I've been studying Spivak's Calculus on Manifolds and I'm really not getting what's behind partitions of unity. Spivak introduces the topic with the following theorem:

Let $A\subset \Bbb R^n$ and let $\mathcal{O}$ be an open cover of $A$. Then there is a collection $\Phi$ of $C^\infty$ functions $\varphi$ defined in an open set containing $A$, with the following properties:

  1. For each $x \in A$ we have $0 \leq \varphi(x) \leq 1$.

  2. For each $x \in A$ there is an open set $V$ containing $x$ such that all but finitely many $\varphi \in \Phi$ are $0$ on $V$.

  3. For each $x \in A$ we have $\sum_{\varphi \in \Phi}\varphi(x)=1$ (by 2 for each $x$ their sum is finite in some open set containing $x$).

  4. For each $\varphi \in \Phi$ there is an open set $U$ in $\mathcal{O}$ such that $\varphi = 0$ outside of some closed set contained in $U$.

The point is that I've heard that partitions of unity are able to transfer local results to global results, and this is of great importance, but I'm not really getting the intution behind this theorem. I mean, why a collection of functions with these four properties is able to do such job?

When I see a theorem/definition, I try to really get the intution behind it: "why we should really think about doing things this way", because I think that this is a good way to understand what we are doing, but with partitions of unity I'm really not getting the idea.

While Spivak uses this just for integration on Calculus on Manifolds for what I've seem, in his Differential Geometry books he starts to use it really more generally to get global results from local ones (obtained with charts).

So, given the great importance of this topic, what's the real intuition behind this theorem and partitions of unity in general?


5 Answers 5


In a few words, the point of partitions of unity is to take functions (or differential forms or vector fields or tensor fields, in general) that are locally defined, bump them off so they're smoothly $0$ outside their domain of definition, and then add them all up to get something globally defined.

For example, suppose you have a surface $S$ in $\mathbb R^3$ that you can locally write as $f=0$, but you don't know how to do so globally. You can cover $S$ with open sets $U_i\subset\mathbb R^3$ on which you have smooth functions $f_i\colon U_i\to\mathbb R$ with $S\cap U_i = \{x\in U_i: f_i(x)=0\}$. Consider $\Phi = \{\phi_i\}$, where $\phi_i$ is supported in $U_i$. Then $f=\sum \phi_if_i$ will define a smooth function with $f=0$ on $S$. If you want $f$ to be zero only on $S$, you can take an additional open sets $U_0 = \mathbb R^3 - S$, set $f_0 = 1$, and throw $\phi_0f_0$ into your sum.

  • 19
    $\begingroup$ Hi Ted: I know this isn't what was originally asked, but for the record it should be said that more hypotheses are needed before you can conclude that this construction yields a function that vanishes only on $S$. First, you have to assume $S$ is closed in $\mathbb R^3$. Second, you need to use the fact that a closed embedded surface in $\mathbb R^3$ separates $\mathbb R^3$ into two components (which I think is nontrivial to prove), and then require that the local defining functions are all positive in the same component. Then you need to set $f_0=1$ on one component and $-1$ on the other. $\endgroup$
    – Jack Lee
    Commented Jun 2, 2015 at 0:11
  • $\begingroup$ You are summing functions with different domains. $\endgroup$
    – Integral
    Commented Jul 30, 2018 at 9:27
  • $\begingroup$ @JackLee Other than showing/assuming $S \subseteq \mathbb{R}^3$ is closed, wouldn't it suffice to just consider $f=\sum_i\phi_if_i^2$ instead? (We may not be able to talk about the components as $\{f<0\}$ and $\{f>0\}$ anymore, but that doesn't seem necessary) $\endgroup$ Commented Jan 4 at 4:38
  • $\begingroup$ @Brian There were lots of things that went unsaid and were sloppy in my glib answer a decade ago. Of course, we need $S$ closed (or else I couldn’t have said $U_0$ was open). We probably want $0$ to beca regular value of $f$ — in which case your trick won’t work. And, as Jack suggested, we want the normal bundle of $S$ to be trivial to complete the task. $\endgroup$ Commented Jan 4 at 5:08
  • 1
    $\begingroup$ @BrianMoehring: Yes, you can do that. And in fact, a modification of the same argument shows that you can find a global smooth function that vanishes on any arbitrary closed subset, not just a surface. (See Thm. 2.29 in my Introduction to Smooth Manifolds, 2nd ed.) When I was worrying about two sides of $S$, I was thinking about trying to find a function for which $S$ is a regualar level set (which, admittedly, Ted Shifrin didn't say was a goal of his construction). $\endgroup$
    – Jack Lee
    Commented Jan 4 at 5:10

Here's what I say when I'm teaching this. These comments are usually spread over several lectures, but I'll say them all at once here.

The first use of partitions of unity is usually to construct integrals over manifolds. For example, let $S$ be a surface in $\mathbb{R}^3$, and say we want to integrate a $2$-form $\omega$ over it (or alternatively, integrate the flux of a vector field across it.)

