Is there such a group?












2














Is there a compact matrix group $Gsubseteq GL(n,mathbb{R})$ such that $|G|$ is countable infinite?






group-theory







share|cite|improve this question
























  • What topology are we using? Would $$left<begin{pmatrix}2&0\0&2end{pmatrix}right>$$ satisfy your conditions? It would form a closed set in the product topology of $mathbb{R^{n^2}}.$
    – Melody
    Dec 3 '18 at 9:15










  • @Melody But it is not bounded.
    – Paul Frost
    Dec 3 '18 at 9:17










  • @Paul Frost True. Nevermind.
    – Melody
    Dec 3 '18 at 9:17








  • 3




    There is no countable infinite compact Hausdorff group. See this: mathoverflow.net/questions/4939/…
    – freakish
    Dec 3 '18 at 9:38








  • 1




    @Servaes Of course. But we know that $det(G)$ is a compact subset of $mathbb{R}$, so it must contain $lim d^n = 0$.
    – Paul Frost
    Dec 3 '18 at 9:43
















2














Is there a compact matrix group $Gsubseteq GL(n,mathbb{R})$ such that $|G|$ is countable infinite?






group-theory







share|cite|improve this question
























  • What topology are we using? Would $$left<begin{pmatrix}2&0\0&2end{pmatrix}right>$$ satisfy your conditions? It would form a closed set in the product topology of $mathbb{R^{n^2}}.$
    – Melody
    Dec 3 '18 at 9:15










  • @Melody But it is not bounded.
    – Paul Frost
    Dec 3 '18 at 9:17










  • @Paul Frost True. Nevermind.
    – Melody
    Dec 3 '18 at 9:17








  • 3




    There is no countable infinite compact Hausdorff group. See this: mathoverflow.net/questions/4939/…
    – freakish
    Dec 3 '18 at 9:38








  • 1




    @Servaes Of course. But we know that $det(G)$ is a compact subset of $mathbb{R}$, so it must contain $lim d^n = 0$.
    – Paul Frost
    Dec 3 '18 at 9:43














2












2








2







Is there a compact matrix group $Gsubseteq GL(n,mathbb{R})$ such that $|G|$ is countable infinite?






group-theory







share|cite|improve this question















Is there a compact matrix group $Gsubseteq GL(n,mathbb{R})$ such that $|G|$ is countable infinite?






group-theory




group-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 4 '18 at 1:12









the_fox

2,47011431




2,47011431










asked Dec 3 '18 at 8:38









yao4015

655




655












  • What topology are we using? Would $$left<begin{pmatrix}2&0\0&2end{pmatrix}right>$$ satisfy your conditions? It would form a closed set in the product topology of $mathbb{R^{n^2}}.$
    – Melody
    Dec 3 '18 at 9:15










  • @Melody But it is not bounded.
    – Paul Frost
    Dec 3 '18 at 9:17










  • @Paul Frost True. Nevermind.
    – Melody
    Dec 3 '18 at 9:17








  • 3




    There is no countable infinite compact Hausdorff group. See this: mathoverflow.net/questions/4939/…
    – freakish
    Dec 3 '18 at 9:38








  • 1




    @Servaes Of course. But we know that $det(G)$ is a compact subset of $mathbb{R}$, so it must contain $lim d^n = 0$.
    – Paul Frost
    Dec 3 '18 at 9:43


















  • What topology are we using? Would $$left<begin{pmatrix}2&0\0&2end{pmatrix}right>$$ satisfy your conditions? It would form a closed set in the product topology of $mathbb{R^{n^2}}.$
    – Melody
    Dec 3 '18 at 9:15










  • @Melody But it is not bounded.
    – Paul Frost
    Dec 3 '18 at 9:17










  • @Paul Frost True. Nevermind.
    – Melody
    Dec 3 '18 at 9:17








  • 3




    There is no countable infinite compact Hausdorff group. See this: mathoverflow.net/questions/4939/…
    – freakish
    Dec 3 '18 at 9:38








  • 1




    @Servaes Of course. But we know that $det(G)$ is a compact subset of $mathbb{R}$, so it must contain $lim d^n = 0$.
    – Paul Frost
    Dec 3 '18 at 9:43
















What topology are we using? Would $$left<begin{pmatrix}2&0\0&2end{pmatrix}right>$$ satisfy your conditions? It would form a closed set in the product topology of $mathbb{R^{n^2}}.$
– Melody
Dec 3 '18 at 9:15




What topology are we using? Would $$left<begin{pmatrix}2&0\0&2end{pmatrix}right>$$ satisfy your conditions? It would form a closed set in the product topology of $mathbb{R^{n^2}}.$
– Melody
Dec 3 '18 at 9:15












@Melody But it is not bounded.
– Paul Frost
Dec 3 '18 at 9:17




@Melody But it is not bounded.
– Paul Frost
Dec 3 '18 at 9:17












@Paul Frost True. Nevermind.
– Melody
Dec 3 '18 at 9:17






@Paul Frost True. Nevermind.
– Melody
Dec 3 '18 at 9:17






3




3




There is no countable infinite compact Hausdorff group. See this: mathoverflow.net/questions/4939/…
– freakish
Dec 3 '18 at 9:38






There is no countable infinite compact Hausdorff group. See this: mathoverflow.net/questions/4939/…
– freakish
Dec 3 '18 at 9:38






1




1




@Servaes Of course. But we know that $det(G)$ is a compact subset of $mathbb{R}$, so it must contain $lim d^n = 0$.
– Paul Frost
Dec 3 '18 at 9:43




