SSymplectic excision
Yael Karshon Xiudi TangJanuary 12, 2021
Abstract
We say that a subset of a symplectic manifold is symplectically (neigh-bourhood) excisable if its complement is symplectomorphic to the ambientmanifold, (through a symplectomorphism that can be chosen to be theidentity outside an arbitrarily small neighbourhood of the subset). Weuse time-independent Hamiltonian flows, and their iterations, to show thatcertain properly embedded subsets of noncompact symplectic manifolds aresymplectically neighbourhood excisable: a ray, a “Cantor brush”, a “boxwith a tail”, and—more generally—epigraphs of lower semi-continuous func-tions; as well as a “ray with two horns”, and—more generally—open-rootedfinite trees.
A couple of years ago, Alan Weinstein circulated the following question, whichappeared in his paper with Christian Blohmann [3, Question 11.2]:Let M be a noncompact symplectic manifold, and let R ⊂ M be the image of a proper embedding of the ray [0 , ∞ ). Is M \ R symplectomorphic to M ? Can a symplectomorphism between them bechosen to be the identity outside a prescribed neighbourhood U of R ?In his earlier paper [12], the author Xiudi Tang proved that such a symplecto-morphism exists if the neighbourhood U is sufficiently big in the following sense:for some ε >
0, there exists a symplectic embedding of B n − ( ε ) × R into U thatmaps R := { } n − × [0 , ∞ ) × { } a r X i v : . [ m a t h . S G ] J a n nto R . Note that such a neighbourhood U has infinite volume. In this paper, weallow U to be arbitrary. This result is strictly stronger; for example, M can havefinite volume. Moreover, we extend this result to more general closed subsets Z ofsymplectic manifolds M .During our final preparations of this paper we learned of Bernd Stratmann’srecent e-print [11]. Stratmann uses time-dependent flows to excise what hecalls “parametrized rays”: closed submanifolds of the form S = [0 , ∞ ) × Q onwhich the two-form ω S comes from a two-form on Q . We use time- independent flows to excise similar sets. Moreover, we use time-independent flows to excisemore general sets; for example, we excise from ( R n , ω can ) the “Cantor brush” { } n − × C × [0 , ∞ ) × { } ( n (cid:62)
2, see Example 5.5) or the “box with a tail”[ − , n − × [0 , ∞ ) ×{ } ∪ { } n − × [ − , ∞ ) ×{ } (see Example 6.8). Furthermore,using iterations of such flows, we excise even more general subsets, for example,a “ray with two horns”, or, more generally, a “tree with a root at infinity” (seeExample 7.1).Our method is a novel variant of the symplectic isotopy extension theorem: weassume that there is a submanifold N of M that contains Z and a Hamiltonianflow ( ψ t ) t on N that sends all the points of Z to infinity in time (cid:54) N \ Z , and we extend it to a Hamiltonian flow on M whose time-1 flow is well defined on M \ Z . The subtle point here is to ensurethat the time-1 flow is well defined on M \ Z .Following preliminary definitions in Section 2, we start with the simplestnontrivial case, of removing a ray R from M , in Section 3. To remove a more generalsubset Z from M , we use a two-step construction: first, we find a hypersurface N containing Z and a null vector field X on N whose time-1 flow excises Z from N ;second, we extend X to a Hamiltonian vector field X F on M whose time-1 flowexcises Z from M . In Section 4, we give the extension of X to X F . In Section 5,we give an explicit construction of the null vector field X when Z is the epigraphof a smooth function over a closed set. Combining the previous two sections, weestablish the Hamiltonian excision of Z from M . In Section 6, we develop a newmethod to excise subsets as epigraphs of lower semi-continuous functions thatare not necessarily smooth. This yields a less explicit but stronger Hamiltonianexcision result than that in the previous section. In Section 7, we exhibit someexamples using the composition of several symplectic excisions. Remark . Polterovich pointed out that a symplectic excision of R from R n arises from the symplectization of a contactomorphism from the punctured stan-dard unit sphere to the standard Euclidean space as a contact version of the2tereographic projection constructed in [5, Proposition 2.13]. In contrast to ourTheorem 3.1, this construction does not give a Hamiltonian flow, and the set wherethis symplectomorphism R n \ R → R n coincides with the identity map (is theopposite ray, which) has empty interior. (cid:5) Acknowledgement
We are grateful to Alan Weinstein, whose question about excising a ray promptedthis project, and to Leonid Polterovich, who pointed out the elegant—thoughnon-localized—solution in the special case of a ray in R n (see Remark 1.1). Theauthor Xiudi Tang thanks Reyer Sjamaar for helpful discussions and thanks NingJiang for the warm hospitality at Wuhan University, where some ideas in thispaper were sparked.This research is partly funded by Natural Sciences and Engineering ResearchCouncil of Canada Discovery Grants RGPIN-2018-05771 and 485904. Definition 2.1.
