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Fiber subbundel of compact subspaces
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Consider the following fiber bundle between smooth manifolds $E'rightarrow B'$ with a fiber $F'$.
Now let $E$ and $ B$ be compact submanifolds of $E'$ and $B'$ respectively, such that $Erightarrow B $ is fiber bundle with a compact fiber $F$.
Is there a fiber bundle $F'rightarrow F$? Is the fiber compact?
differential-geometry algebraic-topology fiber-bundles
asked Jul 14 at 22:27
Ronald
1,5591821
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Consider the following fiber bundle between smooth manifolds $E'rightarrow B'$ with a fiber $F'$.
Now let $E$ and $ B$ be compact submanifolds of $E'$ and $B'$ respectively, such that $Erightarrow B $ is fiber bundle with a compact fiber $F$.
Is there a fiber bundle $F'rightarrow F$? Is the fiber compact?
differential-geometry algebraic-topology fiber-bundles
asked Jul 14 at 22:27
Ronald
1,5591821
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Consider the following fiber bundle between smooth manifolds $E'rightarrow B'$ with a fiber $F'$.
Now let $E$ and $ B$ be compact submanifolds of $E'$ and $B'$ respectively, such that $Erightarrow B $ is fiber bundle with a compact fiber $F$.
Is there a fiber bundle $F'rightarrow F$? Is the fiber compact?
differential-geometry algebraic-topology fiber-bundles
asked Jul 14 at 22:27
Ronald
1,5591821
Consider the following fiber bundle between smooth manifolds $E'rightarrow B'$ with a fiber $F'$.
Now let $E$ and $ B$ be compact submanifolds of $E'$ and $B'$ respectively, such that $Erightarrow B $ is fiber bundle with a compact fiber $F$.
Is there a fiber bundle $F'rightarrow F$? Is the fiber compact?
differential-geometry algebraic-topology fiber-bundles
asked Jul 14 at 22:27
Ronald
1,5591821
asked Jul 14 at 22:27
Ronald
1,5591821
asked Jul 14 at 22:27
Ronald
1,5591821
asked Jul 14 at 22:27
Ronald
1,5591821
1,5591821
add a comment |Â
add a comment |Â
1 Answer
1
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2
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First note that in a fiber bundle with nice enough topological spaces for which the total space is compact, the fiber is automatically compact. One version of "nice enough" is "points are closed in the base". This follows because the fiber is the inverse image of a closed subset of the base, so closed, and a closed subset of a compact space is always compact.
With that out of the way, the answer to your question about bundles $F'rightarrow F$ is, in general, no. Here is an example: begin with the bundle $S^2rightarrow S^2times S^2rightarrow S^2$ where the projection map is projection onto the first factor. Consider the product of equators inside of $S^2times S^2$, with same projection. This is the bundle $S^1rightarrow S^1times S^1times S^1$.
Thus, $F' = S^2$ and $F = S^1$.
Now, I claim there is no bundle $pi:F'rightarrow F$. First, note that the fiber $F''$ must be compact, so it has an Euler characteristic. Then $2 = chi(F') = chi(F)chi(F'') = 0$, a contradiction.
answered Jul 14 at 23:08
Jason DeVito
29.5k473129
Thanks Jason for the nice answer. do you think the statement is still not true even for homogeneous spaces, for example let $G'$ be a Lie group and $G<G'$ and let $E':=G'/H', B':=G'/J'$ and $E:=G/H, B:=G/J$ where $H'<J'$ and $H<J$?
– Ronald
Jul 14 at 23:23
1
I think my answer is already a counter example to that. Let $G' = SO(3)times SO(3), H' = S^1times S^1$, $J' = SO(3)times S^1$, $G = S^1times S^1$, $H = e$, $J =S^1 times e$.
– Jason DeVito
Jul 14 at 23:27
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
First note that in a fiber bundle with nice enough topological spaces for which the total space is compact, the fiber is automatically compact. One version of "nice enough" is "points are closed in the base". This follows because the fiber is the inverse image of a closed subset of the base, so closed, and a closed subset of a compact space is always compact.
