Mixing
Show that the following is not a differentiable submanifold
$begingroup$
I am studying a basic course in differentiable manifolds.I am confused in the following:
Consider $f: mathbb R to mathbb R^2$ given by $f(t)=(t^2,t^3)$.
I have shown that it is a smooth topological embedding onto its image but I do not understand
why is this not differentiable submanifold of $mathbb R^2$
can we make it as an immersed submanifold in $mathbb R^2$
any help would be appericiated...
differential-geometry manifolds smooth-manifolds
edited Jan 17 at 21:30
Henrik
6,03792030
asked Jan 17 at 21:00
user123user123
134
$endgroup$
|
show 6 more comments
$begingroup$
I am studying a basic course in differentiable manifolds.I am confused in the following:
Consider $f: mathbb R to mathbb R^2$ given by $f(t)=(t^2,t^3)$.
I have shown that it is a smooth topological embedding onto its image but I do not understand
why is this not differentiable submanifold of $mathbb R^2$
can we make it as an immersed submanifold in $mathbb R^2$
any help would be appericiated...
differential-geometry manifolds smooth-manifolds
edited Jan 17 at 21:30
Henrik
6,03792030
asked Jan 17 at 21:00
user123user123
134
$endgroup$
$begingroup$
your f is $mathbb R^1 tomathbb R^2$
$endgroup$
– janmarqz
Jan 17 at 21:05
$begingroup$
Oh sorry...i will edit question
$endgroup$
– user123
Jan 17 at 21:06
1
$begingroup$
Welcome to MSE! I have made some changes in your question. Just code, nothing relevant. You can have a look at to see how it changes. Basically I've removed lot of $ symbols.
$endgroup$
– Dog_69
Jan 17 at 21:09
1
$begingroup$
It is not smooth. Do you consider the absolute value function fo be locally Euclidean at the origin?
$endgroup$
– John Douma
Jan 17 at 21:29
1
$begingroup$
See, for example, the answer to this question or this question.
$endgroup$
– Ted Shifrin
Jan 17 at 21:37
|
show 6 more comments
$begingroup$
I am studying a basic course in differentiable manifolds.I am confused in the following:
Consider $f: mathbb R to mathbb R^2$ given by $f(t)=(t^2,t^3)$.
I have shown that it is a smooth topological embedding onto its image but I do not understand
why is this not differentiable submanifold of $mathbb R^2$
can we make it as an immersed submanifold in $mathbb R^2$
any help would be appericiated...
differential-geometry manifolds smooth-manifolds
edited Jan 17 at 21:30
Henrik
6,03792030
asked Jan 17 at 21:00
user123user123
134
$endgroup$
I am studying a basic course in differentiable manifolds.I am confused in the following:
Consider $f: mathbb R to mathbb R^2$ given by $f(t)=(t^2,t^3)$.
I have shown that it is a smooth topological embedding onto its image but I do not understand
why is this not differentiable submanifold of $mathbb R^2$
can we make it as an immersed submanifold in $mathbb R^2$
any help would be appericiated...
differential-geometry manifolds smooth-manifolds
differential-geometry manifolds smooth-manifolds
edited Jan 17 at 21:30
Henrik
6,03792030
asked Jan 17 at 21:00
user123user123
134
edited Jan 17 at 21:30
Henrik
6,03792030
asked Jan 17 at 21:00
user123user123
134
edited Jan 17 at 21:30
Henrik
6,03792030
edited Jan 17 at 21:30
Henrik
6,03792030
edited Jan 17 at 21:30
Henrik
6,03792030
6,03792030
asked Jan 17 at 21:00
user123user123
134
asked Jan 17 at 21:00
user123user123
134
asked Jan 17 at 21:00
user123user123
134
134
$begingroup$
your f is $mathbb R^1 tomathbb R^2$
$endgroup$
– janmarqz
Jan 17 at 21:05
$begingroup$
Oh sorry...i will edit question
$endgroup$
– user123
Jan 17 at 21:06
1
$begingroup$
Welcome to MSE! I have made some changes in your question. Just code, nothing relevant. You can have a look at to see how it changes. Basically I've removed lot of $ symbols.
$endgroup$
– Dog_69
Jan 17 at 21:09
1
$begingroup$
It is not smooth. Do you consider the absolute value function fo be locally Euclidean at the origin?
$endgroup$
– John Douma
Jan 17 at 21:29
1
$begingroup$
See, for example, the answer to this question or this question.
$endgroup$
– Ted Shifrin
Jan 17 at 21:37
|
show 6 more comments
$begingroup$
your f is $mathbb R^1 tomathbb R^2$
$endgroup$
– janmarqz
Jan 17 at 21:05
$begingroup$
Oh sorry...i will edit question
$endgroup$
– user123
Jan 17 at 21:06
1
$begingroup$
Welcome to MSE! I have made some changes in your question. Just code, nothing relevant. You can have a look at to see how it changes. Basically I've removed lot of $ symbols.
