What does determinant bundle of a principal bundle say about the principal bundle
$begingroup$
Let $pi:Prightarrow M$ be a principal $Gl(n,mathbb{R})$ bundle.
Given $xin M$ there is an open set $U$ containing $x$ and a local trivialization $pi^{-1}(U)rightarrow Utimes G$. This gives a cover ${U_alpha}$ of $M$ and local trivializations $pi^{-1}(U_alpha)rightarrow U_alphatimes G$. These in turn give what are called as transition functions $g_{alphabeta}:U_alphacap U_betarightarrow Gl(n,mathbb{R})$. These transition functions determine the principal bundle.
Consider the determinant map $det:Gl(n,mathbb{R})rightarrow mathbb{R}^*=Gl(1,mathbb{R})$. This is smooth. So, is the composition $h_{alphabeta}:U_alphacap U_betarightarrow Gl(n,mathbb{R})rightarrow mathbb{R}^*$. So, we have an open cover ${U_alpha}$ of $M$ and maps $h_{alphabeta}:U_{alphabeta}rightarrow Gl(1,mathbb{R})$ satisfying the cocycle condition.
These maps define $G=Gl(1,mathbb{R})$ bundle on $M$.
We call this the determinant bundle associated to $Prightarrow M$.
I would like to understand what does this determinant bundle say about $Prightarrow M$. In this question I came to know that if determinant bundle is trivial then the bundle is self dual.
Are there any such properties?
differential-geometry principal-bundles
asked Dec 21 '18 at 8:52
Praphulla KoushikPraphulla Koushik
21419
$endgroup$
add a comment |
$begingroup$
Let $pi:Prightarrow M$ be a principal $Gl(n,mathbb{R})$ bundle.
Given $xin M$ there is an open set $U$ containing $x$ and a local trivialization $pi^{-1}(U)rightarrow Utimes G$. This gives a cover ${U_alpha}$ of $M$ and local trivializations $pi^{-1}(U_alpha)rightarrow U_alphatimes G$. These in turn give what are called as transition functions $g_{alphabeta}:U_alphacap U_betarightarrow Gl(n,mathbb{R})$. These transition functions determine the principal bundle.
Consider the determinant map $det:Gl(n,mathbb{R})rightarrow mathbb{R}^*=Gl(1,mathbb{R})$. This is smooth. So, is the composition $h_{alphabeta}:U_alphacap U_betarightarrow Gl(n,mathbb{R})rightarrow mathbb{R}^*$. So, we have an open cover ${U_alpha}$ of $M$ and maps $h_{alphabeta}:U_{alphabeta}rightarrow Gl(1,mathbb{R})$ satisfying the cocycle condition.
These maps define $G=Gl(1,mathbb{R})$ bundle on $M$.
We call this the determinant bundle associated to $Prightarrow M$.
I would like to understand what does this determinant bundle say about $Prightarrow M$. In this question I came to know that if determinant bundle is trivial then the bundle is self dual.
Are there any such properties?
differential-geometry principal-bundles
asked Dec 21 '18 at 8:52
Praphulla KoushikPraphulla Koushik
21419
$endgroup$
add a comment |
$begingroup$
Let $pi:Prightarrow M$ be a principal $Gl(n,mathbb{R})$ bundle.
Given $xin M$ there is an open set $U$ containing $x$ and a local trivialization $pi^{-1}(U)rightarrow Utimes G$. This gives a cover ${U_alpha}$ of $M$ and local trivializations $pi^{-1}(U_alpha)rightarrow U_alphatimes G$. These in turn give what are called as transition functions $g_{alphabeta}:U_alphacap U_betarightarrow Gl(n,mathbb{R})$. These transition functions determine the principal bundle.
Consider the determinant map $det:Gl(n,mathbb{R})rightarrow mathbb{R}^*=Gl(1,mathbb{R})$. This is smooth. So, is the composition $h_{alphabeta}:U_alphacap U_betarightarrow Gl(n,mathbb{R})rightarrow mathbb{R}^*$. So, we have an open cover ${U_alpha}$ of $M$ and maps $h_{alphabeta}:U_{alphabeta}rightarrow Gl(1,mathbb{R})$ satisfying the cocycle condition.
These maps define $G=Gl(1,mathbb{R})$ bundle on $M$.
We call this the determinant bundle associated to $Prightarrow M$.
I would like to understand what does this determinant bundle say about $Prightarrow M$. In this question I came to know that if determinant bundle is trivial then the bundle is self dual.
