, where $G$ is a topological group; so, fixed $G$, the universal fiber $G$-bundle is not the universal principal $G$-bundle.
– Armando j18eos
Jul 26 at 13:25
The latter is just an unwinding of the first, though it's not clear a priori that such an object exists or is unique modulo equivalence. What do you want from an 'intuitive' definition that the first one you cite doesn't have?
– anomaly
Jul 26 at 2:14
The latter is just an unwinding of the first, though it's not clear a priori that such an object exists or is unique modulo equivalence. What do you want from an 'intuitive' definition that the first one you cite doesn't have?
– anomaly
Jul 26 at 2:14
Thanks, "The latter is just an unwinding of the first," -> this 2nd sounds totally a different statement - that has some but little relation to the 1st
– annie heart
Jul 26 at 2:53
Thanks, "The latter is just an unwinding of the first," -> this 2nd sounds totally a different statement - that has some but little relation to the 1st
– annie heart
Jul 26 at 2:53
I think the equivalency follows from there always being an appropriate lifting when the total space is contractible. The universal property has to do with that...
– Chris Custer
Jul 26 at 4:46
I think the equivalency follows from there always being an appropriate lifting when the total space is contractible. The universal property has to do with that...
– Chris Custer
Jul 26 at 4:46
"Fiber $G$-bundles" and "principal $G$-bundles" are different objects, where $G$ is a topological group; so, fixed $G$, the universal fiber $G$-bundle is not the universal principal $G$-bundle.
– Armando j18eos
Jul 26 at 13:25
"Fiber $G$-bundles" and "principal $G$-bundles" are different objects, where $G$ is a topological group; so, fixed $G$, the universal fiber $G$-bundle is not the universal principal $G$-bundle.
– Armando j18eos
Jul 26 at 13:25
add a comment |Â
1 Answer
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The first statement is a bit weaker than the second, since it tells you that for a $G$-bundle $p:E to B$, there exists a "universal bundle" $pi:EG to BG$, where you can find a map $f:B to BG$ so that the pullback $f^*(BG) to B$ is "exactly" the bundle $p:E to B$.
The second statement is telling a a little bit more. For example, it says first of all that the pullback $f^*(BG)$ only depends on the homotopy class of $f:B to BG$.
On the same token, when I said "exactly" above, what I really meant was "up to bundle isomorphism." In other words, for a base space $B$, we can consider $mathrmGbund(B)$ which is the space of all possible $G$ bundles with base space $B$.
The claim is that using the homotopy class of the map $f:B to BG$, we can find each element in $mathrmGbund(B)/sim$, where bundles are identified up to isomorphism.
A fancier way to say this:
Consider the space of maps from $B$ to $BG$ identified up to homotopy. We denote this $[B,BG]$
Consider also the space of vector bundles over $B$ identified up to isomorphism. We denote t his $mathrmGbund/sim$.
Then the assignment of a homotopy class $[f]$ to its pullback $f^*(BG)$, is really well defined map $[B,BG] to mathrmGbund/sim$ given by $[f] mapsto f^*(BG)$.
The claim in the first paragraph is that this map is surjective (for every bundle over $B$, it can be realized as a pullback)
The second paragraph also claims that this is injective, says that the map $[f]$ up to homotopy is essentially unique.
I've really just written everything up into smaller chunks, but I hope that it helps.
answered Jul 26 at 16:18
Andres Mejia
14.5k11243
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
The first statement is a bit weaker than the second, since it tells you that for a $G$-bundle $p:E to B$, there exists a "universal bundle" $pi:EG to BG$, where you can find a map $f:B to BG$ so that the pullback $f^*(BG) to B$ is "exactly" the bundle $p:E to B$.
The second statement is telling a a little bit more. For example, it says first of all that the pullback $f^*(BG)$ only depends on the homotopy class of $f:B to BG$.
On the same token, when I said "exactly" above, what I really meant was "up to bundle isomorphism." In other words, for a base space $B$, we can consider $mathrmGbund(B)$ which is the space of all possible $G$ bundles with base space $B$.
The claim is that using the homotopy class of the map $f:B to BG$, we can find each element in $mathrmGbund(B)/sim$, where bundles are identified up to isomorphism.
A fancier way to say this:
Consider the space of maps from $B$ to $BG$ identified up to homotopy. We denote this $[B,BG]$
Consider also the space of vector bundles over $B$ identified up to isomorphism. We denote t his $mathrmGbund/sim$.
Then the assignment of a homotopy class $[f]$ to its pullback $f^*(BG)$, is really well defined map $[B,BG] to mathrmGbund/sim$ given by $[f] mapsto f^*(BG)$.
The claim in the first paragraph is that this map is surjective (for every bundle over $B$, it can be realized as a pullback)
The second paragraph also claims that this is injective, says that the map $[f]$ up to homotopy is essentially unique.
I've really just written everything up into smaller chunks, but I hope that it helps.
answered Jul 26 at 16:18
Andres Mejia
14.5k11243
add a comment |Â
up vote
3
down vote
The first statement is a bit weaker than the second, since it tells you that for a $G$-bundle $p:E to B$, there exists a "universal bundle" $pi:EG to BG$, where you can find a map $f:B to BG$ so that the pullback $f^*(BG) to B$ is "exactly" the bundle $p:E to B$.
The second statement is telling a a little bit more. For example, it says first of all that the pullback $f^*(BG)$ only depends on the homotopy class of $f:B to BG$.