$endgroup$
– Dog_69
Jan 17 at 21:09
1
$begingroup$
It is not smooth. Do you consider the absolute value function fo be locally Euclidean at the origin?
$endgroup$
– John Douma
Jan 17 at 21:29
1
$begingroup$
See, for example, the answer to this question or this question.
$endgroup$
– Ted Shifrin
Jan 17 at 21:37
$begingroup$
your f is $mathbb R^1 tomathbb R^2$
$endgroup$
– janmarqz
Jan 17 at 21:05
$begingroup$
your f is $mathbb R^1 tomathbb R^2$
$endgroup$
– janmarqz
Jan 17 at 21:05
$begingroup$
Oh sorry...i will edit question
$endgroup$
– user123
Jan 17 at 21:06
$begingroup$
Oh sorry...i will edit question
$endgroup$
– user123
Jan 17 at 21:06
1
1
$begingroup$
Welcome to MSE! I have made some changes in your question. Just code, nothing relevant. You can have a look at to see how it changes. Basically I've removed lot of $ symbols.
$endgroup$
– Dog_69
Jan 17 at 21:09
$begingroup$
Welcome to MSE! I have made some changes in your question. Just code, nothing relevant. You can have a look at to see how it changes. Basically I've removed lot of $ symbols.
$endgroup$
– Dog_69
Jan 17 at 21:09
1
1
$begingroup$
It is not smooth. Do you consider the absolute value function fo be locally Euclidean at the origin?
$endgroup$
– John Douma
Jan 17 at 21:29
$begingroup$
It is not smooth. Do you consider the absolute value function fo be locally Euclidean at the origin?
$endgroup$
– John Douma
Jan 17 at 21:29
1
1
$begingroup$
See, for example, the answer to this question or this question.
$endgroup$
– Ted Shifrin
Jan 17 at 21:37
$begingroup$
See, for example, the answer to this question or this question.
$endgroup$
– Ted Shifrin
Jan 17 at 21:37
|
show 6 more comments
1 Answer
1
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oldest
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$begingroup$
Let $N$ denote the image of $f$. Suppose $N$ were an immersed submanifold of $mathbb{R}^2$ with the standard differentiable structure. Suppose $g:Mto mathbb{R}^2$ is an immersion with $g(M) = N$. Consider the map $p:mathbb{R}^2to mathbb{R}$ given in standard coordinates by $p(x,y) = x^3-y^2$. Of course $p$ is a smooth map, and $pcirc g = 0$, so by the chain rule, $mbox{im}, dg subset ker dp$, and since $ker dp$ has dimension $1$ away from $(0,0)$, we have $mbox{im}, dg$ has dimension at most $1$, and clearly $dim M ge 1$, so $dim M = 1$. Let $ain M$ be such that $g(a) = (0,0)$. Since $g$ is an immersion, $g$ is locally injective at $a$. Let $U$ be an open subset of $M$ containing $a$ on which $g$ is injective with a coordinate chart (diffeomorphism) $phi:U to B_delta(0)subsetmathbb{R}$, the open ball of radius $delta>0$ around $0$ in $mathbb{R}$, with $phi(a) = 0$. Let $h = gcirc phi^{-1}$ and let $t$ be the standard coordinate on $B_delta(0)subsetmathbb{R}$. Note that $h$ is also an injective immersion. Write $h(t) = (h_1(t),h_2(t))$. We have $h_1(t) = h_2(t)^{2/3}$, so
$$h_1'(t) = frac{2}{3}h_2(t)^{-1/3}h_2'(t)$$
for $tne 0$. Since $h_2'$ is continuous, we must have $h_2'(0) = 0$, otherwise $h_1'(t)$ would not exist at $0$. So since $dh_0$ is nonzero, we must have $h_1'(0) ne 0$. Without loss of generality, we assume $h_1'(0) > 0$. Thus there is some $deltageepsilon>0$ so that $h_1$ is increasing in $B_epsilon(0)$. But then for some $-epsilon<t_0<0$ we must have $h_1(t_0)<h_1(0)=0$, and so $h(t_0)notin N$, a contradiction.
