Is my attempt to define the concept “smooth manifold” as a structure satisfying certain axioms correct?

In the lecture notes for a class I'm currently taking, smooth manifold structures are defined as equivalence classes of atlases. However, the issue I'm having is that its not entirely clear (to me) what the smooth structure gives us, beyond the underlying topology. To rectify this problem, I did a bit of snooping around wikipedia, and it seems that there's this concept of a "maximal atlas" which is a bit easier to digest. Still, it would be nice to have a "structure+axioms" definition for a smooth manifold, since this would make it clear what the smooth structure is actually giving us.

Following this line of thought, I toyed with the concept of a maximal atlas for a little while, and ended up writing the following "definition" of a smooth manifold.

Question. Does my definition of a smooth manifold work? If so, how can we show that it is the same as the definition via maximal atlases?

Definition (Prechart). Let $X$ denote a topological space. Then an $n$-prechart on $X$ is a partial function $\varphi : X \rightarrow \mathbb{R}^n$ such that:

• both $\varphi^*\mathbb{R}^n$ and $\varphi_*X$ are open sets
• $\varphi$ induces a homeomorphism $\varphi^*\mathbb{R}^n \rightarrow \varphi_*X$.

Definition (Smooth Manifold Structure). Let $X$ denote a topological space. A smooth $n$-manifold structure on $X$ is a procedure which, for every $n$-prechart $\varphi$ on $X$, returns a decision of yes/no regarding whether $\varphi$ is deemed to be a "chart on the smooth manifold" or not. This procedure is required to satisfy an axiom or two:

1. For all $n$-precharts $\psi$ on $X$, the following are equivalent:

• $\psi$ is a chart
• given any chart $\varphi$ on $X$, we have that $\psi \circ \varphi^{-1}$ is smooth with a smooth inverse (in the usual sense of calculus.)
2. (Optional?) For all $x \in X$, some chart on $X$ is defined at $x$.

I reckon these two axioms might be all we need. The downward implication associated with (1) says that all transition maps are smooth. The upward implication is meant to be a kind of saturation condition. It seems plausible that this is the same as a maximal atlas. Ideas, anyone?

• After rereading I think your axiom is basically a restatement of compatibility for charts. Isn't it? – johndoe Mar 30 '15 at 14:51
• @johndoe, its stronger. One direction says that given $n$-charts $\psi,\varphi,$ if they're both restrictively diffeomorphisms, then so too is $\psi \circ \varphi^{-1}$. This ensures the atlas condition of having smooth transition maps. However, there is a converse direction that is intended to capture atlas maximality. I admit that despite being very precisely worded, nonetheless it is very badly worded; the whole sentence just seems more confusing than it needs to be. – goblin Mar 30 '15 at 14:56
• @johndoe, I've tried rephrasing it for better readability. – goblin Mar 30 '15 at 15:01
• a-ah now I see, yes it makes perfectly sense to me. Honestly for me your "compact" definition does not add much that was not already included in the standard defn via a maximal atlas, but if it helps you... – johndoe Mar 30 '15 at 15:19
• @user86418, yes, I used "restrictively a diffeomorphism" in two different senses, as you rightly point out. I've changed the terminology somewhat, which should hopefully make things clearer. – goblin Mar 31 '15 at 4:36

However, the issue I'm having is that its not entirely clear (to me) what the smooth structure gives us, beyond the underlying topology.

A smooth structure on $X$ tells you what a smooth function $X \to \mathbb{R}$ on the manifold is. You want to know which ones the smooth functions are so you can do calculus to them.

(I might have said "a smooth structure on $X$ tells you what a smooth map $X \to Y$ to another smooth manifold is," but in fact a continuous map $X \to Y$ between smooth manifolds is smooth if and only if smooth functions $Y \to \mathbb{R}$ pull back to smooth functions $X \to \mathbb{R}$, so all you need to know is what a smooth function $X \to \mathbb{R}$ is.)

One way of saying this, which fits quite well into a general yoga of describing various kinds of spaces, is that a smooth structure equips a manifold with a sheaf of rings which for every open set $U$ returns the ring $C^{\infty}(U)$ of smooth functions $U \to \mathbb{R}$. The axiom to be satisfied is that this sheaf is in a suitable sense locally isomorphic to the sheaf of smooth functions on $\mathbb{R}^n$.

The sheaf of smooth functions naturally sits inside the sheaf of continuous functions, and the extra information the smooth structure gives you, beyond the topology, is the information of this interesting subsheaf of the sheaf of continuous functions.

In fact, equipped with this sheaf, a smooth manifold is naturally a locally ringed space. This is a good context in which to view many kinds of spaces; for example, complex manifolds, varieties, and more generally schemes are also naturally certain locally ringed spaces.

• This sounds like a nice idea--anything algebra makes me happy--but how can we recover the smooth structure on $X$ from the subsheaf (am I using that word correctly?) of smooth partial functions into $\mathbb{R}$? – goblin Mar 31 '15 at 3:51
• @goblin: once you know the sheaf of smooth functions, you know what a smooth function on each open subset is, so (if you believe what I said earlier about this being enough to tell you what a smooth map to another smooth manifold is) you know what a diffeomorphism from an open subset to $\mathbb{R}^n$ is, and so you know what charts are. – Qiaochu Yuan Mar 31 '15 at 3:58
• Right. I think I get it, but you seem to be using chart differently to how we've been taught. In class, we learned that a chart on a topological space is an open set $U$ of that space together with homeomorphism of $U$ onto an open subspace of Euclidean space. You seem to be using "chart" differently - in some sense, you're using the phrase "chart on a smooth manifold" to mean "chart on the underlying topological space that is compatible with the smooth structure." Is that right, and if so, is this standard? – goblin Mar 31 '15 at 4:27
• @goblin: yes, that's what "chart on a smooth manifold" means. – Qiaochu Yuan Mar 31 '15 at 4:27