Desvl's blog https://desvl.xyz/icon.png 2020-11-28T00:22:20.481Z https://desvl.xyz/ Desvl Hexo Study Vector Bundle in a Relatively Harder Way - Definition https://desvl.xyz/2020/11/27/vector-bundle-1/ 2020-11-27T10:22:57.000Z 2020-11-28T00:22:20.481Z Motivation

Direction is a considerable thing. For example take a look at this picture (by David Gunderman): The position of the red ball and black ball shows that this triple of balls turns upside down every time they finish one round. This wouldn’t happen if this triple were on a normal band, which can be denoted by $S^1 \times (0,1)$. What would happen if we try to describe their velocity on the Möbius band, both locally and globally? There must be some significant difference from a normal band. If we set some move pattern on balls, for example let them run horizontally or zig-zagly, hopefully we get different set of vectors. those vectors can span some vector spaces as well.

## A Formal Construction

Here and in the forgoing posts, we will try to develop purely formally certain functorial constructions having to do with vector bundles. It may be overly generalized, but we will offer some examples to make it concrete.

Let $M$ be a manifold (of class $C^p$, where $p \geq 0$ and can be set to $\infty$) modeled on a Banach space $\mathbf{E}$. Let $E$ be another topological space and $\pi: E \to M$ a surjective $C^p$-morphism. A vector bundle is a topological construction associated with $M$ (base space), $E$ (total space) and $\pi$ (bundle projection) such that, roughly speaking, $E$ is locally a product of $M$ and $\mathbf{E}$.

We use $\mathbf{E}$ instead of $\mathbb{R}^n$ to include the infinite dimensional cases. We will try to distinguish finite-dimensional and infinite-dimensional Banach spaces here. There are a lot of things to do, since, for example, infinite dimensional Banach spaces have no countable Hamel basis, while the finite-dimensional ones have finite ones (this can be proved by using the Baire category theorem).

Next we will show precisely how $E$ locally becomes a product space. Let $\mathfrak{U}=(U_i)_i$ be an open covering of $M$, and for each $i$, suppose that we are given a mapping

satisfying the following three conditions.

VB 1 $\tau_i$ is a $C^p$ diffeomorphism making the following diagram commutative: where $pr$ is the projection of the first component: $(x,y) \mapsto x$. By restricting $\tau_i$ on one point of $U_i$, we obtain an isomorphism on each fiber $\pi^{-1}(x)$:

VB 2 For each pair of open sets $U_i$, $U_j \in \mathfrak{U}$, we have the map

to be a toplinear isomorphism (that is, it preserves $\mathbf{E}$ for being a topological vector space).

VB 3 For any two members $U_i$, $U_j \in \mathfrak{U}$, we have the following function to be a $C^p$-morphism:

REMARKS. As with manifold, we call the set of 2-tuples $(U_i,\tau_i)_i$ a trivializing covering of $\pi$, and that $(\tau_i)$ are its trivializing maps. Precisely, for $x \in U_i$, we say $U_i$ or $\tau_i$ trivializes at $x$.

Two trivializing coverings for $\pi$ is said to be VB-equivalent if taken together they also satisfy conditions of VB 2 and VB 3. It’s immediate that VB-equivalence is an equivalence relation and we leave the verification to the reader. It is this VB-equivalence class of trivializing coverings that determines a structure of vector bundle on $\pi$. With respect to the Banach space $\mathbf{E}$, we say that the vector bundle has fiber $\mathbf{E}$, or is modeled on $\mathbf{E}$.

