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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.)

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    $\begingroup$ I don't think you're going to find that there are very many smooth functions at all by this definition. $\endgroup$ – Dustan Levenstein Sep 2 '14 at 22:25
  • $\begingroup$ Well, I can find infinitely many: take any positive integer power of $x$. So can you elaborate? $\endgroup$ – symplectomorphic Sep 2 '14 at 22:27
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    $\begingroup$ You said all charts, not just the ones you provided. I presumed you meant all charts as a topological manifold; did you just mean the charts of a given atlas? $\endgroup$ – Dustan Levenstein Sep 2 '14 at 22:28
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    $\begingroup$ Well, a good reason that smooth transition functions are desirable would be that, under the constraint of smooth transition functions, you only need to check one of the charts containing $p$, to determine that a function is smooth at $p$. The idea being that a single chart containing $p$ should include all the local information about $p$ that you need to know. $\endgroup$ – Dustan Levenstein Sep 2 '14 at 22:35
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    $\begingroup$ I can't think of another one off the top of my head, but I'm not sure why you wouldn't consider that to be a good reason? Charts are supposed to encode local information; if they fail at that task, then I don't think they serve their purpose... $\endgroup$ – Dustan Levenstein Sep 2 '14 at 22:41
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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)


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)

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  • $\begingroup$ This doesn't answer the question, at least not in enough detail. The structure I gave on the interval $(-2,2)$ is enough to define a diffeomorphism. At least it seems to me. Is it not? $\endgroup$ – symplectomorphic Sep 2 '14 at 22:47
  • $\begingroup$ I think his last paragraph addresses the question better than you give it credit for. When the desired structure is an encoding of continuity, the transition functions should be continuous. Similarly, when the desired structure is an encoding of smoothness, the transition functions should be smooth. $\endgroup$ – Dustan Levenstein Sep 2 '14 at 22:57
  • $\begingroup$ @Dustan: I don't know what "is an encoding of" means in your comment. I accept Hurkyl's claim that to define a topology the transition functions should be continuous. But I don't then see in what sense you can say that fact is analogous in the smooth case to saying transition functions "should" be smooth. You can't say "in order to define a smooth structure" the transitions should be smooth, because it's precisely the definition of smooth structure I'm asking about. $\endgroup$ – symplectomorphic Sep 3 '14 at 2:09
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    $\begingroup$ @symplec: Your problem is that you claim that both restrictions $\phi_U : (-1, 1) \to \mathbb{R}$ and $\phi_V : (-1, 1) \to \mathbb{R}$ are diffeomorphisms. (that can't be true because it's not surjective, but it's fine to use $(-1, 1)$ in place of $\mathbb{R}$) That would imply $\phi_V \circ \phi_U^{-1}$ is a diffeomorphism from $(-1,1) \to (-1,1)$ where both the domain and range have their usual Euclidean differential structure. But $\phi_V \circ \phi_U^{-1}(x) = \sqrt[3]{x}$ which is clearly not a diffeomorphism, thus our hypothesis that $\phi_U$ and $\phi_V$ are diffoemorphisms is false. $\endgroup$ – user14972 Sep 3 '14 at 4:44
  • $\begingroup$ @Hurkyl: thanks. I follow the counterexample: smoothness of transitions will be necessary for restrictions to be diffeomorphisms. But I'm having trouble seeing that it's sufficient. If both $\phi_U\circ\phi_V^{-1}$ and $\phi_V\circ\phi_U^{-1}$ are smooth, why must the restriction of $\phi_U$ be smooth? A counterexample is $\phi_U=\phi_V=$ something non-smooth. Can you give me a hint for what to do if I assume they aren't equal? $\endgroup$ – symplectomorphic Sep 3 '14 at 13:40
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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.)


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.


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.

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  • $\begingroup$ Per the comments, the original question intended to restrict to charts in a given atlas. $\endgroup$ – Dustan Levenstein Sep 2 '14 at 22:38
  • $\begingroup$ @DustanLevenstein: Thanks; I just saw this after posting my answer! $\endgroup$ – guy-in-seoul Sep 2 '14 at 22:39
  • $\begingroup$ "So it is not the usual notion." I'm aware it isn't the usual definition... I'm basically asking what we want the usual notion for. Telling me it better handles transition functions, as I said in the post, isn't an answer, because it doesn't tell me what I want to use the transitions for. Besides the locality idea (with transitions it is enough to check in one chart rather than all). $\endgroup$ – symplectomorphic Sep 2 '14 at 22:54
  • $\begingroup$ @symplectomorphic: A smooth manifold (one whose atlas has smooth transition functions) is locally diffeomorphic to Euclidean space. This won't typically be true with your notion; you have defined some larger collection of structures, which don't have an easy or standard local characterization in general (as far as I can tell). Incidentally, saying "besides the locality idea" is dismissing the whole notion of local coordinates; that's a pretty big "besides"! $\endgroup$ – guy-in-seoul Sep 2 '14 at 23:06
  • $\begingroup$ @guy-in-seoul: my example is locally diffeomorphic to Euclidean space. can you exhibit a counterexample? & regarding my word "besides": I'm just asking whether there are other reasons; it is a purely logical exercise. I think people are reading too much into the fact that I'm asking what else can be said. I'm doing so not because I'm dissatisfied with the one-chart-should-be-sufficient-to-check-smoothness answer but just because I'm curious about how else the structure I've defined might be pathological. $\endgroup$ – symplectomorphic Sep 3 '14 at 2:05

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