Groups of Diffeomorphisms

Groups of Diﬀeomorphisms

Diplomarbeit zur Erlangung des akademischen Grades Magister rerum naturalium an der Formal- und Naturwissenschaftlichen Fakult¨at der Universit¨atWien

vorgelegt von Stefan Haller

betreut von Univ.-Doz. tit. Ao. Prof. Peter W. Michor

erstellt am Institut f¨urMathematik Strudelhofgasse 4, 1090 Wien

Vorwort

Die vorliegende Arbeit basiert größtenteils auf Arktikeln die von D. B. A. Epstein, J. Mather, R. P. Filipkiewicz und A. Banyaga zwischen 1970 und 1988 publiziert wurden. Das erste Kapitel geht im wesentlichen auf D. B. A Epstein zurück. Hier wird gezeigt, daß die Kommutatorgruppen gewisser Homöomorphismengruppen einfach sind (vgl. Theorem 1.3). Dieses eher allgemein gehaltene Resultat läßtsich auf r r Diffc(M) anwenden, und so hat Diffc(M) eine einfache Kommutatorgruppe (vgl. Corollary◦ 1.17). ◦ In den Kapiteln 2&3 wird ein Resultat von J. Mather präsentiert, das besagt: r Diffc(M) ist einfach, falls r = dim(M) + 1. Eigentlich wird nur gezeigt, daß r ◦ 6 Diffc(M) perfekt ist, und dann das Ergebnis des ersten Kapitels verwendet. Für dim(M) +◦ 2 r ist dies im zweiten Kapitel (vgl. Corollary 2.44), und für1 ≤ ≤ ∞ dim(M)+1 ≤ r dim(M) im dritten Kapitel (vgl. Corollary 3.5) ausgeführt.Ob Diffc (M) auch≤ perfekt ist, ist meines Wissens noch nicht geklärt,der hier präsentierte Beweis◦ versagt jedenfalls völligbei r = dim(M) + 1. Im vierten Kapitel wird gezeigt, daß jeder Gruppenisomorphismus zwischen ge- wissen Diffeomorphismengruppen von einem Diffeomorphismus der zugrundeliegenden Mannigfaltigkeiten induziert wird. Insbesondere stammen isomorphe Diffeo- morphismengruppen von diffeomorphen Mannigfaltigkeiten. Auch hier wird zuerst r ein allgemeineres Resultat bewiesen (vgl. Theorem 4.14), das dann auf Diffc(M) angewandt wird (vgl. Theorem 4.17). Die Beweise dieser Sätzegehen auf R. P. Fil-◦ ipkiewicz und A. Banyaga zurück. Die Methoden und Ergebnisse aus Kapitel 2 & 3 werden dabei nicht benötigt,und daher kann das vierte Kapitel auch unmittelbar nach dem ersten Kapitel gelesen werden. Der Anhang behandelt das “diskrete Analogon” zu Theorem 4.17. Die Mannig- faltigkeiten werden durch endliche Mengen, und die Diffeomorphismengruppen durch Permutationsgruppen ersetzt. Sind die Mengen nicht sechselementig, wird wieder gezeigt, daß ein Gruppenisomorphismus zwischen Permutationsgruppen von einer Bijektion der zugrundeliegenden Mengen induziert wird (vgl. Corollary A.8). Für sechselementige Mengen stimmt das nicht (vgl. Theorem A.10). Im Anhang werden nur elementare gruppentheoretische Methoden verwendet, er kann also unabhängig von den anderen Kapiteln gelesen werden.

iii Danke!

Bei dieser Gelegenheit möchte ich mich bei allen Seminarteilnehmern, ganz besonders aber bei Prof. Peter Michor, fürihre Anregungen und ihr Zuhörenbedanken. Auch meinen Eltern möchte ich fürihre Unterstützungan dieser Stelle einmal ein herzliches “Dankeschön!” aussprechen. Nicht zuletzt gilt mein Dank natürlich auch meiner Schwester Maria und allen meinen Freunden und Freundinnen.

iv Contents

Abstract ...... vi

r r 1. The Simplicity of [Diffc(M) , Diffc(M) ] ...... 1 ◦ ◦ r 2. The Simplicity of Diffc(M) , dim(M) + 2 r ...... 13 ◦ ≤ ≤ ∞ r 3. The Simplicity of Diffc(M) , 1 r dim(M)...... 48 ◦ ≤ ≤ 4. Isomorphisms between groups of diffeomorphisms ...... 58

A. Isomorphisms between permutation groups ...... 71

Bibliography...... 76

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v Abstract

Mainly this work is based on articles that were published by D. B. A. Epstein, J. Mather, R. P. Filipkiewicz and A. Banyaga between 1970 and 1988. The first chapter is essentially due to D. B. A. Epstein. Here it is shown that the commutator groups of certain groups of homeomorphisms are simple (cf. r r Theorem 1.3). This result can be applied to Diffc(M) and thus Diffc(M) has a simple commutator group (cf. Corollary 1.17). ◦ ◦ r In chapter 2 & 3 a result of J. Mather is presented, that says: Diffc(M) is simple, r ◦ provided r = dim(M) + 1. Actually it is shown that Diffc(M) is perfect, then the result of the6 first chapter is used. For dim(M) + 2 r ◦this is carried out in the second chapter (cf. Corollary 2.44), and for 1 ≤r ≤dim( ∞ M) in the third one ≤ ≤ dim(M)+1 (cf. Corollary 3.5). As far as I know it is not yet known whether Diffc (M) is simple as well. However, the proof presented here fails completely in the case◦ r = dim(M) + 1. In the fourth chapter it is shown, that every isomorphism of groups between certain groups of diffeomorphisms is induced by a diffeomorphism of the underlying manifolds. Especially isomorphic groups of diffeomorphisms come from diffeomor- phic manifolds. First the more general result Theorem 4.14 is proved, then it is r n applied to Diffc(R ) (cf. Theorem 4.17). The proofs of these theorems are due to R. P. Filipkiewicz and◦ A. Banyaga. The methods and results of chapter 2 & 3 are not required for chapter 4, and thus it can be read immediately after the first chapter. The appendix deals with the discrete analogon of Theorem 4.17. The manifolds are replaced by finite sets and the groups of diffeomorphisms by groups of permutations. If the sets do not consist of six elements, it is again shown that an isomorphism of groups between two groups of permutations is induced by a bijection of the underlying sets (cf. Corollary A.8). If the sets do consist of six elements this is not true (cf. Theorem A.10). In the appendix only elementary group theoretic methods are used, hence it can be read independently of the other chapters.

vi r r 1. The Simplicity of [Diffc(M) , Diffc(M) ] ◦ ◦ 1.1. Definition. Let G be a group and g, h G. The commutator of g and h is 1 1 ∈ [g, h] := g− h− gh. If E,F are two arbitrary subsets of G then E denotes the h iG subgroup of G generated by E, and we let [E,F ] := [e, f]: e E, f F G.[G, G] is called the commutator subgroup of G, it’s a normalh{ subgroup∈ of G. The∈ }inormalizer 1 of a subgroup U G is NG(U) := g G : g− Ug = U . It’s the largest subgroup of G that contains⊆U as normal subgroup.{ ∈ If N is a normal} subgroup of G we write N ¡ G. A group G is said to be simple iff it has no normal subgroups except e and G. G is called perfect iff [G, G] = G. { } If X is a topological space, and g : X X a homeomorphism then the support of g is → supp(g) := x X : g(x) = x . { ∈ 6 }

1.2. Definition (Epstein’s Axioms). Let X be a paracompact Hausdorff space, a basis of open subsets of X and G a group of homeomorphisms of X. The triple (UG, X, ) is said to satisfy Epstein’s Axioms iff the following holds: U Axiom 1: U = g(U) . ∈ U ⇒ ∈ U Axiom 2: G acts transitively on . U Axiom 3: Let g G, U and let be a covering of X. Then there exists an ∈ ∈ U V ⊆ U integer n, elements g1, . . . , gn G and V1,...,Vn such that g = gn g1, ∈ ∈ V ··· supp(gi) Vi and supp(gi) gi 1 g1(U) = X for 1 i n. ⊆ ∪ − ··· 6 ≤ ≤

1.3. Theorem (Epstein). [3] Let (X, G, ) satisfy Epstein’s Axioms and let 1 = U { } 6 N G be a subgroup such that [G, G] NG(N). Then [G, G] N. ⊆ ⊆ ⊆ Proof. Obviously we may assume G = Id . 6 { } Claim 1. X is connected, and no open subset of X is ﬁnite.

Proof of Claim 1. X is connected, for suppose conversely X = U V where U and V are disjoint open subsets of X. Then := W : W U∪or W V is a covering of X, and hence by Axiom 3, G isW generated{ ∈ by U elements⊆ g G ⊆supported} on W . Therefore G is generated by elements supported either∈ in U or in V . It follows,∈ W that each element of G maps U to U and V to V , but this contradicts Axiom 2 and hence X is connected. Suppose F is a finite open subset of X. Choose x F . Since X is Hausdorff x is an open and closed subset of X, and hence X = ∈x , but this is a contradiction{ } to the assumption G = Id . { } 2 6 { } 1 r r 2 1. The Simplicity of [Diffc(M) , Diffc(M) ] ◦ ◦

Claim 2. Let U, V X be open subsets and let g, h G be homeomorphisms with supp(h) V , U supp(⊆ h) = and g(U) = V . Then ∈ ⊆ ∩ ∅ g 1h 1g on g 1(supp(h)) [g, h] = ( − − − (1.1) h elsewhere

1 Especially we have supp([g, h]) supp(h) g− (supp(h)) V U. ⊆ ∪ ⊆ ∪ 1 1 Proof of Claim 2. On g− (supp(h)) (1.1) follows from g− (supp(h)) supp(h) = , 1 ∩ ∅ on supp(h) it follows from g(supp(h)) supp(h− ) = and elsewhere it’s obvious. 2 ∩ ∅ Claim 3. [G, G] acts transitively on . U

Proof of Claim 3. Let U1,U2 and choose g G with g(U1) = U2 (Axiom 2). ∈ U ∈ We have to ﬁnd g0 [G, G] with g0(U1) = U2. By Axiom 3 there exist n N, ∈ ∈ V1,...,Vn and h1, . . . , hn G with g = hn h1, supp(hi) Vi and ∈ U ∈ ··· ⊆

Ki := supp(hi) hi 1 h1(U1) = X 1 i n. ∪ − ··· 6 ∀ ≤ ≤

Choose Wi with Wi X Ki and gi G with gi(Wi) = Vi (Axiom 2). Since ∈ U ⊆ \ ∈ Wi supp(hi) = we can apply Claim 2 and obtain ∩ ∅ 1 1 1 gi− hi− gi on gi− (supp(hi)) [gi, hi] = ( (1.2) hi elsewhere

1 From gi− (supp(hi)) hi 1 h1(U1) Wi Ki = and (1.2) we get ∩ − ··· ⊆ ∩ ∅ [g , h ] = h i i hi 1 h1(U1) i hi 1 h1(U1) | − ··· | − ··· and thus inductively

[gn, hn] [g1, h1](U1) = [gn, hn] [g2, h2]h1(U1) ··· ··· [G,G] | ∈ {z } . .

= hn h1(U1) = g(U1) = U2. ··· 2

Claim 4. Let U , h G with supp(h) U. Then there exists g N with ∈ U ∈ ⊆ ∈ g U = h U . | | Proof of Claim 4. Since N = Id we can choose α N, α = Id and x X, 6 { } ∈ 6 1 ∈ α(x) = x. Further we choose U with x U and U α− (U) = . Since a paracompact6 Hausdorff spacee is∈ normal U we find∈ e V,W e ∩ suche that∅x V , V W = and V W U. By Axiom 2 there exists g ∈G Uwith g(W ) = V ∈, and ∩ ∅ ∪ ⊆ e ∈ 1 by Claim 3 we find k [G, G] with k(V ) = U. Let h :=e k− hk andeρ := [α, [g, h]]. ∈ e e We have supp(h) V and ρ [N, [G, G]] N, for [G, G] NG(N). From Claime 2 e ⊆ ∈ ⊆ ⊆ r r 1. The Simplicity of [Diffc(M) , Diffc(M) ] 3 ◦ ◦

we obtain [g, h] V = h V and [α, [g, h]] U = [g, h] U , since supp([g, h]) U. Hence e | | e | e | 1 e ⊆ e ρ V = h V . Sincee k [G, G] NG(eN) wee havee g :=e kρk− N ande | e| ∈ ⊆ ∈ 1 1 1 1 g U = kρk− U = khk− U = kk− hkk− U = h U . | | e | | | 2

Claim 5. The orbits of N are the same as the orbits of G. In particular they are dense by Axiom 2.

Proof of Claim 5. Let x X and y G(x). Choose n minimal with respect to ∈ ∈ the condition that there exist g1, . . . , gn G, U1,...,Un with supp(gi) Ui ∈ ∈ U ⊆ and y = gn g1(x). By Claim 4 we obtain hi N with hi U = gi U . Since n is ··· ∈ | i | i minimal gi 1 g1(x) Ui, 1 i n, for otherwise we could omit gi and this would contradict− the··· minimality∈ of n≤. Thus≤ we obtain inductively

y = gn g1(x) = hngn 1 g1(x) ··· . − ··· .

= ... = hn h1(x) N(x). ··· ∈ 2

Claim 6. Let h1, h2 G, V , supp(hi) V . Then [h1, h2] N. ∈ ∈ U ⊆ ∈ 1 1 Proof of Claim 6. Let x X. By Claim 5 we ﬁnd α1, α2 N with x, α1− (x), α2− (x) ∈ 1 ∈ 1 distinct. Now choose U with x U and U, α1− (U), α2− (U) disjoint and choose ∈1 U 1 ∈ g1, g2 G such that x, g1− (x), g2− (x) are distinct elements of U. Further we select ∈ 1 1 V with x V and V , g1− (V ), g2− (V ) disjoint subsets of U. By Claim 3 we ﬁnd e ∈ U ∈ e e e e 1 k [G, G] with k(V ) = V . As we did in Claim 4, we let hi := k− hik. Whereas hi are ∈ e e supported in an arbitrary V , hi are supported in the suitable V . Suppose ∈ U e 1 1 1 e ∈ U1 1 we have [h1, h2] N then [h1, h2] = kk− [h1, h2]kk− = k[k− h1k, k− h2k]k− = e1 e ∈ k[h1, h2]k− N, since [G, G] NG(N). e e ∈ ⊆ So it remains to show [h1, h2] N. Let ρi := [αi, [gi, hi]] [N, [G, G]] N. As e e ∈ e ∈ ⊆ in Claim 4 it follows ρi V = hi V and hence | e | e e [ρ1, ρ2] = [h1, h2] . (1.3) |V |V 1 1 e e ρi X V is supported on gi− (V ) αi− (U). Because of the choice of the sets U and V , | \ e ∪ e supp(eρ1 X V ) supp(ρ2 X V ) = and hence | \ ∩ | \ ∅ e e [ρ1, ρ2] X V = Id X V = [h1, h2] X V . (1.4) | \ | \ e e | \ e e e From equations (1.3) and (1.4) we obtain [h1, h2] = [ρ1, ρ2] N. 2 ∈ Claim 7. There exists a covering of X, such that the following holds: V ⊆ U

V1,V2 ,V1 V2 = = U : V1 V2 U. ∈ V ∩ 6 ∅ ⇒ ∃ ∈ U ∪ ⊆ r r 4 1. The Simplicity of [Diﬀc(M) , Diﬀc(M) ] ◦ ◦

Proof of Claim 7. Since X is paracompact, we can choose Uλ λ Λ , Uλ λ Λ { } ∈ ⊆ U { e } ∈ ⊆ U with Uλ Uλ and such that Uλ λ Λ is a locally ﬁnite cover of X. For every x X ⊆ e { } ∈ ∈ we choose Vx , x Vx, such that Vx intersects only ﬁnitely many Uλ1 ,...Uλnx . e ∈ U ∈ e e We can assume x Uλi Uλi , 1 i nx, by shrinking Vx if necessary. Let Vx , ∈ ⊆ e ≤ ≤ e ∈ U x Vx and ∈ nx V V U x x \ λi ⊆ e ∩ i=1 e

Let := Vx x X . Obviously is a covering of X and . Furthermore we have V { } ∈ V V ⊆ U V ,V Uλ = = V Uλ (1.5) ∈ V ∩ 6 ∅ ⇒ ⊆ e Now take V1,V2 with V1 V2 = . Since Uλ λ Λ is a covering of X we ﬁnd Uλ ∈ V ∩ 6 ∅ { } ∈ intersecting V1 and V2. By (1.5) we obtain V1 Uλ and V2 Uλ. Since Uλ , we are done. ⊆ e ⊆ e e ∈ U 2 Let be the covering constructed in Claim 7, and consider V E := h G : V with supp(h) V . { ∈ ∃ ∈ V ⊆ } E generates G by Axiom 3. Let h1, h2 E and choose V1,V2 with supp(hi) Vi. ∈ ∈ V ⊆ If V1 V2 = then [h1, h2] = Id N. If V1 V2 = then by Claim 7 there exists ∩ ∅ ∈ ∩ 6 ∅ U with V1 V2 U and by Claim 6 we obtain [h1, h2] N too. Summing up we∈ have U ∪ ⊆ ∈ [h1, h2] N h1, h2 E. (1.6) ∈ ∀ ∈ In general [E,E] need not generate [G, G], so we are not ﬁnished yet. 1 Claim 8. g G, h1, h2 E = g[h1, h2]g− N. ∈ ∈ ⇒ ∈

Proof of Claim 8. Choose Wi with supp(hi) Wi for i = 1, 2. We can assume ∈ V ∈ W1 W2 = , for otherwise [h1, h2] = Id. Let U with W1 W2 U. By ∩ 6 ∅ ∈ U ∪ ⊆ Axiom 3 we obtain g1, . . . , gn G, V1,...,Vn with supp(gi) Vi, g = gn g1 ∈ ∈ U ⊆ ··· and Ki := supp(gi) gi 1 g1(U) = X. Now choose Ui with Ui X Ki and ∪ − ··· 6 ∈ U ⊆ \ choose βi G with βi(Ui) = Vi. From Claim 2 we get ∈ 1 1 βi− gi− βi on Ui γi := [βi, gi] = ( (1.7) gi on X Ui \ We will show 1 1 1 g[h1, h2]g− = γn γ1[h1, h2]γ− γ− (1.8) ··· 1 ··· n for we are done then since γi [G, G] NG(N) and [h1, h2] N by (1.6). To see (1.8) ∈ 1⊆ 1 ∈ notice that supp(gi 1 g1[h1, h2]g1− gi− 1) gi 1 g1(U) since supp([h1, h2]) U and hence − ··· ··· − ⊆ − ··· ⊆ 1 1 gi 1 g1[h1, h2]g1− gi− 1 Ui = Id Ui (1.9) − ··· ··· − | | for we have Ui gi 1 g1(U) = by the choice of Ui. Since gi(Ui) = Ui and ∩ − ··· ∅ γi(Ui) = Ui by (1.7) we obtain from (1.9)

1 1 1 1 1 gi g1[h1, h2]g1− gi− Ui = Id Ui = γigi 1 g1[h1, h2]g1− gi− 1γi− Ui . (1.10) ··· ··· | | − ··· ··· − | r r 1. The Simplicity of [Diﬀc(M) , Diﬀc(M) ] 5 ◦ ◦

From (1.10) and since gi X Ui = γi X Ui by (1.7), we have | \ | \ 1 1 1 1 1 gi g1[h1, h2]g1− gi− = γigi 1 g1[h1, h2]g1− gi− 1γi− , ··· ··· − ··· ··· − and thus (1.8) follows inductively. 2 1 Consider S := g[h1, h2]g− : g G, h1, h2 E G. Obviously this is a normal subgroup of G andh{ by Claim 8 we have∈ S N∈. Since}i G is generated by E, G/S ⊆ is generated by hS : h E . Since [h1, h2] S for hi E the generators of G/S commute and hence{ G/S∈is abelian.} Consequently∈ ∈

[g, h]S = [gS, hS] = S g, h G ∀ ∈ and hence [G, G] S N. This completes the proof of Theorem 1.3. 2 ⊆ ⊆ 1.4. Corollary. Let (G, X, ) satisfy Epstein’s Axioms, then [G, G] is simple. U

Proof. Suppose Id = N¡[G, G] then [G, G] NG(N). By Theorem 1.3 [G, G] N and hence [G, G{] is} simple. 6 ⊆ ⊆ 2

1.5. Corollary. Let (G, X, ) satisfy Epstein’s Axioms, then every nontrivial normal subgroup of G contains [G,U G]. Thus [G, G] is the unique minimal normal subgroup of G.

Proof. Let Id = N ¡ G. This implies [G, G] NG(N) = G, and hence by Theorem 1.3{ [G,} G 6] N. ⊆ 2 ⊆ 1.6. Definition. Throughout the text a smooth manifold is a connected, paracompact, separable, boundaryless C∞-manifold, 1 dim(M) < . Suppose M, N are smooth manifolds and 0 r , then C≤r(M,N) denotes∞ the space of Cr- mappings M N equipped with≤ the≤ ∞Whitney topology (cf. Proposition 1.7). Fur- r → r r ther Diff (M) resp. Diffc(M) denotes the group of C - resp. compactly supported r r C -diffeomorphisms of M. If A is a fixed subset of M we let DiffA(M) be the group of Cr-diffeomorphisms supported on A. We equip these groups with the Whitney- or Cr- topology i.e. the initial topology with respect to the inclusions Diffr(M) Cr(M,M), Diffr(M) Cr(M,M) and Diffr (M) Cr(M,M). ⊆ c ⊆ A ⊆ If G is a topological group we denote by Go the component containing Id. Further n n n we let Br (x) := y R : x y < r , the open ball in R with radius r and n { ∈ n k − k } center x, and B := B1 (0), the open unit ball. 1.7. Proposition. We list some facts concerning the Whitney- or Cr-topology. For details see [6] and [9].

(1) Diffr(M) is open in Cr(M,M) for 1 r . ≤ ≤ ∞ (2) The set Embr(M,N) Cr(M,N) of Cr-embeddings of M in N is open for 1 r . ⊆ ≤ ≤ ∞ r r 6 1. The Simplicity of [Diffc(M) , Diffc(M) ] ◦ ◦

r r (3) Let Cprop(M,N) C (M,N) denote the subspace of proper mappings (f : M ⊆ 1 → N is proper iﬀ K N compact f − (K) is compact). Then ⊆ ⇒ Cr(N,P ) Cr (M,N) Cr(M,P ) × prop → (g, h) g h 7→ ◦ is continuous for 0 r . ≤ ≤ ∞ (4) The following mapping is continuous for 0 r : ≤ ≤ ∞ r r r C (M,N1) C (M,N2) C (M,N1 N2) × → × (f1, f2) (f1, f2) 7→

(5) If 0 r and E π M is a ﬁnite dimensional, smooth vector bundle over ≤ ≤r ∞ π → r π M, then Γc(E M), the set of compactly supported C -sections of E M equipped with the→ Whitney-topology, i.e. the initial topology with respect to→ the r π r inclusion Γc(E M) C (M,E), is a locally convex vector space. (compact support is essential→ here,⊆ for otherwise scalar-multiplication won’t be continuous if M is not compact!)

(6) If 0 r s then Cs(M,N) Cr(M,N) is a dense subset. Further if f ≤Cr(M,N≤ )≤, ∞a neighborhood of⊆f Cr(M,N), A M closed, and f is already∈ Cs on aN neighborhood of A, then∈ there exists g ⊆ Cs(M,N) with ∈ N ∩ g A = f A. | | r r r (7) Diff (M) is a topological group for 0 r . Since Diffc(M) and Diffc(M) are subgroups of Diffr(M) equipped with≤ the≤ initial ∞ topology, they are topological◦ groups too.

(8) We have r C∞(M,N) = lim C (M,N), ←− where the projective limit is over all 0 r < and the inclusions Cr(M,N) , Ck(M,N) for r k. ≤ ∞ → ≥

(9) Given a locally finite family of charts Φ = (ui,Ui)i I on M, a family of compact ∈ subsets K = (Ki)i I with Ki Ui, a family of charts Ψ = (vi,Vi)i I on N ∈ ⊂ ∈ with f(Ki) Vi and a family of positive numbers = (i)i I , then we define r ⊆ r ∈ (f;Φ, Ψ, K, ) to be the space of g C (M,N) with g(Ki) Vi i I and N ∈ ⊆ ∀ ∈ k 1 k 1 D (vifu− )(x) D (vigu− )(x) < i k i − i k r i I, x ui(Ki) and 0 k r. The sets (f;Φ, Ψ, K, ) as Φ, Ψ, K, ∀vary∈ form∀ a∈ neighborhood∀ basis≤ of≤ f Cr(M,NN) for 0 r < . ∈ ≤ ∞ r r 1. The Simplicity of [Diffc(M) , Diffc(M) ] 7 ◦ ◦

(10) For 0 r < consider the vector bundle of r-jets J r(M,N) π M and the canonical≤ injective∞ mapping −→

jr :Cr(M,N) , C0(M,J r(M,N)). → If we put the graph topology on C0(M,J r(M,N)) then the Whitney-topology on Cr(M,N) is the topology induced by the inclusion jr. In other words the Whitney-topology is the coarsest topology on Cr(M,N) such that jr becomes continuous.

r (11) If 0 r < and dr is a metric on J (M,N) generating the topology, then a neighborhood≤ ∞ basis for f Cr(M,N) is given by: ∈ r r r (f; r, ) := g C (M,N): dr(j f(x), j g(x)) < (x) N { ∈ } where varies over C(M, (0, )). ∞

1.8. Remark. We have:

r r r r r r Diffc(M) ¡ Diff (M), Diffc(M) ¡ Diff (M) , Diffc(M) ¡ Diffc(M) r ◦ r r ◦ r ◦ ◦ Diffc(M) ¡ Diff (M), Diff (M) ¡ Diff (M) ◦

r 1 Proof. If f, g Diff (M), then supp(gfg− ) = g(supp(f)). Hence if f is com- ∈ 1 pactly supported, then so is gfg− . Now everything follows from the continuity of r r 1 conjg : Diff (M) Diff (M), conjg(f) := gfg− and the fact conjg(Id) = Id r → ∈ Diffc(M) . 2 ◦ 1.9. Definition. Let M be a smooth manifold and 0 r .ACr-diffeotopy of r ≤ ≤ ∞ r M is a C -mapping H : M I M such that Ht := H( , t) Diff (M). H1 is said to be Cr-diffeotopic to the× identity→ through H. The support· of∈ a diffeotopy H is

supp(H) := x M : t I with H(x, t) = x . { ∈ ∃ ∈ 6 }

1.10. Remark. Given a compactly supported, time dependent Cr-vector field X on M, one obtains a compactly supported Cr-diffeotopy H : M I M by integrating × → d X, i.e. the unique mapping H : M I M with H0 = Id and ds tHs(x) = r × → | X(Ht(x), t) is a C -diffeotopy. Conversely, given a compactly supported Cr-diffeotopy H : M I M, one r 1 × → obtains a compactly supported, time dependent C − -vector field H˙ on M given by ˙ d 1 H(x, t) := tHs(H− (x)). Clearly these two constructions are inverse to each other. ds | t 1.11. Lemma. Let M be a connected, smooth manifold, σ : I M a smooth path in M and let U be an open neighborhood of Im(σ) M. Then→ there exists r ⊆ f DiffU∞(M) with f(σ(0)) = σ(1). In particular Diffc(M) acts transitively on M for∈1 r ◦. ◦ ≤ ≤ ∞ r r 8 1. The Simplicity of [Diffc(M) , Diffc(M) ] ◦ ◦

Proof. First we choose a smooth path σ : I M joining x = σ(0) and y = σ(1), such → that σ is an embedding and Im(σ) eU. Hence there exists a C∞-vector field X on ⊆ d X M withe supp(X) U and X(σ(te)) = ds tσ(s). By Remark 1.10 Fl : M I M is ⊆ | X × → a compactly supported C∞-diffeotopye ande thus Fl1 DiffU∞(M) . Finally we have FlX (x) = y since σ : I M is an integral curve of X∈. ◦ 2 1 → e 1.12. Lemma. If 1 r and H : M I M is a compactly supported r ˇ ≤ ≤ ∞ r × → ˇ C -diffeotopy then H : I Diffc(M) is continuous. Especially we have H : I r ˇ → → Diffc(M) since H(0) = Id. ◦ Proof. For finite r this follows easily from the description of the Whitney-topology given in Proposition 1.7(11) and for r = it follows from Proposition 1.7(8) and the universal property of the projective limit.∞ 2

1.13. Proposition. [9] Let M be a smooth manifold, exp : TM M the exponential mapping obtained from a Riemannian metric on M and let 1 →r . Then there ≤1 ≤ ∞π exists a convex, open neighborhood U 0 of the zero section in Γc (TM M) and an 1 → open neighborhood U of Id Diffc (M) such that the map ∈ ◦ r π r ϕ : U 0 Γ (TM M) U Diff (M) ∩ → −→ ∩ σ exp σ 7→ ◦ is well defined and a homeomorphism.