On the same token, when I said "exactly" above, what I really meant was "up to bundle isomorphism." In other words, for a base space $B$, we can consider $mathrmGbund(B)$ which is the space of all possible $G$ bundles with base space $B$.
The claim is that using the homotopy class of the map $f:B to BG$, we can find each element in $mathrmGbund(B)/sim$, where bundles are identified up to isomorphism.
A fancier way to say this:
Consider the space of maps from $B$ to $BG$ identified up to homotopy. We denote this $[B,BG]$
Consider also the space of vector bundles over $B$ identified up to isomorphism. We denote t his $mathrmGbund/sim$.
Then the assignment of a homotopy class $[f]$ to its pullback $f^*(BG)$, is really well defined map $[B,BG] to mathrmGbund/sim$ given by $[f] mapsto f^*(BG)$.
The claim in the first paragraph is that this map is surjective (for every bundle over $B$, it can be realized as a pullback)
The second paragraph also claims that this is injective, says that the map $[f]$ up to homotopy is essentially unique.
I've really just written everything up into smaller chunks, but I hope that it helps.
answered Jul 26 at 16:18
Andres Mejia
14.5k11243
add a comment |Â
up vote
3
down vote
up vote
3
down vote
The first statement is a bit weaker than the second, since it tells you that for a $G$-bundle $p:E to B$, there exists a "universal bundle" $pi:EG to BG$, where you can find a map $f:B to BG$ so that the pullback $f^*(BG) to B$ is "exactly" the bundle $p:E to B$.
The second statement is telling a a little bit more. For example, it says first of all that the pullback $f^*(BG)$ only depends on the homotopy class of $f:B to BG$.
On the same token, when I said "exactly" above, what I really meant was "up to bundle isomorphism." In other words, for a base space $B$, we can consider $mathrmGbund(B)$ which is the space of all possible $G$ bundles with base space $B$.
The claim is that using the homotopy class of the map $f:B to BG$, we can find each element in $mathrmGbund(B)/sim$, where bundles are identified up to isomorphism.
A fancier way to say this:
Consider the space of maps from $B$ to $BG$ identified up to homotopy. We denote this $[B,BG]$
Consider also the space of vector bundles over $B$ identified up to isomorphism. We denote t his $mathrmGbund/sim$.
Then the assignment of a homotopy class $[f]$ to its pullback $f^*(BG)$, is really well defined map $[B,BG] to mathrmGbund/sim$ given by $[f] mapsto f^*(BG)$.
The claim in the first paragraph is that this map is surjective (for every bundle over $B$, it can be realized as a pullback)
The second paragraph also claims that this is injective, says that the map $[f]$ up to homotopy is essentially unique.
I've really just written everything up into smaller chunks, but I hope that it helps.
answered Jul 26 at 16:18
Andres Mejia
14.5k11243
The first statement is a bit weaker than the second, since it tells you that for a $G$-bundle $p:E to B$, there exists a "universal bundle" $pi:EG to BG$, where you can find a map $f:B to BG$ so that the pullback $f^*(BG) to B$ is "exactly" the bundle $p:E to B$.
The second statement is telling a a little bit more. For example, it says first of all that the pullback $f^*(BG)$ only depends on the homotopy class of $f:B to BG$.
On the same token, when I said "exactly" above, what I really meant was "up to bundle isomorphism." In other words, for a base space $B$, we can consider $mathrmGbund(B)$ which is the space of all possible $G$ bundles with base space $B$.
The claim is that using the homotopy class of the map $f:B to BG$, we can find each element in $mathrmGbund(B)/sim$, where bundles are identified up to isomorphism.
A fancier way to say this:
Consider the space of maps from $B$ to $BG$ identified up to homotopy. We denote this $[B,BG]$
Consider also the space of vector bundles over $B$ identified up to isomorphism. We denote t his $mathrmGbund/sim$.
Then the assignment of a homotopy class $[f]$ to its pullback $f^*(BG)$, is really well defined map $[B,BG] to mathrmGbund/sim$ given by $[f] mapsto f^*(BG)$.
The claim in the first paragraph is that this map is surjective (for every bundle over $B$, it can be realized as a pullback)
The second paragraph also claims that this is injective, says that the map $[f]$ up to homotopy is essentially unique.
I've really just written everything up into smaller chunks, but I hope that it helps.
answered Jul 26 at 16:18
Andres Mejia
14.5k11243
edited Jul 26 at 16:27
answered Jul 26 at 16:18
Andres Mejia
14.5k11243
answered Jul 26 at 16:18
Andres Mejia
14.5k11243
answered Jul 26 at 16:18
Andres Mejia
14.5k11243
14.5k11243
add a comment |Â
add a comment |Â
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The latter is just an unwinding of the first, though it's not clear a priori that such an object exists or is unique modulo equivalence. What do you want from an 'intuitive' definition that the first one you cite doesn't have?
– anomaly
Jul 26 at 2:14
Thanks, "The latter is just an unwinding of the first," -> this 2nd sounds totally a different statement - that has some but little relation to the 1st
– annie heart
Jul 26 at 2:53
I think the equivalency follows from there always being an appropriate lifting when the total space is contractible. The universal property has to do with that...
– Chris Custer
Jul 26 at 4:46
"Fiber $G$-bundles" and "principal $G$-bundles" are different objects, where $G$ is a topological group; so, fixed $G$, the universal fiber $G$-bundle is not the universal principal $G$-bundle.
– Armando j18eos
Jul 26 at 13:25