Topological Tools for Manifolds
A construction made in one coordinate chart need not agree with a construction made in another. Partitions of unity let us combine local smooth functions while controlling their supports. The proof requires a cover that is locally finite and smooth functions that vanish outside prescribed neighbourhoods.
We develop these ingredients in that order. Local finiteness makes the eventual sums smooth: near any fixed point, only finitely many terms remain.
Lemma 1. Let \(M\) be a topological manifold. Then, there exists a countable collection of open sets \(\{X_i\}_{i \in \mathbb N}\) such that \(\overline X_j\) is compact and \(\overline X_j \subset X_{j+1}\) for all \(j\), and \(M = \bigcup_j X_j\).
Proof. Every point has arbitrarily small coordinate neighbourhoods with compact closure: choose a Euclidean ball whose closed ball lies inside a chart. From a countable basis, retain the elements whose closures are compact; these still cover \(M\), because a basis element contained in such a relatively compact neighbourhood has compact closure. Enumerate them as \(Y_1,Y_2,\ldots\).
Set \(X_1=Y_1\). Suppose \(X_j\) has compact closure. Choose finitely many basis elements \(Y_{i_1},\ldots,Y_{i_r}\) covering \(\overline X_j\), and put
It is open, contains \(\overline X_j\), and its closure is a closed subset of the compact union \(\overline Y_{j+1}\cup\bigcup_{k=1}^r\overline Y_{i_k}\), hence compact. Inductively \(Y_1\cup\cdots\cup Y_j\subset X_j\), proving exhaustion. The open set \(X_j\) itself need not be compact; it is its closure that is compact. ◻
Remark 2. The countable collection in Lemma 1 is called an exhaustion of \(M\).
Lemma 3 (Existence of Refinements). Let \(M\) be a topological manifold. For any open cover \(\mathscr U = \{U_\alpha\}_{\alpha \in A}\) of \(M\), there exists two countable family of open covers, \(\mathscr V = \{V_j\}_{j\in \mathbb N}\) and \(\mathscr W = \{W_j\}_{j\in \mathbb N}\) such that
\(\overline V_j\) is compact for all \(j\) and \(\overline V_j \subset W_j\),
\(\mathscr W\) is a refinement of \(\mathscr U\), i.e. for all \(j\), there exists some \(\alpha = \alpha(j)\) such that \(W_j \subset U_\alpha\),
\(\mathscr W\) is a locally finite cover, i.e. for any \(p \in M\), there exists a neighbourhood \(W_p\) such that \(W_p \cap W_j \neq \emptyset\) for only finitely many \(j\).
Proof. Take the exhaustion from Lemma 1 and set \(X_0=X_{-1}=\varnothing\). For each \(j\geq1\), the shell \(K_j=\overline X_j\setminus X_{j-1}\) is compact and lies in the open strip \(X_{j+1}\setminus\overline X_{j-2}\). The shells cover \(M\): for a point, take the first \(j\) for which it lies in \(X_j\).
For every \(p\in K_j\), choose \(U_{\alpha(p)}\ni p\). By small coordinate balls, choose open sets \(V_p,W_p\) with
Compactness gives finitely many \(V_p\)'s covering \(K_j\). Keep these pairs for every shell, and enumerate their countable union. Both resulting families cover \(M\), and the closure and refinement properties hold by construction.
To prove local finiteness, choose \(N\) with \(p\in X_N\). If \(j\geq N+2\), a set selected for shell \(j\) is disjoint from \(\overline X_{j-2}\supset X_N\). Thus the neighbourhood \(X_N\) meets only sets from shells \(1,\ldots,N+1\), of which finitely many were selected. This also covers the innermost shell, which must not be omitted. ◻
The refinement gives us small neighbourhoods whose closures remain inside larger ones. We now use smooth coordinates to place a function between each such pair: it will equal one on the smaller set and vanish outside the larger set.
