## Dusa McDuff and Dietmar Salamon

Print publication date: 2017

Print ISBN-13: 9780198794899

Published to Oxford Scholarship Online: June 2017

DOI: 10.1093/oso/9780198794899.001.0001

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# (p.574) Appendix A Smooth maps

Source:
Introduction to Symplectic Topology
Publisher:
Oxford University Press

This appendix clarifies the notion of smooth maps on manifolds with boundary used throughout the book. Section A.1 discusses smooth functions on manifolds with corners. Section A.2 discusses how to extend smooth (homotopies of) embeddings of a closed ball $B⊂Rm$ into a manifold M to smooth (homotopies of) embeddings of $Rm$ into M. This is used in Chapter 3, where the extension results are carried over to the symplectic setting. Section A.3 gives an explicit formula for a smooth function that is used in Section 7.1 for the construction of symplectic forms on blow-ups.

# A.1 Smooth functions on manifolds with corners

Denote by

$Display mathematics$

the right upper quadrant in $Rm$ and by

$Display mathematics$

the set of nonnegative integers.

Definition A.1.1

Let $V⊂Qm$ be an open set in the relative topology. A function $f:Qm∩V→R$ is called smooth if its restriction to $int(Qm)∩V$ is smooth and all its partial derivatives extend to continuous functions on $Qm∩V$.

In Milnor [484] and Guillemin–Pollack [293], a function $f:X→R$, defined on an arbitrary subset $X⊂Rm$, is called smooth if, for every $x∈X$, there is an open neighbourhood $U⊂Rm$ of x and a smooth function $F:U→R$ such that the restriction of F to $X∩U$ agrees with f. The next theorem shows that, for $X=Qm$, their definition of smooth agrees with that in Definition A.1.1.

Theorem A.1.2

Let $f:Qm→R$ be a compactly supported smooth function and let $U⊂Rm$ be an open neighbourhood of Qm. Then there is a compactly supported smooth function $F:Rm→R$ such that

$Display mathematics$

The proof is based on the following lemma.

(p.575) Lemma A.1.3

Let $δ>0$ and let $ak,ck∈R$ be sequences such that

(A.1.1)
$Display mathematics$

for $k∈N0$. Let $β:R→[0,1]$ be a smooth cutoff function such that

(A.1.2)
$Display mathematics$

and define $F:R→R$ by

(A.1.3)
$Display mathematics$

Then F is smooth, vanishes for $|x|≥δ$, and satisfies $F(ℓ)(0)=aℓ$ for $ℓ∈N0$.

Proof:

The right hand side in (A.1.3) vanishes for $|x|≥δ$. It converges absolutely and uniformly because $|xβ(x)|≤1$ for all $x∈R$ and hence, for $k≥1$,

$Display mathematics$

This shows that F is continuous on $(−∞,0]$ and $F(0)=a0$. Differentiate the right hand side of equation (A.1.3) term by term to obtain

(A.1.4)
$Display mathematics$

Use the inequalities $|akβ(ckx)x|≤1$ and $|β′(ckx)ckx|≤max|β′|$ for all $x∈R$ and $k≥2$ to show that the right hand side in equation (A.1.4) converges uniformly and absolutely on $R$. Hence F is continuously differentiable and $F′(0)=a1$.

To examine the higher derivatives of F, define $βk:R→R$ by

$Display mathematics$

for $k∈N0$ and $x∈R$. Then, for every $i∈N$ there is a constant $Ci>0$ such that

$Display mathematics$

for all $k∈N0$ and all $x∈R$. Moreover, $F=∑k=0∞akβkkk!$ and hence

(A.1.5)
$Display mathematics$

where $Gℓ$ is a finite sum, satisfying $Gℓ(0)=aℓ$, and $pj,k,ℓ$ is a polynomial in the functions $βki−1⋅∂xiβk$ for $i=1,…,j$. Thus the functions $pj,k,ℓ$ satisfy a uniform bound, independent of k, and hence the series converges absolutely and uniformly. This shows that F is smooth and $F(ℓ)(0)=aℓ$ for all $ℓ∈N$. This proves Lemma A.1.3.