Proof. It is well known that there exists an open neighborhood V 0 of the image of the zero section in TM, and an open neighborhood V of the diagonal diag(M) M M such that ⊆ × (π,exp) TM V 0 V M M ⊇ −→ ⊆ × is an C∞-diﬀeomorphism. Now let

r π U 0 := σ Γ (TM M): σ(M) V 0 , { ∈ c → ⊆ } U := f Cr(M,M):(Id, f)(M) V , { ∈ c ⊆ }

deﬁne ϕ : U 0 U like above and let → ψ U U 0 → 1 f (π, exp)− (Id, f) 7→ ◦

Clearly everything is well deﬁned, U and U 0 are open, 0 U 0 and Id U To see that ψ(f) is a section of TM π M compute as follows: ∈ ∈ → 1 1 π ψ(f) = π (π, exp)− (Id, f) = pr1 (π, exp) (π, exp)− (Id, f) = Id. ◦ ◦ ◦ ◦ ◦ ◦ Since

1 1 ϕ(ψ(f)) = exp (π, exp)− (Id, f) = pr2 (π, exp) (π, exp)− (Id, f) = f ◦ ◦ ◦ ◦ ◦ r r 1. The Simplicity of [Diﬀc(M) , Diﬀc(M) ] 9 ◦ ◦ and 1 1 ψ(ϕ(σ)) = (π, exp)− (Id, exp σ) = (π, exp)− (π, exp) σ = σ, ◦ ◦ ◦ ◦ ψ is the inverse to ϕ. By Proposition 1.7(3)&(4) both, ϕ and ψ, are continuous. Using Proposition 1.7(1)&(5) we can shrink U and U 0 such that the proposition holds. 2

r 1.14. Corollary. Let M be a smooth manifold and 1 r . Then Diffc(M) is locally contractible and consists of exactly those Cr-diffeomorphisms≤ ≤ ∞ that are dif-◦ feotopic to the identity through a compactly supported Cr-diffeotopy.

r r r Proof. Let f Diffc(M) . Then f : Diffc(M) Diffc(M) , f (g) := fg is a homeomorphism by∈ Proposition◦ 1.7(3).∗ Hence with◦ → the notation◦ from∗ Proposition 1.13 f ϕ : U 0 f (U) is a homeomorphism onto an open neighborhood of f. Since ∗ ◦ → ∗ U 0 is a convex subset of a topological vector space it is contractible, and thus the neighborhood f (U) of f is contractible. ∗ r r By Lemma 1.12 Diffc(M) contains all C -diffeomorphisms that are diffeotopic to the identity through a compactly◦ supported Cr-diffeotopy. Conversely if f U r r ∈ ⊆ Diffc(M) we have a compactly supported C -diffeotopy with H0 = Id and H1 = f ◦ 1 r given by H(x, t) := ϕ(tϕ− (f))(x). Since Diffc(M) is generated by U it remains to show that ’being Cr-diffeotopic to Id’ is compatible◦ with composition. Therefore r consider two compactly supported C -diffeotopies F resp. G with F0 = G0 = Id. Then 1 H(x, t) := F (G− (G(x, 1 t)), t) 1 − r 1 is a compactly supported C -diffeotopy from F1G1− to Id. 2 1.15. Example. Consider M = Rn and the exponential mapping obtained from the usual Riemannian metric on Rn. Via the homeomorphism (cf. Proposition 1.7(3))

π pr Γr(T Rn Rn) = Γr(Rn Rn 1 Rn) ∼= Cr(Rn, Rn) c → ∼ c × → → c σ pr2 σ 7→ ◦ r n n r n n (here σ Cc(R , R ) means: σ C (R , R ) and σ is zero outside a compact set) ∈ ∈ r n n the mapping ϕ defined in Proposition 1.13 is given by ϕ :Cc(R , R ) U 0 U r n ⊇ → ⊆ Diffc(R ) , σ σ + Id. ◦ 7→ r n r n r n We have Diffc(R ) = Diffc(R ) , since for arbitrary f Diffc(R ) we have a r ◦ n n ∈ 1 compactly supported C -diffeotopy H : R [, 1] R , given by H(x, t) := tf( t x). × → 0 n By choosing sufficiently small we can achieve, that H Diffc (R ) and hence there n n 0 ∈ ◦ is a prolongation H : R I R that is a C -diffeotopy from f to Id with Ht = Id 0 × → r r t [0, 2 ]. Since C -mappings are dense in C -mappings and since H is already C ∀ ∈ n 1 r on R ([0, 2 ] [, 1] we find a compactly supported C -diffeotopy H from f to × ∪ r n n f Id, and thus f Diffc(R ) . Moreover, if supp(f) B , then H can be chosen such that supp(H∈) Bn. To◦ see the last assertion let⊆H be a diffeotopyf from f to f ⊆ n f Id like above, choose 1 D R with supp(H) BD(0) and assume that Ht = f ≤ ∈ f ⊆ f r r 10 1. The Simplicity of [Diffc(M) , Diffc(M) ] ◦ ◦

1 t [ , 1]. If we choose λ C∞([0, 1], [1,D]) with λ [0, 1 ] = D and λ(1) = 1, then ∀ ∈ 2 ∈ | 2 1 r n G(x, t) := λ(t) H(λ(t)x, t) is a C -diffeotopy from f to Id and supp(G) B . f ⊆ 1.16. Lemma. [3] Let M be a smooth manifold, 1 r , n = dim(M) and ≤ ≤ ∞ ϕ r(M) := ϕ(Bn): Rn , M Cr embedding . U { → − } r r Then (Diffc(M) ,M, (M)) satisfies Epstein’s Axioms. ◦ U r r Proof. Obviously (Diffc(M) ,M, (M)) satisfies Axiom 1. ◦ U r n To see, that Axiom 2 is satisfied, consider C -embeddings ϕ1, ϕ2 : R , M. We r n n r→ have to find f Diffc(M) with and f(ϕ1(B )) = ϕ2(B ). Since Diffc(M) acts ∈ ◦ ◦n transitively on M we can assume ϕ1(0) = ϕ2(0), and since we can shrink ϕi(B ) by r n elements of Diffc(M) we can additionally assume that ϕi(B ) U, where (u, U) ◦ ⊆ r n is a chart of M centered at ϕ1(0) = ϕ2(0). So it suffices to find f Diffc(R ) with f(ϕ(Bn)) = Bn for all embeddings ϕ : Rn , Rn with ϕ(0) = 0.∈ Besides we◦ → n n may assume that det(Dϕ(0)) > 0, for if we define σ : R R by σ(x1, . . . , xn) = n n → ( x1, x2, . . . , xn), we have ϕ σ(B ) = ϕ(B ) and det(D(ϕ σ)(0)) = det(Dϕ(0)). − ◦ ◦ − Claim 1. There exists a Cr-isotopy from ϕ to Id, i.e. a Cr-mapping H : Rn I Rn n n × → such that Ht := H( , t): R R is an embedding t I, H0 = Id and H1 = ϕ. · → ∀ ∈ r 1 Proof of Claim 1. Consider the following C − -isotopy from Dϕ(0) to ϕ: Rn I Rn × → 1 ϕ(tx) if t = 0 (x, t) ( t 6 7→ Dϕ(0) x if t = 0 · + + r 1 Since Dϕ(0) GL (n, R) and GL (n, R) is connected, we find a C − -isotopy ∈ r 1 from Id to Dϕ(0) and by composing and reparametrizing we obtain a C − -isotopy n n 1 2 G : R I R with Gt = Id t [0, 3 ] and Gt = ϕ t [ 3 , 1]. Finally, we × → r ∀ ∈ n ∀ 1∈ n 3 approximate G by a C -mapping H that equals G on R [0, 4 ] R [ 4 , 1]. This is possible, since G is already Cr on Rn [0, 1 ) Rn ( 2 , 1].× Choosing∪ H× sufficiently × 3 ∪ × 3 near G we can achieve that Ht remain embeddings (cf. Proposition 1.7(2)) and hence H is a Cr-isotopy. 2 r 1 n Now consider the locally defined, time dependent C − -vector field on R given by ˙ d 1 d ˙ ˙ H(x, t) := ds tHs(Ht− (x)). We then have ds tHs(x) = H(Ht(x), t). The domain of H |n | is (H, prI )(R I). Hence we can find a compactly supported, globally defined, time r ×1 ˙ n ˙ ˙ n dependent C − -vector field G on R such that G equals H on (H, prI )(B2 (0) I). ˙ r 1 × Thus, integrating G we obtain a compactly supported C − -diffeotopy G with G0 = r n Id and G Bn(0) I = H Bn(0) I , especially G is C on B2 (0) I. By approximation 2 × 2 × | | r × we find a compactly supported C -diffeotopy G with G Bn I = G Bn I and thus n n n r e n e| × | × G1(B ) = G1(B ) = H1(B ). Since G1 Diffc(R ) , we have verified Axiom 2. ∈ ◦ e e r Claim 2. (cf. (3.1) Lemma in [11]) Let f Diffc(M) , be a cover of M. ∈ ◦ Vr ⊆ U Then there exists an integer n, V1,...,Vn and fi Diffc(M) , 1 i n, with ∈ V ∈ ◦ ≤ ≤ supp(fi) Vi and f = fn f1. ⊆ ··· r r 1. The Simplicity of [Diffc(M) , Diffc(M) ] 11 ◦ ◦

Proof of Claim 2. Choose a partition of unity (λi)i N subordinated to the cover ∈ V and choose Vi with supp(λi) Vi. Let ϕ : U 0 U as in Proposition 1.13 and r∈ V π ⊆ → deﬁne W 0 Γ (TM M) by ⊆ c → n n W := σ U :( λ )σ U , ϕ(( λ )σ)(supp(λ )) V n N . 0 0 X i 0 X i n+1 n+1 { ∈ i=1 ∈ i=1 ⊆ ∀ ∈ }

W 0 is an open neighborhood of the zero section and hence W := ϕ(W 0) is an open r neighborhood of Id Diffc(M) . We can assume that f W , for W generates r ∈ ◦ ∈ Diffc(M) . Since f is compactly supported there exists n N such that ◦ ∈ 1 supp(λi) supp(ϕ− (f)) = i > n. (1.11) ∩ ∅ ∀ i 1 r 1 r Now define gi := ϕ(( j=1 λj)ϕ− (f)) Diffc(M) and let fi := gigi− 1 Diffc(M) , P ◦ ◦ i = 1, . . . , n. Using (1.11) we then have∈ − ∈

n f f = g g = ϕ(( λ )ϕ 1(f)) = f. n 1 n 0 X i − ··· i=1 =Id |{z} 1 | =ϕ−{z(f) }

Further, since gi M supp(λi) = gi 1 M supp(λi) we obtain | \ − | \ 1 supp(fi) = supp(gigi− 1) gi 1(supp(λi)) Vi. − ⊆ − ⊆ This ﬁnishes the proof of Claim 2. 2

r Claim 3. Let U, V , f Diﬀc(M) and supp(f) V . Then there exist r ∈ U ∈ ◦ ⊆ f1, f2 Diﬀc(M) with supp(fi) V , f = f2f1, supp(f1) U = M and supp(f2) ∈ ◦ ⊆ ∪ 6 ∪ f1(U) = M. 6

Proof of Claim 3. Suppose we have neighborhoods Ni of xi Ni M U, i = 1, 2 r ∈ ⊆ \ and f1 Diffc(M) with supp(f1) V , f1 N1 = f N1 and f1 N2 = Id. We then ∈ 1 ◦ ⊆r | | | let f2 := ff1− and obtain f2 Diffc(M) , supp(f2) V , f = f2f1. Since x2 / ∈ ◦ ⊆ ∈ supp(f1) U, we have supp(f1) U = M. Further we have f2f1 N1 = f N = f1 N ∪ ∪ 6 | | 1 | 1 and thus f2 f (N ) = Id. Hence f1(x1) / supp(f2), and since f1(x1) / f1(U) we obtain | 1 1 ∈ ∈ supp(f2) f1(U) = M. So it remains to find xi, Ni and f1 presumed above. ∪ 6 r By Example 1.15 we can choose a C -diffeotopy H : M I M with H0 = Id, × → H1 = f and supp(H) V . In the case n = 1 we think of V as V R via a chart, and let H(x, t) := tf(x)+(1⊆ t)x x V, t I, extended trivially on ⊆M V . We have U = − ∀ ∈ ∈ \ 6 M and thus we find x1 and N1 with x1 N1 M U, such that H(N1 I) M U. ∈ ⊆ \ × 6⊇ \ If n > 1 this is possible since H( x1 I) is compact and M U is open, in the case { } × \ n = 1 the existence of x1 and N1 follows from our special choice of H. Now choose x2 and N2 such that x2 N2 M (U H(N1 I)), λ C∞(M, R) with λ N2 = 0, ˙ ∈ ⊆ d \ ∪ 1 × ∈ ˙ |˙ λ H(N1 I) = 1, let H(x, t) := ds tHs(Ht− (x)) and define G(x, t) := λ(x)H(x, t). | × |r 1 Integration of G˙ gives rise to a C − -diffeotopy G : M I M supported on V × → with Gt N2 = Id t I and G N1 I = H N1 I . Hence we can approximate G by | ∀ ∈ | × | × r r 12 1. The Simplicity of [Diffc(M) , Diffc(M) ] ◦ ◦

r an C -diffeotopy G, supported on V , which equals G on (N1 N2) I, where Ni e fr ∪ f × f are smaller neighborhoods of xi. Thus we have G1 Diffc(M) , supp(G1) V , e ∈ ◦ e ⊆ G1 = Id and G1 = G1 = H1 = f . 2 |N2 |N1 |N1 |N1 |N1 e e r Nowe let f Diffce(M) , e a covere of Me and let U . By Claim 2 we obtain ∈ ◦ Vr ⊆ U ∈ U V1,...,Vn and fi Diffc(M) , 1 i n with supp(fi) Vi and f = fn f1. ∈ V ∈ ◦ ≤ ≤ 1 2 r⊆ ···j Since fi 1 f1(U) we obtain from Claim 3 fi , fi Diffc(M) with supp(fi ) − 2···1 ∈1 U ∈ 2 1 ◦ ⊆ Vi, fi = fi fi , supp(fi ) fi 1 f1(U) = M and supp(fi ) fi fi 1 f1(U) = M. 2 1 ∪ 2 −1 ··· 6 ∪ − ··· 6 Since fi f1 = f f f f we have verified Axiom 3. 2 ··· i i ··· 1 1 1.17. Corollary. Let 1 r and let M be a connected smooth manifold. Then r r ≤ ≤ ∞ r r r [Diffc(M) , Diffc(M) ] is simple, [Diffc(M) , Diffc(M) ] ¡ Diff (M) and every non- ◦ ◦ r ◦ r ◦ r trivial, normal subgroup of Diff (M) contains [Diffc(M) , Diffc(M) ]. Consequently r r ◦ ◦ r [Diffc(M) , Diffc(M) ] is the unique minimal, normal subgroup of Diff (M). The ◦ ◦ r r r r same is true when Diff (M) is replaced by Diffc(M), Diff (M) or Diffc(M) . ◦ ◦ r r Proof. The simplicity of [Diffc(M) , Diffc(M) ] follows immediately from Corol- lary 1.4 and Lemma 1.16. To see the second◦ assertion◦ notice, that if G is an arbitrary 1 1 1 group and H ¡ G then [H,H] ¡ G, since g[h1, h2]g− = [gh1g− , gh2g− ]. By Re- r r r r mark 1.8 we thus obtain [Diffc(M) , Diffc(M) ]¡Diff (M). Let Id = H¡Diff (M). We’ll show ◦ ◦ { } 6 r r H [Diffc(M) , Diffc(M) ] = Id , (1.12) ∩ ◦ ◦ 6 { } for we are done then, since we have

r r r r H [Diffc(M) , Diffc(M) ] ¡ [Diffc(M) , Diffc(M) ] ∩ ◦ ◦ ◦ ◦ r r and [Diffc(M) , Diffc(M) ] is simple. To see (1.12) choose h H, U and r ◦ ◦ ∈ ∈ U f1, f2 Diffc(M) , such that U h(U) = , supp(fi) U and [f1, f2] = Id. We ∈ ◦ r ∩ ∅ r ⊆ r 6 r have gi := [h, fi] H Diffc(M) since H ¡ Diff (M) and Diffc(M) ¡ Diff (M). ∈ ∩ r ◦ r ◦ Hence [g1, g2] H [Diffc(M) , Diffc(M) ], but we have gi U = fi U : U U ∈ ∩ ◦ ◦ | | → (cf. Claim 2 on page 2) and hence [g1, g2] U = [f1, f2] U = Id by the choice of fi. This shows (1.12). | | 6 r r Nothing essential in the proof changes, when Diff (M) is replaced by Diffc(M), r r Diff (M) or Diffc(M) . To see the last case one could also apply Corollary 1.5. 2 ◦ ◦ r 2. The Simplicity of Diffc(M) , dim(M) + 2 r ◦ ≤ ≤ ∞ 2.1. Definition. A modulus of continuity is a continuous, strictly increasing function α : [0, ) R, such that α(0) = 0 and α(tx) tα(x), x [0, ) t 1. ∞ → ≤ ∀ ∈ ∞ ∀ ≥ Consider two metric spaces (X, dX ) and (Y, dY ), a modulus of continuity α, and a mapping f : X Y . f is said to be α-continuous, if there exist C > 0 and > 0, such that →

x1, x2 X, dX (x1, x2) = dY (f(x1), f(x2)) Cα(dX (x1, x2)). ∈ ≤ ⇒ ≤ f is called locally α-continuous if each point has a neighborhood U, such that f U is α-continuous. |

2.2. Remark. Id is a modulus of continuity and a mapping is Id-continuous, iﬀ it is Lipschitz. If 0 < t < s then α(s) = α(t s ) s α(t), hence t t is an increasing t ≤ t 7→ α(t) function, and thus Id :(X, d) (X, d) is α-continuous for all moduli of continuity → α. If α, α0 are moduli of continuity that equal on a neighborhood of 0 then a mapping is α-continuous iﬀ it is α0-continuous.

2.3. Lemma. Let X and Y be two metric spaces, each of them equipped with two metrics which are equivalent in the following sense: there exist positive constants mX , MX , mY , MY , with

mX dX (x1, x2) dX (x1, x2) MX dX (x1, x2) ≤ e ≤ mY dY (y1, y2) dY (y1, y2) MY dY (y1, y2) ≤ e ≤ for all x1, x2 X and y1, y2 Y . Then a mapping f :(X, dX ) (Y, dY ) is α- ∈ ∈ → continuous, iﬀ f :(X, dX ) (Y, dY ) is α-continuous. e → e Proof. Suppose f :(X, dX ) (Y, dY ) is α-continuous and choose C > 0 and > 0 with →

dY (f(x1), f(x2)) Cα(dX (x1, x2)) dX (x1, x2) . ≤ ∀ ≤ We assume mX 1 and let := mX . If dX (x1, x2) we have dX (x1, x2) , ≤ ≤ ≤ hence e e e 1 dY (f(x1), f(x2)) MY Cα(dX (x1, x2)) MY C α(dX (x1, x2)), e ≤ ≤ mX e | C{z:= } e and thus f :(X, dX ) (Y, dY ) is α-continuous. The second implication is proven e → e similarly, but now the constants mY and MX have to be used. 2

2.4. Lemma. Let (X, d) be a metric space and (V, ) a normed vector space. If f, g : X V are α-continuous and λ, µ R then λfk+ · kµg is α-continuous too. → ∈ 13 r 14 2. The Simplicity of Diﬀc(M) , dim(M) + 2 r ◦ ≤ ≤ ∞

Proof. Choose f , g > 0 and Cf ,Cg > 0, such that

f(x1) f(x2) Cf α(d(x1, x2)) d(x1, x2) f k − k ≤ ∀ ≤ g(x1) g(x2) Cgα(d(x1, x2)) d(x1, x2) g, k − k ≤ ∀ ≤ and let := min f , g , C := λ Cf + µ Cg. If d(x1, x2) then we have { } | | | | ≤

f g 2.5. Lemma. Let (X, dX ) (Y, dY ) (Z, dZ ) be mappings of metric spaces, and let α, β be moduli of continuity.→ If f is →α-continuous and g is β-continuous then g f is β α-continuous. ◦ ◦

Proof. We choose f , g > 0 and Cf ,Cg > 1 analogously to the proof of Lemma 2.4, let C := CgCf and choose 0 < f , such that Cf α() g. If dX (x1, x2) then ≤ ≤ ≤

dY (f(x1), f(x2)) Cf α(dX (x1, x2)) Cf α() g ≤ ≤ ≤ hence

dZ ((g f)(x1), (g f)(x2)) Cgβ(dY (f(x1), f(x2))) ◦ ◦ ≤ Cgβ(Cf α(dX (x1, x2))) ≤ CgCf (β α)(dX (x1, x2)). ≤ ◦ and thus Lemma 2.5 is proven. 2

2.6. Remark. From Remark 2.2 and Lemma 2.5 we obtain, that every f :(X, dX ) → (Y, dY ) that is Lipschitz, is α-continuous for all moduli of continuity α. Especially a C1-mapping Rn U f Rm is locally α-continuous for all moduli of continuity α. ⊃ →

2.7. Lemma. Let fi :(X, dX ) (Yi, dYi ) be α-continuous mappings, 1 i n, → n ≤ ≤ and let d((y1, . . . , yn), (z1, . . . , zn)) := dY (yi, zi). Then (f1, . . . , fn):(X, dX ) Pi=1 i → (Y1 Yn, d) is α-continuous. By Remark 2.3 d can be replaced by many common ×· · ·× 2 2 metrics on Y1 Yn (e.g. max dY , . . . , dY or d + + d ). × · · · × { 1 n } q Y1 ··· Yn n Proof. Let := min f , . . . , f , C := Cf , and dX (x1, x2) . Then we can 1 n Pi=1 i estimate { } ≤ n d((f , . . . , f )(x ), (f , . . . , f )(x )) = d (f (x ), f (x )) 1 n 1 1 n 2 X Yi i 1 i 2 i=1 n ( C )α(d (x , x )), X fi X 1 2 ≤ i=1 and this is Lemma 2.7. 2 r 2. The Simplicity of Diﬀc(M) , dim(M) + 2 r 15 ◦ ≤ ≤ ∞

2.8. Lemma. Consider a multi linear mapping B : V1 Vn W of ﬁnite × · · · × → dimensional, normed vector spaces, and α-continuous mappings fi :(X, d) Vi. If → the images of fi are bounded, then B (f1, . . . , fn) is α-continuous. ◦ 1 Proof. Since B is C it is locally Lipschitz, and since K := Im(f1) Im(fn) is × · · · × compact we B K is Lipschitz. Thus by Lemma 2.5 and Lemma 2.7 B (f1, . . . , fn) = | ◦ B K (f1, . . . , fn) is α-continuous. 2 | ◦ 2.9. Lemma. Let f :(X, ) (Y, d) be a continuous map from a compact, convex subset of a normed vectork·k space→ to a metric space. Then there exists a modulus of continuity α, such that f is α-continuous.

Proof. Let β(t) := sup d(f(x1), f(x2)). Since X is compact, β : R R is well x1 x2 t → k − k≤ deﬁned and β < . Obviously β(0) = 0, t1 t2 = β(t1) β(t2) and ∞ ≤ ⇒ ≤

d(f(x1), f(x2)) β( x1 x2 ) x1, x2 X. (2.1) ≤ k − k ∀ ∈ It remains to ﬁnd a modulus of continuity α with β α. First, in order to show that β is continuous, let > 0. For X is compact, f ≤is uniformly continuous, and thus there exists δ > 0 with d(f(x), f(x0)) x x0 δ. ≤ 2 ∀k − k ≤

If t2 t1 δ and x1 x2 t2 we ﬁnd x0 , x0 X with x1 x0 δ, x2 x0 δ | − | ≤ k − k ≤ 1 2 ∈ k − 1k ≤ k − 2k ≤ and x0 x0 t1 (here we used the convexity of X). We then have k 1 − 2k ≤

d(f(x1), f(x2)) d(f(x1), f(x0 )) +d(f(x0 ), f(x0 )) + d(f(x0 ), f(x2)) ≤ 1 1 2 2 | ≤{z2 } | ≤{z2 } + d(f(x0 ), f(x0 )) ≤ 1 2 This shows

sup d(f(x1), f(x2)) sup d(f(x1), f(x2)) t2 t1 δ, x1 x2 t2 − x1 x2 t1 ≤ ∀| − | ≤ k − k≤ k − k≤ and hence β is continuous. Since X is compact there exist C and N, such that β [N, ] C. In the sequel | ∞ ≡ we will construct a sequence βn n N with the following properties: { } ∈ (1) βn : [0, ] [0, ] is continuous, increasing and βn [N, ] C ∞ → ∞ | ∞ ≡ (2) lim βn(0) = 0 n →∞

(3) βn(tx) tβn(x) x [0, ] 1 t ≤ ∀ ∈ ∞ ∀ ≤ (4) β βn ≤ C (5) βn βn 1 0 n , where 0 denotes the sup-norm. k − − k ≤ 2 k · k r 16 2. The Simplicity of Diﬀc(M) , dim(M) + 2 r ◦ ≤ ≤ ∞

Suppose we have such a sequence, then α := lim βn + IdR exists and is continuous, n since the sequence converges uniformly by (5).→∞ From the other properties we obtain, that α is a modulus of continuity, and β α. ≤ So it remains to show the existence of the sequence βn n N. Let β0 : C. There C { } ∈ ≡ exists x1 with β(x1) = 2 . We let C C β1 [x1, ] = β0 [x1, ], β1(x) := + x x [0, x1]. | ∞ | ∞ 2 2x1 ∀ ∈

C Inductively we choose xn with β(xn) = 2n and let C n C βn [xn, ] = βn 1 [xn, ], βn(x) := + x x [0, xn]. ∞ − ∞ n X i | | 2 i=1 2 x1 ∀ ∈ It is easily seen, that this sequence has all the desired properties. 2

2.10. Definition. Let f Cr(U, Rm), where U is an open subset of Rn. Then Drf : U Lr(Rn, Rm) denotes∈ the r-th derivative of f, where Lr(E,F ) denotes the vector→ space of r-linear mappings E E F . We say f is of class Cr,α, iff it is Cr and Drf is locally α-continuous.× · · By · × Remark→ 2.3 this doesn’t depend on r n m ,α the choice of a norm on L (R , R ). C∞ will mean C∞. A mapping of smooth manifolds f : M N is of class Cr,α, iff for every x M and for every chart (V, v) on N with f(x) →V , there exists a chart (U, u) on M∈ with x U, f(U) V and 1 r,α∈ ∈ ⊆ v f u− is C . We define ◦ ◦ Cr,α(M,N) := f : M N : f is Cr,α Cr(M,N) { → } ⊆ equipped with the initial topology with respect to the inclusion. By Remark 2.6 we have Cr+1(M,N) Cr,α(M,N) for all moduli of continuity α and 0 r < . If E π M is a smooth⊂ vector bundle over M we denote by ∀ ≤ ∞ → Γr,α(E π M) := Γr(E π M) Cr,α(M,E) → → ∩ the set of all Cr,α-sections of E π M. → 2.11. Lemma. Let M be a smooth manifold, E π M be a smooth vector bundle over M and V a finite dimensional vector space. Then→ Cr,α(M,V ) and Γr,α(E π M) are vector spaces. →

Proof. Everything follows immediately from Lemma 2.4. 2

2.12. Lemma. Let 1 r , M,N,P be smooth manifolds, g Cr,α(M,N) and f Cr,α(N,P ). Then ≤f g≤ ∞Cr,α(M,P ). ∈ ∈ ◦ ∈ Proof. Obviously we may assume g Cr,α(U, V ) and f Cr,α(V, Rp), where U resp. V are open subsets in Rm resp. Rn.∈ Recall the formulas∈

D(f g) = (Df g) Dg (2.2) ◦ ◦ · r 2. The Simplicity of Diﬀc(M) , dim(M) + 2 r 17 ◦ ≤ ≤ ∞

Dr(f g) = (Drf g) (Dg Dg) + (Df g) Drg ◦ ◦ · × · · · × ◦ · + C (Dif g) (Dj1 g Dji g) (2.3) X i,j1,...,ji 1integers independent of f and g. In the case r = 1, we have Dg and Df g are locally α-continuous (cf. Lemma 2.5). Since the dot in (2.2) is bilinear, we obtain◦ from Lemma 2.8, that D(f g) is locally α-continuous, and hence f g is Cr,α. ◦ In the◦ case r > 1 we use (2.3). By similar arguments one sees, that Drf g, Dg, Df g and Drg are locally α-continuous. Hence by Lemma 2.8 we have (D◦ r g) (Dg◦ Dg) and (Df g) Drg are locally α-continuous, for the dots in◦ (2.3)· are multi× · · · linear. × Each of the◦ remaining· summands of (2.3) is C1 and hence locally α-continuous by Remark 2.6. Thus, by Lemma 2.4, Dr(f g) is locally α-continuous, and hence f g is Cr,α. ◦ 2 ◦ r,α 1 1 2.13. Lemma. If 1 r, f C (M,N), and if f has a C inverse, then f − Cr,α(N,M). ≤ ∈ ∈

Proof. We assume once again, that f Cr,α(U, V ), where U and V are open subsets of Rn. Recall that we have ∈

1 1 D(f − ) = Inv Df f − , (2.4) ◦ ◦ 1 where Inv is the inversion of linear mappings, which is known to be C∞. Since f − r 1 r 1,α r 1,α is C , f − is C − , and Df is C − . So we obtain from (2.4), Lemma 2.12 and 1 r 1,α 1 r,α Lemma 2.5 (r = 1), that D(f − ) is C − , and thus f − is C . 2 2.14. Definition. Let M be a smooth manifold, E π M a smooth vector bundle, α a modulus of continuity, 1 r and K M an→ arbitrary subset. We then we define ≤ ≤ ∞ ⊆ Diffr,α (M) := Diffr (M) Cr,α(M,M) [c][K] [c][K] ∩ Γr,α (E π M) := Γr (E π M) Cr,α(M,E) [c][K] → [c][K] → ∩ and equip each of these spaces with the initial topology. (Here [c][K] means: option- ally c, K or nothing)

r,α 2.15. Lemma. In the situation of Definition 2.14 Diffc (M) is a subgroup of Diffr(M) and Γr,α(TM π M) is a linear subspaces of Γr(TM π M). The homeo- c c → c → morphism ϕ : U 0 U of Proposition 1.13 restricts to a homeomorphism → r,α π r,α ϕ : U 0 Γ (TM M) U Diff (M). ∩ → → ∩ r,α r,α Further Diffc (M) is locally contractible, and Diffc (M) consists exactly of those Cr,α-diffeomorphisms that are compactly supported Cr,α-diffeotopic◦ to Id.

