LECTURE27:FIO-SYMPLECTICCATEGORY1. BASICS ON CATEGORY Category and functor.Definition 1.1 (Category). A category C consists of(1) a class Ob(C) whose elements are called objects,(2) a class Mor(C) of morphisms between the objects, such that.Each morphism f has a source object X E Ob(C) and a target objectY eOb(C). We write f: X -→Y and say “f is a morphism from X toy".We denote all morphismsfromX toYby Mor(X,Y).1. The composition of morphisms f : X → Y and g : Y -→ Z is a morphismgo f : X → Z, such that(a) (associativity) if f:X→Y,g:Y→Zand h:Z→W,thenho(gof)=(hog)of,(b) (identity) for any X e Ob(C), there is an identity morphism Idx :X → X, such that for any morphism f : Z → X and g : X → Y,Idxof=f and goIdx=g.Definition 1.2. A (covariant) functor 3 from a category C to a category D(1) associates to each object X in C an object F(X) in D,(2) associates to each morphism f:X→Y in C a morphism (f):3(X)→3(Y)in D such that thefollowing two conditions hold: 3(Idx) = Ids(x) for every object X in C,. 3(g o f) = 3(g) o3(f) for any morphism f:X-→Y and g:Y-→Z in C.Erample.We have the category of topological spaces TOP,.Ob(TOP) = all topological spaces,: morphisms are continuous maps between topological spaces,and the category of groups GROuP,.Ob(gROuP)=allgroups,·morphisms are group homomorphismsMoreover, we have many functors from TOP to gROuP: n, H", Hn etc whichwere studied extensively in algebraic topology.'t is possible that for some objects X and Y, there is no morphism from X to Y, so thatMor(X,Y) = 0.1
LECTURE 27: FIO – SYMPLECTIC CATEGORY 1. Basics on Category ¶ Category and functor. Definition 1.1 (Category). A category C consists of (1) a class Ob(C) whose elements are called objects, (2) a class Mor(C) of morphisms between the objects, such that • Each morphism f has a source object X ∈ Ob(C) and a target object Y ∈ Ob(C). We write f : X → Y and say “f is a morphism from X to Y ”. We denote all morphisms from X to Y by Mor(X, Y ). 1 • The composition of morphisms f : X → Y and g : Y → Z is a morphism g ◦ f : X → Z, such that (a) (associativity) if f : X → Y, g : Y → Z and h : Z → W, then h ◦ (g ◦ f) = (h ◦ g) ◦ f, (b) (identity) for any X ∈ Ob(C), there is an identity morphism IdX : X → X, such that for any morphism f : Z → X and g : X → Y , IdX ◦ f = f and g ◦ IdX = g. Definition 1.2. A (covariant) functor F from a category C to a category D (1) associates to each object X in C an object F(X) in D, (2) associates to each morphism f :X →Y in C a morphism F(f):F(X)→F(Y ) in D such that the following two conditions hold: • F(IdX) = IdF(X) for every object X in C, • F(g ◦ f) = F(g) ◦ F(f) for any morphism f :X →Y and g :Y →Z in C. Example. We have the category of topological spaces T OP, • Ob(T OP) = all topological spaces, • morphisms are continuous maps between topological spaces, and the category of groups GROUP, • Ob(GROUP) = all groups, • morphisms are group homomorphisms. Moreover, we have many functors from T OP to GROUP: πn, Hn , Hn etc which were studied extensively in algebraic topology. 1 It is possible that for some objects X and Y , there is no morphism from X to Y , so that Mor(X, Y ) = ∅. 1
2LECTURE27:FIO-SYMPLECTICCATEGORYErample. One can define a category M so that. Ob(M) = all smooth manifolds..Mor(X,Y)= diffeomorphisms from X to Y,and a category S so that. Ob(S) = all symplectic manifolds? Mor(M,N)= symplectomorphisms from M to N.Then the map 3 that sends X to (T*X,w+x) and sends a diffeomorphism f toits lifting f is a functor.Unfortunately the categories M and S have an obviousshortage:formostpairsof objects,thereisnomorphism at all!Here is one way to get a more reasonable category M of smooth manifolds:. Ob(M) = all smooth manifolds, Mor(X,Y) = smooth maps from X to Y.But, what is a reasonable way to define a category of symplectic manifolds?I The category S&TS.In many cases, it is natural to define a morphism from X to Y to be a mapfrom X to Y that satisfies specific conditions, so that the composition of morphismscan be defined by composition, and the identity morphism is just the identity map.However, there does exist interesting examples where morphisms are not maps:Erample. Let's define a category S&TS as follows:.Ob(SeTS)=“all"sets (with some restriction to avoid logical issues.). Mor(X,Y) = all subsets of X × Y.Recall that a subset of X × Y is also called a relation. So we are using relation-s instead of mapstodefinemorphisms.And,we candefinethe compositionofmorphismsto bethe composition of relations:.For F EMor(X,Y) and I2 EMor(Y,Z),the composition is(1)F2 o Ti = [(r, z) E X x Z IEy E Y so that (c, y) E Ti, (y, z) ET2.)It is not hard to prove the associativity (T oF2)o =F3 o (2 oFi), and it is alsoclear that the identity morphism is(2)Idx =((r,r)EX xX IrEX)=x.Moreover, any relation I c X × Y has a natural "transpose"(3)TT = ((y,r) I (r,y)) cY× X.Finally if we define a category S&TS of sets by the usual way,. Ob(SeTS) = “all" sets (with some restriction to avoid logical issues)
