Mapping class groups of surfaces with noncompact boundary components
MMapping class groups of surfaces withnoncompact boundary components
Ryan DickmannJanuary 5, 2021
Abstract
We show that the pure mapping class group is uniformly perfect fora certain class of infinite type surfaces with noncompact boundary com-ponents. We then combine this result with recent work in the remainingcases to give a complete classification of the perfect and uniformly perfectpure mapping class groups for infinite type surfaces. We also develop amethod to cut a general surface into simpler surfaces and extend somemapping class group results to the general case.
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Let S be a connected, orientable, and second-countable surface, possiblywith boundary. The mapping class group MCG( S ) is the group of orientationpreserving homeomorphisms of S which fix the boundary of S pointwise. Theelements in this group are considered up to isotopy relative to the boundary.An infinite type surface refers to a surface with π ( S ) not finitely generated,and the MCG( S ) of such surfaces are commonly referred to as big mapping classgroups . These groups have been the recent focus of many papers, but the caseof noncompact boundary components has been largely untouched with only asingle paper known to the author considering such groups [Fab03].The pure mapping class group PMCG( S ) is the subgroup of MCG( S ) con-sisting of elements which fix the ends of S , and PMCG c ( S ) is the subgroup ofcompactly supported elements. Recently George Domat (and the author in onecase) has shown the following. Theorem 1.1. [DD20] Let S be any infinite type surface with only compactboundary components. Then PMCG c ( S ) and PMCG( S ) are not perfect. This partially answered Problem 8 from [APV17]. In the finite type case,it is a well-known result of Powell that pure mapping class groups are perfectfor genus at least 3 [Pow78]. Surprisingly, a new phenomenon occurs when wealso consider surfaces with noncompact boundary components, and even thoughthe general case seems extremely complicated at first glance, it turns out it ispossible to completely classify the surfaces with perfect or uniformly perfect puremapping class groups. A
Disk with Handles will refer to a surface constructedfrom a disk by removing points from the boundary and then attaching infinitelymany handles accumulating to some subset of these points. We say compactboundary components are added to a surface when we delete open balls withdisjoint closures from the interior. We say punctures are added when we removeisolated interior points. See Definition 3.11 for the definition of a degenerateend.
Theorem A.
Let S be an infinite type surface without degenerate ends. • PMCG c ( S ) is uniformly perfect if and only if S is a Disk with Handles. • PMCG c ( S ) is perfect if and only if S is a connected sum of finitely manyDisks with Handles with possibly finitely many punctures or compact bound-ary components added. In [APV17], it was shown for surfaces with only compact boundary com-ponents that PMCG( S ) = PMCG c ( S ) if and only if S has at most one end2ccumulated by genus, and otherwise PMCG( S ) factors as a semidirect productof PMCG c ( S ) with some Z n . See Theorem 6.1 for the precise statement of thistheorem. Once we extend this result to the general case, we immediately get aclassification of the perfect PMCG( S ). A Disk with Handles with exactly oneend will be referred to as a Sliced Loch Ness Monster . Theorem B.
Let S be an infinite type surface without degenerate ends. • PMCG( S ) is uniformly perfect if and only if S is a Sliced Loch Ness Mon-ster. • PMCG( S ) is perfect if and only if S is a Sliced Loch Ness Monster withpossibly finitely many punctures or compact boundary components added. α α α α β β β β γ Figure 1: A Sliced Loch Ness Monster with two infinite collections of curves.The collection { β i } eventually leaves every compact subsurface, but every curvein the collection { α i } intersects an arc γ .For a Sliced Loch Ness Monster, the pure mapping class group and the map-ping class group coincide. Therefore, this also gives new examples of surfaceswith uniformly perfect mapping class groups. These results show there is aninteresting distinction between these mapping class groups and the previouslystudied cases. In particular, the results of Powell and Domat demonstrate a con-sistent behavior for pure mapping class groups of surfaces without noncompactboundary components, but the cases we study demonstrate a more complicatedbehavior. Also many of the tools from the other cases do not easily extend asone would hope, so new techniques need to be discovered.Disks with Handles and Sliced Loch Ness Monsters will be an essential partof this paper. In Section 4 we will show how to cut a Disk with Handles into acollection of Sliced Loch Ness Monsters, so we can use these simpler surfaces asbuilding blocks for a general argument. We can summarize the decompositionresults with the following theorem which is partially inspired by a result in[APV17]. See Section 3 for some of the terminology. This name was chosen because the interior of such a surface is often referred to as the
Loch Ness Monster . The author apologizes for adding to the already out of hand terminology. heorem C. Every Disk with Handles without planar ends can be cut along acollection of disjoint essential arcs into Sliced Loch Ness Monsters.Furthermore, any infinite type surface with infinite genus and no planar endscan be cut along disjoint essential simple closed curves into components whichare either(i) Loch Ness Monsters with k ∈ N ∪{∞} compact boundary components addedpossibly accumulating to the single end. (ii) Disks with Handles with k ∈ N ∪{∞} compact boundary components addedpossibly accumulating to some subset of the ends. Acknowledgements
The author would like to thank his advisor, Mladen Bestvina, for helping himthrough this project. Thanks to George Domat for many helpful conversationsand for writing up the first version of the fragmentation lemma which is criticalto this paper.
In Section 3 we discuss the necessary background including the ClassificationTheorem for orientable noncompact surfaces. The case of compact boundarywas done by Ker´ekj´art´o [Ker23] and Richards [Ric63]. The general case wasdone by Brown and Messer [BM79].In Section 4 we prove Theorem C, and in Section 5 we prove Theorem A.The proof that PMCG c ( S ) is uniformly perfect for a Disk with Handles followsa similar proof for the symmetric group on a countably infinite set [Ore51]. Firstwe use a fragmentation lemma to decompose a map in PMCG c ( S ) into a productof two simpler maps. Then after decomposing the surface into simpler piecesusing Theorem C, we can apply standard tricks to write each of the simplermaps as a product of commutators. In Section 6 we discuss how to extend thework of [APV17] to the general case, proving Theorem B.This work was inspired by the observation that the mapping class groupfor a surface with noncompact boundary components often corresponds to aproper subgroup of the mapping class group of the interior surface. For exam-ple, consider the surface with infinite genus, one end, one noncompact boundarycomponent, and no compact boundary components. It follows from the Classifi-cation Theorem in Section 3 that there is a unique surface with these properties.This is the , and we denote it L s . If we take an in-finite collection of curves { α i } accumulating to the boundary as in Figure 1,then the infinite product of Dehn twists · · · T α T α T α does not correspond toa homeomorphism of L s . To see this take another infinite collection of curves { β i } and an arc γ as shown in the figure. If we delete the boundary of L s , and Here we are using N = { , , , ... } . L be the resulting surface, then the infinite product of twists correspondsto a well-defined homeomorphism T = Π ∞ i =1 T α i ∈ MCG( L ). The inclusion ofsurfaces induces a homomorphism i : MCG( L s ) → MCG( L )but T does not correspond to an element in the image. Assume otherwise, andconflate T with a homeomorphism on L s . Note T ( γ ) intersects all of the β i ,so it follows the image is not compact, a contradiction. This follows a similarargument from Proposition 7.1 in [PV18].We will show that i is injective in Section 