Proof. Using Lemma 2.11, Lemma 2.12 and Lemma 2.13 we obtain the ﬁrst two r,α assertions. To see that ϕ : U 0 U restricts to C mappings as stated above, → r 18 2. The Simplicity of Diﬀc(M) , dim(M) + 2 r ◦ ≤ ≤ ∞

notice that ϕ = exp , exp is C∞ and use Lemma 2.12. The proof of Corollary 1.14 is exactly the same in∗ the Cr,α-case and hence we obtain the remaining assertions of Lemma 2.15 2

r,α n r+1 n 2.16. Example. Since Diffc (R ) is locally contractible and since Diffc (R ) = r+1 n r,α n ◦ Diffc (R ) (cf. Example 1.15) is dense in Diffc (R ) (cf. Proposition 1.7(6)), we r,α n r,α n r,α n r,α n obtain Diffc (R ) = Diffc (R ). Similarly we have DiffBn (R ) = DiffBn (R ). ◦ ◦ 2.17. Lemma. Let M be a connected, smooth manifold and let 1 r . Then ≤ ≤ ∞ Diffr(M) = Diffr,α(M) , (2.5) c [ c ◦ α ◦ where the union is taken over all moduli of continuity α.

Proof. This is trivial for r = , and hence we may assume r < . One inclusion ( ) is obvious, so it remains to∞ show the other one. ∞ ⊇ r r Let f Diffc(M) and let (M) be the covering from Lemma 1.16. By Claim 2 ∈ ◦ U r r on page 10 we obtain m N, U1,...,Um (M) and f1, . . . , fm Diffc(M) ∈ ∈ U ∈ ◦ with supp(fi) Ui and f = fm f1. Using the charts ϕi belonging to Ui, we have ∈1 r n ··· r fi := ϕi fi ϕi− DiffBn (R ) (cf. Example 1.15). Hence there exist C -diffeotopies ◦ e ◦ ◦ ∈ n Hi, 1 i m, supported on B with Hi( , 1) = fi. ≤ ≤ · e Dr(H): Rn I Lr(Rn I, Rn) × → × is continuous and supported on the convex, compact set Bn I. Thus, using r × Lemma 2.9, we obtain moduli of continuity αi, such that D Hi is αi-continuous, m 1 i m. Now let α := i=1 αi. This is again a modulus of continuity and r≤ ≤ P r,α D Hi is α-continuous 1 i m. Consequently H is a C -diffeotopy and hence r,α n ∀ ≤ ≤r,α r,α fi DiffBn (R ) . So fi Diffc (M) and thus f Diffc (M) . 2 e ∈ ◦ ∈ ◦ ∈ ◦ 2.18. Remark. In the next Chapter we will need the following improvement: Lemma 2.17 remains valid, if the union is only taken over all moduli of continuity satisfying α(tx) √tα(x) t 1 x [0, ). ≤ ∀ ≥ ∀ ∈ ∞ To see this let α := √α, where α is the modulus of continuity constructed in Lemma 2.17. Thene α is a modulus of continuity and e α(tx) = α(tx) tα(x) = √tα(x) t 1 x [0, ). q ≤ q ∀ ≥ ∀ ∈ ∞ e e r Since α α in a neighborhood of 0 D Hi are locally α-continuous, hence fi ≤ ∈ r,α n e r,α e e DiffBn (R ) , and thus f Diffc (M) . e ◦ ∈ e ◦ 2.19. Theorem (Mather, Epstein). [7] [4] Let M be a smooth manifold, n = r,α dim(M), n + 2 r and let α be a modulus of continuity. Then Diffc (M) is perfect, i.e. equals≤ its≤ own ∞ commutator group. ◦ r 2. The Simplicity of Diffc(M) , dim(M) + 2 r 19 ◦ ≤ ≤ ∞

It will take us the rest of this chapter to prove this theorem. We start with:

2.20. Deﬁnition. Let ϕ : G G0 be a homomorphism of groups. We deﬁne → H1(G) := G/[G, G], and let H1(ϕ): H1(G) H1(G0) be the induced homomor- → phism. H1 is a covariant functor from the category of groups to the category of abelian groups.

n 2.21. Lemma. Let M be a smooth manifold, ϕ : R , M a C∞-embedding, 1 r , and α a modulus of continuity. We define homomorphisms→ of groups r,≤[α] ≤ ∞r,[α] n r,[α] τϕ : Diffc (R ) Diffc (M) by ◦ → ◦ 1 n ϕ f ϕ− on ϕ(R ) f ( ◦ ◦ 7→ Id elsewhere

r,[α] r,[α] n r,[α] Then H1(τϕ ): H1(Diffc (R ) ) H1(Diffc (M) ) is surjective. ◦ → ◦ r,[α] Proof. Clearly everything is well defined, and τϕ is a homomorphism of groups. r,[α] r,α Let f Diffc (M) . By Claim 2 on page 10, which is proven similarly in the C - ∈ ◦ r r,[α] case, we obtain m N, V1,...,Vm (M) and f1, . . . , fm Diffc (M) , with ∈ ∈ U ∈ ◦ supp(fi) Vi and f = fm f1. Choose g1, . . . , gm Diffc∞(M) with gi(Vi) n ⊆ 1··· r,[α] ∈ ◦ ⊆ ϕ(R ). Thus we have gifig− Im(τ ) (cf. Example 2.16), and hence i ∈ ϕ 1 1 r,[α] [f] = [fm] [f1] = [gmfmg− ] ... [g1f1g− ] Im(H1(τ )). ··· m 1 ∈ ϕ 2

2.22. Remark. By Lemma 2.21 it suffice to show Theorem 2.19 in the case M = Rn. r,[α] n r,[α] For then we have H1(Diffc (R ) ) = e , and hence using the surjectivity of τϕ r,[α] ◦ { } r,[α] H1(Diffc (M) ) = e , but this is the perfectness of Diffc (M) . ◦ { } n ◦ Moreover it suffices to find a compact set K R with K◦ = and ⊆ 6 ∅ r,[α] n r,[α] n r,[α] n DiffK (R ) [Diffc (R ) , Diffc (R ) ]. ◦ ⊆ ◦ ◦ r,[α] n r,[α] For then given f Diffc (R ) and a compactly supported C -diffeotopy F from ∈ ◦ n 1 f to Id, we choose g Diffc∞(R ) with g(supp(F )) K◦ and obtain gfg− r,[α] n ∈ 1◦ r,[α] ⊆ 1 ∈ DiffK (R ) since (x, t) g(F (g− (x), t)) is a C -diffeotopy from gfg− to Id, supported on◦ K. Hence 7→

1 r,[α] n r,[α] n r,[α] n gfg− DiffK (R ) [Diffc (R ) , Diffc (R ) ], ∈ ◦ ⊆ ◦ ◦ r,[α] n and thus Diffc (R ) is perfect since ◦ 1 1 r,[α] n [f] = [g][f][g]− = [gfg− ] = [Id] H1(Diffc (R ) ). ∈ ◦

2.23. Definition. Let A 1. Choose ρA C∞(R, [0, 1]) with supp(ρA) = [ 2A ≥ ∈ n − − 1, 2A + 1] and ρA [ 2A,2A] 1. We definee ρA :C∞(R , [0, 1]) by ρA(ex1, . . . , xn) := | − ≡ e r 20 2. The Simplicity of Diffc(M) , dim(M) + 2 r ◦ ≤ ≤ ∞

n ρA(x1) ρA(xn). Then supp(ρA) = [ 2A 1, 2A + 1] and ρA [ 2A,2A]n 1. Let − ···ρA∂i n − − | ≡ τei,A := Fl1 e Diffc∞(R ) , and ∈ ◦ n ϕi,A n x R : xj < 2A + 1, j = i R { ∈ | | 6 } −→ ϕ (x , . . . , x ) := FlρA∂i (x ,..., 0, . . . , x ). i,A 1 n xi 1 n n Here ∂i denotes the unit vector field on R in the direction of the ith coordinate, X and Flt denotes the flow of the vector field X. Further we denote by Ti the unit ∂i translation in the direction of the ith coordinate, i.e. Ti := Fl1 . 2.24. Lemma. In this situation we have

(1) ϕi,A( x1 R xn ) x1 R xn . { } × · · · × × · · · × { } ⊆ { } × · · · × × · · · × { } (2) ϕi,A [ 2A,2A]n = Id | − n (3) ϕi,A Diﬀ∞(dom(ϕi,A), ( 2A 1, 2A + 1) ) ∈ − −

(4) ϕi,A∗ (ρA∂i) = ∂i

(5) τi,A ϕi,A = ϕi,A Ti. ◦ ◦ 1 n (6) pri ϕ− (x) pri(x) x ( 2A 1, 2A + 1) | ◦ i,A | ≥ | | ∀ ∈ − − i 1 n i (7) pri ϕi,A pr = pri ϕi,A x [ 2A, 2A] − R [ 2A, 2A] − ◦ ◦ i ◦ ∀ ∈ − × × − n f n n Here pri : R R and pr : R R are given by pri(x1, . . . , xn) := xi and → i → pri(x1, . . . , xn) := (0, . . . , xif,..., 0). f Proof. (1) and (2) are obvious. To see (3), recall that ϕi,A is C∞, supp(ρA) = n [ 2A 1, 2A + 1] , and thus by the uniqueness of integral curves ϕi,A : dom(ϕi,A) (−2A− 1, 2A + 1)n is bijective. → − − ∂ϕi,A = ∂ FlρA∂i (x ,..., 0, . . . , x ) xi 1 n ∂xi x ∂xi x = (ρ ∂ )(FlρA∂i (x ,..., 0, . . . , x )) = ρ (ϕ (x))∂ (2.6) A i xi 1 n A i,A i

∂ From (2.6) we obtain (pri ϕi,A) = pri(ρA(ϕi,A(x))∂i) = ρA(ϕi,A(x)) and thus, ∂xi x ◦ since prj ϕi,A = prj, j = i, we have det(Dϕi,A)(x)) = ρA(ϕi,A(x)) = 0. Hence ◦ 6 n 6 ϕi,A Diﬀ∞(dom(ϕi,A), ( 2A 1, 2A + 1) ). (4) follows from (2.6) and (5) is an ∈ − − immediate consequence of (4). From 0 ρA 1 we obtain (6). Because of the ≤ ≤ i 1 n i special choice of ρA we have ρA pr = ρA on [ 2A, 2A] − R [ 2A, 2A] − and ◦ i − × × − this implies (7). f 2

2.25. Definition. Let U be an open subset of Rn, 0 r < , α a modulus of continuity and let f Cr(U, Rm). Then we define ≤ ∞ ∈ r f r := sup D f(x) k k x U k k ≤ ∞ ∈ Drf(x) Drf(y) f r,α := sup k − k k k x=y U α( x y ) ≤ ∞ 6 ∈ k − k r 2. The Simplicity of Diffc(M) , dim(M) + 2 r 21 ◦ ≤ ≤ ∞ where Drf(x) denotes the norm of the multi linear mapping Drf(x) Lr(Rn, Rm), i.e. k k ∈ r D f(x) (v1, . . . , vr) Drf(x) = sup k · k n k k vj R v1 vr ∈ k k · · · k k 0 if 1 r and D f(x) = f(x) . Further we let µr(f) := f Id r, µr,α(f) := ≤ k k k k k − k f Id r,α and Mr(f) := max µ1(f), . . . , µr(f) . Finally we define the distance of twok − nonemptyk sets A, B Rn {to be } ⊆ dist(A, B) := inf x y . a A,b B ∈ ∈ k − k

2.26. Lemma. We list some useful properties of µr,[α] and r,[α], we’ll need in the sequel. k · k

n r m (1) µr(f) = f r r 2, U R open and f C (U, R ) k k ∀ ≥ ∀ ⊆ ∀ ∈ n 1 m (2) µ1(f) f 1 + 1, f 1 µ1(f) + 1 U R open and f C (U, R ) ≤ k k k k ≤ ∀ ⊆ ∀ ∈ n r m (3) µr,α(f) = f r,α α, r 1, U R open and f C (U, R ) k k ∀ ∀ ≥ ∀ ⊆ ∀ ∈ r+1 n m n r (4) Let 0 r < and f C (R , R ). If there exists y0 R with D f(y0) = 0 then≤ ∞ ∈ ∈ n r f r sup distx, y R : D f(y) = 0 f r+1. k k ≤ x Rn { ∈ } k k ∈ n r If there exists y0 R with D (f Id)(y0) = 0 then ∈ − n r µr(f) sup distx, y R : D (f Id)(y) = 0 µr+1(f). ≤ x Rn { ∈ − } ∈ r n m n r (5) Let 0 r < , f C (R , R ), suppose there exists y0 R with D f(y0) = ≤ ∞ ∈ n r ∈ 0, C := supx Rn distx, y R : D f(y) = 0 < , and choose N N with ∈ { ∈ } ∞ ∈ α( C ) 1. Then N ≤ f r N f r,α. k k ≤ k k n r n If there exists y0 R with D f(y0) = 0, C0 := supx Rn distx, y R : ∈ ∈ { ∈ r C0 D (f Id)(y) = 0 < and α( N ) 1 then − } ∞ 0 ≤

µr(f) N 0µr,α(f). ≤ (6) Let 0 r < , f Cr+1(Rn, Rm) and suppose there exists a closed subset ≤n ∞ ∈r A R such that D f A is constant. Then ⊆ | 1 f r,α 1 + 2 sup dist(x, A) f r+1. k k ≤ α(1) x Rn k k ∈ r If D (f Id) A is constant then − | 1 µr,α(f) 1 + 2 sup dist(x, A)µr+1(f). ≤ α(1) x Rn ∈ r 22 2. The Simplicity of Diﬀc(M) , dim(M) + 2 r ◦ ≤ ≤ ∞

Proof. (1), (2) and (3) follow immediately from IdRn 1 = 1 and IdRn r = 0 if r 2. To see (4) let x Rn, and choose z Rnkwith kDrf(z) = 0k and xk z = ≥ ∈ ∈ k − k dist x, y Rn : Drf(y) = 0 . Then we have { ∈ } Drf(x) = Drf(x) Drf(z) k k k 1 − k r+1 = Z D f(tx + (1 t)z) (z x)dt k 0 − · − k f r+1 x z , ≤ k k k − k and thus we get

r n r f r = sup D f(x) sup distx, y R : D f(y) = 0 f r+1. k k x Rn k k ≤ x Rn { ∈ } k k ∈ ∈ The second assertion in (4) is an immediate consequence of the ﬁrst one. Too see (5) let x Rn, choose z Rn with Drf(z) = 0 and x z = dist x, y ∈ ∈ k − k { ∈ n r x z R : D f(y) = 0 . and let xi := z + i − , i = 0,...,N. Then } N r r r D f(x) = D f(xN ) D f(x0) k k k − k N r r D f(xi) D f(xi 1) − X k − k ≤ i=1 α( xi xi 1 ) k − − k N f r,α, ≤ k k x z C where we used α( xi xi 1 ) = α( k − k ) α( ) 1 for the ﬁrst inequality. Since k − − k N ≤ N ≤ N doesn’t depend on x we obtain f r N f r,α. The second assertion of (5) is now obvious. k k ≤ k k Let f and A as assumed in (6). If x, y Rn with x y 1 then we can estimate as follows ∈ k − k ≤

r r D f(x) D f(y) f r+1 x y f r+1 k − k k k k − k k k , (2.7) α( x y ) ≤ α( x y ) ≤ α(1) k − k k − k t since t α(t) is increasing (cf. Remark 2.2). If x y > 1 we choose x0, y0 A 7→ k − k r r ∈ with x x0 = dist(x, A) and y y0 = dist(x, A). Since D f(x0) = D f(y0) we obtaink − k k − k

r r r r r r D f(x) D f(y) D f(x) D f(x0) + D f(y0) D f(y) k − k k − k k − k α( x y ) ≤ α(1) k − k f r+1 k k ( x x0 + y y0 ) ≤ α(1) k − k k − k

2 f r+1 k k sup dist(z, A) (2.8) ≤ α(1) z Rn ∈ Estimations (2.7) and (2.8) prove the ﬁrst assertion of (6), the second follows immediately. 2 r 2. The Simplicity of Diﬀc(M) , dim(M) + 2 r 23 ◦ ≤ ≤ ∞

2.27. Remark. Let K Rn be a closed subset satisfying ⊆ n CK := sup distx, R K < , x K \ ∞ ∈ let α be a modulus of continuity, 0 r < and let f be a diﬀeomorphism supported on K. Then, by Lemma 2.26, there≤ exists∞ a constant 0 < C < , depending only on ∞ CK and α with

µr(f) Cµr+1(f) µr(f) Cµr,α(f) µr,α(f) Cµr+1(f), ≤ ≤ ≤ provided f is of suﬃciently high diﬀerentiability class.

2.28. Deﬁnition. An admissible polynomial is a polynomial whose coeﬃcients are nonnegative and which has no constant or linear terms.

2.29. Lemma.

1 (1) If f1, . . . , fk, 1 k, are C -mappings between open subsets of Euclidean spaces, ≤ such that the composition f1 fk is defined, then ··· k 1 − µ1(f1 fk) k sup µ1(fi)1 + sup µ1(fi) . ··· ≤ 1 i k 1 i k ≤ ≤ ≤ ≤ (2) If r 2 then there exists an admissible polynomial F of one variable with ≥ r(k 1) − µr(f1 fk) k sup µr(fi)1 + sup µ1(fi) + F ( sup Mr 1(fi)) ··· ≤ 1 i k 1 i k 1 i k − ≤ ≤ ≤ ≤ ≤ ≤ r for all C -mappings f1, . . . , fk between open subsets of Euclidean spaces, such that the composition f1 fk is defined. ··· 1 n 1 (3) If f Diff (R ) and µ1(f) then ∈ ≤ 2 1 µ1(f − ) 2µ1(f) ≤

(4) If r 2, then there exists an admissible polynomial F 0 of one variable, such that ≥ 1 r+1 µr(f − ) µr(f)1 + 2µ1(f) + F 0(Mr 1(f)) ≤ − r n 1 f Diﬀ (R ) with µ1(f) . ∀ ∈ ≤ 2

Proof. We will prove (1) by induction on k. For k = 1 this is obvious, so suppose (1) is proven for k 1. Then we have −

D(f1 fk 1fk)(x) Id = k ··· − − k = D(f1 fk 1)(fk(x))Dfk(x) Id k ··· − − k D(f1 fk 1)(fk(x)) IdDfk(x) + Dfk(x) Id ≤ k ··· − − k k − k r 24 2. The Simplicity of Diﬀc(M) , dim(M) + 2 r ◦ ≤ ≤ ∞

µ1(f1 fk 1) fk 1 +µ1(fk) ≤ ··· − k k 1+µ1(fk) ≤| {z } k 2 − (k 1) sup µ1(fi)1 + sup µ1(fi) 1 + µ1(fk) + µ1(fk) ≤ − 1 i k 1 1 i k 1 ≤ ≤ − ≤ ≤ − k 1 − (k 1) sup µ1(fi)1 + sup µ1(fi) + sup µ1(fi) ≤ − 1 i k 1 i k 1 i k ≤ ≤ ≤ ≤ ≤ ≤ k 1 − k sup µ1(fi)1 + sup µ1(fi) ≤ 1 i k 1 i k ≤ ≤ ≤ ≤ and thus (1) is proven. We will prove (2) by induction on k too. For k = 1 it is again obvious and hence we assume it is yet shown for all i k 1. Then, using (2.3) and (1), we obtain ≤ −

r D (f1 fk)(x) k ··· k ≤ r f1 fk 1 r fk 1 + f1 fk 1 1 fk r ≤ k ··· − k k k k ··· − k k k + Ci,j1,...,ji f1 fk 1 i fk j1 ... fk ji X − 1

f1 fk 1 i = k ··· − k = µi(f1 fk 1) − ··· i(k 1) − (i 1) sup µi(fl)1 + sup µ1(fl) + F3( sup Mi 1(fl)) ≤ − 1 l k 1 1 l k 1 1 l k 1 − ≤ ≤ − ≤ ≤ − ≤ ≤ − i(k 1) − (i 1) sup Mr 1(fl)1 + sup Mr 1(fl) + F3( sup Mr 1(fl)), ≤ − 1 l k − 1 l k − 1 l k − ≤ ≤ ≤ ≤ ≤ ≤ r 2. The Simplicity of Diﬀc(M) , dim(M) + 2 r 25 ◦ ≤ ≤ ∞ and thus F really doesn’t have any constant or linear terms. If f Diﬀ1(Rn) and ∈ µ1(f) < 1 we have m (Df(x)) 1 = ∞ (Df(x) Id) − X m=0 − − and thus

1 1 1 1 µ1(f − ) = sup D(f − Id)(x) = sup (Df)− (f − (x)) Id x Rn k − k x Rn k − k ∈ ∈ ∞ 1 m = sup (Df(f − (x)) Id) x Rn k X − − k ∈ m=1 ∞ sup (Df(x) Id) m ≤ X x Rn k − k m=1 ∈ ∞ m 1 µ1(f) µ1(f) = 1 = 2µ1(f), ≤ X 1 µ1(f) − 1 µ1(f) ≤ m=1 − − 1 provided µ1(f) . We will prove (4) by induction on r. From (2.3) we get ≤ 2 r 1 r 1 1 r 1 r 1 0 = D (ff − ) = (D f f − )D(f − ) + (Df f − )D (f − ) ◦ ◦ + C (Dif) Dj1 (f 1) Dji (f 1) X i,j1,...,ji − − 1

:=R 1 | {zf,f− } and thus

r 1 1 r 1 1 r 1 D (f − ) = D(f − )(D f f − )D(f − ) D(f − )R 1 . (2.9) − ◦ − f,f − 1 If r = 2 and µ1(f) we obtain from (2.9) and (3) ≤ 2 2 1 1 1 2 D (f − )(x) f − 1 f 2 f − k k ≤ k k k k k k1 1 3 3 µ2(f) 1 + µ1(f − ) µ2(f) 1 + 2µ1(f) , ≤ ≤ this shows (4) in the case r = 2. Now suppose the formula in (4) above is proven for all j r 1. Using once again (2.9) and (3) we get ≤ − r 1 D (f − )(x) k k ≤ f 1 f f 1 r + f 1 C f f 1 f 1 − 1 r − 1 − 1 X i,j1,...,ji i − j1 − ji ≤ k k k k k k k k 1

where F 0 is an admissible polynomial of one variable (here we used that ji < r, and that there exists at least one jl 2). 2 0 ≥ 2.30. Lemma. Let 1 r < , α a modulus of continuity and let K Rn be a closed subset, satisfying≤ ∞ ⊆ n CK := sup distx, R K < . x Rn \ ∞ ∈ (1) There exist δ > 0 and C > 0 depending on n, r, α and CK , such that

µr,α(f g) µr,α(f) + µr,α(g) + Cµr,α(f)µr,α(g) ◦ ≤ r,α n 1 1 f, g Diff (R ) with µr,α(f), µr,α(g) δ and D (f Id) Rn K = D (g ∀ ∈ ≤ − | \ − Id) Rn K 0. | \ ≡ (2) Given λ > 1 there exists δ > 0 depending on n, r, α and CK , such that 1 µr,α(f − ) λµr,α(f) ≤ r,α n 1 f Diff (R ) with µr,α(f) δ and D (f Id) Rn K 0. ∀ ∈ ≤ − | \ ≡ Proof. In view of (2.3) we take f, g Cr,α(Rn, Rn) and estimate ∈ x i j1 ji (D f g)(D g D g) k ◦ × · · · × y k α( x y ) ≤ k x− k i (D f g) Dif(g(y)) x k ◦ y k jl jl D g(x) + k k D g ≤ α( x y ) Y k k α( x y ) Y k y k 1 l i 1 l i k − k ≤ ≤ k − k ≤ ≤ x jl α( g(x) g(y) ) D 0 g µ (f) g + f k y k 2 g i,α k − k Y jl i Y jl ≤ α( x y ) 1 l i k k k k α( x y ) 1 l i k k k − k ≤ ≤ k − k l≤=l≤ 6 0 α( g 1 x y ) i 1 µi,α(f) k k k − k g j + 2 − f iµj ,α(g) g j Y l l0 Y l ≤ α( x y ) 1 l i k k k k 1 l i k k k − k ≤ ≤ l≤=l≤ 6 0 i 1 µi,α(f) µ1(g) + 1 g j + 2 − f iµj ,α(g) g j (2.10) Y l l0 Y l ≤ 1 l i k k k k 1 l i k k ≤ ≤ l≤=l≤ 6 0 where 1 l0 i can be chosen arbitrarily. Together with (2.3) and Lemma 2.26 this yields≤ ≤ r r 1 r 1 µr,α(f g) µr,α(f) µ1(g) + 1 g + 2 − f rµ1,α(g) g − ◦ ≤ k k1 k k k k1 + µ1,α(f) µ1(g) + 1 g r + f 1µr,α(g) k k k k + C µ (f)(µ (g) + 1) g X i,j1,...,ji i,α 1 Y jl 2 i r 1,1 ji 1 l i k k ≤j +≤ −+j =≤r ≤ ≤ 1 ··· i +2i 1 f µ (g) g − i j1,α Y jl k k 2 l i k k ≤ ≤ µr,α(f) + µr,α(g) + C1µr,α(f)µr,α(g) ≤ r 2. The Simplicity of Diffc(M) , dim(M) + 2 r 27 ◦ ≤ ≤ ∞ provided µr,α(f), µr,α(g) are sufficiently small. Here C1 is a constant, depending only on CK , n, r and α. This shows (1) of Lemma 2.30. In order to prove (2) we first assume r = 1. Then if µ1,α(f) is sufficiently small we have µ1(f) < 1 (cf. Lemma 2.26) and thus

m (Df) 1(x) = ∞ (Df(x) Id) . − X m=0 − − Hence

1 1 1 D(f − )(x) D(f − )(y) µ1,α(f − ) = sup k − k x=y Rn α( x y ) 6 ∈ k − k x 1 1 (Df)− f − 1 1 k ◦ y k α( f − (x) f − (y) ) = sup 1 1 k − k x=y Rn α( f − (x) f − (y) ) α( x y ) 6 ∈ k x − k k − k 1 (Df)− 1 k y k α( f − 1 x y ) sup sup k k k − k ≤ x=y Rn α( x y ) x=y Rn α( x y ) 6 ∈ k − k 6 ∈ k − k 1 µ1(f )+1 √λ | ≤ −{z ≤ } m m ∞ (Df(x) Id) (Df(y) Id) √λ sup k − − − k ≤ x=y Rn X α( x y ) 6 ∈ m=1 k − k x m 1 m k 1 k k=0− (Df(x) Id) − − Df (Df(y) Id) √λ ∞ P k − kk y kk − k ≤ X α( x y ) m=1 k − k √λ ∞ m µ (f)m 1 µ (f) λµ (f) X 1 − 1,α 1,α ≤ m=1 · ≤