Let (
M, ω ) be a symplectic manifold, and let Z be a subsetof M . A symplectic excision of Z from M is a symplectomorphism ϕ from M \ Z to M . The subset Z of ( M, ω ) is symplectically neighbourhood excisable if forevery neighbourhood U of Z in M there exists a symplectic excision of Z from M that restricts to the identity map on some neighbourhood of M \ U in M \ Z .It is Hamiltonian excisable if there is a Hamiltonian flow on M whose forward-trajectories that start in Z stay in Z as long as they are defined and whose time-1map is a symplectic excision of Z from M . It is time-independent Hamiltonianexcisable if the Hamiltonian flow can be chosen to be time-independent. (cid:5) Remark . If Z is symplectically excisable from ( M, ω ) and is non-empty, then Z is closed in M and M is non-compact. More generally, if M is a topologicalmanifold and Z is a non-empty subset and M \ Z is homeomorphic to M , then Z is closed in M and M is non-compact. (cid:5) Remark . If Z is Hamiltonian excisable from ( M, ω ), then Z is symplecticneighbourhood excisable from ( M, ω ). Indeed, for any neighbourhood U of Z in M , we can multiply the Hamiltonian function in Definition 2.1 by a smooth bumpfunction that is supported in U and is equal to 1 near Z . (cid:5)
3e denote by X F the Hamiltonian vector field of a smooth function F : M → R on a symplectic manifold ( M, ω ). We use the convention X F ⌟ ω = d F . Removing a ray from R n Consider R n with coordinates ( x , y , . . . , x n , y n ), with the standard symplecticform ω can := d x ∧ d y + . . . + d x n ∧ d y n , and, in it, consider the ray R := { } n − × [0 , ∞ ) × { } . Theorem 3.1. R is time-independent Hamiltonian excisable from ( R n , ω can ) .Proof. Because there is a symplectomorphism from ( − , × R to R that takes[0 , × { } to [0 , ∞ ) × { } , (for example, take the cotangent lift of the diffeomor-phism t t/ (1 − t ) from ( − ,
1) to R ,) it is enough to show that R := { } n − × [0 , × { } is time-independent Hamiltonian excisable from M := R n − × ( − , × R . We claim the following stronger result than is required. There exists a smoothfunction F : M → R , with Hamiltonian vector field X F , whose flow domain D ⊂ M × R is given by D ∩ ( { z } × R ) = { z } × ( S ( z ) , T ( z ))for all z ∈ M , such that• the forward flow of X F preserves R ;• T > M \ R ;• T (cid:54) R ;• S < − M . 4his claim completes the proof because the time-1 Hamiltonian flow of F is thena symplectomorphism from M \ R to M .We will now prove this claim. Suppose n (cid:62)
2. We write points of M as z = ( p ; x n , y n ) with p = ( x , y , . . . , x n − , y n − ) . Let ε ∈ (0 ,
1) and let h : [ − ε, → (0 , ∞ )be a smooth strictly decreasing function that converges to 0 at 1. Let U := n x n ∈ ( − ε, , | p | + y n < h ( x n ) o . Then | p | + y n (cid:54) h ( − ε ) on U .Fix a smooth function χ : M → [0 , U and is equal to 1 in a neighbourhood of R . Assume thatd χ = 0 wherever χ = 0; this can be achieved, for instance, by replacing χ by ρ ◦ χ where ρ ( s ) = 3 s − s .Let F ( z ) := 1 − x n | p | + 1 − x n χ ( z ) y n . Then X F (0; x n ,
0) = χ (0; x n , ∂ x n for all x n (cid:62)
0, so the forward flow of X F preserves R .Fix c >
0. Because the function F is the product of y n with a function thattakes values in [0 ,
1] and is supported in U , if | F ( z ) | (cid:62) c >
0, then z ∈ U and | y n | (cid:62) c . From the definition of U , this further implies that x n ∈ [ − ε,
1) and h ( x n ) (cid:62) c . Because h : [ − ε, → R > approaches 0, these inequalities imply that x n ∈ [ − ε, b ] for some b ∈ [ − ε, {| F | (cid:62) c } is compact, as it isclosed in M and contained in { x n ∈ [ − ε, b ] , | p | + y n (cid:54) h ( − ε ) } .So F | M \ F − (0) : M \ F − (0) → R \ { } is a proper map. In particular, all the non-zero level sets of F are compact.We will now calculate S ( z ) and T ( z ) explicitly for each z ∈ M .5 n { y n = 0 } : At each point z = ( p ; x n , y n ) with y n = 0, if F ( z ) = 0, then F − ( F ( z )) is compact, and if F ( z ) = 0, then ( χ ( z ) = 0; by the choice of χ also d χ | z = 0; and so) X F ( z ) = 0. We conclude that, for each such a point z ,the trajectory of X F through z is defined for all times, and so S ( z ) = −∞ and T ( z ) = ∞ . On { y n = 0 } : At each point z with y n = 0, we have X F ( p ; x n ,
0) = 1 − x n | p | + 1 − x n χ ( p ; x n , ∂ x n . Since X F is proportional to ∂ x n and vanishes for x < − ε , we have S ( z ) = −∞ forany point z with y n = 0. It remains to calculate T ( z ) for such z . On { y n = 0 } ∩ { p = 0 } : At each point z = ( p ; x n , y n ) with y n = 0 and p = 0, wehave | p | >
0. By the comparison1 − x n | p | + 1 − x n χ ( p ; x n , (cid:54) − x n | p | + 1 − x n and the completeness of the vector field − x n b +1 − x n ∂ x n on ( − ,
1) for b >
0, we have T ( z ) = ∞ . On { y n = 0 } ∩ { p = 0 } : At each point z = ( p ; x n , y n ) with y n = 0 and p = 0,which we write as z = (0; x n , X F (0; x n ,
0) = χ (0; x n , ∂ x n . Because χ (0; x n ,
0) = 1 on the set { x n (cid:62) } , we have T ( z ) (cid:54) x n (cid:62) T ( z ) = 1when x n = 0. Because the vector field is a non-negative multiple of ∂ x n , thefunction x n T ( z ) is weakly decreasing; becuase the vector field is a positivemultiple of ∂ x n near x n = 0, we have T ( z ) > x n <
0. So T ( z ) (cid:54) x n (cid:62) T ( z ) (cid:54) z ∈ R , and that S ( z ) = −∞ for all z ∈ M . This justifies our claim for n (cid:62) n = 1. Let U be an open neighbourhood of R in M where the x -coordinate is bounded away from − c > U with | y | (cid:62) c is relatively compact in M . Fix asmooth function χ : M → [0 ,
1] that is supported in U and is equal to 1 in aneighbourhood of R . Let F ( x , y ) := χ ( x , y ) y . On one hand, for any c > x , y ) ∈ M , the inequality | F ( x , y ) | (cid:62) c impliesthat ( x , y ) ∈ U and | y | (cid:62) c . Then F | M \ F − (0) : M \ F − (0) → R \ { } is a6roper map. On the other hand, the vector field X F ( x ,
0) = χ ( x , ∂ x for all x ∈ ( − , T ( x , (cid:54) x (cid:62) S ( x ,
0) = −∞ forany x ∈ ( − , S ( z ) = −∞ and T ( z ) = ∞ for all z = ( x , y )with y = 0.We have justified our claim for any dimension and our proof is complete.The analogous result of [12] only guarantees that, for any ε >
0, there is asymplectomorphism R n → R n \ R that is the identity outside B n − ( ε ) × W ( ε ),where W ( ε ) := B ( ε ) ∪ { ( x, y ) ∈ R | x > √ ε , x | y | < ε } . (This implies the resultthat we quoted in the introduction.) This result cannot be used to symplecticallyexcise R from M if M has finite volume. In contrast, Theorem 3.1 allows us toremove a ray from any symplectic manifold, as we now show. Applications
We begin with the following corollary of Theorem 3.1, which answers Weinsteinand Blohmann’s question about excising rays from arbitrary symplectic manifolds:
Corollary 3.2.