6. Note we have just shown that i is not surjective, but it requires more work to show that MCG( L s ) and MCG( L )are not abstractly isomorphic. Once we are done though, this will follow fromTheorem A. Here we summarize the Classification Theorems from [Ric63] and [BM79]starting with the case of compact boundary. We briefly review the necessaryterminology. We always let a surface refer to a connected, orientable, andsecond-countable 2-manifold. We will assume subsurfaces are connected unlessstated otherwise. A complementary domain of a surface S is a subsurface whichis the closure of some component of S \ K for a compact subsurface K . Definition 3.1. An exhausting sequence for a surface S is a sequence of sub-surfaces { U i } such that the following properties hold: • U i +1 ⊂ U i for all i . • (cid:84) ∞ i =1 U i = ∅ . • Each U i is a complementary domain.Two such sequences { U i } and { U (cid:48) i } are considered equivalent if for any i thereexists a j with U j ⊂ U (cid:48) i , and vice versa. This defines an equivalence relation onthe set of exhausting sequences, and an equivalence class is referred to as an end of the surface. The ends space of S is the collection of all equivalence classes,and it is denoted E( S ).For a given subsurface U , let U (cid:63) be the set of ends such that there is arepresentative sequence eventually contained within U . We now equip E ( S )with a basis generated by sets of the form U (cid:63) ranging over all subsurfaces U suchthat U is a complementary domain. This basis gives a topology on E ( S ) whichis totally disconnected, second-countable, compact, and Hausdorff (see [AS60]5hapter 1, Proposition 36 and 37). Topological spaces with these properties arealways homeomorphic to a closed subset of the Cantor set.We say an end is accumulated by genus if there is a representative sequence { U i } so every U i has infinite genus. We denote the set of ends accumulatedby genus as E ∞ ( S ). An end is planar if there is a representative sequence sosome U i is homeomorphic to a subset of the plane. The space of planar ends isexactly E ( S ) \ E ∞ ( S ). We say an end is isolated if it is isolated in the topologyon the space of ends. Isolated planar ends are referred to as punctures .When we consider surfaces with compact boundary, there is the followingClassification Theorem. Theorem 3.2 (Classification of Surfaces (compact ∂ ), [Ric63]) . A surface S with compact boundary is determined up to homeomorphism by the quadruple ( g, b, E( S ) , E ∞ ( S )) , where g ∈ N ∪ {∞} is the genus of S , b ∈ N is the numberof boundary components, E ( S ) is a totally disconnected, second-countable, com-pact Hausdorff space, and E ∞ ( S ) is a closed subset of E ( S ) which is nonemptyif and only if g is infinite. The sets E ( S ) and E ∞ ( S ) are considered up tohomeomorphism as a pair. Now we summarize the ideas for the general case. The previous definitionsall apply to a general surface without adaptation, but we need more informationto capture the new possibilities.For a surface with infinitely many compact boundary components, we mustrecord the ends which are accumulated by these components. We refer to theseas ends accumulated by boundary , and we denote the space of these ends by E ∂ ( S ). This can be precisely defined in a similar manner to accumulated bygenus.Let ˆ ∂S be the union of the noncompact boundary components of a surface S .Each element of ˆ ∂S has two ends which correspond to possibly equal elementsof E ( S ). Let E ( ˆ ∂S ) be the set of ends of the components of ˆ ∂S . This is justa discrete space with two ends for each component. Let v : E ( ˆ ∂S ) → E ( S ) bethe function that takes an end of a noncompact boundary component to theend of the surface to which it corresponds. We refer to E ( ˆ ∂S ) as the set ofboundary ends . The elements of E ( ˆ ∂S ) are referred to as boundary ends , but wewill conflate definitions and use this to also refer to ends of E ( S ) in the imageof v . An end in E ( S ) is said to be an interior end otherwise. If a boundary endin S is isolated from the other ends, then we refer to it as a boundary puncture .Let e : E ( ˆ ∂S ) → π ( ˆ ∂S ) be the map that takes an end to the correspondingnoncompact boundary component. If we fix an orientation on S , then for anarbitrary component p ∈ π ( ˆ ∂S ) we may distinguish the right and left ends of e − ( p ). An orientation of E ( ˆ ∂S ) is a subset O ⊂ E ( ˆ ∂S ) that contains exactlythe right ends for the given orientation. Now we have all of the necessaryinformation to determine a general surface. We collect this in the followingdiagram: 6 ( ˆ ∂S ) E ( ˆ ∂S ) E ( S ) O E ∞ ( S ) E ∂ ( S ) e v The unlabeled arrows are the inclusion maps.
Definition 3.3. (Surface Diagram) A surface diagram is a diagram of the aboveform with the following properties: • E ( S ) is totally disconnected, second-countable, compact Hausdorff space. E ∞ ( S ) and E ∂ ( S ) are closed subsets of E ( S ). • E ( ˆ ∂S ) and π ( ˆ ∂S ) are countable and discrete spaces. O is some subset of E ( ˆ ∂S ). • The map e is two-to-one and onto. • If q ∈ E ( ˆ ∂S ), let q (cid:48) denote the other element of e − ( e ( q )). If V is anyopen and closed subset of E ( S ), then(i) S ( V ) = { q ∈ E ( ˆ ∂S ) | v ( q ) ∈ V, v ( q (cid:48) ) / ∈ V } is a finite set with an evennumber of elements.(ii) S ( V ) ∩ O has the same number of elements as S ( V ) ∩ ( E ( ˆ ∂S ) \ O ).(iii) If V ∩ E ∞ ( S ) = ∅ , then for any finite subset E of π ( ˆ ∂S ) there is apartition V = V ∪ · · · ∪ V k such that, for each i = 1 , ..., k , V i is open,and S E ( V i ) = S ( V i ) ∪ ( e − ( E ) ∩ v − ( V i ))has zero or two elements.We consider surface diagrams to be homeomorphic when there are homeo-morphisms between each of the sets which commute with the arrows.See [BM79] for the proof that every surface has a surface diagram, and theconstruction of a surface from a given diagram. The surface diagram definitionis quite technical, so we will develop a practical method for interpreting thisdata in the next section. Now we are ready to state the general ClassificationTheorem. Theorem 3.4. (Classification of Surfaces, [BM79]) A surface S is determinedup to homeomorphism by a surface diagram (considered up to homeomorphism),the number of genus which is infinite if and only if E ∞ ( S ) (cid:54) = ∅ , and the numberof compact boundary components which is infinite if and only if E ∂ ( S ) (cid:54) = ∅ . Since the general case is vastly more complicated, we give a few illustrativeexamples. 7 xample 3.5.
See Figure 2. The two surfaces shown have homeomorphicends spaces E ( S ) = E ∞ ( S ) = ω · E ( S ), but the lower surface does not.Figure 2: Nonhomeomorphic surfaces with homeomorphic doubles. Example 3.6.
Take an annulus and from each boundary component removea point and a sequence accumulating to the point monotonically. There is achoice whether both sequences converge in the same direction or not, and thisgives two nonhomeomorphic surfaces. These surfaces have homeomorphic endspaces E ( S ) = ω · Definition 3.7. (Disk with Handles) A
Disk with Handles is a surface whichcan be constructed by taking a disk, removing a closed embedded subset P ofthe Cantor set from the boundary, and then attaching infinitely many handlesaccumulating to some subset of P . Remark 3.8.
Let D be a disk with boundary points removed, and S a Diskwith Handles constructed from D . When we attach a sequence of handles to D , it is possible the two corresponding sequences of open balls accumulate todifferent points in E ( D ). This joins these ends into a single end of E ( S ).We also want to consider a more specific class of surfaces. Definition 3.9. (Loch Ness Monsters) A
Loch Ness Monster refers to theunique surface with one end, infinite genus, and empty boundary. A
SlicedLoch Ness Monster is any of the surfaces with one end, infinite genus, no com-pact boundary components, but at least one noncompact boundary component.Equivalently, a Sliced Loch Ness Monster is a Disk with Handles with one end.Note it is possible for a Sliced Loch Ness to have infinitely many boundarycomponents. We sometimes refer to an n -Sliced Loch Ness to emphasize thenumber of boundary components.Figure 3: A visualization of the 2-Sliced Loch NessIn order to help visualize these surfaces, we give the following construction. Construction 3.10.
