This post was translated from Korean by LLM (Kimi). The translation may contain errors or awkward sentences. The Korean original is the source of truth.
We defined the notion of a topological manifold in topology; in this series of posts we deal with differentiable manifolds, and in particular smooth manifolds.
Notation
Since we will frequently work with coordinate systems of dimension \(m\), we fix the following notation. For \(\mathbb{R}^m\), we denote the \(i\)-th projection \(\pr_i\) by \(r^i\). Similarly, for any set \(X\) and function \(f:X\rightarrow\mathbb{R}^m\), the \(i\)-th component function of \(f\) is defined by the formula \(f^i=r^i\circ f\).
Now let \(f\) be a function from \(\mathbb{R}^m\) to \(\mathbb{R}\). Then we define the partial derivative of \(f\) with respect to its \(i\)-th component by the formula
\[\frac{\partial}{\partial r^i}\bigg|_t f=\frac{\partial f}{\partial r^i}\bigg|_t=\lim_{h\rightarrow 0}\frac{f(t^1,\ldots, t^{i-1}, t^i+h, t^{i+1},\ldots, t^m)-f(t^1,\ldots, t^m)}{h}\]As in the notation above, following [Lee] we write the \(i\)-th component as \(x^i\) rather than \(x_i\).
If each component function is \(k\)-times differentiable and the result is continuous, we say the function \(f\) is \(C^k\). For example, saying a function \(f:\mathbb{R}^2\rightarrow\mathbb{R}\) is \(C^2\) means that all the following partial derivatives
\[\frac{\partial^2 f}{\partial x^2},\quad\frac{\partial^2 f}{\partial x\partial y},\quad\frac{\partial^2 f}{\partial y\partial x},\quad\frac{\partial^2 f}{\partial y^2}\]exist and are continuous. If a function \(f\) is \(C^k\) for every natural number \(k\), we call it \(C^\infty\).
Differentiable Manifolds
Unlike general topological spaces, a topological manifold looks locally like \(\mathbb{R}^n\), so we can import the notion of differentiation defined there to \(M\). This is possible because differentiability is essentially a local property.
Definition 1 Let a topological manifold \(M\) be given. For \(0\leq k\leq\infty\), coordinate charts \((U,\varphi)\) and \((V,\psi)\) are \(C^k\)-compatible if both transition maps
\[\psi\circ\varphi^{-1}:\varphi(U\cap V)\rightarrow\psi(U\cap V),\qquad\varphi\circ\psi^{-1}:\psi(U\cap V)\rightarrow\varphi(U\cap V)\]are \(C^k\). A collection \(\mathcal{A}=\{(U_\lambda, \varphi_\lambda)\}_{\lambda\in\Lambda}\) of \(C^k\)-compatible charts satisfying \(M=\bigcup U_\lambda\) is called a \(C^k\)-atlas.
Among \(C^k\)-atlases defined on \(M\), an atlas that is maximal with respect to inclusion is called a \(C^k\)-differentiable structure, and in this case \(M\) is called a \(C^k\)-differentiable manifold. In the special case \(k=\infty\), this structure is called a smooth differentiable manifold or more simply a differentiable manifold.
The reason we think of a maximal atlas as giving a differentiable structure in this definition is that it is entirely possible for two non-maximal atlases to give essentially the same differentiable structure. For example, \(\mathbb{R}\) has the \(C^\infty\)-atlas
\[\mathcal{A}=\{(\mathbb{R}, \id_\mathbb{R})\}\]but also another atlas
\[\mathcal{A}'=\{((-\infty, 1), \id_{(-\infty, 1)}), ((-1, \infty),\id_{(-1,\infty)})\}\]However, as we will see in Proposition 3, since a maximal atlas containing any given atlas is uniquely determined, in an essential sense this is not such a big difference.
On the other hand, to understand an object in mathematics it suffices to know the functions defined on it. Henceforth we assume all manifolds are smooth differentiable manifolds.
Definition 2 Consider a manifold \(M\) and a point \(p\in M\). A function \(f\) defined on a suitable open neighborhood of \(p\) is \(C^\infty\) at \(p\) if for some coordinate chart \((U,\varphi)\) containing \(p\), the function \(f\circ\varphi^{-1}:U'\rightarrow \mathbb{R}\) is \(C^\infty\) at the point \(\varphi(p)\).
Suppose another coordinate chart \((V,\psi)\) is defined on another open neighborhood of \(p\). If \(f\circ\varphi^{-1}\) is \(C^\infty\) at \(\varphi(p)\) but \(f\circ\psi^{-1}\) is not at \(\psi(p)\), then this definition would not be well-defined. However, on \(\psi(U\cap V)\) we have
\[f\circ\psi^{-1}=(f\circ\varphi^{-1})\circ(\varphi\circ\psi^{-1})\]so \(f\circ\psi^{-1}\) is \(C^\infty\) at \(\psi(p)\). By a similar argument one can show the following.
