Are Hausdorff measures on the real line Haar measures for some locally compact topology?
$begingroup$
For $0leq dleq 1$, let $lambda_d$ be the $d$-dimensional Hausdorff measure on $mathbb{R}$. Note that it is translation-invariant. Does there exist a locally compact topology $mathscr{T}_d$ on $mathbb{R}$, finer than the usual topology and compatible with the (additive) group structure (i.e., $+$ and $-$ are continuous), such that $lambda_d$ is, up to some normalization, the Haar measure for $(mathbb{R},+,mathscr{T}_d)$?
(For $d=1$ the usual topology provides a positive answer. For $d=0$ the discrete topology does. So the question is whether we can do something in between.)
Bonus points if $mathscr{T}_d$ can somehow be made "canonical".
gn.general-topology measure-theory topological-groups haar-measure hausdorff-dimension
asked Feb 15 at 15:16
Gro-TsenGro-Tsen
9,740234100
$endgroup$
add a comment |
$begingroup$
For $0leq dleq 1$, let $lambda_d$ be the $d$-dimensional Hausdorff measure on $mathbb{R}$. Note that it is translation-invariant. Does there exist a locally compact topology $mathscr{T}_d$ on $mathbb{R}$, finer than the usual topology and compatible with the (additive) group structure (i.e., $+$ and $-$ are continuous), such that $lambda_d$ is, up to some normalization, the Haar measure for $(mathbb{R},+,mathscr{T}_d)$?
(For $d=1$ the usual topology provides a positive answer. For $d=0$ the discrete topology does. So the question is whether we can do something in between.)
Bonus points if $mathscr{T}_d$ can somehow be made "canonical".
gn.general-topology measure-theory topological-groups haar-measure hausdorff-dimension
asked Feb 15 at 15:16
Gro-TsenGro-Tsen
9,740234100
$endgroup$
add a comment |
$begingroup$
For $0leq dleq 1$, let $lambda_d$ be the $d$-dimensional Hausdorff measure on $mathbb{R}$. Note that it is translation-invariant. Does there exist a locally compact topology $mathscr{T}_d$ on $mathbb{R}$, finer than the usual topology and compatible with the (additive) group structure (i.e., $+$ and $-$ are continuous), such that $lambda_d$ is, up to some normalization, the Haar measure for $(mathbb{R},+,mathscr{T}_d)$?
(For $d=1$ the usual topology provides a positive answer. For $d=0$ the discrete topology does. So the question is whether we can do something in between.)
Bonus points if $mathscr{T}_d$ can somehow be made "canonical".
gn.general-topology measure-theory topological-groups haar-measure hausdorff-dimension
asked Feb 15 at 15:16
Gro-TsenGro-Tsen
9,740234100
$endgroup$
For $0leq dleq 1$, let $lambda_d$ be the $d$-dimensional Hausdorff measure on $mathbb{R}$. Note that it is translation-invariant. Does there exist a locally compact topology $mathscr{T}_d$ on $mathbb{R}$, finer than the usual topology and compatible with the (additive) group structure (i.e., $+$ and $-$ are continuous), such that $lambda_d$ is, up to some normalization, the Haar measure for $(mathbb{R},+,mathscr{T}_d)$?
(For $d=1$ the usual topology provides a positive answer. For $d=0$ the discrete topology does. So the question is whether we can do something in between.)
Bonus points if $mathscr{T}_d$ can somehow be made "canonical".
gn.general-topology measure-theory topological-groups haar-measure hausdorff-dimension
gn.general-topology measure-theory topological-groups haar-measure hausdorff-dimension
asked Feb 15 at 15:16
Gro-TsenGro-Tsen
9,740234100
asked Feb 15 at 15:16
Gro-TsenGro-Tsen
9,740234100
asked Feb 15 at 15:16
Gro-TsenGro-Tsen
9,740234100
asked Feb 15 at 15:16
Gro-TsenGro-Tsen
9,740234100
asked Feb 15 at 15:16
Gro-TsenGro-Tsen
9,740234100
9,740234100
add a comment |
add a comment |
1 Answer
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The answer is NO because the Euclidean and the discrete topologies are the unique locally compact group topologies on $mathbb R$, which are stronger that the Euclidean topology of the real line.
