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I am wondering whether or not the graph a polynomial (or perhaps any basic analytic function) is

  • a manifold
  • a smooth manifold

I believe that a function such as $x^2$ indeed gives rise to a manifold, though perhaps non-compact or non-connected. And if it is a manifold (with one coordinate map), then wouldn't it be smooth, because $x^2$ itself is smooth?

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    $\begingroup$ What do you mean by a function to be a manifold exactly ? Its graph ? In that case (of the polynomials), the answer is yes, and it is diffeomorphic to R $\endgroup$
    – Anthony
    Commented Aug 23, 2020 at 15:59
  • $\begingroup$ Yes exactly, meaning $(-\infty, \infty)$ is your topology and $f(x)=x^2$ is your coordinate map. Thus $((-\infty, \infty), x^2)$ is your coordinate chart (and Atlas). $\endgroup$
    – Tug Witt
    Commented Aug 23, 2020 at 19:10

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Just because of purism, $$(-\infty,+\infty)$$ is not a topology, but rather a set that you have to endow with some topology (in your case, it is the "narural" normed topology).

Then, my previous comment holds : for any smooth function $f:\mathbb{R}\to\mathbb{R}$, its graph $$\mathcal{G}f=\{(x,f(x)),\;x\in\mathbb{R}\}$$ is a smooth manifold, diffeomorphic to $\mathbb{R}$ (and you have an embedding of $\mathcal{G}f$ in $\mathbb{R}^2$ given by the usual inclusion). Polynomials are just a special example.

However, $f$ need not be a homeomorphism, which means the atlas isn't what you'd hope for ; $f$ is not a chart map. An atlas could be composed of the one map $$\varphi:\mathcal{G}f\to\mathbb{R}$$ defined as $\varphi(x,y)=x$.

So the role that $f$ plays is that it is a transition map : from $\mathbb{R}$ to $\mathcal{G}f$ by : $$x\mapsto(x,f(x)).$$

At last, the graph being a connected topological space, the manifold is therefore connected (but that's also a corollary of $\mathcal{G}f\cong\mathbb{R}$, also giving us all the homotopy and homology groups of $\mathcal{G}f$ for instance, that is a much much stronger statement).


Aside : A similar statement holds in lower regularities : if $f$ is only assumed to be $\mathscr{C}^k$, then $\mathcal{G}f$ is a $\mathscr{C}^k$-manifold (transition maps are required to have the regularity $\mathscr{C}^k$), $\mathscr{C}^k$-diffeomorphic to $\mathbb{R}$ (in the case $k=0$, understand homeomorphic). There is an interesting theory of non-smooth manifolds (e.g. the tangent bundle is endowed with a $\mathscr{C}^{k-1}$-manifold structure).

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  • $\begingroup$ So you're saying that, instead of considering $f$ as your coordinate chart, $\varphi$ is actually the full map between the graph $\mathcal{G}f$ and the output space, $\mathbb{R}^2$ - thus the reason $f$ is actually a transition map is because its mapping between $x$ values (change of coordinates) on $\mathcal{G}f$? $\endgroup$
    – Tug Witt
    Commented Aug 23, 2020 at 22:27
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    $\begingroup$ What do you mean by $\mathbb{R}^2$ to be the output space ? The embedding $\mathcal{G}f\subset\mathbb{R}^2$ means that $\mathcal{G}f$ lives naturally in $\mathbb{R}^2$. However, it is still a one-dimensional manifold, meaning that your open sets in your atlas are homeomorphic to $\mathbb{R}$. Neither $f$ nor $\varphi$ link $\mathcal{G}f$ to $\mathbb{R}^2$, but rather with $\mathbb{R}$. The only connection between $\mathcal{G}f$ and $\mathbb{R}^2$ is the embedding (the inclusion of the underlying sets). $\endgroup$
    – Anthony
    Commented Aug 25, 2020 at 18:17

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