Proposition 3 Let a \(C^k\)-atlas \(\mathcal{A}\) on a topological manifold \(M\) be given. Then there exists a unique maximal \(C^k\)-atlas containing \(\mathcal{A}\). Therefore any \(C^k\)-atlas \(\mathcal{A}\) defines a unique \(C^k\)-differentiable structure on \(M\).
Proof
Define \(\mathcal{A}'\) by the formula
\[\mathcal{A}'=\{(V,\psi)\mid\psi\circ\varphi_\lambda^{-1}, \varphi_\lambda\circ\psi^{-1}\text{ are $C^k$ for all $\varphi_\lambda\in\mathcal{A}$}\}\]Then \(\mathcal{A}'\) contains \(\mathcal{A}\), and thus covers \(M\) with coordinate charts. On the other hand, if \((V,\psi)\) and \((V',\psi')\) are elements of \(\mathcal{A}'\) and \(V\cap V'\neq\emptyset\), then the transition map
\[\psi'\circ\psi^{-1}:\psi(V\cap V')\rightarrow\psi'(V\cap V')\]is \(C^k\). For any \(p\in\psi(V\cap V')\), choosing \((U,\varphi)\in\mathcal{A}\) with \(p\in U\), on \(U\cap V\cap V'\) we have
\[\psi'\circ\psi^{-1}=(\psi'\circ\varphi^{-1})\circ(\varphi\circ\psi^{-1})\]so \(\psi'\circ\psi^{-1}\) is \(C^k\) at \(p\). Since \(p\) was arbitrary, this shows that \(\psi'\circ\psi^{-1}\) is \(C^k\). Of course, reversing the roles of \((V,\psi)\) and \((V',\psi')\) shows that the reverse transition map is also \(C^k\).
By definition \(\mathcal{A}'\) is clearly a maximal \(C^k\)-atlas, and its uniqueness is easily verified.
Example 4 On the real numbers \(\mathbb{R}\), consider the two atlases
\[\mathcal{A}_1=\{(\mathbb{R},\id_\mathbb{R})\},\qquad \mathcal{A}_2=\{(\mathbb{R}, x\mapsto x^3)\}\]These are atlases consisting of a single chart, so they are trivially \(C^\infty\). By the preceding Proposition 3, there exists a differentiable structure containing each of them. However, they are not equal, because the two charts \((\mathbb{R},\id_\mathbb{R})\) and \((\mathbb{R}, x\mapsto x^3)\) are not \(C^\infty\)-compatible. (While \(x\mapsto x^3\) is a \(C^\infty\) function, its inverse \(x\mapsto x^{1/3}\) is not.)
However, although the two atlases in Example 4 do not give the same differentiable structure, they give differentiable structures that are diffeomorphic to each other.
Smooth Partition of Unity
We showed that a continuous partition of unity exists on any topological manifold, but when dealing with differentiable manifolds a merely continuous partition of unity is of little help. For example, if we multiply an arbitrary \(C^\infty\) function by a partition of unity that is only continuous, the degree of differentiability of this function would immediately deteriorate.
Therefore we need to construct a smooth partition of unity, and for this it suffices to prove the following lemma.
Lemma 5 (\(C^\infty\) Urysohn Lemma) Let real numbers \(a'<a<b<b'\) be given. Then there exists a \(C^\infty\) function \(\psi:\mathbb{R}\rightarrow[0,1]\) that equals \(1\) on \([a,b]\) and \(0\) outside \((a',b')\).
Proof
We may assume without loss of generality that \(a'=-2,a=-1,b=1,b'=2\). First define a function \(f\) by
\[f(t)=\begin{cases}e^{-1/t}&t>0\\0&t\leq 0\end{cases}\]Then in particular \(f\) is always non-negative and is \(C^\infty\). Now define
\[g(t)=\frac{f(t)}{f(t)+f(1-t)}\]Then \(g\) is likewise always non-negative, its value is always less than or equal to \(1\), and in particular it is identically \(1\) when \(t\geq 1\) and identically \(0\) when \(t\leq 0\). Therefore it suffices to define \(\psi\) by the formula
\[\psi(t)=g(t+2)g(2-t)\]Using the above \(C^\infty\) Urysohn lemma in place of the general Urysohn lemma, we can construct a smooth partition of unity on a differentiable manifold.
References
[Lee] John M. Lee. Introduction to Smooth Manifolds, Graduate texts in mathematics, Springer, 2012
[War] Frank W. Warner. Foundations of Differentiable Manifolds and Lie Groups, Graduate texts in mathematics, Springer, 2013
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