Yamabe's theorem proof
$begingroup$
*I'm trying to make the proof of Yamabe's Theorem that says that an arcwise connected subgroup of a Lie group G is a Lie subgroup of G.
I found the proof in Goto's article (https://www.ams.org/journals/proc/1969-020-01/S0002-9939-1969-0233923-X/S0002-9939-1969-0233923-X.pdf) and in the book "Structure and Geometry of Lie Groups", from Joachim Hilgert and Karl-Hermann Neeb. In booth of them, they use the following lemma to conclude the proof of Yamabe's theorem:
Blockquote: Lemma: If $H$ is an arcwise connected topological subgroup of $G$, then
for each identity neighborhood $U$ of $1$ in $G$, the arc-component $U_a$ of $Ucap H$
generates $H$.
On the first one, Goto just says that the proof of this lemma is trivial. On the second one, they make a proof:
Blockquote: Let $H_U ⊆ H$ denote the subgroup generated by $U_a$. Further, let $hin H$
and $gamma : [0,1] to H$ be a continuous path from $1$ to $h$. We consider the set $$S := {t ∈ [0, 1] : γ(t) ∈ H_U}.$$
Then the subset $gamma^{−1}(U)$ of $S$ is a neighborhood of zero. For any $t_0 ∈ S$, there exists an $epsilon > 0$ with $γ(t_0)^{−1}γ(t) ∈ U_a$
for $|t−t_0| ≤ ε$. This implies that $S$ is an open subset of $[0, 1]$. If $t_1 in [0, 1]setminus S$, then we also find an $epsilon > 0$ with
$gamma(t_1)^{−1}gamma(t) in U_a subseteq H_U$
for $|t − t_1| ≤ epsilon$, so that $gamma(t_1) ∈ H_U$ leads to $gamma(t) ∈ H_U$ for $|t − t_1| ≤ epsilon$. Thus $[0, 1] setminus S$ is also open in $[0, 1]$, and now the connectedness of $[0, 1]$ implies that
$S = [0, 1]$, hence $h = gamma(1) in H_U$.
Apparently, where I emphasis added, they used that $U_a$ is a neighborhood of $1$ in $H$. But how can I guarantee this if I don't have that $H$ is a locally connected space? Is there some result that makes this real for topological subgroups of a Lie group? I tryed to proof that but I coudn't. I'd be grateful if anyone had any tips.*
lie-groups topological-groups path-connected
asked Dec 8 '18 at 18:52
Jéssica BuzattoJéssica Buzatto
112
$endgroup$
add a comment |
$begingroup$
*I'm trying to make the proof of Yamabe's Theorem that says that an arcwise connected subgroup of a Lie group G is a Lie subgroup of G.
I found the proof in Goto's article (https://www.ams.org/journals/proc/1969-020-01/S0002-9939-1969-0233923-X/S0002-9939-1969-0233923-X.pdf) and in the book "Structure and Geometry of Lie Groups", from Joachim Hilgert and Karl-Hermann Neeb. In booth of them, they use the following lemma to conclude the proof of Yamabe's theorem:
Blockquote: Lemma: If $H$ is an arcwise connected topological subgroup of $G$, then
for each identity neighborhood $U$ of $1$ in $G$, the arc-component $U_a$ of $Ucap H$
generates $H$.
On the first one, Goto just says that the proof of this lemma is trivial. On the second one, they make a proof:
Blockquote: Let $H_U ⊆ H$ denote the subgroup generated by $U_a$. Further, let $hin H$
and $gamma : [0,1] to H$ be a continuous path from $1$ to $h$. We consider the set $$S := {t ∈ [0, 1] : γ(t) ∈ H_U}.$$
Then the subset $gamma^{−1}(U)$ of $S$ is a neighborhood of zero. For any $t_0 ∈ S$, there exists an $epsilon > 0$ with $γ(t_0)^{−1}γ(t) ∈ U_a$
for $|t−t_0| ≤ ε$. This implies that $S$ is an open subset of $[0, 1]$. If $t_1 in [0, 1]setminus S$, then we also find an $epsilon > 0$ with
$gamma(t_1)^{−1}gamma(t) in U_a subseteq H_U$
for $|t − t_1| ≤ epsilon$, so that $gamma(t_1) ∈ H_U$ leads to $gamma(t) ∈ H_U$ for $|t − t_1| ≤ epsilon$. Thus $[0, 1] setminus S$ is also open in $[0, 1]$, and now the connectedness of $[0, 1]$ implies that
$S = [0, 1]$, hence $h = gamma(1) in H_U$.
Apparently, where I emphasis added, they used that $U_a$ is a neighborhood of $1$ in $H$. But how can I guarantee this if I don't have that $H$ is a locally connected space? Is there some result that makes this real for topological subgroups of a Lie group? I tryed to proof that but I coudn't. I'd be grateful if anyone had any tips.*
lie-groups topological-groups path-connected
asked Dec 8 '18 at 18:52
Jéssica BuzattoJéssica Buzatto
112
$endgroup$
add a comment |
$begingroup$
*I'm trying to make the proof of Yamabe's Theorem that says that an arcwise connected subgroup of a Lie group G is a Lie subgroup of G.
