Gluing (Principal) Bundles?
0
$begingroup$
Say $X $ is a manifold, and let $U$ and $V$ be two disjoint submanifolds. Then given two Lie groups $H_1$ and $H_2$ we can fiber $U$ with $H_1$ fibers and similarly with $V$ and $H_2$. It seems to me that one can "lift" the fibration to one of $X$ in a naive way: pick a group $G$ which contains both of the $H_i$ groups, fiber $G$ over $X$ such that the induced injections into the submanifolds map into the original fibrations by gluing cosets. This seems plausibly kosher by adding a few lines of math, even if the lift is far from unique. My question is the reverse of this, given two principal bundles is there a nice (canonical) way of gluing these two together?
general-topology differential-geometry differential-topology
asked Jan 9 at 20:47
user614057user614057
1
$endgroup$
add a comment |
0
$begingroup$
Say $X $ is a manifold, and let $U$ and $V$ be two disjoint submanifolds. Then given two Lie groups $H_1$ and $H_2$ we can fiber $U$ with $H_1$ fibers and similarly with $V$ and $H_2$. It seems to me that one can "lift" the fibration to one of $X$ in a naive way: pick a group $G$ which contains both of the $H_i$ groups, fiber $G$ over $X$ such that the induced injections into the submanifolds map into the original fibrations by gluing cosets. This seems plausibly kosher by adding a few lines of math, even if the lift is far from unique. My question is the reverse of this, given two principal bundles is there a nice (canonical) way of gluing these two together?
general-topology differential-geometry differential-topology
asked Jan 9 at 20:47
user614057user614057
1
$endgroup$
$begingroup$
I am confused by the claim "we can fiber $U$ with $H_1$ fibers and similarly with $V$ and $H_2$." Are you claiming that such a fibering is possible for any arbitrary disjoint submanifolds $U, V$ of $X$? In fact, could you kindly explain precisely what you mean by "we can fiber $U$ with $H_1$ fibers"?
$endgroup$
– Jesse Madnick
Jan 9 at 20:56
$begingroup$
These Lie groups sound arbitrary. Are they?
$endgroup$
– Randall
Jan 9 at 21:05
$begingroup$
Jesse the rough picture I have in mind is as follows: let's assume X is a compact topological (Hausdorff) manifold and choose an open set $U$ of X. We can regard $U$ as a manifold itself. Then we can forget about X and construct the trivial bundle $U times T$ where $T$ is the torus viewed as two copies of $U(1)$. Then if we choose another open set $V$ we can construct $V times U(1)$. To me it seems that we can do a similar operation to $X$ in such a way that restrictions behave nicely.
$endgroup$
– user614057
Jan 9 at 21:42
1
$begingroup$
In general, it's not even clear that every manifold $X$ having $U$ as a (not necessarily open) submanifold will admit an $H_1$-structure. For example, suppose $X$ is a non-orientable $n$-dimensional manifold, $U$ is an orientable (let's say open here) submanifold, and $H_1=SO(n)$.
$endgroup$
– Ted Shifrin
Jan 9 at 22:06
$begingroup$
I see. So, when you say "fiber $U$ with $H_1$ fibers," you actually mean "construct a bundle over $U$ with $H_1$ fibers." Your usage is at odds with standard terminology: usually, one describes the total space the bundle as being fibered with $H_1$ fibers, not the base $U$. Regardless, now that I understand your meaning properly, I'm not convinced that the "plausibly kosher" claim is true.
$endgroup$
– Jesse Madnick
Jan 9 at 23:31
add a comment |
0
0
0
$begingroup$
Say $X $ is a manifold, and let $U$ and $V$ be two disjoint submanifolds. Then given two Lie groups $H_1$ and $H_2$ we can fiber $U$ with $H_1$ fibers and similarly with $V$ and $H_2$. It seems to me that one can "lift" the fibration to one of $X$ in a naive way: pick a group $G$ which contains both of the $H_i$ groups, fiber $G$ over $X$ such that the induced injections into the submanifolds map into the original fibrations by gluing cosets. This seems plausibly kosher by adding a few lines of math, even if the lift is far from unique. My question is the reverse of this, given two principal bundles is there a nice (canonical) way of gluing these two together?
general-topology differential-geometry differential-topology
asked Jan 9 at 20:47
user614057user614057
1
$endgroup$
Say $X $ is a manifold, and let $U$ and $V$ be two disjoint submanifolds. Then given two Lie groups $H_1$ and $H_2$ we can fiber $U$ with $H_1$ fibers and similarly with $V$ and $H_2$. It seems to me that one can "lift" the fibration to one of $X$ in a naive way: pick a group $G$ which contains both of the $H_i$ groups, fiber $G$ over $X$ such that the induced injections into the submanifolds map into the original fibrations by gluing cosets. This seems plausibly kosher by adding a few lines of math, even if the lift is far from unique. My question is the reverse of this, given two principal bundles is there a nice (canonical) way of gluing these two together?
general-topology differential-geometry differential-topology
general-topology differential-geometry differential-topology
asked Jan 9 at 20:47
user614057user614057
1
asked Jan 9 at 20:47
user614057user614057
1
asked Jan 9 at 20:47
user614057user614057
1
asked Jan 9 at 20:47
user614057user614057
1
asked Jan 9 at 20:47
user614057user614057
1
1
$begingroup$
I am confused by the claim "we can fiber $U$ with $H_1$ fibers and similarly with $V$ and $H_2$." Are you claiming that such a fibering is possible for any arbitrary disjoint submanifolds $U, V$ of $X$? In fact, could you kindly explain precisely what you mean by "we can fiber $U$ with $H_1$ fibers"?
