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Introduction to Symplectic Topology$

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

(p.574) Appendix A Smooth maps

Source:
Introduction to Symplectic Topology
Author(s):

Dusa McDuff

Dietmar Salamon

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 BRm 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

Qmx=(x1,,xm)Rm|xi0 for i=1,,m

the right upper quadrant in Rm and by

N0N{0}

the set of nonnegative integers.

Definition A.1.1

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

In Milnor [484] and Guillemin–Pollack [293], a function f:XR, defined on an arbitrary subset XRm, is called smooth if, for every xX, there is an open neighbourhood URm of x and a smooth function F:UR such that the restriction of F to XU 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:QmR be a compactly supported smooth function and let URm be an open neighbourhood of Qm. Then there is a compactly supported smooth function F:RmR such that

F|Qm=f,supp(F)U.

The proof is based on the following lemma.

(p.575) Lemma A.1.3

Let δ>0 and let ak,ckR be sequences such that

(A.1.1)
ckδ1+|ak|

for kN0. Let β:R[0,1] be a smooth cutoff function such that

(A.1.2)
β(x)=1,for |x|1/3,0,for |x|2/3,

and define F:RR by

(A.1.3)
F(x)k=0ak(β(ckx)x)kk!forxR.

Then F is smooth, vanishes for |x|δ, and satisfies F()(0)=a for 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 xR and hence, for k1,

|β(ckx)x|1ckδ1,|akβ(ckx)x||ak|ck1.

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)
F(x)=k=1ak(β(ckx)x)k1(k1)!(β(ckx)+β(ckx)ckx).

Use the inequalities |akβ(ckx)x|1 and |β(ckx)ckx|max|β| for all xR and k2 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:RR by

βk(x)β(ckx)x

for kN0 and xR. Then, for every iN there is a constant Ci>0 such that

βk(x)i1didxiβk(x)Ci

for all kN0 and all xR. Moreover, F=k=0akβkkk! and hence

(A.1.5)
F()=G+j=1k=akβkk(k+j1)!pj,k,,

where G is a finite sum, satisfying G(0)=a, and pj,k, is a polynomial in the functions βki1xiβk for i=1,,j. Thus the functions pj,k, satisfy a uniform bound, independent of k, and hence the series converges absolutely and uniformly. This shows that F is smooth and F()(0)=a for all 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:QmR be a compactly supported smooth function, let URm be an open neighbourhood of Qm, and let n{1,,m}. Then there exists a compactly supported smooth function Fn:Rn×QmnR such that

Fn|Qm=f,supp(Fn)U.

Choose constants

ckδ1+sup(x2,,xm)Qm11kf(0,x2,,xk)

for kN0 and define the function F1:R×Qm1R by F1|Qmf and

F1(x)k=01kf(0,x2,,xm)k!(β(ckx1)x1)k

for x1<0 and (x2,,xm)Qm1. 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 Fn1 has been constructed. Choose constants

ckδ1+sup(x1,,xn1)Rn1(xn+1,,xm)QmnnkFn1(x1,,xn1,0,xn,,xm)

and define Fn:Rn×QmnR by F|Rn1×Qmn+1Fn1 and

Fn(x)k=0nkFn1(x1,,xn1,0,xn,,xm)k!(β(ckxn)xn)k

for (x1,,xn1)Rn1, xn<0, and (xn,,xm)Qmn. 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:Uαϕα(Uα) (called coordinate charts) onto open sets ϕα(Uα)Qm (in the relative topology induced by Rm) such that the transition maps ϕβαϕβϕα1:ϕα(UαUβ)ϕβ(UαUβ) 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

{Uα,ϕα}α.

A function f:MR is called smooth if the composition

fαfϕα1:ϕα(Uα)R

is smooth in the sense of Definition A.1.1 for every α‎. A function f:MRn is called smooth if its coordinate functions fi:MR are smooth for i=1,,n. A function f:MN with values in a manifold N is called smooth if its composition with an embedding ι:NRn 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 pM, choose indices α,β such that pUαUβ, and define the coordinate vectors x,yQm by

x=(x1,,xm)ϕα(p),y=(y1,,ym)ϕβ(p).