Take a strip R × [ − , n,
0) for n ∈ Z \ { } . Now identify pairsof boundary components centered at ( ± n, n -Sliced Loch Ness for any finite n by taking a disk with n points removed9rom the boundary and attaching tubes to join all of the ends. For the infinitecase, we can take a disk with any infinite embedded subset of the Cantor setremoved from the boundary and attach tubes to join every end as before. Bythe Classification Theorem, no matter what set of points we remove we alwaysget the ∞ -Sliced Loch Ness. Let Homeo + ∂ ( S ) be the group of orientation preserving homeomorphisms of asurface S which fix the boundary pointwise. The mapping class group MCG( S )is defined to be MCG( S ) = Homeo + ∂ ( S ) / ∼ where two homeomorphisms are equivalent if they are isotopic relative to theboundary of S . We will often conflate a mapping class group element with arepresentative homeomorphism. We equip Homeo + ∂ ( S ) with the compact-opentopology, which induces a quotient topology on MCG( S ). We equip subgroups ofMCG( S ) with the subspace topology. The pure mapping class group PMCG( S )is the subgroup of MCG( S ) consisting of elements which fix the ends of S . Noteif every end of a surface is a boundary end, then PMCG( S ) = MCG( S ).We say f ∈ MCG( S ) is compactly supported if f has a representative thatis the identity outside of a compact subsurface of S . The subgroup consist-ing of compactly supported mapping classes is denoted PMCG c ( S ). Note anycompactly supported mapping class is in the subgroup PMCG( S ). Definition 3.11. (Degenerate ends) Notice removing a proper closed subset ofpoints from the boundary of a surface does not change the underlying mappingclass group. We refer to the resulting ends as degenerate . More generally,this will refer to boundary ends with a representative sequence { U i } such thatsome U i is homeomorphic to a disk with boundary points removed. It may beconvenient in some cases to only work with homeomorphism types of surfacesup to filling in the degenerate ends. We will allow these ends except when statedotherwise.Now we review the definition of a handle shift. Let Σ be the surface obtainedby gluing handles onto R × [ − ,
1] periodically with respect to the map ( x, y ) (cid:55)→ ( x + 1 , y ). We refer to this surface as a Strip with Genus . For some (cid:15) >
0, let σ : R × [ − , → R × [ − ,
1] be the map determined by setting σ ( x, y ) = ( x + 1 , y ) for ( x, y ) ∈ R × [ − (cid:15), − (cid:15) ] ,σ ( x, y ) = ( x, y ) for ( x, y ) ∈ R × {− , } , and interpolating continuously on R × [ − , − (cid:15) ] ∪ R × [1 − (cid:15), σ . A homeomorphism h : S → S is a handle shift if there exists a10roper embedding ι : Σ → S such that h ( x ) = (cid:40) ( ι ◦ σ ◦ ι − )( x ) if x ∈ ι (Σ) x otherwise . The embedding ι is required to be proper, so it induces a map ˆ ι : E (Σ) → E ∞ ( S ). A handle shift h then has an attracting and repelling end denoted by h + and h − , respectively. A simple closed curve in a surface S is the image of a topological embedding S (cid:44) → S . A simple closed curve is trivial if it is isotopic to a point; it is peripheralif it is either isotopic to a boundary component or bounds a once-punctured disk.We will often refer to a simple closed curve as just a curve.An arc in S is a topological embedding α : I (cid:44) → S , where I is the closedunit interval, and α ( ∂I ) ⊂ ∂S . We consider all isotopies between arcs to berelative to ∂I ; i.e., the isotopies are not allowed to move the endpoints. An arcis trivial if it is isotopic to an arc whose image is completely contained in ∂S ;it is peripheral if it bounds a disk with single boundary puncture.A curve or arc is essential when it is not trivial nor peripheral; it is separatingif its complement is disconnected and nonseparating otherwise. We will oftenconflate a curve or arc with its isotopy class. All curves and arcs will be assumedto be essential unless stated otherwise.By cutting along a collection of curves or arcs, we mean removing disjointopen regular neighborhoods of each of the curves or arcs. Throughout this paper,we will conflate the complement of a curve or arc with this cut surface. We willalso occasionally conflate the complement of a subsurface with its closure. In this section, we prove Theorem C along with several other decompositionresults. This is crucial for the proof of the main theorem since a general sur-face can be extremely complicated. We also want an approachable method forvisualizing a surface from the surface diagram data. Our work builds off theBrown-Messer classification [BM79] with some inspiration from [PM07]. Theargument in the latter paper is incomplete, but provides a more intuitive ap-proach. We precisely define some of the ideas from this paper.The main idea is to study what happens when we remove the boundary ofa surface S . Deleting a compact boundary component leaves a puncture whichcorresponds to an isolated end of S o , the interior of S . Deleting the noncompactboundary components is more complicated as there could be several boundaryends getting sent to a single end. 11e show that we can think of the noncompact boundary components as be-ing grouped together into chains, and that removing the boundary componentsfrom a chain sends all of these boundary ends to a single end of S o . Then forthe final decomposition theorem, we show we can cut a surface along curves soeach resulting component has at most one boundary chain. Finally we representthe components with boundary chains as disks with points removed from theboundary, with handles attached or additional boundary components added, sothe boundary chain corresponds to the boundary of the modified disk. First we want to precisely define the map on ends spaces induced by deletingthe boundary. A ray in a surface S refers to a proper embedding of a half-closedinterval into S . Choose some end in p ∈ E ( S ), and let α be a ray in S whose endcorresponds to p . Since α is also a ray in S o , it determines some end q ∈ E ( S o ) . Note different rays corresponding to p always give the same q . Otherwise, wecan find two disjoint complementary domains in S o which each contain the tailof one of the rays. After adding in the boundary to get complementary domainsfor S , we see that the tails of the rays must correspond to different ends of S .Therefore, we have a well-defined canonical map π : E ( S ) → E ( S o ) Proposition 4.1.
The map π is continuous.Proof. Let U (cid:63) ⊆ E ( S o ) be a basis element defined by some complementarydomain U in S o . This gives a complementary domain U S in S after adding inthe boundary, and so it defines a basis element U (cid:63)S ⊆ E ( S ). We are done afternoting that π − ( U (cid:63) ) = U (cid:63)S .Now recall the definition of a surface diagram (Definition 3.3). Let B ( S ) bethe image of v , the map which sends ends of noncompact boundary componentsto ends of S . Recall that we will conflate definitions, and refer to B ( S ) as theset of boundary ends of E ( S ).Now we use the map π to define a boundary chain. Intuitively, this can bethought of as a set of boundary ends and boundary components which can berealized in the surface as a circle with points removed. Definition 4.2. (Boundary Chains) A boundary chain of a surface S is a subsetof E ( S ) of the form π − ( p ) where p ∈ π ( B ( S )). The collection of all such setsis denoted by C ( S ) and is referred to as the set of boundary chains for S .Occasionally, we will conflate definitions and use boundary chain to refer to theunion of noncompact boundary components with ends in the chain.We can use π to define an equivalence relation on B ( S ) for which C ( S ) isthe resulting quotient. After equipping B ( S ) with the subspace topology, C ( S )inherits a quotient topology. Note π is injective on E ( S ) \ B ( S ). The set ofboundary chains exactly records the noninjectivity of π on B ( S ).12 emark 4.3. Since there are countably many boundary components in a sur-face, π ( B ( S )) is countable. Remark 4.4.
The subset B ( S ) ⊆ E ( S ) is not necessarily closed. For exampletake a once-punctured sphere, remove infinitely many open balls with disjointclosures accumulating to the puncture, and then remove a single point fromeach of the resulting boundary components. It is not necessarily open eitheras in the case of a disk with a point removed from the boundary with interiorpunctures added accumulating to the boundary end. By Proposition 4.1, eachboundary chain is a closed subset. Then by Remark 4.3, B ( S ) is the countableunion of closed subsets. Example 4.5.
Consider any Disk with Handles S . The interior of S is the LochNess Monster Surface. Every boundary end of S gets sent by π to the singleend of the Loch Ness, so any Disk with Handles has a single boundary chain.This last statement has a partial converse which provides a more intuitiveway to think about a boundary chain. Lemma 4.6.
Every surface S with infinite genus, one boundary chain, and onlyboundary ends is a Disk with Handles possibly with compact boundary compo-nents added. Before we prove this we first need a few facts.
Proposition 4.7.
Let S be a noncompact surface without boundary. Then thefollowing are equivalent:(i) There exists a compact exhaustion { S i } of S such that each ∂S i has asingle component.(ii) S has exactly one end.Proof. Since the complementary regions of a compact exhaustion can be usedto build exhausting sequences, and the ends space is independent of this choiceof a compact exhaustion, the first condition immediately implies the second.Assuming the second condition, S is either a finite type surface with one punc-ture, or the Loch Ness Monster. In either case, we can directly construct thedesired exhaustion. Proposition 4.8.
Let S be a noncompact surface with no compact boundarycomponents and no interior ends. Then the following are equivalent:(i) There exists a compact exhaustion { S i } of S such that each ∂S i has asingle component.(ii) S has exactly one boundary chain. roof. Suppose the first condition holds. To get the second condition it suf-fices to show that the interior of S has a single end. Remove open regularneighborhoods of the boundary from each S i , shrinking the neighborhoods aswe increase i so we get a compact exhaustion for the interior. Each subsurfacein this exhaustion has one boundary component, so we are done by the aboveproposition.Now suppose the second condition holds, and let { K i } be a compact exhaus-tion of the interior of S such that each ∂K i has a single boundary component.Let N be an open regular neighborhood of the boundary chain. Note if we set S i = K i ∩ ( S \ N ), then we get a compact exhaustion { S i } for S \ N .We want to modify the K i so the resulting S i each have a single boundarycomponent. First remove subsurfaces from { K i } so K ∩ N (cid:54) = ∅ . Now isotope ∂K so it is transverse to ∂N , and each component of K ∩ N is a bigon.Now we proceed inductively. Remove some subsurfaces from the exhaustion so K i contains the previously modified K i − , and isotope ∂K i in S \ K i − so itsposition with N is as above. We can ensure the bigons exhaust the interior of N , so the modified sequence { K i } is an exhaustion of the interior of S . Nowsince S i is the result of removing disjoint bigons from K i , we conclude that each S i has one boundary component. We are now done since S \ N is homeomorphicto S .Now we are ready to prove the above lemma. Proof of Lemma 4.6.