$endgroup$
– Jesse Madnick
Jan 9 at 20:56
$begingroup$
These Lie groups sound arbitrary. Are they?
$endgroup$
– Randall
Jan 9 at 21:05
$begingroup$
Jesse the rough picture I have in mind is as follows: let's assume X is a compact topological (Hausdorff) manifold and choose an open set $U$ of X. We can regard $U$ as a manifold itself. Then we can forget about X and construct the trivial bundle $U times T$ where $T$ is the torus viewed as two copies of $U(1)$. Then if we choose another open set $V$ we can construct $V times U(1)$. To me it seems that we can do a similar operation to $X$ in such a way that restrictions behave nicely.
$endgroup$
– user614057
Jan 9 at 21:42
1
$begingroup$
In general, it's not even clear that every manifold $X$ having $U$ as a (not necessarily open) submanifold will admit an $H_1$-structure. For example, suppose $X$ is a non-orientable $n$-dimensional manifold, $U$ is an orientable (let's say open here) submanifold, and $H_1=SO(n)$.
$endgroup$
– Ted Shifrin
Jan 9 at 22:06
$begingroup$
I see. So, when you say "fiber $U$ with $H_1$ fibers," you actually mean "construct a bundle over $U$ with $H_1$ fibers." Your usage is at odds with standard terminology: usually, one describes the total space the bundle as being fibered with $H_1$ fibers, not the base $U$. Regardless, now that I understand your meaning properly, I'm not convinced that the "plausibly kosher" claim is true.
$endgroup$
– Jesse Madnick
Jan 9 at 23:31
add a comment |
$begingroup$
I am confused by the claim "we can fiber $U$ with $H_1$ fibers and similarly with $V$ and $H_2$." Are you claiming that such a fibering is possible for any arbitrary disjoint submanifolds $U, V$ of $X$? In fact, could you kindly explain precisely what you mean by "we can fiber $U$ with $H_1$ fibers"?
$endgroup$
– Jesse Madnick
Jan 9 at 20:56
$begingroup$
These Lie groups sound arbitrary. Are they?
$endgroup$
– Randall
Jan 9 at 21:05
$begingroup$
Jesse the rough picture I have in mind is as follows: let's assume X is a compact topological (Hausdorff) manifold and choose an open set $U$ of X. We can regard $U$ as a manifold itself. Then we can forget about X and construct the trivial bundle $U times T$ where $T$ is the torus viewed as two copies of $U(1)$. Then if we choose another open set $V$ we can construct $V times U(1)$. To me it seems that we can do a similar operation to $X$ in such a way that restrictions behave nicely.
$endgroup$
– user614057
Jan 9 at 21:42
1
$begingroup$
In general, it's not even clear that every manifold $X$ having $U$ as a (not necessarily open) submanifold will admit an $H_1$-structure. For example, suppose $X$ is a non-orientable $n$-dimensional manifold, $U$ is an orientable (let's say open here) submanifold, and $H_1=SO(n)$.
$endgroup$
– Ted Shifrin
Jan 9 at 22:06
$begingroup$
I see. So, when you say "fiber $U$ with $H_1$ fibers," you actually mean "construct a bundle over $U$ with $H_1$ fibers." Your usage is at odds with standard terminology: usually, one describes the total space the bundle as being fibered with $H_1$ fibers, not the base $U$. Regardless, now that I understand your meaning properly, I'm not convinced that the "plausibly kosher" claim is true.
$endgroup$
– Jesse Madnick
Jan 9 at 23:31
$begingroup$
I am confused by the claim "we can fiber $U$ with $H_1$ fibers and similarly with $V$ and $H_2$." Are you claiming that such a fibering is possible for any arbitrary disjoint submanifolds $U, V$ of $X$? In fact, could you kindly explain precisely what you mean by "we can fiber $U$ with $H_1$ fibers"?
$endgroup$
– Jesse Madnick
Jan 9 at 20:56
$begingroup$
I am confused by the claim "we can fiber $U$ with $H_1$ fibers and similarly with $V$ and $H_2$." Are you claiming that such a fibering is possible for any arbitrary disjoint submanifolds $U, V$ of $X$? In fact, could you kindly explain precisely what you mean by "we can fiber $U$ with $H_1$ fibers"?
