Is the image of $f: R to S^1 times S^1, f(t) = (e^{it}, e^{sqrt 2it})$ a submanifold?
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Let $f: R to S^1 times S^1$ be $f(t) = (e^{it}, e^{sqrt 2it})$,is $f(mathbb R)$ a submanifold? By easy verification, $f$ is an injective immersion. Then if I want to show that the image is a submanifold, I will want to show that $f$ is a smooth embedding. It remains to show that the inverse is continuous. Can the inverse be written down explicitly?
differential-topology
asked Dec 10 '18 at 22:44
pennypenny
635
$endgroup$
add a comment |
$begingroup$
Let $f: R to S^1 times S^1$ be $f(t) = (e^{it}, e^{sqrt 2it})$,is $f(mathbb R)$ a submanifold? By easy verification, $f$ is an injective immersion. Then if I want to show that the image is a submanifold, I will want to show that $f$ is a smooth embedding. It remains to show that the inverse is continuous. Can the inverse be written down explicitly?
differential-topology
asked Dec 10 '18 at 22:44
pennypenny
635
$endgroup$
1
$begingroup$
The image is not a sub-manifold. I have an argument which you may not like: first you notice that $f$ is a homomorphism (from the additive $mathbb{R}$ to the multiplicative $S^1times S^1$). So the image is a subgroup of $S^1times S^1$. If it is a manifold then it is a Lie group, hence closed in $S^1times S^1$. But in this case the image is dense. I believe you can find another argument but If you want to prove that it is a sub-manifold I think you will fail...
$endgroup$
– Yanko
Dec 10 '18 at 23:04
$begingroup$
I am not familiar with Lie groups, so I guess I need to find a way to show that the image is a closed subset.
$endgroup$
– penny
Dec 10 '18 at 23:10
$begingroup$
The only thing you have to show is that $f(mathbb{R})$ is dense in $S^1 times S^1$. A dense subset can never be a submanifold (consider a submanifold chart).
$endgroup$
– Paul Frost
Dec 11 '18 at 0:21
$begingroup$
Thak makes sense. Thank you! Could you maybe write something as an answer so that I can accept it.@PaulFrost
$endgroup$
– penny
Dec 11 '18 at 4:12
add a comment |
$begingroup$
Let $f: R to S^1 times S^1$ be $f(t) = (e^{it}, e^{sqrt 2it})$,is $f(mathbb R)$ a submanifold? By easy verification, $f$ is an injective immersion. Then if I want to show that the image is a submanifold, I will want to show that $f$ is a smooth embedding. It remains to show that the inverse is continuous. Can the inverse be written down explicitly?
differential-topology
asked Dec 10 '18 at 22:44
pennypenny
635
$endgroup$
Let $f: R to S^1 times S^1$ be $f(t) = (e^{it}, e^{sqrt 2it})$,is $f(mathbb R)$ a submanifold? By easy verification, $f$ is an injective immersion. Then if I want to show that the image is a submanifold, I will want to show that $f$ is a smooth embedding. It remains to show that the inverse is continuous. Can the inverse be written down explicitly?
differential-topology
differential-topology
asked Dec 10 '18 at 22:44
pennypenny
635
asked Dec 10 '18 at 22:44
pennypenny
635
asked Dec 10 '18 at 22:44
pennypenny
635
asked Dec 10 '18 at 22:44
pennypenny
635
asked Dec 10 '18 at 22:44
pennypenny
635
635
1
$begingroup$
The image is not a sub-manifold. I have an argument which you may not like: first you notice that $f$ is a homomorphism (from the additive $mathbb{R}$ to the multiplicative $S^1times S^1$). So the image is a subgroup of $S^1times S^1$. If it is a manifold then it is a Lie group, hence closed in $S^1times S^1$. But in this case the image is dense. I believe you can find another argument but If you want to prove that it is a sub-manifold I think you will fail...
$endgroup$
– Yanko
Dec 10 '18 at 23:04
$begingroup$
I am not familiar with Lie groups, so I guess I need to find a way to show that the image is a closed subset.
$endgroup$
– penny
Dec 10 '18 at 23:10
$begingroup$
The only thing you have to show is that $f(mathbb{R})$ is dense in $S^1 times S^1$. A dense subset can never be a submanifold (consider a submanifold chart).