First cap any compact boundary components of S withdisks. Let { S i } be a compact exhaustion of S given by Proposition 4.7 suchthat each ∂S i has one component. Now we can find an infinite sequence ofnonseparating curves in S such that cutting along the curves gives a connectedsurface with no genus and one boundary chain. See Figure 4 for an example. Tosee this note we can cut each S i along a collection of curves into a surface with nogenus and one boundary component, and we can ensure each collection extendsto subsequent collections. The desired collection of curves is then the increasingunion of these collections. Cut S along these curves, and cap the resultingboundary components with disks. Now by the Classification of Surfaces andProposition 4.7, the interior of S is homeomorphic to a once-punctured sphere.If we fill in the ends of S , we get a compact surface which must be homeomorphicto a disk. We are then done after reversing the above steps. Lemma 4.9.
Every infinite type surface S without boundary and without planarends can be cut along a collection of disjoint essential simple closed curves intoLoch Ness Monsters with k ∈ N ∪ {∞} compact boundary components added.Proof. This was first shown in [APV17] as a tool to prove Theorem 6.1. SeeSection 6 for this argument. We provide a different proof which gives us morecontrol over the ends of the components in the cut surface. Recall that the ends14igure 4: A surface satisfying the conditions of Lemma 4.6 with a collection ofcurves that cuts it into a surface with zero genus and one boundary chain.space E ( S ) is homeomorphic to a closed subset of the Cantor set. Let T besome locally finite tree with E ( T ) homeomorphic to E ( S ). We can think of S as a thickened version of T with genus added accumulating to every end. Forsimplicity, we will assume T has no vertices of valence one.We may write T as a union of rays { R i } where for each distinct R i and R j , R i ∩ R j is empty or a single vertex. To see this enumerate a countable densesubset { x i } of E ( T ), and fix some basepoint vertex v . Begin by letting R bethe ray from v to x , then let R be the ray from v to x with the interior of theoverlap with R deleted. Continue in this manner to build the desired collection { R i } . Since T has no vertices of valance one, this will exhaust the entire tree.Associate each R i with a Loch Ness Monster L i . Let n i ∈ N ∪ {∞} bethe number of vertices in R i ∩ (cid:83) j (cid:54) = i R j . For each i , remove n i open balls withdisjoint closures from L i with the balls accumulating to the single end when n i isinfinite. Associate each boundary component of L i with a vertex in R i ∩ (cid:83) j (cid:54) = i R j ,and attach the boundary components of distinct L i , L j when these componentscorrespond to the same vertex of T . Let S (cid:48) be the resulting surface, and let { α i } be the collection of curves in S (cid:48) corresponding to the attached boundarycomponents.Now E ( T ) and E ( S (cid:48) ) are homeomorphic. This requires showing a corre-spondence between a compact exhaustion of T and an exhaustion of S (cid:48) . Oneapproach is to subdivide T , then write it as a union of stars of the vertices fromthe original tree. Then associate each star with an n -holed torus where n is thenumber of edges in the star. The stars and the tori can then be attached tobuild compact exhaustions for T and S (cid:48) , respectively. By the Classification ofSurfaces, S (cid:48) is homeomorphic to S . Cutting S (cid:48) along the α i gives componentswhich are Loch Ness Monsters, so we are done.The argument from this lemma will be referenced often in the followingproofs. By decomposing a tree T into rays we mean writing T as a union ofrays which are either disjoint or intersect one another at a single vertex. The ends space of a tree is defined analogously to the ends space of a surface. For locallyfinite trees, the ends space is always homeomorphic to a closed subset of the Cantor set.
Lemma 4.10.
Every surface S can be cut along a collection of disjoint es-sential simple closed curves into components with at most one boundary chain.Additionally, we may assume the components with boundary chains have onlyboundary ends.Proof. Recall that by the definition of a boundary chain, two boundary ends p, q ∈ B ( S ) are in the same boundary chain if and only if π ( q ) = π ( p ). Supposewe can cut S o along curves into one-ended components such that each end in π ( B ( S )) corresponds to one of the components. Now when we cut S along thesesame curves, each component of the cut surface has at most one boundary chain.Since π maps interior ends outside of π ( B ( S )), we get the last statement of thelemma. Therefore, it suffices to decompose S o in this manner.Following the proof of Lemma 4.9, represent S o by a tree T with no valanceone vertices. Fix a base vertex v , and let T (cid:48) be the union of rays from v toan end in π ( B ( S )). By Remark 4.3, π ( B ( S )) is countable, so enumerate theelements of π ( B ( S )) as a sequence { x i } . As in the proof of Lemma 4.9, we canuse an inductive process to decompose T (cid:48) into a collection of rays { R i } whereeach element of { x i } corresponds to the end of one of the rays. Then we candecompose the remainder of T into rays. Now we follow the proof of Lemma4.9 to decompose S o as desired. Lemma 4.11.
Every Disk with Handles S without planar ends can be cut alonga collection of disjoint essential arcs into Sliced Loch Ness Monsters. roof. Let D be a disk with points removed from the boundary used to con-struct S . Note we may realize D as a closed neighborhood of a tree T properlyembedded in C . As before we will assume this tree has no valence one vertices.Recall from Remark 3.8 that the handles may be attached in a way that joinsends of D together. By similar reasoning to Proposition 4.1 and the remarksat the beginning of Section 4, the process of attaching handles determines awell-defined continuous quotient map q : E ( D ) → E ( S )By Classification of Surfaces, two Disks with Handles without planar ends arehomeomorphic when there is a homeomorphism between the base disks whichrespects the quotient maps induced by attaching the handles.We argue by analogy to the proof of Lemma 4.9. First suppose that q isinjective. Decompose T into rays { R i } , and then associate each ray with a1-Sliced Loch Ness. Attach these surfaces along arcs { α i } on their boundariesaccording to the incidences of the R i in T . This gives a Disk with Handleswith a base disk homeomorphic to D and an injective quotient map, so it ishomeomorphic to S . It then follows that we can cut S into 1-Sliced Loch NessMonsters. However if q is not injective, we need to be a little more careful. Let U = { p ∈ E ( S ) | | q − ( p ) | ≥ } Enumerate a countable dense subset { x i } of U . Now when we decompose T into rays, first choose rays that exhaust a dense subset of each q − ( x i ). Thendecompose the remainder of T to exhaust a dense subset of the entire endsspace. Let { R i } be the resulting rays. As before, associate each R i with a diskwith one boundary puncture D i , and attach the D i along arcs { α i } to get abase disk homeomorphic to D .Choose some x i , and consider the subset of rays with an end correspondingto an element of q − ( x i ). Attach infinitely many handles to the union of therespective D i in order to join the boundary ends of the D i into a single end. Sim-ilar to Construction 3.10, this gives n -Sliced Loch Ness Monsters where n ≥ x i . Now for the remaining rays attach handles tothe corresponding disks to get 1-Sliced Loch Ness Monsters. Attaching handlesin this manner to the base disk gives an equivalence relation on E ( D ) whichagrees with the equivalence relation given by q on a dense subset. Therefore bycontinuity and the above remarks, this construction gives a surface homeomor-phic to S . Now we are done since cutting this surface along the α i gives SlicedLoch Ness Monsters.Now we combine everything thus far. Proof of Theorem C.
The first statement of this theorem is Lemma 4.11. Wecan apply Lemma 4.10 to decompose a general surface into components with at One approach is to take a triangulation of D , and then build T from a spanning tree ofthe dual 1-skeleton. Remark 4.12.
If we allow planar ends, then similar decomposition resultshold. When decomposing a Disk with Handles as in Lemma 4.11, we may getdisks with one boundary puncture, and when decomposing a surface withoutboundary as in Lemma 4.9, we may get singly punctured spheres with compactboundary components added. We may also need to allow peripheral curves andarcs.