Are there any such properties?
differential-geometry principal-bundles
asked Dec 21 '18 at 8:52
Praphulla KoushikPraphulla Koushik
21419
$endgroup$
Let $pi:Prightarrow M$ be a principal $Gl(n,mathbb{R})$ bundle.
Given $xin M$ there is an open set $U$ containing $x$ and a local trivialization $pi^{-1}(U)rightarrow Utimes G$. This gives a cover ${U_alpha}$ of $M$ and local trivializations $pi^{-1}(U_alpha)rightarrow U_alphatimes G$. These in turn give what are called as transition functions $g_{alphabeta}:U_alphacap U_betarightarrow Gl(n,mathbb{R})$. These transition functions determine the principal bundle.
Consider the determinant map $det:Gl(n,mathbb{R})rightarrow mathbb{R}^*=Gl(1,mathbb{R})$. This is smooth. So, is the composition $h_{alphabeta}:U_alphacap U_betarightarrow Gl(n,mathbb{R})rightarrow mathbb{R}^*$. So, we have an open cover ${U_alpha}$ of $M$ and maps $h_{alphabeta}:U_{alphabeta}rightarrow Gl(1,mathbb{R})$ satisfying the cocycle condition.
These maps define $G=Gl(1,mathbb{R})$ bundle on $M$.
We call this the determinant bundle associated to $Prightarrow M$.
I would like to understand what does this determinant bundle say about $Prightarrow M$. In this question I came to know that if determinant bundle is trivial then the bundle is self dual.
Are there any such properties?
differential-geometry principal-bundles
differential-geometry principal-bundles
asked Dec 21 '18 at 8:52
Praphulla KoushikPraphulla Koushik
21419
asked Dec 21 '18 at 8:52
Praphulla KoushikPraphulla Koushik
21419
asked Dec 21 '18 at 8:52
Praphulla KoushikPraphulla Koushik
21419
asked Dec 21 '18 at 8:52
Praphulla KoushikPraphulla Koushik
21419
asked Dec 21 '18 at 8:52
Praphulla KoushikPraphulla Koushik
21419
21419
add a comment |
add a comment |
2 Answers
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$begingroup$
The determinant bundle is trivial if and only if the bundle is orientable.
The determinant bundle is a principal $mathbb{R}$-bundle, it is trivial if and only if it has a $(mathbb{R}^+,times)$-reduction since $mathbb{R}^+$ is contractible, this is equivalent that $M$ is orientable, in in fact the obstruction of the triviality of the determinant bundle is the 0-stiefel-Whithney class.
An alternative description of the first Stiefel-Whitney class
answered Dec 21 '18 at 9:38
Tsemo AristideTsemo Aristide
58.3k11445
$endgroup$
$begingroup$
reference please
$endgroup$
– Praphulla Koushik
Dec 21 '18 at 9:39
$begingroup$
math.stackexchange.com/questions/379820/…
$endgroup$
– Tsemo Aristide
Dec 21 '18 at 9:44
add a comment |
$begingroup$
The answer of the question you link to refers to bundles of rank two, it is not true in general. Also, the staement in there refers to the associated vector bundle (corresponding to the obvious representation of $GL(n,mathbb R)$ on $mathbb R^n$ rather than to the principal bundle itself. Calling this associated bundle $E$, there is a simple characterization of the fact that the determinant bundle of $P$ is trivial, namely that $E$ is orientable. The point here is that the determinant of a matrix can be interpreted at the induced action on the top exterior power $Lambda ^nmathbb R^{n*}$. This easily implies that the determinant bundle of $P$ is the frame bundle of the line bundle $Lambda^n E^*$ so triviality of these two bundles is equivalent. But triviality of $Lambda^n E^*$ is well known to be equivalent to the fact that it has a global non-vanishng section and thus to orientability of $E$.
answered Dec 21 '18 at 9:39
Andreas CapAndreas Cap
11.2k923
$endgroup$
$begingroup$
Sir, can you suggest some reference which discusses about determinant bundle.
$endgroup$
– Praphulla Koushik
Dec 21 '18 at 9:42
add a comment |
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2 Answers
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2 Answers
2
active
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$begingroup$
The determinant bundle is trivial if and only if the bundle is orientable.
The determinant bundle is a principal $mathbb{R}$-bundle, it is trivial if and only if it has a $(mathbb{R}^+,times)$-reduction since $mathbb{R}^+$ is contractible, this is equivalent that $M$ is orientable, in in fact the obstruction of the triviality of the determinant bundle is the 0-stiefel-Whithney class.
