Milnor's definition of smooth manifold

Question

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?

Answer 1

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.

Answer 2

What Milnor describes is more often defined as a "sub-manifold of $\mathbb{R}^n$.

It is quite natural to see that such an object is a smooth manifold in the usual definition, by taking for $(U_{\alpha})$ all the neighbourhoods of the points $x$ in $M$ given by Milnor's definition.

The other implication is not that easy and is known as Whitney embedding theorem : every manifold $M$ of dimension $m$ (with the usual definition) can be embedded in $\mathbb{R}^{2m}$, thus providing as diffeomorphism between M and the image of $M$ in $\mathbb{R}^{2m}$, the latter being a submanifold of $\mathbb{R}^{2m}$ (i.e. a "manifold" with Milnor definition).

An easier version of Whitney theorem is : every manifold $M$ of dimension $m$ (with the usual definition) can be embedded in somme $\mathbb{R}^{N}$. Here is a proof when M is a compact manifold : Let $(U_{i},\phi_{i})_{1 \leq i \leq p}$ be a finite number of charts of our manifold M (possible because $M$ is assumed to be compact), and let $(\rho_{i})$ be a partition of unity subordinate to the open cover $(U_{i})$. Then the map

$f : \begin{array}{ccc} M& \to & (\mathbb{R}^{m})^p \times \mathbb{R}^p \\ x &\mapsto & (\rho_{1}(x)\phi_{1}(x),\cdots,\rho_{p}(x)\phi_{p}(x),\;\;\;\rho_{1}(x),\rho_{2}(x),\cdots,\rho_{p}(x)) \end{array}$

is an embedding of $M$ to $\mathbb{R}^{(m+1)p}$.

1. It is clearly smooth ;
2. It is injective : if $f(x)=f(y)$, then there is a common $i$ such that $\rho_i(x)=\rho_i(y) \neq 0$, and then we get $\phi_i(x)=\phi_i(y)$, therefore $x=y$.
3. Its differential is injective everywhere. For $x \in M$, there is an $i$ such that $x \in U_i$. If $d_xf[u] = 0$, then $d_x(\rho_i\phi_i)[u]=0$ and $d_x(\rho_i)[u]=0$. Then we get $\rho_i(x).d_x\phi_i[u]=0$, and as $x \in U_i$, we get $d_x\phi_i[u] =0$. As $\phi_i$ is a diffeomorphism, we get $u=0$ and thus $d_xf$ is injective.

When $M$ is not compact, the same kind of ideas can be used but a bit more is needed, namely tranversality.

Answer 3

Milnor defines what is usually called a submanifold of $\mathbb R^n$. This is a more special approach than considering "abstract" smooth manifolds which are Hausdorff and second countable topological spaces endowed with an atlas having smooth transition functions.

Clearly all subspaces of $\mathbb R^n$ are Hausdorff and second countable. The essential question is this:

• Are Milnor's manifolds special cases of abstract smooth manifolds? More precisely, do they have a "natural" atlas with smooth transition functions?

The answer is yes.

On p.1 Milnor extends the "classic" concept of a smooth map between open subsets $U, V$ of Euclidean spaces to maps $f : X \to Y$ between arbitrary subsets $X \subset \mathbb R^n$ and $Y \subset \mathbb R^m$ by requiring that for all $x \in X$ there exists an open neigborhood $U$ of $x$ in $\mathbb R^n$ and a smooth extension (in the classic sense) $f_U : U \to \mathbb R^m$ of $f \mid_{U \cap M}$. A diffeomorphism $f : X \to Y$ is then a homeomorphism such that $f, f^{-1}$ are smooth.

For the moment let us reserve the word "smooth map" for a map between open subsets of Euclidean spaces which is smooth in the classic sense and use the phrase "$\mu$-smooth map" for the generalized case. We also use the phrase "$\mu$-diffeomorphism" to distinguish it from a classic diffeomorphism.

Let us make the following observations:

• If $X$ is an open subset of $\mathbb R^n$ and $Y$ is an open subset of $\mathbb R^m$, the $f : X \to Y$ is $\mu$-smooth if and only if it is smooth.

• If $X$ is an open subset of $\mathbb R^n$, then $f : X \to Y \subset \mathbb R^m$ is $\mu$-smooth if and only if $\bar f : X \stackrel{f}{\to} Y \hookrightarrow \mathbb R^m$ is smooth.