Next we shall give some motivations of each condition. Each pair $(U_i,\tau_i)$ determines a local product of ‘a part of the manifold’ and the model space, on the latter of which we can deploy the direction with ease. This is what VB 1 tells us. But that’s far from enough if we want our vectors fine enough. We do want the total space $E$ to actually be able to qualify our requirements. As for VB 2, it is ensured that using two different trivializing maps will give the same structure of some Banach spaces (with equivalent norms). According to the image of $\tau_{ix}$, we can say, for each point $x \in X$, which can be determined by a fiber $\pi^{-1}(x)$ (the pre-image of $\tau_{ix}$), can be given another Banach space by being sent via $\tau_{jx}$ for some $j$. Note that $\pi^{-1}(x) \in E$, the total space. In fact, VB 2 has an equivalent alternative:

VB 2’ On each fiber $\pi^{-1}(x)$ we are given a structure of Banach space as follows. For $x \in U_i$, we have a toplinear isomorphism which is in fact the trivializing map:

As stated, VB 2 implies VB 2’. Conversely, if VB 2’ is satisfied, then for open sets $U_i$, $U_j \in \mathfrak{U}$, and $x \in U_i \cap U_j$, we have $\tau_{jx} \circ \tau_{ix}^{-1}:\mathbf{E} \to \mathbf{E}$ to be an toplinear isomorphism. Hence, we can consider VB 2 or VB 2’ as the refinement of VB 1.

In finite dimensional case, one can omit VB 3 since it can be implied by VB 2, and we will prove it below.

(Lemma) Let $\mathbf{E}$ and $\mathbf{F}$ be two finite dimensional Banach spaces. Let $U$ be open in some Banach space. Let

be a $C^p$-morphism such that for each $x \in U$, the map

given by $f_x(v)=f(x,v)$ is a linear map. Then the map of $U$ into $L(\mathbf{E},\mathbf{F})$ given by $x \mapsto f_x$ is a $C^p$-morphism.

PROOF. Since $L(\mathbf{E},\mathbf{F})=L(\mathbf{E},\mathbf{F_1}) \times L(\mathbf{E},\mathbf{F_2}) \times \cdots \times L(\mathbf{E},\mathbf{F_n})$ where $\mathbf{F}=\mathbf{F_1} \times \cdots \times \mathbf{F_n}$, by induction on the dimension of $\mathbf{F}$ and $\mathbf{E}$, it suffices to assume that $\mathbf{E}$ and $\mathbf{F}$ are toplinearly isomorphic to $\mathbb{R}$. But in that case, the function $f(x,v)$ can be written $g(x)v$ for some $g:U \to \mathbb{R}$. Since $f$ is a morphism, it follows that as a function of each argument $x$, $v$ is also a morphism, Putting $v=1$ shows that $g$ is also a morphism, which finishes the case when both the dimension of $\mathbf{E}$ and $\mathbf{F}$ are equal to $1$, and the proof is completed by induction. $\blacksquare$

To show that VB 3 is implied by VB 2, put $\mathbf{E}=\mathbf{F}$ as in the lemma. Note that $\tau_j \circ \tau_i^{-1}$ maps $U_i \cap U_j \times \mathbf{E}$ to $\mathbf{E}$, and $U_i \cap U_j$ is open, and for each $x \in U_i \cap U_j$, the map $(\tau_j \circ \tau_i^{-1})_x=\tau_{jx} \circ \tau_{ix}^{-1}$ is toplinear, hence linear. Then the fact that $\varphi$ is a morphism follows from the lemma.

## Examples

### Trivial bundle

Let $M$ be any $n$-dimensional smooth manifold that you are familiar with, then $pr:M \times \mathbb{R}^n \to M$ is actually a vector bundle. Here the total space is $M \times \mathbb{R}^n$ and the base is $M$ and $pr$ is the bundle projection but in this case it is simply a projection. Intuitively, on a total space, we can determine a point $x \in M$, and another component can be any direction in $\mathbb{R}^n$, hence a vector.

We need to verify three conditions carefully. Let $(U_i,\varphi_i)_i$ be any atlas of $M$, and $\tau_i$ is the identity map on $U_i$ (which is naturally of $C^p$). We claim that $(U_i,\tau_i)_i$ satisfy the three conditions, thus we get a vector bundle.