Theorem 4 (Existence of Smooth Bump Functions on Compact Sets on Manifolds). Let \(M\) be a differentiable manifold, \(A \subset M\) be a compact subset, and \(U \subset M\) be an open set that contains \(A\). Then, there exists a bump function \(\varphi \in \mathscr C^{\infty}_c (M)\) such that \(0 \leq \varphi \leq 1\), \(\varphi \equiv 1\) on \(A\), and \(\operatorname{spt} \varphi \subset U\).
The construction starts on the real line with a function that is flat at the origin. Rescaling it gives a cutoff on an interval, and a radial version gives the Euclidean bump functions needed in charts.
Lemma 5. The function \(f : \mathbb R \to \mathbb R\) defined by
is smooth.
Proof. For every integer \(r\geq0\), the one-sided limit
follows by substituting \(t=1/x\): the power-series bound \(e^t\geq t^{r+1}/(r+1)!\) gives \(t^re^{-t}\leq(r+1)!/t\to0\). This is a right-hand limit; the formula \(e^{-1/x}\) is not used for negative \(x\).
For \(x>0\), induction gives
where differentiation yields the exact recurrence
Thus \(p_k\) is a polynomial of degree at most \(k\). To extend the derivatives through zero, suppose the \(k\)-th derivative is zero at zero and on the negative half-line. Its right difference quotient at zero is \(x^{-2k-1}e^{-1/x}p_k(x)\), which tends to zero by the preceding estimate. The left quotient is zero. Hence \(f^{(k+1)}(0)=0\), and the formula for \(f^{(k+1)}\) tends to zero from the right as well, proving continuity. Starting at \(k=0\) completes induction and proves smoothness. ◻
Lemma 6 (Existence of Bump Function on \(\mathbb R\)). Given any \(a, b \in \mathbb R\) such that \(a < b\), there exists a smooth function \(g : \mathbb R \to \mathbb R\) such that \(g \equiv 1\) on \(x \leq a\), \(g(x) \in (0,1)\) for \(x \in (a,b)\), and \(g \equiv 0\) on \(x \geq b\).
Proof. With \(f\) from Lemma 5, set
The denominator is positive: the inequalities \(b-x\leq0\) and \(x-a\leq0\) cannot both hold when \(a<b\). Thus \(g\) is a smooth quotient. For \(x\leq a\), the second term vanishes and the first is positive, so \(g=1\); for \(x\geq b\), the first vanishes and the second is positive, so \(g=0\). In between, both terms are positive, giving \(0<g<1\). ◻
Lemma 7 (Euclidean bump function). For \(0\leq a<b\), there is a smooth \(h:\mathbb R^n\to[0,1]\) equal to one on \(\overline B_a(0)\), strictly between zero and one when \(a<|x|<b\), and zero for \(|x|\geq b\).
Proof. Apply Lemma 6 with thresholds \(a^2<b^2\), and set \(h(x)=g(|x|^2)\). The squared norm is smooth everywhere, so this composition is smooth even at the origin, including when \(a=0\). The three value ranges follow from the corresponding ranges of \(|x|^2\). Its support is the closed ball of radius \(b\), hence compact. ◻
We can now transfer the Euclidean construction to charts. Compactness lets us use finitely many chart neighbourhoods to cover the set on which the cutoff must equal one.
Proof of Theorem 4. For every \(q\in A\), choose a chart \(\varphi_q:U_q\to\mathbb R^n\) with \(q\in U_q\subset U\), translate its coordinates so that \(\varphi_q(q)=0\), and choose \(0<a_q<b_q\) with \(\overline B_{b_q}(0)\subset\varphi_q(U_q)\). Pull the bump function of Lemma 7 back to \(U_q\) and extend it by zero outside \(U_q\). Call the result \(h_q\). Its support is contained in the compact set \(\varphi_q^{-1}(\overline B_{b_q})\subset U_q\); near every point outside this compact set it is identically zero, which proves smoothness of the extension across the chart boundary. Moreover \(h_q=1\) on \(\varphi_q^{-1}(B_{a_q})\).