2 LECTURE 27: FIO – SYMPLECTIC CATEGORY Example. One can define a category Mf so that • Ob(Mf) = all smooth manifolds, • Mor(X, Y ) = diffeomorphisms from X to Y , and a category Se so that • Ob(Se) = all symplectic manifolds, • Mor(M, N) = symplectomorphisms from M to N. Then the map F that sends X to (T ∗X, ωT ∗X) and sends a diffeomorphism f to its lifting ˜f is a functor. Unfortunately the categories Mf and Se have an obvious shortage: for most pairs of objects, there is no morphism at all! Here is one way to get a more reasonable category M of smooth manifolds: • Ob(M) = all smooth manifolds, • Mor(X, Y ) = smooth maps from X to Y . But, what is a reasonable way to define a category of symplectic manifolds? ¶ The category SET S. In many cases, it is natural to define a morphism from X to Y to be a map from X to Y that satisfies specific conditions, so that the composition of morphisms can be defined by composition, and the identity morphism is just the identity map. However, there does exist interesting examples where morphisms are not maps: Example. Let’s define a category SET S as follows: • Ob(SET S) = “all” sets (with some restriction to avoid logical issues.) • Mor(X, Y ) = all subsets of X × Y . Recall that a subset of X × Y is also called a relation. So we are using relations instead of maps to define morphisms. And, we can define the composition of morphisms to be the composition of relations: • For Γ1 ∈ Mor(X, Y ) and Γ2 ∈ Mor(Y, Z), the composition is (1) Γ2 ◦ Γ1 = {(x, z) ∈ X × Z | ∃y ∈ Y so that (x, y) ∈ Γ1,(y, z) ∈ Γ2.} It is not hard to prove the associativity (Γ3 ◦ Γ2) ◦ Γ1 = Γ3 ◦ (Γ2 ◦ Γ1), and it is also clear that the identity morphism is (2) IdX = {(x, x) ∈ X × X | x ∈ X} = ∆X. Moreover, any relation Γ ⊂ X × Y has a natural “transpose” (3) ΓT = {(y, x) | (x, y) ∈ Γ} ⊂ Y × X. Finally if we define a category SET S ^ of sets by the usual way, • Ob(SET S ^) = “all” sets (with some restriction to avoid logical issues)
3LECTURE27:FIO-SYMPLECTICCATEGORY. Mor(X, Y) = maps from X to Y,then one can embed SETS into STs by a functor that maps any set to itselfandmapsanymap f :X-→Y to itsgraphTF=((r, f(r)) [rEx)In other words, the category SeTS enlarges the categorySeTs Categorical “points".In the category S&TS (and in all examples above), there is a distinguishedobject: the one point set “"pt". In a category with such a distinguished object “pt",one can define for every object X in this category the “points" of X:Definition 1.3. A categorical point of X is a morphism I:pt-→XNote that every morphism I : X → Y induces a map I* from “points" of X to"points" of Y by composition.Erample. In the categories STS, M, M, S, the categorical points are the pointsin the usual sense. However, in the category STS, a categorical point of a set Xis a subset of X.2.LINEAR SYMPLECTIC CATEGORY The category Cs.WecandefinethecategoryCstobe. Ob(CS)= symplectic vector spaces. Mor((V,2v), (W, 2w) = linear symplectomorphisms from (V,2v) to (W,w).Obviously this is not very interesting, because the set of morphisms betweentwo symplectic vector spaces is non-empty only if they have the same dimension.In what follows we will define a slightly larger category, CS, in which the set ofmorphisms between any two symplectic vector spaces is non-empty.However weprefer not to use all relations as morphisms as in the example SeTs above -wewant ourmorphisms to begeometrically interesting.For simplicitywewill abbreviate (V,2y)to V,and abbreviate (V,-2v)to V-By this way, V × W- means the symplectic vector space (V @ W,2y @ (-2w)).We have shown in lecture 26 that the graph of any linear symplectomorphism fromV to W is a Lagrangian subspace of V × W-.On the other hand, V × W- hasmore Lagrangian subspaces than those arise from the graphs of linear symplecto-morphisms. Moreover, for the existence of Lagrangian subspace in V × W-we don'trequire dim V = dim W. Note that a Lagrangian subspace of V × W- is a specialsubset of V × W, and thus a special relation from V to W
