# Milnor's definition of smooth manifold

In Milnor's book "Topology from a differential viewpoint" on page one he defines a smooth manifold to be a subset $M \subset \mathbb R^n$ which is locally diffeomorphic to some open subset of $\mathbb R^k$, i.e. every point $x \in M$ has a neighborhood $U \subset \mathbb R^n$ such that $U \cap M = V$ for some open $V \subset \mathbb R^k$ . The usual definition I know is that a smooth manifold is a (hausdorff and second countable) topological space $M$ together with an open cover $\{U_{\alpha}\}$ and homeomorphisms $f_{\alpha} : U_{\alpha} \rightarrow V_{\alpha}$ such that $V_{\alpha} \subset \mathbb R^k$ are open and the transition functions $f_{\beta}f_{\alpha}^{-1}$ are smooth (where defined).

Question: How does the usual definition of smooth manifold imply Milnor's definition of smooth manifold?

• Maybe you mean the other way around? – Hoot Jun 28 '16 at 15:37

You seem to be bothered by the fact that in Milnor's definition one talks of smooth maps, whereas in the chart definition, the $f_\alpha$ are only homeomorphisms. However, once we require the transition functions to be smooth, we can actually view the $f_\alpha$ as being smooth as well.
By the implicit function theorem, a $k$-manifold smoothly embedded in Euclidean $n$-space will be the graph of a smooth vector-valued function over a suitable coordinate $k$-plane in $\mathbb{R}^n$. This implies the smoothness of the transition functions. Conversely, of you have a smooth manifold (in the sense of smooth transition functions, etc.) then Whitney gives you a smooth embedding in Euclidean space.
• so if I have a smooth $k$ manifold $M$ by whitneys theorem I get a smooth embedding $M \rightarrow R^{2k}$. How does this imply that for each point $x\in M$ there is a open neighbourhood $x\in U \subset R^{2k}$ such that $U\cap M$ is diffeomorphic to an open subset of $R^k$. – math635 Jun 28 '16 at 15:50
• That's just a compactness argument. First you choose a coordinate neighborhood which exists by the hypothesis that $M$ is a manifold. In this neighborhood you choose a ball containing the point. As far as points outside this coordinate chart, by compactness they are going to be at a distance separated from zero from the center of the ball, and then we choose a smaller radius to make sure we eliminate any risk of points outside the chosen coordinate chart. – Mikhail Katz Jun 28 '16 at 16:18
• Im not quiet following. We can choose open subset $x\in U \subset R^k$ such that $U \cap M$ is homeomorphic to an open subset of $R^k$. Now we take a ball $B$ which is contained in $U$ and contains $x$. Then $B$ is homeomorphic to an open subset of $R^k$ but why can we shrink $B$ to get an diffeomorphism? – math635 Jun 28 '16 at 16:37
• No, the shrinking was meant to address a different problem namely the one you raised: how do we show that a neighborhood of the point is actually diffeomorphic to a $k$-disk in the coordinate chart, and guarantee that there are no other points. That's the step that involves shrinking the radius. As far as the construction of diffeomorphism is concerned, that's the implicit function theorem. – Mikhail Katz Jun 28 '16 at 16:43
• yes thats what I mean. A priory we only know that the neighborhood is homeomorphic to an open subset of $R^k$ and not diffeomorphic. I don't see why shrinking helps with this problem.( I can't edit my second comment but there is a mistake, I meant: we can choose an open subset $x \in U \subset R^n$) – math635 Jun 28 '16 at 21:30