With that out of the way, the answer to your question about bundles $F'rightarrow F$ is, in general, no. Here is an example: begin with the bundle $S^2rightarrow S^2times S^2rightarrow S^2$ where the projection map is projection onto the first factor. Consider the product of equators inside of $S^2times S^2$, with same projection. This is the bundle $S^1rightarrow S^1times S^1times S^1$.
Thus, $F' = S^2$ and $F = S^1$.
Now, I claim there is no bundle $pi:F'rightarrow F$. First, note that the fiber $F''$ must be compact, so it has an Euler characteristic. Then $2 = chi(F') = chi(F)chi(F'') = 0$, a contradiction.
answered Jul 14 at 23:08
Jason DeVito
29.5k473129
Thanks Jason for the nice answer. do you think the statement is still not true even for homogeneous spaces, for example let $G'$ be a Lie group and $G<G'$ and let $E':=G'/H', B':=G'/J'$ and $E:=G/H, B:=G/J$ where $H'<J'$ and $H<J$?
– Ronald
Jul 14 at 23:23
1
I think my answer is already a counter example to that. Let $G' = SO(3)times SO(3), H' = S^1times S^1$, $J' = SO(3)times S^1$, $G = S^1times S^1$, $H = e$, $J =S^1 times e$.
– Jason DeVito
Jul 14 at 23:27
add a comment |Â
up vote
2
down vote
accepted
First note that in a fiber bundle with nice enough topological spaces for which the total space is compact, the fiber is automatically compact. One version of "nice enough" is "points are closed in the base". This follows because the fiber is the inverse image of a closed subset of the base, so closed, and a closed subset of a compact space is always compact.
With that out of the way, the answer to your question about bundles $F'rightarrow F$ is, in general, no. Here is an example: begin with the bundle $S^2rightarrow S^2times S^2rightarrow S^2$ where the projection map is projection onto the first factor. Consider the product of equators inside of $S^2times S^2$, with same projection. This is the bundle $S^1rightarrow S^1times S^1times S^1$.
Thus, $F' = S^2$ and $F = S^1$.
Now, I claim there is no bundle $pi:F'rightarrow F$. First, note that the fiber $F''$ must be compact, so it has an Euler characteristic. Then $2 = chi(F') = chi(F)chi(F'') = 0$, a contradiction.
answered Jul 14 at 23:08
Jason DeVito
29.5k473129
Thanks Jason for the nice answer. do you think the statement is still not true even for homogeneous spaces, for example let $G'$ be a Lie group and $G<G'$ and let $E':=G'/H', B':=G'/J'$ and $E:=G/H, B:=G/J$ where $H'<J'$ and $H<J$?
– Ronald
Jul 14 at 23:23
1
I think my answer is already a counter example to that. Let $G' = SO(3)times SO(3), H' = S^1times S^1$, $J' = SO(3)times S^1$, $G = S^1times S^1$, $H = e$, $J =S^1 times e$.
– Jason DeVito
Jul 14 at 23:27
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
First note that in a fiber bundle with nice enough topological spaces for which the total space is compact, the fiber is automatically compact. One version of "nice enough" is "points are closed in the base". This follows because the fiber is the inverse image of a closed subset of the base, so closed, and a closed subset of a compact space is always compact.
With that out of the way, the answer to your question about bundles $F'rightarrow F$ is, in general, no. Here is an example: begin with the bundle $S^2rightarrow S^2times S^2rightarrow S^2$ where the projection map is projection onto the first factor. Consider the product of equators inside of $S^2times S^2$, with same projection. This is the bundle $S^1rightarrow S^1times S^1times S^1$.
Thus, $F' = S^2$ and $F = S^1$.