$endgroup$
– Paul Frost
Dec 11 '18 at 0:21
$begingroup$
Thak makes sense. Thank you! Could you maybe write something as an answer so that I can accept it.@PaulFrost
$endgroup$
– penny
Dec 11 '18 at 4:12
add a comment |
1
$begingroup$
The image is not a sub-manifold. I have an argument which you may not like: first you notice that $f$ is a homomorphism (from the additive $mathbb{R}$ to the multiplicative $S^1times S^1$). So the image is a subgroup of $S^1times S^1$. If it is a manifold then it is a Lie group, hence closed in $S^1times S^1$. But in this case the image is dense. I believe you can find another argument but If you want to prove that it is a sub-manifold I think you will fail...
$endgroup$
– Yanko
Dec 10 '18 at 23:04
$begingroup$
I am not familiar with Lie groups, so I guess I need to find a way to show that the image is a closed subset.
$endgroup$
– penny
Dec 10 '18 at 23:10
$begingroup$
The only thing you have to show is that $f(mathbb{R})$ is dense in $S^1 times S^1$. A dense subset can never be a submanifold (consider a submanifold chart).
$endgroup$
– Paul Frost
Dec 11 '18 at 0:21
$begingroup$
Thak makes sense. Thank you! Could you maybe write something as an answer so that I can accept it.@PaulFrost
$endgroup$
– penny
Dec 11 '18 at 4:12
1
1
$begingroup$
The image is not a sub-manifold. I have an argument which you may not like: first you notice that $f$ is a homomorphism (from the additive $mathbb{R}$ to the multiplicative $S^1times S^1$). So the image is a subgroup of $S^1times S^1$. If it is a manifold then it is a Lie group, hence closed in $S^1times S^1$. But in this case the image is dense. I believe you can find another argument but If you want to prove that it is a sub-manifold I think you will fail...
$endgroup$
– Yanko
Dec 10 '18 at 23:04
$begingroup$
The image is not a sub-manifold. I have an argument which you may not like: first you notice that $f$ is a homomorphism (from the additive $mathbb{R}$ to the multiplicative $S^1times S^1$). So the image is a subgroup of $S^1times S^1$. If it is a manifold then it is a Lie group, hence closed in $S^1times S^1$. But in this case the image is dense. I believe you can find another argument but If you want to prove that it is a sub-manifold I think you will fail...
$endgroup$
– Yanko
Dec 10 '18 at 23:04
$begingroup$
I am not familiar with Lie groups, so I guess I need to find a way to show that the image is a closed subset.
$endgroup$
– penny
Dec 10 '18 at 23:10
$begingroup$
I am not familiar with Lie groups, so I guess I need to find a way to show that the image is a closed subset.
$endgroup$
– penny
Dec 10 '18 at 23:10
$begingroup$
The only thing you have to show is that $f(mathbb{R})$ is dense in $S^1 times S^1$. A dense subset can never be a submanifold (consider a submanifold chart).
$endgroup$
– Paul Frost
Dec 11 '18 at 0:21
$begingroup$
The only thing you have to show is that $f(mathbb{R})$ is dense in $S^1 times S^1$. A dense subset can never be a submanifold (consider a submanifold chart).
$endgroup$
– Paul Frost
Dec 11 '18 at 0:21
$begingroup$
Thak makes sense. Thank you! Could you maybe write something as an answer so that I can accept it.@PaulFrost
$endgroup$
– penny
Dec 11 '18 at 4:12
$begingroup$
Thak makes sense. Thank you! Could you maybe write something as an answer so that I can accept it.@PaulFrost
$endgroup$
– penny
Dec 11 '18 at 4:12
add a comment |
1 Answer
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Let us prove that $f(mathbb{R})$ is dense in $T = S^1 times S^1$. This will show that $f(mathbb{R})$ is not a submanifold of $T$ (since a submanifold $S$ of a manifold $M$ is never dense).
It is a well-known result that any infinite subgroup of $S^1$ (endowed with complex multplication) is dense. See for examaple Subgroup of the unit circle under complex multiplication.
Consider $(x,y) in T$ and $epsilon > 0$. Choose $t in mathbb{R}$ such that $e^{it} = x$. Then for each $t_k = t + 2kpi$, $k in mathbb{Z}$, we also have $e^{it_k} = x$.