answered Jan 18 at 2:15
cspruncsprun
1,54828
$endgroup$
add a comment |
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1 Answer
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$begingroup$
Let $N$ denote the image of $f$. Suppose $N$ were an immersed submanifold of $mathbb{R}^2$ with the standard differentiable structure. Suppose $g:Mto mathbb{R}^2$ is an immersion with $g(M) = N$. Consider the map $p:mathbb{R}^2to mathbb{R}$ given in standard coordinates by $p(x,y) = x^3-y^2$. Of course $p$ is a smooth map, and $pcirc g = 0$, so by the chain rule, $mbox{im}, dg subset ker dp$, and since $ker dp$ has dimension $1$ away from $(0,0)$, we have $mbox{im}, dg$ has dimension at most $1$, and clearly $dim M ge 1$, so $dim M = 1$. Let $ain M$ be such that $g(a) = (0,0)$. Since $g$ is an immersion, $g$ is locally injective at $a$. Let $U$ be an open subset of $M$ containing $a$ on which $g$ is injective with a coordinate chart (diffeomorphism) $phi:U to B_delta(0)subsetmathbb{R}$, the open ball of radius $delta>0$ around $0$ in $mathbb{R}$, with $phi(a) = 0$. Let $h = gcirc phi^{-1}$ and let $t$ be the standard coordinate on $B_delta(0)subsetmathbb{R}$. Note that $h$ is also an injective immersion. Write $h(t) = (h_1(t),h_2(t))$. We have $h_1(t) = h_2(t)^{2/3}$, so
$$h_1'(t) = frac{2}{3}h_2(t)^{-1/3}h_2'(t)$$
for $tne 0$. Since $h_2'$ is continuous, we must have $h_2'(0) = 0$, otherwise $h_1'(t)$ would not exist at $0$. So since $dh_0$ is nonzero, we must have $h_1'(0) ne 0$. Without loss of generality, we assume $h_1'(0) > 0$. Thus there is some $deltageepsilon>0$ so that $h_1$ is increasing in $B_epsilon(0)$. But then for some $-epsilon<t_0<0$ we must have $h_1(t_0)<h_1(0)=0$, and so $h(t_0)notin N$, a contradiction.
answered Jan 18 at 2:15
cspruncsprun
1,54828
$endgroup$
add a comment |
$begingroup$
Let $N$ denote the image of $f$. Suppose $N$ were an immersed submanifold of $mathbb{R}^2$ with the standard differentiable structure. Suppose $g:Mto mathbb{R}^2$ is an immersion with $g(M) = N$. Consider the map $p:mathbb{R}^2to mathbb{R}$ given in standard coordinates by $p(x,y) = x^3-y^2$. Of course $p$ is a smooth map, and $pcirc g = 0$, so by the chain rule, $mbox{im}, dg subset ker dp$, and since $ker dp$ has dimension $1$ away from $(0,0)$, we have $mbox{im}, dg$ has dimension at most $1$, and clearly $dim M ge 1$, so $dim M = 1$. Let $ain M$ be such that $g(a) = (0,0)$. Since $g$ is an immersion, $g$ is locally injective at $a$. Let $U$ be an open subset of $M$ containing $a$ on which $g$ is injective with a coordinate chart (diffeomorphism) $phi:U to B_delta(0)subsetmathbb{R}$, the open ball of radius $delta>0$ around $0$ in $mathbb{R}$, with $phi(a) = 0$. Let $h = gcirc phi^{-1}$ and let $t$ be the standard coordinate on $B_delta(0)subsetmathbb{R}$. Note that $h$ is also an injective immersion. Write $h(t) = (h_1(t),h_2(t))$. We have $h_1(t) = h_2(t)^{2/3}$, so
$$h_1'(t) = frac{2}{3}h_2(t)^{-1/3}h_2'(t)$$
for $tne 0$. Since $h_2'$ is continuous, we must have $h_2'(0) = 0$, otherwise $h_1'(t)$ would not exist at $0$. So since $dh_0$ is nonzero, we must have $h_1'(0) ne 0$. Without loss of generality, we assume $h_1'(0) > 0$. Thus there is some $deltageepsilon>0$ so that $h_1$ is increasing in $B_epsilon(0)$. But then for some $-epsilon<t_0<0$ we must have $h_1(t_0)<h_1(0)=0$, and so $h(t_0)notin N$, a contradiction.