LECTURE 27: FIO – SYMPLECTIC CATEGORY 3 • Mor(X, Y ) = maps from X to Y , then one can embed SET S ^ into SET S by a functor F that maps any set to itself and maps any map f : X → Y to its graph Γf = {(x, f(x)) | x ∈ X}. In other words, the category SET S enlarges the category SET S ^. ¶ Categorical “points”. In the category SET S (and in all examples above), there is a distinguished object: the one point set “pt”. In a category with such a distinguished object “pt”, one can define for every object X in this category the “points” of X: Definition 1.3. A categorical point of X is a morphism Γ : pt → X. Note that every morphism Γ : X → Y induces a map Γ∗ from “points” of X to “points” of Y by composition. Example. In the categories SET S ^, M, Mf, Se, the categorical points are the points in the usual sense. However, in the category SET S, a categorical point of a set X is a subset of X. 2. Linear Symplectic Category ¶ The category LS. We can define the category LSf to be • Ob(LSf ) = symplectic vector spaces , • Mor((V, ΩV ),(W, ΩW )) = linear symplectomorphisms from (V, ΩV ) to (W, ΩW ). Obviously this is not very interesting, because the set of morphisms between two symplectic vector spaces is non-empty only if they have the same dimension. In what follows we will define a slightly larger category, LS, in which the set of morphisms between any two symplectic vector spaces is non-empty. However we prefer not to use all relations as morphisms as in the example SET S above – we want our morphisms to be geometrically interesting. For simplicity we will abbreviate (V, ΩV ) to V , and abbreviate (V, −ΩV ) to V −. By this way, V × W− means the symplectic vector space (V ⊕ W, ΩV ⊕ (−ΩW )). We have shown in lecture 26 that the graph of any linear symplectomorphism from V to W is a Lagrangian subspace of V × W−. On the other hand, V × W− has more Lagrangian subspaces than those arise from the graphs of linear symplectomorphisms. Moreover, for the existence of Lagrangian subspace in V ×W− we don’t require dim V = dim W. Note that a Lagrangian subspace of V × W− is a special subset of V × W, and thus a special relation from V to W
4LECTURE27:FIO-SYMPLECTICCATEGORYDefinition 2.1.A Lagrangian subspace F of V x W- is called a linear canonicalrelation from V to W.Using this language we can define the category CS as.Ob(Cs)= symplecticvectorspaces,. Mor(V,W) = linear canonical relations from V to W.It remains to prove that CS is a category, i.e. the composition of two linear canonicalrelations is a linear canonical relation. (One also need to prove the associativity,but let's ignore this.) Before we prove this let's first admit that s is a categoryand describe its categorical points:The distinguished object “pt" is just the 0-dimensional vector space [o]. So for any symplectic vector space V, its categoricalpoints are the linear canonical relations F C fol × V-.They are of course in one-to-one correspondence with the Lagrangian subspaces of V. In other words, in thecategoryLS.Categorical points of V-Lagrangian subspaces of V.A. Weinstein gave a physical interpretation of this: According to Heisenberg's uncer-tainty principle, if you specify the position of a quantum particle, i.e. if you insiststhat it lie on the Lagrangian submanifold, c, = ai,1 ≤ i≤ n, of R2n = T*Rn, youforfeitall hope of knowing about its momentums.Similar statement for specifyingthe momentum. In other words, in quantum mechanics, points of R2n are irrelevant.Instead, the Lagrangians, i.e. the categorical points, are relevant. Composition of linear canonical relations.Let i C U × V- and 2 C V × W- be two linear canonical relations. Theircomposition is defined in analogy to the composition of relations in SeTS, i.e.(4) T, oFi =f(u, w) EU x W |u E V such that (u,) eTi, (u,w) eI2)Our goal is to proveTheorem 2.2. I2 oIi C U x W- is a linear canonical relation.Before we prove this, we will first prove a special case where U =[O], i.e.Theorem 2.3. Suppose F C V × W- is a linear canonical relation and A C V aLagrangian subspace. Then T(A) is a Lagrangian subspace of W.In other words, linear canonical relations“"maps"“points"to“points”Proof. If we denote Ti : V × W V and 2 : V× W→ W be the projections, thenT(A) = 2((A × W)nF) = 2(元-(A)T):Sor(A)isavectorsubspace of W.Itremains toprove(1) F(A) is isotropic in W.(2) dimI(A) =dim W