For VB 1 things are clear: since $pr^{-1}(U_i)=U_i \times \mathbb{R}^n$, the diagram is commutative. Each fiber $pr^{-1}(x)$ is essentially $(x) \times \mathbb{R}^n$, and still, $\tau_{jx} \circ \tau_{ix}^{-1}$ is the identity map between $(x) \times \mathbb{R}^n$ and $(x) \times \mathbb{R}^n$, under the same Euclidean topology, hence VB 2 is verified, and we have no need to verify VB 3.

### Möbius band

First of all, imagine you have embedded a circle into a Möbius band. Now we try to give some formal definition. As with quotient topology, $S^1$ can be defined as

where $I$ is the unit interval and $0 \sim_1 1$ (identifying two ends). On the other hand, the infinite Möbius band can be defined by

where $(0,v) \sim_2 (1,-v)$ for all $v \in \mathbb{R}$ (not only identifying two ends of $I$ but also ‘flips’ the vertical line). Then all we need is a natural projection on the first component:

And the verification has few difference from the trivial bundle. Quotient topology of Banach spaces follows naturally in this case, but things might be troublesome if we restrict ourself in $\mathbb{R}^n$.

### Tangent bundle of the sphere

The first example is relatively rare in many senses. By $S^n$ we mean the set in $\mathbb{R}^{n+1}$ with

and the tangent bundle can be defined by

where, of course, $\mathbf{x} \in S^n$ and $\mathbf{y} \in \mathbb{R}^{n+1}$. The vector bundle is given by $pr:TS^n \to S^n$ where $pr$ is the projection of the first factor. This total space is of course much finer than $M \times \mathbb{R}^n$ in the first example. Each point in the manifold now is associated with a tangent space $T_x(M)$ at this point.

More generally, we can define it in any Hilbert space $H$, for example, $L^2$ space:

where

The projection is natural:

But we will not cover the verification in this post since it is required to introduce the abstract definition of tangent vectors. This will be done in the following post.

## There are still many things remain undiscovered

We want to study those ‘vectors’ associated to some manifold both globally and locally. For example we may want to describe the tangent line of some curves at some point without heavily using elementary calculus stuff. Also, we may want to describe the vector bundle of a manifold globally, for example, when will we have a trivial one? Can we classify the manifold using the behavior of the bundle? Can we make it a little more abstract, for example, consider the class of all isomorphism bundles? How do one bundle transform to another? But to do this we need a big amount of definitions and propositions.

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The Big Three Pt. 6 - Closed Graph Theorem with Applications https://desvl.xyz/2020/11/13/big-3-pt-6/ 2020-11-13T14:45:17.000Z 2020-11-19T07:31:31.670Z (Before everything: elementary background of topology and vector spaces, Banach spaces, is assumed.)

## A surprising result of Banach spaces

We can define several relations between two norms. Suppose we have a topological vector space $X$ and two norms $\lVert \cdot \rVert_1$ and $\lVert \cdot \rVert_2$. One says $\lVert \cdot \rVert_1$ is weaker than $\lVert \cdot \rVert_2$ if there is $K>0$ such that $\lVert x \rVert_1 \leq K \lVert x \rVert_2$ for all $x \in X$. Two norms are equivalent if each is weaker than the other (trivially this is a equivalence relation). The idea of stronger and weaker norms is related to the idea of the “finer” and “coarser” topologies in the setting of topological spaces.

So what about their limit of convergence? Unsurprisingly this can be verified with elementary $\epsilon-N$ arguments. Suppose now $\lVert x_n - x \rVert_1 \to 0$ as $n \to 0$, we immediately have

for some large enough $n$. Hence $\lVert x_n - x \rVert_2 \to 0$ as well. But what about the converse? We give a new definition of equivalence relation between norms.

(Definition) Two norms $\lVert \cdot \rVert_1$ and $\lVert \cdot \rVert_2$ of a topological vector space are compatible if given that $\lVert x_n - x \rVert_1 \to 0$ and $\lVert x_n - y \rVert_2 \to 0$ as $n \to \infty$, we have $x=y$.