Domat has shown for surfaces with compact boundary components and atleast two ends accumulated by genus that PMCG c ( S ) is not perfect [DD20].In the appendix of this paper, the author and Domat use the Birman ExactSequence to extend this to the case with one end accumulated by genus. On theother hand, Calegari has shown that the mapping class group of the sphere minusa Cantor set is uniformly perfect [Cal09]. Now we want to show many surfaceswith noncompact boundary components have uniformly perfect mapping classgroups.First we need to extend a result of Patel-Vlamis to the general case, sincewe will use this implicitly throughout the proof of Theorem A. Theorem 5.1. [PV18] For any infinite type surface S with only compact bound-ary components and at most one end accumulated by genus, PMCG c ( S ) =PMCG( S ) . This result was originally stated for compact boundary, but the proof in[PV18] also works when there are infinitely many compact boundary compo-nents. The argument uses pants decompositions which we can construct withoutadaptation when there are only compact boundary components. Pants decom-positions are more tedious to define in the general case, so we will not use them.Instead, we give a modified proof using a more general compact exhaustion. Tosimplify our arguments we will assume surfaces do not have degenerate ends(Definition 3.11) for the entirety of Section 5.
Theorem 5.2.
For any infinite type surface S with at most one end accumulatedby genus, PMCG c ( S ) = PMCG( S ) .Proof. Let f ∈ PMCG( S ) be an arbitrary element. We want to find a sequence { f i } of elements of PMCG c ( S ) such that f i → f in the compact-open topology.Let { S i } be an exhaustion of S by essential finite type surfaces. It will sufficeto show that there is always some compactly supported f i which agrees with f S i . We can assume that the complementary domains of each S i are infinitetype.Note the orbit of any curve in S under PMCG( S ) is determined, up toisotopy, by the partition it determines on E ( S ), the partition it determineson the compact boundary components of S , and the topological type of thecomplementary domains. The orbit of an arc, up to isotopy, is determined bythe same properties and the endpoints of the arc. This is also true for curvesand arcs in any surface.Fix some S i , and let n be large enough so f ( S i ) ⊂ S n . Let { α k } be thecomponents of ∂S i \ ∂S . First suppose α k is a separating curve or arc. Since S has at most one end accumulated by genus, S \ α k has one component U withfinite genus. Increase n if necessary so S n ∩ U contains all of this genus. Note f ( α k ) and α k determine the same partition on E ( S n ) and the same partitionon the compact boundary components of S n .Let V = S n ∩ U and W = S n ∩ f ( U ). Since S n contains all the genus of U ,we must have that V and W have the same genus. It follows that α k and f ( α k )have homeomorphic complementary domains. Now if α k is nonseparating, then α k and f ( α k ) are both nonseparating in S n , and so the complementary domainsare homeomorphic in this case too. Therefore, we can find some g ∈ PMCG( S n )which takes f ( α k ) to α k . We can also require gf to fix the orientation of α k when it is a nonseparating curve.Now we build a compactly supported element f i which approximates f on S i . Start by finding some g ∈ PMCG( S n ) which takes f ( α ) to α . Now findsome g ∈ PMCG( S n \ α ) which takes g f ( α ) to α . Repeat this process tofind a sequence of compactly supported elements { g k } so g = · · · g g g sendseach f ( α k ) to α k . Also choose the g k so gf fixes the orientation of each α k . Nowfinally we have that g ∈ PMCG c ( S ) sends f ( S i ) to S i . Then let h = ( gf ) | S i , so f i = g − h agrees with f on S i . The main tool for the proof of Theorem A is a fragmentation lemma thatallows us write a map in PMCG c ( S ) as a product of two simpler maps. Thisis based on fragmentation results from [EK71] and [Man19], and was originallyformulated by Domat in the case without boundary. Here we provide a proofthat works in the general case. Lemma 5.3 (Fragmentation) . Let S be any infinite type surface and f ∈ PMCG c ( S ) . There exist two sequences of compact subsurfaces { K i } and { C i } ,with each sequence consisting of pairwise disjoint surfaces, and g, h ∈ PMCG c ( S ) such that(i) supp( g ) ⊆ (cid:83) i C i and supp( h ) ⊆ (cid:83) i K i ,(ii) f = hg .Proof. Fix a compact exhaustion { S i } of S , and begin by setting K (cid:48) = S .Choose some n large enough so f ( K (cid:48) ) ⊂ S n , and then set K = S n . Now there19xists some φ ∈ PMCG( K ) such that φ f fixes ∂K (cid:48) . Let ψ = φ f | K (cid:48) ∈ PMCG( K (cid:48) )Then ψ − φ ∈ PMCG( K ) and ψ − φ f fixes K (cid:48) . Let g = ψ − φ . Next let K (cid:48) , . . . , K (cid:48) j be the components of some S n \ S n − where n is large enough sothe following properties hold for each 2 ≤ i ≤ j :(1) f ( K (cid:48) i ) is disjoint from K ,(2) Some curves/arcs in ∂K (cid:48) i \ ∂S and ∂K \ ∂S together bound an essentialcompact subsurface C i − .Now we run the same argument as before to get new elements g , . . . , g j con-tained in PMCG( K ) , . . . , PMCG( K j ), respectively, with all of the K i disjoint.Our choices for the new K i will be the components of some S n \ S m where n and m are chosen so f ( K (cid:48) i ) ⊂ S n \ S m , for each 2 ≤ i ≤ j . Then g j · · · g g f fixes K (cid:48) i for i = 1 , . . . , j .Continue this process to exhaust the surface and obtain a sequence of el-ements g i and compact subsurfaces K i and C i . Note each of the g i are com-pactly supported with disjoint supports so the product · · · g g g converges to g ∈ PMCG c ( S ). Set g = gf and h = g − , so that f = hg . Now supp( h ) ⊆ (cid:83) i K i as desired. We also have (cid:83) i C i = S \ (cid:83) i K (cid:48) i and g = gf fixes each of the K (cid:48) i which shows that supp( g ) ⊆ (cid:83) i C i .Figure 6: Example of one of maps produced via fragmentation on a surfacewith two boundary chains (bold lines). The blue shaded regions represent the K i before we modify them. Remark 5.4.
A critical observation is that some of the compact subsurfaceswe get from fragmentation can be modified. Say K is a compact subsurfacewhose boundary is composed of alternating essential arcs in S and arcs in ∂S .20et f ∈ MCG( K ), and conflate f with a representative homeomorphism. Since f fixes ∂K we can assume after an isotopy that f is the identity in an openregular neighborhood N of ∂K , so f ∈ MCG( K (cid:48) ) where K (cid:48) = K \ N . Theboundary of K (cid:48) is then a union of essential simple closed curves in S .Modifying the subsurfaces in this manner may turn a surface which separatesinto one that does not. For example, the rightmost two subsurfaces shown inFigure 6 can be modified to be nonseparating. This idea can be extended asfollows. Lemma 5.5.
Suppose S is a Disk with Handles. Let g be map given by frag-mentation, and { K i } the corresponding sequence of compact subsurfaces. Wecan assume each K i has single boundary component, and S \ ∪ i K i is homeo-morphic to S with compact boundary components added accumulating to somesubset of the ends.Proof. Recall fragmentation depends on a given choice of a compact exhaustion { S i } . We can choose our exhaustion so each ∂S i has one component composedof alternating essential arcs in S and arcs in ∂S . When we run fragmentationon this type of exhaustion, each component of ∂K i intersects ∂S . We can thenmodify the K i as in the above remark so ∂K i is a union of essential simpleclosed curves. See Figure 7 for an example. In the case of fragmentation on the2-Sliced Loch Ness (see Figure 3 and Construction 3.10), this process will oftengive K i with two boundary components, and in general this could can give anypositive number of boundary components.Now note that none of the K i bound a common subsurface. We can checkfrom the proof of Lemma 5.3 that S \ ∪ i K i has infinite genus. It follows that S \ ∪ i K i is a Disk with Handles with compact boundary components added.We can also assume an end in S \ ∪ i K i is accumulated by genus if and only ifthe corresponding end of S is accumulated by genus. Therefore, we have thesecond condition of the lemma.For any K i with multiple boundary components, connect the componentstogether with disjoint arcs { α k } in S \ ∪ i K i . Now enlarge K i by adding in aclosed regular neighborhood of ∂K i ∪ (cid:83) k α k . Repeat this for every K i mak-ing sure the new subsurfaces are all disjoint. Now we have the first conditionof the lemma. Note this proof applies to either of the two maps given fromfragmentation. Lemma 5.6.