Now, I claim there is no bundle $pi:F'rightarrow F$. First, note that the fiber $F''$ must be compact, so it has an Euler characteristic. Then $2 = chi(F') = chi(F)chi(F'') = 0$, a contradiction.
answered Jul 14 at 23:08
Jason DeVito
29.5k473129
First note that in a fiber bundle with nice enough topological spaces for which the total space is compact, the fiber is automatically compact. One version of "nice enough" is "points are closed in the base". This follows because the fiber is the inverse image of a closed subset of the base, so closed, and a closed subset of a compact space is always compact.
With that out of the way, the answer to your question about bundles $F'rightarrow F$ is, in general, no. Here is an example: begin with the bundle $S^2rightarrow S^2times S^2rightarrow S^2$ where the projection map is projection onto the first factor. Consider the product of equators inside of $S^2times S^2$, with same projection. This is the bundle $S^1rightarrow S^1times S^1times S^1$.
Thus, $F' = S^2$ and $F = S^1$.
Now, I claim there is no bundle $pi:F'rightarrow F$. First, note that the fiber $F''$ must be compact, so it has an Euler characteristic. Then $2 = chi(F') = chi(F)chi(F'') = 0$, a contradiction.
answered Jul 14 at 23:08
Jason DeVito
29.5k473129
answered Jul 14 at 23:08
Jason DeVito
29.5k473129
answered Jul 14 at 23:08
Jason DeVito
29.5k473129
answered Jul 14 at 23:08
Jason DeVito
29.5k473129
29.5k473129
Thanks Jason for the nice answer. do you think the statement is still not true even for homogeneous spaces, for example let $G'$ be a Lie group and $G<G'$ and let $E':=G'/H', B':=G'/J'$ and $E:=G/H, B:=G/J$ where $H'<J'$ and $H<J$?
– Ronald
Jul 14 at 23:23
1
I think my answer is already a counter example to that. Let $G' = SO(3)times SO(3), H' = S^1times S^1$, $J' = SO(3)times S^1$, $G = S^1times S^1$, $H = e$, $J =S^1 times e$.
– Jason DeVito
Jul 14 at 23:27
add a comment |Â
Thanks Jason for the nice answer. do you think the statement is still not true even for homogeneous spaces, for example let $G'$ be a Lie group and $G<G'$ and let $E':=G'/H', B':=G'/J'$ and $E:=G/H, B:=G/J$ where $H'<J'$ and $H<J$?
– Ronald
Jul 14 at 23:23
1
I think my answer is already a counter example to that. Let $G' = SO(3)times SO(3), H' = S^1times S^1$, $J' = SO(3)times S^1$, $G = S^1times S^1$, $H = e$, $J =S^1 times e$.
– Jason DeVito
Jul 14 at 23:27
Thanks Jason for the nice answer. do you think the statement is still not true even for homogeneous spaces, for example let $G'$ be a Lie group and $G<G'$ and let $E':=G'/H', B':=G'/J'$ and $E:=G/H, B:=G/J$ where $H'<J'$ and $H<J$?
– Ronald
Jul 14 at 23:23
Thanks Jason for the nice answer. do you think the statement is still not true even for homogeneous spaces, for example let $G'$ be a Lie group and $G<G'$ and let $E':=G'/H', B':=G'/J'$ and $E:=G/H, B:=G/J$ where $H'<J'$ and $H<J$?
– Ronald
Jul 14 at 23:23
1
1
I think my answer is already a counter example to that. Let $G' = SO(3)times SO(3), H' = S^1times S^1$, $J' = SO(3)times S^1$, $G = S^1times S^1$, $H = e$, $J =S^1 times e$.
– Jason DeVito
Jul 14 at 23:27
I think my answer is already a counter example to that. Let $G' = SO(3)times SO(3), H' = S^1times S^1$, $J' = SO(3)times S^1$, $G = S^1times S^1$, $H = e$, $J =S^1 times e$.
– Jason DeVito
Jul 14 at 23:27
add a comment |Â
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