But now the map $g : mathbb{Z} to S^1, g(k) = e^{2sqrt 2 pi i k}$, is an injective group homomorphism. Hence its image is an infinite subgroup of $S^1$ and therefore dense in $S^1$. Choose $k$ such that $lvert x g(k) - y rvert < epsilon$ (note that $mu_x : S^1 to S^1, mu_x(z) = xz$, is a homeomorphism, hence $mu_x g(mathbb{Z})$ is dense in $S^1$). Then
$$e^{sqrt 2i t_k} = e^{sqrt 2i t} e^{2sqrt 2pi i k} = xg(k)$$
and
$$lvert e^{sqrt 2i t_k} - y rvert < epsilon .$$
This show that each neighborhood of $(x,y)$ in $T$ contains an element of $f(mathbb{R})$.
answered Dec 11 '18 at 20:53
Paul FrostPaul Frost
10.3k3933
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$begingroup$
Let us prove that $f(mathbb{R})$ is dense in $T = S^1 times S^1$. This will show that $f(mathbb{R})$ is not a submanifold of $T$ (since a submanifold $S$ of a manifold $M$ is never dense).
It is a well-known result that any infinite subgroup of $S^1$ (endowed with complex multplication) is dense. See for examaple Subgroup of the unit circle under complex multiplication.
Consider $(x,y) in T$ and $epsilon > 0$. Choose $t in mathbb{R}$ such that $e^{it} = x$. Then for each $t_k = t + 2kpi$, $k in mathbb{Z}$, we also have $e^{it_k} = x$.
But now the map $g : mathbb{Z} to S^1, g(k) = e^{2sqrt 2 pi i k}$, is an injective group homomorphism. Hence its image is an infinite subgroup of $S^1$ and therefore dense in $S^1$. Choose $k$ such that $lvert x g(k) - y rvert < epsilon$ (note that $mu_x : S^1 to S^1, mu_x(z) = xz$, is a homeomorphism, hence $mu_x g(mathbb{Z})$ is dense in $S^1$). Then
$$e^{sqrt 2i t_k} = e^{sqrt 2i t} e^{2sqrt 2pi i k} = xg(k)$$
and
$$lvert e^{sqrt 2i t_k} - y rvert < epsilon .$$
This show that each neighborhood of $(x,y)$ in $T$ contains an element of $f(mathbb{R})$.
answered Dec 11 '18 at 20:53
Paul FrostPaul Frost
10.3k3933
$endgroup$
add a comment |
$begingroup$
Let us prove that $f(mathbb{R})$ is dense in $T = S^1 times S^1$. This will show that $f(mathbb{R})$ is not a submanifold of $T$ (since a submanifold $S$ of a manifold $M$ is never dense).
It is a well-known result that any infinite subgroup of $S^1$ (endowed with complex multplication) is dense. See for examaple Subgroup of the unit circle under complex multiplication.
Consider $(x,y) in T$ and $epsilon > 0$. Choose $t in mathbb{R}$ such that $e^{it} = x$. Then for each $t_k = t + 2kpi$, $k in mathbb{Z}$, we also have $e^{it_k} = x$.
But now the map $g : mathbb{Z} to S^1, g(k) = e^{2sqrt 2 pi i k}$, is an injective group homomorphism. Hence its image is an infinite subgroup of $S^1$ and therefore dense in $S^1$. Choose $k$ such that $lvert x g(k) - y rvert < epsilon$ (note that $mu_x : S^1 to S^1, mu_x(z) = xz$, is a homeomorphism, hence $mu_x g(mathbb{Z})$ is dense in $S^1$). Then
$$e^{sqrt 2i t_k} = e^{sqrt 2i t} e^{2sqrt 2pi i k} = xg(k)$$
and
$$lvert e^{sqrt 2i t_k} - y rvert < epsilon .$$
This show that each neighborhood of $(x,y)$ in $T$ contains an element of $f(mathbb{R})$.
answered Dec 11 '18 at 20:53
Paul FrostPaul Frost
10.3k3933
$endgroup$
add a comment |
$begingroup$
Let us prove that $f(mathbb{R})$ is dense in $T = S^1 times S^1$. This will show that $f(mathbb{R})$ is not a submanifold of $T$ (since a submanifold $S$ of a manifold $M$ is never dense).
