Is the set $\{(x,y)\in \mathbb R^2 : y=|x|\} \subseteq \mathbb R^2 $ a smooth manifold ?
I can only see that if it is a smooth manifold then it is of dimension $1$ . Please help . Thanks in advance
Is $\{(x,y)\in \mathbb R^2 : y=|x|\} \subseteq \mathbb R^2 $ a smooth manifold ?
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multivariable-calculus
differential-geometry
smooth-manifolds
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0I think the ( C, $p_x$) is a chart of the curve C where $p_x$ is the map on the projection to the x axis. – 2017-02-12
1 Answers
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If considered as a subset of $\mathbb{R}^2$ then no it's not. The point $(0,0)$ does not have a neighbourhood for which the definition (any of the equivalent definintions, like being the graph of a smooth function or the image of a smooth map with nonzero derivative) would apply.
To see this assume you have a smooth curve with nonzero derivative parametrizing the set. Near $x=0$ it's easy to see that the left hand derivative is equal to the right hand derivative, so the derivative cannot be continuous.
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0but why is the derivative not continuous ? say $f(t)=|t^5|$ is $C^1$ ... – 2017-02-12
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0@SaunDev Continuity of the derivative alone is not sufficient. It has to be nonzero as well. The problem arises from the simultaneous requirement for the derivativ of being nonzero and continuous. – 2017-02-12