Let ( M, ω ) be a symplectic manifold, and let γ : [0 , ∞ ) → M be a proper embedding with image R . Then R is time-independent Hamiltonianexcisable from M .Proof. By Theorem 3.1, there exists a smooth function F : R n → R with Hamil-tonian flow whose forward-trajectories that start in R stay in R and whosetime-1 flow is a symplectomorphism of R n \ R with R n . By the Weinsteinsymplectic tubular neighbourhood theorem, there exists a symplectic embedding ψ : ( U , ω can ) → ( M, ω ) from an open neighbourhood U of R to an open neigh-bourhood U of R in M , sending R to R . Because R is closed in M , there exists asmooth function χ : M → [0 ,
1] supported in U that equals 1 in a neighbourhoodof R in M . Then the function F := χ · ( F ◦ ψ − ) : M → R is smooth, and itgenerates a Hamiltonian flow whose forward-trajectories that start in R stay in R and whose time-1 map is a symplectomorphism of M \ R with M .The original motivation for Weinstein and Blohmann’s question was to showthat every exact symplectic form has a nowhere vanishing primitive. We show thisin Theorem 3.5 below, after the preliminary Lemma 3.4.7 emark . Blohmann and Weinstein’s paper [3] contains a statement of Lemma 3.4and an idea for the proof of Theorem 3.5. Stratmann’s paper [10] contains proofsof Lemma 3.4 and Theorem 3.5 that do not rely on the excision of a ray. (cid:5)
Lemma 3.4.
Any exact -form on a smooth manifold has a primitive whose zeroset is discrete.Proof. Let ω = d θ be an exact 2-form on a smooth manifold M . Let φ =( x , . . . , x n ) : M → R n be a smooth embedding. The function F : M × R n → T ∗ M,F ( x, s ) = θ ( x ) + n X i =1 s i d x x i , where s = ( s , . . . , s n ), is transverse to the zero section 0 T ∗ M of T ∗ M . By theTransversality Theorem (see, for instance, [7]), we deduce that F ( · , s ) : M → T ∗ M is transverse to 0 T ∗ M for almost every s ∈ R n . Fix such an s , and let ρ := P ni =1 s i x i : M → R . Then the zero set of the one-form β := θ + d ρ = F ( · , s ) ∈ Ω ( M ) is a discrete set of points in M . Because d β = d θ = ω , the one-form β is aprimitive of ω . Theorem 3.5.
Any exact symplectic form has a nowhere vanishing primitive.Proof.
Let (
M, ω ) be an exact symplectic manifold. By Lemma 3.4, ω has aprimitive β whose zeroes are isolated. Choose such a β .We construct an exhaustion of M by a sequence of compact subsets ( K j ) ∞ j =1 such that for each j ∈ N , any point in K j +1 \ K j can be joined by a smooth pathin M \ K j to a point in M \ K j +1 . This can be achieved by taking the unions ofregular superlevel sets of an exhaustion function for M with the bounded connectedcomponents of their complements, as was done in [6, 8, 9, 10].Let ( z i ) i (cid:62) , be the (finite or countable) set of zeroes of β . We now construct foreach i a properly embedded ray R i starting with z i such that the rays are pairwisedisjoint and each point of M has a neighbourhood that meets only finitely many ofthe rays. Our construction is step-by-step inside M \ K j for j = 1 , , . . . . For any j ∈ N , we draw a smooth path in M \ K j from each z i or endpoint of a previouspath that is contained in K j +1 \ K j to a point in M \ K j +1 , such that all the newpaths are disjoint from each other, from previous paths, and from all the points z i . Moreover, we arrange the paths to have non-zero velocity, and whenever weextend an earlier path, we arrange that the concatenated path will be smooth.8et ( U i ) i (cid:62) be pairwise disjoint open neighbourhoods of ( R i ) i (cid:62) , such that eachpoint of M has a neighbourhood that meets only finitely many of the sets U i . ByCorollary 3.2, for each i , there is a smooth function F i : M → R supported in U i whose time-one Hamiltonian flow, ϕ i , excises R i from M . Let R := S i (cid:62) R i , let U := S i (cid:62) U i , let F := P i (cid:62) F i , and let ϕ : M \ R → M be the composition of( ϕ i ) i (cid:62) . Then F is well defined and smooth, it is supported in U , and the time-1flow of X F is ϕ . Let α := ( ϕ − ) ∗ ( β | M \ R ). Then we haved α = d(( ϕ − ) ∗ ( β | M \ R )) = ( ϕ − ) ∗ ( ω | M \ R ) = ω, so α is a primitive of ω . Moreover, α has no zeroes, because β has no zeroes in M \ R . In this section, we construct symplectic vector fields on symplectic manifolds thatextend null vector fields on submanifolds.
Definition 4.1. A presymplectic form is a closed two-form. A presymplectic man-ifold is a manifold S equipped with a presymplectic form ω S . On a presymplecticmanifold ( S, ω S ), a null vector field is a vector field X that is everywhere in thenull space ( T S ) ω S of the presymplectic form. (cid:5) On a presymplectic manifold (
S, ω S ), every null vector field X is presymplectic ,in the sense that L X ω S = 0; moreover, it is Hamiltonian in the sense that X ⌟ ω S is exact.A function on a topological space vanishes at infinity if all of the positiveepigraphs of its absolute value are compact. This is equivalent to being properaway from the zero locus, as a map to R \ { } . Any smooth manifold admits apositive smooth function vanishing at infinity, for instance, as the reciprocal of apositive exhaustion function. Theorem 4.2.
