Mixing
Index of covering in covering group
$begingroup$
In wikipedia I read:
In mathematics, a covering group of a topological group $H$ is a covering space $G$ of $H$ such that $G$ is a topological group and the covering map $p : G rightarrow H$ is a continuous group homomorphism. The map $p$ is called the covering homomorphism. A frequently occurring case is a double covering group, a topological double cover in which $H$ has index 2 in $G$; examples include the Spin groups, Pin groups, and metaplectic groups.
In this definition $H$ is not necessary a subgroup of $G$ (so "$H$ has index 2 in $G$" doesn't make sense in the usual way). So my question is: what is "the index of $H$ in $G$" in this context?
general-topology group-theory covering-spaces
edited Jan 7 at 15:24
user1729
16.9k64085
asked Dec 29 '18 at 13:11
asvasv
2841211
$endgroup$
|
show 4 more comments
$begingroup$
In wikipedia I read:
In mathematics, a covering group of a topological group $H$ is a covering space $G$ of $H$ such that $G$ is a topological group and the covering map $p : G rightarrow H$ is a continuous group homomorphism. The map $p$ is called the covering homomorphism. A frequently occurring case is a double covering group, a topological double cover in which $H$ has index 2 in $G$; examples include the Spin groups, Pin groups, and metaplectic groups.
In this definition $H$ is not necessary a subgroup of $G$ (so "$H$ has index 2 in $G$" doesn't make sense in the usual way). So my question is: what is "the index of $H$ in $G$" in this context?
general-topology group-theory covering-spaces
edited Jan 7 at 15:24
user1729
16.9k64085
asked Dec 29 '18 at 13:11
asvasv
2841211
$endgroup$
$begingroup$
It's just a speculation, perhaps by saying covering map they mean that $p^{-1}(H)$ is homeomorphic to a disjoint union of spaces homeomorphic to $H$. If so the index is the number of such spaces.
$endgroup$
– Yanko
Dec 29 '18 at 13:16
$begingroup$
Maybe is $[G:p^{-1}(H)]=2$
$endgroup$
– asv
Dec 29 '18 at 14:36
$begingroup$
that's less likely. If you consider $G=Htimes mathbb{Z}/2mathbb{Z}$ with the map $p(h,a) = h$ then $p^{-1}(H)=G$ so by your definition the index is $1$ which in my opinion makes less sense than saying that $p^{-1}(H) = Htimes{0}bigsqcup Htimes{1}$ and so the index is $2$ (As the usual index).
$endgroup$
– Yanko
Dec 29 '18 at 14:38
$begingroup$
Maybe $[G:p^{-1}(H)]$ is always 1 because, if I'm not wrong, $p$ is surjective by definition then $p^{-1}(H)=G$. Here: en.wikipedia.org/wiki/Covering_space says $p$ is surjective.
$endgroup$
– asv
Dec 29 '18 at 14:42
2
$begingroup$
The question currently has 4 votes to close. I cannot understand why this is, apart from perhaps that the question is not initially clear. I have tried to make the question clearer (although I am unsure if my edit improve readability - anyone should feel free to edit it!).
$endgroup$
– user1729
Jan 7 at 15:26
|
show 4 more comments
$begingroup$
In wikipedia I read:
In mathematics, a covering group of a topological group $H$ is a covering space $G$ of $H$ such that $G$ is a topological group and the covering map $p : G rightarrow H$ is a continuous group homomorphism. The map $p$ is called the covering homomorphism. A frequently occurring case is a double covering group, a topological double cover in which $H$ has index 2 in $G$; examples include the Spin groups, Pin groups, and metaplectic groups.
