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A smooth manifold in $\mathbb{R}^2$ is locally the graph of a $C^1$ function. Consider the graph of $f(x)=x^{1/3}$. Since $f$ is not differentiable at zero, we are in trouble. However, this subset is the graph of $x=y^3$, which IS differentiable at zero.

So is this a manifold? I think so, but I remain confused on the general situation. Is the point that "being a manifold" is a topological property, so it doesn't matter which variable, x or y, is used to locally parametrize the graph?

Any generally illuminating comments will be appreciated!

Wouter Zeldenthuis
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“Being a manifold” is indeed a topological property. However, “being a smooth manifold” is not. Anyhow, in this case it does matter which variable you use.

There are several equivalent definitions of smooth submanifolds of a Euclidean space. One of them requires that the set can be parametrized by some set of coordinates locally: In your example, using the $y$ coordinate works for this purpose. That the $x$ coordinate doesn't is then immaterial.

Another definition requires the submanifold to be locally the zero set of a smooth (possibly vector valued) function, where $0$ is not a critical value of this function. In your example, $x-y^3$ satisfies this requirement.

Harald Hanche-Olsen
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