I found the proof in Goto's article (https://www.ams.org/journals/proc/1969-020-01/S0002-9939-1969-0233923-X/S0002-9939-1969-0233923-X.pdf) and in the book "Structure and Geometry of Lie Groups", from Joachim Hilgert and Karl-Hermann Neeb. In booth of them, they use the following lemma to conclude the proof of Yamabe's theorem:
Blockquote: Lemma: If $H$ is an arcwise connected topological subgroup of $G$, then
for each identity neighborhood $U$ of $1$ in $G$, the arc-component $U_a$ of $Ucap H$
generates $H$.
On the first one, Goto just says that the proof of this lemma is trivial. On the second one, they make a proof:
Blockquote: Let $H_U ⊆ H$ denote the subgroup generated by $U_a$. Further, let $hin H$
and $gamma : [0,1] to H$ be a continuous path from $1$ to $h$. We consider the set $$S := {t ∈ [0, 1] : γ(t) ∈ H_U}.$$
Then the subset $gamma^{−1}(U)$ of $S$ is a neighborhood of zero. For any $t_0 ∈ S$, there exists an $epsilon > 0$ with $γ(t_0)^{−1}γ(t) ∈ U_a$
for $|t−t_0| ≤ ε$. This implies that $S$ is an open subset of $[0, 1]$. If $t_1 in [0, 1]setminus S$, then we also find an $epsilon > 0$ with
$gamma(t_1)^{−1}gamma(t) in U_a subseteq H_U$
for $|t − t_1| ≤ epsilon$, so that $gamma(t_1) ∈ H_U$ leads to $gamma(t) ∈ H_U$ for $|t − t_1| ≤ epsilon$. Thus $[0, 1] setminus S$ is also open in $[0, 1]$, and now the connectedness of $[0, 1]$ implies that
$S = [0, 1]$, hence $h = gamma(1) in H_U$.
Apparently, where I emphasis added, they used that $U_a$ is a neighborhood of $1$ in $H$. But how can I guarantee this if I don't have that $H$ is a locally connected space? Is there some result that makes this real for topological subgroups of a Lie group? I tryed to proof that but I coudn't. I'd be grateful if anyone had any tips.*
lie-groups topological-groups path-connected
asked Dec 8 '18 at 18:52
Jéssica BuzattoJéssica Buzatto
112
$endgroup$
*I'm trying to make the proof of Yamabe's Theorem that says that an arcwise connected subgroup of a Lie group G is a Lie subgroup of G.
I found the proof in Goto's article (https://www.ams.org/journals/proc/1969-020-01/S0002-9939-1969-0233923-X/S0002-9939-1969-0233923-X.pdf) and in the book "Structure and Geometry of Lie Groups", from Joachim Hilgert and Karl-Hermann Neeb. In booth of them, they use the following lemma to conclude the proof of Yamabe's theorem:
Blockquote: Lemma: If $H$ is an arcwise connected topological subgroup of $G$, then
for each identity neighborhood $U$ of $1$ in $G$, the arc-component $U_a$ of $Ucap H$
generates $H$.
On the first one, Goto just says that the proof of this lemma is trivial. On the second one, they make a proof:
Blockquote: Let $H_U ⊆ H$ denote the subgroup generated by $U_a$. Further, let $hin H$
and $gamma : [0,1] to H$ be a continuous path from $1$ to $h$. We consider the set $$S := {t ∈ [0, 1] : γ(t) ∈ H_U}.$$
Then the subset $gamma^{−1}(U)$ of $S$ is a neighborhood of zero. For any $t_0 ∈ S$, there exists an $epsilon > 0$ with $γ(t_0)^{−1}γ(t) ∈ U_a$
for $|t−t_0| ≤ ε$. This implies that $S$ is an open subset of $[0, 1]$. If $t_1 in [0, 1]setminus S$, then we also find an $epsilon > 0$ with
$gamma(t_1)^{−1}gamma(t) in U_a subseteq H_U$
for $|t − t_1| ≤ epsilon$, so that $gamma(t_1) ∈ H_U$ leads to $gamma(t) ∈ H_U$ for $|t − t_1| ≤ epsilon$. Thus $[0, 1] setminus S$ is also open in $[0, 1]$, and now the connectedness of $[0, 1]$ implies that
$S = [0, 1]$, hence $h = gamma(1) in H_U$.
Apparently, where I emphasis added, they used that $U_a$ is a neighborhood of $1$ in $H$. But how can I guarantee this if I don't have that $H$ is a locally connected space? Is there some result that makes this real for topological subgroups of a Lie group? I tryed to proof that but I coudn't. I'd be grateful if anyone had any tips.*
lie-groups topological-groups path-connected
lie-groups topological-groups path-connected
asked Dec 8 '18 at 18:52
Jéssica BuzattoJéssica Buzatto
112
asked Dec 8 '18 at 18:52
Jéssica BuzattoJéssica Buzatto
112
edited Dec 8 '18 at 19:12
Jéssica Buzatto
asked Dec 8 '18 at 18:52
Jéssica BuzattoJéssica Buzatto
112
asked Dec 8 '18 at 18:52
Jéssica BuzattoJéssica Buzatto
112
asked Dec 8 '18 at 18:52
Jéssica BuzattoJéssica Buzatto
112
112
add a comment |
add a comment |
1 Answer
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You are right: $U_a$ is not necessarily a neighbourhood of $1$ in $H$, and $H$ is not necessarily locally connected.