4 LECTURE 27: FIO – SYMPLECTIC CATEGORY Definition 2.1. A Lagrangian subspace Γ of V × W− is called a linear canonical relation from V to W. Using this language we can define the category LS as • Ob(LS) = symplectic vector spaces, • Mor(V, W) = linear canonical relations from V to W. It remains to prove that LS is a category, i.e. the composition of two linear canonical relations is a linear canonical relation. (One also need to prove the associativity, but let’s ignore this.) Before we prove this let’s first admit that LS is a category and describe its categorical points: The distinguished object “pt” is just the 0- dimensional vector space {0}. So for any symplectic vector space V , its categorical points are the linear canonical relations Γ ⊂ {0} × V −. They are of course in oneto-one correspondence with the Lagrangian subspaces of V . In other words, in the category LS, Categorical points of V = Lagrangian subspaces of V . A. Weinstein gave a physical interpretation of this: According to Heisenberg’s uncertainty principle, if you specify the position of a quantum particle, i.e. if you insists that it lie on the Lagrangian submanifold, xi = ai , 1 ≤ i ≤ n, of R 2n = T ∗R n , you forfeit all hope of knowing about its momentums. Similar statement for specifying the momentum. In other words, in quantum mechanics, points of R 2n are irrelevant. Instead, the Lagrangians, i.e. the categorical points, are relevant. ¶ Composition of linear canonical relations. Let Γ1 ⊂ U × V − and Γ2 ⊂ V × W− be two linear canonical relations. Their composition is defined in analogy to the composition of relations in SET S, i.e. (4) Γ2 ◦ Γ1 = {(u, w) ∈ U × W | ∃v ∈ V such that (u, v) ∈ Γ1,(v, w) ∈ Γ2} Our goal is to prove Theorem 2.2. Γ2 ◦ Γ1 ⊂ U × W− is a linear canonical relation. Before we prove this, we will first prove a special case where U = {0}, i.e. Theorem 2.3. Suppose Γ ⊂ V × W− is a linear canonical relation and Λ ⊂ V a Lagrangian subspace. Then Γ(Λ) is a Lagrangian subspace of W. In other words, linear canonical relations “maps” “points” to “points”. Proof. If we denote π1 : V × W → V and π2 : V × W → W be the projections, then Γ(Λ) = π2((Λ × W) ∩ Γ) = π2(π −1 1 (Λ) ∩ Γ). So Γ(Λ) is a vector subspace of W. It remains to prove (1) Γ(Λ) is isotropic in W. (2) dim Γ(Λ) = 1 2 dim W
5LECTURE27:FIO-SYMPLECTICCATEGORYThe proof of (1): Suppose wi, w2 E T(A), then by definition, there exists vi, 2 E Aso that % = (ui, w;) e I,i = 1, 2. For simplicity we will denote = 2v @ (-Nw).Since AV andr cV×W-areboth Lagrangians,2v(u1, V2) = 0, 2(1, 2) = 0.On the other hand.(2(1, 2) = 2v(1, V2) - w(w1, W2)Itfollows2w(wi,w2)=0.The proof of (2): Let H = π-'(A) = A× W c V ×W, then H? = A×[0]. ConsiderthemapQ=20Lnr:HnTV×W-W,then Im(α) = 2(Hn) =T(A), andker(α) = (u, 0) Hn IE A) = Hn.Similarly if we let + be the mapT:HXF-VXW,((h,)-h -,then Im() = H +I. It follows(Im()? = H"nI = ker(α).In particular, dim ker(α) = dim coker(), and thusdim Im(α) = dim H nF- dim coker()To calculate the right hand side, we consider the short exact sequence0→HnH×V×W→coker()→0.The exactness (which is easy to check) impliesdim HnF+ dim V × W = dim H ×I+ dim coker()Itfollowsdim HnF- dimcoker() = dimH +dimI- dim V-dim W dim W2口Proof of Theorem 2.2. We let V = U xV- and W - U × W-. Then we can identifyV×W-= (U×U-) × (V×W-)-.Since A := Ii is a Lagrangian subspace of V, andI := Au × I2 c V×W-is a linear canonical relation, Theorem 2.3 implies thatI(A) = [(u, w) E U × W I (u, u) eFi such that (u, u, u, w) eF) =I, oIiis a Lagrangian subspace of w.口
LECTURE 27: FIO – SYMPLECTIC CATEGORY 5 The proof of (1): Suppose w1, w2 ∈ Γ(Λ), then by definition, there exists v1, v2 ∈ Λ so that γi = (vi , wi) ∈ Γ, i = 1, 2. For simplicity we will denote Ω = ΩV ⊕ (−ΩW ). Since Λ ⊂ V and Γ ⊂ V × W− are both Lagrangians, ΩV (v1, v2) = 0, Ω(γ1, γ2) = 0. On the other hand, Ω(γ1, γ2) = ΩV (v1, v2) − ΩW (w1, w2). It follows ΩW (w1, w2) = 0. The proof of (2): Let H = π −1 1 (Λ) = Λ×W ⊂ V ×W, then HΩ = Λ×{0}. Consider the map α = π2 ◦ ιH∩Γ : H ∩ Γ ,→ V × W → W, then Im(α) = π2(H ∩ Γ) = Γ(Λ), and ker(α) = {(v, 0) ∈ H ∩ Γ | v ∈ Λ} = H Ω ∩ Γ. Similarly if we let τ be the map τ : H × Γ → V × W, (h, γ) 7→ h − γ, then Im(τ ) = H + Γ. It follows (Im(τ ))Ω = H Ω ∩ Γ = ker(α). In particular, dim ker(α) = dim coker(τ ), and thus dim Im(α) = dim H ∩ Γ − dim coker(τ ). To calculate the right hand side, we consider the short exact sequence 0 → H ∩ Γ ∆ ,→ H × Γ τ→ V × W → coker(τ ) → 0. The exactness (which is easy to check) implies dim H ∩ Γ + dim V × W = dim H × Γ + dim coker(τ ). It follows dim H ∩ Γ − dim coker(τ ) = dim H + dim Γ − dim V − dim W = 1 2 dim W. Proof of Theorem 2.2. We let Ve = U ×V − and Wf = U ×W−. Then we can identify Ve × Wf− = (U × U −) × (V × W−) −. Since Λ := Γ1 is a Lagrangian subspace of Ve, and Γ := ∆U × Γ2 ⊂ Ve × Wf− is a linear canonical relation, Theorem 2.3 implies that Γ(Λ) = {(u, w) ∈ U × W | ∃(u, v) ∈ Γ1 such that (u, u, v, w) ∈ Γ} = Γ2 ◦ Γ1 is a Lagrangian subspace of Wf.