(p.576) Proof of Theorem A.1.2:

We prove the following by induction on n.

Claim. Let $f:Qm→R$ be a compactly supported smooth function, let $U⊂Rm$ be an open neighbourhood of Qm, and let $n∈{1,…,m}$. Then there exists a compactly supported smooth function $Fn:Rn×Qm−n→R$ such that

$Display mathematics$

Choose constants

$Display mathematics$

for $k∈N0$ and define the function $F1:R×Qm−1→R$ by $F1|Qm≔f$ and

$Display mathematics$

for $x1<0$ and $(x2,…,xm)∈Qm−1$. By Lemma A.1.3 this function is smooth. It is supported in U whenever $δ>0$ is chosen sufficiently small. Let $n∈{2,…,m}$ and suppose $Fn−1$ has been constructed. Choose constants

$Display mathematics$

and define $Fn:Rn×Qm−n→R$ by $F|Rn−1×Qm−n+1≔Fn−1$ and

$Display mathematics$

for , and $(xn,…,xm)∈Qm−n$. By Lemma A.1.3, Fn satisfies the requirements of the claim and this proves Theorem A.1.2.

We now briefly discuss the notion of a manifold with corners. In the following we will mostly be interested in smooth functions whose domain is either a closed ball B, i.e. a manifold with boundary, or is the product $B×[0,1]$, which is a manifold with corners of codimension 2. For an in-depth discussion of manifolds with corners and their tangent bundles, see Joyce [351] and the references cited therein.

Definition A.1.4

A manifold with corners (of dimension m) is a second countable Hausdorff topological space M, equipped with an open cover ${Uα}α$ and a collection of homeomorphisms $ϕα:Uα→ϕα(Uα)$ (called coordinate charts) onto open sets $ϕα(Uα)⊂Qm$ (in the relative topology induced by $Rm$) such that the transition maps $ϕβα≔ϕβ∘ϕα−1:ϕα(Uα∩Uβ)→ϕβ(Uα∩Uβ)$ are smooth in the sense of Definition A.1.1. The collection of coordinate charts ${Uα,ϕα}α$ is called an atlas.

(p.577) Here is the corresponding notion of smooth map.

Definition A.1.5

Let M be an m-manifold with corners and with atlas

$Display mathematics$

A function $f:M→R$ is called smooth if the composition

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is smooth in the sense of Definition A.1.1 for every α‎. A function $f:M→Rn$ is called smooth if its coordinate functions $fi:M→R$ are smooth for $i=1,…,n$. A function $f:M→N$ with values in a manifold N is called smooth if its composition with an embedding $ι:N→Rn$ is smooth.

For these to be sensible definitions, we need to know that the corner structure of M, i.e. the stratification of its boundary into pieces of different dimension, is preserved by the coordinate changes. This is not true in the topological category; for example, Q2 is homeomorphic to the half space $Q1×R$. However, the next exercise shows that it is true in the smooth category.

Exercise A.1.6

Let M be an m-manifold with corners with atlas ${Uα,ϕα}α.$ Let $p∈M$, choose indices $α,β$ such that $p∈Uα∩Uβ$, and define the coordinate vectors $x,y∈Qm$ by

$Display mathematics$

Prove that $#i|xi=0=#j|yj=0$.

Exercise A.1.6 shows that there is a well-defined map

$Display mathematics$

which assigns to each point $p∈M$ the number of components of the vector $ϕα(p)∈Qm$ that vanish for some, and hence every, index α‎ such that $p∈Uα$. For $k∈{0,1,…,m}$ define the set

$Display mathematics$

Thus

$Display mathematics$

The boundary of M is the set

$Display mathematics$

and the interior of M is the set

$Display mathematics$

Exercise A.1.7

Prove that $∂kM$ is an $(m−k)$-manifold with corners.