An alternative description of the first Stiefel-Whitney class
answered Dec 21 '18 at 9:38
Tsemo AristideTsemo Aristide
58.3k11445
$endgroup$
$begingroup$
reference please
$endgroup$
– Praphulla Koushik
Dec 21 '18 at 9:39
$begingroup$
math.stackexchange.com/questions/379820/…
$endgroup$
– Tsemo Aristide
Dec 21 '18 at 9:44
add a comment |
$begingroup$
The determinant bundle is trivial if and only if the bundle is orientable.
The determinant bundle is a principal $mathbb{R}$-bundle, it is trivial if and only if it has a $(mathbb{R}^+,times)$-reduction since $mathbb{R}^+$ is contractible, this is equivalent that $M$ is orientable, in in fact the obstruction of the triviality of the determinant bundle is the 0-stiefel-Whithney class.
An alternative description of the first Stiefel-Whitney class
answered Dec 21 '18 at 9:38
Tsemo AristideTsemo Aristide
58.3k11445
$endgroup$
$begingroup$
reference please
$endgroup$
– Praphulla Koushik
Dec 21 '18 at 9:39
$begingroup$
math.stackexchange.com/questions/379820/…
$endgroup$
– Tsemo Aristide
Dec 21 '18 at 9:44
add a comment |
$begingroup$
The determinant bundle is trivial if and only if the bundle is orientable.
The determinant bundle is a principal $mathbb{R}$-bundle, it is trivial if and only if it has a $(mathbb{R}^+,times)$-reduction since $mathbb{R}^+$ is contractible, this is equivalent that $M$ is orientable, in in fact the obstruction of the triviality of the determinant bundle is the 0-stiefel-Whithney class.
An alternative description of the first Stiefel-Whitney class
answered Dec 21 '18 at 9:38
Tsemo AristideTsemo Aristide
58.3k11445
$endgroup$
The determinant bundle is trivial if and only if the bundle is orientable.
The determinant bundle is a principal $mathbb{R}$-bundle, it is trivial if and only if it has a $(mathbb{R}^+,times)$-reduction since $mathbb{R}^+$ is contractible, this is equivalent that $M$ is orientable, in in fact the obstruction of the triviality of the determinant bundle is the 0-stiefel-Whithney class.
An alternative description of the first Stiefel-Whitney class
answered Dec 21 '18 at 9:38
Tsemo AristideTsemo Aristide
58.3k11445
edited Dec 21 '18 at 9:43
answered Dec 21 '18 at 9:38
Tsemo AristideTsemo Aristide
58.3k11445
answered Dec 21 '18 at 9:38
Tsemo AristideTsemo Aristide
58.3k11445
answered Dec 21 '18 at 9:38
Tsemo AristideTsemo Aristide
58.3k11445
58.3k11445
$begingroup$
reference please
$endgroup$
– Praphulla Koushik
Dec 21 '18 at 9:39
$begingroup$
math.stackexchange.com/questions/379820/…
$endgroup$
– Tsemo Aristide
Dec 21 '18 at 9:44
add a comment |
$begingroup$
reference please
$endgroup$
– Praphulla Koushik
Dec 21 '18 at 9:39
$begingroup$
math.stackexchange.com/questions/379820/…
$endgroup$
– Tsemo Aristide
Dec 21 '18 at 9:44
$begingroup$
reference please
$endgroup$
– Praphulla Koushik
Dec 21 '18 at 9:39
$begingroup$
reference please
$endgroup$
– Praphulla Koushik
Dec 21 '18 at 9:39
$begingroup$
math.stackexchange.com/questions/379820/…
$endgroup$
– Tsemo Aristide
Dec 21 '18 at 9:44
$begingroup$
math.stackexchange.com/questions/379820/…
$endgroup$
– Tsemo Aristide
Dec 21 '18 at 9:44
add a comment |
$begingroup$
The answer of the question you link to refers to bundles of rank two, it is not true in general. Also, the staement in there refers to the associated vector bundle (corresponding to the obvious representation of $GL(n,mathbb R)$ on $mathbb R^n$ rather than to the principal bundle itself. Calling this associated bundle $E$, there is a simple characterization of the fact that the determinant bundle of $P$ is trivial, namely that $E$ is orientable. The point here is that the determinant of a matrix can be interpreted at the induced action on the top exterior power $Lambda ^nmathbb R^{n*}$. This easily implies that the determinant bundle of $P$ is the frame bundle of the line bundle $Lambda^n E^*$ so triviality of these two bundles is equivalent. But triviality of $Lambda^n E^*$ is well known to be equivalent to the fact that it has a global non-vanishng section and thus to orientability of $E$.
answered Dec 21 '18 at 9:39
Andreas CapAndreas Cap
11.2k923
$endgroup$
$begingroup$
Sir, can you suggest some reference which discusses about determinant bundle.
