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# Elementary Algebra MATH M0070

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EQUIVARIANT ORIENTATIONS AND THOM ISOMORPHISMS JP MAY CONTENTS 1 The fundamental groupoid and categorical de nitions 3 2 Functors to the category of Gspaces over orbits 5 3 Coherent families of connected covers 7 4 Orientability of spherical G brations 8 5 Coherent families of Thom classes and Thom isomorphisms 10 6 Orientations in ordinary equivariant cohomology l3 7 Concluding remarks 16 References 16 Let G be a compact Lie group and let Ea be an ROGgraded cohomology theory on Gspaces We shall explain a sensible way to think about orientations and the Thom isomorphism theorem in the theory Ea offering an alternative to the approach given by Costenoble and Waner in Both approaches generalize the restricted theory given by Lewis and myself in 15 and both grew out of joint work of Costenoble Waner7 and myself In the study of nonequivariant bundles and their orientations an innocuous rst step is to assume that the base space is path connected The analogous equivariant assumption is that the Gset of components of the base space is a single orbit but this assumption doesn t get us very far There is an entirely satisfactory theory of equivariant Thom isomorphisms and Poincare duality under the much more stringent hypothesis that the base space X be G connected7 in the sense that each XH is nonempty and path connected This is developed in detail in 15 HI 6 and X 5 The basic problem7 then is to generalize that theory to a less restricted class of base spaces The obvious approach is to parametrize changes of ber representation on the fundamental groupoid However doing this directly leads to a fairly complicated and hard to compute7 generalization of equivariant cohomology 5 67 77 8 We seek a variant approach that allows us to work within the framework of ROGgraded cohomology theories7 so that we can apply rather than generalize the preexisting theory of 15 Our essential idea is that7 to obtain a satisfactory general theory7 it seems reason able to give up the idea that the Thom isomorphism must be a single isomorphism Rather we shall de ne it to be an appropriate family of isomorphisms More pre cisely7 we shall de ne an EEorientation of a Gvector bundle7 or more generally of a spherical G bration7 to be a suitably coherent family of cohomology classes Each class in the family will determine a Thom isomorphism7 and these isomorphisms will be nicely related If the base space is G connected7 then all of these Thom 1 2 JP MAY classes and Thom isomorphisms will be determined by one member of the family In general7 all members of the family will be determined by choosing members of the family indexed on components of xed point spaces X which do not contain any K xed points for K larger than H We begin in Sections 173 by de ning the fundamental groupoid 71X discussing functors de ned on it and constructing the family 0 Hconnected covers77 of a Gspace X This construction can be expected to have other uses in equivariant al gebraic topology Many arguments in algebraic topology begin with the statement e may assume without loss of generality that X is connected Our Hconnected covers give a tool that often allows us to give the same start to equivariant argu ments However7 the reader is warned that there are some prices to be paid7 beyond the intrinsic complexity Probably the most signi cant is that the Hconnected cover of a nite G CW complex will in general be in nite dimensional For this reason I have not yet succeeded in obtaining a satisfactory treatment of Poincare duality that starts from the Thom isomorphism theorem given here We de ne the notion of an orientable spherical G bration in Section 4 This does not depend on our Hconnected covers7 but we use these in our de nition of an EEorientation of a spherical G bration in Section 5 Our Thom isomorphism theorem in E cohomology follows directly from the de nition and the work in 15 We specialize to ordinary ROGgraded cohomology with Burnside ring coef cients in Section 6 Nonequivariantly7 orientability as de ned topologically is equivalent to cohomological orientability with integer coef cients We prove that perhaps the most natural topological notion of equivariant orientability is equivalent to cohomological orientability with Burnside ring coef cients We brie y mention other examples in Section The theory here was suggested by joint work with Steve Costenoble and Stefan Waner 5 and later Igor Kriz that began in the late 1980 s and is still largely unpublished In that work we take the more direct approach7 confronting head on the problem of parametrizing the change of ber representations of a Gvector bundle by the equivariant fundamental groupoid of the base space Rather than giving up the idea that an orientation is a single cohomology class7 we construct more complicated cohomology theories in which a single class can encode all the complexity That approach has been exploited in a series of papers on this and related topics by Costenoble and Waner 6 77 87 97 10 A fully coherent theory of orientation will require a comparison of that approach to the one given here Although Costenoble Kriz