Now we define smooth functions between manifolds. This obviously an necessary definition in order to do any calculus. The approach is to define differentiability in terms of differentiability via charts, which reduces the problem to smoothness in Euclidean space which we already understand. We will also introduce tangent vectors and differentials of smooth maps.
Let \(M, N\) be \(C^{\infty}\) manifolds with maximal atlases \(\{\varphi_{\alpha} : U_{\alpha} \subseteq M \to V_{\alpha} \subseteq \mathbb{R}^m\}\) for \(M\) and \(\{\psi_k : W_k \subseteq N \to Z_k \subseteq \mathbb{R}^n\}\) for \(N\).
A \(C^{\infty}\) function is a function \(f : M \to N\) such that for ever \(\alpha, k\), \(\psi_k \circ f \circ \varphi_{\alpha}^{-1}\) is \(C^{\infty}\) on it's domain.
A function \(f\) between manifolds is thus \(C^{\infty}\) if the local chart representation, \(\psi_k \circ f \circ \varphi_{\alpha}^{-1}\) of \(f\) is \(C^{\infty}\). Note that \(\psi_k \circ f \circ \varphi_{\alpha}^{-1}\) is defined on the domain \[ V_{\alpha,k} := (\varphi_{\alpha}^{-1})^{\ast} (f^{\ast} (W_k)) = \varphi_{\alpha}(f^{\ast}(W_k)) \subseteq V_{\alpha} \subseteq \mathbb{R}^n \] and takes values in \(\mathbb{R}^m\).
The domain \(U_{\alpha,k}\) is an open subset of \(\mathbb{R}^n\) since \(\varphi_{\alpha}^{-1}\) and \(f\) are continuous, while \(W_k\) is open. It's possible that for a given \(\alpha, k\), the domain \(U_{\alpha,k} = \emptyset\). That is, \(f\) maps \(U_{\alpha}\) into the complement of \(W_k\). As usual, in this case that the composition \(\psi_k^{-1} \circ f \circ \varphi_{\alpha}\) is vacuously smooth.
Thus since \(\psi_k \circ f \circ \varphi_{\alpha}^{-1}\) is a map from an open set of Euclidean space to another Euclidean space, we already have a well defined notion of when \(\psi_k \circ f \circ \varphi_{\alpha}^{-1}\) is \(C^{\infty}\).
Let \(M = \mathbb{R}\) and \(N = \mathbb{S}^1 = \{x^2 + y^2 =1\}\). Then the identity function \(\operatorname{Id} : M \to \mathbb{R}\) is a chart covering \(M\) and for charts covering \(N\) we take \[ \psi_{\pm}^{-1} : \theta \in (0, 2\pi) \mapsto \pm (\cos \theta, \sin \theta) \] That is the charts are the inverses of the local parametrisations \(\psi_{\pm}^{-1}\). We saw this cover in a previous example.
The domain of \(\psi_+\) is the range of \(\psi_+^{-1}\) which is the open set \(U_+ = \mathbb{S}^1 \backslash \{(1, 0)\}\) while the domain of \(\psi_-\) is the range of \(\psi_-^{-1}\) which is the open set \(U_- = \mathbb{S}^1 \backslash \{(-1, 0)\}\).
We claim that the function \(f : M \to N\) defined by \(f(x) = (\cos x, \sin x)\) is \(C^{\infty}\). To verify this claim, we compute \[ \psi_{\pm} \circ f \circ \operatorname{Id}^{-1} (x) = \psi_{\pm} (\cos x, \sin x). \] In the case of \(\psi_+\), this is defined for \(x\) such that \((\cos x, \sin x) \in U_+\). That is for \(x \in \mathbb{R}\backslash 2\pi \mathbb{Z}\). For such \(x\), we have \[ \varphi_+^{-1}(\theta) = (\cos \theta, \sin \theta) = (\cos x, \sin x) \] if and only if \(\theta \equiv x \operatorname{mod} 2\pi\). That is, for \(x \in (2\pi n, 2\pi(n+1))\), \[ \psi_+ \circ f \circ \operatorname{Id}^{-1} (x) = x - 2\pi n \in (0, 2\pi) \] is \(C^{\infty}\). The case of \(\psi_-\) is similar.
Show that \(\psi_- \circ f \circ \operatorname{Id}^{-1}\) is \(C^{\infty}\).
You may object to the example since we didn't show that \(\psi \circ f \circ \varphi^{-1}\) is \(C^{\infty}\) for every chart \(\psi\) for \(\mathbb{S}^1\) and every chart \(\varphi\) for \(\mathbb{R}\). We only showed it for the open covers \(\{\psi_{\pm}\}\) for \(\mathbb{S}^1\) and \(\{\operatorname{Id}\}\) for \(\mathbb{R}\). The next result says that this is sufficient.
A function \(f : M \to N\) is \(C^{\infty}\) if and only if there are covers \(\{\varphi_{\alpha} : U_{\alpha} \to V_{\alpha}\}\) and \(\{\psi_k : W_k \to Z_k\}\) for \(N\) such that \(\psi_k \circ f \circ \varphi_{\alpha}^{-1}\) is \(C^{\infty}\).
