Question about Lie subgroups and homeomorphisms
$begingroup$
I am reading about Lie groups and came across the definition of a Lie subgroup, which is simply an injective homomorphism of Lie groups H $hookrightarrow$G. A comment is made that H may not be homeomorphic to its iamge. with the example f:$mathbb{Z} rightarrow S^1$, where n $mapsto e^{in}$. The image is dense, which I understand.
Is the reason the example is not a homeomorphism because f$^{-1}$ is not continuous? For instance, there is no neighborhood of (1,0) whose preimage is a neighborhood of 0?
This example can be generalized to the torus if we think of the identification of the unit square and have a line through the origin with irrational slope. It seems this is an important example and I would like to be sure that I understand it. Thanks for any help and insight.
general-topology lie-groups
asked Jan 2 at 15:28
Joel PereiraJoel Pereira
83719
$endgroup$
add a comment |
$begingroup$
I am reading about Lie groups and came across the definition of a Lie subgroup, which is simply an injective homomorphism of Lie groups H $hookrightarrow$G. A comment is made that H may not be homeomorphic to its iamge. with the example f:$mathbb{Z} rightarrow S^1$, where n $mapsto e^{in}$. The image is dense, which I understand.
Is the reason the example is not a homeomorphism because f$^{-1}$ is not continuous? For instance, there is no neighborhood of (1,0) whose preimage is a neighborhood of 0?
This example can be generalized to the torus if we think of the identification of the unit square and have a line through the origin with irrational slope. It seems this is an important example and I would like to be sure that I understand it. Thanks for any help and insight.
general-topology lie-groups
asked Jan 2 at 15:28
Joel PereiraJoel Pereira
83719
$endgroup$
$begingroup$
$f$ is not surjective, for example $e^{ipi}in S^1$ is not in the image of $f$.
$endgroup$
– mouthetics
Jan 2 at 16:09
$begingroup$
Sorry, I meant homeomorphic onto its image. I have edited.
$endgroup$
– Joel Pereira
Jan 2 at 16:12
add a comment |
$begingroup$
I am reading about Lie groups and came across the definition of a Lie subgroup, which is simply an injective homomorphism of Lie groups H $hookrightarrow$G. A comment is made that H may not be homeomorphic to its iamge. with the example f:$mathbb{Z} rightarrow S^1$, where n $mapsto e^{in}$. The image is dense, which I understand.
Is the reason the example is not a homeomorphism because f$^{-1}$ is not continuous? For instance, there is no neighborhood of (1,0) whose preimage is a neighborhood of 0?
This example can be generalized to the torus if we think of the identification of the unit square and have a line through the origin with irrational slope. It seems this is an important example and I would like to be sure that I understand it. Thanks for any help and insight.
general-topology lie-groups
asked Jan 2 at 15:28
Joel PereiraJoel Pereira
83719
$endgroup$
I am reading about Lie groups and came across the definition of a Lie subgroup, which is simply an injective homomorphism of Lie groups H $hookrightarrow$G. A comment is made that H may not be homeomorphic to its iamge. with the example f:$mathbb{Z} rightarrow S^1$, where n $mapsto e^{in}$. The image is dense, which I understand.
Is the reason the example is not a homeomorphism because f$^{-1}$ is not continuous? For instance, there is no neighborhood of (1,0) whose preimage is a neighborhood of 0?
This example can be generalized to the torus if we think of the identification of the unit square and have a line through the origin with irrational slope. It seems this is an important example and I would like to be sure that I understand it. Thanks for any help and insight.
general-topology lie-groups
general-topology lie-groups
asked Jan 2 at 15:28
Joel PereiraJoel Pereira
83719
asked Jan 2 at 15:28
Joel PereiraJoel Pereira
83719
edited Jan 2 at 16:12
Joel Pereira
asked Jan 2 at 15:28
Joel PereiraJoel Pereira
83719
asked Jan 2 at 15:28
Joel PereiraJoel Pereira
83719
asked Jan 2 at 15:28
Joel PereiraJoel Pereira
83719
83719
$begingroup$
$f$ is not surjective, for example $e^{ipi}in S^1$ is not in the image of $f$.
$endgroup$
– mouthetics
Jan 2 at 16:09
$begingroup$
Sorry, I meant homeomorphic onto its image. I have edited.
$endgroup$
– Joel Pereira
Jan 2 at 16:12
add a comment |
$begingroup$
$f$ is not surjective, for example $e^{ipi}in S^1$ is not in the image of $f$.
$endgroup$
– mouthetics
Jan 2 at 16:09
$begingroup$
Sorry, I meant homeomorphic onto its image. I have edited.
$endgroup$
– Joel Pereira
Jan 2 at 16:12
$begingroup$
$f$ is not surjective, for example $e^{ipi}in S^1$ is not in the image of $f$.