Waner7 and I sketched out such a comparison some years ago the details have yet to be worked out It is a pleasure to thank Costenoble Waner7 and Kriz for numerous discussions of this material This paper is a very small token of thanks to Mel Rothenberg7 my colleague and friend for the last 32 years I wish I had a better paper to offer since this one should have a sign on it saying speculative may not be useful but it is in one of the areas that Mel has pioneered eg 18 and that I have in part learned from him It has been a privilege to work with him all these years to help make Chicago a thriving center of topology EQUIVARIANT ORIENTATIONS AND THOM ISOMORPHISMS 3 1 THE FUNDAMENTAL GROUPOID AND CATEGORICAL DEFINITIONS We here recall our preferred de nition of 77X and give some categorical lan guage that will help us de ne structures in terms of it An equivalent de nition appears in 117 107 and a de nition in terms of Moore loops is given in 177 App We assume that G is a compact Lie group7 and we only consider closed subgroups Let 0G denote the topological category of orbit spaces G H and Gmaps between them where H runs through the closed subgroups of G Let h 39g be the homotopy category of jg Of course7 0G h 39g if G is nite The following observation describes the structure of h 39g for general compact Lie groups G Recall that if a GH H GK is a G map with aeH 9K then g ng C K Lemma 11 Let j a H B be a Ghomotopy between Gmaps GH H GK Then j factors as the composite ofa and a homotopy c GH X I H GH such that CeH t ctH where 00 e and the 0 specify apath in the identity component of the centralizer GgH of H in G Proof Let jeHt gtK Since we can lift this path in GK to a path in G starting at go we may assume that the 9 specify a path in G Now g1Hgt C K for all t so we can de ne d H X I H K by dht g1hgt Since the adjoint d I H MapH K is a path through homomorphisms the MontgomeryZippen Theorem 4 381 implies that there are elements 19 E K such that ho e and dht klgalhggkt De ne C K H Homgangg K by Chh k lh k The image of C may be identi ed with K L where L is the subgroup of elements 19 such that Ce is the inclusion of gangg in K It follows that C is a bundle over its image We may regard d as a path in Homgangg K and we can lift it to a path 19 I H K with 190 5 Thus we may assume that the let specify a path in K Now de ne ct gtklggl Then jeHt CtggK co e and the C are in GgH7 as desired D De nition 12 Let X be a Gspace De ne the fundamental groupoid 77X as follows Its objects are the pairs GHx where H C G and x E XH we think of this pair as the G map x GH H X that sends eH to x The morphisms GHx H GK y are the equivalence classes ago of pairs ao1 where a GH H GK is a Gmap and 01 GH X I H X is a G homotopy from x to yoa Here two such pairs a 01 and 0 of are equivalent if there are Ghomotopies jzazo and920120 suchthat ka0t xa and ka1t yojat for a E GH and t E I If G is nite then a 0 and j is constant Composition is evident De ne a functor a 8X 77X H h 39g by sending GH x to GH and sending aw to the homotopy class a of a A Gmap f X H Y induces a functor 77X H 77Y such that 8y 0 f 8X A Ghomotopy h f 2 f induces a natural isomorphism h H To be precise about orientability and orientations we need some abstract def initions and constructions7 which are taken from joint work with Costenoble and Waner The rst encodes the formal structure of the fundamental groupoid De nition 13 A groupoid over a small category 33 is a small category together with a functor a H 33 that satis es the following properties For an object I 4 JP MAY of 53 the ber W5 is the subcategory of consisting of the objects and morphisms that 8 maps to b and id5 i For each object 5 of 53 W5 is either empty or a groupoid in the sense that each of its morphisms is an isomorphism ii Source lifting For each object y E and each morphism B a H 8y in 53 there is an object at E such that 830 a and a morphism y x H y in such that 87 Divisibility For each pair of morphisms y x H y and y x H y in and each morphism B 830 H 8 in 33 such that 87 87 0 3 there is a morphism 5 x H x in such that 86 B and y o 5 y V Remark 14 We say that has unique divisibility if the morphism 5 asserted to exist in iii is unique This holds for fundamental groupoids when G is nite7 but not when G is a general compact Lie group When it holds7 is exactly a cate gorie br e en groupoides77 over 33 as de ned by Grothendieck 137 p 166 De nitions 15 Let be a groupoid over i is skeletal over 33 if each ber W5 is skeletal has a single object in each isomorphism class of objects ii is faithful over 33 if a is faithful injective on hom sets iii is discrete over 33 if each W5 is discrete has only identity morphisms Lemma 16 is discrete over 33 and only it is skeletal and faithful over Proof Clearly ifW is skeletal and faithful7 then it is discrete IfW is discrete7 then it is clearly skeletal and it is faithful by Remark 17ii below 1 Remarks 17 Let be a groupoid over i If is skeletal over 53 then divisibility implies that the object at asserted to exist in the source lifting property is unique If is skeletal and faithful over 53 then the morphism asserted to exist in the source lifting property is also unique If is faithful over 53 then it is uniquely divisible ii