6LECTURE27:FIO-SYMPLECTICCATEGORY3.SYMPLECTIC“CATEGORY"The“category"s.Motivated by the category CS of symplectic vector spaces, one can define acategory S of symplectic manifolds via.Ob(S)= symplecticmanifolds,? Mor(M, N) = canonical relations from M to N,where, as in the linear case, we use the notionDefinition 3.1. A canonical relation from a symplectic manifold M to a symplecticmanifold N is a Lagrangian submanifold of M × N-, where N-= (N, -wn).BADNEWS:Given canonical relationsiEMixM,andI2EM2×M,theircompositionI2oImayfail to bea smooth submanifold, and asaresult,fail tobea canonical relation fromMi to Ms!As a consequence, S is NOT a true category. Transversal/Clean intersection conditions.Recall from manifold theory the following useful criteria for a submanifold:Definition 3.2. Suppose X, Z are smooth manifolds and Y a submanifold of Z.Let f :XZ be a smooth map.We say f intersects Y transuersally if(5)Im(dfp) + Tf(p)Y = Tf(p)Z, Vp E f-1(Y).Theorem 3.3. If f intersects Y transversally, then f-1(Y) is a smooth submanifoldof X whose tangent space at p E f-1(Y) is(6)Tp(f-1(Y)) = (dfp)-1(Tr(p)Y).Therearetwospecial cases of this definition/theorem:. Suppose X, Y smooth submanifolds of Z. We say X and Y intersect transver-sally in Z if : X Z intersect Y transversally. Equivalently:(7)T,Z=TX+TY,VpEXnYIn this case, X nY is a smooth submanifold of z whose tangent space is(8)T,(XnY)=T,XnT,Y.Suppose fi:X,→Z, i=1,2 are smooth maps.We say fi and f intersecttransversally if theproduct map fi×f2:Xi×X2→Z×Z intersects withthe diagonal z =((z,z)Iz E Z) transversally.In this case, the fiberproduct(9)F= (fi × f2)-1(△z) =[(1,2) [fi(c1) = f2(r2))is a smooth submanifold of Xi × X2 whose tangent space at r = (ri, r2) E Fis(10)TF = [(u1, V2) / Us E Tr,X, (df1)rn(i) = (df2)r2(u2))
6 LECTURE 27: FIO – SYMPLECTIC CATEGORY 3. Symplectic “Category” ¶ The “category” S. Motivated by the category LS of symplectic vector spaces, one can define a category S of symplectic manifolds via • Ob(S) = symplectic manifolds, • Mor(M, N) = canonical relations from M to N, where, as in the linear case, we use the notion Definition 3.1. A canonical relation from a symplectic manifold M to a symplectic manifold N is a Lagrangian submanifold of M × N −, where N − = (N, −ωN ). BAD NEWS: Given canonical relations Γ1 ∈ M1 × M− 2 and Γ2 ∈ M2 × M− 3 , their composition Γ2 ◦ Γ1 may fail to be a smooth submanifold, and as a result, fail to be a canonical relation from M1 to M3! As a consequence, S is NOT a true category. ¶ Transversal/Clean intersection conditions. Recall from manifold theory the following useful criteria for a submanifold: Definition 3.2. Suppose X, Z are smooth manifolds and Y a submanifold of Z. Let f : X → Z be a smooth map. We say f intersects Y transversally if (5) Im(dfp) + Tf(p)Y = Tf(p)Z, ∀p ∈ f −1 (Y ). Theorem 3.3. If f intersects Y transversally, then f −1 (Y ) is a smooth submanifold of X whose tangent space at p ∈ f −1 (Y ) is (6) Tp(f −1 (Y )) = (dfp) −1 (Tf(p)Y ). There are two special cases of this definition/theorem: • Suppose X, Y smooth submanifolds of Z. We say X and Y intersect transversally in Z if ι : X ,→ Z intersect Y transversally. Equivalently: (7) TpZ = TpX + TpY, ∀p ∈ X ∩ Y. In this case, X ∩ Y is a smooth submanifold of Z whose tangent space is (8) Tp(X ∩ Y ) = TpX ∩ TpY. • Suppose fi : Xi → Z, i = 1, 2 are smooth maps. We say f1 and f2 intersect transversally if the product map f1 × f2 : X1 × X2 → Z × Z intersects with the diagonal ∆Z = {(z, z) | z ∈ Z} transversally. In this case, the fiber product (9) F = (f1 × f2) −1 (∆Z) = {(x1, x2) | f1(x1) = f2(x2)} is a smooth submanifold of X1 ×X2 whose tangent space at x = (x1, x2) ∈ F is (10) TxF = {(v1, v2) | vi ∈ TxiXi ,(df1)x1 (v1) = (df2)x2 (v2)}