√λ | ≤{z } provided µ1,α(f) is small (cf. Lemma 2.26 and Lemma 2.29(3)). Now suppose r 2. Recall formula (2.9) and look at ≥

x 1 r 1 1 r D(f − )(D f f − )(D(f − )) k ◦ y k = α( x y ) 1 k − k 1 D(f − )(x) D(f − )(x) 1 r k − k f r f − ≤ α( x y ) k k k k1 k r − k1 r 1 1 D f(f − (x)) D f(f − (x)) 1 r + f − 1 k − k f − k k α( x y ) k k1 k − k r D(f 1)(x) D(f 1)(y) 1 − − + f − 1 f r k − k k k k k α( x y ) k − k 1 1 1 1 r 1 r+1 α(f − (x) f − (y)) µ1,α(f − ) f r f − + f − µr,α(f)k − k ≤ k k k k1 k k1 α( x y ) r 1 k − k 1 1 1 − + f − 1 f rµ1,α(f − ) 2 f − 1 k k k k k k r 28 2. The Simplicity of Diﬀc(M) , dim(M) + 2 r ◦ ≤ ≤ ∞

1 1 r 1 r+1 1 µ1,α(f − ) f r f − + f − µr,α(f)(µ1 f − ) + 1 ≤ k k k k1 k k1 r 1 1 r 1 +2 − f − f rµ1,α(f − ) k k1k k λ 1 1 + − ! µr,α(f) (2.11) ≤ 2

>1 | {z }

provided µr,α(f) is suﬃciently small (cf. Lemma 2.29(3) and Lemma 2.26). Further we have

x 1 D(f )R 1 − f,f − k y k α( x y ) ≤ k − k x x 1 1 D(f ) R 1 (x) D(f )(y) R 1 − f,f − − f,f − k y kk k + k kk y k ≤ α( x y ) α( x y ) k − k k − k µ (f 1) C f f 1 1,α − X i,j1,...,ji i Y − jl ≤ 10 | {z } provided µr,α(f) is suﬃciently small (cf. Lemma 2.29(3)&(4) and Lemma 2.26). Us- ing (2.9) (2.11) and (2.12) we obtain δ > 0 depending on n, r, α and CK with

r 1 r 1 1 D (f − )(x) D (f − )(y) µr,α(f − ) = sup k − k x=y Rn α( x y ) 6 ∈ k − k λ 1 λ 1 1 + − ! µr,α(f) + − ! µr,α(f) = λµr,α(f) ≤ 2 2

r n 1 f Diﬀ (R ) with µr,α(f) δ and D (f Id) Rn K 0. 2 ∀ ∈ ≤ − | \ ≡ 2.31. Corollary. Let K Rn be a compact, convex subset, 1 r < and let α be a modulus of continuity.⊆ Then ≤ ∞

(f, g) f g r,α 7→ k − k r,α n r,α is a metric on DiffK (R ). The induced topology is called C -topology. It is finer r r+1 r+1 n r,α n than the C -topology and coarser than the C -topology on DiffK (R ). DiffK (R ) equipped with the Cr,α-topology is a connected, topological group. r 2. The Simplicity of Diffc(M) , dim(M) + 2 r 29 ◦ ≤ ≤ ∞

r,α n r Proof. Let f, g DiﬀK (R ). Since D (f g) is locally α-continuous, f g Rn K 0 and since K is∈ compact, the set − − | \ ≡

Drf(x) Drf(y) (k − k : x, y Rn) α x y ∈ k − k is bounded, and thus f g r,α < . Obviously f g r,α = g f r,α and k − k ∞ k r,α− nk k − k f h r,α f g r,α + g h r,α for all f, g, h DiffK (R ). Now suppose f, g k −r,α k n ≤ k − k k − k r ∈ ∈ DiffK (R ) and f g r,α = 0. Then D (f g) is constant, and since f g Rn K = 0 k − k − − | \ this yields f = g, and thus we have shown r,α is a metric. By Lemma 2.26(5) the induced topology is finer than the Cr-topologyk · − · k and by Lemma 2.26(6) it is coarser r+1 r+1 n r,α n than the C -topology on DiffK (R ). If f, f 0, g DiffK (R ) we obtain from (2.3) and (2.10) ∈

fg f 0g r,α = (f f 0) g r,α = µr,α((f f 0) g) k − k k − ◦ k − ◦ ≤ C µ (f f ) 1 + µ (g) g X i,j1,...,ji i,α 0 1 Y jl ≤ 1 i r,1 jl − 1 l i k k j ≤+≤ +j≤=r ≤ ≤ 1 ··· i +2i 1 f f µ (g) g − 0 i j1,α Y jl k − k 2 l i k k ≤ ≤ r+1 r 1 − C f f 0 r,α 1 + µr,α(g) + f f 0 r,αµr,α(g) 1 + µr,α(g) ! ≤ k − k k − k where C is a constant depending on n, r, α and CK (cf. Lemma 2.26). Thus r,α n r,α n r,α comp( , g) : DiffK (R ) DiffK (R ) is continuous with respect to C -topology · r,α n → for all g DiffK (R ). Similarly one shows that comp(g, ) is continuous for all g Diffr,α∈(Rn). By Lemma 2.30(1) comp is continuous at (Id,· Id), and hence ∈ K

1 1 comp = comp(f0, ) comp( , g0) comp comp(f − , ) comp( , g− ) · ◦ · ◦ ◦ 0 · × · 0 is continuous at (f0, g0). Since f0 and g0 were arbitrary this yields: comp is continuous with respect to Cr,α-topology. 1 From Lemma 2.30(2) we obtain immediately inv : f f − is continuous at Id. r,α n 7→ If f0 Diff (R ) we have ∈ K 1 1 inv = comp(f − , ) inv comp( , f − ) 0 · ◦ ◦ · 0 r,α n r,α and inv is continuous at f0. Consequently DiffK (R ) together with the C -topology is a topological group, and it remains to show its connectedness. Suppose f Diffr,α(Rn) is C1 near Id. Then ∈ K Rn I H Rn × → (x, t) Ht(x) := tf(x) + (1 t)x 7→ − r 30 2. The Simplicity of Diffc(M) , dim(M) + 2 r ◦ ≤ ≤ ∞

is a Cr,α-diﬀeotopy to Id, supported on K. Further we have

r r D (Ht Ht0 )(x) D (Ht Ht0 )(y) Ht Ht0 r,α = sup k − − − k k − k x=y Rn α( x y ) 6 ∈ r k − k r (t t0)D f(x) (t t0)D f(y) = sup k − − − k x=y Rn α( x y ) 6 ∈ k − k t t0 f r,α ≤ | − |k k ˇ r,α n r,α and thus H : I DiffK (R ) is continuous with respect to C -topology. Since r+1 n → r n r+1 n DiffK (R ) is dense in DiffK (R ) (cf. Proposition 1.7(6)) and since DiffK (R ) is r,α n connected (cf. Example 1.15) this shows that DiffK (R ) is connected. 2

i 1 1 n i 1 1 2.32. Definition. We let Ci := R − S R − , where we think of S as S = R/Z. n × × n n 1 ∼ Let πi : R Ci be the covering projection, and let pi : R R − resp. pi : Ci n 1 → → → R − be the projections, which omit the ith coordinate.e Clearly pi πi = pi. n ◦ The mapping πi : R Ci gives us a preferred system of coordinatese in a → neighborhood of any point of Ci. The transition mappings between these coordinate r r systems are all translations and hence if r 1 and f C (Ci,Ci) then D f : r n n ≥ ∈ Ci L (R , R ) is defined independently of the choice of one of these preferred → charts. Thus the (semi)norms introduced in Definition 2.25 do make sense on Ci too, provided r 1. ≥ n 1 Let A 1. If supp(f) [ 2A, 2A] and µ0(f) 2 , we define Γi,A(f): Ci Ci ≥ ⊆ − n ≤ → by: Given θ Ci we choose x R with πi(x) = θ and pri(x) < 2A. Then we ∈ ∈ N − choose N N such that pri((Tif) (x)) > 2A, and let ∈ N Γi,A(f)(θ) := πi((Tif) (x)), which is independent of the choice of N and x, since supp(f) [ 2A, 2A]n. ⊆ − 1 n 2.33. Lemma. [7] [4] There exists a neighborhood U of Id Diffc (R ) such that ∈ ◦

(1) Γi,A(Id) = Id A 1 1 i n ∀ ≥ ∀ ≤ ≤

(2) Let 2A ai 2, 1 i n, and supp(f) [ a1, a1] [ an, an]. Then ≥ ≥ ≤ ≤ 1 ⊆ − × · · · × − supp(f) [ a1, a1] S [ an, an]. ⊆ − × · · · × × · · · × − (3) Let α be a modulus of continuity, r 1, A 1 and 1 i n. Then ≥ ≥ ≤ ≤ r,[α] n r,[α] Γi,A : Diﬀ[ 2A,2A]n (R ) U Diﬀ[ 2A,2A]i 1 S1 [ 2A,2A]n i (Ci) − ∩ → − − × × − − ◦ is continuous with respect to Cr-topologies.

(4) If r 1, A 1 and if α is a modulus of continuity, then there exists δ > 0, depending≥ on≥n, r, α and A, such that

µr,α(Γi,A(f)) 9Aµr,α(f) ≤ r 2. The Simplicity of Diﬀc(M) , dim(M) + 2 r 31 ◦ ≤ ≤ ∞

r,α n 1 i n, f Diﬀ[ 2A,2A]n (R ) U with µr,α(f) δ, and ∀ ≤ ≤ ∀ ∈ − ∩ ≤

µr,α(Γi,A(f)) 9µr,α(f) ≤ r,α n 1 i n, f Diﬀ[ 2,2]i [ 2A,2A]n i (R ) U with µr,α(f) δ. ∀ ≤ ≤ ∀ ∈ − × − − ∩ ≤ (5) If A 1 and r 2 then there exists an admissible polynomial of one variable F , such≥ that ≥

18Ar µr(Γi,A(f)) 18Aµr(f)(1 + µ1(f)) + F (Mr 1(f)) (2.13) ≤ − r n f Diﬀ[ 2A,2A]n (R ) U, and ∀ ∈ − ∩ 18A µ1(Γi,A(f)) 18Aµ1(f)(1 + µ1(f)) (2.14) ≤ 1 n f Diﬀ[ 2A,2A]n (R ) U. ∀ ∈ − ∩

r n 1 Proof. (1) and (2) are obvious. If f Diff[ 2A,2A](R ) , then (Γi,A(f))− is con- ∈ − n ◦ structed similar to Γi,A(f), but now choose x R with pri(x) > 2A, N such that N 1 ∈ N pri((Tif)− (x)) < 2A, and let Γi,A(f)− (θ) := πi((Tif)− (x)). − Since composition is continuous Γi,A is continuous. If we can choose U, such r,[α] n that Diff[ 2A,2A]n (R ) U is connected 1 r , A 1 and for all moduli of − ∩ ∀ ≤ ≤ ∞ ∀ ≥ continuity α, then (3) follows immediately. So we choose 0 < 1 , such that ≤ 2 1 n µ0(f) < , µ1(f) < = f Diff (R ), ⇒ ∈ and let 1 n U := f Diff (R ): µ0(f), µ1(f) < . { ∈ c } r,[α] n If f Diff[ 2A,2A]n (R ) U, it is easily seen that H(x, t) := tf(x) + (1 t)x is ∈ − ∩ − r,[α] r,[α] n r,[α] n a C -diffeotopy in Diff[ 2A,2A]n (R ) U to Id, and thus Diff[ 2A,2A]n (R ) U is connected. To see (4), we− first show the∩ following − ∩ Sublemma. Given N N there exists δ > 0, depending on n, r, α, A and N, such that ∈ N µr,α((Tif) ) (N + 1)µr,α(f) ≤ r,α n f Diff[ 2A,2A]n (R ) with µr,α(f) δ. ∀ ∈ − ≤

Proof of the Sublemma. By Lemma 2.30(1) there exist δ1 > 0 and C1 > 0, such that

µr,α(fg) µr,α(f) + µr,α(g) + C1µr,α(f)µr,α(g) (2.15) ≤ r,α n 1 1 f, g Diﬀ (R ) with µr,α(f), µr,α(g) δ1, D (f Id) Rn K = D (g Id) Rn K = ∀0, where∈ K := [ 2A N, 2A + N]n. Now≤ choose− > |0 with\ 1 + (1− + )| + ...\ + N 1 − − δ1 (1 + ) − N + 1, and let δ := min δ1, , . From µr,α(f) = µr,α(Tif) and N+1 C1 induction on≤m we get { }

m m 1 µr,α((Tif) ) (1 + (1 + ) + ... + (1 + ) − )µr,α(f) ≤ r 32 2. The Simplicity of Diﬀc(M) , dim(M) + 2 r ◦ ≤ ≤ ∞

r,α n 1 m N and f Diff[ 2A,2A]n (R ) with µr,α(f) δ. In the case m = N this is ∀ ≤ ≤ ∀ ∈ − ≤ N N 1 µr,α((Tif) ) (1 + (1 + ) + ... + (1 + ) − )µr,α(f) (N + 1)µr,α(f) ≤ ≤ r,α n f Diff[ 2A,2A]n (R ) with µr,α(f) δ. 2 ∀ ∈ − ≤ Choose N N with 8A + 3 > N > 8A + 1 and use the Sublemma above to obtain ∈ N µr,α(Γi,A(f)) µr,α((Tif) ) (N + 1)µr,α(f) 9Aµr,α(f) ≤ ≤ ≤ r,α n f Diff[ 2A,2A]n (R ) U with µr,α(f) δ, and ∀ ∈ − ∩ ≤ N 8 µr,α(Γi,A(f)) µr,α((Tif) ) = µr,α((Tif) ) 9µr,α(f) ≤ ≤ r,α n f Diff[ 2,2]i [ 2A,2A]n i (R ) U with µr,α(f) δ. ∀ ∈ − × − − ∩ ≤ To see (5) choose an integer N with 8A + 1 < N < 8A + 3. Since µr(Ti) = 0 for r 1, we obtain from Lemma 2.29(1) ≥ N 2N 1 µ1(Γi,A(f)) = µ1((Tif) ) 2Nµ1(f)(1 + µ1(f)) − ≤ 18A 18Aµ1(f)(1 + µ1(f)) , ≤ and this is (2.14). To see (2.13) we estimate

N r(2N 1) µr(Γi,A(f)) = µr((Tif) ) 2Nµr(f)(1 + µ1(f)) − + F (Mr 1(f)) − ≤ 18Ar 18Aµr(f)(1 + µ1(f)) + F (Mr 1(f)), ≤ − where F is the admissible polynomial of Lemma 2.29(2). This ﬁnishes the proof of (5). 2

1 1 2.34. Deﬁnition. Consider the following S -action S Ci Ci × →

β (x1, . . . , ϑ, . . . , xn) := (x1, . . . , β + ϑ, . . . , xn) · r,[α] r,[α] and let Gi denote the group of equivariant C -diﬀeomorphisms of Ci,

r,[α] r,[α] 1 G := f Diﬀ (Ci): f(β θ) = β f(θ) β S θ Ci . i { ∈ · · ∀ ∈ ∀ ∈ }

r,[α] n 1 n 1 2.35. Remark. If f Gi then there exists a mapping f 0 : R − R − with ∈ 1 r,[α] → pi f = f 0 pi, for f is S -equivariant. f 0 is C , since there exists a global cross ◦ n ◦1 1 1 r,[α] section R − Ci. The inverse of f 0 is given by (f 0)− = (f − )0 and is C by a → r,[α] n 1 similar argument. Thus f 0 Diff (R − ). ∈ 2.36. Lemma (Mather). [7] Let α be a modulus of continuity, 1 r and A 1 n ≤ ≤ ∞ ≥ 1. Then there exists a neighborhood U 0 of Id Diffc (R ) depending on n, r, α and A, r,∈[α] n 1 such that the following holds: If f, g Diff[ 2A,2A]n (R ) U 0 and Γi,A(f)Γi,A(g)− ∈ − ∩ ∈ r,[α] r,[α] n Gi then τi,Af and τi,Ag are conjugated in Diffc (R ) . ◦ r 2. The Simplicity of Diffc(M) , dim(M) + 2 r 33 ◦ ≤ ≤ ∞

Proof. First we assume U 0 U (cf. Lemma 2.33), thus Γi,A is well defined on n ⊆ n n U 0. Now we define Λ : R R by: Given x R choose N N such that N → N ∈ N ∈ 1 pri((Tif)− (x)) < 2A and let Λ(x) := (Tig) (Tif)− (x). Since µ0(f) 2 such an N exists and since supp(− f), supp(g) [ 2A, 2A]n, Λ doesn’t depend on the≤ choice of N. Locally N can be chosen constant,⊆ so− we have Λ Cr,[α](Rn, Rn). The inverse of ∈ N Λ is constructed similarly, but now choose N N such that pri((Ti g)− (x)) < 2A 1 N N ∈ r,[α]◦ n − and set Λ− (x) := (Tif) (Tig)− (x). Hence we obtain Λ Diff (R ). Notice that Λ depends on i, A, f and g, but we omit the indices, since∈ Λ is used solely in this proof. Claim 1. We have

1 (1) ΛTifΛ− = Tig

n (2) Λ = Id on x R : j = i : prj(x) 2A or pri(x) 2A { ∈ ∃ 6 | | ≥ ≤ − } 1 n 1 (3) Γi,A(g)(Γi,A(f))− πi = πi Λ on x R : pri(x) > 2A + ◦ ◦ { ∈ 2 }

n N Proof of Claim 1. To see (1), let x R and N N such that pri((Tig)− (x)) < 2A. Then we have ∈ ∈ − 1 N N ΛTifΛ− (x) = ΛTif(Tif) (Tig)− (x) N+1 (N+1) N+1 N = (Tig) (Tif)− (Tif) (Tig)− (x)

= Tig(x).

(2) is an immediate consequence of supp(f), supp(g) [ 2A, 2A]n. In order to proof ⊆ − (3), we have to shrink U 0, such that

N N pri((Tig) (Tif)− (x)) > 2A (2.16)

n 1 x R with pri(x) > 2A + 2 , f, g U 0 and N N. To achieve this, just require ∀ ∈ 1 1 ∀ ∈ n ∀ ∈ 1 µ0(h) < 2 16A for all h U 0. Now let x R with pri(x) > 2A+ 2 and choose N N N ∈ ∈ 1 N∈ with pri((Tif)− (x)) < 2A. We then have Γi,A(f)− (πi(x)) = πi((Tif)− (x)) − 1 N N and hence we obtain from (2.16) Γi,A(g)Γi,A(f)− (πi(x)) = πi((Tig) (Tif)− (x)) = πi(Λ(x)). 2 Let x ( 2A 1, 2A + 1)n. Using Claim 1(1)&(2), supp(f), supp(g) [ 2A, 2A]n and Lemma∈ − 2.24− we get ⊆ −

1 1 1 1 1 1 1 1 (ϕi,AΛϕi,A− )τi,Af(ϕi,AΛϕi,A− )− (x) = ϕi,AΛϕi,A− ϕi,ATiϕi,A− ϕi,Afϕi,A− ϕi,AΛ− ϕi,A− (x) 1 1 1 = ϕi,AΛTifΛ− ϕi,A− = ϕi,ATigϕi,A− (x) 1 1 = ϕi,ATiϕi,A− ϕi,Agϕi,A− = τi,Ag(x).

1 r,[α] n Suppose ϕi,AΛϕi,A− could be extended to a map λ Diffc (R ) . We are finished ∈ ◦ then, for τi,A Rn ( 2A 1,2A+1)n = Id, f Rn [ 2A,2A]n = g Rn [ 2A,2A]n = Id and hence 1 | \ − − n | \ − | \ − λτi,Afλ− = τi,Ag holds on R . r 34 2. The Simplicity of Diffc(M) , dim(M) + 2 r ◦ ≤ ≤ ∞

1 By Remark 2.35 and Lemma 2.33(3) Γi,A(g)Γi,A(f)− induces a mapping Λ10 r,[α] n 1 1 ∈ Diﬀ[ 2A,2A]n 1 (R − ) , with Λ10 pi = pi Γi,A(g)Γi,A(f)− (cf. Example 1.15 and − ◦ − r,[α]◦ ◦ n 1 n 1 Example 2.16). Choose a C -diﬀeotopy Λ0 : R − R R − , supported on n 1 × → 1 [ 2A, 2A] − with Λt0 = Id t 0 and Λt0 = Λ10 t 1. Since Γi,A(g)Γi,A(f)− − 1 ∀ ≤ r,[α] ∀ ≥ n n is S -equivariant we can lift it to an C -mapping Λ: R R , i.e. πi Λ = 1 e → ◦ e Γi,A(g)Γi,A(f)− πi. By Claim 1(3) we may assume ◦ Λ i 1 1 n i = Λ i 1 1 n i . (2.17) R − (2A+ , ) R − R − (2A+ , ) R − e| × 2 ∞ × | × 2 ∞ × r,[α] n 1 Claim 2. There exists a mapping β C (R − , R) with e ∈ (β pi)∂i pri Λ = pri + β pi = pri Fl1 ◦ ◦ e e ◦ ◦ e e n 1 e and supp(β) [ 2A, 2A] − . e ⊆ − Proof of Claim 2. Consider the R-action on Rn given by

(t, x) tei + x, 7→ n 1 1 where ei denotes the i-th unit vector in R . Since Γi,A(g)Γi,A(f)− ) is S -equivariant we have 1 πi(Λ(tei + x)) = Γi,A(g)Γi,A(f)− (πi(tei + x)) e 1 = Γi,A(g)Γi,A(f)− (π(t) πi(x)) 1· = π(t) Γi,A(g)Γi,A(f)− (πi(x)) · = π(t) πi(Λ(x)) = πi(tei + Λ(x)), · e e 1 where π : R S ∼= R/Z. Hence t Λ(tei + x) and t tei + Λ(x) are both lifts → 7→ e 7→ e of the same path. Since they coincide for t = 0 we obtain Λ(tei + x) = tei + Λ(x) t R, x Rn and thus Λ is equivariant with respect toe the R-action above.e It ∀ ∈ ∀ ∈ e follows, that priΛ pri is constant on the R-orbits, and hence there exists a map n 1 e − r,[α] n n 1 β : R − R with pri Λ pri = β pi. β is C since pi : R R − admits → ◦ − ◦ → e e (β pi)∂i e e e e (β pi)∂i global cross sections. Since Fl1 ◦ (x) = x + β(pi(x))∂i, we obtain pri Fl1 ◦ = e e e ◦ e e pri + β pi. e e ◦ n 1 n 1 n 1 To see,e that supp(β) [ 2A, 2A] − , let x R − [ 2A, 2A] − and choose n e ⊆ − 1 ∈ \ − x R with pi(x) = x and pri(x) > 2A + . From (2.17) and Claim 1(2) we get ∈ 2 e e e e pri(x) + β(pi(x)) = pri(Λ(x)) = pri(Λ(x)) = pri(x) e e e e e e e e and hence β(x) = β(pi(x)) = 0. 2 e e e e n n 1 Using β and Λ0 we now define the extension λ : R R of ϕi,AΛϕi,A− , which will e → conjugate τi,Af and τi,Ag. Claim 3. The following mapping λ : Rn Rn is well defined and Cr,[α]. → 1 n ϕi,AΛϕ− on ( 2A 1, 2A + 1)  i,A − −  ins (Λ p , pr Flρi,A∂i ) on Ri 1 (2A + 1 , ) Rn i λ :=  i 3+20 A pri i i − 2 − − ◦ ◦ β pi × ∞ × e ◦  Id e e otherwise  r 2. The Simplicity of Diffc(M) , dim(M) + 2 r 35 ◦ ≤ ≤ ∞

n 1 n where insi : R − R R , insi(x, t) := (x1, . . . , xi 1, t, xi, . . . , xn 1). Moreover × i 1 → n −1 − supp(λ) [ 2A, 2A] − [ 2A, 2A + 3] [ 2A, 2A] − . ⊆ − × − × − Proof. Clearly λ is Cr,[α], provided it is well deﬁned. So we only have to show

p ϕ Λϕ 1 = Λ p i i,A i,A− 3+20 A pri i ◦ − ◦ e 1 ρi,A∂i e pri ϕi,AΛϕi,A− = pri Fl β pi ◦ ◦ ◦ e i 1 1 en i hold on ( 2A, 2A) − (2A + 2 , 2A + 1) ( 2A, 2A) − , the rest is obvious from − n ×1 × − n n 1 supp(β) [ 2A, 2A] − , supp(ρA) [ 2A 1, 2A + 1] , supp(Λ0) [ 2A, 2A] − , Claime 1(2)⊆ and− Lemma 2.24. ⊆ − − ⊆ − 1 From Lemma 2.24(1) we get piϕi,A = piϕi,A− = pi, and hence we have e e e 1 1 1 pi ϕi,AΛϕi,A− = pi Λϕi,A− = pi Λϕi,A− ◦ ◦ ◦ e e = Λe p ϕ 1 e= Λ p = Λ p , 10 i i,A− 10 i 3+20 A pri i ◦ ◦ ◦ − ◦ e e e i 1 1 n i on ( 2A, 2A) − (2A + 2 , 2A + 1) ( 2A, 2A) − , where we used Lemma 2.24(6) and− (2.17) for the× second equality. × − Since β pi is constant on the integral curves of the vector ﬁeld ρi,A∂i, we have ◦ e e ρi,A∂i (β pi)ρi,A∂i n Fl (x) = Flt ◦ (x) x R t R (β pi)(x)t ◦ e e ∀ ∈ ∀ ∈ e e and thus

1 ρ ∂ (β p )ρ ∂ (ϕ− )∗(β pi)∂i i,A i i i,A i i,A ◦ Fl (x) = Fl1 ◦ (x) = Fl1 (x) β pi(x) e e ◦ e e e e 1 (β pi)∂i (β pi)∂i 1 = (ϕi,A− )∗Fl1 ◦ (x) = ϕi,AFl1 ◦ ϕi,A− (x), e e e e where we used Lemma 2.24(4) to get

1 1 1 (ϕ− )∗(β pi)∂i = (β pi ϕ− )(ϕ− )∗∂i = (β pi)ρi,A∂i i,A ◦ ◦ ◦ i,A i,A ◦ e e e e e e From Lemma 2.24(6)&(7) and Claim 2 we obtain

1 1 1 pri ϕi,AΛϕi,A− = pri ϕi,AΛϕi,A− = pri ϕi,A pri Λϕi,A− ◦ ◦ e ◦ ◦ ◦ e (β pi)∂i 1 f (β pi)∂i 1 = pri ϕi,A pri Fl1 ◦ ϕi,A− = pri ϕi,AFl1 ◦ ϕi,A− ◦ ◦ ◦ e e ◦ e e ρi,A∂i f = priFl β pi ◦ e i 1 1e n i holds on ( 2A, 2A) − (2A + , 2A + 1) ( 2A, 2A) − . 2 − × 2 × − r,[α] n n Clearly λ x R :pri(x)<2A+1 Diff ( x R : pri(x) < 2A + 1 ). Further, since |{ ∈ } ∈ { ∈ } λ n = ins (Λ p , pr ), λ is bijective. We have det(Dλ(x)) = x R :xi 2A+1 i 3+20 A pri i i |{ ∈ ≥ } − ◦ det(DΛ (p (x))) = 0 if pr (x) e 2A + 1. Thus we obtain λ Diffr,[α](Rn) (cf. 3+20 A xi i i c − 6 ≥ ∈ ◦ Example 1.15 ande Example 2.16), and the proof of Lemma 2.36 is finished. 2 r 36 2. The Simplicity of Diffc(M) , dim(M) + 2 r ◦ ≤ ≤ ∞