If \(f\) is \(C^{\infty}\), then \(\psi_k \circ f \circ \varphi_{\alpha}^{-1}\) is \(C^{\infty}\) for all charts, hence is \(C^{\infty}\) for the charts of a cover.
Conversely suppose \(\psi_k \circ f \circ \varphi_{\alpha}^{-1}\) is \(C^{\infty}\) for charts from a cover. We need to show that for any chart (not necessarily from the covers) \(\psi : W \to Z\) for \(N\) and any chart \(\varphi : U \to V\) for \(M\) \(\psi \circ f \circ \varphi^{-1}\) is \(C^{\infty}\). Now \(U\) is covered by the open sets \(\{U \cap U_{\alpha}\}\) and \(W\) is covered by the open sets \(\{W \cap W_k\}\). To show \(\psi \circ f \circ \varphi^{-1}\) is \(C^{\infty}\) it therefore suffices to show that \(\psi \circ f \circ \varphi^{-1}\) is \(C^{\infty}\) on \(\varphi(f^{\ast}(W \cap W_k) \cap (U \cap U_{\alpha}))\) for each \(\alpha, k\).
For this, on \(\varphi(f^{\ast}(W \cap W_k) \cap (U \cap U_{\alpha}))\) we have
\begin{align*} \psi \circ f \circ \varphi^{-1} &= \psi \circ \psi_k^{-1} \circ \psi_k \circ f \circ \varphi_{\alpha}^{-1} \circ \varphi_{\alpha} \circ \varphi^{-1} \\ &= (\psi \circ \psi_k^{-1}) \circ (\psi_k \circ f \circ \varphi_{\alpha}^{-1}) \circ (\varphi_{\alpha} \circ \varphi^{-1}). \end{align*}The outer parenthesised terms are transition functions, hence \(C^{\infty}\). The inner parenthesised term is the expression of \(f\) in the charts of the cover which is \(C^{\infty}\) by assumption. Thus the entire composition is \(C^{\infty}\).
The lemma is very useful in that it means we don't have to check \(C^{\infty}\) for every chart, just on whatever covers are convenient. This is what we did in the example.
The map \(\pi : \mathbb{S}^n \to \mathbb{RP}^n\) defined by \[ \pi(x) = \operatorname{span}(\{x\}) \] is \(C^{\infty}\).
Cover \(\mathbb{S}^n\) by the charts \[ \tilde{\varphi}_i: x \in \tilde{U}_i \mapsto \frac{1}{x_i}\hat{x}_i \] where \(\tilde{U}_i = \{x_i \neq 0\} \cap \mathbb{S}^n\) and \(\hat{x}_i\) denotes \(x\) with the \(i\)'th coordinate removed - for example, \(\hat{x}_2 = (x_1, x_3, \dots, x_{n+1})\).
Cover \(\mathbb{RP}^n\) by the charts \[ \varphi_i : V \in U_i := \pi(\tilde{U}_i) \mapsto \tilde{\varphi}_i(x) \] where \(V = \pi(x)\).
Recall that \(\varphi_i \circ \pi = \tilde{\varphi}_i\). Then \[ \varphi_i \circ \pi \circ \tilde{\varphi}_j^{-1} = \tilde{\varphi_i} \circ \tilde{\varphi}_j^{-1} \] is a transition function on \(\mathbb{S}^n\) hence is \(C^{\infty}\) for every \(i,j\).
Let us now consider the compatibility of smoothness of Euclidean sub-manifolds and smoothness of the ambient \(\mathbb{R}^n\).
Let \(M \subseteq \mathbb{R}^n\) be a \(k\) dimensional embedded sub-manifold. Let \(F : \mathbb{R}^n \to N\) be a \(C^{\infty}\) function. Then \[ f = F|_M : M \to N \] is \(C^{\infty}\).
The assumption is that for every chart \(\psi\) for \(N\) we have \(\psi \circ F\) is \(C^{\infty}\). We need to show that for a cover by charts \(\{\varphi_{\alpha}\}\) for \(M\), the functions \(\psi \circ F \circ \varphi_{\alpha}^{-1}\) is \(C^{\infty}\). For this, cover \(M\) by charts \(\varphi_{\alpha}\) where \(\varphi_{\alpha}^{-1}\) is a local parametrisation. By definition each \(\varphi_{\alpha}^{-1}\) is \(C^{\infty}\) hence \(\psi \circ F \circ \varphi_{\alpha}^{-1} = (\psi \circ F) \circ \varphi_{\alpha}^{-1}\) is a composition of \(C^{\infty}\) functions, hence is \(C^{\infty}\).
Restricting the coordinate functions \(x \mapsto x_i = \pi_i(x)\) to embedded sub-manifolds \(M\) yields smooth functions \(M \to \mathbb{R}\) for \(i =1, \dots, n\).
Let \(M \subseteq \mathbb{R}^n\) be an embedded sub-manifold.