Suppose S is a connected sum of finitely many Disks with Handles.Let g be a map given by fragmentation, and { K i } the corresponding sequence ofcompact subsurfaces. We can assume the following:(i) K separates S into Disks with Handles with one compact boundary com-ponent added.(ii) For the remaining K i , ∂K i is a single essential simple closed curve. K Figure 7: Example of fragmentation on a 1-Sliced Loch Ness. The blue arcscorrespond to the compact exhaustion used for the fragmentation. The K i correspond to the modified subsurfaces containing the support of one of themaps from fragmentation. (iii) Each component of S \∪ i K i is a Disk with Handles with compact boundarycomponents added accumulating to some subset of the ends.Proof. Suppose S is a connected sum of n Disks with Handles. Choose ourexhaustion for fragmentation so each ∂S i has n components, each correspondingto one of the boundary chains, composed of alternating essential arcs in S andarcs in ∂S . Following the proof of the fragmentation lemma, we first set K = S n for some n . Then modifying K as in Remark 5.4 gives the first condition ofthe lemma. We get the remaining conditions by following the proof of Lemma5.5 for each component of S \ K . Note this only gives the conditions for one ofthe fragmentation maps. For the other map, we do not get the first and thirdconditions until we enlarge some of the compact subsurfaces. We could alsostate a version of this lemma with different conditions for the other map, butthat will not be necessary for the following proofs. First we use fragmentation along with standard commutator tricks to showevery element of PMCG c ( S ) can be written as a product of two commutatorswhen S is a Sliced Loch Ness Monster. Then we will show the same for anyDisk with Handles by applying Lemma 4.11. Finally we extend to the remainingcases using Lemma 4.10. During the upcoming proofs, we are implicitly usingthe fact that PMCG c ( S ) = PMCG( S ) = MCG( S )when S is a Sliced Loch Ness Monster. Lemma 5.7.
PMCG c ( S ) is uniformly perfect when S is a Disk with Handles.Proof. Let g be a map given by fragmentation on a general f ∈ PMCG c ( S ), andlet { K i } be the corresponding sequence of compact subsurfaces. First consider22he case when our surface is the 1-Sliced Loch Ness Monster, L s . By Lemma 5.5we may assume each ∂K i has one component and the complement of ∪ i K i ishomeomorphic to L s with infinitely many compact boundary components addedaccumulating to the single end.Realize L s as the closed upper half plane with a handles attached inside an (cid:15) -ball at every integer point, and let h be the map ( x, y ) → ( x + 1 , y ) extendedto the attached handles and isotoped in a neighborhood of the boundary to bethe identity. Now we can assume using Classification of Surfaces (after replacing g with a conjugate) that the K i are contained inside the vertical strip boundedby the lines x = ± . Letting a = Π k ≥ h k gh − k we can now write g = ha − h − a = [ h, a − ]See Figure 8 for a visualization. It now follows that we can write any f ∈ PMCG c ( S ) as the product of two commutators. K K K Figure 8: The last step for showing PMCG c ( L s ) is uniformly perfect. Thesupport of the element a is shown in blue.Next we extend this to any n -Sliced Loch Ness Monster, L ns . First we needa model of this surface that works with the above method. Take a copy of L s with the above half plane model, and denote it by T . Now take the disjointunion with n − L s realized in any way. Attach handles from T to each additional copy of L s to join all the ends into a single end. Whenremoving open balls from T in the process of attaching these handles, choosethe open balls to be below the line y = . Similar to Construction 3.10, thisyields a surface homeomorphic to L ns which we use as our model. Let T (cid:48) ⊂ L ns be the subsurface corresponding to the area of T above the line y = . Nowwe can use Lemma 5.5 and the Classification Theorem as before to assume thatthe K i are contained above the attached handles within a vertical strip of T (cid:48) .We then let h be the map which acts as the previous shift map on T (cid:48) and fixesthe remainder of L ns . Proceed as before to show g = [ h, a − ] . S is any Disk with Handles. After applying Lemma 5.5, we canassume S \ ∪ i K i is homeomorphic to S with infinitely many compact boundarycomponents added accumulating to some subset of the ends. Using a slightvariation of Lemma 4.11 where we allow the Disks with Handles to have compactboundary components added, we can cut S along a collection of disjoint arcs { α i } which miss the K i so the components of the cut surface are Sliced LochNess Monsters. Give the components the models discussed in the previousparagraphs, and apply the Classification of Surfaces argument to assume each K i is contained within a vertical strip within its respective component. Let { h i } be the collection of plane shift maps for each component analogous tothe previous paragraphs. Since the supports of the h i are disjoint we have awell-defined product h = Π i h i , and then we can show g = [ h, a − ] as before. Lemma 5.8.
PMCG c ( S ) is perfect when S is a connected sum of finitely manyDisk with Handles with possibly finitely many punctures or compact boundarycomponents added.Proof. Let g be a map given by fragmentation on a general f ∈ PMCG c ( S ),and let { K i } be the corresponding compact subsurfaces. First suppose S has nopunctures or compact boundary components. When fragmenting in this case,we get supports with boundary components that are curves which separate ends(see the two leftmost subsurfaces from Figure 6). If a map is supported withinone of these subsurfaces, then we cannot move the support off of itself as we didin the other cases. This is commonly referred to as a nondisplacable subsurface.Apply Lemma 5.6 so we can assume the K i have the desired properties. Wecan assume K has genus at least 3 by replacing it with a connected compactsurface containing K and more of the K i . Now g = g g where g ∈ PMCG( K )and g ∈ PMCG( S \ K ) . The classic result of Powell [Pow78] tells us we canwrite g as a product of commutators. By the method in the previous lemma,we can write g as a single commutator. It follows every element in PMCG c ( S )can be written as a product of commutators.The cases with finitely many punctures and compact boundary componentsare done similarly. To consider the cases with punctures, we need to slightlymodify the fragmentation process by replacing a compact exhaustion with anexhaustion of finite type surfaces. Then depending on the number of boundarychains, we use a modification of either Lemma 5.5 or Lemma 5.6 such that K includes the boundary components and punctures.These lemmas complete the forward implications from Theorem A, so nowwe discuss why the other directions hold. For all infinite type surfaces withonly compact boundary components, PMCG c ( S ) is not perfect by the work ofDomat. His proof relies on finding a particular sequence of disjoint essentialannuli. Then he shows some multitwist about the core curves of these annulicannot be written as a product of commutators. His work can be summarizedwith the following theorem. For the statement of this theorem, a nondisplaceable24urface in S refers to an essential subsurface K disjoint from the noncompactboundary components of S such that f ( K ) ∩ K (cid:54) = ∅ for all f ∈ PMCG c ( S ).Note a subsurface K is nondisplaceable if it separates ends; i.e., if S \ K isdisconnected and induces a partition of E ( S ) into two sets. A subsurface is alsonondisplaceable if some component of S \ K is a finite type subsurface containinga compact boundary component of ∂S . Theorem 5.9. [DD20] Let S be an infinite type surface such that there existsan infinite sequence of disjoint nondisplacable essential annuli that eventuallyleaves every compact subsurface. Then PMCG c ( S ) is not perfect. This condition holds whenever there are interior ends of S accumulated bygenus, except in the case of the Loch Ness Monster which was handled separatelyin the appendix of [DD20]. It also holds if there are infinitely many planar endsor infinitely many compact boundary components. By using Lemma 4.10, wesee this condition holds whenever there are infinitely many boundary chains aswell. The only cases that remain are exactly the surfaces from Lemma 5.7 andLemma 5.8.Finally we must explain why PMCG c ( S ) is not uniformly perfect when S has more than one boundary chain, any planar ends, or any compact boundarycomponents. We will only sketch the details since the main ideas here aretaken from [DD20]. The issue in these cases is that there is some essentialcurve α which is nondisplaceable under the action of PMCG c ( S ). Either takea curve which separates ends or bounds a finite type subsurface containing acompact component of ∂S . The orbit of α can then be used to build a Bestvina-Bromberg-Fujiwara projection complex (see [BBF19]) on which PMCG c ( S ) actsby isometries. This complex is quasi-isometric to a tree, and the Dehn twistabout α is a WWPD element. An adaptation of a construction of Brooks[Bro81] then gives a quasimorphism from PMCG c ( S ) to R which is unboundedon { T nα } ∞ n =1 . Combining this with the fact that homogeneous quasimorphismsare bounded on commutators, we see PMCG c ( S ) cannot be uniformly perfect. In the case of surfaces with only compact boundary components, it is knownthat PMCG( S ) factors as a semidirect product containing PMCG c ( S ) as one ofthe factors. Theorem 6.1 (Aramayona-Patel-Vlamis [APV17], Corollary 6) . Let S be aninfinite type surface with compact boundary components. PMCG( S ) = PMCG c ( S ) (cid:111) H where H ∼ = Z n − when there is a finite number n > of ends of S accumulatedby genus, H ∼ = Z ∞ when there are infinitely many ends accumulated by genus, nd H trivial otherwise. Furthermore, H is generated by pairwise commutinghandle shifts. Here Z ∞ refers to the direct product of a countably infinite number of copiesof Z . Although many of the results of Aramayona-Patel-Vlamis are stated forthe case of compact boundary, the proofs all apply to surfaces with only compactboundary components.In order to extend this result, we will also need to extend a well-known factabout maps between mapping class groups. Recall the definition of a degenerateend (Definition 3.11). We say a boundary chain of a surface is degenerate whenevery end in the chain is degenerate. After filling in degenerate ends, degeneratechains become compact boundary components. Similar to a Dehn twist abouta compact boundary component, we can also speak of a Dehn twist about adegenerate chain. Lemma 6.2.