It is a well-known result that any infinite subgroup of $S^1$ (endowed with complex multplication) is dense. See for examaple Subgroup of the unit circle under complex multiplication.
Consider $(x,y) in T$ and $epsilon > 0$. Choose $t in mathbb{R}$ such that $e^{it} = x$. Then for each $t_k = t + 2kpi$, $k in mathbb{Z}$, we also have $e^{it_k} = x$.
But now the map $g : mathbb{Z} to S^1, g(k) = e^{2sqrt 2 pi i k}$, is an injective group homomorphism. Hence its image is an infinite subgroup of $S^1$ and therefore dense in $S^1$. Choose $k$ such that $lvert x g(k) - y rvert < epsilon$ (note that $mu_x : S^1 to S^1, mu_x(z) = xz$, is a homeomorphism, hence $mu_x g(mathbb{Z})$ is dense in $S^1$). Then
$$e^{sqrt 2i t_k} = e^{sqrt 2i t} e^{2sqrt 2pi i k} = xg(k)$$
and
$$lvert e^{sqrt 2i t_k} - y rvert < epsilon .$$
This show that each neighborhood of $(x,y)$ in $T$ contains an element of $f(mathbb{R})$.
answered Dec 11 '18 at 20:53
Paul FrostPaul Frost
10.3k3933
$endgroup$
Let us prove that $f(mathbb{R})$ is dense in $T = S^1 times S^1$. This will show that $f(mathbb{R})$ is not a submanifold of $T$ (since a submanifold $S$ of a manifold $M$ is never dense).
It is a well-known result that any infinite subgroup of $S^1$ (endowed with complex multplication) is dense. See for examaple Subgroup of the unit circle under complex multiplication.
Consider $(x,y) in T$ and $epsilon > 0$. Choose $t in mathbb{R}$ such that $e^{it} = x$. Then for each $t_k = t + 2kpi$, $k in mathbb{Z}$, we also have $e^{it_k} = x$.
But now the map $g : mathbb{Z} to S^1, g(k) = e^{2sqrt 2 pi i k}$, is an injective group homomorphism. Hence its image is an infinite subgroup of $S^1$ and therefore dense in $S^1$. Choose $k$ such that $lvert x g(k) - y rvert < epsilon$ (note that $mu_x : S^1 to S^1, mu_x(z) = xz$, is a homeomorphism, hence $mu_x g(mathbb{Z})$ is dense in $S^1$). Then
$$e^{sqrt 2i t_k} = e^{sqrt 2i t} e^{2sqrt 2pi i k} = xg(k)$$
and
$$lvert e^{sqrt 2i t_k} - y rvert < epsilon .$$
This show that each neighborhood of $(x,y)$ in $T$ contains an element of $f(mathbb{R})$.
answered Dec 11 '18 at 20:53
Paul FrostPaul Frost
10.3k3933
answered Dec 11 '18 at 20:53
Paul FrostPaul Frost
10.3k3933
answered Dec 11 '18 at 20:53
Paul FrostPaul Frost
10.3k3933
answered Dec 11 '18 at 20:53
Paul FrostPaul Frost
10.3k3933
10.3k3933
add a comment |
add a comment |
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1
$begingroup$
The image is not a sub-manifold. I have an argument which you may not like: first you notice that $f$ is a homomorphism (from the additive $mathbb{R}$ to the multiplicative $S^1times S^1$). So the image is a subgroup of $S^1times S^1$. If it is a manifold then it is a Lie group, hence closed in $S^1times S^1$. But in this case the image is dense. I believe you can find another argument but If you want to prove that it is a sub-manifold I think you will fail...
$endgroup$
– Yanko
Dec 10 '18 at 23:04
$begingroup$
I am not familiar with Lie groups, so I guess I need to find a way to show that the image is a closed subset.
$endgroup$
– penny
Dec 10 '18 at 23:10
$begingroup$
The only thing you have to show is that $f(mathbb{R})$ is dense in $S^1 times S^1$. A dense subset can never be a submanifold (consider a submanifold chart).
$endgroup$
– Paul Frost
Dec 11 '18 at 0:21
$begingroup$
Thak makes sense. Thank you! Could you maybe write something as an answer so that I can accept it.@PaulFrost
$endgroup$
– penny
Dec 11 '18 at 4:12