LECTURE4:SYMPLECTOMORPHISMSCONTENTS11.Thegroupof Symplectomorphisms4SymplectomorphismsasLagrangian submanifolds2.73.Generatingfunctions94.Thebilliards1.THEGROUPOF SYMPLECTOMORPHISMS Diffeomorphisms v.s. vector fields.Let M be a smooth manifold and Vect(M) the set of all smooth vectors on MIt is well known that for any X,Y e Vect(M), the Lie bracket[X,Y]= XY -YX E Vect(M)and is bilinear, anti-symmetric and satisfies the Jacobi identity[[X, Y], Z] + [Y, Z], X] +[Z, X],Y]= 0.As a consequence, (Vect(M), [, J) is an (infinitely dimensional) Lie algebra. Whatis the corresponding “Lie group"?Well, any smooth vector field X e Vect(M) generates (at least locally) a one-parameter subgroup of diffeomorphisms of MPt=exp(tX) : M →M, pt(r)=(t),where is the integral curve of X starting at r. Conversely, given any one param-eter subgroup pt of diffeomorphisms on M, one gets a smooth vector field viadpt(r).X(c) =dtlt=0So, the group of diffeomorphisms.Diff(M) = (: M → M I is a diffeomorphism),is the “Lie group" whose Lie algebra is Vect(M).1
LECTURE 4: SYMPLECTOMORPHISMS Contents 1. The group of Symplectomorphisms 1 2. Symplectomorphisms as Lagrangian submanifolds 4 3. Generating functions 7 4. The billiards 9 1. The group of Symplectomorphisms ¶ Diffeomorphisms v.s. vector fields. Let M be a smooth manifold and Vect(M) the set of all smooth vectors on M. It is well known that for any X, Y ∈ Vect(M), the Lie bracket [X, Y ] = XY − Y X ∈ Vect(M) and is bilinear, anti-symmetric and satisfies the Jacobi identity [[X, Y ], Z] + [[Y, Z], X] + [[Z, X], Y ] = 0. As a consequence, (Vect(M), [·, ·]) is an (infinitely dimensional) Lie algebra. What is the corresponding “Lie group”? Well, any smooth vector field X ∈ Vect(M) generates (at least locally) a oneparameter subgroup of diffeomorphisms of M ρt = exp(tX) : M → M, ρt(x) = γ X x (t), where γ X x is the integral curve of X starting at x. Conversely, given any one parameter subgroup ρt of diffeomorphisms on M, one gets a smooth vector field via X(x) = d dt t=0 ρt(x). So, the group of diffeomorphisms, Diff(M) = {ϕ : M → M | ϕ is a diffeomorphism}, is the “Lie group” whose Lie algebra is Vect(M). 1
2LECTURE4:SYMPLECTOMORPHISMS The group of symplectomorphisms.Now let (M,w)be a symplectic manifold andSymp(M,w) = [ : M -→ M I is a symplectomorphism)thegroup of symplectomorphisms of (M,w).This is a“closed"subgroup ofDiff(M).Both Symp(M,w) and Diff(M) are large in the sense that they are infinitely dimensional.Erample.Wehave seen inlecture2that anydiffeomorphism:X→Xlifts toasymplectomorphism :T*X →T*X via(r,s) = ((r), (dp)-1(s)).It is easyto check2p1=2oi.Soweget a natural group monomorphismDiff(X) -→ Symp(T*X,Wcan).Erample. Here is another subgroup of Symp(T*X,wean): For any β e '(X), letG : T*X T*X be the diffeomorphismGp(r,E) = (r,E+βr).Lemma 1.1. Gg is a symplectomorphism if and only if β is closed.Proof. We shall prove Ggacan - Qcan = n*β, which implies the conclusion. Recallthe reproducing property of acan: Qcan is the unique 1-form on T*X such that forany μE2'(X), sα =μ. It follows$Ggacan=(GgoSu)"acan=su+βacan=μ+β=stn*β+μ.