By the uniqueness of limit, we see if two norms are equivalent, then they are compatible. And surprisingly, with the help of the closed graph theorem we will discuss in this post, we have

(Theorem 1) If $\lVert \cdot \rVert_1$ and $\lVert \cdot \rVert_2$ are compatible, and both $(X,\lVert\cdot\rVert_1)$ and $(X,\lVert\cdot\rVert_2)$ are Banach, then $\lVert\cdot\rVert_1$ and $\lVert\cdot\rVert_2$ are equivalent.

This result looks natural but not seemingly easy to prove, since one find no way to build a bridge between the limit and a general inequality. But before that, we need to elaborate some terminologies.

## Preliminaries

(Definition) For $f:X \to Y$, the graph of $f$ is defined by

If both $X$ and $Y$ are topological spaces, and the topology of $X \times Y$ is the usual one, that is, the smallest topology that contains all sets $U \times V$ where $U$ and $V$ are open in $X$ and $Y$ respectively, and if $f: X \to Y$ is continuous, it is natural to expect $G(f)$ to be closed. For example, by taking $f(x)=x$ and $X=Y=\mathbb{R}$, one would expect the diagonal line of the plane to be closed.

(Definition) The topological space $(X,\tau)$ is an $F$-space if $\tau$ is induced by a complete invariant metric $d$. Here invariant means that $d(x+z,y+z)=d(x,y)$ for all $x,y,z \in X$.

A Banach space is easily to be verified to be a $F$-space by defining $d(x,y)=\lVert x-y \rVert$.

(Open mapping theorem) See this post

By definition of closed set, we have a practical criterion on whether $G(f)$ is closed.

(Proposition 1) $G(f)$ is closed if and only if, for any sequence $(x_n)$ such that the limits

exist, we have $y=f(x)$.

In this case, we say $f$ is closed. For continuous functions, things are trivial.

(Proposition 2) If $X$ and $Y$ are two topological spaces and $Y$ is Hausdorff, and $f:X \to Y$ is continuous, then $G(f)$ is closed.

Proof. Let $G^c$ be the complement of $G(f)$ with respect to $X \times Y$. Fix $(x_0,y_0) \in G^c$, we see $y_0 \neq f(x_0)$. By the Hausdorff property of $Y$, there exists some open subsets $U \subset Y$ and $V \subset Y$ such that $y_0 \in U$ and $f(x_0) \in V$ and $U \cap V = \varnothing$. Since $f$ is continuous, we see $W=f^{-1}(V)$ is open in $X$. We obtained a open neighborhood $W \times U$ containing $(x_0,y_0)$ which has empty intersection with $G(f)$. This is to say, every point of $G^c$ has a open neighborhood contained in $G^c$, hence a interior point. Therefore $G^c$ is open, which is to say that $G(f)$ is closed. $\square$ REMARKS. For $X \times Y=\mathbb{R} \times \mathbb{R}$, we have a simple visualization. For $\varepsilon>0$, there exists some $\delta$ such that $|f(x)-f(x_0)|<\varepsilon$ whenever $|x-x_0|<\delta$. For $y_0 \neq f(x_0)$, pick $\varepsilon$ such that $0<\varepsilon<\frac{1}{2}|f(x_0)-y_0|$, we have two boxes ($CDEF$ and $GHJI$ on the picture), namely

and

In this case, $B_2$ will not intersect the graph of $f$, hence $(x_0,y_0)$ is an interior point of $G^c$.

The Hausdorff property of $Y$ is not removable. To see this, since $X$ has no restriction, it suffices to take a look at $X \times X$. Let $f$ be the identity map (which is continuous), we see the graph

is the diagonal. Suppose $X$ is not Hausdorff, we reach a contradiction. By definition, there exists some distinct $x$ and $y$ such that all neighborhoods of $x$ contain $y$. Pick $(x,y) \in G^c$, then all neighborhoods of $(x,y) \in X \times X$ contain $(x,x)$ so $(x,y) \in G^c$ is not a interior point of $G^c$, hence $G^c$ is not open.