In this definition $H$ is not necessary a subgroup of $G$ (so "$H$ has index 2 in $G$" doesn't make sense in the usual way). So my question is: what is "the index of $H$ in $G$" in this context?
general-topology group-theory covering-spaces
edited Jan 7 at 15:24
user1729
16.9k64085
asked Dec 29 '18 at 13:11
asvasv
2841211
$endgroup$
In wikipedia I read:
In mathematics, a covering group of a topological group $H$ is a covering space $G$ of $H$ such that $G$ is a topological group and the covering map $p : G rightarrow H$ is a continuous group homomorphism. The map $p$ is called the covering homomorphism. A frequently occurring case is a double covering group, a topological double cover in which $H$ has index 2 in $G$; examples include the Spin groups, Pin groups, and metaplectic groups.
In this definition $H$ is not necessary a subgroup of $G$ (so "$H$ has index 2 in $G$" doesn't make sense in the usual way). So my question is: what is "the index of $H$ in $G$" in this context?
general-topology group-theory covering-spaces
general-topology group-theory covering-spaces
edited Jan 7 at 15:24
user1729
16.9k64085
asked Dec 29 '18 at 13:11
asvasv
2841211
edited Jan 7 at 15:24
user1729
16.9k64085
asked Dec 29 '18 at 13:11
asvasv
2841211
edited Jan 7 at 15:24
user1729
16.9k64085
edited Jan 7 at 15:24
user1729
16.9k64085
edited Jan 7 at 15:24
user1729
16.9k64085
16.9k64085
asked Dec 29 '18 at 13:11
asvasv
2841211
asked Dec 29 '18 at 13:11
asvasv
2841211
asked Dec 29 '18 at 13:11
asvasv
2841211
2841211
$begingroup$
It's just a speculation, perhaps by saying covering map they mean that $p^{-1}(H)$ is homeomorphic to a disjoint union of spaces homeomorphic to $H$. If so the index is the number of such spaces.
$endgroup$
– Yanko
Dec 29 '18 at 13:16
$begingroup$
Maybe is $[G:p^{-1}(H)]=2$
$endgroup$
– asv
Dec 29 '18 at 14:36
$begingroup$
that's less likely. If you consider $G=Htimes mathbb{Z}/2mathbb{Z}$ with the map $p(h,a) = h$ then $p^{-1}(H)=G$ so by your definition the index is $1$ which in my opinion makes less sense than saying that $p^{-1}(H) = Htimes{0}bigsqcup Htimes{1}$ and so the index is $2$ (As the usual index).
$endgroup$
– Yanko
Dec 29 '18 at 14:38
$begingroup$
Maybe $[G:p^{-1}(H)]$ is always 1 because, if I'm not wrong, $p$ is surjective by definition then $p^{-1}(H)=G$. Here: en.wikipedia.org/wiki/Covering_space says $p$ is surjective.
$endgroup$
– asv
Dec 29 '18 at 14:42
2
$begingroup$
The question currently has 4 votes to close. I cannot understand why this is, apart from perhaps that the question is not initially clear. I have tried to make the question clearer (although I am unsure if my edit improve readability - anyone should feel free to edit it!).
$endgroup$
– user1729
Jan 7 at 15:26
|
show 4 more comments
$begingroup$
It's just a speculation, perhaps by saying covering map they mean that $p^{-1}(H)$ is homeomorphic to a disjoint union of spaces homeomorphic to $H$. If so the index is the number of such spaces.
$endgroup$
– Yanko
Dec 29 '18 at 13:16
$begingroup$
Maybe is $[G:p^{-1}(H)]=2$
$endgroup$
– asv
Dec 29 '18 at 14:36
$begingroup$
that's less likely. If you consider $G=Htimes mathbb{Z}/2mathbb{Z}$ with the map $p(h,a) = h$ then $p^{-1}(H)=G$ so by your definition the index is $1$ which in my opinion makes less sense than saying that $p^{-1}(H) = Htimes{0}bigsqcup Htimes{1}$ and so the index is $2$ (As the usual index).
$endgroup$
– Yanko
Dec 29 '18 at 14:38
$begingroup$
Maybe $[G:p^{-1}(H)]$ is always 1 because, if I'm not wrong, $p$ is surjective by definition then $p^{-1}(H)=G$. Here: en.wikipedia.org/wiki/Covering_space says $p$ is surjective.