What we would probably do in practice is break $S$ up into patches $S = \bigcup U_i$ with parametrizations $f_i : P_i \to U_i$ by various open sets $P_i \subset \mathbb{R}^2$, pull the differential form back across the parametrization and integrate on each $P_i$. We then need the patches $U_i$ to cover $S$ up to measure $0$. For example, if $S$ is the unit sphere, we might use a single patch in spherical coordinates with $P = (-\pi, \pi) \times (-\pi/2, \pi/2)$ and $f(\theta, \phi) = (\cos \theta \cos \phi, \sin \theta \cos \phi, \sin \phi)$. Alternatively we might parameterize the northern and southern hemispheres separately, taking $P_1 = P_2 = \{ (u,v) : u^2+v^2<1 \}$, with $f_1(u,v) = (u,v,\sqrt{1-u^2-v^2})$ and $f_2(u,v) = (u,v,-\sqrt{1-u^2-v^2})$. Or we might use a stereographic projection: $P = \mathbb{R}^2$, $f(u,v) = \left( \tfrac{2u}{1+u^2+v^2}, \tfrac{2v}{1+u^2+v^2}, \tfrac{1-u^2-v^2}{1+u^2+v^2} \right)$.

This is exactly how to compute integrals in practice. But if we use it as our definition in theory, it becomes messy -- we have to talk about the combinatorics of how the patches fit together, and our integrands will have discontinuities at the boundaries of the patches. We also might convert a compactly supported integrand to a noncompactly supported one -- look at the example of a stereographic projection above -- or convert a bounded integrand to an unbounded one -- look at the example with the square root.

A partition of unity allows us to blend from one patch to another more smoothly. For example, instead of saying that every point is either exactly in the northern hemisphere, or exactly in the southern hemisphere, we have two functions $\phi_1+\phi_2$ with $\phi_1+\phi_2=1$, where $\phi_1$ measures how much we will count the point toward the northern integral, and $\phi_2$ measures how much we will count the point toward the southern integral. This gives integrals that are much worse for hand computation, but have cleaner theoretical properties.

Incidentally, I believe this should be better for machine Monte Carlo integration. That is to say, suppose I want to integrate a $2$-form $\omega$ over the sphere $S^2 \subset \mathbb{R}^3$. One approach would be to parametrize the northern hemisphere and southern hemisphere separately, pulling $\omega$ back to forms supported on discs in two copies of $\mathbb{R}^2$, with discontinuities at the boundary of the disc, and compute these integrals by Monte Carlo. Alternatively, I could use a continuous partition of unity supported on the open sets $z<0.1$ and $z>-0.1$, and pull back by stereographic coordinates to slightly larger discs; my integrands would then be continuous. I believe Monte Carlo integration usually prefers continuous integrands, as that way it is not important to determine exactly which side of the discontinuits a sample random point lies on.

Later uses of partitions of unity are also often of the form "I would like to chop my manifold into pieces, but that is too discontinuous an operation." For example, let's show that every short exact sequence of vector bundles splits. Let $0 \to A \to B \to C \to 0$ be the short exact sequence and let $X$ be the manifold. One would like to cut $X$ into pieces $U_i$ where the bundles are trivial and write down a section $\sigma_i : C \to B$ on each $U_i$. But gluing the $\sigma_i$ together is not continuous. If instead we take a partition of unity $\phi_i$ and write $\sigma = \sum \phi_i \sigma_i$, then $\sigma$ is a smoother version of gluing the $\sigma_i$, and it is continuous.

My mental metaphor for a partition of unity is feathering out paint. If your paint stops abruptly at the edge of the brush stroke, it will leave a visible line even once you paint the wall next to it. Instead, you need to smear out the edge of your stroke so it thins out gradually. I haven't tried bringing a paint can into class though yet!


The idea behind a number of proofs is as follows.

  • We want to prove theorem "A" for certain functions $f$.

  • If theorem is true for two functions $f_1$ and $f_2$ , then it is true for $f_1 + f_2$.

  • The theorem is true for the same class of functions locally. For example, the space might be a manifold and the theorem is true for compactly supported functions.

  • So you have the theorem true for a covering of the space; and you then construct a partition of unity $\varphi_i$ subordinate to this cover. The theorem being true for each $\varphi_i f$ being compactly supported, it is true for their sum $f$, this sum being locally finite; so at each point it is a finite sum.


An example: Assume you want to prove Gauss' theorem for a complicated bounded "sponge" $\Omega$ in ${\mathbb R}^3$. You are not willing to describe this sponge in detail, but you can guarantee that at each point $p\in\partial\Omega$ one can draw a cube $C$ with center $p$ and side-length $2h$, such that introducing $(x,y,z)$-coordinates in $C$ with $p=(0,0,0)$ one has $$\Omega\cap C=\{(x,y,z)\in C\>|\>z>\psi(x,y)\}$$ for some $C^1$-function $\psi$.

Setting up a (finite!) partition of unity $(\phi_\iota)_{\iota\in I}$ where the support of each $\phi_\iota$ is either completely contained in $\Omega$ or in one of these cubes allows one to do the "fine computation" for the proof of Gauss' theorem in a single cube and then sum it all up in one big sweep.


It is not entirely clear from your question whether you wish to develop intuition for why such a thing as a partition of unity would exist, or rather why it would be useful. If it is the former, perhaps the following elementary comment may help: consider the constant function on $\mathbb{R}$ identically equal to $1$. Then the function can be decomposed into a sum of indicator functions of the intervals $[n,n+1)$ where $n$ ranges over all the integers. A partition of unity is similar except that the functions being summed are expected to be smooth, which turns out not to be much more difficult to achieve.


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