Let S be any surface, and Σ an essential subsurface. The natu-ral homomorphism i : PMCG(Σ) → PMCG( S ) is injective when the followingholds:(i) No compact component of ∂ Σ bounds a once-punctured disk.(ii) No two compact components of ∂ Σ bound an annulus.(iii) There are no degenerate chains in Σ such that each boundary componentof the chain bounds an upper half plane. Our proof will rely on the Alexander method for infinite type surfaces. Thecase of compact boundary components was done in [HMV19].
Definition 6.3.
A stable Alexander system is a collection of essential simpleclosed curves and essential arcs Γ in a surface S such that the following prop-erties hold: • The elements in Γ are in pairwise minimal position. • For distinct α i , α j ∈ Γ, we have that α i is not isotopic to α j . • For all distinct α i , α j , α k ∈ Γ, at least one of the following sets is empty: α i ∩ α j , α i ∩ α k , α j ∩ α k . • The collection Γ fills S ; i.e., it cuts S into disks and once-punctured disks. • Every f ∈ Homeo + ( S, ∂S ) that preserves the isotopy class of each elementof Γ is isotopic to the identity.
Theorem 6.4. (Alexander method) For any infinite type surface S , there existsa stable Alexander system Γ . roof. We will assume the empty boundary case. For a surface S with noncom-pact boundary components, embed it in the natural way inside the double, dS .Let Γ be a stable Alexander system for dS .For an arbitrary γ ∈ Γ, isotope it to be transverse and in minimal positionwith ∂S so γ ∩ S is either a curve or a union of arcs in S . Let Γ (cid:48) be thecollection of all curves and arcs formed in this manner. After possibly removingnonessential curves or arcs or repeated occurrences of isotopy classes, Γ (cid:48) is astable Alexander system for S .We will also need an additional fact about arcs in an infinite type surface.For only this lemma and its proof, we allow all isotopies of arcs to move theendpoints along the boundary. Lemma 6.5.
Let S be an infinite type surface with nonempty boundary and nodegenerate ends, and α an essential arc in S . There exists a collection of curves Γ disjoint from α so the following holds: If β is an arc with endpoints on thesame boundary components as α , and β can be isotoped to be disjoint from everycurve in Γ , then α and β are isotopic.Proof. Let { S i } be a compact exhaustion of S where each S i has enough genusand boundary components so it contains essential simple closed curves. Deletethe first few subsurfaces in the exhaustion so α is an essential arc of each S i . Firstsuppose α is nonseparating in S i . Now we can find a finite collection of curvesΓ i which fills S i \ α , so the complementary regions are disks, once-punctureddisks, and annuli. When α is separating in S i , it is possible it bounds an annulusor a pair of pants. Then let Γ i be a collection which fills the other component.If the compact component is a pair of pants, add the curve bounding the twoboundary components not containing α to Γ i . For all other cases, we just let Γ i be a collection which fills both components of S i \ α . Let Γ = (cid:83) i Γ i .Suppose β is any arc as in the statement of the lemma, and isotope it tobe disjoint from each curve of Γ. Choose some i large enough so S i containsboth α and β . Also increase i so α and β have endpoints on the same boundarycomponents of S i . Observe that the complementary domain of Γ i in S i whichcontains α is a pair of pants. It is standard fact that there is a unique arc, up toisotopy, between any two boundary components of a pair of pants (see [FM12]Proposition 2.2). It follows β must be isotopic to α in S i . Since this holds forall sufficiently large i , we see that β is isotopic to α in S . Remark 6.6.
The assumption on degenerate ends is necessary since otherwisethere might be essential arcs which bound a disk with boundary points removed.These can be isotoped to be disjoint from any curve. In fact, an essential arcwith this property must bound a disk in any compact subsurface containing it.It follows that such an arc bounds a disk with boundary points removed. Ifwe fill in degenerate ends, then these arcs become trivial. It may be useful insome cases to extend the definition of trivial/peripheral to include arcs or curveswhich are trivial/peripheral in the surface after degenerate ends are filled in.27 roof of Lemma 6.2.
The last condition is similar to the first condition in thesense that it prevents Dehn twists from being in the kernel; in this case, Dehntwists about degenerate chains. We give a proof following Farb-Margalit [FM12].Let f ∈ PMCG(Σ) be in the kernel, and conflate it with a representativehomeomorphism. We extend f by the identity to a homeomorphism whichrepresents i ( f ). Let Γ be an Alexander system for Σ.Let α be any essential simple closed curve in Σ. Since i ( f ) is isotopic tothe identity and i ( f ) agrees with f on Σ, we have that f ( α ) is isotopic to α in S . Let K ⊂ S be a compact essential subsurface which contains this isotopy.If K can be isotoped to be contained within Σ then we are done, so assumeotherwise. Now after isotoping ∂K and ∂ Σ to be transverse and in minimalposition, K ∩ ∂ Σ is a union of arcs in K . Since f ( α ) and α are disjoint fromthese arcs, it follows from a standard fact of isotopies in the compact case thatthere is an isotopy from f ( α ) to α missing the arcs. Therefore, we have that f fixes the isotopy class of every curve in Σ.Note we can remove all arcs of the type in Remark 6.6 from Γ and still havea stable Alexander system. Let α be one of the remaining arcs in Γ. To simplifythe remainder of the argument, we fill in any degenerate ends of Σ. By Lemma6.5, we can find a collection of curves in Σ such that f ( α ) is isotopic to α , by anisotopy possibly moving the endpoints, if it can be isotoped to miss each curvein the collection. This last condition holds since f fixes the isotopy class ofevery curve. Now we can assume by an isotopy not moving the endpoints that f ( α ) agrees with α outside of an open collar neighborhood N of the boundarycomponents. Since Γ descends to a stable Alexander system for S \ N , wecan apply the Alexander method to S \ N to show f is supported in N . Thecomponents of N are annuli and strips R × [ − , f is topologicallygenerated by Dehn twists in the annuli. By the given conditions, we must nowhave that f is isotopic to the identity, since otherwise i ( f ) would be nontrivial. Remark 6.7.
Note deleting a noncompact boundary component is topologicallythe same as attaching an upper half plane to the component. Therefore, we canextend the above proof to show that homomorphisms such as the one discussedin Section 2 are injective. In particular, we will still have injectivity as long aswe do not delete any degenerate chains or compact boundary components.
First we will give a proof of Theorem 6.1, and then explain how to extendit to the general case. We say a handle shift h cuts a curve α when h + and h − are on opposite sides of α . Let S be a surface with only compact boundarycomponents. A principal exhaustion of S is an exhaustion of S by finite typesubsurfaces such that the following conditions hold for all i :(i) Each complementary domain of S i is an infinite type surface.28ii) Each component of ∂S i is separating.Now we state a few results from [APV17] which we will assume for the follow-ing proofs. Let H sep ( S, Z ) denote the subgroup of first homology of a surfacegenerated by classes that can be represented by separating curves on the surface. Lemma 6.8. ([APV17], Lemma 4.2) Let S be a surface with only compactboundary components. Given a principal exhaustion { S i } of S there exists abasis of H sep ( S, Z ) comprised of curves in the boundary of the S i . Lemma 6.9. ([APV17], Proposition 3.3) Suppose S is a surface with onlycompact boundary components. Then we have the following:
1. There is an injection φ from H sep ( S, Z ) to H (PMCG( S ) , Z ), thought ofas the group of all homomorphisms from PMCG( S ) to Z .2. Let α be a curve representing an element in H sep ( S, Z ). The homomor-phism φ ( α ) : PMCG( S ) → Z sends a handle shift h to a nonzero elementif and only if it cuts α , and it sends any map in PMCG c ( S ) to 0. We canassume φ ( α ) sends a given handle shift cutting α to 1. Proof of Theorem 6.1.