LECTURE27:FIO-SYMPLECTICCATEGORY7Remark. Of course the most important thing for us is the conclusions in the abovetheorems,notthe conditions.This motivates thefollowing definition:Definition 3.4.LetX,X,,Y,Z besmooth manifolds, and f,f;smoothmaps.(1) We say f : X → Z intersects Y C Z cleanly if f-1(Y) is a smooth submani-fold of Y and for all pE f-1(Y), T,(f-1(Y)) = (df,)-1(Tr(p)Y).(2) We say X, Y c Z intersect cleanly in Z if t: X Z intersect Y cleanly(3) We say fi and f2 intersect cleanly if the product map fi × f2 intersects withthe diagonal △z C Z × Z cleanly.Note that for (2) and (3),the tangent spacearegiven by (8)and (10)respectivelyT Composition of canonical relations under cleanness condition.Now suppose M; are symplectic and I; C M; × Mi+i are canonical relations.Considertheprojectionsπ:Fi→M2,(m,m2)→m2p:I2→M2, (m2,m3)→m2Theorem 3.5. If and p intersect cleanly, then F, oFi is an immersed canonicalrelations in Mi × M3.Proof. Since and p intersect cleanly, the fiber productF = (π × p)-1(M2) ~ [(m1, m2, m3) / (m1,m2) EI1, (m2,m3) ET2)is a submanifold of Mi × M2 × M3 whose tangent space at m = (mi, m2, m3) equalsTmF = {(1, 2, V3) / Vi ETm,Mi, (1, V2) E T(m1,m2)F1, (2, U3) E T(m2,m3)I2)Let t:F Mi × M2 × M3 be the inclusion, andK: Mi×M2 × M3-→Mi × M3,(m1,m2,m3)- (mi,m3)be the“projection onto the first and third component"map.Then by definition,T2 0T1= Im(K0t)Note that d(kot)m is just the projection mapd(ot)m : TmF-T(m1,m3)(M × M3), (U1, V2,V3) - (U1, U3)This has two implications:.By definitionT(mi,ms)(Im(r ot)) = Im(d(k o t)m) = T(m2,ms)T2 o T(m1.m2)Fisa linear canonical relation in Tm,Mi×TmsM3..kot is a constant rankmap, so itsmage F2ois an immersed submanifoldofMixM3.口It follows that F2 o Fi is an immersed Lagrangian submanifold in Mi × Mg
LECTURE 27: FIO – SYMPLECTIC CATEGORY 7 Remark. Of course the most important thing for us is the conclusions in the above theorems, not the conditions. This motivates the following definition: Definition 3.4. Let X, Xi , Y, Z be smooth manifolds, and f, fi smooth maps. (1) We say f : X → Z intersects Y ⊂ Z cleanly if f −1 (Y ) is a smooth submanifold of Y and for all p ∈ f −1 (Y ), Tp(f −1 (Y )) = (dfp) −1 (Tf(p)Y ). (2) We say X, Y ⊂ Z intersect cleanly in Z if ι : X ,→ Z intersect Y cleanly. (3) We say f1 and f2 intersect cleanly if the product map f1 × f2 intersects with the diagonal ∆Z ⊂ Z × Z cleanly. Note that for (2) and (3), the tangent space are given by (8) and (10) respectively. ¶ Composition of canonical relations under cleanness condition. Now suppose Mi are symplectic and Γi ⊂ Mi × M− i+1 are canonical relations. Consider the projections π : Γ1 → M2, (m1, m2) 7→ m2 ρ : Γ2 → M2, (m2, m3) 7→ m2. Theorem 3.5. If π and ρ intersect cleanly, then Γ2 ◦ Γ1 is an immersed canonical relations in M1 × M− 3 . Proof. Since π and ρ intersect cleanly, the fiber product F = (π × ρ) −1 (∆M2 ) ' {(m1, m2, m3) | (m1, m2) ∈ Γ1,(m2, m3) ∈ Γ2} is a submanifold of M1 × M2 × M3 whose tangent space at m = (m1, m2, m3) equals TmF = {(v1, v2, v3) | vi ∈ TmiMi ,(v1, v2) ∈ T(m1,m2)Γ1,(v2, v3) ∈ T(m2,m3)Γ2}. Let ι : F ,→ M1 × M2 × M3 be the inclusion, and κ : M1 × M2 × M3 → M1 × M3, (m1, m2, m3) 7→ (m1, m3) be the “projection onto the first and third component” map. Then by definition, Γ2 ◦ Γ1 = Im(κ ◦ ι). Note that d(κ ◦ ι)m is just the projection map d(κ ◦ ι)m : TmF → T(m1,m3)(M1 × M3), (v1, v2, v3) 7→ (v1, v3). This has two implications: • By definition, T(m1,m3)(Im(κ ◦ ι)) = Im(d(κ ◦ ι)m) = T(m2,m3)Γ2 ◦ T(m1,m2)Γ1 is a linear canonical relation in Tm1M1 × Tm3M3. • κ ◦ ι is a constant rank map, so its mage Γ2 ◦ Γ1 is an immersed submanifold of M1 × M3. It follows that Γ2 ◦ Γ1 is an immersed Lagrangian submanifold in M1 × M− 3 .