$endgroup$
– Jesse Madnick
Jan 9 at 20:56
$begingroup$
These Lie groups sound arbitrary. Are they?
$endgroup$
– Randall
Jan 9 at 21:05
$begingroup$
These Lie groups sound arbitrary. Are they?
$endgroup$
– Randall
Jan 9 at 21:05
$begingroup$
Jesse the rough picture I have in mind is as follows: let's assume X is a compact topological (Hausdorff) manifold and choose an open set $U$ of X. We can regard $U$ as a manifold itself. Then we can forget about X and construct the trivial bundle $U times T$ where $T$ is the torus viewed as two copies of $U(1)$. Then if we choose another open set $V$ we can construct $V times U(1)$. To me it seems that we can do a similar operation to $X$ in such a way that restrictions behave nicely.
$endgroup$
– user614057
Jan 9 at 21:42
$begingroup$
Jesse the rough picture I have in mind is as follows: let's assume X is a compact topological (Hausdorff) manifold and choose an open set $U$ of X. We can regard $U$ as a manifold itself. Then we can forget about X and construct the trivial bundle $U times T$ where $T$ is the torus viewed as two copies of $U(1)$. Then if we choose another open set $V$ we can construct $V times U(1)$. To me it seems that we can do a similar operation to $X$ in such a way that restrictions behave nicely.
$endgroup$
– user614057
Jan 9 at 21:42
1
1
$begingroup$
In general, it's not even clear that every manifold $X$ having $U$ as a (not necessarily open) submanifold will admit an $H_1$-structure. For example, suppose $X$ is a non-orientable $n$-dimensional manifold, $U$ is an orientable (let's say open here) submanifold, and $H_1=SO(n)$.
$endgroup$
– Ted Shifrin
Jan 9 at 22:06
$begingroup$
In general, it's not even clear that every manifold $X$ having $U$ as a (not necessarily open) submanifold will admit an $H_1$-structure. For example, suppose $X$ is a non-orientable $n$-dimensional manifold, $U$ is an orientable (let's say open here) submanifold, and $H_1=SO(n)$.
$endgroup$
– Ted Shifrin
Jan 9 at 22:06
$begingroup$
I see. So, when you say "fiber $U$ with $H_1$ fibers," you actually mean "construct a bundle over $U$ with $H_1$ fibers." Your usage is at odds with standard terminology: usually, one describes the total space the bundle as being fibered with $H_1$ fibers, not the base $U$. Regardless, now that I understand your meaning properly, I'm not convinced that the "plausibly kosher" claim is true.
$endgroup$
– Jesse Madnick
Jan 9 at 23:31
$begingroup$
I see. So, when you say "fiber $U$ with $H_1$ fibers," you actually mean "construct a bundle over $U$ with $H_1$ fibers." Your usage is at odds with standard terminology: usually, one describes the total space the bundle as being fibered with $H_1$ fibers, not the base $U$. Regardless, now that I understand your meaning properly, I'm not convinced that the "plausibly kosher" claim is true.
$endgroup$
– Jesse Madnick
Jan 9 at 23:31
add a comment |
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$begingroup$
I am confused by the claim "we can fiber $U$ with $H_1$ fibers and similarly with $V$ and $H_2$." Are you claiming that such a fibering is possible for any arbitrary disjoint submanifolds $U, V$ of $X$? In fact, could you kindly explain precisely what you mean by "we can fiber $U$ with $H_1$ fibers"?
$endgroup$
– Jesse Madnick
Jan 9 at 20:56
$begingroup$
These Lie groups sound arbitrary. Are they?
$endgroup$
– Randall
Jan 9 at 21:05
$begingroup$
Jesse the rough picture I have in mind is as follows: let's assume X is a compact topological (Hausdorff) manifold and choose an open set $U$ of X. We can regard $U$ as a manifold itself. Then we can forget about X and construct the trivial bundle $U times T$ where $T$ is the torus viewed as two copies of $U(1)$. Then if we choose another open set $V$ we can construct $V times U(1)$. To me it seems that we can do a similar operation to $X$ in such a way that restrictions behave nicely.
$endgroup$
– user614057
Jan 9 at 21:42
1
$begingroup$
In general, it's not even clear that every manifold $X$ having $U$ as a (not necessarily open) submanifold will admit an $H_1$-structure. For example, suppose $X$ is a non-orientable $n$-dimensional manifold, $U$ is an orientable (let's say open here) submanifold, and $H_1=SO(n)$.
$endgroup$
– Ted Shifrin
Jan 9 at 22:06
$begingroup$
I see. So, when you say "fiber $U$ with $H_1$ fibers," you actually mean "construct a bundle over $U$ with $H_1$ fibers." Your usage is at odds with standard terminology: usually, one describes the total space the bundle as being fibered with $H_1$ fibers, not the base $U$. Regardless, now that I understand your meaning properly, I'm not convinced that the "plausibly kosher" claim is true.
$endgroup$
– Jesse Madnick
Jan 9 at 23:31