Also, as an immediate consequence, every affine algebraic variety in $\mathbb{C}^n$ and $\mathbb{R}^n$ is closed with respect to Euclidean topology. Further, we have the Zariski topology $\mathcal{Z}$ by claiming that, if $V$ is an affine algebraic variety, then $V^c \in \mathcal{Z}$. It’s worth noting that $\mathcal{Z}$ is not Hausdorff (example?) and in fact much coarser than the Euclidean topology although an affine algebraic variety is both closed in the Zariski topology and the Euclidean topology.

## The closed graph theorem

After we have proved this theorem, we are able to prove the theorem about compatible norms. We shall assume that both $X$ and $Y$ are $F$-spaces, since the norm plays no critical role here. This offers a greater variety but shall not be considered as an abuse of abstraction.

(The Closed Graph Theorem) Suppose

(a) $X$ and $Y$ are $F$-spaces,

(b) $f:X \to Y$ is linear,

(c) $G(f)$ is closed in $X \times Y$.

Then $f$ is continuous.

In short, the closed graph theorem gives a sufficient condition to claim the continuity of $f$ (keep in mind, linearity does not imply continuity). If $f:X \to Y$ is continuous, then $G(f)$ is closed; if $G(f)$ is closed and $f$ is linear, then $f$ is continuous.

Proof. First of all we should make $X \times Y$ an $F$-space by assigning addition, scalar multiplication and metric. Addition and scalar multiplication are defined componentwise in the nature of things:

The metric can be defined without extra effort:

Then it can be verified that $X \times Y$ is a topological space with translate invariant metric. (Potentially the verifications will be added in the future but it’s recommended to do it yourself.)

Since $f$ is linear, the graph $G(f)$ is a subspace of $X \times Y$. Next we quote an elementary result in point-set topology, a subset of a complete metric space is closed if and only if it’s complete, by the translate-invariance of $d$, we see $G(f)$ is an $F$-space as well. Let $p_1: X \times Y \to X$ and $p_2: X \times Y \to Y$ be the natural projections respectively (for example, $p_1(x,y)=x$). Our proof is done by verifying the properties of $p_1$ and $p_2$ on $G(f)$.

For simplicity one can simply define $p_1$ on $G(f)$ instead of the whole space $X \times Y$, but we make it a global projection on purpose to emphasize the difference between global properties and local properties. One can also write $p_1|_{G(f)}$ to dodge confusion.

Claim 1. $p_1$ (with restriction on $G(f)$) defines an isomorphism between $G(f)$ and $X$.

For $x \in X$, we see $p_1(x,f(x)) = x$ (surjectivity). If $p_1(x,f(x))=0$, we see $x=0$ and therefore $(x,f(x))=(0,0)$, hence the restriction of $p_1$ on $G$ has trivial kernel (injectivity). Further, it’s trivial that $p_1$ is linear.

Claim 2. $p_1$ is continuous on $G(f)$.

For every sequence $(x_n)$ such that $\lim_{n \to \infty}x_n=x$, we have $\lim_{n \to \infty}f(x_n)=f(x)$ since $G(f)$ is closed, and therefore $\lim_{n \to \infty}p_1(x_n,f(x_n)) =x$. Meanwhile $p_1(x,f(x))=x$. The continuity of $p_1$ is proved.

Claim 3. $p_1$ is a homeomorphism with restriction on $G(f)$.

We already know that $G(f)$ is an $F$-space, so is $X$. For $p_1$ we have $p_1(G(f))=X$ is of the second category (since it’s an $F$-space and $p_1$ is one-to-one), and $p_1$ is continuous and linear on $G(f)$. By the open mapping theorem, $p_1$ is an open mapping on $G(f)$, hence is a homeomorphism thereafter.

Claim 4. $p_2$ is continuous.

This follows the same way as the proof of claim 2 but much easier since we have no need to care about $f$.

Now things are immediate once one realizes that $f=p_2 \circ p_1|_{G(f)}^{-1}$, and hence $f$ is continuous. $\square$

## Applications

Before we go for theorem 1 at the beginning, we drop an application on Hilbert spaces.