That $M \subset \mathbb R^n$ is a smooth manifold of dimension $k$ means that for each $x \in M$ there exists a pair $(W,\phi)$ consisting of an open neigborhood $W$ of $x$ in $M$ and a $\mu$-diffeomorphism $\phi : W \to U$ to an open subset $U \subset\mathbb R^k$. This means in particular that $M$ is a topological manifold (i.e. is locally Euclidean) and that the above $(W,\phi)$ form an atlas $\mathcal A$ for $M$. We shall prove that $\mathcal A$ is a smooth atlas, that is, $M$ has a natural structure of an "abstract" smooth manifold.

So let $(W_i,\phi_i) \in \mathcal A$ with $\phi_i : W_i \to U_i \subset \mathbb R^k$. Consider the transition function $$\phi_{12} : \phi_1(W_1 \cap W_2) \stackrel{\phi_1^{-1}}{\to} W_1 \cap W_2 \stackrel{\phi_2}{\to} \phi_2(W_1 \cap W_2) $$ and a point $\xi \in \phi_1(W_1 \cap W_2)$.

Since $\phi_2$ is $\mu$-smooth, $x = \phi_1^{-1}(\xi)$ admits an open neigborhood $V_2$ in $\mathbb R^n$ and a smooth extension $\phi_2^* : V_2 \to \mathbb R^k$ of $\phi_2 \mid_{V_2 \cap M}$. Since $\phi_2^*(x) = \phi_2(x) \in \phi_2(W_1 \cap W_2)$, the continuity of $\phi_2^*$ shows that we may w.lo.g. assume that $\phi_2^*(V_2) \subset \phi_2(W_1 \cap W_2)$ (shrink $V_2$ if necessary). Thus we may regard $\phi_2^*$ as a smooth map $$\phi_2^* : V_2 \to \phi_2(W_1 \cap W_2) .$$

Since $\phi_1^{-1}$ is $\mu$-smooth, our above observation says that $\psi : U_1 \stackrel{\phi_1^{-1}}{\to} W_1 \hookrightarrow \mathbb R^n$ is smooth. Since $\psi$ is continuous, there exists an open neighborhood $U \subset U_1$ of $\xi$ such that $\psi(U) \subset V_2$. Thus the composition $$\tau : U \stackrel{\psi}{\to} V_2 \stackrel{\phi_2^*}{\to} \mathbb R^k$$ is smooth. By construction $$\tau = \phi_{12} \mid_U$$ which shows that $\phi_{12} \mid_U$ is smooth. But smoothness is local property, therefore $\phi_{12}$ is smooth.

Remark.

Although submanifolds are special cases of manifolds, focussing on submanifolds is not really a restriction. This follows from the fact that each abstract smooth manifold is diffeomorphic to a submanifold of some $\mathbb R^n$.

Submanifolds have benefits for motivational purposes. For example, the concept of the tangent space at a point of a submanifold is fairly intuitive. See The motivation for a tangent space and Equivalent definition of a tangent space?

For alternative definitions of submanifolds see Equivalences of definitions of submanifolds in $\mathbb{R^n}$. Note that Milnor's definition implies characterization $(2)$. In fact, let $x \in M$ and $\phi : W \to U$ be a $\mu$-diffeomorphism from an open neigborhood $W$ of $x$ in $M$ to an open subset $U \subset\mathbb R^k$. W.l.o.g. we may assume that $\phi(x) = 0$ (otherwise compose $\phi$ with a translation on $\mathbb R^k$). Let $$f : U \stackrel{\phi^{-1}}{\to} W \hookrightarrow \mathbb R^n .$$ This is a smooth map. Since $\phi$ is $\mu$-smooth, there exist an open neighborhood $V$ of $x$ in $\mathbb R^n$ and a smooth extension $\phi_V : V \to \mathbb R^k$ of $\phi \mid_{V \cap M}$. We have $$\phi_V \circ f \mid_{f^{-1}(V)} = id .$$ Since both $\phi_V$ and $f \mid_{f^{-1}(V)}$ are smooth, we get $$D\phi_V \mid_{x} \circ Df_0 = id$$ which implies that $Df_0$ has rank $k$.