1 r,[α] 2.37. Remark. We used the condition Γi,A(g)Γi,A(f)− Gi only to extend 1 1 ∈ ϕi,AΛϕi,A− . Now suppose ‘conversely’ ϕi,AΛϕi,A− extends to a continuous mapping λ defined on [ 2A 1, 2A + 1]n. Then we will show − − n piΛ(x0) = piΛ(x) x, x0 R : pi(x) = pi(x0), (2.18) e e ∀ ∈ e 1 e 1 1 e e and hence Γi,A(g)Γi,A(f)− maps S -orbits to S -orbits. Recall that we defined Λ 1 r,[α] 1 e without using Γi,A(g)Γi,A(f)− Gi . Since Λ is a lift of Γi,A(g)Γi,A(f)− r,[α] ∈ e ∈ Diffc (Ci) we have n Λ(nei + x) = nei + Λ(x) x R n Z. e e ∀ ∈ ∀ ∈ If x, x0 dom(ϕi,A) we thus obtain ∈ piΛ(x) = lim piϕi,A(nei + Λ(x)) = lim piϕi,AΛ(nei + x) n n e →∞ e →∞ e e e 1 e = lim pi ϕi,AΛϕ− ϕi,A(nei + x) = piλ lim ϕi,A(nei + x) = n i,A n →∞ e e →∞ | {zλ } = piλ lim ϕi,A(nei + x0) = ... = piΛ(x0), n e →∞ e e since λ is continuous on [ 2A 1, 2A + 1]n. Elsewhere (2.18) is obvious. − − 2.38. Lemma (Mather). [7] [4] Let A 1 and let α be a modulus of continuity. n ≥ 1 n Then there exists a neighborhood UA of Id Diff[ 2A,2A]n (R ) and mappings ∈ − ◦ n 1 n Ψi,A : UA Diff[ 2A,2A]n (R ) , 1 i n −→ − ◦ ≤ ≤ with the following properties:

(1) Ψi,A(Id) = Id (2)

n r,[α] n r,[α] n Ψi,A : UA Diﬀ[ 2,2]i 1 [ 2A,2A]n (i 1) (R ) Diﬀ[ 2,2]i [ 2A,2A]n i (R ) ∩ − − × − − − → − × − − ◦ is continuous with respect to Cr-topologies.

n r,[α] n r,[α] n (3) f UA Diﬀ (R ) = [f] = [Ψi,A(f)] H1(Diﬀc (R ) ) ∈ ∩ ⇒ ∈ ◦ (4) There exists δ > 0 depending on n, r, α, A and C > 1 depending on n, r,α but independent of A with

µr,α(Ψi,A(f)) CAµr,α(f) ≤ n r,α n f U Diff (R ) with µr,α(f) δ. ∀ ∈ A ∩ ≤ (5) If r 2 then there exists a constant K > 1, independent of r, A, and an admissible≥ polynomial of one variable F , depending on n, r, α and A, such that r µr(Ψi,A(f)) K Aµr(f) + F (Mr 1(f)) ≤ − f Diffr(Rn) U n. ∀ ∈ ∩ A r 2. The Simplicity of Diffc(M) , dim(M) + 2 r 37 ◦ ≤ ≤ ∞

n Proof. First we assume UA U (cf. Lemma 2.33), and thus Γi,A(f) is defined for all n n ⊆ 1 f U . Given f U we define g C (Ci,Ci) by ∈ A ∈ A ∈ g(x1, . . . , ϑ, . . . , xn) := Γi,A(f)(x1,..., 0, . . . , xn) + (0, . . . , ϑ, . . . , 0). (2.19) Using Proposition 1.7(3) and Lemma 2.33 one sees that g depends continuously on f, n 1 and thus we can shrink U , such that g Diff i 1 1 n i (Ci) . Notice, A [ 2A,2A] − S [ 2A,2A] − 1 ∈ − × × − ◦ that g Gi and g θ Ci:pri(θ)=0 = Γi,A(f) θ Ci:pri(θ)=0 (in fact g is unique with ∈ |{ ∈ } 1 |{ ∈ 1 } these properties). Further let h := g− Γi,A(f) Diff i 1 1 n i (Ci) . [ 2A,2A] − S [ 2A,2A] − ∈ − 2 × × − ◦ If pr (x1, . . . , ϑ, . . . , xn) = (x1,..., 0, . . . , xn) we have D (pr ) = 0 and g Id = i i − (Γi,Af(f) Id) pr . From (2.3) we get f − ◦ i r f r r D (g Id) = D (Γi,A(f) Id) pr Dpr r 1, − − ◦ i i ∀ ≥ thus f f r µr(g) Γi,A(f) Id r pr = µr(Γi,A(f)), ≤ k − k k ik1 f=1 | {z } and x r D (Γi,A(f) Id) priDpri k − ◦ y k µr,α(g) = sup f f x=y Ci α(dist(x, y)) 6 ∈ x r D (Γi,A(f) Id) pri pri 1 k − ◦ y kk k sup f f ≤ x=y Ci α(dist(x, y)) 6 ∈ x r D (Γi,A(f) Id) pri k − ◦ y k sup f µr,α(Γi,A(f)). ≤ x=y Ci α(dist(x, y)) ≤ 6 ∈ Choosing δ1 > 0 sufficiently small (cf. Lemma 2.33(4) and Lemma 2.30), we get 1 µr,α(h) = µr,α(g− Γi,A(f)) 1 1 µr,α(g− ) + µr,α(Γi,A(f)) + C1µr,α(Γi,A(f)) µr,α(g− ) ≤ 1 | ≤{z } 2µr,α(g) + µr,α(Γi,A(f)) + 2µr,α(g) ≤ 5µr,α(Γi,A(f)) ≤ n r,α n n 1 f UA Diff (R ) with µr,α(f) δ1. If we shrink UA such that µ1(Γi,A(f)) 2 ∀ ∈ ∩ 1 n ≤ ≤ and µ1(f) for all f U , we can use (2.14), Lemma 2.29(1)&(3) and µr(g) ≤ A ∈ A ≤ µr(Γi,A(f)) to estimate 1 µ1(h) = µ1(g− Γi,A(f)) 1 1 2 µ1(g− ) +µ1(Γi,A(f)) 1 + µ1(g− ) +µ1(Γi,A(f)) ≤ 2µ1(Γi,A(f)) 2µ1(Γi,A(f)) ≤ | {z } ≤ | {z } 6µ1(Γi,A(f)) 1 + 3 µ1(Γi,A(f)) ≤ 1 | ≤{z2 } 18A 15 18Aµ1(f) 1 + µ1(f) K1Aµ1(f), ≤ · ≤ 1 | ≤{zA } r 38 2. The Simplicity of Diffc(M) , dim(M) + 2 r ◦ ≤ ≤ ∞

1 A where K1 is a constant, independent of A, since (1+ ) is bounded on (0, ). From A ∞ Lemma 2.29(4) we obtain an admissible polynomial of one variable F1 with

1 r+1 µr(g− ) µr(g)1 + 2µ1(g) + F1(Mr 1(g)) ≤ − r+1 2 µr(Γi,A(f)) + F1(Mr 1(Γi,A(f))). (2.20) ≤ − Thus there exists an admissible polynomial F2, depending on r, such that

1 r+1 Mr(g− ) 2 Mr(Γi,A(f)) + F2(Mr 1(Γi,A(f))) (2.21) ≤ − (cf. Lemma 2.29(3)). Using (2.20), (2.21) and Lemma 2.29(2) we obtain

1 µr(h) = µr(g− Γi,A(f)) 1 1 r 2 µr(g− ) + µr(Γi,A(f)) 1 + µ1(g− ) + µ1(Γi,A(f)) ≤ 1 1 | ≤{z } | ≤{z2 } 1 +F3(Mr 1(g− ) + Mr 1(Γi,A(f))) − − r K4 µr(Γi,A(f)) + F4(Mr 1(Γi,A(f))) ≤ − Together with (2.13), this yields an admissible polynomial F5 of one variable and a constant K5, independent of r and A, such that

r µr(h) K5 Aµr(f) + F5(Mr 1(f)). ≤ − A 1 A Here we used again, that (1 + µ1(f)) (1 + A ) is bounded on (0, ). Summing up we have ≤ ∞ (1) f U n Diﬀr,α(Rn) = g Gr,α ∈ A ∩ ⇒ ∈ i

(2) h θ Ci:pri(θ)=0 = Id |{ ∈ } (3) f = Id = h = Id ⇒ n r,[α] n r,[α] (4) f UA Diﬀ[ 2,2]i 1 [ 2A,2A]n (i 1) (R ) = h Diﬀ[ 2,2]i 1 S1 [ 2A,2A]n i (Ci) ∈ ∩ − − − ⇒ ∈ − − × × − − ◦ and h depends− continuously× − on f with respect to Cr-topologies.

(5) There exists δ1 > 0 depending on n, r, α and A, such that

µr,α(h) 5µr,α(Γi,A(f)) (2.22) ≤ r,α n n f Diﬀ (R ) U with µr,α(f) δ1 ∀ ∈ ∩ A ≤ (6) There exists a constant K5, independent of r and A, and an admissible polynomial of one variable F5, such that

µ1(h) K5Aµ1(f) (2.23) ≤ and

r µr(h) K5 Aµr(f) + F5(Mr 1(f)). (2.24) ≤ − r 2. The Simplicity of Diﬀc(M) , dim(M) + 2 r 39 ◦ ≤ ≤ ∞

1 n Using (2) above we can lift h to a mapping h Diﬀ i 1 n i (R ) [ 2A,2A] − R [ 2A,2A] − with e ∈ − × × − ◦

n h πi = πi h and h x R :pri(x) Z = Id. ◦ ◦ e e|{ ∈ ∈ } Notice that h depends continuously on h, µr(h) = µr(h) and µr,α(h) = µr,α(h). Now e e e choose a periodic function ξ C∞(R, [0, 1]) of period 1 with ∈

ξ [m 1 ,m+ 1 ] 0, ξ [m+ 2 ,m+ 4 ] 1 m Z | − 6 6 ≡ | 6 6 ≡ ∀ ∈ and let 1 h := (ξ pri)(h Id) + Id, h+ := hh− . e− ◦ e − e ee− Thus h i 1 1 1 n i = Id and h+ i 1 2 4 n i = Id, m Z. If R − [m ,m+ ] R − h(R [m+ ,m+ ] R ) e−| × − 6 6 × e | − × 6 6 × − ∀ ∈ n i e1 5 7 n i i 1 we choose UA suﬃciently small then h(R − [m + 12 , m + 12 ] R − ) R − 2 4 n i e × × ⊆ ×n [m + 6 , m + 6 ] R − and h+ Ri 1 [m+ 5 ,m+ 7 ] Rn i = Id, m Z. By shrinking UA × | − × 12 12 × − ∀ ∈ e 1 n we may further assume h Diﬀ i 1 n i (R ) . In order to estimate [ 2A,2A] − R [ 2A,2A] − e± ∈ − × × − ◦ µr,α(h ) in terms of µr,α(h) we start with e± e x r D ((ξ pri)(h Id)) k ◦ − y k µr,α(h ) = sup e e− x=y Rn α( x y ) 6 ∈ k − k x k r k r D (ξ pri)D − (h Id) r k ◦ − y k sup e ≤ x=y Rn X k α( x y ) 6 ∈ k=0 k x− k k r k r r D (ξ pri) D − (h Id)(x) ( k ◦ y kk − k sup e ≤ X k x=y Rn α( x y ) k=0 6 ∈ k − k x k r k D (ξ pri)(y) D − (h Id) k ◦ kk − y k) + sup e x=y Rn α( x y ) 6 ∈ k − k r r ξ pri k,αµr k(h) + ξ pri kµr k,α(h) X n − − o ≤ k=0 k k ◦ k e k ◦ k e

C2µr,α(h) ≤ e where C2 is a constant independent of A. If µr,α(h) is small we thus have e 1 µr,α(h− ) 2µr,α(h ) e− ≤ e− (cf. Lemma 2.30), and hence there exists a constant C3 > 0 independent of A, such that

1 µr,α(h+) = µr,α(hh− ) − e ee 1 1 µr,α(h) + µr,α(h− ) + C4µr,α(h) µr,α(h− ) ≤ e e− e e− 1 | ≤{z } (1 + 2C2)µr,α(h) (1 + 2C2)µr,α(h) C3µr,α(Γi,A(f)) ≤ e ≤ ≤ r 40 2. The Simplicity of Diﬀc(M) , dim(M) + 2 r ◦ ≤ ≤ ∞ provided µr,α(f) is small. For the last inequality we used (2.22). Further we have

µr(h ) = (ξ pri)(h Id) r e− kr ◦ e − k r k r k sup D (ξ pri)(x) D − (h Id)(x) ≤ X k x Rn k ◦ kk − k k=0 ∈ e r r ξ pri kµr k(h) X − ≤ k=0 k k ◦ k e r 1 − r = µr(h) + ξ pri) kµr k(h) + ξ pri) r µ0(h) X − e k=1 k k ◦ k e k ◦ k e µ1(h) |≤ {z } e µr(h) + KMr 1(h), ≤ f − where K is a constant depending on r. Here we used µr(h) = µr(h) if r 1 and ≥ µ (h) fµ (h), since h Id is periodic with period 1, and eh Id i 1 n i = 0. 0 1 R − Z R − Togethere ≤ withe (2.23)e and− (2.24) and Lemma 2.29(4) this yieldse − | × ×

r µr(h ) K6 Aµr(f) + F6(Mr 1(f)) e± ≤ − where F6 is an admissible polynomial of one variable, and K6 is a constant independent of r and A. Summing up we have

(1) f = Id = h = Id ⇒ e±

(2) h = h+h e e e− (3) If m Z then ∈

h i 1 1 1 n i = Id, h+ i 1 5 7 n i = Id R − [m ,m+ ] R − R − [m+ ,m+ ] R − e−| × − 6 6 × e | × 12 12 ×

n r,[α] n r,[α] n (4) f UA Diﬀ[ 2,2]i 1 [ 2A,2A]n (i 1) (R ) h Diﬀ[ 2,2]i 1 R [ 2A,2A]n i (R ) ∈ ∩ − − × − − − ⇒ e± ∈ − − × × − − ◦ and h depend continuously on f with respect to Cr-topologies. e±

(5) There exist δ3 > 0 depending on n, r, α, A and C3 > 0 depending on n, r, α but independent of A, such that

µr,α(h ) C3µr,α(Γi,A(f)) (2.25) e± ≤ n r,α n f U Diﬀ (R ) with µr,α(f) δ3. ∀ ∈ A ∩ ≤

(6) There exists a constant K6 independent of r and A and an admissible polynomial of one variable F6 such that

r µr(h ) K6 Aµr(f) + F6(Mr 1(f)) (2.26) e± ≤ − r 2. The Simplicity of Diﬀc(M) , dim(M) + 2 r 41 ◦ ≤ ≤ ∞

We let

n n 1 3 E := x R : 1 pri(x) 0 ,E+ := x R : pri(x) − { ∈ − ≤ ≤ } { ∈ 2 ≤ ≤ 2} and deﬁne Ψi,A(f) by

n Ψi,A(f) E+ = h+ E+ , Ψi,A(f) E = h E , Ψi,A R (E E+) = Id. | e | | − e−| − | \ −∪ (1) and (2) of Lemma 2.38 follow immediately. To see (3) notice that there exist 1 h Diff (Ci) with h πi = πi h . Since h+h = h+h = h we get h+h = h, ± ∈ ± ◦ ◦ e± g− e e− e − hence Γi,A(Ψi,A(f)) = h+h = h and thus − 1 1 r,[α] Γi,A(f)Γi,A(Ψi,A(f))− = Γi,A(f)h− = g G ∈ i n r,[α] n n f U Diff (R ). By shrinking U we may assume f, Ψi,A(f) U 0 (cf. ∀ ∈ A ∩ A ∈ Lemma 2.36), and we can apply Lemma 2.36. Consequently τi,Af and τi,AΨi,A(f) r,[α] n r,[α] n are conjugated in Diffc (R ) and thus [f] = [Ψi,A(f)] H1(Diffc (R ) ). (2.25), Lemma 2.33(4) and the obvious◦ fact ∈ ◦

µr,α(Ψi,A(f)) µr,α(h ) + µr,α(h+) ≤ e− e together yield (4) of Lemma 2.38. Similarly we obtain (5) of Lemma 2.38 from

µr(Ψi,A(f)) = max µr(h ), µr(h+) { − } and (2.26). 2 n For the following construction we fix gA Diffc∞(R ) with gA [ 2,2]n = A Id. ∈ ◦ | − · n 1 n 2.39. Corollary. For A 1 there exists a neighborhood VA of Id Diff[ 2,2]n (R ) such that ≥ ∈ − ◦

n 1 n ϑf,A : VA Diff[ 2,2]n (R ) → − ◦ 1 h ψn,A . . . ψ1,A(gAfhg− ) 7→ A is well defined f V n, and has the following properties: ∀ ∈ A (1) If α is a modulus of continuity, 1 r and f V n Diffr,[α](Rn) then ≤ ≤ ∞ ∈ A ∩ n r,[α] n r,[α] n ϑf,A V n Diffr,[α](Rn) : VA Diff (R ) Diff[ 2,2]n (R ) | A ∩ ∩ → − ◦ is continuous with respect to Cr-topologies.

n r,[α] n r,[α] n (2) f, h VA Diﬀ (R ) = [ϑf,A(h)] = [fh] H1(Diﬀc (R ) ) ∈ ∩ ⇒ ∈ ◦

1 Proof. Since Ψi,A are continuous (cf. Lemma 2.38) and since gAfhgA− depends contin- n 1 n uously on (f, h), there exists a neighborhood VA of Id Diff[ 2,2]n (R ) , such that 1 n ∈ − ◦ n ψn,A . . . ψ1,A(gAfhg− ) is defined f, h V . Since supp(fh) [ 2, 2] we have A ∀ ∈ A ⊆ − r 42 2. The Simplicity of Diffc(M) , dim(M) + 2 r ◦ ≤ ≤ ∞

1 n supp(gAfhgA− ) [ 2A, 2A] and thus (1) follows immediately from Lemma 2.38(2). From Lemma 2.38(3)⊆ − we obtain

1 1 r,[α] n [ϑf,A(h)] = [Ψn,A Ψ1,A(gAfhgA− )] = [gAfhgA− ] = [fh] H1(Diﬀc (R ) ). ··· ∈ ◦ 2

2.40. Lemma. Let α be a modulus of continuity and suppose n + 2 r < . Then ≤ ∞ there exists A0 1 and 0 > 0, such that ≥ n µr,α(f), µr,α(h) = f, h V , µr,α(ϑf,A (h)) ≤ ⇒ ∈ A0 0 ≤ r,α n 0 < 0 and f, h Diﬀ[ 2,2]n (R ). ∀ ≤ ∀ ∈ − n 1 r+n Proof. Since n + 2 r we can choose A0 1 such that 3C A0− 1, where C is the constant in Lemma≤ 2.38(4). Using Lemma≥ 2.26(4)&(5) one sees,≤ that there exists 0 > 0, such that

r,α n n f Diﬀ[ 2,2]n (R ), µr,α(f) 0 = f VA0 . ∈ − ≤ ⇒ ∈

By Lemma 2.30(1) we can shrink 0 such that

µr,α(fh) µr,α(f) + µr,α(h) + C1µr,α(f) µr,α(h) 3, ≤ ≤ 1 | ≤{z } r,α n 0 < 0 and f, h Diﬀ[ 2,2]n (R ) with µr,α(f), µr,α(h) . In order to estimate ∀ ≤ 1 ∀ ∈ − ≤ µr,α(gA0 fhgA−0 ), we ﬁrst calculate 1 Dr(g fhg 1) = Dr (A Id)fh( Id) A0 A−0 0 · A0 · r 1 1 1 = D − A0 (D(fh) ( Id)) · ◦ A0 · A0 r 2 2 1 1 = D − (D (fh) ( Id)) ◦ A0 · A0 1 r r 1 = ... = A0− D (fh) Id. (2.27) ◦ A0 Thus we have

r 1 r 1 1 D (gA0 fhgA−0 )(x) D (gA0 fhgA−0 )(y) µr,α(gA0 fhgA−0 ) = sup k − k x=y Rn α( x y ) 6 ∈ k − k Dr(fh)( 1 x) Dr(fh)( 1 y) 1 r A0 A0 = A0− sup k 1 − 1 k x=y Rn α(A0 x y ) 6 ∈ k A0 − A0 k Dr(fh)( 1 x) Dr(fh)( 1 y) 1 r A0 A0 A0− sup k 1 − 1 k ≤ x=y Rn α( x y ) 6 ∈ k A0 − A0 k 1 r 1 r A − µr,α(fh) 3A − , ≤ 0 ≤ 0 r 2. The Simplicity of Diﬀc(M) , dim(M) + 2 r 43 ◦ ≤ ≤ ∞

r,α n 0 < 0 and f, h Diﬀ[ 2,2]n (R ) with µr,α(f), µr,α(h) . By shrinking ∀ ≤ ∀ ∈ − ≤ 0 once again we may assume 0 δ, where δ is the constant in Lemma 2.38(4) ≤ i 1 r+i corresponding to C above. Since A0,C 1 we have 3C A0− 0 δ for 0 i n, and hence using Lemma 2.38(4) we obtain≥ inductively ≤ ≤ ≤

1 µr,α(ϑf,A0 (h)) = µr,α(Ψn,A0 Ψ1,A0 (gA0 fhgA−0 )) n n ··· 1 C A0 µr,α(gA0 fhgA−0 ) ≤ n n 1 r C A 3A − ≤ 0 0 ≤ 1 | ≤{z } r,α n 0 < 0 and f, h Diﬀ[ 2,2]n (R ) with µr,α(f), µr,α(h) . 2 ∀ ≤ ∀ ∈ − ≤ 2.41. Lemma. Let α be a modulus of continuity, 1 r < and let a > 0. If > 0 is suﬃciently small, then ≤ ∞

r,α n L := h Diff[ a,a]n (R ): µr,α(h) { ∈ − ≤ } equipped with the Cr-topology has the fixed-point property, i.e. every continuous mapping L L has at least one fixed point. → r n n Proof. Recall that C[ a,a]n (R , R ), r is a Banach space, and look at the subspace − k · k

r,α n n r n n L0 := h C[ a,a]n (R , R ): h r,α C[ a,a]n (R , R ), r. { ∈ − k k ≤ } ⊆ − k · k r n n r Since the topology induced from r on C[ a,a]n (R , R ) is exactly the C -topology k · k − (cf. Lemma 2.26(4)), h h Id is a homeomorphism from L onto L0, provided > 0 is small (cf. Example 1.15).7→ − r n n Next we show L0 is closed in C[ a,a]n (R , R ), r. If (fi)i N L0 is a sequence − k·k ∈ ∈ r n n r r that converges to f C[ a,a]n (R , R ), r, then limi D fi(x) = D f (x), n ∞ ∈ − k · k n →∞ ∞ x R . Since fi L0 we have for arbitrary x = y R ∀ ∈ ∈ 6 ∈ r r D fi(x) D fi(y) k − k fi r,α , i N, α( x y ) ≤ k k ≤ ∀ ∈ k − k and thus Drf (x) Drf (y) k ∞ − ∞ k . α( x y ) ≤ k − k So we have f r,α , hence f L0 and thus L0 is closed. Next considerk ∞k the≤ linear mapping∞ ∈

r n n T 0 n r n n C[ a,a]n (R , R ), r C[ a,a]n (R ,L (R , R )), 0 − k · k −→ − k · k h Drh 7→ T is an isometry since

r T h 0 = sup D h(x) = h r. k k x Rn k k k k ∈ r 44 2. The Simplicity of Diﬀc(M) , dim(M) + 2 r ◦ ≤ ≤ ∞

0 n r n n Now look at L00 := T (L0). L00 is closed in C[ a,a]n (R ,L (R , R )), 0, for L0 is − k · k closed and T is an isometry. From Lemma 2.26(5) one obtains immediately that L00 is bounded. Further we have Drh(x) Drh(y) Drh(x) Drh(y) = k − k α( x y ) k − k α( x y ) k − k k − k | ≤{z } α( x y ), ≤ k − k n n x, y R h L00, and hence L00 is equicontinuous. Since [ a, a] is compact, we ∀ ∈ ∀ ∈ − can apply Ascoli-Arzela’s theorem, and so L00 is relative compact. But L00 is closed and hence L0 ∼= L00 is compact. Since L0 is a convex, compact subset of a Banach space we obtain by the fixed- point theorem of Schauder-Tychonoff, that every continuous map L0 L0 has a fixed → point. Since L ∼= L0 the same is true for L. 2 Proof of Theorem 2.19 in the case n + 2 r < . By Remark 2.22 we only have to show ≤ ∞

r,α n r,α n r,α n r,α n Diff[ 2,2]n (R ) [Diffc (R ) , Diffc (R ) ] Diff[ 2,2]n (R ) (2.28) − ⊆ ◦ ◦ ⊆ −

Given n, r and α, we fix A0 and 0 by Lemma 2.40, and we choose 0 < 0 such that ≤ r,α n L := h Diff[ 2,2]n (R ): µr,α(h) { ∈ − ≤ } has the fixed-point property (cf. Lemma 2.41). By Lemma 2.40 and Corollary 2.39(1) we have L V n , and for all f L ⊆ A0 ∈

ϑf,A L r,α n 0 | r,α n Diﬀ[ 2,2]n (R ) L L Diﬀ[ 2,2]n (R ) − ⊇ −→ ⊆ − r is continuous with respect to C -topology. Hence for every f L there exists hf L ∈ ∈ with ϑf,A0 (hf ) = hf . Together with Corollary 2.39(2) this yields

r,α n [f][hf ] = [fhf ] = [ϑf,A(hf )] = [hf ] H1(Diffc (R ) ), ∈ ◦ r,α n hence [f] = [Id] H1(Diffc (R ) ) for all f L, and thus we have shown ∈ ◦ ∈ r,α n r,α n L [Diffc (R ) , Diffc (R ) ]. ⊆ ◦ ◦ r,α n r,α Diff[ 2,2]n (R ) equipped with the C -topology is a connected topological group (cf. − r,α n Corollary 2.31), and L is a neighborhood of Id. So L generates Diff[ 2,2]n (R ), and this yields (2.28). − 2

2.42. Lemma. If A 1 and r 2, then there exists a constant K > 1, independent of A and r, and an admissible≥ polynomial≥ F depending on n, r, α and A, such that

n+1 r rn µr(ϑf,A(h)) A − K µr(f) + µr(h) + F (Mr 1(f) + Mr 1(h)) ≤ − − r 2. The Simplicity of Diﬀc(M) , dim(M) + 2 r 45 ◦ ≤ ≤ ∞

r n n n 1 f, h Diﬀ (R ) VA , where VA is the C -neighborhood of Id deﬁned in Corol- ∀lary 2.39.∈ ∩

Proof. From (2.27) we obtain

1 1 r r 1 1 r µr(gAfhgA− ) = sup A − D (fh)( x) = A − µr(fh) x Rn k A k ∈ 1 r r = 2A − (µr(f) + µr(h))(1 + µ1(f) + µ1(h))

+F1(Mr 1(f) + Mr 1(h)), − − where F1 is the admissible polynomial in Lemma 2.29(2). Using Lemma 2.38(5) we obtain inductively

n rn 1 1 µr(ϑf,A(h)) A K1 µr(gAfhgA− ) + F2(Mr 1(gAfhgA− )), ≤ − where F2 is an admissible polynomial of one variable. Recall that µ1( ) is bounded n · on VA . Thus we have

1 r+n rn µr(ϑf,A(h)) A − K2 µr(f) + µr(h) + F3(Mr 1(f) + Mr 1(h)) ≤ − − f, h Diﬀr(Rn) V n, but this is Lemma 2.42. 2 ∀ ∈ ∩ A

2.43. Lemma. Let a > 0, i0 N and let bi , bi +1,... be a sequence of positive ∈ 0 0 numbers. If bi0 is suﬃciently small then

n L := h Diff[∞a,a]n (R ): µi(h) bi i0 i { ∈ − ≤ ∀ ≤ } equipped with the Whitney C∞-topology has the fixed-point property, i.e. every continuous mapping L L has at least one fixed point. → Proof. Consider

n n n n L0 := h C[∞a,a]n (R , R ): h i bi i0 i C[∞a,a]n (R , R ), ( i)i N. { ∈ − k k ≤ ∀ ≤ } ⊆ − k · k ∈ n n Since the topology induced from ( i)i N on C[∞a,a]n (R , R ) is exactly the Whitney- k·k ∈ − C∞-topology h h Id is a homeomorphism from L onto L0, provided bi0 > 0 is suﬃciently small7→ (cf.− Example 1.15 and Lemma 2.26(4)). L0 is closed, since the (semi-)norms i are continuous. For i N look at the linear mapping k · k ∈

n n Si 0 n i n n C[∞a,a]n (R , R ), ( i)i N C[ a,a]n (R ,L (R , R )), 0 − k · k ∈ −→ − k · k h Dih 7→

Si is continuous i N, since ∀ ∈ i Sih 0 = sup D h(x) = h i. (2.29) k k x Rn k k k k ∈ r 46 2. The Simplicity of Diﬀc(M) , dim(M) + 2 r ◦ ≤ ≤ ∞

Further Si(L0) is bounded. If i n + 2 this is obvious from (2.29) and the definition ≥ of L0, if i < n + 2 one additionally needs the estimation in Lemma 2.26(4). Since we have i i Si(h)(x) Si(h)(y) = D h(x) D h(y) h i+1 x y k − k k − k ≤ k k k − k Si(L0) is equicontinuous i N. Hence, by a theorem of Ascoli-Arzela, Si(L0) is 0 ∀ ∈ n i n n relative compact in C[ a,a]n (R ,L (R , R )), 0 . Let (hi)i N be an arbitrary ∈ − k · k 1 sequence in L0. Since S1(L0) is relative compact there exists a subsequence (hi )i N 1 0 n 1 n n ∈ of (hi)i N, such that (S1(hi ))i N converges in C[ a,a]n (R ,L (R , R )), 0 . Now, ∈ ∈ − 2 1 k·k since S2(L0) is relative compact, we obtain a subsequence (hi )i N of (hi )i N, such that 2 0 n 2 n n ∈ ∈ (S2(hi ))i N converges in C[ a,a]n (R ,L (R , R )), 0 . We continue inductively ∈ −j j 1 k · k j and obtain subsequences (hi )i N of (hi− )i N, such that (Sj(hi ))i N converges in ∈ ∈ ∈ 0 n j n n i C[ a,a]n (R ,L (R , R )), 0 , j N. Now consider the sequence (hi)i N. Then ∈ − i k · k ∀ ∈ i (Sj(hi))i N converges for all j, and hence by (2.29) (hi)i N is a Cauchy sequence with ∈ n n ∈ respect to j on C∞(R , R ), j N. Since L0 is closed subspace of a Fréchet i k · k ∀ ∈ space, (hi)i N converges in L0 and thus L0 is compact. Since L0 is a convex and compact subset∈ of a Fréchet space we obtain by a theorem of Schauder-Tychonoff that every continuous map L0 L0 has at least one fixed point. Since L ∼= L0 the same is true for L. → 2

Proof of Theorem 2.19 in the case r = . Recall that the constant K in Lemma 2.42 ∞ n (n+2)n is independent of A and deﬁne A0 := max K , 4K . If r n + 2 we thus have { } ≥ n r (n+2) n+1 r rn n+1 (n+2) n(n+2) K − 1 A0 − K = A0 − K . A0 ≤ 4 1 n(n+2) 1 A− K | 0 {z ≤ 4 } 1 | ≤{z } n n n By Lemma 2.26(4) we can choose n+2 > 0, such that f VA0 , f Diﬀ[∞2,2] (R ) ∈ ∀ ∈ − with µn+2(f) n+2. Using again the estimation of Lemma 2.26(4) and Lemma 2.42 we obtain ≤

n+1 r rn µn+2(ϑf,A (h)) A − K (µn+2(f) + µn+2(h)) 0 ≤ 0 1 2n+2 | ≤{z4 } | ≤ {z } + F (Mn+1(f) + Mn+1(h)) n+2, ≤ n+2 | ≤ {z2 }

n f, h Diﬀ[∞2,2]n (R ) with µn+2(f), µn+2(h) n+2, provided n+2 is chosen suﬃ- ciently∀ ∈ small.− Here it is essential, that F doesn’t≤ have any linear or constant terms.