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

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

ν:X{1,,m}

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

kMpM|ν(p)k.

Thus

mMm1M2M1M0M=M.

The boundary of M is the set

M1M,

and the interior of M is the set

int(M)MM.

Exercise A.1.7

Prove that kM is an (mk)-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:RmM 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]×RmM of ψ‎, whose image has compact closure, such that Ψ(t,):RmM is an embedding for every t.

  3. (iii) Let ψ:[0,1]×B(λ)M be as in (ii) and let Ψ0,Ψ1:RmM 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 λ0.

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

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

  3. (iii) Let f:[0,1]×B(λ)R be a smooth function and let F0,F1:RmR 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 xB(λ), define fx:(,0]R by

fx(s)f(esx)

for s0. Choose a sequence of real numbers ck such that ck1+|fx(k)(0)| for all kN0 and all xB(λ). Define F:RmR by F|B(λ)f and

F(esx)k=1fx(k)(0)k!(β(cks)s)k

for xB(λ) and s0. 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]×RmR is any compactly supported smooth extension of f and let F0,F1:RmR 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

F(t,x)G(t,x)+β(t)(F0(x)G(0,x))+(1β(t))(F1(x)G(1,x))

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 URn be a tubular neighbourhood of the image of the embedding and denote by π:UM 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

ρ:(0,)(0,λ+ε)

such that ρ(r)=r for 0<rλ. Then the map Ψ:RmM, defined by

Ψ(x)ϕρ(|x|)x|x|

for xRm{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:ΩM, 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

ϕtϕ(t,):ΩtxRm|(t,x)ΩM

is a embedding for every t. Now choose a smooth function (0,1)(0,):tεt such that B(λ+εt)Ωt for all t, and εt tends to infinity for t0 and t1. Then choose a smooth function

0t1{t}×(0,λ+εt)(0,):(t,r)ρt(r)

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]×RmM, defined by

Ψ(t,x)ϕtρt(|x|)x|x|,0t1,

for xRm{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

(0,)2×(0,1)R:(r,λ,ε)fλ,ε(r)

such that

(A.3.1)
fλ,ε(r)=λ2+r2,for rε,r,for rλ+ε,fλ,ε(r)>0

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 rε and f(r)=r for r1+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)
αc(t)=0,for t[0,c],t+1,for t[1,c3],0αc(t)t+3,0cαc=c22.

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

(A.3.3)
ϕ(t)=0,for t1/2,t+1,for t3/4,ψ(t)=ψ(t),ψ|[1/2,1]1,

and define

(A.3.4)
αc(t)ϕ(t),for 0t1,t+1,for 1tc3,cϕ(c2t),for c3tc2,c2(1ψ(t+1c)),for c2tc.

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

1c3αc=c222c,c3c2αc=c01ϕ,c2cαc=c,

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

(0,)×(0,1)R:(r,ε)βε(r)

such that

(A.3.5)
βε(r)=1+r2,for rε,r2,for r1+ε,βε(r)>0

for all ε(0,1) and all r>0.

Let αc be as in Step 1 and define

(A.3.6)
βε(r)1+r2ε20rα4/ε4sεεds

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

βε(r)=2rε2α4/ε4rεε2rε24rε1=ε2.

This shows that βε(r)>0 for εr1+ε and βε(r)=2r for r1+ε. Moreover, it follows from equation (A.3.2) in Step 1 that

βε(1+ε)=1+(1+ε)2ε2ε1+εα4/ε4sεεds =1+(1+ε)2ε2804/εα4/εtdt =(1+ε)2.

Hence βε(r)=r2 for r1+ε. This proves Step 2.

Let βε be as in Step 2. Then the function

fλ,ε(r)λβε/λ(r/λ)

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