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Is there a Borel-measurable projection to a closed subgroup
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Suppose that $G$ is a compact metrizable group and let $H$ be a closed subgroup of $G$.
Is it true that there must exists a Borel-measurable projection map $p:Grightarrow H$ with the property that for every $hin H$ and $gin G$ one has that $p(gh)=p(g)cdot h$? (In particular $p(h)=h$).
measure-theory topological-groups
asked Aug 2 at 16:28
Yanko
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Suppose that $G$ is a compact metrizable group and let $H$ be a closed subgroup of $G$.
Is it true that there must exists a Borel-measurable projection map $p:Grightarrow H$ with the property that for every $hin H$ and $gin G$ one has that $p(gh)=p(g)cdot h$? (In particular $p(h)=h$).
measure-theory topological-groups
asked Aug 2 at 16:28
Yanko
3,463620
3
You are supposed to do some work. See if you can find a selection theorem to help for this ... en.wikipedia.org/wiki/List_of_selection_theorems
– GEdgar
Aug 2 at 18:34
@GEdgar Thank you. I will look up for that!
– Yanko
Aug 2 at 22:47
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Suppose that $G$ is a compact metrizable group and let $H$ be a closed subgroup of $G$.
Is it true that there must exists a Borel-measurable projection map $p:Grightarrow H$ with the property that for every $hin H$ and $gin G$ one has that $p(gh)=p(g)cdot h$? (In particular $p(h)=h$).
measure-theory topological-groups
asked Aug 2 at 16:28
Yanko
3,463620
Suppose that $G$ is a compact metrizable group and let $H$ be a closed subgroup of $G$.
Is it true that there must exists a Borel-measurable projection map $p:Grightarrow H$ with the property that for every $hin H$ and $gin G$ one has that $p(gh)=p(g)cdot h$? (In particular $p(h)=h$).
measure-theory topological-groups
asked Aug 2 at 16:28
Yanko
3,463620
asked Aug 2 at 16:28
Yanko
3,463620
asked Aug 2 at 16:28
Yanko
3,463620
asked Aug 2 at 16:28
Yanko
3,463620
3,463620
3
You are supposed to do some work. See if you can find a selection theorem to help for this ... en.wikipedia.org/wiki/List_of_selection_theorems
– GEdgar
Aug 2 at 18:34
@GEdgar Thank you. I will look up for that!
– Yanko
Aug 2 at 22:47
add a comment |Â
3
You are supposed to do some work. See if you can find a selection theorem to help for this ... en.wikipedia.org/wiki/List_of_selection_theorems
– GEdgar
Aug 2 at 18:34
@GEdgar Thank you. I will look up for that!
– Yanko
Aug 2 at 22:47
3
3
You are supposed to do some work. See if you can find a selection theorem to help for this ... en.wikipedia.org/wiki/List_of_selection_theorems
– GEdgar
Aug 2 at 18:34
You are supposed to do some work. See if you can find a selection theorem to help for this ... en.wikipedia.org/wiki/List_of_selection_theorems
– GEdgar
Aug 2 at 18:34
@GEdgar Thank you. I will look up for that!
– Yanko
Aug 2 at 22:47
@GEdgar Thank you. I will look up for that!
– Yanko
Aug 2 at 22:47
add a comment |Â
1 Answer
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The answer is a yes! (Thanks to GEdgar who pointed out the relation with "selection theorems")
In the book "A Course on Borel sets" page 186 they proved a claim that implies the following:
Theorem: Let $H,G$ be as in the question. Then there exists a "cross-section" $s:G/Hrightarrow G$ such that $scirc p = id$ where $p:Grightarrow G/H$ the quotient map $p(g)=gH$.
Proof: Let $Pi = a_i H : a_iH text is a coset of H $ then by the theorem in the book there exists a set Borel set $S$ such that $Scap a_i H = s_i$.
This defines a (Borel) map $s:G/Hrightarrow G$ such that $s(u) = s_i$ if $uin a_i H$. (The map is Borel because $s^-1(U)=s^-1(Ucap S) = p(Ucap S)$ and quotient map is open so it sends Borel to Borel), which proves the theorem.
This theorem leads to our result, as one can can define a map $varphi:Grightarrow H$ by $varphi(g) = gcdot (s(p(g))^-1$.
Since for every $hin H$ one has that $p(gh)=p(g)$ we have that $varphi(gh)=varphi(g)cdot h$.
This also answers positively my (now obviously related) other question:
$Gcong G/Htimes H$ measureably
answered yesterday
Yanko
3,463620
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
The answer is a yes! (Thanks to GEdgar who pointed out the relation with "selection theorems")
In the book "A Course on Borel sets" page 186 they proved a claim that implies the following:
Theorem: Let $H,G$ be as in the question. Then there exists a "cross-section" $s:G/Hrightarrow G$ such that $scirc p = id$ where $p:Grightarrow G/H$ the quotient map $p(g)=gH$.
Proof: Let $Pi = a_i H : a_iH text is a coset of H $ then by the theorem in the book there exists a set Borel set $S$ such that $Scap a_i H = s_i$.
