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?

  • 1
    $\begingroup$ Maybe you mean the other way around? $\endgroup$
    – Hoot
    Jun 28, 2016 at 15:37

3 Answers 3


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.

  • $\begingroup$ 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 $R^k$. $\endgroup$
    – math635
    Jun 28, 2016 at 15:50
  • $\begingroup$ 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. $\endgroup$ Jun 28, 2016 at 16:18
  • $\begingroup$ 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? $\endgroup$
    – math635
    Jun 28, 2016 at 16:37
  • $\begingroup$ 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. $\endgroup$ Jun 28, 2016 at 16:43
  • $\begingroup$ 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 $x \in U \subset R^n$) $\endgroup$
    – math635
    Jun 28, 2016 at 21:30

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.


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.


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 $k$.


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