# (p.578) A.2 Extension

We now show that ball embeddings have smooth extensions. Let $B(λ)⊂Rm$ denote the closed ball of radius $λ$.

Theorem A.2.1 (Smooth embedding extension)

Let M be a smooth manifold without boundary and fix a constant $λ>0$.

1. (i) Every embedding $ψ:B(λ)→M$ extends to an embedding $Ψ:Rm→M$ whose image has compact closure.

2. (ii) Let $ψ:[0,1]×B(λ)→M$ be a smooth map such that $ψ(t,⋅):B(λ)→M$ is an embedding for every t. Then there is a smooth extension $Ψ:[0,1]×Rm→M$ of ψ‎, whose image has compact closure, such that $Ψ(t,⋅):Rm→M$ is an embedding for every t.

3. (iii) Let $ψ:[0,1]×B(λ)→M$ be as in (ii) and let $Ψ0,Ψ1:Rm→M$ be embeddings, whose images have compact closure, such that $Ψi|B(λ)=ψ(i,⋅)$ for $i=0,1$. Then the extension Ψ‎ in (ii) can be chosen such that $Ψ(i,⋅)=Ψi$ for $i=0,1$.

The proof is based on the following lemma.

Lemma A.2.2 (Smooth extension)

Fix a constant $λ≥0$.

1. (i) Every smooth function $f:B(λ)→R$ extends to a compactly supported smooth function $F:Rm→R$.

2. (ii) Every smooth function $f:[0,1]×B(λ)→R$ extends to a compactly supported smooth function $F:[0,1]×Rm→R$.

3. (iii) Let $f:[0,1]×B(λ)→R$ be a smooth function and let $F0,F1:Rm→R$ be compactly supported smooth functions such that $Fi|B(λ)=f(i,⋅)$ for $i=0,1$. Then the extension F in (ii) can be chosen such that $F(i,⋅)=Fi$ for $i=0,1$.

Proof:

Assertion (i) is trivial for $λ=0$ (take any smooth cutoff function). To prove it for $λ>0$ choose a smooth cutoff function $β:[0,∞)→[0,1]$ that satisfies (A.1.2). For $x∈∂B(λ)$, define $fx:(−∞,0]→R$ by

$Display mathematics$

for $s≤0$. Choose a sequence of real numbers ck such that $ck≥1+|fx(k)(0)|$ for all $k∈N0$ and all $x∈∂B(λ)$. Define $F:Rm→R$ by $F|B(λ)≔f$ and

$Display mathematics$

for $x∈∂B(λ)$ and $s≥0$. Then F is smooth by Lemma A.1.3 and this proves (i). The same formula proves (ii) because of the smooth dependence of F on f. To prove (iii), assume that $G:[0,1]×Rm→R$ is any compactly supported smooth extension of f and let $F0,F1:Rm→R$ be compactly supported smooth functions (p.579) such that $Fi|B(λ)=f(i,⋅)$ for $i=0,1$. Let $β:[0,1]→[0,1]$ be a smooth cutoff function that satisfies (A.1.2). Then the function

$Display mathematics$

satisfies (iii). This proves Lemma A.2.2.

Proof of Theorem A.2.1:

Choose any embedding of M (or of an open subset of M that contains the image of f and has compact closure) into some Euclidean space $Rn$. Let $U⊂Rn$ be a tubular neighbourhood of the image of the embedding and denote by $π:U→M$ the projection. Use part (i) of Lemma A.2.2 to construct any smooth extension of ψ‎ to a function with values in $Rn$ and compose it with π‎ to obtain a smooth extension $ϕ:B(λ+ε)→M$ for some $ε>0$. Shrinking $ε$, if necessary, we may assume without loss of generality that ϕ‎ is an embedding. Now choose a diffeomorphism

$Display mathematics$

such that $ρ(r)=r$ for $0. Then the map $Ψ:Rm→M$, defined by

$Display mathematics$

for $x∈Rm∖{0}$ and $Ψ(0)≔ϕ(0)$, is an embedding and agrees with ψ‎ on $B(λ)$. The same argument, using part (ii) of Lemma A.2.2, proves assertion (ii).