Let ( M, ω ) be a symplectic manifold with a codimension onesubmanifold ( N, ω N ) . Let Z be a subset of N that is closed in M . Let X be anull vector field on ( N, ω N ) that is non-vanishing on Z . Then there are smoothfunctions χ : M → [0 , and F : M → R such that χ = 1 on Z , X F | N = χX , andevery trajectory of X F that intersects M \ N is defined for all times. roof. Because X is a null vector field, α := X ⌟ ω is a section of the conormalbundle T ∗ M/N . We view α as a function on the normal bundle T M/N := T M | N /T N that is linear on each fibre. By the tubular neighbourhood theorem, there exist anopen neighbourhood W of N in M and a diffeomorphism (cid:15) : T M/N → W that identifies the zero section with N and whose differential along the zero sectioninduces the identity map on the normal bundle. Let H := α ◦ (cid:15) − : W → R . Thend H | N = α , and X H | N = X .Let π : W → N be the composition of (cid:15) − : W → T M/N with projection map T M/N → N . By the construction of H , and since N has codimension 1,for any z ∈ W \ N , H ( z ) = 0 if and only if X ( π ( z )) = 0 . (4.1)Let H be a positive smooth function on M vanishing at infinity. Because X is non-vanishing on Z and H vanishes on N , the set U := { z ∈ W | X ( π ( z )) = 0 and H ( z ) < H ( z ) } is an open neighbourhood of Z in M . Let χ : M → [0 ,
1] be a smooth functionthat is supported in U and is equal to 1 on Z . Moreover, suppose that d χ = 0wherever χ = 0; this can be achieved, for instance, by replacing χ by ρ ◦ χ where ρ ( s ) = 3 s − s .The function F := χH : M → R satisfies X F | N = χX H | N = χX (where the first equality is because H vanisheson N ). We will now show that, on the subset M \ N , we have dF = 0 whenever F = 0.Suppose that z ∈ M \ N and F ( z ) = 0. We claim that χ ( z ) = 0. Indeed, if z U , then χ ( z ) = 0 because the support of χ is contained in U . And if z ∈ U ,then z ∈ W \ N and X ( π ( z )) = 0 by the choice of U ; by (4.1) we have H ( z ) = 0;since ( χH )( z ) = F ( z ) = 0, we have χ ( z ) = 0. This proves our claim that χ ( z ) = 0.By the choice of χ , we then have that dχ | z = 0, and so dF | z = 0.Since | F | = | χH | (cid:54) | U H | < H , the function F vanishes at infinity, since H does. Then F is proper away fromthe zero locus, i.e., F | { F =0 } : { F = 0 } → R \ { } is proper. In particular, all the10on-zero level sets of F are compact. Then for any z ∈ M \ N , either F ( z ) = 0and F − ( F ( z )) is compact, or F ( z ) = 0 and X F ( z ) vanishes. This shows thatevery trajectory of X F that intersects M \ N is defined for all times. Throughout this section, let I := ( − , . We recall the solution theory of an autonomous ordinary differential equationof first order on I . Consider a non-negative smooth function v : I → [0 , ∞ ) andthe ordinary differential equation on I d x d t = v ( x ) . (5.1)Let γ = γ t ,x be the maximal solution to (5.1) with the initial condition ( t , x ).By Barrow’s formula [1], t − t = Z γ ( t ) x d ξv ( ξ ) if v ( x ) > γ ( t ) = x if v ( x ) = 0 . The flow of v , Φ v : D v → I, given by Φ v ( t, x ) := γ ,x ( t ) , is defined on the flow domain of v , which is an open subset D v of R × I of theform D v = { ( t, x ) ∈ R × I | S v ( x ) < t < T v ( x ) } for some upper semi-continuous function S v : I → [ −∞ , T v : I → (0 , ∞ ];we call these functions, respectively, the backward time function and the forwardtime function . By Barrow’s formula, T v ( x ) = Z x d ξv ( ξ ) ∈ (0 , ∞ ] if v ( ξ ) > ξ ∈ [ x, , ∞ if v ( ξ ) = 0 for some ξ ∈ [ x, , S v . In particular, if v ( x ) = 0 for all x close to the left endpoint − I , then, for all x ∈ I , S v ( x ) = −∞ . Now we consider a model with parameters.Recall that I = ( − , u : I × [ − , × [0 , × I → [0 , ∞ ) by u ( a, b, c ; x ) := χ a ( x )(1 − b ) 1 − x − x + c , (5.2)where χ a : I → [0 , a ∈ I , is the non-decreasing smooth function χ a ( x ) := − < x (cid:54) a −
12 ;exp( − x − ( a − / )exp( − x − ( a − / ) + exp( − a − x ) if a − (cid:54) x (cid:54) a ;1 if a (cid:54) x < . Consider the initial value problem d γ d t = u ( a, b, c ; γ ) ,γ (0) = x ;let γ = γ a,b,c ; x : ( S u ( a, b, c ; x ) , T u ( a, b, c ; x )) → I be its maximal solution curve. Theset D u := { ( a, b, c ; t, x ) ∈ I × [ − , × [0 , × R × I | S u ( a, b, c ; x ) < t < T u ( a, b, c ; x ) } is open in I × [ − , × [0 , × R × I , and the flow of u ∂∂x ,Φ u : D u → I, Φ u ( a, b, c ; t, x ) := γ a,b,c ; x ( t ) , is smooth.We calculate D u explicitly. The backward time is always −∞ : S u ( a, b, c ; x ) = −∞ for all ( a, b, c ; x ) . (5.3)If x (cid:54) a − or b = 1, the forward time T u ( a, b, c ; x ) is ∞ . Otherwise, a − < x < b <
1, and we have T u ( a, b, c ; x ) = Z x d ξu ( a, b, c ; ξ ) = Z x − b ) χ a ( ξ ) c − ξ ! d ξ. a (cid:54) x <
1, then T u ( a, b, c ; x ) = " ξ − b + c − b ) ln (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ξ − ξ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) x = − x − b , c = 0; ∞ , c > . (5.4)If a − < x < a then T u ( a, b, c ; x ) = − a − b + Z ax ξ − ( a − / − a − ξ )1 − b d ξ, c = 0; ∞ , c > . (5.5)Define the function µ : I × [ − , → [ − ,
1) by T u ( a, b, µ ( a, b )) = 1 . By (5.4) and (5.5) we conclude that µ is a smooth function increasing in b suchthat µ ( a, b ) = b for b (cid:62) a and µ ( a, b ) is greater than both b and a − for b < a .Then, for all ( a, b, c ; x ) ∈ I × [ − , × [0 , × I , T u ( a, b, c ; x ) (cid:54) b < , c = 0 , and µ ( a, b ) (cid:54) x. (5.6) Lemma 5.1.
Let B be a smooth manifold, let C ⊆ B be a closed subset, and let λ : C → ( − , be a smooth function. Then there exists a vector field on B × I , ofthe form v ( p, x ) ∂∂x with v ( p, x ) (cid:62) , whose forward time function T v and backwardtime function S v satisfy T v ( p, x ) (cid:54) if and only if p ∈ C and λ ( p ) (cid:54) x < and S v ( p, x ) = −∞ for all ( p, x ) ∈ B × I. Proof.