$endgroup$
– mouthetics
Jan 2 at 16:09
$begingroup$
$f$ is not surjective, for example $e^{ipi}in S^1$ is not in the image of $f$.
$endgroup$
– mouthetics
Jan 2 at 16:09
$begingroup$
Sorry, I meant homeomorphic onto its image. I have edited.
$endgroup$
– Joel Pereira
Jan 2 at 16:12
$begingroup$
Sorry, I meant homeomorphic onto its image. I have edited.
$endgroup$
– Joel Pereira
Jan 2 at 16:12
add a comment |
1 Answer
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$begingroup$
Yes, $f^{-1}:mathrm{Im}(f)longrightarrow mathbb{Z}$ isn't continuous. For, ${ n}$ is open in $mathbb{Z}$ but ${e^{in}}$ is not open in $mathrm{Im}(f)$. If ${e^{in}}$ were open in $mathrm{Im}(f)$ then ${e^{in}}=mathrm{Im}(f)cap U$ for some open subset $Usubset S^1$ (a posteriori $Uneq {e^{in}}$ as a singleton is not open in $S^1$). But this is impossible since $mathrm{Im}(f)$ is dense in $S^1$.
answered Jan 2 at 18:49
moutheticsmouthetics
52137
$endgroup$
add a comment |
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1 Answer
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$begingroup$
Yes, $f^{-1}:mathrm{Im}(f)longrightarrow mathbb{Z}$ isn't continuous. For, ${ n}$ is open in $mathbb{Z}$ but ${e^{in}}$ is not open in $mathrm{Im}(f)$. If ${e^{in}}$ were open in $mathrm{Im}(f)$ then ${e^{in}}=mathrm{Im}(f)cap U$ for some open subset $Usubset S^1$ (a posteriori $Uneq {e^{in}}$ as a singleton is not open in $S^1$). But this is impossible since $mathrm{Im}(f)$ is dense in $S^1$.
answered Jan 2 at 18:49
moutheticsmouthetics
52137
$endgroup$
add a comment |
$begingroup$
Yes, $f^{-1}:mathrm{Im}(f)longrightarrow mathbb{Z}$ isn't continuous. For, ${ n}$ is open in $mathbb{Z}$ but ${e^{in}}$ is not open in $mathrm{Im}(f)$. If ${e^{in}}$ were open in $mathrm{Im}(f)$ then ${e^{in}}=mathrm{Im}(f)cap U$ for some open subset $Usubset S^1$ (a posteriori $Uneq {e^{in}}$ as a singleton is not open in $S^1$). But this is impossible since $mathrm{Im}(f)$ is dense in $S^1$.
answered Jan 2 at 18:49
moutheticsmouthetics
52137
$endgroup$
add a comment |
$begingroup$
Yes, $f^{-1}:mathrm{Im}(f)longrightarrow mathbb{Z}$ isn't continuous. For, ${ n}$ is open in $mathbb{Z}$ but ${e^{in}}$ is not open in $mathrm{Im}(f)$. If ${e^{in}}$ were open in $mathrm{Im}(f)$ then ${e^{in}}=mathrm{Im}(f)cap U$ for some open subset $Usubset S^1$ (a posteriori $Uneq {e^{in}}$ as a singleton is not open in $S^1$). But this is impossible since $mathrm{Im}(f)$ is dense in $S^1$.
answered Jan 2 at 18:49
moutheticsmouthetics
52137
$endgroup$
Yes, $f^{-1}:mathrm{Im}(f)longrightarrow mathbb{Z}$ isn't continuous. For, ${ n}$ is open in $mathbb{Z}$ but ${e^{in}}$ is not open in $mathrm{Im}(f)$. If ${e^{in}}$ were open in $mathrm{Im}(f)$ then ${e^{in}}=mathrm{Im}(f)cap U$ for some open subset $Usubset S^1$ (a posteriori $Uneq {e^{in}}$ as a singleton is not open in $S^1$). But this is impossible since $mathrm{Im}(f)$ is dense in $S^1$.
answered Jan 2 at 18:49
moutheticsmouthetics
52137
answered Jan 2 at 18:49
moutheticsmouthetics
52137
answered Jan 2 at 18:49
moutheticsmouthetics
52137
answered Jan 2 at 18:49
moutheticsmouthetics
52137
52137
add a comment |
add a comment |
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$begingroup$
$f$ is not surjective, for example $e^{ipi}in S^1$ is not in the image of $f$.
$endgroup$
– mouthetics
Jan 2 at 16:09
$begingroup$
Sorry, I meant homeomorphic onto its image. I have edited.
$endgroup$
– Joel Pereira
Jan 2 at 16:12