Let $T$ be a bounded operator on the Hilbert space $L_2([0,1])$ so that if $\phi \in L_2([0,1])$ is a continuous function so is $T\phi$. Then the restriction of $T$ to $C([0,1])$ is a bounded operator of $C([0,1])$.

Now we go for the identification of norms. Define

i.e. the identity map between two Banach spaces (hence $F$-spaces). Then $f$ is linear. We need to prove that $G(f)$ is closed. For the convergent sequence $(x_n)$

we have

Hence $G(f)$ is closed. Therefore $f$ is continuous, hence bounded, we have some $K$ such that

By defining

we see $g$ is continuous as well, hence we have some $K’$ such that

Hence two norms are weaker than each other.

## The series

Since there is no strong reason to write more posts on this topic, i.e. the three fundamental theorems of linear functional analysis, I think it’s time to make a list of the series. It’s been around half a year.

• Walter Rudin, Functional Analysis
• Peter Lax, Functional Analysis
• Jesús Gil de Lamadrid, Some Simple Applications of the Closed Graph Theorem
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Partition of Unity on Different Manifolds (Part 1. Introduction) https://desvl.xyz/2020/11/08/Partition-of-Unity-on-Different-Manifolds/ 2020-11-08T15:25:01.000Z 2020-11-08T15:25:01.182Z An application of partition of unity

Partition of unity builds a bridge between local properties and global properties. A nice example is the Stokes’ theorem on manifolds.

Suppose $\omega$ is a $(n-1)$-form with compact support on a oriented manifold $M$ of dimension $n$ and if $\partial{M}$ is given the induced orientation, then

This theorem can be proved in two steps. First, by Fubini’s theorem, one proves the identity on $\mathbb{R}^n$ and $\mathbb{H}^n$. Second, for the general case, let $(U_\alpha)$ be an oriented atlas for $M$ and $(\rho_\alpha)$ a partition of unity to $(U_\alpha)$, one naturally writes $\omega=\sum_{\alpha}\rho_\alpha\omega$. Since $\int_M d\omega=\int_{\partial M}\omega$ is linear with respect to $\omega$, it suffices to prove it only for $\rho_\alpha\omega$. Note that the support of $\rho_\alpha\omega$ is contained in the intersection of supports of $\rho_\alpha$ and $\omega$, hence a compact set.

On the other hand, $U_\alpha$ is diffeomorphic to either $\mathbb{R}^n$ or $\mathbb{H}^n$, it is immediate that

Which furnishes the proof for the general case.

As is seen, to prove a global thing, we do it locally. If you have trouble with these terminologies, never mind. We will go through this right now (in a more abstract way however). If you are familiar with them however, fell free to skip.

## Prerequisites

### Manifold (of finite or infinite dimension)

Throughout, we use bold letters like $\mathbf{E}$, $\mathbf{F}$ to denote Banach spaces. We will treat Euclidean spaces as a case instead of our restriction. Indeed since Banach spaces are not necessarily of finite dimension, our approach can be troublesome. But the benefit is a better view of abstraction.

Let $X$ be a set. An atlas of class $C^p$ ($p \geq 0$) on $X$ is a collection of pairs $(U_i,\varphi_i)$ where $i$ ranges through some indexing set, satisfying the following conditions:

AT 1. Each $U_i$ is a subset of $X$ and $\bigcup_{i}U_i=X$.

AT 2. Each $\varphi_i$ is a bijection of $U_i$ onto an open subset $\varphi_iU_i$ of some Banach space $\mathbf{E}_i$ and for any $i$ and $j$, $\phi_i(U_i \cap U_j)$ is open in $E_i$.

AT 3. The map

is a $C^p$-isomorphism for all $i$ and $j$.

One should be advised that isomorphism here does not come from group theory, but category theory. Precisely speaking, it’s the isomorphism in the category $\mathfrak{O}$ whose objects are the continuous maps of Banach spaces and whose morphisms are the continuous maps of class $C^p$.