口So by reproducing property again we see Ggacan - r*β = acan:As a consequence, we get another group monomorphismZ'(X) -→ Symp(T*X,wean),where z'(X) is the space of closed 1-forms on X. Symplectic vector fields.Question: What is the “Lie algebra" of Symp(M,w)? Well, since Symp(M,w)is a “closed" subgroup of Diff(M), its Lie algebra should be a Lie subalgebra of(Vect(M),,D).Let's try to find the condition for a smooth vector to be"symplec-tic"Asymplecticvector field X shouldbea vector field whose flowIptl consists ofsymplectomorphisms.In otherwords p,w=wfor all t.Itfollowsd0：dipiw=piCxw.Definition1.2.Asmoothvector fieldX iscalled symplecticifCxw=0Since w is closed, the Cartan's magic formula implies
2 LECTURE 4: SYMPLECTOMORPHISMS ¶ The group of symplectomorphisms. Now let (M, ω) be a symplectic manifold and Symp(M, ω) = {ϕ : M → M | ϕ is a symplectomorphism} the group of symplectomorphisms of (M, ω). This is a “closed” subgroup of Diff(M). Both Symp(M, ω) and Diff(M) are large in the sense that they are infinitely dimensional. Example. We have seen in lecture 2 that any diffeomorphism ϕ : X → X lifts to a symplectomorphism ˜ϕ : T ∗X → T ∗X via ϕ˜(x, ξ) = (ϕ(x),(dϕ∗ x ) −1 (ξ)). It is easy to check ϕ^2 ◦ ϕ1 = ˜ϕ2 ◦ ϕ˜1. So we get a natural group monomorphism Diff(X) → Symp(T ∗X, ωcan). Example. Here is another subgroup of Symp(T ∗X, ωcan): For any β ∈ Ω 1 (X), let Gβ : T ∗X → T ∗X be the diffeomorphism Gβ(x, ξ) = (x, ξ + βx). Lemma 1.1. Gβ is a symplectomorphism if and only if β is closed. Proof. We shall prove G∗ βαcan − αcan = π ∗β, which implies the conclusion. Recall the reproducing property of αcan: αcan is the unique 1-form on T ∗X such that for any µ ∈ Ω 1 (X), s ∗ µα = µ. It follows s ∗ µG ∗ βαcan = (Gβ ◦ sµ) ∗αcan = s ∗ µ+βαcan = µ + β = s ∗ µπ ∗β + µ. So by reproducing property again we see G∗ βαcan − π ∗β = αcan. As a consequence, we get another group monomorphism Z 1 (X) → Symp(T ∗X, ωcan), where Z 1 (X) is the space of closed 1-forms on X. ¶ Symplectic vector fields. Question: What is the “Lie algebra” of Symp(M, ω)? Well, since Symp(M, ω) is a “closed” subgroup of Diff(M), its Lie algebra should be a Lie subalgebra of (Vect(M), [·, ·]). Let’s try to find the condition for a smooth vector to be “symplectic”. A symplectic vector field X should be a vector field whose flow {ρt} consists of symplectomorphisms. In other words ρ ∗ tω = ω for all t. It follows 0 = d dtρ ∗ tω = ρ ∗ tLXω. Definition 1.2. A smooth vector field X is called symplectic if LXω = 0. Since ω is closed, the Cartan’s magic formula implies
3LECTURE4:SYMPLECTOMORPHISMSLemma 1.3. A vector field X on (M,w) is symplectic if and only if txw is closed.The set of all symplectic vector fields on (M,w) is denoted by Vect(M,w).Weneed to check that if X,Y eVect(M,w),so is[X,Y].In other words, (Vect(M,w),[,)is a Lie sub algebra of (Vect(M), [, l). This follows fromLemma 1.4. If X,Y are symplectic, tx,yjw = d(-w(X,Y)..Proof. By the definition of exterior differential, for any Z E Vect(M),0 = (dw)(X,Y,Z) = X(w(Y, Z))-Y(w(X, Z))+Z(w(X,Y))-w([X,Y], Z)+w([X, Z),Y)-w([Y, Z],X)0 = (dixw)(Y,Z) = Y(w(X,Z)) - Z(w(X,Y)) -w(X,[Y,Z),0 = (dtyw)(X, Z) = X(w(Y, Z)) - Z(w(Y,X)) -w(Y,[X, Z)Comparing thethree equationswe conclude-Z(w(X, Y)) - w([X,Y], Z) = 0,or in other words,(x,rw = d(-w(X,Y)口Remark. So the bracket [X,Y] of two symplectic vector fields is better than beinga symplectic vector field: tx,yjw is not only closed, but in fact eract! Hamiltonian vector fields.Definition 1.5. A vector field X is Hamiltonian if txw is exact.Remark. According to the lemma above, the space of all Hamiltonian vector fieldsis an ideal of Vect(M,w).So if X is hamiltonian, then there exists a smoothfunction f e Co(M) so thattxw = df. Conversely, since w is non-degenerate, for any f e Co(M), there is aunique vector field X on M so thatix,w = df.We will call X, the Hamiltonian vector field associated to the Hamiltonian functionf. The flow generated by Xf is called the Hamiltonian flow associated to f.Remark.AssumeMis connected.Thentwo smooth functions definethe sameHamiltonian vector field if and only if they differ by a constant. So as a vector spacethe space of Hamiltonian vector fields is isomorphic to C(M)/R, which can alsobeidentifiedwithC(M)o = (f E C(M) / /. f(r)da = 0.)if M is compact
LECTURE 4: SYMPLECTOMORPHISMS 3 Lemma 1.3. A vector field X on (M, ω) is symplectic if and only if ιXω is closed. The set of all symplectic vector fields on (M, ω) is denoted by Vect(M, ω). We need to check that if X, Y ∈ Vect(M, ω), so is [X, Y ]. In other words, (Vect(M, ω), [·, ·]) is a Lie sub algebra of (Vect(M), [·, ·]). This follows from Lemma 1.4. If X, Y are symplectic, ι[X,Y ]ω = d(−ω(X, Y )). Proof. By the definition of exterior differential, for any Z ∈ Vect(M), 0 = (dω)(X, Y, Z) = X(ω(Y, Z))−Y (ω(X, Z))+Z(ω(X, Y ))−ω([X, Y ], Z)+ω([X, Z], Y )−ω([Y, Z], X) 0 = (dιXω)(Y, Z) = Y (ω(X, Z)) − Z(ω(X, Y )) − ω(X, [Y, Z]), 0 = (dιY ω)(X, Z) = X(ω(Y, Z)) − Z(ω(Y, X)) − ω(Y, [X, Z]). Comparing the three equations we conclude −Z(ω(X, Y )) − ω([X, Y ], Z) = 0, or in other words, ι[X,Y ]ω = d(−ω(X, Y )). Remark. So the bracket [X, Y ] of two symplectic vector fields is better than being a symplectic vector field: ι[X,Y ]ω is not only closed, but in fact exact!. ¶ Hamiltonian vector fields. Definition 1.5. A vector field X is Hamiltonian if ιXω is exact. Remark. According to the lemma above, the space of all Hamiltonian vector fields is an ideal of Vect(M, ω). So if X is hamiltonian, then there exists a smooth function f ∈ C ∞(M) so that ιXω = df. Conversely, since ω is non-degenerate, for any f ∈ C ∞(M), there is a unique vector field Xf on M so that ιXfω = df. We will call Xf the Hamiltonian vector field associated to the Hamiltonian function f. The flow generated by Xf is called the Hamiltonian flow associated to f. Remark. Assume M is connected. Then two smooth functions define the same Hamiltonian vector field if and only if they differ by a constant. So as a vector space the space of Hamiltonian vector fields is isomorphic to C ∞(M)/R, which can also be identified with C ∞(M)0 = {f ∈ C ∞(M) | Z M f(x)dx = 0.} if M is compact.