By the definition of a smooth function on a subset of a manifold, andbecause the subset C of B is closed, λ is the restriction to C of a smooth function b : B → ( − , a : B → I be a smooth function such that a ( p ) (cid:54) b ( p ) forall p ; for example, we can take a ( p ) = − b ( p )2 . Because C is closed, it is the zerolocus of a smooth function c : B → [0 ,
1] (see for instance [4]). We set v ( p, x ) := u ( a ( p ) , b ( p ) , c ( p ); x )13ith the function u ( a, b, c ; x ) of (5.2). Then for all ( p, x ) ∈ B × I , we have T v ( p, x ) = T u ( a ( p ) , b ( p ) , c ( p ); x ) and S v ( p, x ) = S u ( a ( p ) , b ( p ) , c ( p ); x ). By (5.3), S v ( p, x ) = −∞ for all ( p, x ). By (5.6), for all ( p, x ) ∈ B × I , T v ( p, x ) (cid:54) b ( p ) < , c ( p ) = 0 , and µ ( a ( p ) , b ( p )) (cid:54) x. Because a ( p ) (cid:54) b ( p ), we have µ ( a ( p ) , b ( p )) = b ( p ); the condition c ( p ) = 0 meansthat p ∈ C , which implies that b ( p ) = λ ( p ); because x ∈ I , we have x <
1; so T v ( p, x ) (cid:54) p ∈ C and λ ( p ) (cid:54) x < . Corollary 5.2.
Let ( B, ω B ) be a presymplectic manifold, and equip N := B × I with the presymplectic structure ω N := ω B ⊕ , namely, the pullback of ω B by theprojection to B . Let C ⊆ B be a closed subset, let λ : C → ( − , be a smoothfunction, and let Z be the epigraph of λ in N : Z := { ( p, x ) ∈ B × I | p ∈ C and x (cid:62) λ ( p ) } . Then the vector field of Lemma 5.1 is a null vector field, and the time- map ofthe flow that it generates is a presymplectomorphism N \ Z → N. Proof.
Let X = v ( p, x ) ∂∂x be the vector field of Lemma 5.1, and let ( ψ t ) t ∈ R be theflow that it generates. Each ψ t is a diffeomorphism between open subsets of N . Thedomain of ψ t is { z ∈ N | T v ( z ) > t } , and the range of ψ t is { z ∈ N | S v ( z ) < − t } .Because S v ( z ) = −∞ for all z , the range of ψ t is all of N . Because T v ( z ) (cid:54) z ∈ Z , the domain of ψ is N \ Z . The vector field X is a null vector field becauseit is a multiple of ∂∂x (where we write coordinates on B × I as ( p, x )) and ω N is thepullback of a two-form on B . It follows that each ψ t is a presymplectomorphism.So ψ is a presymplectomorphism from N \ Z to N .We now prove a symplectic excision result for closed subsets of codimension atleast one. As before, let I = ( − , Theorem 5.3.
Let ( M, ω ) be a n -dimensional symplectic manifold, let ( B, ω B ) bea (2 n − -dimensional symplectic manifold, and let ψ : B × I → M be an embeddingsuch that ψ ∗ ω = ω B ⊕ . Let λ : C → ( − , be a smooth function on a closedsubset C of B . Let Z be the image under ψ of the epigraph { ( p, x ) ∈ B × I | x ∈ C and x (cid:62) λ ( p ) } . Suppose that Z is closed in M . Then Z is time-independentHamiltonian excisable from ( M, ω ) . roof. We use ψ to identify N := B × I with its image in M ; we write ω N for thepullback of ω to N , which is equal to ω B ⊕ X = v ( p, x ) ∂∂x be a vector field on N as in Lemma 5.1, so that its forwardtime function T v and backward time function S v satisfy T v ( z ) (cid:54) z ∈ Z and S v ( z ) = −∞ for all z ∈ N. Note that X is a null vector field on ( N, ω N ) that is non-vanishing on Z . ByTheorem 4.2, there are smooth functions χ : M → [0 ,
1] and F : M → R , such that χ = 1 on Z , X F | N = χ X , and every trajectory of X F that starts on M \ N isdefined for all times.We now analyse the dynamical behaviour of X F . Let S F and T F be itsforward and backward time functions. By the choice of F , for all z ∈ M \ N we have S F ( z ) = −∞ and T F ( z ) = ∞ . Because the restriction of X F to N is a multiple of X by a function with values in [0 , z ∈ N we have S F ( z ) (cid:54) S v ( z ) < T v ( z ) (cid:54) T F ( z ). Finally, because Z is invariant under the forwardflow of X and is closed in M and X F | Z = v , for all z ∈ Z we have T v ( z ) = T F ( z ).We conclude that T F ( z ) (cid:54) z ∈ Z and S F ( z ) = −∞ for all z ∈ M. The time-1 map of the flow generated by X F is then a symplectomorphism from M \ Z to M . Remark . Theorem 3.1, about excising the ray { } n − × [0 , ∞ ) × { } from( M, ω ) = ( R n , dx ∧ dy + . . . + dx n ∧ dy n ), also follows from Theorem 5.3. Tosee this, take N ⊂ M to be given by the vanishing of the y n coordinate, take( B, ω B ) = ( R n − , dx ∧ dy + . . . + dx n − ∧ dy n − ), identify B × I with N througha diffeomorphism I → R in the x n coordinate, and take C = { } n − and λ = 0. (cid:5) It is interesting to note that we can excise subsets with complicated topology:
Example 5.5.
The “Cantor brush” Z := { } n − × C × [0 , ∞ ) × { } , where n (cid:62) C is the Cantor set, is Hamiltonian excisable from ( R n , ω can ); see Figure 5.1.To see this, note that there is a symplectomorphism of ( − , × R with R thattakes [0 , × { } to [0 , ∞ ) × { } , and apply Theorem 5.3 with the function λ = 0on C . (cid:71) n \ ( C × ray) R n Figure 5.1: Removing the Cantor brush.
In Section 5, we excise the epigraph Z in B × ( − ,
1) of a smooth function λ : C → ( − ,
1] over a closed subset C of B . We can view this same Z as theepigraph in B × ( − ,
1) of the function B × ( − ,
1] that is obtained from λ bydefining it to be equal to 1 outside C . This function might not be smooth, but itis lower semi-continuous.Can we excise epigraphs of more general functions?Such a function must be lower semi-continuous. Indeed, as we noted inRemark 2.2, the subset that we excise must be closed, and the epigraph of afunction is closed if and only if the function is lower semi-continuous.In this section we excise the epigraphs of arbitrary lower semi-continuousfunctions. The proof relies on the following smooth version of a theorem of Baire(see [2]). Lemma 6.1.