We further assume that n+2 satisﬁes the condition on bi0 in Lemma 2.43. n Sublemma. If f Diﬀ[∞2,2]n (R ) and µn+2(f) n+2, then there exist f,i > 0, n + 3 i < , such∈ that − ≤ ≤ ∞

µi(h) f,i n + 2 i < = µi(ϑf,A (h)) f,i n + 2 i < , ≤ ∀ ≤ ∞ ⇒ 0 ≤ ∀ ≤ ∞ r 2. The Simplicity of Diﬀc(M) , dim(M) + 2 r 47 ◦ ≤ ≤ ∞ where f,n+2 := n+2.

Proof of the Sublemma. We will choose f,i inductively. Suppose we have chosen f,n+2, . . . , f,i 1 with −

µj(h) f,j n + 2 j i 1 = µj(ϑf,A (h)) f,j n + 2 j i 1. ≤ ∀ ≤ ≤ − ⇒ 0 ≤ ∀ ≤ ≤ −

Then there exists a constant ai > 0 depending on f, i and f,n+2, . . . , f,i 1, such that n − F (Mi 1(h) + Mi 1(f)) ai for all h Diﬀ[∞2,2]n (R ) with µj(h) f,j, n + 2 j − − ≤ ∈ − ≤ ≤ ≤ i 1. If we let f,i := µi(f) + 2ai, Lemma 2.42 yields − n+1 i in µi(ϑf,A0 (h)) A0 − K (µi(f) + µi(h)) + F (Mi 1(h) + Mi 1(f)) ≤ − − 1 2µi(f)+2ai ai | ≤{z4 } | ≤ {z } | ≤{z } 1 3 µi(f) + ai f,i ≤ 2 2 ≤ n for all h Diﬀ[∞2,2]n (R ) with µj(h) f,j, n + 2 j i. 2 ∈ − ≤ ≤ ≤ Next consider

n Lf := h Diﬀ[∞2,2]n (R ): µi(h) f,i n + 2 i < . { ∈ − ≤ ∀ ≤ ∞}

From the Sublemma above and Lemma 2.39(1) we obtain, that ϑf,A0 restricts to a continuous mapping

ϑf,A L n 0 | f n Diff[∞2,2]n (R ) Lf Lf Diff[∞2,2]n (R ), − ⊇ −→ ⊆ − n f Diff[∞2,2]n (R ) with µn+2(f) n+2. By Lemma 2.43 there exists hf Lf with ∀ ∈ − ≤ ∈ ϑf,A0 (hf ) = hf . From Lemma 2.39(2), we get

n [f][hf ] = [fhf ] = [ϑf,A(hf )] = [hf ] H1(Diffc∞(R ) ), ∈ ◦ n n hence [f] = [Id] H1(Diffc∞(R ) ) for all f Diff[∞2,2]n (R ) with µn+2(f) n+2, and thus ∈ ◦ ∈ − ◦ ≤

n n n f Diff[∞2,2]n (R ): µn+2(f) n+2 [Diffc∞(R ) , Diffc∞(R ) ]. { ∈ − ≤ } ⊆ ◦ ◦ n But the set on the left side generates Diff[∞2,2]n (R ) and consequently − ◦ n n n Diff[∞2,2]n (R ) [Diffc∞(R ) , Diffc∞(R ) ]. − ⊆ ◦ ◦ By Remark 2.22 this is all we have to show. This completes the proof of Theo- rem 2.19. 2

2.44. Corollary. Let M be a connected, smooth manifold, n = dim(M), n + 2 r ≤ r . Then Diffc(M) is perfect and hence simple by Corollary 1.17. ≤ ∞ ◦ Proof. This is an immediate consequence of Theorem 2.19 and Lemma 2.17. 2 r 3. The Simplicity of Diffc(M) , 1 r dim(M) ◦ ≤ ≤ 3.1. Lemma (Mather). [8] Let A 1 and let α be a modulus of continuity. Then n ≥ 1 n there exists a neighborhood UA of Id Diff[ 2A,2A]n (R ) and mappings e ∈ − ◦ n 1 n Ψi,A : UA Diff[ 2A,2A]n (R ) 1 i n e e → − ◦ ≤ ≤ with the following properties:

(1) Ψi,A(Id) = Id e (2)

n r,α n r,α n Ψi,A : UA Diﬀ[ 2,2]i [ 2A,2A]n i (R ) Diﬀ[ 2,2]i 1 [ 2A,2A]n (i 1) (R ) e e ∩ − × − − → − − × − − − ◦ is continuous with respect to Cr-topologies.

n r,α n r,α n (3) f UA Diff (R ) = [f] = [Ψi,A(f)] H1(Diffc (R ) ) ∈ e ∩ ⇒ e ∈ ◦ (4) There exists δ > 0 depending on n, r, α, A and C > 0 depending on n, r, α, but independent of A, with C µr,α(Ψi,A(f)) µr,α(f) e ≤ A n r,α n f UA Diff[ 2,2]i [ 2A,2A]n i (R ) with µr,α(f) δ. ∀ ∈ e ∩ − × − − ≤

n Proof. As we did in the proof of Lemma 2.38 we choose UA sufficiently small, such n 1 e that for f U we can define g Diff i 1 1 n i (Ci) by A [ 2A,2A] − S [ 2A,2A] − ∈ e ∈ − × × − ◦ g(x1, . . . , ϑ, . . . , xn) := Γi,A(f)(x1,..., 0, . . . , xn) + (0, . . . , ϑ, . . . , 0).

1 1 Further we let h := g− Γi,A(f) Diﬀ i 1 1 n i (Ci) , and obtain simi- [ 2A,2A] − S [ 2A,2A] − larly to the proof of Lemma 2.38:∈ − × × − ◦

n r,α n r,α (1) f UA Diﬀ (R ) = g Gi ∈ e ∩ ⇒ ∈

(2) h θ Ci:pri(θ)=0 = Id |{ ∈ } (3) f = Id = h = Id ⇒ n r,α n r,α (4) f U Diﬀ i n i (R ) = h Diﬀ i 1 1 n i (Ci) and A [ 2,2] [ 2A,2A] − [ 2,2] − S [ 2A,2A] − h depends∈ e ∩ continuously− × − on f with respect⇒ ∈ to Cr-topologies.− × × − ◦

(5) There exists δ1 > 0 depending on n, r, α and A, such that

µr,α(h) 5µr,α(Γi,A(f)) (3.1) ≤ r,α n n f Diﬀ (R ) UA with µr,α(f) δ1 ∀ ∈ ∩ e ≤ 48 r 3. The Simplicity of Diﬀc(M) , 1 r dim(M) 49 ◦ ≤ ≤

1 n Using (2) above we may lift h to a mapping h Diﬀ i 1 n i (R ) [ 2A,2A] − R [ 2A,2A] − with e ∈ − × × − ◦

n h πi = πi h, h x R :pri(x) Z = Id. (3.2) ◦ ◦ e e|{ ∈ ∈ } Let a be the least half-integer with 2A a, and for j N deﬁne − ≤ ∈ j n E := x R : a + 3j 3 pri(x) a + 3j 2 − { ∈ − ≤ ≤ − } j n 3 1 E := x R : a + 3j pri(x) a + 3j . + { ∈ − 2 ≤ ≤ − 2}

1 Further let B be the greatest integer such that a + 3B 2 2A and notice that A 1 1 2 − ≤ B B B 2 . So we have at least A disjoint stripes E ,E+,E ,...,E ,E+ , all contained ≥ i 1 n i − − − in R − [ 2A, 2A] R − . Now we define × − × 1 h1 := (h Id) + Id e B e − 1 1 h2 := (hh1− Id) + Id e B 1 ee − . − . 1 1 1 hj := (hh1− hj− 1 Id) + Id e B (j 1) ee ··· e − − . − − . 1 1 hB := (hh−1 hB− 1 Id) + Id e ee ··· e − − n 1 n By shrinking U we may arrange h1,..., hB Diff i 1 n i (R ) . In A [ 2A,2A] − R [ 2A,2A] − e e e ∈ − × × − ◦ order to estimate µr,α(hj) in terms of µr,α(h) we need the following Sublemma, since the estimations in Lemmae 2.30 do not sufficee for this purpose. Sublemma. Let 0 < λ < 1, 1 r < , α a modulus of continuity and let r,α n ≤ ∞ f Diff[ 2A,2A]i 1 R [ 2A,2A]n i (R ). If g := λ(f Id) + Id then there exists C0 > 0 ∈ − − × × − − − and δ0 > 0 with 1 2 µr,α(fg− ) = (1 λ)µr,α(f) + C0µr,α(f) . − provided µr,α(f) δ0. ≤ Proof of the Sublemma. First consider the case r = 1. We have Dg = λ(Df Id)+Id, − hence µ1α(g) = λµ1,α(f) and

m (Dg) 1(x) = ∞ λ(Df(x) Id) − X m=0 − −

1 provided µ1(f) < λ , but this can be arranged by choosing δ0 suﬃciently small (cf. Lemma 2.26). Hence we have

1 D(fg− )(g(x)) = 1 = Df(x)(Dg)− (x) r 50 3. The Simplicity of Diﬀc(M) , 1 r dim(M) ◦ ≤ ≤

m = Id + Df(x) Id ∞ λ(Df(x) Id) X − m=0 − − 1 m = Id + (1 λ) Df(x) Id + (1 ) ∞ λ(Df(x) Id) X − − − λ m=2 − − and thus for suﬃciently small µ1,α(f)

1 µ1,α(fg− ) ≤ 1 1 α( x y ) D(fg− )(g(x)) D(fg− )(g(y)) sup k − k sup k − k ≤ x=y Rn α( g(x) g(y) ) x=y Rn α( x y ) 6 ∈ k − k 6 ∈ k − kx α( g 1 x y ) (1 λ)(Df Id) − 1 ( k − − y k sup k k k − k sup ≤ x=y Rn α( x y ) x=y Rn α( x y ) 6 ∈ k − k 6 ∈ k − k x m 1 (λ(Df Id)) ∞ k − y k) +(1 ) sup − λ X x=y Rn α( x y ) m=2 6 ∈ k − k 1 µ (g 1) + 1 ((1 λ)µ (f) + (1 ) ∞ mλmµ (f)( µ (f) )m 1) 1 − 1,α X 1,α 1 − ≤ − − λ m=2 C µ (f) 2C µ1,α(g) 10 1,α | ≤ 10{z } ≤ | {z }

C0 µ1,α(f) + 1 ((1 λ)µ1,α(f) ≤ 2 − 1 + (1 ) ∞ (m + 2)λm+2(C )m+1µ (f)m µ (f)2) X 10 1,α 1,α − λ m=0 C | ≤{z30 } 2 (1 λ)µ1,α(f) + C0µ1,α(f) ≤ − r r Now consider the case r > 1. Then we have D g = λD f, µr(g) = λµr(f), µr,α(g) = λµr,α(f) and from (2.3) we get

r 1 r 1 1 r 1 r 1 D f g− = (D f g− )D(g− ) + (Df g− )D (g− ) + R 1 (3.3) ◦ ◦ ◦ f,g−

Similarly to (2.10) there exists a constant C50 such that we have

R 1 (x) R 1 (y) f,g− f,g− 1 2 k − k C0 µr,α(f)µr,α(g− ) C0 µr,α(f) (3.4) α( x y ) ≤ 4 ≤ 5 k − k

provided µr,α(f) is suﬃciently small. Further we have

x r 1 1 r 1 r 1 (D f g− )(D(g− )) + (Df g− )D (g− ) k ◦ ◦ y k α( x y ) ≤ x k − k r x r 1 1 r 1 r 1 (D f g− ) D(g− )(x) Id + D f g− D g g− k ◦ y − k k ◦ − ◦ y k ≤ α( x y ) k − k r 3. The Simplicity of Diﬀc(M) , 1 r dim(M) 51 ◦ ≤ ≤

x r x r 1 1 1 r 1 D f(g− (y))D(g− ) + (Df g− ) D (g− )(x) +k y k k ◦ y k α( x y ) k − k x r 1 1 r 1 D g g− + Df(g− (y))D (g− ) +k ◦ y k α( x y ) k − k 1 1 r 1 µr,α(f) µ1(g− ) + 1 µ1(g− ) + (1 λ)µr,α(f) µ1(g− ) + 1 ≤ − 1 1 r 1 1 1 + f rµ1,α(g− )(2 g− ) − + µr,α(f)µ1(g− ) + 1 g− r k k k k x k k r 1 1 r 1 D g g− + Df(g− (y))D (g− ) +k ◦ y k α( x y ) k − k x r 1 1 r 1 D g g− + Df(g− (y))D (g− ) 2 k ◦ y k (1 λ)µr,α(f) + C0 µr,α(f) + (3.5) ≤ − 6 α( x y ) k − k for suﬃciently small µr,α(f). Here we frequently used the estimations in Lemma 2.26. From (2.3) we get

r 1 1 r 1 1 r 1 D (g− ) = D(g− )(D (g) g− )D(g− ) D(g− )R 1 . − ◦ − g,g− Similarly to (3.4) we have

R 1 (x) R 1 (y) g,g− g,g− 1 2 k − k C0 µr,α(g)µr,α(g− ) C0 µr,α(f) α( x y ) ≤ 7 ≤ 8 k − k and

1 1 2 R 1 (x) Ci,j ,...,j g i g− j g− j C0 µr,α(f) , g,g− X 1 i 1 i 9 k k ≤ 2 i r,1 jl k k k k · · · k k ≤ j ≤+≤ +j≤=r 1 ··· i provided µr,α(f) is small. Further we have

x r 1 1 r 1 D g g− + Df(g− (y))D (g− ) k ◦ y k = α( x y ) k − k x r 1 1 1 r 1 1 r D g g + Df(g (y))D(g ) (D g g )D(g ) R 1 − − − − − g,g− = k ◦ ◦ − y k α( x y ) k −x k 1 r 1 Id Df(g− (y))D g g− +k − ◦ y k ≤ α( x y ) k − k x 1 1 r 1 Df(g− (y))(Id D(g− ))(D g g− ) +k − ◦ y k α( x y ) k − k r x 1 1 r 1 1 Df(g− (y))D(g− )(D g g− )Id D(g− ) +k ◦ − y k α( x y ) k − k r 52 3. The Simplicity of Diﬀc(M) , 1 r dim(M) ◦ ≤ ≤

x 1 1 Df(g (y))D(g )R 1 − − g,g− +k y k α( x y ) k − k 1 1 µ1(f)µr,α(g) µ1(g− ) + 1 + f 1µ1,α(g− ) g r ≤ k k k k 1 1 1 1 r + f 1µ1,α(g− )µr,α(g) µ1(g− ) + 1 + f 1µ1,α(g− ) g rµ1(g− ) k k k k k k 1 1 1 r + f 1 g− 1µr,α(g) µ1(g− ) + 1 µ1(g− ) k k k k 1 1 1 r 1 + f 1 g− 1 g rrµ1,α(g− )µ1(g− ) − k k k k k1 k 2 1 2 + f 1µ1,α(g− )C90 µr,α(f) + f 1 g− 1C80 µr,α(f) k k 2 k k k k C0 µr,α(f) (3.6) ≤ 10 provided µr,α(f) is suﬃciently small. (3.3), (3.4), (3.5) and (3.6) together yield

r 1 r 1 1 D (fg− )(x) D (fg− )(y) µr,α(fg− ) = sup k − k x=y Rn α( x y ) 6 ∈ k − k 2 (1 λ)µr,α(f) + (C0 + C0 + C0 ) µr,α(f) , ≤ − 5 6 10 | C{z0:= } and this completes the proof of the Sublemma. 2 If 1 j B we have ≤ ≤ 1 1 1 1 1 1 µr,α(hj) = hh1− hj− 1 Id r,α = µr,α(hh1− hj− 1). B (j 1)k ··· − − k B (j 1) ··· − e − − ee e − − ee e 1 1 1 Using this and the Sublemma above (f = hh1− hj− 2, λ = B (j 2) and g = hj 1) we obtain ee ··· e − − − e −

1 1 1 µr,α(hj) = µr,α(hh1− hj− 1) B (j 1) ··· − e − − ee e 1 B (j 1) ( 1 1 1 1 2) − − µr,α(hh1− hj− 2) +C0µr,α(hh1− hj− 2) ≤ B (j 1) B (j 2) ··· − ··· − − − − − ee e ee e =(B (j 2))µr,α(hj 1) | − − {z − } 2 e B (j 2) C 0 2 = µr,α(hj 1) + − − µr,α(hj 1) − B (j 1) − e − − e for all 2 j B, and thus inductively ≤ ≤ 2 µr,α(hj) µr,α(h1) + C2µr,α(h1) e ≤ e e 1 C2 = µr,α(h) + µr,α(h) µr,α(h) B e B e e 1 | ≤{z } 2 4 20 = µr,α(h) µr,α(h) µr,α(Γi,A(f)) B e ≤ A ≤ A A provided µr,α(f) is suﬃciently small. For the last inequalities we used B 2 and (3.1). Summing up we have ≥ r 3. The Simplicity of Diﬀc(M) , 1 r dim(M) 53 ◦ ≤ ≤

(1) f = Id = hj = Id 1 j B ⇒ e ∀ ≤ ≤

(2) h = hB h1 e e ··· e n r,α n r,α n (3) f UA Diﬀ[ 2,2]i [ 2A,2A]n i (R ) = hj Diﬀ[ 2,2]i 1 R [ 2A,2A]n i (R ) , − − − ◦ ∈ e ∩ − × − ⇒ e ∈ r− × × − and hj depend continuously on f with respect to C -topologies. e

(4) There exists δ2 > 0 depending on n, r, α, A such that

20 µr,α(hj) µr,α(Γi,A(f)) 1 j B (3.7) e ≤ A ≤ ≤

n r,α n f UA Diﬀ (R ) with µr,α(f) δ2. ∀ ∈ e ∩ ≤

Now choose a periodic function ξ C∞(R, [0, 1]) of period 1 with ∈

ξ [m 1 ,m+ 1 ] 0, ξ [m+ 2 ,m+ 4 ] 1 m Z | − 6 6 ≡ | 6 6 ≡ ∀ ∈ and let

1 hj, := (ξ pri)(hj Id) + Id, hj,+ := hjhj,− 1 j B. e − ◦ e − e e e − ≤ ≤ n r,α n If UA is chosen suﬃciently small, then hj, Diﬀ[ 2A,2A]i 1 R [ 2A,2A]n i (R ) . In e e ± ∈ − − × × − − ◦ order to estimate µr,α(hj, ) in terms of µr,α(hj) we start with e ± e

µr,α(hj, ) = − e r r D ((ξ pri)(hj Id))(x) D ((ξ pri)(hj Id))(y) = sup k ◦ e − − ◦ e − k x=y Rn α( x y ) 6 ∈ k − k r k r k k r k r D (ξpri)D − (hj Id)(x) D (ξpri)D − (hj Id)(y) sup k e − − e − k ≤ x=y Rn X k α( x y ) 6 ∈ k=0 k − k r r Dk(ξ pr )(x) Dk(ξ pr )(y) Dr k(h Id)(x) ( i i − j sup k ◦ − ◦ kk e − k ≤ X k x=y Rn α( x y ) k=0 6 ∈ k − k Dk(ξ pr )(y) Dr k(h Id)(x) Dr k(h Id)(y) i − j − j ) + sup k ◦ kk e − − e − k x=y Rn α( x y ) 6 ∈ k − k r r ξ pri k,αµr k(hj) + ξ pri kµr k,α(hj) X n − − o ≤ k=0 k k ◦ k e k ◦ k e

C3µr,α(hj) ≤ e where C3 is a constant independent of A. If µr,α(hj) is small we thus have e 1 µr,α(hj,− ) 2µr,α(hj, ) e − ≤ e − r 54 3. The Simplicity of Diﬀc(M) , 1 r dim(M) ◦ ≤ ≤

(cf. Lemma 2.30(2)), and hence there exists a constant C4 > 0 independent of A, such that

1 µr,α(hj,+) = µr,α(hjhj,− ) − e e e 1 1 µr,α(hj) + µr,α(hj,− ) + C5µr,α(hj) µr,α(hj,− ) ≤ e e − e e − 1 | ≤{z } C4 (1 + 2C3)µr,α(hj) µr,α(Γi,A(f)), ≤ e ≤ A

provided µr,α(f) is small. For the last inequality we used (3.7). Summing up we have

(1) f = Id = hj, = Id 1 j B ⇒ e ± ∀ ≤ ≤ (2) h = hB,+hB, hB 1,+hB 1, h1,+h1, e e e −e − e − − ··· e e − (3) If m Z and 1 j B then ∈ ≤ ≤

hj, i 1 1 1 n i = Id hj,+ i 1 5 7 n i = Id R − [m ,m+ ] R − R − [m+ ,m+ ] R − e −| × − 6 6 × e | × 12 12 ×

n r,α n r,α n (4) f UA Diﬀ[ 2,2]i [ 2A,2A]n i (R ) = hj, Diﬀ[ 2,2]i 1 R [ 2A,2A]n i (R ) , − ± − − ◦ ∈ e ∩ − × − ⇒ e ∈ −r × × − and hj, depend continuously on f with respect to C -topologies. e ± (5) There exists δ4 > 0 depending on n, r, α, A and C4 > 0 depending on n, r, α but independent of A, such that

C4 µr,α(hj, ) µr,α(Γi,A(f)) 1 j B (3.8) e ± ≤ A ≤ ≤ n r,α n f UA Diﬀ (R ) with µr,α(f) δ4. ∀ ∈ e ∩ ≤ Using (3) above we may deﬁne Ψi,A(f) by e

Ψi,A(f) Ej = hj,+ Ej , Ψi,A(f) Ej = hj, Ej 1 j B + + − e | e | e | − e | − ≤ ≤ Ψi,A(f) Rn B (Ej Ej ) = Id j=1 + e | \S −∪

From (2) above and and (3.2) we get Γi,A(Ψi,a(f)) = h (cf. Lemma 2.33) and consequently e

1 1 r,α n r,α n Γi,A(Ψi,A(f))Γi,A(f)− = hh− g = g Gi f UA Diff (R ). e ∈ ∀ ∈ e ∩ n By shrinking UA if necessary we can apply Lemma 2.36, hence τi,AΨi,A(f) and τi,Af e r,α n e are conjugated in Diffc (R ) and thus ◦ r,α n n r,α n [Ψi,A(f)] = [f] H1(Diffc (R ) ) f UA Diff (R ). e ∈ ◦ ∀ ∈ e ∩ (4) of Lemma 3.1 follows immediately from (3.8), Lemma 2.33(4) and

µr,α(Ψi,A(f)) 2 max µr,α(hj,+), µr,α(hj, ) . 1 j B − e ≤ ≤ ≤ { e e } r 3. The Simplicity of Diﬀc(M) , 1 r dim(M) 55 ◦ ≤ ≤

The remaining assertions of Lemma 3.1 are now obvious. 2

n 3.2. Corollary. Let A 1 and let gA Diffc∞(R ) with gA [ 2,2]n = A Id (cf. page ≥ ∈ n ◦ 1 | − n · 41). Then there exists a neighborhood VA of Id Diff[ 2A,2A]n (R ) such that e ∈ − ◦ n 1 n ϑf,A : VA Diff[ 2A,2A]n (R ) → − ◦ e e 1 h Ψ1,A Ψn,A(gA− fhgA) 7→ e ··· e n is well defined f VA , and the following holds: ∀ ∈ e n r,α n (1) If α is a modulus of continuity, 1 r and f VA Diff (R ) then ≤ ≤ ∞ ∈ e ∩ n r,α n r,α n ϑf,A : VA Diff (R ) Diff[ 2A,2A]n (R ) e e ∩ → − ◦ is continuous with respect to Cr-topologies.

n r,α n r,α n (2) f, h VA Diﬀ (R ) [ϑf,A(h)] = [fh] H1(Diﬀc (R ) ) ∈ e ∩ ⇒ e ∈ ◦

n Proof. Similarly to the proof of Corollary 2.39 we ﬁnd VA such that ϑf,A(h) is well n n e e n deﬁned for all f, h VA . If f, h VA then supp(fh) [ 2A, 2A] and thus 1 ∈ en ∈ e ⊆ − supp(gA− fhgA) [ 2, 2] . Now (1) follows immediately from Lemma 3.1(2). Using Lemma 3.1(3) we⊆ obtain−

1 1 r,α n [ϑf,A(h)] = [Ψ1,A Ψn,A](gA− fhgA)] = [gA− fhgA] = [fh] H1(Diﬀc (R ) ) e e ··· e ∈ ◦ and thus Corollary 3.2 is proved. 2

