One dimensional topological subgroups of the torus
$begingroup$
I was recently asked to characterize Lie subgroups of the torus $mathbb{S}^1timesmathbb{S}^1$. I found the one-dimensional case more difficult than I imagined, having to appeal to either Lie algebras or covering space theory to show they have "constant slope" in the lift. In particular, I made heavy use of smoothness.
I naively thought that a one-dimensional connected topological subgroup of the plane would be a linear subspace. Is this wrong?
lie-groups topological-groups
asked Dec 18 '18 at 21:06
RyanRyan
358214
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$begingroup$
I was recently asked to characterize Lie subgroups of the torus $mathbb{S}^1timesmathbb{S}^1$. I found the one-dimensional case more difficult than I imagined, having to appeal to either Lie algebras or covering space theory to show they have "constant slope" in the lift. In particular, I made heavy use of smoothness.
I naively thought that a one-dimensional connected topological subgroup of the plane would be a linear subspace. Is this wrong?
lie-groups topological-groups
asked Dec 18 '18 at 21:06
RyanRyan
358214
$endgroup$
add a comment |
$begingroup$
I was recently asked to characterize Lie subgroups of the torus $mathbb{S}^1timesmathbb{S}^1$. I found the one-dimensional case more difficult than I imagined, having to appeal to either Lie algebras or covering space theory to show they have "constant slope" in the lift. In particular, I made heavy use of smoothness.
I naively thought that a one-dimensional connected topological subgroup of the plane would be a linear subspace. Is this wrong?
lie-groups topological-groups
asked Dec 18 '18 at 21:06
RyanRyan
358214
$endgroup$
I was recently asked to characterize Lie subgroups of the torus $mathbb{S}^1timesmathbb{S}^1$. I found the one-dimensional case more difficult than I imagined, having to appeal to either Lie algebras or covering space theory to show they have "constant slope" in the lift. In particular, I made heavy use of smoothness.
I naively thought that a one-dimensional connected topological subgroup of the plane would be a linear subspace. Is this wrong?
lie-groups topological-groups
lie-groups topological-groups
asked Dec 18 '18 at 21:06
RyanRyan
358214
asked Dec 18 '18 at 21:06
RyanRyan
358214
asked Dec 18 '18 at 21:06
RyanRyan
358214
asked Dec 18 '18 at 21:06
RyanRyan
358214
asked Dec 18 '18 at 21:06
RyanRyan
358214
358214
add a comment |
add a comment |
1 Answer
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$begingroup$
$mathbb{R}^2$ has some extremely weird connected subgroups, and as far as I know no one has ever classified them. (I am not sure how you define one-dimensionality if you don't want to assume it's a Lie subgroup). See, for instance, this paper: https://www.jstor.org/stable/2046253?seq=1#page_scan_tab_contents.
It is true that path-connected subgroups of $mathbb{R}^2$ are the linear subspaces. This is implied by Yamabe's theorem (although maybe one can come up with a simpler proof). For a reference I recommend Structure and Geometry of Lie Groups by Hilgert and Neeb. This is in particular true for Lie subgroups, but maybe it's an assumption which would be more to your liking, since it seems you were bothered by relying too heavily on smoothness. The proof, however, naturally does use smoothness, so maybe that's not good enough.
I hope this answers your question.
answered Dec 19 '18 at 20:53
CronusCronus
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1 Answer
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$begingroup$
$mathbb{R}^2$ has some extremely weird connected subgroups, and as far as I know no one has ever classified them. (I am not sure how you define one-dimensionality if you don't want to assume it's a Lie subgroup). See, for instance, this paper: https://www.jstor.org/stable/2046253?seq=1#page_scan_tab_contents.
It is true that path-connected subgroups of $mathbb{R}^2$ are the linear subspaces. This is implied by Yamabe's theorem (although maybe one can come up with a simpler proof). For a reference I recommend Structure and Geometry of Lie Groups by Hilgert and Neeb. This is in particular true for Lie subgroups, but maybe it's an assumption which would be more to your liking, since it seems you were bothered by relying too heavily on smoothness. The proof, however, naturally does use smoothness, so maybe that's not good enough.
I hope this answers your question.
answered Dec 19 '18 at 20:53
CronusCronus
1,108518
$endgroup$
add a comment |
$begingroup$
$mathbb{R}^2$ has some extremely weird connected subgroups, and as far as I know no one has ever classified them. (I am not sure how you define one-dimensionality if you don't want to assume it's a Lie subgroup). See, for instance, this paper: https://www.jstor.org/stable/2046253?seq=1#page_scan_tab_contents.
It is true that path-connected subgroups of $mathbb{R}^2$ are the linear subspaces. This is implied by Yamabe's theorem (although maybe one can come up with a simpler proof). For a reference I recommend Structure and Geometry of Lie Groups by Hilgert and Neeb. This is in particular true for Lie subgroups, but maybe it's an assumption which would be more to your liking, since it seems you were bothered by relying too heavily on smoothness. The proof, however, naturally does use smoothness, so maybe that's not good enough.
I hope this answers your question.
answered Dec 19 '18 at 20:53
CronusCronus
1,108518
$endgroup$
add a comment |
$begingroup$
$mathbb{R}^2$ has some extremely weird connected subgroups, and as far as I know no one has ever classified them. (I am not sure how you define one-dimensionality if you don't want to assume it's a Lie subgroup). See, for instance, this paper: https://www.jstor.org/stable/2046253?seq=1#page_scan_tab_contents.
It is true that path-connected subgroups of $mathbb{R}^2$ are the linear subspaces. This is implied by Yamabe's theorem (although maybe one can come up with a simpler proof). For a reference I recommend Structure and Geometry of Lie Groups by Hilgert and Neeb. This is in particular true for Lie subgroups, but maybe it's an assumption which would be more to your liking, since it seems you were bothered by relying too heavily on smoothness. The proof, however, naturally does use smoothness, so maybe that's not good enough.
I hope this answers your question.
answered Dec 19 '18 at 20:53
CronusCronus
1,108518
$endgroup$
$mathbb{R}^2$ has some extremely weird connected subgroups, and as far as I know no one has ever classified them. (I am not sure how you define one-dimensionality if you don't want to assume it's a Lie subgroup). See, for instance, this paper: https://www.jstor.org/stable/2046253?seq=1#page_scan_tab_contents.
It is true that path-connected subgroups of $mathbb{R}^2$ are the linear subspaces. This is implied by Yamabe's theorem (although maybe one can come up with a simpler proof). For a reference I recommend Structure and Geometry of Lie Groups by Hilgert and Neeb. This is in particular true for Lie subgroups, but maybe it's an assumption which would be more to your liking, since it seems you were bothered by relying too heavily on smoothness. The proof, however, naturally does use smoothness, so maybe that's not good enough.
I hope this answers your question.
answered Dec 19 '18 at 20:53
CronusCronus
1,108518
answered Dec 19 '18 at 20:53
CronusCronus
1,108518
answered Dec 19 '18 at 20:53
CronusCronus
1,108518
answered Dec 19 '18 at 20:53
CronusCronus
1,108518
1,108518
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add a comment |
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