defining smooth functions on manifolds *without* smooth chart transitions

Question

Let $M$ be a topological manifold, covered by an atlas of charts ${(U,\phi_U)}$ (which are homeomorphisms into Euclidean space), and let $p\in M$. Say a function $f:M\to\mathbb{R}$ is smooth at $p$ if $f\circ\phi^{-1}$ is smooth for all charts in the atlas that contain $p$. (Note that this definition says nothing about smooth chart transitions on the domain where two charts overlap.)

This definition is enough to pick out a collection of smooth functions on $M$. So it must be that we don't need the apparatus of smooth atlases, i.e. atlases composed of charts with smooth transitions, just to talk about the collection of smooth functions on a manifold.

So what do we need the extra structure for? In other words, why is this a bad definition? What do we want/need it do that it does not allow us to do? (Note that "it doesn't allow us to talk about smooth chart transitions," by itself, is not an answer: it doesn't explain what we need that condition for.)

For a concrete example, take $M$ to be the interval $(-2,2)\subset\mathbb{R}$ with charts $$U=(-2,1),\phi_U(x)=x\,\,\,\text{(so $\phi_U^{-1}=x$)}$$ and $$V=(-1,2),\phi_V(x)=\sqrt[3]{x}\,\,\,\text{(so $\phi_V^{-1}=x^3$)}$$ This is a topological manifold equipped with an atlas. And the definition above allows us to say which functions are smooth at, say, $x=0$ on $M$. For example, $f(x)=\sqrt[3]{x}$ is not smooth at $x=0$ according to the definition, because it fails to be smooth using the chart $U$. However, $f(x)=x$ is smooth in both charts.

EDIT: I realize that one reason this definition is bad is that it requires us to check smoothness of $f$ in all charts in the atlas, whereas smooth transitions allow us to check only in one chart. I'm not dissatisfied with that response as a basis for rejecting this definition; I'm just interested in what this structure does entail, and how else it might be pathological or deviate from the usual definition. (I'm basically trying to understand how to pick out the usual notion of smoothness as requiring an atlas with smooth transitions from alternative definitions of smoothness. This is one alternative.)

Answer 1

Unless I've misunderstood what you've written, there are no such non-zero functions $f$. E.g. if $M = \mathbb R$, then you're asking for functions $f: \mathbb R \to \mathbb R$ such that $f\circ \phi^{-1}$ is smooth for all homeomorphisms $\phi$ of $\mathbb R$; the only such functions $f$ are constant functions. (Note that, replacing $\phi$ by $\phi^{-1}$ --- which takes homeos to homeos --- it is the same to ask that $f \circ \phi$ is smooth for all homeos $\phi$, which might be easier to think about.)

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Your example computation is wrong, or at least misleading. For example, there is another chart $W = (-2,2)$ with $\phi_W (x) = x^3$, so that $\phi_W^{-1} = x^{1/3}$, and if we take $f(x) = x$, then $f \circ \phi_W^{-1}$ is not smooth.

You happened to choose two charts for which it is smooth, but in this third chart it is not. Thus $f(x) = x$ is not a smooth function in your definition. Just as in the case $M = \mathbb R$, the only smooth functions on any open subset of $M$ (in your definition) will be constant functions.

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Added: I see from looking at the comments that I did misunderstandd your definition; you plan to choose some atlas and use that to define smooth functions. Now, as pointed out in comments, if the transition functions are not smooth then your notion of "smooth" cannot be tested in just one chart around $p$. So it does't seem particularly useful to me.

E.g. if you take $M = (-2,2)$ and let the atlas be your $U$ and $V$, then smooth functions on $M$ turn out to be just smooth functions on $M$ in the usual sense.

But if you take the atlas to be your $U$ and my $W$ (or even just my $W$ alone), then they are functions on $M$ such that $f(x^{1/3})$ is smooth, which is to say, functions of the form $g(x^3)$, where $g$ is a smooth function (in the usual sense) on $(-8,8)$.

Ultimately, the kind of functions your definition cuts out depends on the behaviour of the transition functions. In the case of the usual definition of a smooth manifold, where these are diffeos, we get the usual notion of smoothness, which can be checked in any chart around a point. In your case, we get some notion that depends highly on the particular nature of the transition functions. So it is not the usual notion of smoothness, and exactly what it is depends very much on the choice of atlas.

Answer 2

The reason we have charts on a topological manifold is so that if we pick a chart $U \to \mathbb{R}^n$ that is a continuous function, we can say that the topological space $U$ is homeomorphic to the topological space $\mathbb{R}^n$, so any questions about the topology of $U$ can be moved to questions about the topology of Euclidean space, which we understand very well.

For differentiable manifolds, we want charts that make $U$ diffeomorphic to $\mathbb{R}^n$, so that any questions about the differentiable structure on $U$ can be moved to questions about the differentiable structure of Euclidean space, which we understand well.

For smooth manifolds, we want $U$ to be isomorphic to $\mathbb{R}^n$ in the appropriate sense as well.

This is why we don't consider all charts on a differentiable manifold or a smooth manifold.

Of course, for this to make sense, we need a way to be able to define what it means for a map to be diffeomorphic. We know how to define diffeomorphisms of subspaces of Euclidean space. The transition map idea is a device to let us take that definition and use it to define what a diffeomorphism is from a subspace of our manifold. (and happily, that turns out to be all we need to make the definition work)

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It may be a fun exercise to show the analogous fact for topological manifolds. Suppose you didn't even bother defining a topology on the set $M$: you just had an atlas of charts that are just ordinary bijective functions. Try using the atlas to define a topology on $M$. You'll find that you need the transition maps to be continuous, and that's all you need.

(EDIT: you also need the domain and image of the transition maps to be open)