$endgroup$
– asv
Dec 29 '18 at 14:42
2
$begingroup$
The question currently has 4 votes to close. I cannot understand why this is, apart from perhaps that the question is not initially clear. I have tried to make the question clearer (although I am unsure if my edit improve readability - anyone should feel free to edit it!).
$endgroup$
– user1729
Jan 7 at 15:26
$begingroup$
It's just a speculation, perhaps by saying covering map they mean that $p^{-1}(H)$ is homeomorphic to a disjoint union of spaces homeomorphic to $H$. If so the index is the number of such spaces.
$endgroup$
– Yanko
Dec 29 '18 at 13:16
$begingroup$
It's just a speculation, perhaps by saying covering map they mean that $p^{-1}(H)$ is homeomorphic to a disjoint union of spaces homeomorphic to $H$. If so the index is the number of such spaces.
$endgroup$
– Yanko
Dec 29 '18 at 13:16
$begingroup$
Maybe is $[G:p^{-1}(H)]=2$
$endgroup$
– asv
Dec 29 '18 at 14:36
$begingroup$
Maybe is $[G:p^{-1}(H)]=2$
$endgroup$
– asv
Dec 29 '18 at 14:36
$begingroup$
that's less likely. If you consider $G=Htimes mathbb{Z}/2mathbb{Z}$ with the map $p(h,a) = h$ then $p^{-1}(H)=G$ so by your definition the index is $1$ which in my opinion makes less sense than saying that $p^{-1}(H) = Htimes{0}bigsqcup Htimes{1}$ and so the index is $2$ (As the usual index).
$endgroup$
– Yanko
Dec 29 '18 at 14:38
$begingroup$
that's less likely. If you consider $G=Htimes mathbb{Z}/2mathbb{Z}$ with the map $p(h,a) = h$ then $p^{-1}(H)=G$ so by your definition the index is $1$ which in my opinion makes less sense than saying that $p^{-1}(H) = Htimes{0}bigsqcup Htimes{1}$ and so the index is $2$ (As the usual index).
$endgroup$
– Yanko
Dec 29 '18 at 14:38
$begingroup$
Maybe $[G:p^{-1}(H)]$ is always 1 because, if I'm not wrong, $p$ is surjective by definition then $p^{-1}(H)=G$. Here: en.wikipedia.org/wiki/Covering_space says $p$ is surjective.
$endgroup$
– asv
Dec 29 '18 at 14:42
$begingroup$
Maybe $[G:p^{-1}(H)]$ is always 1 because, if I'm not wrong, $p$ is surjective by definition then $p^{-1}(H)=G$. Here: en.wikipedia.org/wiki/Covering_space says $p$ is surjective.
$endgroup$
– asv
Dec 29 '18 at 14:42
2
2
$begingroup$
The question currently has 4 votes to close. I cannot understand why this is, apart from perhaps that the question is not initially clear. I have tried to make the question clearer (although I am unsure if my edit improve readability - anyone should feel free to edit it!).
$endgroup$
– user1729
Jan 7 at 15:26
$begingroup$
The question currently has 4 votes to close. I cannot understand why this is, apart from perhaps that the question is not initially clear. I have tried to make the question clearer (although I am unsure if my edit improve readability - anyone should feel free to edit it!).
$endgroup$
– user1729
Jan 7 at 15:26
|
show 4 more comments
1 Answer
1
active
oldest
votes
$begingroup$
This community wiki solution is intended to clear the question from the unanswered queue.
As Moishe Cohen pointed out in his comment, the phrase "A frequently occurring case is a double covering group, a topological double cover in which $H$ has index $2$ in $G$" appeared for the first time in the January 2009 revision of the Wikipedia article and it is wrong.
In general $H$ is not a subgroup of $G$, and it does not even embed as a subgroup into $G$. Therefore it does not make sense to speak about the index of $H$ in $G$.
Only in the trivial case $G = H times mathbb{Z}_2$ there would be a reasonable interpretation.
