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While trying to better understand why we use partitions of unity I stumbled across this post. There are a few points I do not understand:

This is exactly how to compute integrals in practice. But if we use it as our definition in theory, it becomes messy

Why exactly does it become messy? The definition pointed at in the post seems to be the following:

Let $\{(U_i,\phi_i)\}$ be an atlas for $S$ up to a null set consisting of disjoint charts. We define the integral of $\omega$ over $M$ as $$\int_S\omega := \sum_i\int_{U_i}\omega|_{U_i} := \sum_i\int_{V_i} (\phi_i^{-1})^*\omega|_{U_i}$$ where, for the integrands in the sum to be well-defined, we need the guarantee that

  • $(\phi_i^{-1})^*\omega|_{U_i}$ is compactly supported in some rectangle $V_i\supseteq\phi_i(U_i)$.
  • $(\phi_i^{-1})^*\omega|_{U_i} = f\ dx^1\wedge dx^2$ for some bounded function $f$.

Is there anything else we would need to make this definition proper?

If I understand the author of the post correctly, each of the atlases he suggests for the unit sphere has their own issues. What exactly are these? As far as I can tell these are:

  • Spherical coordinates: as $\phi(S) = \mathbb{R}^2$, our pullback form fails to be compactly supported. However, could we not 'patch' our previous definition so as to include a limit process as follows: $$\int_{U_i}\omega|_{U_i} := \lim_{r\to\infty}\int_{B(r)}(\phi^{-1})^*\omega|_{\phi^{-1}(B(r))}\ ?$$ where $B(r)$ is, say, the box $[-r,r]^2$. I've seen similar ideas used to attribute a value to integrals such as $$\int_0^1\frac{1}{\sqrt{x}}dx := \int_{(0,1)}\frac{1}{\sqrt{x}}dx$$ as opposed to, although possible, to define them in terms of partitions of unity.

  • By northern and southern hemispheres: the atlas is given by $$\Big\{\Big(S\cap\{x^3<0\},\phi_1\Big),\Big(S\cap\{x^3>0\},\phi_2\Big)\Big\} \ \ \text{where} \ \ \phi_i:(x,y,z)\mapsto (x,y) \ \ \text{ and } \ \ \phi_i^{-1}:(u,v)\mapsto \left(u,v,\pm\sqrt{1-u^2-v^2}\right).$$

Here the issue is that $f$ would become unbounded (why?):

convert a bounded integrand to an unbounded one -- look at the example with the square root.

  • Spherical coordinates: here I simply fail to see what the issue would be. What is it? Perhaps that 'our integrands will have discontinuities at the boundaries of the patches'? If so, how would this cause trouble?
Arctic Char
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Sam
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    To me, the other post is trying to say that you should use an open covering of coordinate charts, and hence the need of partition of unity, because they have to overlap. It is not a good idea to partition the manifold into pieces which don't overlap but are not open. In practice, you do the latter method of chopping, but in theory building, you have to do the open covering, and again partition of unity. – Three aggies Jan 12 '24 at 02:15
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    I view partitions of unity as a technical tool that makes certain proofs easier. They can also be useful when working with, say, Čech cohomology on a manifold. Otherwise, they are not practical when computing and understanding the geometric properties of integrals. I prefer to define an integral by decomposing a manifold into regions that are parameterized by rectangles and overlap only along their boundaries. The big catch, however, is that you need the highly nontrivial theorem that any smooth manifold can be decomposed this way. Even then I prefer to assume this theorem and go from there. – Deane Jan 18 '24 at 00:51

1 Answers1

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Thinking more about your question, I start to see that you have a very strong point. I have to agree that none of these charts is problematic for the purpose of doing integrals.

To me, the story is that using integrals to motivate partition of unity is not a very strong case, since integrals don't care about a subset with measure 0 (say a lower dimensional submanifold or a finite union of them). At the end of this post, I will mention some other cases for partition of unity (the original post also has some more).

So your formula of $$ \int_S \omega = \sum_i \int_{V_i} (\phi^{-1}_i)^*\omega|_{U_i}, $$ is totally valid, as long as you are sure that $S\backslash \bigcup_i U_i$ has measure 0.

Here is a second point that works in favor of your argument. You can't make the integral $\int_{V_i} (\phi^{-1}_i)^*\omega|_{U_i}$ converge by your choice, they are determined to converge or not by the original $S$ and $\omega$. (I am talking about Lebesgue integral, so convergence is absolute convergence -- there is no conditional convergence.)

What I am saying is that if you have two charts $(U_1,\phi_1)$ and $(U_2,\phi_2)$, then $$ \int_{\phi_1(U_1\cap U_2)} (\phi^{-1}_1)^*\omega|_{U_1\cap U_2} = \int_{\phi_2(U_1\cap U_2)} (\phi^{-1}_2)^*\omega|_{U_1\cap U_2}. $$ This follows from the change of variables formula, since $\phi_2\circ \phi_1^{-1}: \phi_1(U_1\cap U_2)\to \phi_2(U_1\cap U_2) $ is a diffeomorephism, and $$ (\phi_2\circ \phi_1^{-1})^*\big((\phi^{-1}_2)^*\omega|_{U_1\cap U_2}\big)=(\phi^{-1}_1)^*\omega|_{U_1\cap U_2}. $$

In this sense, there is nothing you can do to make the integral converge. Either it does or it doesn't.

That also means that yes, in the stereographic projection case, although $\phi(U)={\mathbb R}^2$ is not compact, you still have the integral convergent if the original integral is convergent. You can compute the integral using a limit procedure as you proposed for improper integrals, or I don't even think Lebesgue integral cares about if the domain is compact. Such kind of integrals over the whole space actually happen often, when you do integrals over, say, ${\mathbb C}P^1$ and you would use a coordinate chart ${\mathbb C}\subset {\mathbb C}P^1$.

I think the same goes for the hemisphere coordinates and the spherical coordinates. The square function is not $C^1$ at $u^2+v^2=1$, but I believe everything will still be fine if it were fine originally.

Now back to situations where you genuinely need partition of unity. These questions would require you to consider the whole manifold with nothing left out. Then you have to use partition of unity for an open covering since you know more locally and you want to produce a global result. Here are some examples.

  1. Existence of a Riemannian metric on any manifold.
  2. The Mayer-Vietoris short exact sequence for de Rham cohomology $$ 0\to \Omega^k(X)\to \Omega^k(U)\oplus \Omega^k(V)\to \Omega^k(U\cap V)\to 0. $$
  3. Even constructing an inward vector field on the boundary of a manifold.

The list goes on, but you get the idea.

Three aggies
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