Let B be a manifold, and let λ : B → (0 , be a lower semi-continuousfunction. Then there exists a strictly increasing sequence of positive smoothfunctions λ n : B → R , for n ∈ N , whose supremum is λ .Proof. Fix a metric d ( · , · ) on B . We will construct a strictly increasing sequenceof smooth functions λ n : B → R that have the following property. For each n ∈ N and y ∈ B , λ ( y ) > λ n ( y ) (cid:62) (1 − n ) inf (cid:26) λ ( x ) | d ( x, y ) < n (cid:27) . (6.1)16his property implies that the supremum of the sequence is λ . Indeed, fix y ∈ B , and let ε >
0. Because λ ( y ) is positive, the set { x | λ ( x ) > (1 − ε λ ( y ) } contains y ; because the function λ is lower semi-continuous, this set is a neigh-bourhood of y in B . Let N ∈ N be such that d ( x, y ) < N implies that x is in thisneighbourhood and such that N < ε . Then, for any n (cid:62) N , λ ( y ) > λ n ( y ) (cid:62) λ N ( y ) (cid:62) (1 − N ) inf { λ ( x ) | d ( x, y ) < N } (cid:62) (1 − ε λ ( y ) > (1 − ε ) λ ( y ) . Since λ ( y ) is positive and ε > n →∞ λ n ( y ) = λ ( y ).So far, λ n is positive for all n (cid:62)
2, and λ (cid:62)
0. Replacing λ by ( λ + λ )yields a sequence of positive functions as required.We now construct a sequence λ n with the above property. For n = 1, we define λ ( y ) := 0 for all y ∈ B . Arguing recursively, fix n (cid:62)
2, and let λ n − : B → R bea smooth function such that λ ( y ) > λ n − ( y ) for all y ∈ B . We will construct asmooth function λ n : B → R that satisfies (6.1) and such that λ n ( y ) > λ n − ( y ) forall y ∈ B .For each x ∈ B , let c ( x, n ) be a real number such that λ ( x ) > c ( x, n ) > max { λ n − ( x ) , (1 − n ) λ ( x ) } , and let U ( x, n ) := { y ∈ B | d ( y, x ) < n and λ ( y ) > c ( x, n ) > λ n − ( y ) } . Because λ ( x ) > c ( x, n ) > λ n − ( x ) (by the choice of c ( x, n )), the set U ( x, n )contains x ; because λ n − is (smooth, hence) continuous and λ is lower semi-continuous, the set U ( x, n ) is open. So U ( x, n ) is a neighbourhood of x in B .For each x ∈ B and y ∈ U ( x, n ), λ ( y ) > c ( x, n ) > λ n − ( y ) . (6.2)Also, because c ( x, n ) > (1 − n ) λ ( x ) (by the choice of c ( x, n )) and d ( y, x ) < n (because y ∈ U ( x, n )), c ( x, n ) > (1 − n ) inf { λ ( z ) | d ( y, z ) < n } . (6.3)17et { ρ n,i } i (cid:62) be a smooth partition of unity with supp ρ n,i ⊆ U ( x i , n ) for all i .Define λ n ( y ) := X i (cid:62) ρ n,i c ( x i , n ) . Then λ n is smooth. For each y ∈ B , from the inequalities (6.2) and (6.3) where x runs over those x i such that ρ n,i ( y ) = 0, we obtain λ ( y ) > λ n ( y ) > λ n − ( y ) and λ n ( y ) > (1 − n ) inf { λ ( z ) | d ( y, z ) < n } . So λ n > λ n − , and λ n satisfies (6.1).We spell out an easy lemma: Lemma 6.2.
Let B be a manifold, and let f : B → (0 , be a continuous function.Then there exists a smooth function g : B → (0 , such that > g > f .Proof. For each x ∈ B , let c ( x ) be a real number such that 1 > c ( x ) > f ( x ), andlet U ( x ) := { y ∈ B | c ( x ) > f ( y ) } . By the choice of c ( x ), the set U ( x ) contains x ;because f is continuous, the set U ( x ) is open. So U ( x ) is a neighbourhood of x in B . Let { ρ i } i (cid:62) be a smooth partition of unity with supp ρ i ⊆ U ( x i ). Take g ( y ) := X i (cid:62) ρ i ( y ) c ( x i ) . We continue with a couple of preparatory lemmas.
Lemma 6.3.
There exists a smooth family of functions u a,b,τ : (0 , → (0 , , parametrized by triples of real numbers a, b, τ such that < a < b < and τ > ,such that, for each such a, b, τ , the following holds. • The function u a,b,τ is equal to outside the interval ( a, b ) . • The flow of the vector field u a,b,τ ( x ) ∂∂x on (0 , takes a to b in time b − a + τ . roof. Let u a,b,τ ( x ) := b − a R − exp( − − ξ ) d ξ b − a R − exp( − − ξ ) d ξ + τ exp( − ( b − a ) x − a )( b − x ) )for x ∈ ( a, b ), and u a,b,τ ( x ) := 1 for x ( a, b ). By Barrow’s formula, the flow ofthe vector field u a,b,τ ( x ) ∂∂x takes a to b in time Z ba d ξu a,b,τ ( ξ ) = b − a + τ R ba exp( − ( b − a ) ξ − a )( b − ξ ) ) d ξ b − a R − exp( − − ξ ) d ξ . With the change of variable ξ = a + b + η b − a , this becomes . . . = b − a + τ R − exp( − − η ) b − a d η b − a R − exp( − − ξ ) d ξ = b − a + τ. Fix a manifold B . Consider the product B × (0 , p, x )with p ∈ B and x ∈ (0 , Lemma 6.4.