$endgroup$
add a comment |
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$begingroup$
This community wiki solution is intended to clear the question from the unanswered queue.
As Moishe Cohen pointed out in his comment, the phrase "A frequently occurring case is a double covering group, a topological double cover in which $H$ has index $2$ in $G$" appeared for the first time in the January 2009 revision of the Wikipedia article and it is wrong.
In general $H$ is not a subgroup of $G$, and it does not even embed as a subgroup into $G$. Therefore it does not make sense to speak about the index of $H$ in $G$.
Only in the trivial case $G = H times mathbb{Z}_2$ there would be a reasonable interpretation.
$endgroup$
add a comment |
$begingroup$
This community wiki solution is intended to clear the question from the unanswered queue.
As Moishe Cohen pointed out in his comment, the phrase "A frequently occurring case is a double covering group, a topological double cover in which $H$ has index $2$ in $G$" appeared for the first time in the January 2009 revision of the Wikipedia article and it is wrong.
In general $H$ is not a subgroup of $G$, and it does not even embed as a subgroup into $G$. Therefore it does not make sense to speak about the index of $H$ in $G$.
Only in the trivial case $G = H times mathbb{Z}_2$ there would be a reasonable interpretation.
$endgroup$
add a comment |
$begingroup$
This community wiki solution is intended to clear the question from the unanswered queue.
As Moishe Cohen pointed out in his comment, the phrase "A frequently occurring case is a double covering group, a topological double cover in which $H$ has index $2$ in $G$" appeared for the first time in the January 2009 revision of the Wikipedia article and it is wrong.
In general $H$ is not a subgroup of $G$, and it does not even embed as a subgroup into $G$. Therefore it does not make sense to speak about the index of $H$ in $G$.
Only in the trivial case $G = H times mathbb{Z}_2$ there would be a reasonable interpretation.
$endgroup$
This community wiki solution is intended to clear the question from the unanswered queue.
As Moishe Cohen pointed out in his comment, the phrase "A frequently occurring case is a double covering group, a topological double cover in which $H$ has index $2$ in $G$" appeared for the first time in the January 2009 revision of the Wikipedia article and it is wrong.
In general $H$ is not a subgroup of $G$, and it does not even embed as a subgroup into $G$. Therefore it does not make sense to speak about the index of $H$ in $G$.
Only in the trivial case $G = H times mathbb{Z}_2$ there would be a reasonable interpretation.
edited Jan 7 at 14:23
community wiki
2 revs
Paul Frost
add a comment |
add a comment |
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$begingroup$
It's just a speculation, perhaps by saying covering map they mean that $p^{-1}(H)$ is homeomorphic to a disjoint union of spaces homeomorphic to $H$. If so the index is the number of such spaces.
$endgroup$
– Yanko
Dec 29 '18 at 13:16
$begingroup$
Maybe is $[G:p^{-1}(H)]=2$
$endgroup$
– asv
Dec 29 '18 at 14:36
$begingroup$
that's less likely. If you consider $G=Htimes mathbb{Z}/2mathbb{Z}$ with the map $p(h,a) = h$ then $p^{-1}(H)=G$ so by your definition the index is $1$ which in my opinion makes less sense than saying that $p^{-1}(H) = Htimes{0}bigsqcup Htimes{1}$ and so the index is $2$ (As the usual index).
$endgroup$
– Yanko
Dec 29 '18 at 14:38
$begingroup$
Maybe $[G:p^{-1}(H)]$ is always 1 because, if I'm not wrong, $p$ is surjective by definition then $p^{-1}(H)=G$. Here: en.wikipedia.org/wiki/Covering_space says $p$ is surjective.
$endgroup$
– asv
Dec 29 '18 at 14:42
2
$begingroup$
The question currently has 4 votes to close. I cannot understand why this is, apart from perhaps that the question is not initially clear. I have tried to make the question clearer (although I am unsure if my edit improve readability - anyone should feel free to edit it!).
$endgroup$
– user1729
Jan 7 at 15:26