checking $C^infty$-manifold analytically
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We look at the set $C : =lbrace (x,y) in mathbb{R}^2 mid y^2 = x^2+x^3 rbrace$.
We want to know if $C$ is a $C^infty$-manifold. When looking at the plot it is easy to see that $(0,0)$ is the problem here but we are interested in solving this analytically.
With $lbrace (x,y) in mathbb{R}^2 mid y^3 = x^2 rbrace$ we can see that a projection $phi$ onto the $x$-axis is homeomorph but $phi^{-1}$ is not continuously differentiable. So there is no chart.
How would you argument on $C$ while dont have any idea how the set looks?
calculus manifolds
asked Nov 22 at 14:58
Arjihad
378111
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We look at the set $C : =lbrace (x,y) in mathbb{R}^2 mid y^2 = x^2+x^3 rbrace$.
We want to know if $C$ is a $C^infty$-manifold. When looking at the plot it is easy to see that $(0,0)$ is the problem here but we are interested in solving this analytically.
With $lbrace (x,y) in mathbb{R}^2 mid y^3 = x^2 rbrace$ we can see that a projection $phi$ onto the $x$-axis is homeomorph but $phi^{-1}$ is not continuously differentiable. So there is no chart.
How would you argument on $C$ while dont have any idea how the set looks?
calculus manifolds
asked Nov 22 at 14:58
Arjihad
378111
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
We look at the set $C : =lbrace (x,y) in mathbb{R}^2 mid y^2 = x^2+x^3 rbrace$.
We want to know if $C$ is a $C^infty$-manifold. When looking at the plot it is easy to see that $(0,0)$ is the problem here but we are interested in solving this analytically.
With $lbrace (x,y) in mathbb{R}^2 mid y^3 = x^2 rbrace$ we can see that a projection $phi$ onto the $x$-axis is homeomorph but $phi^{-1}$ is not continuously differentiable. So there is no chart.
How would you argument on $C$ while dont have any idea how the set looks?
calculus manifolds
asked Nov 22 at 14:58
Arjihad
378111
We look at the set $C : =lbrace (x,y) in mathbb{R}^2 mid y^2 = x^2+x^3 rbrace$.
We want to know if $C$ is a $C^infty$-manifold. When looking at the plot it is easy to see that $(0,0)$ is the problem here but we are interested in solving this analytically.
With $lbrace (x,y) in mathbb{R}^2 mid y^3 = x^2 rbrace$ we can see that a projection $phi$ onto the $x$-axis is homeomorph but $phi^{-1}$ is not continuously differentiable. So there is no chart.
How would you argument on $C$ while dont have any idea how the set looks?
calculus manifolds
calculus manifolds
asked Nov 22 at 14:58
Arjihad
378111
asked Nov 22 at 14:58
Arjihad
378111
asked Nov 22 at 14:58
Arjihad
378111
asked Nov 22 at 14:58
Arjihad
378111
asked Nov 22 at 14:58
Arjihad
378111
378111
add a comment |
add a comment |
1 Answer
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It won't be a manifold although if you remove the point $(0,0)$ it will be a 1-dimensional manifold.
When $x$ is very close to 0, the $x^2$ term dominates, and so the set looks like the graph of $y=pm x$, a cross, which is not locally homeomorphic to an interval of $mathbb{R}$.
For points away from $(0,0)$ you can use the regular level set theorem (aka the preimage theorem).
answered Nov 22 at 22:37
user25959
1,559816
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1 Answer
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1 Answer
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active
oldest
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oldest
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active
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up vote
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down vote
It won't be a manifold although if you remove the point $(0,0)$ it will be a 1-dimensional manifold.
When $x$ is very close to 0, the $x^2$ term dominates, and so the set looks like the graph of $y=pm x$, a cross, which is not locally homeomorphic to an interval of $mathbb{R}$.
For points away from $(0,0)$ you can use the regular level set theorem (aka the preimage theorem).
answered Nov 22 at 22:37
user25959
1,559816
add a comment |
up vote
0
down vote
It won't be a manifold although if you remove the point $(0,0)$ it will be a 1-dimensional manifold.
When $x$ is very close to 0, the $x^2$ term dominates, and so the set looks like the graph of $y=pm x$, a cross, which is not locally homeomorphic to an interval of $mathbb{R}$.
For points away from $(0,0)$ you can use the regular level set theorem (aka the preimage theorem).
answered Nov 22 at 22:37
user25959
1,559816
add a comment |
up vote
0
down vote
up vote
0
down vote
It won't be a manifold although if you remove the point $(0,0)$ it will be a 1-dimensional manifold.
When $x$ is very close to 0, the $x^2$ term dominates, and so the set looks like the graph of $y=pm x$, a cross, which is not locally homeomorphic to an interval of $mathbb{R}$.
For points away from $(0,0)$ you can use the regular level set theorem (aka the preimage theorem).
answered Nov 22 at 22:37
user25959
1,559816
It won't be a manifold although if you remove the point $(0,0)$ it will be a 1-dimensional manifold.
When $x$ is very close to 0, the $x^2$ term dominates, and so the set looks like the graph of $y=pm x$, a cross, which is not locally homeomorphic to an interval of $mathbb{R}$.
For points away from $(0,0)$ you can use the regular level set theorem (aka the preimage theorem).
answered Nov 22 at 22:37
user25959
1,559816
answered Nov 22 at 22:37
user25959
1,559816
answered Nov 22 at 22:37
user25959
1,559816
answered Nov 22 at 22:37
user25959
1,559816
1,559816
add a comment |
add a comment |
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