# Question about definition of a smooth manifolds (do all transition maps have to be smooth)?

I need to clear up my confusion on the definition of a smooth manifold. So we say that $M$ is a smooth manifold (of dimension $n$), if $M$ is Hausdorff and if every $x \in M$ is contained in a neighborhood $U$ that's homeomorphic to an n-ball (the pair $\phi, U$ is called a chart), and if two such charts $\phi_1, U_1$, and $\phi_2, U_2$ overlap, then

$$\phi_2 \circ \phi_1^{-1} : \phi_1(U_1 \cap U_2) \to \phi_2(U_1 \cap U_2)$$

is a smooth map.

But I remember my professor proving that a certain space was a smooth manifold by merely finding an atlas (an open covering of the space by charts) such that the above holds. But according to the definition I wrote, this would be insufficient. Can anyone clear up my confusion?

• Why would that be insufficient? – Potato Oct 13 '12 at 17:43
• Well, consider two atlases $A1$ and $A2$. It could be that all transition maps within $A1$ and within $A2$ are smooth, but if two charts (one from A1 and one from A2) overlap, then that transition map may not be smooth, and the definition I wrote(which may be wrong), indicates that any transition map must be smooth. – Jorge Oct 13 '12 at 17:47
• A manifold is a pairing of a topological space with an atlas. Topological spaces can support many incompatible manifold structures. There is not always a canonical choice. Your professor has taken a topological space and given it a smooth structure, so you get a manifold. Whether or not it is compatible with some other atlas doesn't really matter. – Potato Oct 13 '12 at 17:50
• My preferred definition of a smooth manifold is that of using sheaves. We consider a ringed space $(M, \mathcal{O}_M)$, where $M$ is a Hausdorff space and $\mathcal{O}_M$ is a subsheaf of the sheaf of germs of continuous functions on $M$. Let $V$ be an open subset of $\mathbb{R}^n$. Let $\mathcal{O}_V$ be the sheaf of germs of smooth functions on $V$. The ringed space $(V, \mathcal{O}_V)$ is called a model space. If every point $p$ of $M$ has an open neighborhood $U$ such that $(U, \mathcal{O}_M|U)$ is isomorphic to a model space, $(M, \mathcal{O}_M)$ is called a smooth manifold. – Makoto Kato Oct 13 '12 at 19:23

## 2 Answers

I think your definition of a manifold is missing something. A topological manifold is a Hausdorff space $M$ such that every $x\in M$ has a neighbourhood $U$ that is homeomorphic to some open subset of $\mathbb R^n$. (Usually we also assume paracompactness.) The pair $(\phi,U)$ where $\phi$ is the homeomorphism I mentioned is called a chart.

Now the main point: to specify a smooth structure on $M$ we have to specify which charts we consider to be smooth. So, a smooth manifold is a topological manifold together with a set of smooth charts, which is a subset of all charts. And it is these smooth charts that we require to be smoothly compatible, i.e. for any two smooth charts $(\phi,U)$, $(\psi,V)$, we require that $$\psi\circ\phi^{-1}:\phi(U\cap V)\to\psi(U\cap V)$$ is smooth. A collection of such smooth charts that covers $M$ is called a smooth atlas.

(The situation is a bit similar to the one in the definition of a topological space: a topological space is a space in which we specify which subsets are open. A smooth manifold is a manifold in which we specify which charts are smooth.)

For precisely this reason, some prefer to define a manifold as its underlying topological space together with (a) a maximal atlas, i.e. one which contains all compatible atlases, or (b) an equivalence class of atlases. It's more or less the same thing either way (IIRC the equivalence relation is determined by maximal atlases).