answered Jan 18 at 2:15
cspruncsprun
1,54828
$endgroup$
add a comment |
$begingroup$
Let $N$ denote the image of $f$. Suppose $N$ were an immersed submanifold of $mathbb{R}^2$ with the standard differentiable structure. Suppose $g:Mto mathbb{R}^2$ is an immersion with $g(M) = N$. Consider the map $p:mathbb{R}^2to mathbb{R}$ given in standard coordinates by $p(x,y) = x^3-y^2$. Of course $p$ is a smooth map, and $pcirc g = 0$, so by the chain rule, $mbox{im}, dg subset ker dp$, and since $ker dp$ has dimension $1$ away from $(0,0)$, we have $mbox{im}, dg$ has dimension at most $1$, and clearly $dim M ge 1$, so $dim M = 1$. Let $ain M$ be such that $g(a) = (0,0)$. Since $g$ is an immersion, $g$ is locally injective at $a$. Let $U$ be an open subset of $M$ containing $a$ on which $g$ is injective with a coordinate chart (diffeomorphism) $phi:U to B_delta(0)subsetmathbb{R}$, the open ball of radius $delta>0$ around $0$ in $mathbb{R}$, with $phi(a) = 0$. Let $h = gcirc phi^{-1}$ and let $t$ be the standard coordinate on $B_delta(0)subsetmathbb{R}$. Note that $h$ is also an injective immersion. Write $h(t) = (h_1(t),h_2(t))$. We have $h_1(t) = h_2(t)^{2/3}$, so
$$h_1'(t) = frac{2}{3}h_2(t)^{-1/3}h_2'(t)$$
for $tne 0$. Since $h_2'$ is continuous, we must have $h_2'(0) = 0$, otherwise $h_1'(t)$ would not exist at $0$. So since $dh_0$ is nonzero, we must have $h_1'(0) ne 0$. Without loss of generality, we assume $h_1'(0) > 0$. Thus there is some $deltageepsilon>0$ so that $h_1$ is increasing in $B_epsilon(0)$. But then for some $-epsilon<t_0<0$ we must have $h_1(t_0)<h_1(0)=0$, and so $h(t_0)notin N$, a contradiction.
answered Jan 18 at 2:15
cspruncsprun
1,54828
$endgroup$
Let $N$ denote the image of $f$. Suppose $N$ were an immersed submanifold of $mathbb{R}^2$ with the standard differentiable structure. Suppose $g:Mto mathbb{R}^2$ is an immersion with $g(M) = N$. Consider the map $p:mathbb{R}^2to mathbb{R}$ given in standard coordinates by $p(x,y) = x^3-y^2$. Of course $p$ is a smooth map, and $pcirc g = 0$, so by the chain rule, $mbox{im}, dg subset ker dp$, and since $ker dp$ has dimension $1$ away from $(0,0)$, we have $mbox{im}, dg$ has dimension at most $1$, and clearly $dim M ge 1$, so $dim M = 1$. Let $ain M$ be such that $g(a) = (0,0)$. Since $g$ is an immersion, $g$ is locally injective at $a$. Let $U$ be an open subset of $M$ containing $a$ on which $g$ is injective with a coordinate chart (diffeomorphism) $phi:U to B_delta(0)subsetmathbb{R}$, the open ball of radius $delta>0$ around $0$ in $mathbb{R}$, with $phi(a) = 0$. Let $h = gcirc phi^{-1}$ and let $t$ be the standard coordinate on $B_delta(0)subsetmathbb{R}$. Note that $h$ is also an injective immersion. Write $h(t) = (h_1(t),h_2(t))$. We have $h_1(t) = h_2(t)^{2/3}$, so
$$h_1'(t) = frac{2}{3}h_2(t)^{-1/3}h_2'(t)$$
for $tne 0$. Since $h_2'$ is continuous, we must have $h_2'(0) = 0$, otherwise $h_1'(t)$ would not exist at $0$. So since $dh_0$ is nonzero, we must have $h_1'(0) ne 0$. Without loss of generality, we assume $h_1'(0) > 0$. Thus there is some $deltageepsilon>0$ so that $h_1$ is increasing in $B_epsilon(0)$. But then for some $-epsilon<t_0<0$ we must have $h_1(t_0)<h_1(0)=0$, and so $h(t_0)notin N$, a contradiction.
answered Jan 18 at 2:15
cspruncsprun
1,54828
answered Jan 18 at 2:15
cspruncsprun
1,54828
answered Jan 18 at 2:15
cspruncsprun
1,54828
answered Jan 18 at 2:15
cspruncsprun
1,54828
1,54828
add a comment |
add a comment |
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$begingroup$
your f is $mathbb R^1 tomathbb R^2$
$endgroup$
– janmarqz
Jan 17 at 21:05
$begingroup$
Oh sorry...i will edit question
$endgroup$
– user123
Jan 17 at 21:06
1
$begingroup$
Welcome to MSE! I have made some changes in your question. Just code, nothing relevant. You can have a look at to see how it changes. Basically I've removed lot of $ symbols.
$endgroup$
– Dog_69
Jan 17 at 21:09
1
$begingroup$
It is not smooth. Do you consider the absolute value function fo be locally Euclidean at the origin?
$endgroup$
– John Douma
Jan 17 at 21:29
1
$begingroup$
See, for example, the answer to this question or this question.
$endgroup$
– Ted Shifrin
Jan 17 at 21:37