4LECTURE4:SYMPLECTOMORPHISMSErample. On the 2-sphere S? with symplectic form w = doAdz, if we take f(z,0) = zthe height function, then Xf = , and the Hamiltonian flow is the rotation aboutthe vertical axis:pt(z,0) = (z, + t).Erample. On (T2,dede2) the vector fields and are both symplectic but80notHamiltonian.We will return to Hamiltonian vector fields later. Hamiltonian symplectomorphisms.Recall that an isotopy is a family of diffeomorphisms pt so that po = Id. If eachPtis a symplectomorphism,we call theisotopya symplectic isotopy.It is easytosee that a symplectic isotopy is generated bya family of symplecticvector fields X,withddtPt = Xi(pt)IfeachX,isnot only symplectic,butinfactHamiltonian,thenwecall theisotopya Hamiltonianisotopy.Definition 1.6. A symplectomorphism is called Hamiltonian symplectomorphismif there exists a Hamiltonian isotopy pt such that pi =p.The space of Hamiltonian symplectomorphisms is denoted by Ham(M,w). Itturns out that Ham(M,w) is a normal subgroup of Symp(M,w) whose Lie algebrais the algebra of all Hamiltonian vector fields.Since any function (modulo constants)gives a family of Hamiltonian symplec-tomorphisms, we see that the group of symplectomorphisms is huge.2.SYMPLECTOMORPHISMS AS LAGRANGIAN SUBMANIFOLDSLagrangian submanifolds v.s.symplectomorphisms.Weinstein's symplectic creed:EVERYTHINGISALAGRANGIANSUBMANIFOLD!In what follows we shall study symplectomorphisms according to this creed.Let (Mi,wi) and (M2,w2) be 2n dimensional symplectic manifolds and let pr, :Mi× M2 →M, be the projection.We have seen in lecture 2 that for any nonzerorealnumbersA1and>2,Aipr'wi+>2praw2isasymplecticformontheproductM =Mi × M2. In particular, one has two important symplectic forms:.the product symplectic form w=priwi+prsw2,.the twisted product form =priwi-przw2
4 LECTURE 4: SYMPLECTOMORPHISMS Example. On the 2-sphere S 2 with symplectic form ω = dθ∧dz, if we take f(z, θ) = z the height function, then Xf = ∂ ∂θ , and the Hamiltonian flow is the rotation about the vertical axis: ρt(z, θ) = (z, θ + t). Example. On (T 2 , dθ1 ∧ dθ2) the vector fields ∂ ∂θ1 and ∂ ∂θ1 are both symplectic but not Hamiltonian. We will return to Hamiltonian vector fields later. ¶ Hamiltonian symplectomorphisms. Recall that an isotopy is a family of diffeomorphisms ρt so that ρ0 = Id. If each ρt is a symplectomorphism, we call the isotopy a symplectic isotopy. It is easy to see that a symplectic isotopy is generated by a family of symplectic vector fields Xt with d dtρt = Xt(ρt). If each Xt is not only symplectic, but in fact Hamiltonian, then we call the isotopy a Hamiltonian isotopy. Definition 1.6. A symplectomorphism ϕ is called Hamiltonian symplectomorphism if there exists a Hamiltonian isotopy ρt such that ρ1 = ϕ. The space of Hamiltonian symplectomorphisms is denoted by Ham(M, ω). It turns out that Ham(M, ω) is a normal subgroup of Symp(M, ω) whose Lie algebra is the algebra of all Hamiltonian vector fields. Since any function (modulo constants) gives a family of Hamiltonian symplectomorphisms, we see that the group of symplectomorphisms is huge. 2. Symplectomorphisms as Lagrangian submanifolds ¶ Lagrangian submanifolds v.s. symplectomorphisms. Weinstein’s symplectic creed: EVERYTHING IS A LAGRANGIAN SUBMANIFOLD! In what follows we shall study symplectomorphisms according to this creed. Let (M1, ω1) and (M2, ω2) be 2n dimensional symplectic manifolds and let pri : M1 × M2 → Mi be the projection. We have seen in lecture 2 that for any nonzero real numbers λ1 and λ2, λ1pr∗ 1ω1 + λ2pr∗ 2ω2 is a symplectic form on the product M = M1 × M2. In particular, one has two important symplectic forms: • the product symplectic form ω = pr∗ 1ω1 + pr∗ 2ω2, • the twisted product form ω˜ = pr∗ 1ω1 − pr∗ 2ω2