Also, by setting $\tau_X=(U_i)_i$, we see $\tau_X$ is a topology, and $\varphi_i$ are topological isomorphisms. Also, we see no need to assume that $X$ is Hausdorff unless we start with Hausdorff spaces. Lifting this restriction gives us more freedom (also sometimes more difficulty to some extent though).

For condition AT 2, we did not require that the vector spaces be the same for all indexes $i$, or even that they be toplinearly isomorphic. If they are all equal to the same space $\mathbf{E}$, then we say that the atlas is an $\mathbf{E}$-atlas.

Suppose that we are given an open subset $U$ of $X$ and a topological isomorphism $\phi:U \to U’$ onto an open subset of some Banach space $\mathbb{E}$. We shall say that $(U,\varphi)$ is compatible with the atlas $(U_i,\varphi_i)_i$ if each map $\varphi\circ\varphi^{-1}$ is a $C^p$-isomorphism. Two atlas are said to be compatible if each chart of one is compatible with other atlas. It can be verified that this is a equivalence relation. An equivalence relation of atlases of class $C^p$ on $X$ is said to define a structure of $C^p$-manifold on $X$. If all the vector spaces $\mathbf{E}_i$ in some atlas are toplinearly isomorphic, we can find some universal $\mathbf{E}$ that is equal to all of them. In this case, we say $X$ is a $\mathbf{E}$-manifold or that $X$ is modeled on $\mathbf{E}$.

As we know, $\mathbb{R}^n$ is a Banach space. If $\mathbf{E}=\mathbb{R}^n$ for some fixed $n$, then we say that the manifold is $n$-dimensional. Also we have the local coordinates. A chart

is given by $n$ coordinate functions $\varphi_1,\cdots,\varphi_n$. If $P$ denotes a point of $U$, these functions are often written

or simply $x_1,\cdots,x_n$.

### Topological prerequisites

Let $X$ be a topological space. A covering $\mathfrak{U}$ of $X$ is locally finite if every point $x$ has a neighborhood $U$ such that all but a finite number of members of $\mathfrak{U}$ do not intersect with $U$ (as you will see, this prevents some nonsense summation). A refinement of a covering $\mathfrak{U}$ is a covering $\mathfrak{U}’$ such that for any $U’ \in \mathfrak{U}’$, there exists some $U \in \mathfrak{U}$ such that $U’ \subset U$. If we write $\mathfrak{U} \leq \mathfrak{U}’$ in this case, we see that the set of open covers on a topological space forms a direct set.

A topological space is paracompact if it is Hausdorff, and every open covering has a locally finite open refinement. Here follows some examples of paracompact spaces.

1. Any compact Hausdorff space.
2. Any CW complex.
3. Any metric space (hence $\mathbb{R}^n$).
4. Any Hausdorff Lindelöf space.
5. Any Hausdorff $\sigma$-compact space

These are not too difficult to prove, and one can easily find proofs on the Internet. Below are several key properties of paracompact spaces.

If $X$ is paracompact, then $X$ is normal. (Proof here)

Let $X$ be a paracompact (hence normal) space and $\mathfrak{U}=(U_i)$ a locally finite open cover, then there exists a locally finite open covering $\mathfrak{V}=(V_i)$ such that $\overline{V_i} \subset U_i$. (Proof here. Note the axiom of choice is assumed.

One can find proofs of the following propositions on Elements of Mathematics, General Topology, Chapter 1-4 by N. Bourbaki. It’s interesting to compare them to the corresponding ones of compact spaces.

Every closed subspace $F$ of a paracompact space $X$ is paracompact.

The product of a paracompact space and a compact space is paracompact.

Let $X$ be a locally compact paracompact space. Then every open covering $\mathfrak{R}$ of $X$ has a locally finite open refinement $\mathfrak{R}’$ formed of relatively compact sets. If $X$ is $\sigma$-compact then $\mathfrak{R}’$ can be taken to be countable.