$endgroup$
– Praphulla Koushik
Dec 21 '18 at 9:42
add a comment |
$begingroup$
The answer of the question you link to refers to bundles of rank two, it is not true in general. Also, the staement in there refers to the associated vector bundle (corresponding to the obvious representation of $GL(n,mathbb R)$ on $mathbb R^n$ rather than to the principal bundle itself. Calling this associated bundle $E$, there is a simple characterization of the fact that the determinant bundle of $P$ is trivial, namely that $E$ is orientable. The point here is that the determinant of a matrix can be interpreted at the induced action on the top exterior power $Lambda ^nmathbb R^{n*}$. This easily implies that the determinant bundle of $P$ is the frame bundle of the line bundle $Lambda^n E^*$ so triviality of these two bundles is equivalent. But triviality of $Lambda^n E^*$ is well known to be equivalent to the fact that it has a global non-vanishng section and thus to orientability of $E$.
answered Dec 21 '18 at 9:39
Andreas CapAndreas Cap
11.2k923
$endgroup$
$begingroup$
Sir, can you suggest some reference which discusses about determinant bundle.
$endgroup$
– Praphulla Koushik
Dec 21 '18 at 9:42
add a comment |
$begingroup$
The answer of the question you link to refers to bundles of rank two, it is not true in general. Also, the staement in there refers to the associated vector bundle (corresponding to the obvious representation of $GL(n,mathbb R)$ on $mathbb R^n$ rather than to the principal bundle itself. Calling this associated bundle $E$, there is a simple characterization of the fact that the determinant bundle of $P$ is trivial, namely that $E$ is orientable. The point here is that the determinant of a matrix can be interpreted at the induced action on the top exterior power $Lambda ^nmathbb R^{n*}$. This easily implies that the determinant bundle of $P$ is the frame bundle of the line bundle $Lambda^n E^*$ so triviality of these two bundles is equivalent. But triviality of $Lambda^n E^*$ is well known to be equivalent to the fact that it has a global non-vanishng section and thus to orientability of $E$.
answered Dec 21 '18 at 9:39
Andreas CapAndreas Cap
11.2k923
$endgroup$
The answer of the question you link to refers to bundles of rank two, it is not true in general. Also, the staement in there refers to the associated vector bundle (corresponding to the obvious representation of $GL(n,mathbb R)$ on $mathbb R^n$ rather than to the principal bundle itself. Calling this associated bundle $E$, there is a simple characterization of the fact that the determinant bundle of $P$ is trivial, namely that $E$ is orientable. The point here is that the determinant of a matrix can be interpreted at the induced action on the top exterior power $Lambda ^nmathbb R^{n*}$. This easily implies that the determinant bundle of $P$ is the frame bundle of the line bundle $Lambda^n E^*$ so triviality of these two bundles is equivalent. But triviality of $Lambda^n E^*$ is well known to be equivalent to the fact that it has a global non-vanishng section and thus to orientability of $E$.
answered Dec 21 '18 at 9:39
Andreas CapAndreas Cap
11.2k923
answered Dec 21 '18 at 9:39
Andreas CapAndreas Cap
11.2k923
answered Dec 21 '18 at 9:39
Andreas CapAndreas Cap
11.2k923
answered Dec 21 '18 at 9:39
Andreas CapAndreas Cap
11.2k923
11.2k923
$begingroup$
Sir, can you suggest some reference which discusses about determinant bundle.
$endgroup$
– Praphulla Koushik
Dec 21 '18 at 9:42
add a comment |
$begingroup$
Sir, can you suggest some reference which discusses about determinant bundle.
$endgroup$
– Praphulla Koushik
Dec 21 '18 at 9:42
$begingroup$
Sir, can you suggest some reference which discusses about determinant bundle.
$endgroup$
– Praphulla Koushik
Dec 21 '18 at 9:42
$begingroup$
Sir, can you suggest some reference which discusses about determinant bundle.
$endgroup$
– Praphulla Koushik
Dec 21 '18 at 9:42
add a comment |
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