8LECTURE27:FIO-SYMPLECTICCATEGORYFact from manifold theory: the image of the constant rank map k o i is anembedded submanifold if k o t is proper and its level sets are connected.(For aproof, see the appendix to this lecture.) SoTheorem 3.6.Suppose and p intersect cleanly.In addition suppose kot is properand for any (mi,m3) e 2 oFi, ( ot)-1(mi,m3) is connected. Then I2 oFi is acanonical relation in Mi × M3:The category M v.s. the“"category"S.Recall that the category Mof smoothmanifolds consists. Ob(M) = all smooth manifolds,. Mor(X,Y)= smooth maps from X to Y.We also learned that any object X in M corresponds to an object M = T*X in S.Now suppose f : X-→Y is a morphism, i.e. a smooth map. Then its graph X isa submanifold of Xi x X2, associated to which we have a Lagrangian submanifoldN*Xf of T*Xi × T*X2. If we let be the involutiona : T*Xi × T*X2 -→T*Xi × T*X2,(r,s,y,n) -→ (a,S,y, -n),then If = o(N*X) is a canonical relation. So we get an embedding (a covariant"functor")of category M into“category"S:.X-M =T*X.. (f : X -→Y) - If =o(N*X)). Composition of a sympectomorphism with a canonical relation.Suppose Fi E Mi × M2 is a canonical relation and I, is the graph of a symplec-tomorphism from M2 to M3.Thendp(m2,ma) : T(m2,ms)F2 → Tm, M2is surjective. It follows that for each m = (mi, m2, m2, m3) E I × I2, the image ofd(π× p)m contains all vectors of the form (O, w) e Tm,M2 × Tm,M2. In other words.π×pintersects the diagonal M,transversally.So thefiber product Fis a smoothsubmanifold.Moreover,themapk is given byk:(mi,m2,g(m2))-(mi,g(m2)),where g : M2 → Mg is the symplectomorphism. Since g is one-to-one, so is k. Itfollows that 2 o T1 is a canonical relation. Similar argument holds if Fi is definedby a symplectomorphism. In conclusion, we showedTheorem3.7.The composition of the graph ofa symplectomorphism with a canon-ical relation is again a canonical relation.In particular, we see that the composition of a canonical relation F e M × N-with the diagonal △m is again a canonical relation
8 LECTURE 27: FIO – SYMPLECTIC CATEGORY Fact from manifold theory: the image of the constant rank map κ ◦ ι is an embedded submanifold if κ ◦ ι is proper and its level sets are connected. (For a proof, see the appendix to this lecture.) So Theorem 3.6. Suppose π and ρ intersect cleanly. In addition suppose κ◦ι is proper and for any (m1, m3) ∈ Γ2 ◦ Γ1, (κ ◦ ι) −1 (m1, m3) is connected. Then Γ2 ◦ Γ1 is a canonical relation in M1 × M− 3 . ¶ The category M v.s. the “category” S. Recall that the category M of smooth manifolds consists • Ob(M) = all smooth manifolds, • Mor(X, Y ) = smooth maps from X to Y . We also learned that any object X in M corresponds to an object M = T ∗X in S. Now suppose f : X → Y is a morphism, i.e. a smooth map. Then its graph Xf is a submanifold of X1 × X2, associated to which we have a Lagrangian submanifold N∗Xf of T ∗X1 × T ∗X2. If we let σ be the involution σ : T ∗X1 × T ∗X2 → T ∗X1 × T ∗X2, (x, ξ, y, η) 7→ (x, ξ, y, −η), then Γf = σ(N∗Xf ) is a canonical relation. So we get an embedding (a covariant “functor”) of category M into “category” S: • X 7→ M = T ∗X. • (f : X → Y ) 7→ Γf = σ(N∗Xf ). ¶ Composition of a sympectomorphism with a canonical relation. Suppose Γ1 ∈ M1 × M2 is a canonical relation and Γ2 is the graph of a symplectomorphism from M2 to M3. Then dρ(m2,m3) : T(m2,m3)Γ2 → Tm2M2 is surjective. It follows that for each m = (m1, m2, m2, m3) ∈ Γ1 × Γ2, the image of d(π ×ρ)m contains all vectors of the form (0, w) ∈ Tm2M2 ×Tm2M2. In other words, π × ρ intersects the diagonal ∆M2 transversally. So the fiber product F is a smooth submanifold. Moreover, the map κ is given by κ : (m1, m2, g(m2)) 7→ (m1, g(m2)), where g : M2 → M3 is the symplectomorphism. Since g is one-to-one, so is κ. It follows that Γ2 ◦ Γ1 is a canonical relation. Similar argument holds if Γ1 is defined by a symplectomorphism. In conclusion, we showed Theorem 3.7. The composition of the graph of a symplectomorphism with a canonical relation is again a canonical relation. In particular, we see that the composition of a canonical relation Γ ∈ M × N − with the diagonal ∆M is again a canonical relation