@Servaes Of course. But we know that $det(G)$ is a compact subset of $mathbb{R}$, so it must contain $lim d^n = 0$.
– Paul Frost
Dec 3 '18 at 9:43










1 Answer
1






active

oldest

votes


















4














So the answer is negative. In general there is no countably infinite and compact Hausdorff group. See this:



https://mathoverflow.net/questions/4939/is-there-a-compact-group-of-countably-infinite-cardinality



To add a bit to the answer (so that it isn't completely naked ;) ): every compact subgroup of $GL_n(mathbb{R})$ is obviously contained in the preimage $det^{-1}({-1,1})$ which is the largest compact subgroup of $mathbb{R}^*$. But we can do better. The orthogonal group $O(n)$ is a maximal subgroup of $GL_n(mathbb{R})$ among compact subgroups (see this). Therefore every compact subgroup of $GL_n(mathbb{R})$ is a subconjugate of $O(n)$ which is a consequence of the Cartan-Iwasawa-Malcev theorem.



In particular if you wish to analyze compact subgroups of $GL_n(mathbb{R})$ then it is enough to look at subgroups of $O(n)$.






share|cite|improve this answer























  • thank you very much
    – yao4015
    Dec 4 '18 at 1:09











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









4














So the answer is negative. In general there is no countably infinite and compact Hausdorff group. See this:



https://mathoverflow.net/questions/4939/is-there-a-compact-group-of-countably-infinite-cardinality



To add a bit to the answer (so that it isn't completely naked ;) ): every compact subgroup of $GL_n(mathbb{R})$ is obviously contained in the preimage $det^{-1}({-1,1})$ which is the largest compact subgroup of $mathbb{R}^*$. But we can do better. The orthogonal group $O(n)$ is a maximal subgroup of $GL_n(mathbb{R})$ among compact subgroups (see this). Therefore every compact subgroup of $GL_n(mathbb{R})$ is a subconjugate of $O(n)$ which is a consequence of the Cartan-Iwasawa-Malcev theorem.



In particular if you wish to analyze compact subgroups of $GL_n(mathbb{R})$ then it is enough to look at subgroups of $O(n)$.






share|cite|improve this answer























  • thank you very much
    – yao4015
    Dec 4 '18 at 1:09
















4














So the answer is negative. In general there is no countably infinite and compact Hausdorff group. See this:



https://mathoverflow.net/questions/4939/is-there-a-compact-group-of-countably-infinite-cardinality



To add a bit to the answer (so that it isn't completely naked ;) ): every compact subgroup of $GL_n(mathbb{R})$ is obviously contained in the preimage $det^{-1}({-1,1})$ which is the largest compact subgroup of $mathbb{R}^*$. But we can do better. The orthogonal group $O(n)$ is a maximal subgroup of $GL_n(mathbb{R})$ among compact subgroups (see this). Therefore every compact subgroup of $GL_n(mathbb{R})$ is a subconjugate of $O(n)$ which is a consequence of the Cartan-Iwasawa-Malcev theorem.



In particular if you wish to analyze compact subgroups of $GL_n(mathbb{R})$ then it is enough to look at subgroups of $O(n)$.






share|cite|improve this answer























  • thank you very much
    – yao4015
    Dec 4 '18 at 1:09














4












4








4






So the answer is negative. In general there is no countably infinite and compact Hausdorff group. See this:



https://mathoverflow.net/questions/4939/is-there-a-compact-group-of-countably-infinite-cardinality



To add a bit to the answer (so that it isn't completely naked ;) ): every compact subgroup of $GL_n(mathbb{R})$ is obviously contained in the preimage $det^{-1}({-1,1})$ which is the largest compact subgroup of $mathbb{R}^*$. But we can do better. The orthogonal group $O(n)$ is a maximal subgroup of $GL_n(mathbb{R})$ among compact subgroups (see this). Therefore every compact subgroup of $GL_n(mathbb{R})$ is a subconjugate of $O(n)$ which is a consequence of the Cartan-Iwasawa-Malcev theorem.



In particular if you wish to analyze compact subgroups of $GL_n(mathbb{R})$ then it is enough to look at subgroups of $O(n)$.






share|cite|improve this answer














So the answer is negative. In general there is no countably infinite and compact Hausdorff group. See this:



https://mathoverflow.net/questions/4939/is-there-a-compact-group-of-countably-infinite-cardinality



To add a bit to the answer (so that it isn't completely naked ;) ): every compact subgroup of $GL_n(mathbb{R})$ is obviously contained in the preimage $det^{-1}({-1,1})$ which is the largest compact subgroup of $mathbb{R}^*$. But we can do better. The orthogonal group $O(n)$ is a maximal subgroup of $GL_n(mathbb{R})$ among compact subgroups (see this). Therefore every compact subgroup of $GL_n(mathbb{R})$ is a subconjugate of $O(n)$ which is a consequence of the Cartan-Iwasawa-Malcev theorem.



In particular if you wish to analyze compact subgroups of $GL_n(mathbb{R})$ then it is enough to look at subgroups of $O(n)$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 3 '18 at 10:02

























answered Dec 3 '18 at 9:57









freakish

11.5k1629




11.5k1629












  • thank you very much
    – yao4015
    Dec 4 '18 at 1:09


















  • thank you very much
    – yao4015
    Dec 4 '18 at 1:09
















thank you very much
– yao4015
Dec 4 '18 at 1:09




thank you very much
– yao4015
Dec 4 '18 at 1:09


















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