First assume S is a surface without any planar ends orcompact boundary components. The case of at most one end accumulated bygenus was done in [PV18], so assume S has at least two ends accumulated bygenus. Let { α i } be a collection of curves forming a basis for H sep ( S, Z ). Nowcut S along each of the α i . Each separating curve in the cut surface boundsa compact surface, since otherwise the collection of curves above would notform a basis. Therefore, each component of the cut surface is a Loch Ness with k ∈ N ∪ {∞} compact boundary components added. Each component Z of thecut surface can be modeled as R with k open disks removed along the horizontalaxis and handles attached periodically and vertically above each removed disk.Let Y be the surface obtained from [ − , × [0 , ∞ ) ⊂ R by attaching a handleinside a small neighborhood about each interior integer point. We can properlyembed k disjoint copies of Y into Z so each copy of [ − , × { } ⊂ Y is mappedto a different boundary component of Z .Now we paste all of the components back together to form the original surface S . We can choose the embeddings of Y above so the union of their imagesis a collection of disjoint Strips with Genus. This then gives a collection ofhandle shifts { h i } where each h i cuts only α i . By the above lemma, we havehomomorphisms φ ( α i ) : PMCG( S ) → Z such that φ ( α i ) sends h i to 1 and everyother h j to 0. Let H be the subgroup topologically generated by the { h i } . Sinceall of the h i commute, H is a direct product of countably many copies of Z .The product map φ = Π ni =1 φ ( α i ) gives a homomorphism from PMCG( S ) to H .Then by the above lemma, we have a split exact sequence:1 PMCG c ( S ) PMCG( S ) H φs s is inclusion. The cases of surfaces with planar ends and compactboundary components are done similarly. When there are planar ends, we choosehandle shifts which miss the planar ends. Then we get the desired semidirectproduct using the forgetful map between pure mapping class groups.The general case is a corollary of this result by capping the boundary chainsof a surface S with noncompact boundary components. Using Lemma 4.10, wecut S along curves into surfaces with one boundary chain. Then by Lemma4.6, we can build these components from disks with points removed from theboundary. Next we cap off each boundary chain by attaching a copy of thecorresponding disk with points removed from the boundary, and we let (cid:101) S be theresulting surface. As an example, note that capping the boundary chain of anySliced Loch Ness gives the Loch Ness. The inclusion of a surface in the cappedoff surface induces a map on the ends spaces which preserves ends accumulatedby genus and planar ends. Note there is a natural homomorphism i : PMCG( S ) → PMCG( (cid:101) S )induced by inclusion, and i is injective by Lemma 6.2. Theorem 6.10.
Theorem 6.1 holds for any infinite type surface S .Proof. Recall the case of at most one end accumulated by genus was done inTheorem 5.2. Assume S is a surface with noncompact boundary components,without planar ends or compact boundary components, and with at least twoends accumulated by genus. Let i be the homomorphism between pure mappingclass groups as above. Note (cid:101) S has the same number of ends accumulated bygenus as S . By Theorem 6.1, there is a split exact sequence as above with (cid:101) S in the place of S . Recall H is the subgroup topologically generated by disjointhandle shifts { h i } , and s is the inclusion map. It suffices to show each of the h i can be chosen to be inside i (PMCG( S )), because then by injectivity of i we geta split exact sequence:1 PMCG c ( S ) PMCG( S ) H φ ◦ ii − ◦ s As mentioned above, we can cut a surface along a collection of curves { α i } soeach component of the cut surface has one boundary chain. Then each of thesecomponents can be represented as Disks with Handles. We can piece togetherfinite type exhaustions on the components to get an exhaustion { S i } for S . Wecan choose the exhaustion so ∂S i \ ∂S is always composed of separating curvesand arcs with endpoints on boundary components of the same chain. Also wecan assume the exhaustion satisfies the first condition in the definition of aprincipal exhaustion.Now we modify this exhaustion to get a principal exhaustion of (cid:101) S . For everyarc β k in ∂S i \ ∂S , there is a corresponding arc γ k in the attached disk which,30igure 9: A surface (blue) embedded in the capped off surface. The disks in thepicture are identified with each other by the vertical symmetry. The red curvesare created by closing up arcs in the central surface. The blue and purple curvesare separating curves which replace the red curves.together with β k , closes up to a curve α k . The α k together with the curvesin ∂S i \ ∂S bound a compact subsurface K i ⊂ (cid:101) S . Then { K i } is a compactexhaustion for (cid:101) S which is not necessarily principal, but we can modify it so itbecomes principal. Let U be a complementary domain of K , and then connecteach component of ∂U together with disjoint arcs in U ∩ S . Now enlarge K by adding a closed regular neighborhood in U of the arcs and the boundarycomponents, then repeat this for each complementary domain. See Figure 9 foran example. Now remove some subsurfaces from the exhaustion so K ⊂ K ,and then repeat the above process for K . Continue in this manner to get aprincipal exhaustion.Now we sketch the final details. Find a homology basis { α i } of H sep ( (cid:101) S, Z )comprised of curves that are boundary components for surfaces in the aboveexhaustion. Then we cut (cid:101) S along these curves, and we get components whichare Loch Ness Monsters with compact boundary components added. Next webuild the subgroup H by taking the group topologically generated by disjointhandle shifts h i where each h i cuts α i and no other curve in the basis. In thispart of the proof there is a great deal of choice for how to embed these strips;in particular, we can assume the strips are contained in S . The remaining casesare done similarly to the proof of Theorem 6.1. References [APV17] J. Aramayona, P. Patel, and N. G. Vlamis. The first integral coho-mology of pure mapping class groups. arXiv preprint arXiv:1711.03132 ,2017. 31AS60] Lars Ahlfors and Leo Sario.
Riemann surfaces . Princeton UniversityPress, 1960.[BBF19] Mladen Bestvina, Kenneth Bromberg, and Koji Fujiwara. Proper ac-tions on finite products of quasi-trees. arXiv preprint arXiv:1905.10813 ,2019.[BM79] Edward M. Brown and Robert Messer. The classification of two-dimensional manifolds.
Transactions of the American MathematicalSociety , 255:377–402, 1979.[Bro81] Robert Brooks. Some remarks on bounded cohomology. In
Riemannsurfaces and related topics: Proceedings of the 1978 Stony Brook Con-ference , volume 97, pages 53–63, 1981.[Cal09] Calegari, Danny. Big mapping class groups and dy-namics. https://lamington.wordpress.com/2009/06/22/big-mapping-class-groups-and-dynamics/ , 2009. [Online; Ac-cessed on 11/30/2020].[DD20] George Domat and Ryan Dickmann. Big pure mapping class groups arenever perfect. arXiv preprint arXiv:2007.14929 , 2020.[EK71] Robert D Edwards and Robion C Kirby. Deformations of spaces ofimbeddings.
Annals of Mathematics , pages 63–88, 1971.[Fab03] Paul Fabel. The mapping class group of a disk with infinitely manyholes. arXiv preprint arXiv:math/0303042 , 2003.[FM12] Benson Farb and Dan Margalit.
A Primer on Mapping Class Groups .Princeton University Press, 2012.[HMV19] Jes´us Hern´andez Hern´andez, Israel Morales, and Ferr´an Valdez. Thealexander method for infinite-type surfaces.
The Michigan MathematicalJournal , 68(4):743–753, 2019.[Ker23] B´ela Ker´ekj´art´o.
Vorlesungen uber Topologie . J. Springer, 1923.[Man19] Kathryn Mann. Automatic continuity for homeomorphism groups ofnoncompact manifolds. arXiv preprint arXiv:2003.01173 , 2019.[Ore51] Oystein Ore. Some remarks on commutators.
Proceedings of the Amer-ican Mathematical Society , 2(2):307–341, 1951.[PM07] A. O. Prishlyak and K. I. Mischenko. Classification of noncompactsurfaces with boundary.
Methods of Functional Analysis and Topology ,13(1):62–66, 2007.[Pow78] Jerome Powell. Two theorems on the mapping class group of a surface.
Proceedings of the American Mathematical Society , 68(3):347–350, 1978.32PV18] Priyam Patel and Nicholas Vlamis. Algebraic and topological prop-erties of big mapping class groups.
Algebraic & Geometric Topology ,18(7):4109–4142, 2018.[Ric63] Ian Richards. On the classification of noncompact surfaces.