Let f, a, b be real valued smooth functions on B such that < f < a < b < . Then there exists a vector field on B × (0 , of the form X | ( p,x ) = v ( p, x ) ∂∂x , with < v ( p, x ) (cid:54) , with the following properties. • The forward time function T X of X satisfies T X ( p, x ) (cid:54) if and only if x (cid:62) f ( p ) . • The vector field X coincides with ∂∂x on some neighbourhood of the set { ( p, x ) | f ( p ) (cid:54) x (cid:54) a ( p ) or x (cid:62) b ( p ) } .Proof. Take v ( p, x ) := u a ( p ) ,b ( p ) ,f ( p ) ( x ) , where u a,b,τ is the function from Lemma 6.3 with a = a ( p ), b = b ( p ), τ = f ( p ).19 emma 6.5. Let f, h, b, c be real valued smooth functions on B such that < f < h < b < c < . Let X be a vector field on B × (0 , of the form X | ( p,x ) = v ( p, x ) ∂∂x , with < v ( p, x ) (cid:54) , with the following properties. • The forward time function T X of X satisfies T X ( p, x ) (cid:54) if and only if x (cid:62) f ( p ) . • The vector field X coincides with ∂∂x on some neighbourhood of the set { ( p, x ) | x (cid:62) b ( p ) } .Then there exists a vector field X on B × (0 , of the form X | ( p,x ) = v ( p, x ) ∂∂x , with v ( p, x ) (cid:62) v ( p, x ) > , with the following properties. • The forward time function T X of X satisfies T X ( p, x ) (cid:54) if and only if x (cid:62) h ( p ) . • The vector field X coincides with X on some neighbourhood of the set { ( p, x ) | < x (cid:54) b ( p ) } and coincides with ∂∂x on some neighbourhood of theset { ( p, x ) | x (cid:62) c ( p ) } .Proof. By Barrow’s formula, the flow of X takes each point ( p, f ( p )) to the point( p, h ( p )) in time τ ( p ) := Z h ( p ) f ( p ) dxv ( p, x ) . The function τ : B → (0 ,
1) is smooth. The forward time function T X of X satisfies T X ( p, h ( p )) = 1 − τ ( p ) . v ( p, x ) := v ( p, x ) if 0 < x (cid:54) b ( p ) ,u b ( p ) ,c ( p ) ,τ ( p ) if b ( p ) < x (cid:54) c ( p ) , c ( p ) < x < , where u b,c,τ is as in Lemma 6.3. Theorem 6.6.
Let λ : B → (0 , be a lower semi-continuous function. Then there exists a vector field X on B × (0 , of the form v ( p, x ) ∂∂x , with (cid:54) v ( p, x ) (cid:54) , whose forward and backward time functions T X and S X satisfy T X ( p, x ) (cid:54) if and only if x (cid:62) λ ( p ) and S X = −∞ everywhere . Proof.
Lemma 6.1 implies that there exists a strictly increasing sequence of positivesmooth functions f n : B → (0 ,
1) for n ∈ N , < f < f < f < . . . , such that λ = sup n ∈ N f n . There also exists a strictly increasing sequence of smooth functions g n : B → (0 , , g < g < g < g < . . . such that f n < g n − for all n ∈ N and sup n ∈ N g n ≡ . Indeed, for n = 0, apply Lemma 6.2 with the continuous function f to finda function g : B → (0 ,
1) such that f < g <
1. Arguing recursively, givena smooth function g n − : B → (0 , { g n − , f n +1 , − n } to find a function g n : B → (0 ,
1) such that1 > g n > max { g n − , f n +1 , − n } .We will now construct, recursively, a sequence of smooth vector fields X n on B × (0 ,
1) of the form X n | ( p,x ) = v n ( p, x ) ∂∂x , parametrized by n ∈ N , with the following properties.(i) For all n (cid:62) p, x , v n − ( p, x ) (cid:62) v n ( p, x ) > . (ii) The forward time function T X n of each X n satisfies T X n ( p, x ) (cid:54) x (cid:62) f n ( p ) . (iii) For each n (cid:62)
2, the function v n coincides with the function v n − on aneighbourhood of the set { ( p, x ) | x (cid:54) g n − ( p ) } and is equal to 1 on aneighbourhood of the set { ( p, x ) | x (cid:62) g n ( p ) } .Here is the construction. We obtain X by applying Lemma 6.4 to the functions f = f , a = g , b = g . Given X n − , we obtain X n by applying Lemma 6.5 to thevector field X := X n − and to the functions f := f n − , h := f n , b := g n − , c := g n .Because sup g n = 1, the union of the open sets O n := { ( x, p ) | x < g n ( p ) } is all of B × (0 , v ∞ ( p, x ) on B × (0 ,
1) that, for each n , coincides with v n on O n . We define X ∞ := v ∞ ( p, x ) ∂∂x . We now show that that, for each ( p, x ) ∈ B × (0 , T X ∞ ( p, x ) > x < λ ( p ) . First, suppose that x < λ ( p ). Because sup f n = λ , there exists an n such that x < f n ( p ). Fix such an n . By Property (ii), T X n ( p, x ) >
1. Property (i) implies22hat v n ( p, · ) (cid:62) v ∞ ( p, · ) > x, T X ∞ ( p, x ) (cid:62) T X n ( p, x ). Putting thesetogether, we obtain T X ∞ ( p, x ) > T X ∞ ( p, x ) >
1. Let ( p, y ) be the image of ( p, x ) under thetime-1 flow of X ∞ . Because sup g n = 1 and y <
1, there exists an n such that y < g n ( p ). Fix such an n . Because on the set { ( p, x ) | x < g n ( p ) } the vectorfield X ∞ coincides with X n , the time-1 flow of X n also takes ( p, x ) to ( p, y ). So T X n ( p, x ) >
1. By Property (ii), x < f n ( p ). This implies that x < λ ( p ).To finish, take X := χ ( p, x ) X ∞ where χ ( p, x ) = 0 if 0 < x (cid:54) f ( p ) and χ ( p, x ) = 1 if x (cid:62) f ( p ).We now use these results to excise epigraphs of functions that are not necessarilysmooth. As we noted in Remark 2.2, the subset that we excise must be closed.Recall that the epigraph of a function is closed if and only if the function is lowersemi-continuous. Theorem 6.7.
