Are Hausdorff measures on the real line Haar measures for some locally compact topology?

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For $0leq dleq 1$, let $lambda_d$ be the $d$-dimensional Hausdorff measure on $mathbbR$. Note that it is translation-invariant. Does there exist a locally compact topology $mathscrT_d$ on $mathbbR$, 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 $(mathbbR,+,mathscrT_d)$?



(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 $mathscrT_d$ can somehow be made "canonical".










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    12












    $begingroup$


    For $0leq dleq 1$, let $lambda_d$ be the $d$-dimensional Hausdorff measure on $mathbbR$. Note that it is translation-invariant. Does there exist a locally compact topology $mathscrT_d$ on $mathbbR$, 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 $(mathbbR,+,mathscrT_d)$?



    (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 $mathscrT_d$ can somehow be made "canonical".










    share|cite|improve this question









    $endgroup$














      12












      12








      12


      1



      $begingroup$


      For $0leq dleq 1$, let $lambda_d$ be the $d$-dimensional Hausdorff measure on $mathbbR$. Note that it is translation-invariant. Does there exist a locally compact topology $mathscrT_d$ on $mathbbR$, 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 $(mathbbR,+,mathscrT_d)$?



      (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 $mathscrT_d$ can somehow be made "canonical".










      share|cite|improve this question









      $endgroup$




      For $0leq dleq 1$, let $lambda_d$ be the $d$-dimensional Hausdorff measure on $mathbbR$. Note that it is translation-invariant. Does there exist a locally compact topology $mathscrT_d$ on $mathbbR$, 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 $(mathbbR,+,mathscrT_d)$?



      (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 $mathscrT_d$ can somehow be made "canonical".







      gn.general-topology measure-theory topological-groups haar-measure hausdorff-dimension






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      asked Feb 15 at 15:16









      Gro-TsenGro-Tsen

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      9,802234101




















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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.






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            1 Answer
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            $begingroup$

            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.






            share|cite|improve this answer









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              15












              $begingroup$

              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.






              share|cite|improve this answer









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                15












                15








                15





                $begingroup$

                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.






                share|cite|improve this answer









                $endgroup$



                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.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Feb 15 at 16:32









                Taras BanakhTaras Banakh

                17.1k13495




                17.1k13495



























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