Choose finitely many of these inner neighbourhoods covering \(A\), say those indexed by \(q_1,\ldots,q_N\). Define
Every factor lies in \([0,1]\), so \(0\leq\varphi\leq1\). At each point of \(A\), one factor is zero, giving \(\varphi=1\). Outside the finite union of the compact supports, every \(h_{q_i}\) is zero, so \(\varphi=0\). Thus \(\varphi\) is smooth, compactly supported, and its support lies in \(U\). If \(A=\varnothing\), take \(\varphi=0\). The finite product also avoids reversing the cutoff values in the final composition. ◻
A partition of unity requires one further step. Starting from nonnegative bump functions whose positive sets cover the manifold, divide each function by their sum. The denominator is positive everywhere, and local finiteness allows us to differentiate the sum.
Proposition 8 (Partition of unity). For any open cover \(\{U_\alpha\}_{\alpha\in A}\) of a smooth manifold \(M\), there are smooth functions \(f_\alpha:M\to[0,1]\) such that the family of their supports is locally finite, \(\operatorname{spt}f_\alpha\subset U_\alpha\), and \(\sum_\alpha f_\alpha=1\). Some \(f_\alpha\) may be identically zero.
Proof. Use Lemma 3 to obtain countable covers \(V_j,W_j\) with \(\overline V_j\) compact, \(\overline V_j\subset W_j\subset U_{\alpha(j)}\), and \(\{W_j\}\) locally finite. By Theorem 4 choose \(0\leq\varphi_j\leq1\), equal to one on \(\overline V_j\), with compact support in \(W_j\). The sum
is smooth because every point has a neighbourhood on which only finitely many summands are nonzero. It satisfies \(\Phi\geq1\), since the sets \(V_j\) cover \(M\). Therefore \(\psi_j=\varphi_j/\Phi\) is smooth, \(0\leq\psi_j\leq1\), and \(\sum_j\psi_j=1\).
Group terms by the original cover index:
These sums may have infinitely many terms globally, but they are locally finite and hence smooth. Nonnegativity and the total-sum identity give \(0\leq f_\alpha\leq1\) and \(\sum_\alpha f_\alpha=1\). The supports of the \(\psi_j\)'s form a locally finite family. Lemma 9 shows that each union \(\bigcup_{\alpha(j)=\alpha}\operatorname{spt}\psi_j\) is closed, so it contains \(\operatorname{spt}f_\alpha\) and lies in \(U_\alpha\). Near any point only finitely many original \(\psi_j\)'s can be nonzero; their finitely many labels \(\alpha(j)\) are the only possible supports of \(f_\alpha\) meeting a sufficiently small neighbourhood. This verifies local finiteness after regrouping. ◻
Lemma 9 (Properties of Locally Finiteness). Let \(M\) be a topological manifold, and \(\mathscr X\) be a locally finite collection of subsets of \(M\). Then,
\(\{\overline X : X \in \mathscr X\}\) is locally finite, and
\(\overline{\bigcup_{X \in \mathscr X} X} = \bigcup_{X \in \mathscr X} \overline X\).
Proof. Choose an open neighbourhood \(U\) of \(p\) meeting only \(X_1,\ldots,X_N\). If \(U\cap\overline X\ne\varnothing\), the openness of \(U\) and the definition of closure imply \(U\cap X\ne\varnothing\). Thus the same neighbourhood meets only those finitely many closures, proving part (a).
The inclusion \(\bigcup_X\overline X\subset\overline{\bigcup_X X}\) always holds. For the reverse, take \(p\notin\bigcup_X\overline X\), and choose \(U\) as above. Then
is an open neighbourhood of \(p\) disjoint from every \(X\). Hence \(p\notin\overline{\bigcup_X X}\). Equivalently, the locally finite union of the closed sets \(\overline X\) is closed. Keeping the intersection with \(U\) is necessary to exclude the other members of the family. ◻
Reading: Carmo (1992), Riemannian geometry.