Let ( M, ω ) be a n -dimensional symplectic manifold. Let ( B, ω B ) be a (2 n − -dimensional symplectic manifold, let λ : B → (0 , be a lowersemi-continuous function, and let U ⊆ B × I be an open neighbourhood of theepigraph Z := { ( p, x ) ∈ B × I | x (cid:62) λ ( p ) } . Let ψ : U → M be an embedding such that ψ ∗ ω = ω B ⊕ . Suppose that Z := ψ ( Z ) is closed in M . Then Z is time-independent Hamiltonian excisable from ( M, ω ) .Proof. Let X be a vector field on B × (0 ,
1) as in Theorem 6.6, so that X := v ( p, x ) ∂∂x with v : B × (0 , → [0 , T X ( p, x ) (cid:54) x (cid:62) λ ( p ) and S X = −∞ everywhere . Let χ : B × (0 , → [0 ,
1] be a smooth function that is supported in U and isequal to 1 on Z . Then X := χ X | U is a null vector field on U whose forwardand backward time functions satisfy T X ( p, x ) (cid:54) x (cid:62) λ ( p ) and S X = −∞ everywhere . Applying Theorem 4.2 to the codimension one submanifold N := ψ ( U ) of M , with its subset Z that is closed in M and the null vector field Y := ψ ∗ X
23n (
N, ω N ) that is non-vanishing on Z , we obtain a smooth function F : M → R whose Hamiltonian vector field X F has the following properties. X F | N = χY for some χ : M → [0 ,
1] such that χ = 1 on Z , and every trajectory of X F thatintersects M \ N is defined for all times.We now analyse the dynamical behaviour of X F . Let S F and T F be itsforward and backward time functions. By the choice of F , for all z ∈ M \ N we have S F ( z ) = −∞ and T F ( z ) = ∞ . Because the restriction of X F to N is a multiple of Y by a function with values in [0 , z ∈ N we have S F ( z ) (cid:54) S Y ( z ) < T Y ( z ) (cid:54) T F ( z ); since S Y ( z ) = −∞ , also S F ( z ) = −∞ .Finally, because Z is invariant under the forward flow of Y and is closed in M and X F | Z = Y , for all z ∈ Z we have T v ( z ) = T F ( z ). We conclude that T F ( z ) (cid:54) z ∈ Z, and S F ( z ) = −∞ for all z ∈ M. The time-1 map of the flow generated by X F is then a symplectomorphism from M \ Z to M . Example 6.8.
The “box with a tail” Z := [ − , n − × [0 , ∞ ) × { } ∪ { } n − × [ − , ∞ ) × { } is Hamiltonian excisable from ( R n , ω can ) when n (cid:62)
2; see Figure 6.1. Indeed, Z is the epigraph (times { } ) of the lower semi-continuous function λ : R n − → ( −∞ , ∞ ] that takes the { } n − to −
1, that takes [ − , n − \ { } n − to 0, andthat takes R n − \ [ − , n − to ∞ . Identify R n with B × (0 , × R with B = R n − ,so that Z becomes the epigraph Z (times { } ) of a new lower semi-continuousfunction λ : B → (0 , (cid:71) R n \ (box ∪ ray) R n Figure 6.1: Removing a box with a tail.24
Symplectic excision in stages
Recall that a tree is a connected acyclic graph with no cycles. We define an open-rooted tree as a tree with one of its nodes removed. The ray [0 , ∞ ) is anexample of an open-rooted tree; so is the “ray with two horns”,[0 , ∞ ) × { } ∪ { ( x, y ) | − (cid:54) y (cid:54) x = −| y |} . A singleton, a closed line segment, and the “double Y”,[ − , × { } ∪ { ( x, y ) | − (cid:54) y (cid:54) | x | = 1 + | y |} , are examples of trees. A tree, or an open-rooted tree, is finite if it has finitelymany vertices. R n \ T T R n \ T T R n \ R R R n Figure 7.1: Removing an open-rooted tree.
Example 7.1.
Any properly embedded open-rooted finite tree in a symplecticmanifold is symplectically neighbourhood excisable; see Figure 7.1.
Proof.
We prove this by induction on the number of vertices in the tree.25et (
M, ω ) be a symplectic manifold, and let T be a properly embedded open-rooted finite tree in M . Let U be a neighbourhood of T in M . Fix a leaf z of T ,and let R z be the branch connecting z that is closed at z and open at the otherend. Then• T := T \ R z is a properly embedded open-rooted tree in M with one lessvertex, and U is a neighbourhood of T in M .• R z is a properly embedded ray in M := M \ T , and U z := U \ T is aneighbourhood of R z in M .By Corollary 3.2, R z is Hamiltonian excisable from M , so there exists (seeRemark 2.3) a symplectomorphism ϕ z from M \ R z (= M \ T ) to M that is theidentity on some neighbourhood V z of M \ U z (= M \ U ) in M \ R z (= M \ T ): ϕ z : M \ T → M , identity on V z , , M \ U ⊂ V z ⊂ M \ T. By the induction hypothesis, T is symplectically neighbourhood excisable from M ,so there exists a symplectomorphism ϕ z from M \ T (= M ) to M that is theidentity on some neighbourhood V z of M \ U in M \ T (= M ): ϕ z : M → M , identity on V z , , M \ U ⊂ V z ⊂ M . The composition ϕ := ϕ z ◦ ϕ z is a symplectomorphism from M \ T to M that isthe identity on the intersection V z ∩ ϕ − z ( V z ) . It remains to show that this intersection contains M \ U . Because V z and V z contain M \ U , it is enough to show ϕ − z ( M \ U ) contains M \ U . Since ϕ − z ( M \ U ) =( M \ T ) \ ϕ − z ( U ) and T is contained in U , this is equivalent to showing that ϕ − z ( U )is contained in U . Indeed, suppose that x ∈ ϕ − z ( U ). That is, x ∈ M \ T and ϕ z ( x ) ∈ U . Seeking a contradiction, suppose that x is not in U . Then ϕ z ( x ) = x .So ϕ z ( x ) is also not in U , which contradicts our assumption on x .Here is another interesting question: given a compact subset K of a symplecticmanifold ( M, ω ), is ( M \ K, ω ) symplectomorphic to M with several puncturedpoints? Example 7.2 (Retracting a tree) . Let (
M, ω ) be a symplectic manifold, and let T be an embedded finite tree in M . Then for any neighbourhood U of T in M and point z in T there is a symplectomorphism from M \ T to M \ { z } that isthe identity on some neighbourhood of M \ U in M \ T ; see Figure 7.2.26 roof. Note that T \ { z } is a disjoint union of finitely many open-rooted trees T j ,indexed by j ∈ { , . . . , k } for some k ∈ N , each of which is properly embedded in M \ { z } . Let U j , for j = 1 , . . . , k , be disjoint open neighbourhoods of the open-rooted trees T j in U \{ z } . By Example 7.1, for each j there is a symplectomorphismfrom M \ T j to M \ { z } that is the identity on some neighbourhood of M \ U j in M \ T j . The composition of these symplectomorphisms is a symplectomorphismfrom M \ T to M \ { z } that is the identity on some neighbourhood of M \ U in M \ T . M \ T M \ [ − , M \ pt Figure 7.2: Retracting a tree.
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International Mathematics Research Notices , 2020,https://doi.org/10.1093/imrn/rnaa007. rnaa007.Xiudi Tang [email protected]
Department of Mathematical and Computational SciencesUniversity of Toronto Mississauga3359 Mississauga RoadMississauga, ON L5L 1C6, Canada
Yael Karshon [email protected]