This defines a (Borel) map $s:G/Hrightarrow G$ such that $s(u) = s_i$ if $uin a_i H$. (The map is Borel because $s^-1(U)=s^-1(Ucap S) = p(Ucap S)$ and quotient map is open so it sends Borel to Borel), which proves the theorem.
This theorem leads to our result, as one can can define a map $varphi:Grightarrow H$ by $varphi(g) = gcdot (s(p(g))^-1$.
Since for every $hin H$ one has that $p(gh)=p(g)$ we have that $varphi(gh)=varphi(g)cdot h$.
This also answers positively my (now obviously related) other question:
$Gcong G/Htimes H$ measureably
answered yesterday
Yanko
3,463620
add a comment |Â
up vote
0
down vote
The answer is a yes! (Thanks to GEdgar who pointed out the relation with "selection theorems")
In the book "A Course on Borel sets" page 186 they proved a claim that implies the following:
Theorem: Let $H,G$ be as in the question. Then there exists a "cross-section" $s:G/Hrightarrow G$ such that $scirc p = id$ where $p:Grightarrow G/H$ the quotient map $p(g)=gH$.
Proof: Let $Pi = a_i H : a_iH text is a coset of H $ then by the theorem in the book there exists a set Borel set $S$ such that $Scap a_i H = s_i$.
This defines a (Borel) map $s:G/Hrightarrow G$ such that $s(u) = s_i$ if $uin a_i H$. (The map is Borel because $s^-1(U)=s^-1(Ucap S) = p(Ucap S)$ and quotient map is open so it sends Borel to Borel), which proves the theorem.
This theorem leads to our result, as one can can define a map $varphi:Grightarrow H$ by $varphi(g) = gcdot (s(p(g))^-1$.
Since for every $hin H$ one has that $p(gh)=p(g)$ we have that $varphi(gh)=varphi(g)cdot h$.
This also answers positively my (now obviously related) other question:
$Gcong G/Htimes H$ measureably
answered yesterday
Yanko
3,463620
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The answer is a yes! (Thanks to GEdgar who pointed out the relation with "selection theorems")
In the book "A Course on Borel sets" page 186 they proved a claim that implies the following:
Theorem: Let $H,G$ be as in the question. Then there exists a "cross-section" $s:G/Hrightarrow G$ such that $scirc p = id$ where $p:Grightarrow G/H$ the quotient map $p(g)=gH$.
Proof: Let $Pi = a_i H : a_iH text is a coset of H $ then by the theorem in the book there exists a set Borel set $S$ such that $Scap a_i H = s_i$.
This defines a (Borel) map $s:G/Hrightarrow G$ such that $s(u) = s_i$ if $uin a_i H$. (The map is Borel because $s^-1(U)=s^-1(Ucap S) = p(Ucap S)$ and quotient map is open so it sends Borel to Borel), which proves the theorem.
This theorem leads to our result, as one can can define a map $varphi:Grightarrow H$ by $varphi(g) = gcdot (s(p(g))^-1$.
Since for every $hin H$ one has that $p(gh)=p(g)$ we have that $varphi(gh)=varphi(g)cdot h$.
This also answers positively my (now obviously related) other question:
$Gcong G/Htimes H$ measureably
answered yesterday
Yanko
3,463620
The answer is a yes! (Thanks to GEdgar who pointed out the relation with "selection theorems")
In the book "A Course on Borel sets" page 186 they proved a claim that implies the following:
Theorem: Let $H,G$ be as in the question. Then there exists a "cross-section" $s:G/Hrightarrow G$ such that $scirc p = id$ where $p:Grightarrow G/H$ the quotient map $p(g)=gH$.
Proof: Let $Pi = a_i H : a_iH text is a coset of H $ then by the theorem in the book there exists a set Borel set $S$ such that $Scap a_i H = s_i$.
This defines a (Borel) map $s:G/Hrightarrow G$ such that $s(u) = s_i$ if $uin a_i H$. (The map is Borel because $s^-1(U)=s^-1(Ucap S) = p(Ucap S)$ and quotient map is open so it sends Borel to Borel), which proves the theorem.
This theorem leads to our result, as one can can define a map $varphi:Grightarrow H$ by $varphi(g) = gcdot (s(p(g))^-1$.
Since for every $hin H$ one has that $p(gh)=p(g)$ we have that $varphi(gh)=varphi(g)cdot h$.
This also answers positively my (now obviously related) other question:
$Gcong G/Htimes H$ measureably
answered yesterday
Yanko
3,463620
answered yesterday
Yanko
3,463620
answered yesterday
Yanko
3,463620
answered yesterday
Yanko
3,463620
3,463620
add a comment |Â
add a comment |Â
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3
You are supposed to do some work. See if you can find a selection theorem to help for this ... en.wikipedia.org/wiki/List_of_selection_theorems
– GEdgar
Aug 2 at 18:34
@GEdgar Thank you. I will look up for that!
– Yanko
Aug 2 at 22:47