LECTURE27:FIO-SYMPLECTICCATEGORY94.APPENDIXInthisappendixweproveTheorem 4.1. Let f : M N be a constant rank proper map and each level set isconnected. Then the image of f is an embedded submanifold on N.Proof. We know that the image of a constant rank map is an immersed submanifold.In other words, for any r E M, there exist open neighborhoods Uof r in M andV(r) of f(r) in N such that f(U)is an embedded submanifold of N.In whatfollows we try to find open neighborhood Vr() C Vf(r) of f(α) in N such thatf(M)n Vr(a) = f(Ua)n Vr(n),from which we conclude that f(M) is an embedded submanifold of NWewillproceedbycontradiction.Weshallneedthefollowingfact:Fact: If f(r') = f(r), then for any open neighborhood Ur and Ur,there exists open neighborhood UrC U and UrC Ur so thatf(Ur) = f(Ur)Proof. We define an equivalence relation on f-1(f(r)) by ~ ' if the property in the statement holds for r and r.Then by local structure theorem of constant rank map, each equiva-lence class is open in f-1(f(r). Since the complement of any equiv-alence class is a union of equivalence classes, each equivalence classis also closed in f-1(f(r)).Thus by the connectedness of f-1(f()),口there is only one equivalence class.Now suppose no such Vr(a) exists. Then there exists yk = f(rk) E f(X) but yk f(U)such thatyk→f(r).Sincef isproperand(yk,f(r)) is compact,rk'slies ina compact set in M and thus have a convergent subsequence k, -→ r' e f-1(f(r)).According to the fact we just proved, there exists open neighborhood Ur c Ur andUrso that f(U)=f(Ur). It follows that for k large enough,ykEf(U)=f(U.)Cf(U),口a contradiction
LECTURE 27: FIO – SYMPLECTIC CATEGORY 9 4. Appendix In this appendix we prove Theorem 4.1. Let f : M → N be a constant rank proper map and each level set is connected. Then the image of f is an embedded submanifold on N. Proof. We know that the image of a constant rank map is an immersed submanifold. In other words, for any x ∈ M, there exist open neighborhoods Ux of x in M and Vf(x) of f(x) in N such that f(Ux) is an embedded submanifold of N. In what follows we try to find open neighborhood Vef(x) ⊂ Vf(x) of f(x) in N such that f(M) ∩ Vef(x) = f(Ux) ∩ Vef(x) , from which we conclude that f(M) is an embedded submanifold of N. We will proceed by contradiction. We shall need the following fact: Fact: If f(x 0 ) = f(x), then for any open neighborhood Ux and Ux0, there exists open neighborhood Uex ⊂ Ux and Uex0 ⊂ Ux0 so that f(Uex) = f(Uex0). Proof. We define an equivalence relation on f −1 (f(x)) by x ∼ x 0 if the property in the statement holds for x and x 0 . Then by local structure theorem of constant rank map, each equivalence class is open in f −1 (f(x)). Since the complement of any equivalence class is a union of equivalence classes, each equivalence class is also closed in f −1 (f(x)). Thus by the connectedness of f −1 (f(x)), there is only one equivalence class. Now suppose no such Vef(x) exists. Then there exists yk = f(xk) ∈ f(X) but yk 6∈ f(Ux) such that yk → f(x). Since f is proper and {yk, f(x)} is compact, x k ’s lies in a compact set in M and thus have a convergent subsequence xki → x 0 ∈ f −1 (f(x)). According to the fact we just proved, there exists open neighborhood Uex ⊂ Ux and Uex0 so that f(Uex) = f(Uex0). It follows that for k large enough, yk ∈ f(Uex0) = f(Uex) ⊂ f(Ux), a contradiction.