3.3. Lemma. Let 1 r n and let α be a modulus of continuity. If r = n we ≤ ≤ further assume α(tx) √tα(x) x R and t 1. Then there exists A0 1 and ≤ ∀ ∈ ∀ ≥ ≥ 0 > 0 with

n µr,α(f), µr,α(h) f, h VA0 and µr,α(ϑf,A0 (h)) . ≤ ⇒ ∈ e e ≤ r,α n 0 < and f, h Diﬀ n (R ). 0 [ 2A0,2A0] ∀ ≤ ∀ ∈ −

Proof. First, using the assumption on r, we choose A0 1 with ≥ 1 n r n n 2 3C A − 1 if r < n and 3C A− 1 if r = n, 0 ≤ 0 ≤ where C is the constant of Lemma 3.1(4). Choosing 0 > 0 sufficiently small we n r,α n may arrange that f V for all f Diff n (R ), with µ (f) . From A0 [ 2A0,2A0] r,α 0 ∈ e ∈ − ≤ 1 a calculation similar to (2.27) we obtain (notice that the rôlesof gA0 and gA−0 are reversed) r 1 r 1 r D (g− fhgA ) = A − (D (fh) (A0 Id)), A0 0 0 ◦ · r 56 3. The Simplicity of Diffc(M) , 1 r dim(M) ◦ ≤ ≤

r,α n for all f, h Diﬀ n (R ), and thus [ 2A0,2A0] ∈ − r 1 r 1 1 D (gA−0 fhgA0 )(x) D (gA−0 fhgA0 )(y) µr,α(gA−0 fhgA0 ) = sup k − k x=y Rn α( x y ) 6 ∈ r k − kr r 1 D (fh)(A0x) D (fh)(A0y) = A0− sup k 1 − k x=y Rn α( A0x A0y ) 6 ∈ A0 r k r − k r D (fh)(A0x) D (fh)(A0y) A0 sup k − k if 1 r < n  n α( A0x A0y )  x=y R k − k ≤  6 ∈1 r r  r 1+ 2 D (fh)(A0x) D (fh)(A0y) ≤ A − sup k − k if r = n 0 α( A0x A0y ) x=y Rn k − k  6 ∈  r A0µr,α(fh) if 1 r < n ( 1 ≤ r 2 ≤ A0− µr,α(fh) if r = n

Using Lemma 3.1(4) and Lemma 2.30 we shrink 0, such that the following estimation holds:

1 µr,α(ϑf,A0 (h)) = µr,α(Ψ1,A0 Ψn,A0 (gA−0 fhgA0 )) e en ··· e C 1 µr,α(gA−0 fhgA0 ) ≤ A0 n r n C A0− µr,α(fh) if 1 r < n ( 1 ≤ n r 2 n ≤ C A0− − µr,α(fh) if r = n n r n C A0− 3 if 1 r < n ( 1 ≤ ) n 2 ≤ C A0− 3 if r = n ≤

r,α n 0 < and f, h Diff n (R ) with µ (f), µ (h) . 2 0 [ 2A0,2A0] r,α r,α ∀ ≤ ∀ ∈ − ≤ 3.4. Theorem (Mather). [8] Let M be a connected, smooth manifold, 1 r dim(M) and let α be a modulus of continuity. If r = dim(M) we further assume≤ that≤ r,α α(tx) √tα(x), x [0, ) and t 1. Then Diffc (M) is perfect. ≤ ∀ ∈ ∞ ∀ ≥ ◦ Proof. By Remark 2.22 it suffices to show

r,α n r,α n r,α n Diff n (R ) [Diff (R ) , Diff (R ) ], (3.9) [ 2A0,2A0] c c − ⊆ ◦ ◦

where A0 is the constant of Lemma 3.3. Analogously to the proof of Theorem 2.19 (cf. page 44) choose 0 < 0 such that ≤ r,α n L := h Diﬀ n (R ): µ (h) [ 2A0,2A0] r,α { ∈ − ≤ } has the ﬁxed-point property (cf. Lemma 2.41). By Corollary 3.2(1) and Lemma 3.3 n we have L VA0 , and for all f L ⊂ e ∈

ϑf,A L r,α n 0 | r,α n Diff n (R ) L Diff n (R ) [ 2A0,2A0] [ 2A0,2A0] − ⊇e−→ ⊆ − r 3. The Simplicity of Diffc(M) , 1 r dim(M) 57 ◦ ≤ ≤ is continuous with respect to Cr-topologies. Hence for every f L there exists ∈ hf L with ϑf,A0 (hf ) = hf . From Corollary 3.2(2) we get ∈ e r,α n [f][hf ] = [fhf ] = [ϑf,A0 (hf )] = [hf ] H1(Diffc (R ) ) e ∈ ◦ r,α n r,α n r,α n and thus [f] = [Id] H1(Diffc (R ) ). So f [Diffc (R ) , Diffc (R ) ] for all f ∈ r,α ◦ n ∈ ◦ ◦ ∈ L and since L generates Diff n (R ) (cf. Corollary 2.31) this yields (3.9). 2 [ 2A0,2A0] − 3.5. Corollary. Let M be a connected, smooth manifold and let 1 r dim(M). r,α ≤ ≤ Then Diffc (M) is perfect and hence simple by Corollary 1.17. ◦ Proof. This is an immediate consequence of Theorem 3.4, Lemma 2.17 and Re- mark 2.18. 2 4. Isomorphisms between groups of diffeomorphisms

4.1. Definition (Property P). Let M be a smooth, connected manifold and let G be a group of diffeomorphisms, i.e. G Diffr(M) for some 1 r . We say G has Property P (path transitivity) iff⊆ for every smooth path ≤σ : I≤ ∞M and for every open neighborhood U of Im(σ) there exists g G with supp(g)→ U and g(σ(0)) = σ(1). ∈ ⊆ If x M we denote by SxG the isotropy- or stabilizer subgroup at x, i.e. ∈ SxG := g G : g(x) = x { ∈ } and for g G we let ∈ Fix(g) := x M : g(x) = x . { ∈ } If we speak of a group of diffeomorphisms Gr(M) we will always assume that M is a smooth, connected manifold, 1 r and Gr(M) Diffr(M) is a subgroup. ≤ ≤ ∞ ⊆ 4.2. Lemma. Let Gr(M) be a group of diffeomorphisms that has Property P. If r r f, g G (M) SxG (M) and f(x), g(x) are contained in the same component of M ∈x then \ \{ } r r f (SxG (M))g(SxG (M)) (4.1) ∈ r r Especially SxG (M) is a maximal subgroup of G (M). Proof. Since f(x), g(x) are contained in the same component of M x there exists a smooth path connecting f(x) = x and g(x) = x, that doesn’t meat\{x.} By Property P r 6 6 1 r we obtain h SxG (M) with h(f(x)) = g(x). Hence g− hf SxG (M) and thus r ∈ r ∈ f (SxG (M))g(SxG (M)). ∈ r r In order to show that SxG (M) is a maximal subgroup of G (M) notice first that r r r r SxG (M) = G (M) by Property P. Now let SxG (M) H G (M) and choose 6 r r ⊂ ⊆r h H SxG (M). We have to show H = G (M), but since SxG (M) H it suffices ∈ \ r r r r ⊂ to show G (M) SxG (M) H. If f G (M) SxG (M) then either f(x), h(x) 1 \ ⊆ ∈ \ { } or f(x), h− (x) is contained in one component of M x and we can apply (4.1). Hence{ we obtain} \{ } r 1 r f SxG (M) h± SxG (M) H, ∈ ∈ H H H | ∈{z } |∈{z} | ∈{z } r r and thus G (M) SxG (M) H. 2 \ ⊆ 4.3. Lemma. [1] [5] Let Gr(M) and Gs(N) be groups of diffeomorphisms having Property P, and let Φ: Gr(M) Gs(N) be an isomorphism of groups. Notice that we do not consider any topology→ on Gr(M) and Gs(N) and hence do not require Φ to be continuous. Assume further that there exists x0 M and y0 N with r s ∈ ∈ Φ(Sx0 G (M)) = Sy0 G (N). Then there exists a unique homeomorphism ϕ : M N inducing Φ, i.e. →

1 r Φ(f) = ϕ f ϕ− f G (M) (4.2) ◦ ◦ ∀ ∈ 58 4. Isomorphisms between groups of diﬀeomorphisms 59

Proof. First we define ϕ : M N: given x M, we choose g Gr(M) with → ∈ ∈ g(x0) = x (Property P) and let ϕ(x) := Φ(g)(y0). Too see, that this is well defined r 1 r take another g0 G (M) with g0(x0) = x. Then g− g0 Sx0 G (M) and hence 1 ∈ 1 s 1 ∈ Φ(g)− Φ(g0) = Φ(g− g0) Sy0 G (N). But then Φ(g)− Φ(g0)(y0) = y0 and thus ∈ 1 s r 1 Φ(g0)(y0) = Φ(g)(y0). Using Φ− : G (N) G (M) one defines analogously ϕ− : N M. Clearly this is the inverse of ϕ.→ To check (4.2) take f Gr(M), x M → r ∈ ∈ and g G (M) with g(x0) = x. Then (fg)(x0) = f(x) and thus ∈

Φ(f)(ϕ(x)) = Φ(f)Φ(g)(y0) = Φ(fg)(y0) = ϕ(f(x)). Since x and f were arbitrary we have shown Φ(f) ϕ = ϕ f for all f Gr(M), but this is (4.2). ◦ ◦ ∈ Now suppose there exists another bijection ψ : M N inducing Φ. Then 1 r → ψ− ϕ : M M is bijective and for all f G (M) we have ◦ → ∈ 1 1 1 1 1 (ψ− ϕ)f(ψ− ϕ)− = ψ− Φ(f) ψ = Φ− (Φ(f)) = f. (4.3) ◦ ◦ ◦ ◦ 1 If ϕ = ψ then there exists x M with ψ− (ϕ(x)) = x and by Property P we 6 r ∈ 1 6 1 can choose f G (M) with f(x) = x and f(ψ− (ϕ(x))) = ψ− (ϕ(x)). Since 1 ∈ 6 f(ψ− (ϕ(x))) = x we have 6 1 1 1 1 1 (ψ− ϕ)f(ψ− ϕ)− (ψ− (ϕ(x))) = (ψ− ϕ)(f(x)) ◦ ◦ 1 ◦ 1 = ψ− (ϕ(x)) = f(ψ− (ϕ(x))) 6 1 1 1 and hence (ψ− ϕ)f(ψ− ϕ)− = f, a contradiction to (4.3). So ϕ = ψ and we have proven the◦ uniqueness◦ of ϕ, stated6 above. 1 It remains to show, that ϕ and ϕ− are continuous. Therefore look at

r M := M Fix(f): f G (M) U { \ ∈ s } N := N Fix(g): g G (N) . U { \ ∈ }

Obviously M and N consist of open sets, and by Property P every neighborhood U U of x M contains an open neighborhood M Fix(f) of x. Thus M is a basis of the ∈ \ U topology on M. Similarly N is a basis of the topology on N. We have U 1 x Fix(Φ(f)) Φ(f)(x) = x ϕ f ϕ− (x) = x ∈ ⇔ 1 ⇔ ◦ ◦ ϕ− (x) Fix(f) x ϕ(Fix(f)), (4.4) ⇔ ∈ ⇔ ∈ r 1 and thus Fix(Φ(f)) = ϕ(Fix(f)) for all f G (M). Hence ϕ and ϕ− induce ∈ bijections between M and N and are thus continuous. 2 U U 4.4. Lemma. Let Gr(M) and Gs(N) be groups of diﬀeomorphisms, which have Property P, and let Φ: Gr(M) Gs(N) be an isomorphism of groups. Suppose → there exists y0 N and a closed, nonempty, proper subset A of M (i.e. A M, A = and A =∈M), such that ⊆ 6 ∅ 6 1 s g(A) = A g F := Φ− (Sy G (N)). ∀ ∈ 0 60 4. Isomorphisms between groups of diﬀeomorphisms

r Then A A◦ = x and F = SxG (M). \ { }

Proof. Since M is connected we have A A◦ = . Suppose A A◦ = x for some x M. Then, since g is continuous\ and 6g(A∅) = A for all g\ F , we{ } have ∈ r s ∈ g(x) = x, g F , and hence F SxG (M). By Lemma 4.2 Sy0 G (N) is a maximal ∀ ∈s ⊆1 s r subgroup of G (N), so F = Φ− (Sy0 G (N)) is a maximal subgroup of G (M), and r thus F = SxG (M). Now suppose A A◦ contains at least two distinct points x1 and x2. We then \ choose U1 and U2, neighborhoods of x1 and x2, with U1 U2 = . Using Property P r ∩ ∅ we find gi G (M), i = 1, 2 with supp(gi) Ui and gi(xi) / A. Hence gi / F ∈ s ⊆ ∈ 1 ∈ and thus Φ(gi) / Sy0 G (N). If necessary we replace g1 by g1− and obtain gi r ∈ ∈ G (M) F , such that Φ(g1)(y0), Φ(g2)(y0) is contained in one component of N y0 . \ { } r \{ } Further we let g3 := g2g1. Then g3 G (M) F and Φ(gi)(y0): i = 1, 2, 3 ∈ \ { } is contained in one component of N y0 . From Lemma 4.2 we obtain Φ(gi) s s \{ } ∈ (Sy0 G (N))Φ(g3)(Sy0 G (N)) and thus gi F g3F for i = 1, 2. Next define Ai := 1 ∈ A g− (M A) for i = 1, 2, 3 and notice that Ai Ui, A3 = A1 A2 and x2 A2 = . ∩ i \ ⊆ ∪ ∈ 6 ∅ 1 Claim 1. If i 1, 2 and t g− F gi F , then t(Ai) = A3. ∈ { } ∈ 3 ∩ 1 Proof of Claim 1. Since t F we have t(A) = A, and since t g3− F gi we find f F 1 ∈ ∈ ∈ with t = g− fgi. From g3t(Ai) = fgi(Ai) f(M A) = M A, we obtain 3 ⊆ \ \ 1 t(Ai) A g− (M A) = A3, (4.5) ⊆ ∩ 3 \ 1 1 1 and since git− (A3) = f − g3(A3) f − (M A) = M A we have ⊆ \ \ 1 1 t− (A3) A g− (M A) = Ai. (4.6) ⊆ ∩ i \

Combining (4.5) and (4.6) we get t(Ai) = A3 2

1 Claim 2. There exist si, ti g− F gi F for i = 1, 2 with s1s2 = t2t1. ∈ 3 ∩ s Proof of Claim 2. Using Property P we may choose z N and s10 , s20 , t10 G (N) with ∈ ∈

1 1 1 s20 (y0) = y0, s20 Φ(g2− )(y0) = Φ(g3− )(y0), s20 (z) = Φ(g1− )(y0) 1 1 1 t10 (y0) = y0, t10 Φ(g1− )(y0) = Φ(g3− )(y0), t10 (z) = Φ(g2− )(y0)

1 This is possible, even if dim(N) = 1 since Φ(gi− )(y0): i = 1, 2, 3 is contained in one { 1 } s component of N y0 . We let s10 := t10 and t20 := s10 s20 t10− . Then si0 , ti0 Sy0 G (N), 1\{ } s ∈ and s0 , t0 Φ(g− )(Sy G (N))Φ(gi) since we have i i ∈ 3 0 1 1 Φ(g3)si0 Φ(gi− )(y0) = Φ(g3)Φ(g3− )(y0) = y0 1 1 Φ(g3)t10 Φ(g1− )(y0) = Φ(g3)Φ(g3− )(y0) = y0 1 1 1 Φ(g3)t20 Φ(g2− )(y0) = Φ(g3)s10 s20 t10− Φ(g2− )(y0)

= Φ(g3)s10 s20 (z) 1 = Φ(g3)Φ(g3− )(y0) = y0. 4. Isomorphisms between groups of diﬀeomorphisms 61

1 1 1 s If we let ti := Φ− (ti0 ) and si := Φ− (si0 ), Claim 2 follows from Φ− (Sy0 G (N)) = F . 2 Using Claim 1 we obtain

s1s2(A2) = s1(A3) s1(A1) = A3 ⊇ t2t1(A1) = t2(A3) t2(A2) = A3, ⊇ but since A1 A2 = and A3 = , this is a contradiction to the bijectivity of ∩ ∅ 6 ∅ s1s2 = t2t1. So A A◦ contains exactly one point, and we are done. 2 \ 4.5. Remark. In the situation of Lemma 4.4 assume additionally dim(M) 2. r ≥ Then A = x , for if x = x0 A we find f SxG (M) = F with f(x0) / A (Property P),{ a} contradiction6 to∈f(A) = A for all∈ f F . If dim(M) = 1, however,∈ ∈ this is not true. For example take M = R, A = [0, ), x = 0 and G = Diffc∞(R) . ∞ ◦ Then f(A) = A for all f SxG. ∈ 4.6. Definition. If Gr(M) is a group of diffeomorphisms, we denote by

Gr(M) := Gr(M) Diﬀr(M) c ∩ c the subgroup of all compactly supported Gr(M)-diﬀeomorphisms. If U M is an arbitrary subset we let ⊆

Gr (M) := Gr(M) Diffr (M), U ∩ U the subgroup of all Gr(M)-diffeomorphisms supported on U.A Gr(M)-diffeotopy is r r a C -mapping H : M I M, such that Ht := H( , t) G (M) and H0 = Id. By Gr(M) we denote all× G→r(M)-diffeomorphisms, that· are ∈Gr(M)-diffeotopic to Id. r ◦ r r r By Gc(M) we will always mean (Gc(M)) . Hence f Gc(M) iff f is Gc(M)- ◦ r r ◦ ∈ ◦ diffeotopic to Id. Similarly GU (M) = (GU (M)) . Like we did in the first Chapter, we let ◦ ◦ ϕ r(M) := ϕ(Bn): Rn , M Cr-embedding . U { → } 4.7. Remark. Notice that Gr(M) is a subgroup of Gr(M), for if F,H : M I I r ◦ × → are G (M)-diffeotopies with F0 = H0 = Id then

1 (x, t) Ft(H1 t(H1− (x))) 7→ − r 1 r is a G (M)-diﬀeotopy from F1H− to Id. Further, if g G (M) then 1 ∈ 1 (x, t) g− (Ft(g(x))) 7→ r 1 is a G (M)-diﬀeotopy from g− F1g to Id. Thus, analogously to Remark 1.8 we have

r r r r r r Gc(M) ¡ G (M),Gc(M) ¡ G (M) ,Gc(M) ¡ Gc(M) r ◦ r r ◦ r ◦ ◦ Gc(M) ¡ G (M),G (M) ¡ G (M) ◦ for all groups of diﬀeomorphisms Gr(M). 62 4. Isomorphisms between groups of diﬀeomorphisms

4.8. Remark. Since we do not consider any topology on Gr(M) the definition of Gr(M) is different to the one given in Definition 1.6. But by Corollary 1.14 they coincide◦ in the case Gr(M) = Diffr(M). More precisely, for U r(M) we have: ∈ U r r r r r G (M) Gc(M) Gc(M) GU (M) GU (M) r r r ◦ r r ◦ Diff (M) Diffc(M) Diffc(M) DiffU (M) DiffU (M) r r r ◦ r r ◦ Diffc(M) Diffc(M) Diffc(M) DiffU (M) DiffU (M) r r r ◦ r r ◦ Diffc(M) Diffc(M) Diffc(M) DiffU (M) DiffU (M) r ◦ r ◦ r ◦ r ◦ r ◦ DiffU (M) DiffU (M) DiffU (M) DiffU (M) DiffU (M) r r r ◦ r r ◦ DiffU (M) DiffU (M) DiffU (M) DiffU (M) DiffU (M) ◦ ◦ ◦ ◦ ◦

4.9. Definition (Property L). A group of diffeomorphisms Gr(M) is said to have Property L (locality), iff for every subgroup F Gr(M) and for every open cover r r r ⊆ r r (M) with [GU (M) ,GU (M) ] F U we have [Gc(M) ,Gc(M) ] F . U ⊆ U ◦ ◦ ⊆ ∀ ∈ U ◦ ◦ ⊆ 4.10. Lemma. Let Gr(M), Gs(N) be groups of diffeomorphisms, which have Prop- r r erty P and Property L. Assume further, that Gc(M) and Gc(N) are nonabelian, and let Φ: Gr(M) Gs(N) be an isomorphism of groups.◦ For every◦ y N we let → ∈ r r r 1 s y := U (M):[GU (M) ,GU (M) ] Φ− (SyG (N)) C { ∈ U ◦ ◦ ⊆ }

and Cy := M U y U. Then we have \ S ∈C

(1) Cy is closed in M

1 s (2) f(Cy) = Cy f Φ− (SyG (N)) ∀ ∈

(3) Cy = 6 ∅

1 s Proof. (1) is obvious, since y consists of open sets. To see (2) take f Φ− (SyG (N)) C r ∈ and U y. Then f(U) (M) and ∈ C ∈ U r r r 1 r 1 [Gf(U)(M) ,Gf(U)(M) ] = [fGU (M) f − , fGU (M) f − ] ◦ ◦ ◦ ◦ r r 1 = f[GU (M) ,GU (M) ]f − ◦ ◦ 1 s φ− (SyG (N)). ⊆

So we have f(U) y, and hence f induces a bijection on y. Thus ∈ C C f(C ) = f(M U) = M f( U) = M U = C . y [ [ [ y \ U y \ U y \ U y ∈C ∈C ∈C

In order to prove (3), suppose conversely Cy = . Hence y is an open cover of M with the following property: ∅ C

r r 1 s [GU (M) ,GU (M) ] Φ− (SyG (N)) U y. ◦ ◦ ⊆ ∀ ∈ C 4. Isomorphisms between groups of diﬀeomorphisms 63

If z N we may choose g Gs(N) with g(y) = z (Property P), and let f := 1 ∈ r ∈ r Φ− (g) G (M). Clearly (f(U))U y is an open cover of M by sets of (M), and ∈ ∈C U for all U y we have ∈ C r r r 1 r 1 [Gf(U)(M) ,Gf(U)(M) ] = [fGU (M) f − , fGU (M) f − ] ◦ ◦ ◦ ◦ r r 1 = f[GU (M) ,GU (M) ]f − ◦ ◦ 1 s 1 fΦ− (SyG (N))f − ⊆ 1 s 1 = Φ− (gSyG (N)g− ) 1 s = Φ− (SzG (N)). Hence, using Property L of Gr(M), we obtain

r r 1 s [Gc(M) ,Gc(M) ] Φ− (SzG (N)) z N ◦ ◦ ⊆ ∀ ∈ and thus

r r 1 s 1 s [Gc(M) ,Gc(M) ] Φ− (SzG (N)) = Φ− ( SzG (N)) = IdM ◦ ◦ \ \ ⊆ z N z N { } ∈ ∈ r This is a contradiction, since Gc(M) is nonabelian. 2 ◦ 4.11. Definition (Property B). Let Gr(M) be a group of diffeomorphisms. Gr(M) is said to have Property B (ball), iff U = ϕ(Bn) r(M) there exists f Gr(M) with Fix(f) = (M U) ϕ(0) . ∀ ∈ U ∈ \ ∪ { } 4.12. Lemma. Let Gr(M) and Gs(N) be groups of diffeomorphisms having Prop- erty P and Property B. If Φ: Gr(M) Gs(N) is an isomorphism of groups, then 1→ s y N there exists Id = f F := Φ− (SyG (N)), such that Fix(f)◦ = . ∀ ∈ 6 ∈ 6 ∅ n Proof. Suppose conversely Fix(f)◦ = , Id = f F . Choose xi = ϕi(0) ϕi(B ) = r ∅ ∀ 6 ∈ r ∈ Ui (M), i = 1,..., 4, such that Ui Uj = i = j. Since G (M) has Property B, ∈ U r ∩ ∅ ∀ 6 we may choose gi G (M) such that Fix(gi) = (M Ui) xi . Since Fix(gi)◦ = ∈ s \ ∪ { } 6 ∅1 we have gi / F , hence Φ(gi) / SyG (N) and thus we may assume (replace gi by g− ∈ ∈ i if necessary) that Φ(gi)(y): i = 1,..., 4 is contained in one component of N y . { } \{ } Further let g5 := g3g4 and U5 := U3 U4. Similarly to Claim 2 on page 60, we find 1 ∪ si, ti g− F gi F , i = 1, 2, with s1s2 = t2t1. ∈ 5 ∩ 1 Sublemma. Let i 1, 2 and let t g− F gi F . Then M U5 t(Ui). ∈ { } ∈ 5 ∩ \ ⊆ 1 Proof of the Sublemma. Suppose conversely M U5 t(Ui). Then t− (M U5) 1 \ 6⊆ \1 ∩ (M Ui) = . Since t g− F gi there exists s F with sg5t = gi. If x t− (M \ 6 ∅ ∈ 5 ∈ ∈ \ U5) (M Ui) we have st(x) = sg5t(x) = gi(x) = x. Hence ∩ \ 1 = t− (M U5) (M Ui) Fix(st)◦ ∅ 6 \ ∩ \ ⊆ 1 1 and thus st = Id, since st F . But then t− g5t = gi, and thus Fix(gi) = t− (Fix(g5)). ∈ So Fix(gi) and Fix(g5) are homeomorphic, but

Fix(h5) = (M (U3 U4)) x3, x4 \ ∪ ∪ { } 64 4. Isomorphisms between groups of diﬀeomorphisms

has two discrete points, and

Fix(gi) = (M Ui) xi \ ∪ { } has only one discrete point, a contradiction. 2

Applying the Sublemma to si and ti above, we obtain

s1s2(U2) s1(M U5) s1(U1) M U5, ⊇ \ ⊇ ⊇ \ and t2t1(U1) t2(M U5) t2(U2) M U5. ⊇ \ ⊇ ⊇ \ But this is a contradiction, since M U5 = , U1 U2 = and s1s2 = t2t1 is bijective. \ 6 ∅ ∩ ∅ 2

4.13. Lemma. Let Gr(M) and Gs(N) be groups of diﬀeomorphisms which have Property P and Property B, and let Φ: Gr(M) Gs(N) be an isomorphism of → groups. For y N consider the set Cy := M U y U, where ∈ \ S ∈C r r r 1 s y := U (M):[GU (M) ,GU (M) ] Φ− (SyG (N)) , C { ∈ U ◦ ◦ ⊆ }

we have introduced in Lemma 4.10 and for x M let Dx := N V x V , where ∈ \ S ∈D s s s r x := V (N):[GV (N) ,GV (N) ] Φ(SxG (M)) . D { ∈ U ◦ ◦ ⊆ } Then either there exists y0 N with Cy = M or there exists x0 M with Dx = N. ∈ 0 6 ∈ 0 6 1 s Proof. Choose y0 N. Then by Lemma 4.12 there exists Id = f0 Φ− (Sy G (N)) ∈ 6 ∈ 0 with A := Fix(f0)◦ = . We have y0 B := Fix(Φ(f0)), thus B = , and B = N 6 ∅ ∈ r 6 ∅ 6 since f0 = Id. Now consider the following two subgroups of G (M): 6 1 s H := Φ− ( g G (N): g(B) = B ) 1 { ∈ s } K := Φ− ( g G (N): B Fix(g) ) { ∈ ⊆ } We then have: (1) K ¡ H

(2) f0 K and thus K,H = Id ∈ 6 { } 1 s (3) K Φ− (Sy G (N)) ⊆ 0 (4) Gr (M) H A ⊆ r Everything is obvious, except (4). If f G (M) we have ff0 = f0f since supp(f) ∈ A ∩ supp(f0) A (M A) = . Hence ⊆ ∩ \ ∅ 1 Φ(f)(B) = Φ(f)(Fix(Φ(f0))) = Fix(Φ(f)Φ(f0)Φ(f)− ) 1 = Fix(Φ(ff0f − )) = Fix(Φ(f0)) = B, and thus f H. We now∈ distinguish between two possibilities: 4. Isomorphisms between groups of diﬀeomorphisms 65

Case a) A Fix(k) k K ⊆ ∀ ∈

Case b) k K x0 A : k(x0) = x0 ∃ ∈ ∃ ∈ 6 s In case a) we choose x0 A and V (N) with V N B. This is possible since ∈ s ∈ U ⊆ \ B = N and B is closed. If g GV (N) then Fix(g) N V N (N B) = B, 6 1 ∈1 ⊇ \ ⊇ 1 \ \ r hence Φ− (g) K and thus Φ− (g)(x0) = x0, since x0 A. So Φ− (g) Sx0 G (M) ∈ r ∈ ∈ and thus g Φ(Sx G (M)), Therefore we have shown ∈ 0 s s s r [GV (N) ,GV (N) ] GV (N) Φ(Sx0 G (M)), ◦ ◦ ⊆ ⊆ and consequently V x0 , that is Dx0 = N. So it remains to prove∈ D Lemma 4.136 under the additional assumption of case b), i.e. there exists k K and x0 A with k(x0) = x0. Using Property P we ﬁnd r ∈ ∈ 6 1 1 g1, g2 G (M) with x0 = gi(x0), gi(k(x0)) = k(x0) and g− (x0) = g− (x0). If we let ∈ A 6 1 6 2 ki := [gi, k] for i = 1, 2, we obtain