We prove (iii). The same argument as in the proof of (i) and (ii), using part (iii) of Lemma A.2.2, gives rise to a smooth map $ϕ:Ω→M$, defined on an open neighbourhood $Ω⊂[0,1]×Rm$ of $({(0,1)}×Rm)∪([0,1]×B(λ))$, such that $ϕ|[0,1]×B(λ)=ψ$, $ϕ(i,⋅)=Ψi$ for $i=0,1$, the image of ϕ‎ has compact closure, and the map

$Display mathematics$

is a embedding for every t. Now choose a smooth function $(0,1)→(0,∞):t↦εt$ such that $B(λ+εt)⊂Ωt$ for all t, and $εt$ tends to infinity for $t→0$ and $t→1$. Then choose a smooth function

$Display mathematics$

such that $ρt:(0,∞)→(0,λ+εt)$ is a diffeomorphism satisfying $ρt|(0,λ]=id$ for all t, and $ρ0=ρ1=id$. Then the function $Ψ:[0,1]×Rm→M$, defined by

$Display mathematics$

for $x∈Rm∖{0}$ and $Ψt(0)≔ϕt(0)$, satisfies the requirements of (iii). This proves Theorem A.2.1.

# (p.580) A.3 Construction of a smooth function

The following construction is used in Section 7.1.

Lemma A.3.1

There exists a smooth function

$Display mathematics$

such that

(A.3.1)
$Display mathematics$

for all $ε∈(0,1)$ and all $λ,r>0$

Consider first the case $λ=1$ and fix a constant $ε>0$. Then it is a standard problem in analysis to construct a function $f1,ε:(0,∞)→R$ that satisfies (A.3.1). Moreover, the set of all such functions is a convex open set in the set of all smooth functions $f:(0,∞)→R$ that satisfy the constraints $f(r)=1+ε2$ for $r≤ε$ and $f(r)=r$ for $r≥1+ε$. By varying $ε∈(0,1)$ and using this convexity, one can prove the existence of a smooth function $(r,ε)↦f1,ε(r)$ that satisfies (A.3.1) for $λ=1$, and then, by varying $λ$ similarly, for all $λ>0$. Below we construct the functions $fλ,ε$ explicitly.

Exercise A.3.2

Complete the details of the above proof.

Proof of Lemma A.3.1

The explicit construction of $fλ,ε$ requires two preparatory steps.

Step 1. There is a smooth function $R×(4,∞)→R:(t,c)↦αc(t)$ such that

(A.3.2)
$Display mathematics$

Let $ϕ:[0,1]→[0,2]$ and $ψ:[−1,1]→[−1,1]$ be smooth functions such that

(A.3.3)
$Display mathematics$

and define

(A.3.4)
$Display mathematics$

The map $(c,t)↦αc(t)$ is well-defined and smooth and satisfies the first two equations in (A.3.2). Moreover,

$Display mathematics$

and hence $αc$ also satisfies the integral condition in (A.3.2). This proves Step 1.

(p.581) Step 2. There exists a smooth function

$Display mathematics$

such that

(A.3.5)
$Display mathematics$

for all $ε∈(0,1)$ and all $r>0$.

Let $αc$ be as in Step 1 and define

(A.3.6)
$Display mathematics$

for $0<ε<1$ and $r>0$. If $0 then the integral vanishes and so $βε(r)=1+r2$. Moreover, since $α4/ε(t)≤t+3$ for every $t∈R$, we have

$Display mathematics$

This shows that $βε′(r)>0$ for $ε≤r≤1+ε$ and $βε′(r)=2r$ for $r≥1+ε$. Moreover, it follows from equation (A.3.2) in Step 1 that

$Display mathematics$

Hence $βε(r)=r2$ for $r≥1+ε$. This proves Step 2.

Let $βε$ be as in Step 2. Then the function

$Display mathematics$

satisfies the requirements of Lemma A.3.1. (p.582)