### Partition of unity

A partition of unity (of class $C^p$) on a manifold $X$ consists of an open covering $(U_i)$ of $X$ and a family of functions

satisfying the following conditions:

PU 1. For all $x \in X$ we have $\phi_i(x) \geq 0$.

PU 2. The support of $\psi_i$ is contained in $U_i$.

PU 3. The covering is locally finite

PU 4. For each point $x \in X$ we have

The sum in PU 4 makes sense because for given point $x$, there are only finite many $i$ such that $\psi_i(x) >0$, according to PU 3.

A manifold $X$ will be said to admit partition of unity if it is paracompact, and if, given a locally finite open covering $(U_i)$, there exists a partition of unity $(\psi_i)$ such that the support of $\psi_i$ is contained in $U_i$.

### Bump function

This function will be useful when dealing with finite dimensional case.

For every integer $n$ and every real number $\delta>0$ there exist maps $\psi_n \in C^{\infty}(\mathbb{R}^n;\mathbb{R})$ which equal $1$ on $B(0,1)$ and vanish in $\mathbb{R}^n\setminus B(1,1+\delta)$.

Proof. It suffices to prove it for $\mathbb{R}$ since once we proved the existence of $\psi_1$, then we may write

Consider the function $\phi: \mathbb{R} \to \mathbb{R}$ defined by

The reader may have seen it in some analysis course and should be able to check that $\phi \in C^{\infty}(\mathbb{R};\mathbb{R})$. Integrating $\phi$ from $-\infty$ to $x$ and divide it by $\lVert \phi \rVert_1$ (you may have done it in probability theory) to obtain

it is immediate that $\theta(x)=0$ for $x \leq a$ and $\theta(x)=1$ for $x \geq b$. By taking $a=1$ and $b=(1+\delta)^2$, our job is done by letting $\psi_1(x)=1-\theta(x^2)$. Considering $x^2=|x|^2$, one sees that the identity about $\psi_n$ and $\psi_1$ is redundant. $\square$

In the following blog posts, we will generalize this to Hilbert spaces.

## Is partition of unity ALWAYS available?

Of course this is desirable. But we will give an example that sometimes we cannot find a satisfying partition of unity.

Let $D$ be a connected bounded open set in $\ell^p$ where $p$ is not an even integer. Assume $f$ is a real-valued function, continuous on $\overline{D}$ and $n$-times differentiable in $D$ with $n \geq p$. Then $f(\overline{D}) \subset \overline{f(\partial D)}$.

(Corollary) Let $f$ be an $n$-times differentiable function on $\ell^p$ space, where $n \geq p$, and $p$ is not an even integer. If $f$ has its support in a bounded set, then $f$ is identically zero.

It follows that for $n \geq p$, $C^n$ partitions of unity do not exists whenever $p$ is not an even integer. For example,e $\ell^1[0,1]$ does not have a $C^2$ partition of unity. It is then our duty to find that under what condition does the desired partition of unity available.

## Existence of partition of unity

Below are two theorems about the existence of partitions of unity. We are not proving them here but in the future blog post since that would be rather long. The restrictions on $X$ are acceptable. For example $\mathbb{R}^n$ is locally compact and hence the manifold modeled on $\mathbb{R}^n$.

Let $X$ be a manifold which is locally compact Hausdorff and whose topology has a countable base. Then $X$ admits partitions of unity

Let $X$ be a paracompact manifold of class $C^p$, modeled on a separable Hilbert space $E$, then $X$ admits partitions of unity (of class $C^p$)

## References

• N. Bourbaki, Elements of Mathematics
• S. Lang, Fundamentals of Differential Geometry
• M. Berger, Differential Geometry: Manifolds, Curves, and Surfaces
• R. Bonic and J. Frampton, Differentiable Functions on Certain Banach Spaces
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Stirling公式的几种经典估计 https://desvl.xyz/2020/10/18/stirling-approximation/ 2020-10-18T07:06:33.000Z 2020-11-09T06:12:20.607Z Stirling公式

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