1 1 1 1 1 ki(x0) = g− k− gik(x0) = g− k− k(x0) = g− (x0) = x0, (4.7) i i i 6 r and ki K, since k K ¡ H and gi GA(M) H. From (4.7) we get x0 = ki(x0), ∈ ∈ 1 ∈ ⊆ 6 k1(x0) = k2(x0), and thus x0 = ki− (x0) since ki are bijections. Using the continuity of 6 r 6 1 ki we ﬁnd U (M) with x0 U A and such that ki(U) U = , ki− (U) U = ∈ U ∈ ⊆ r 1 1 ∩ ∅ ∩ ∅ and k1(U) k2(U) = . If h1, h2 GU (M) then [hi− , ki− ] K for i = 1, 2, since 1 ∩ 1 ∅ r ∈ r ∈ k− K ¡ H and h− G (M) G (M) H. Further i ∈ i ∈ U ⊆ A ⊆ 1 1 kihi− ki− on ki(U) 1 1  [hi− , ki− ] =  hi on U Id otherwise  for i = 1, 2, and

1 1 h− h− h1h2 on U  1 2  1 1 1 1  Id on k1(U)  [[h1− , k1− ], [h2− , k2− ]] =   = [h1, h2], Id on k2(U)  Id otherwise    1 s r and thus [h1, h2] K Φ− (Sy0 G (N)). Since h1, h2 GU (M) were arbitrary, we have shown ∈ ⊆ ∈

r r r r 1 s [GU (M) ,GU (M) ] [GU (M),GU (M)] Φ− (Sy0 G (N)). ◦ ◦ ⊆ ⊆

So U y and thus Cy = M. 2 ∈ C 0 0 6 4.14. Theorem (Filipkiewicz, Banyaga). [5] [1] Let Gr(M) and Gs(N) be groups of diﬀeomorphisms that have Property P, Property B, Property L and assume that r s r s Gc(M) , Gc(N) are nonabelian. If Φ: G (M) G (N) is an isomorphism of ◦ ◦ → 66 4. Isomorphisms between groups of diﬀeomorphisms

groups (we do not assume that Φ is continuous with respect to any topology), then there exists an unique homeomorphism ϕ : M N with → 1 r Φ(f) = ϕ f ϕ− f G (M). ◦ ◦ ∀ ∈

1 Proof. By Lemma 4.13 there exists y0 N with Cy = M (replace Φ by Φ− if ∈ 0 6 necessary). From Lemma 4.10 we obtain that Cy0 is a nonempty, closed subset of M, 1 s invariant under Φ− (Sy G (N)). So we can apply Lemma 4.4 and get x0 M with 0 ∈ 1 s r Φ− (Sy0 G (N)) = Sx0 G (M), and thus, by Lemma 4.3, we are ﬁnished. 2

4.15. Lemma. Let be a cover of Bn by open sets of Rn and let 0 < r 1. Then U n ≤ there exist m N, U1,...,Um and fi, gi DiffU∞(R ) such that ∈ ∈ U ∈ i ◦ n n [fm, gm] [f1, g1](B ) B (0) ··· ⊆ r Proof. We fix and let U R := r (0, 1] : Lemma 4.15 holds for r . { ∈ } Obviously R is an interval containing 1, and thus we are done if we show r0 := n n inf R = 0. Suppose conversely r0 > 0 and choose > 0, such that x B , B (x) is n ∀ ∈ contained in some U . Now choose x1, . . . , xk ∂B (0) with ∈ U ∈ r0 k n n ∂B (0) B (xi) r0 [ 2 ⊆ i=1 n and choose gi DiffB∞n(x )(R ) such that ∈ i ◦ n n n 1 n n gi(B (xi)) R Br0+ (0) and gi− (B (xi)) Br0 (0). 2 ⊆ \ 2 2 ⊆ − 2 n By the choice of there exist U1,...,Uk with B (xi) Ui and hence gi n ∈ U ⊆ ∈ DiffU∞(R ) . Further we choose 0 < 0 < with i ◦ 2 k n n n B (0) B (0) B (xi) r0+0 r0 0 [ 2 \ − ⊆ i=1 n and f DiffB∞n (0) Bn (0)(R ) with r + r ◦ ∈ 0 0 \ 0− 0 1 n n f − (B (0)) B (0). r0+ 0 r0 0 2 ⊆ − 2 n By Claim 2 on page 10 we find m N and f Diff∞n (R ) with f = f f . j B (xi(j)) 1 m ∈ ∈ 2 ◦ ··· n Hence fj DiffU∞ (R ) and we have ∈ i(j) ◦ 1 n fj− on B (xi(j))  2 1 1 n hj := [fj, gi(j)] =  g− fjgi on g− (B (xi(j))) .  i(j) i(j) 2  Id otherwise  4. Isomorphisms between groups of diffeomorphisms 67

Especially n n 1 n hm h1(Br0 (0)) Br0 (0) = f − (Br0 (0)). (4.8) ··· − 2 ⊆ − 2 − 2 n Since supp(fj) B (xi(j)) we have ⊆ 2 1 hj Bn (0) Bn (0) = f − Bn (0) Bn (0) r + r j r + r | 0 2 \ 0− 2 | 0 2 \ 0− 2 and thus 1 hm h1 Bn (0) Bn (0) = f − Bn (0) Bn (0). (4.9) r + r r + r ··· | 0 2 \ 0− 2 | 0 2 \ 0− 2 (4.8) and (4.9) together yield

n 1 n n hm h1(B (0)) f − (B (0)) B (0), r0+ 0 r0+ 0 r0 0 ··· 2 ⊆ 2 ⊆ − 2 a contradiction to r0 = inf R. 2 4.16. Lemma. [5] Let M be a connected, smooth manifold, 1 dim(M) < , r r ≤r r ∞ 1 r and let G (M) be one of the groups Diff (M), Diffc(M), Diffc(M) . ≤ ≤r ∞ r ◦ Then Gc(M) is nonabelian and G (M) has Property P, Property B and Property L. ◦ r Proof. Obviously Gc(M) is nonabelian (cf. Remark 4.8). From Lemma 1.11 we r r ◦ r obtain Diffc(M) G (M) has Property P, and hence G (M) has it too. Given U = n r ◦ ⊆ ϕ(B ) (M) we choose λ C∞([0, ), [0, )) with λ(x) = 0 x [1, ) and ∈ U ∈ ∞2 ∞ n ⇔X ∈ ∞ n consider the vector field X(x) := λ( x ) x on R . Then f := Fl1 Diff∞n (R ) k k · ∈ B ◦ and Fix(f) = (Rn Bn) 0 . If we let \ ∪ { } 1 n ϕ f ϕ− on ϕ(R ) f := ( ◦ ◦ e Id otherwise

r r r then f Diﬀc(M) G (M) and Fix(f) = (M U) ϕ(0) . Thus G (M) has Propertye ∈ B. ◦ ⊆ e \ ∪ { } r r r In order to show that G (M) has Property L, recall that GU (M) = DiﬀU (M) for all U r(M) (cf. Remark 4.8). Let F be a subgroup of Gr(M) and◦ let r(◦M) be an∈ open U cover of M, such that U ⊆ U

r r r r [DiffU (M) , DiffU (M) ] = [GU (M) ,GU (M) ] F U . ◦ ◦ ◦ ◦ ⊆ ∀ ∈ U Now consider the following subgroup of Diffr(M):

r r r H := [DiffV (M) , DiffV (M) ]: V (M) Diffr(M) h ◦ ◦ ∈ U i Obviously we have r r Id = H [Diffc(M) , Diffc(M) ], { } 6 ⊆ ◦ ◦ r r r r and H ¡ Diff (M), for if V (M), h1, h2 DiffV (M) and f Diff (M) then f(V ) r(M) and thus ∈ U ∈ ◦ ∈ ∈ U 1 1 1 r 1 r 1 f[h1, h2]f − = [fh1f − , fh2f − ] [fDiffV (M) f − , fDiffV (M) f − ] H. ∈ ◦ ◦ ⊆ =[Diffr (M) ,Diffr (M) ] | f(V ) {z◦ f(V ) ◦ } 68 4. Isomorphisms between groups of diffeomorphisms

r r Since [Diffc(M), Diffc(M)] is simple (cf. Corollary 1.17) we obtain r r H = [Diffc(M) , Diffc(M) ], ◦ ◦ and it thus remains to show

r r r [DiﬀV (M) , DiﬀV (M) ] F V (M), (4.10) ◦ ◦ ⊆ ∀ ∈ U for then by Remark 4.8

r r r r [Gc(M) ,Gc(M) ] = [Diffc(M) , Diffc(M) ] = H F. ◦ ◦ ◦ ◦ ⊆ In order to prove (4.10) let V = ϕ(Bn) r(M). Then, since covers M, 1 n ∈ U U (ϕ− (U))U is an open cover of R , and we can apply Lemma 4.15. Hence there ∈U n exist U0,...,Um and fi, gi Diffϕ∞ 1(U )(R ) with ∈ U ∈ − i ◦ n 1 [fm, gm] [f1, g1](B ) ϕ− (U0). ··· ⊆ If we let 1 n ϕ fi ϕ− on ϕ(R ) fi := ( ◦ ◦ e Id otherwise ϕ g ϕ 1 on ϕ(Rn) g := ( i − i Id◦ ◦ otherwise e we obtain f , g Diffr (M) , h := [f , g ] [f , g ] F , h(V ) U and thus i i Ui m m 1 1 0 e ∈ ◦ e e ··· e ∈ e ⊆ ∈ U er r e r e r [DiffV (M) , DiffV (M) ] [Diff 1 (M) , Diff 1 (M) ] ◦ ◦ ⊆ h− (U0) ◦ h− (U0) ◦ 1 e r re = h− [DiffU0 (M) , DiffU0 (M) ]h F. e ◦ ◦ e ⊆ This completes the proof of Lemma 4.16. 2

4.17. Theorem (Filipkiewicz). [5] Let M and N be connected, smooth manifolds, r r r r s let G (M) be one of the groups Diff (M), Diffc(M), Diffc(M) and let G (N) be s s s r s ◦ Diff (N), Diffc(N) or Diffc(N) . If Φ: G (M) G (N) is an isomorphism of groups (we do not assume that Φ◦ is continuous with→ respect to any topologies) then r = s and there exists an unique Cr-diffeomorphism ϕ : M N inducing Φ, i.e. → 1 r Φ(f) = ϕ f ϕ− f G (M). ◦ ◦ ∀ ∈

Proof. By Lemma 4.16 we can apply Theorem 4.14 and obtain an unique homeomorphism ϕ : M N inducing Φ. Hence it remains to show r = s and ϕ is a Cr diffeomorphism→ . − Sublemma. (cf. p. 77-78 in [2]) If m0 M and n = dim(M) then there exists an n ∈ C∞-embedding ι : R , M with ι(0) = m0, → l : Rn M M × → ι(ι 1(m) + x) on ι(Rn) (x, m) ( − 7→ m otherwise 4. Isomorphisms between groups of diﬀeomorphisms 69

n is a C∞-action of the Lie group (R , +) on M, and lx := l(x, ) Diﬀc∞(M) for all x Rn. · ∈ ◦ ∈

Proof of the Sublemma. Choose λ Diff∞([0, 1), [0, )) with ∈ ∞ 1 t 0 t 3 λ(t) := ( 1 ≤ ≤ ., e (1 t)2 2 t < 1 − 3 ≤ n n and define σ Diff∞(B , R ) by ∈ σ : Bn Rn → λ( x ) x k k x 7→ x · k k n n If y R we denote by ∂y the vector field constant y on R and define a vector field ∈ n Xy on R by σ (∂ ) on Bn X (x) := ( ∗ y . y 0 otherwise

An elementary calculation, using the special choice of λ, shows that Xy is C∞ for all n n y R . Choose a C∞-chart ψ on M with ψ(m0) = 0 and Im(ψ) = R . Then the vector∈ ﬁeld ψ∗(Xy) on dom(ψ) Xy := ( g 0 otherwise

Xy n is C∞ and Fl1 Diﬀc∞(M) for all y R . On the other side we have f ∈ ◦ ∈

ψ∗σ∗(∂y) 1 n Xy ( Fl1 (m) on (σ ψ)− (R ) Fl1 (m) = 0 ◦ f Fl1(m) otherwise ((σ ψ) Fl(∂y))(m) on (σ ψ) 1(Rn) = ( ∗ 1 − m ◦ otherwise◦ (σ ψ) 1((σ ψ)(m) + y) on (σ ψ) 1(Rn) = ( − − , m ◦ ◦ otherwise◦

1 and thus ι := (σ ψ)− is a C∞-embedding with ι(0) = m0 and l deﬁned above is ◦ C∞. Further we have l0 = IdM , and

1 1 n ι(ι− (ι(ι− (m) + y)) + x) m ι(R ) lx(ly(m)) = ( ∀ ∈ lx(m) otherwise ι(ι 1(m) + y + x) m ι(Rn) = ( − m otherwise∀ ∈

= lx+y(m),

n and thus l is a C∞-action of (R , +) on M. 2 70 4. Isomorphisms between groups of diﬀeomorphisms

Let m0 M and consider the C∞-action l constructed in the Sublemma above. Since φ is a homeomorphism∈

l : Rn N N e × → 1 (x, n) ϕ(lx(ϕ− (n))) 7→ n n s is a continuous action of (R , +) on N. If we fix x R , then lx = Φ(lx) G (N), r r ∈ e ∈ since lx Diffc(M) G (M). Hence, by a Theorem of Montgomery and Zippin (chapter∈ 5.2 of [10])◦l ⊆is a Cs-action. We have e 1 lx(ϕ(m0)) = ϕ(lx(ϕ− (ϕ(m0)))) = ϕ(lx(m0)) = ϕ(ι(x)) e n s 1 and thus ϕ ι : R N is C . But since ι− is a chart around m0, and since m0 was ◦ → s 1 1 r arbitrary in M we obtain ϕ is C . Replacing Φ by Φ− we get ϕ− is C . Let p be maximal in N 0, with ϕ is a Cp-diffeomorphism. What we have done above shows min r, s∪ { p∞}. Now suppose r = s and assume without loss of generality r < s. Then{ }p ≤ < min r, s = r, for6 otherwise ϕ would induce an isomorphism Gr(M) Gs(N) Diffr{(N)}= Gs(N). But then p < min r, s p, a contradiction. So we must→ have r∩= s p, and6 thus we have finished the{ proof} ≤ of Theorem 4.17. 2 ≤ Appendix A. Isomorphisms between permutation groups One could ask whether Theorem 4.17 remains valid if we replace the manifolds M, N r s by finite sets X, Y and Diffc(M), Diffc(N) by Σ(X), Σ(Y ) (cf. Definition A.3). The answer is ‘yes’, except in the case X = 6 (cf. Corollary A.8 and Theorem A.10). The existence of this exceptional case| made| me present this elementary group theoretical result here. For references see [12]. A.1. Definition. If G is a group we denote by Aut(G) the group of automorphisms of G, i.e. the group of isomorphisms G G. If g G we let conjg Aut(G) be 1 → ∈ ∈ given by conjg(h) := ghg− . The group of inner automorphisms of G is

Inn(G) := conj : g G . { g ∈ } For an arbitrary subset A G the centralizer of A in G is ⊆

CG(A) := g G : ga = ag a A , { ∈ ∀ ∈ } and the center of G is C(G) := CG(G). If X is a set we denote by X the cardinality of X. G is called the order of the group G. If G < and g G| then| the minimal integer| n| 1 with gn = e is called the order of|g.| For∞ a subgroup∈ H G we denote by G/H ≥the set of right cosets of H in G. The index of H in G is G/H⊆ . | | A.2. Remark. If G is a group, g G and ϕ Aut(G) we have ∈ ∈ 1 1 1 1 (ϕconjgϕ− )(x) = ϕ(gϕ− (x)g− ) = ϕ(g)xϕ(g)− = conjϕ(g)(x), and thus Inn(G) ¡ Aut(G). So Out(G) := Aut(G)/Inn(G) is as well a group. CG(A) is a subgroup of G and C(G) ¡ G. The mapping G Inn(G): g conjg is a surjective homomorphism with kernel C(G). → 7→

A.3. Deﬁnition. If X is a set, we denote by Σ(X) the group of permutations of X, i.e. the group of bijective mappings X X. Especially we let Σn := Σ( 1, . . . , n ). In analogy to Deﬁnition 4.1 we denote→ by { }

SxΣ(X) := π Σ(X): π(x) = x { ∈ } the isotropy group of x X. If a1, . . . , ar 1, . . . , n with ai = aj for i = j we ∈ ∈ { } 6 6 denote by (a1, . . . , ar) the permutation in Σn given by

ai ai+1 1 i r 1 7→ ∀ ≤ ≤ − ar a1 m 7→ m otherwise, 7→ 71 72 A. Isomorphisms between permutation groups

and call it an r-cycle. 2-cycles are also called transpositions. Two cycles (a1, . . . , ar) and (b1, . . . , bs) are called disjoint iﬀ a1, . . . , ar b1, . . . , bs = . { } ∩ { } ∅

A.4. Remark. It is well known that Σn = n! and that every permutation in Σn can be written as a composition of disjoint| cycles.| Moreover (1, i): i = 2, . . . , n = h{ }iΣn Σn, and hence every permutation can be written as composition of transpositions. Such a representation is not unique, but if π = τr τ1 = σs σ1, where τi, σj are transpositions, then r s (mod 2). If r is even resp.··· odd we··· call π an even resp. ≡ odd permutation and let sgn(π) := 1 resp. sgn(π) := 1. sgn : Σn ( 1, ) is a − → ± · homomorphism, and it is onto iﬀ n 2. We call An := ker(sgn), the alternating ≥ group of order n. We have An ¡ Σn and An = 3 cycles . h{ − }iΣn If n 3 and Id = σ Σn then there exists a transposition which does not ≥ 6 ∈ 1 commute with σ (let τ := (i, j) where σ(i) = i and j = i, σ(i); then τστ − (i) = 1 6 6 6 τστ − (j) = σ(i)) and thus C(Σn) = Id . { }

A.5. Lemma. Let Φ Aut(Σn) and assume that Φ maps transpositions to transpo- ∈ sitions. Then Φ Inn(Σn). ∈

Proof. Since Φ((1, 2)) is a transposition there exist a1 = a2 1, . . . , n with 6 ∈ { } Φ((1, 2)) = (a1, a2). Similarly there exist b = a3 1, . . . , n with Φ((1, 3)) = (b, a3). Since the order of an element of a group is6 invariant∈ { under} isomorphisms, and since

(b, a3)(a1, a2) = Φ((1, 3))Φ((1, 2)) = Φ((1, 3)(1, 2)) = Φ((1, 2, 3))

is of order 3, a1, a2 and b, a3 must have exactly one element in common, say { } { } a1 = b. Analogously we ﬁnd a1, . . . , an 1, . . . , n with Φ((1, i)) = (a1, ai) for ∈ { } i = 2, . . . , n. Since Φ is bijective we have ai = aj for i = j and hence we may deﬁne 6 6 ϕ Σn by ϕ(i) := ai. We then have ∈ 1 Φ((1, i)) = (a1, ai) = ϕ(1, i)ϕ− = conjϕ((1, i))

and thus Φ = conj Inn(Σn), since (1, i): i = 2, . . . , n generates Σn. 2 ϕ ∈ { }

A.6. Theorem. If n = 6 then Inn(Σn) = Aut(Σn). 6

Proof. Let Φ Aut(Σn) and let τ Σn be a transposition. Counting the elements ∈ ∈ of Σn that commute with τ one obtains

CΣ ( τ ) = (n 2)! 2. | n { } | − ·

Clearly, since Φ Aut(Σn) the order of Φ(τ) is as well 2, and thus Φ(τ) can be written as a composition∈ of disjoint 2-cycles, i.e.

Φ(τ) = (a1, b1) (ar, br) ··· n where a1, b1, . . . , ar, br are all distinct, and 1 r . We then have ≤ ≤ 2 r CΣ ( Φ(τ) ) = (n 2r)!2 r!, | n { } | − A. Isomorphisms between permutation groups 73

but since Φ Aut(Σn) we must have ∈ r (n 2)! 2 = CΣ ( τ ) = CΣ ( Φ(τ) ) = (n 2r)!2 r!. (A.1) − · | n { } | | n { } | − If r = 1 (A.1) holds n 2r. If r = 2 there is no 2r n N, satisfying (A.1). If r = 3 then n = 6 is∀ the≥ unique solution of (A.1) with≤ 2r ∈ n N. If r 4 then there is no 2r n N satisfying (A.1), for if 2r n and r≤ 4∈ then ≥ ≤ ∈ ≤ ≥ (n 2r + 1) (n 2r + 2) (n 3) (n 2) − − ··· − − ≥ 2 2(r 1) | ≥{z } |≥ {z− } (n 2r + 1)(n 2r + 3) (n 3) 2 4 6 2(r 1) ≥ − − ··· − · · · ··· − 2r 3 =2r 1(r 1)! |≥ {z− } | −{z − } r 1 (2r 3)2 − (r 1)! ≥ r −1 − r 1 > r2 − (r 1)! = 2 − r!, − and thus

(n 2)! 2 = (n 2r)! 2 (n 2r + 1) (n 2r + 2) (n 2) − · − · · r −1 · − r ··· − > (n 2r)! 2 2 − r! = (n 2r)!2 r!, − · · − a contradiction to (A.1). Since we have assumed n = 6 we obtain r = 1 and hence Φ maps transpositions to transpositions. So we can6 apply Lemma A.5 and obtain Aut(Σn) = Inn(Σn). 2

A.7. Corollary to the proof of Theorem A.6. If Φ Aut(Σ6) then Φ maps transpositions either to transpositions or to compositions∈ of exactly three disjoint transpositions.

Proof. If n = 6 we found r = 1 or r = 3 (cf. proof of Theorem A.6), but this is Corollary A.7. 2

A.8. Corollary. Let X and Y be two ﬁnite, nonempty sets, and let Φ : Σ(X) Σ(Y ) be an isomorphism of groups. Then X = Y and, if we assume additionally→ X = 6, there exists a bijection ϕ : X Y| inducing| | | Φ, i.e. | | 6 → 1 Φ(π) = ϕ π ϕ− π Σ(X). ◦ ◦ ∀ ∈ Moreover ϕ is unique, provided X 3. | | ≥ Proof. Since X ! = Σ(X) = Σ(Y ) = Y !, we obtain X = Y , and we may choose a bijection ψ|: X| |Y . Now| consider| | the| | isomorphism| of| groups| | Ψ : Σ(X) Σ(Y ) → 1 1 → given by Ψ(π) := ψ π ψ− . We then have Ψ− Φ Aut(Σ(X)), and hence by ◦ ◦ 1 ∈ Theorem A.6 there exists σ Σ(X) with Ψ− Φ = conjσ. Thus, for every π Σ(X) we have ∈ ∈

1 1 1 1 Φ(π) = Ψ(conj (π)) = Ψ(σπσ− ) = ψ σπσ− ψ− = ϕ π ϕ− , σ ◦ ◦ ◦ ◦ 74 A. Isomorphisms between permutation groups where ϕ := ψ σ : Σ(X) Σ(Y ). ◦ → 1 Suppose ϕ0 : X Y is another bijection inducing Φ. Then ϕ− ϕ0 Σ(X) and → ◦ ∈ conjϕ 1 ϕ = IdΣ(X), for − ◦ 0 1 1 1 1 1 π = Φ− (Φ(π)) = Φ− (ϕ0 π ϕ0− ) = (ϕ− ϕ0)π(ϕ0− ϕ) = conjϕ 1 ϕ (π). ◦ ◦ ◦ ◦ − ◦ 0

Since X 3, Σ(X) has trivial center (cf. Remark A.4) and hence conj 1 = ϕ− ϕ0 | | ≥ 1 ◦ IdΣ(X) yields ϕ− ϕ0 = IdX (cf. Remark A.2). Thus φ is unique. 2 ◦

A.9. Lemma. If n = 4 then Id , An and Σn are all normal subgroups of Σn. 6 { }

Proof. Assume n 3 and let Id = N ¡ Σn. Now choose Id = σ N and ≥ { } 6 6 ∈ a transposition τ Σn, that does not commute with σ (cf. Remark A.4). We then ∈ have Id = ρ := [σ, τ] N, since N ¡Σn, and ρ is a composition of two transpositions, 6 1 1 ∈ namely σ− τ − σ and τ. Thus ρ is either a 3-cycle or a composition of two disjoint transpositions. Conjugating ρ we obtain that N either contains all 3-cycles or N contains all compositions of two disjoint transpositions. In the second case we must have n 5 by assumption. So given any 3-cycle (a, b, c) we ﬁnd (d, e) disjoint to (a, b, c).≥ Then (a, b, c) = (a, c)(d, e) (d, e)(a, b),

N N | ∈{z } | ∈{z } and thus, even in the second case, N contains all 3-cycles. Since the 3-cycles generate An we have An N and consequently N = An or N = Σn, since An Σn has index 2. ⊆ ⊆ 2

A.10. Theorem. We have

Out(Σ6) := Aut(Σ6)/Inn(Σ6) ∼= Z/2Z, and hence Corollary A.8 does not hold in the case X = 6. | |

Proof. Suppose there exist Φ, Ψ Aut(Σ6) Inn(Σ6). Then, by Corollary A.7 and Lemma A.5, Φ and Ψ map transpositions∈ to\ compositions of exactly three disjoint 6 1 6 4 2 transpositions. In Σ6 there are 2 = 15 transpositions and 3! 2 2 2 = 15 1 compositions of three disjoint transpositions. Hence Ψ− Φ maps transpositions to 1 transpositions and thus, by Lemma A.5, we obtain Ψ− Φ Inn(Σ6). Summing up we have ∈ Aut(Σ6)/Inn(Σ6) 2. | | ≤ We ﬁnish the proof by showing Aut(Σ6) Inn(Σ6) = . Consider Σ5 and denote by \ 6 ∅ S5 the set of those subgroups of Σ5, that have order 5. Since every subgroup in Σ5 of order 5 consists of Id and four 5-cycles, and since U V = Id U = V S5, 5!{ } ∩4! { } ∀ 6 ∈ and since there are 5 = 4! 5-cycles in Σ5, we get S5 = 4 = 6. Now consider the following homomorphism: | |

φ Σ5 Σ(S5) → 1 π (U πUπ− U S5) 7→ 7→ ∀ ∈ A. Isomorphisms between permutation groups 75

Since A5 (123) / ker(φ) ¡ Σ5, we obtain from Lemma A.9, ker(φ) = Id and 3 ∈ { } thus φ is injective. So H := φ(Σ5) is a subgroup of index 6, but H is no isotropy group, since Σ5 acts (via φ) transitively on S5. Now consider M := Σ(S5)/H and the homomorphism

Ψ Σ(S5) Σ(M) → π (σH πσH σ Σ(S5)). 7→ 7→ ∀ ∈ Then M = 6 and ker(Ψ) H. From Lemma A.9 we obtain ker(Ψ) = Id , and so| Ψ| is an isomorphism.⊆ Hence Ψ(H) Σ(M) is a subgroup of index 6,{ that} ⊆ ﬁxes H M and thus Ψ(H) = SH Σ(M). Using isomorphisms Σ(S5) = Σ6 = Σ(M) ∈ ∼ ∼ induced from bijections S5 = 1,..., 6 = M we obtain Ψ0 Aut(Σ6) and a subgroup ∼ { } ∼ ∈ H0 Σ6, such that Ψ0(H0) is an isotropy group, but H0 is not. So Ψ0 can’t be inner, ⊆ and thus Out(Σ6) = 2. 2 | | Bibliography

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76 Lebenslauf

Stefan Haller, geboren am 27. September 1971 in Wien Nationalit¨at: Osterreich¨

1978-1982 Oﬀentliche¨ Volksschule in Wien

1982-1990 Mathematisches Realgymnasium (Bundesgymnasium und Bundesreal- gymnasium, Wien 21), Matura am 7. Juni 1990 ab 1990 Studium der Mathematik an der Formal- und Naturwissenschaftlichen Fakult¨atder Universit¨atWien