However, since $U$ is open, for any $t_0in S$ there exists $varepsilon >0$ such that for all $tin [0,1]$ with $|t-t_0|<varepsilon$ one has $gamma(t_0)^{-1}gamma(t)in U$. Since the image of $gamma$ lies in $H$, we also have $gamma(t_0)^{-1}gamma(t)in Ucap H$. It is, moreover, in the same arc-component (in $Ucap H$) as the element $1$ (since it is connected to it by the path $$smapsto gamma(s)^{-1}gamma(t)$$ where $s$ goes from $t$ to $t_0$), so $gamma(t_0)^{-1}gamma(t)$ is in $U_a$.
answered Dec 11 '18 at 9:41
CronusCronus
1,088517
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
You are right: $U_a$ is not necessarily a neighbourhood of $1$ in $H$, and $H$ is not necessarily locally connected.
However, since $U$ is open, for any $t_0in S$ there exists $varepsilon >0$ such that for all $tin [0,1]$ with $|t-t_0|<varepsilon$ one has $gamma(t_0)^{-1}gamma(t)in U$. Since the image of $gamma$ lies in $H$, we also have $gamma(t_0)^{-1}gamma(t)in Ucap H$. It is, moreover, in the same arc-component (in $Ucap H$) as the element $1$ (since it is connected to it by the path $$smapsto gamma(s)^{-1}gamma(t)$$ where $s$ goes from $t$ to $t_0$), so $gamma(t_0)^{-1}gamma(t)$ is in $U_a$.
answered Dec 11 '18 at 9:41
CronusCronus
1,088517
$endgroup$
add a comment |
$begingroup$
You are right: $U_a$ is not necessarily a neighbourhood of $1$ in $H$, and $H$ is not necessarily locally connected.
However, since $U$ is open, for any $t_0in S$ there exists $varepsilon >0$ such that for all $tin [0,1]$ with $|t-t_0|<varepsilon$ one has $gamma(t_0)^{-1}gamma(t)in U$. Since the image of $gamma$ lies in $H$, we also have $gamma(t_0)^{-1}gamma(t)in Ucap H$. It is, moreover, in the same arc-component (in $Ucap H$) as the element $1$ (since it is connected to it by the path $$smapsto gamma(s)^{-1}gamma(t)$$ where $s$ goes from $t$ to $t_0$), so $gamma(t_0)^{-1}gamma(t)$ is in $U_a$.
answered Dec 11 '18 at 9:41
CronusCronus
1,088517
$endgroup$
add a comment |
$begingroup$
You are right: $U_a$ is not necessarily a neighbourhood of $1$ in $H$, and $H$ is not necessarily locally connected.
However, since $U$ is open, for any $t_0in S$ there exists $varepsilon >0$ such that for all $tin [0,1]$ with $|t-t_0|<varepsilon$ one has $gamma(t_0)^{-1}gamma(t)in U$. Since the image of $gamma$ lies in $H$, we also have $gamma(t_0)^{-1}gamma(t)in Ucap H$. It is, moreover, in the same arc-component (in $Ucap H$) as the element $1$ (since it is connected to it by the path $$smapsto gamma(s)^{-1}gamma(t)$$ where $s$ goes from $t$ to $t_0$), so $gamma(t_0)^{-1}gamma(t)$ is in $U_a$.
answered Dec 11 '18 at 9:41
CronusCronus
1,088517
$endgroup$
You are right: $U_a$ is not necessarily a neighbourhood of $1$ in $H$, and $H$ is not necessarily locally connected.
However, since $U$ is open, for any $t_0in S$ there exists $varepsilon >0$ such that for all $tin [0,1]$ with $|t-t_0|<varepsilon$ one has $gamma(t_0)^{-1}gamma(t)in U$. Since the image of $gamma$ lies in $H$, we also have $gamma(t_0)^{-1}gamma(t)in Ucap H$. It is, moreover, in the same arc-component (in $Ucap H$) as the element $1$ (since it is connected to it by the path $$smapsto gamma(s)^{-1}gamma(t)$$ where $s$ goes from $t$ to $t_0$), so $gamma(t_0)^{-1}gamma(t)$ is in $U_a$.
answered Dec 11 '18 at 9:41
CronusCronus
1,088517
edited Dec 11 '18 at 10:17
answered Dec 11 '18 at 9:41
CronusCronus
1,088517
answered Dec 11 '18 at 9:41
CronusCronus
1,088517
answered Dec 11 '18 at 9:41
CronusCronus
1,088517
1,088517
add a comment |
add a comment |
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