The reason is that $mathbb R$ endowed with such topology $tau$ is a locally compact abelian topological group without small subgroups, so is a Lie group (by the Gleason-Mongomery-Zippin Theorem). Since $(mathbb R,tau)$ admits a continuous injective map into $mathbb R$, it has dimension $le 1$. If the dimension of the Lie group $(mathbb R,tau)$ is 1, then it is (locally) homeomorphic to $mathbb R$. If the Lie group $(mathbb R,tau)$ has dimension zero, then it is discrete.
answered Feb 15 at 16:32
Taras BanakhTaras Banakh
16.8k13495
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1 Answer
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1 Answer
1
active
oldest
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active
oldest
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active
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votes
$begingroup$
The answer is NO because the Euclidean and the discrete topologies are the unique locally compact group topologies on $mathbb R$, which are stronger that the Euclidean topology of the real line.
The reason is that $mathbb R$ endowed with such topology $tau$ is a locally compact abelian topological group without small subgroups, so is a Lie group (by the Gleason-Mongomery-Zippin Theorem). Since $(mathbb R,tau)$ admits a continuous injective map into $mathbb R$, it has dimension $le 1$. If the dimension of the Lie group $(mathbb R,tau)$ is 1, then it is (locally) homeomorphic to $mathbb R$. If the Lie group $(mathbb R,tau)$ has dimension zero, then it is discrete.
answered Feb 15 at 16:32
Taras BanakhTaras Banakh
16.8k13495
$endgroup$
add a comment |
$begingroup$
The answer is NO because the Euclidean and the discrete topologies are the unique locally compact group topologies on $mathbb R$, which are stronger that the Euclidean topology of the real line.
The reason is that $mathbb R$ endowed with such topology $tau$ is a locally compact abelian topological group without small subgroups, so is a Lie group (by the Gleason-Mongomery-Zippin Theorem). Since $(mathbb R,tau)$ admits a continuous injective map into $mathbb R$, it has dimension $le 1$. If the dimension of the Lie group $(mathbb R,tau)$ is 1, then it is (locally) homeomorphic to $mathbb R$. If the Lie group $(mathbb R,tau)$ has dimension zero, then it is discrete.
answered Feb 15 at 16:32
Taras BanakhTaras Banakh
16.8k13495
$endgroup$
add a comment |
$begingroup$
The answer is NO because the Euclidean and the discrete topologies are the unique locally compact group topologies on $mathbb R$, which are stronger that the Euclidean topology of the real line.
The reason is that $mathbb R$ endowed with such topology $tau$ is a locally compact abelian topological group without small subgroups, so is a Lie group (by the Gleason-Mongomery-Zippin Theorem). Since $(mathbb R,tau)$ admits a continuous injective map into $mathbb R$, it has dimension $le 1$. If the dimension of the Lie group $(mathbb R,tau)$ is 1, then it is (locally) homeomorphic to $mathbb R$. If the Lie group $(mathbb R,tau)$ has dimension zero, then it is discrete.
answered Feb 15 at 16:32
Taras BanakhTaras Banakh
16.8k13495
$endgroup$
The answer is NO because the Euclidean and the discrete topologies are the unique locally compact group topologies on $mathbb R$, which are stronger that the Euclidean topology of the real line.
The reason is that $mathbb R$ endowed with such topology $tau$ is a locally compact abelian topological group without small subgroups, so is a Lie group (by the Gleason-Mongomery-Zippin Theorem). Since $(mathbb R,tau)$ admits a continuous injective map into $mathbb R$, it has dimension $le 1$. If the dimension of the Lie group $(mathbb R,tau)$ is 1, then it is (locally) homeomorphic to $mathbb R$. If the Lie group $(mathbb R,tau)$ has dimension zero, then it is discrete.
answered Feb 15 at 16:32
Taras BanakhTaras Banakh
16.8k13495
answered Feb 15 at 16:32
Taras BanakhTaras Banakh
16.8k13495
answered Feb 15 at 16:32
Taras BanakhTaras Banakh
16.8k13495
answered Feb 15 at 16:32
Taras BanakhTaras Banakh
16.8k13495
16.8k13495
add a comment |
add a comment |
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