5LECTURE4:SYMPLECTOMORPHISMSNow let f : Mi → M2 be a diffeomorphism, then its graphF= [(r, f(r) IrE Mi)isa2ndimensionalsubmanifoldofthe4ndimensionalmanifoldMi×M2Theorem 2.1.f is a symplectomorphism ifand only ifF, is Lagrangian withrespecttow.Proof.Let t:If→M be the inclusion and :Mi→Ifbe the obvious diffeomor-phism, thenIf is Lagrangian t*w =0*w=0**(priwi - prw2) = 01-f*2=0fisasymplectomorphism口 Lift of smooth maps as Lagrangian submanifolds.In particular, suppose Mi = T*Xi and M2 = T*X2 be cotangent bundles andwi=-dai,w2=-da2 thecanonical symplectic forms.Then M=Mi×M2=T*X,whereX=X,×X2.Moreover,thecanonical1-form onM=T*X isQ=ai@Q2,sothe product symplectic form w = wi @ w2 on M = T*X is the canonical symplecticform. Let02 : M2 → M2, (c,) - (r, -E)Then oα2=-α2, and thus ow2=-w2.It follows theorem thatProposition 2.2.If f:Mi-→M2 is a diffeomorphism, then f is a symplectomor-phism if and only if Ta2of is a Lagrangian submanifold of (M,w).Now suppose g : Xi → X, is a diffeomorphism. As we have seen, g lifts to asymplectomorphismg:Mi→M2.Recall thatg(r,s) = (y,n) y=g(r),E=(dgr)nAs a consequence,Fg = {(r,s,y,n) [y=g(r),E=-(dgr)*n)is a Lagrangian submanifold of M =T*X. Here is another way to see this: LetXg = ((r,g(r) I rEXi) C X = Xi x X2be the graph of g, then Fg = N*Xg. In fact, we have a more general theorem:Theorem 2.3. Let Xi,X2 be arbitrary smooth manifolds and g : XiX2 a smoothmap, then the set F defined as above is eractly N*Xg, and thus a Lagrangiansubmanifold of M = T*X
LECTURE 4: SYMPLECTOMORPHISMS 5 Now let f : M1 → M2 be a diffeomorphism, then its graph Γf = {(x, f(x) | x ∈ M1} is a 2n dimensional submanifold of the 4n dimensional manifold M1 × M2. Theorem 2.1. f is a symplectomorphism if and only if Γf is Lagrangian with respect to ω˜. Proof. Let ι : Γf ,→ M be the inclusion and γ : M1 → Γf be the obvious diffeomorphism, then Γf is Lagrangian ⇐⇒ ι ∗ω˜ = 0 ⇐⇒ γ ∗ ι ∗ω˜ = 0 ⇐⇒ γ ∗ ι ∗ (pr∗ 1ω1 − pr∗ 2ω2) = 0 ⇐⇒ ω1 − f ∗ω2 = 0 ⇐⇒ f is a symplectomorphism. ¶ Lift of smooth maps as Lagrangian submanifolds. In particular, suppose M1 = T ∗X1 and M2 = T ∗X2 be cotangent bundles and ω1 = −dα1, ω2 = −dα2 the canonical symplectic forms. Then M = M1×M2 = T ∗X, where X = X1×X2. Moreover, the canonical 1-form on M = T ∗X is α = α1⊕α2, so the product symplectic form ω = ω1 ⊕ ω2 on M = T ∗X is the canonical symplectic form. Let σ2 : M2 → M2,(x, ξ) 7→ (x, −ξ). Then σ ∗ 2α2 = −α2, and thus σ ∗ 2ω2 = −ω2. It follows theorem that Proposition 2.2. If f : M1 → M2 is a diffeomorphism, then f is a symplectomorphism if and only if Γσ2◦f is a Lagrangian submanifold of (M, ω). Now suppose g : X1 → X2 is a diffeomorphism. As we have seen, g lifts to a symplectomorphism ˜g : M1 → M2. Recall that g˜(x, ξ) = (y, η) ⇐⇒ y = g(x), ξ = (dgx) ∗ η. As a consequence, Γ σ g˜ = {(x, ξ, y, η) | y = g(x), ξ = −(dgx) ∗ η} is a Lagrangian submanifold of M = T ∗X. Here is another way to see this: Let Xg = {(x, g(x)) | x ∈ X1} ⊂ X = X1 × X2 be the graph of g, then Γσ g˜ = N∗Xg. In fact, we have a more general theorem: Theorem 2.3. Let X1, X2 be arbitrary smooth manifolds and g : X1 → X2 a smooth map, then the set Γ σ g˜ defined as above is exactly N∗Xg, and thus a Lagrangian submanifold of M = T ∗X.