By divisibility any two morphisms x H y of over the same morphism of 33 differ by precomposition with an automorphism of at over the identity mor phism of Thus is faithful over 33 if and only if the only automorphisms in each W5 are identity maps iii If y x H y is a morphism of such that 87 is an isomorphism7 then 7 is an isomorphism7 as we see by application of divisibility to the equality a ya y 1 id If every endomorphism of every object of 33 is an isomor phism7 as holds in jg then every endomorphism of every object of is an isomorphism Construction 18 Let be a groupoid over We construct the discrete groupoid over 33 associated to i Construct a faithful groupoid with the same objects as by setting Ime H dx The quotient functor H is the universal map from into a faithful groupoid over ii Construct a skeletal subgroupoid W of by choosing one object in each isomorphism class of objects of W5 for each object 5 of 33 and letting W be EQUIVARIANT ORIENTATIONS AND THOM ISOMORPHISMS 5 the resulting full subcategory of The inclusion b H is an adjoint equivalence of categories over 53 its left inverse is a retraction p obtained from any choice of isomorphisms from each object of each W5 to an object of We call 5 a skeleton of v By Remark 17iii passage from to creates no new isomorphisms so that we can make the same choices of objects for and for when forming skeleta Then W78 We call this category the discrete groupoid over 33 associated to Lemma 19 Joyal A discrete groupoid b39 over determines and is determined by an associated contravai iant functor F 33 H Sets Proof Given 539 de ne F as follows For an object 5 of B Fb is the set of objects of 53 For a morphism B a H b of 33 and an object y of 53 is the unique object of Wu that is the source of a map covering Given F de ne as follows Its objects are pairs 5 y where b is an object of 33 and y E Fb A morphism a x H b y is a morphism B a H b of 33 such that x Composition and the functor a H 53 are evident D Now return to the fundamental groupoid Notations 110 Let 770 X denote the discrete groupoid over h 39g associated to the fundamental groupoid 71X The quotient functor 77X H 77Xa identi es aw and ozw whenever a 0 so that functors de ned on 77X factor through 77Xa if their values on morphisms are independent of paths The cate gory 770X is obtained from 77Xa by choosing one point in each component of each xed point space The notation 770X is justi ed since the associated con travariant functor h 39g H Sets can be identi ed with the evident functor that sends an orbit GH to the set of components 770XH 2 FUNCTORS TO THE CATEGORY OF G SPACES OVER ORBITS In the next section we show how to construct a system of interrelated H connected covers77 associated to a given Gspace X The interrelationships will be encoded in terms of functors de ned on 71X We describe the target category and the abstract nature of the functors we will be concerned with in this section Let lt2 be the category of compactly generated weak Hausdorff spaces and let G W be the category of G spaces De nition 21 De ne Glt2jlg to be the category of Gspaces over Gm bits The objects of this category are G maps X X H GH and the morphisms are commutative diagrams of Gmaps f X H Y X i G H H G K There is an evident notion of a homotopy between such morphisms and a resulting homotopy category hG W quotG Let a hGlt2jlg H h 39g be the evident augmen tation functor it forgets the G spaces and remembers the Gorbits 6 JP MAY We think of a Gmap X X H GH as having the total space X and base space GH although we do not require X to be a bration Let V X 1eH C X Then V is an Hspace and the action of G on X induces a Gmap 5 G X V H X that is easily veri ed to be a bijection To avoid pointset pathology we agree to restrict attention to completely regular eg normal Gspaces For such X 5 is a homeomorphism of G spaces The following remark gives a more concrete but less canonical description of the category hGlt2jlg Remark 22 For a commutative diagram of Gmaps G X V Hf G gtltK W X 11 GH Hagt GK with aeH 9K de ne f V H W by g 1fv Then f is an Hmap where H acts on W by hw g lhgw and 23 an mm for j E G We call f the Gmap associated to the pair g Suppose that maps f0 and f1 over homotopic Gmaps a0 and al are associated to pairs f0gg and f1gl Then f0a0 and f1a1 are homotopic if and only if there is a path ft connecting f0 to f1 in the space of Hmaps V H W The point is that by Lemma 11 the homotopy a0 2 a1 can be written in the form ateH CtggK where C is a path in GgH starting at e and the conjugate Haction on W is then the same throughout the homotopy De nition 24 Let be a groupoid over h 39g A ltzquot space is a functor Y H hGlt2jlg over h 39g Given a map A H b of groupoids over h 39g a map c5 Y H Y from a ltitfspace Y to a ltquotrmspace Y is a natural transformation zYHY voverhquotG There is a less conceptual but perhaps more easily understood version of this def inition in terms of our concrete description of G Wj39g We shall work throughout in terms of this alternative version Lemma 25 A ltzquot space Y determines and is determined by the following data i An Hspace Zx for each object as in the ber KgH of ii A homotopy class of H maps Z yg Zx H Zy for each morphism y x H y of and element 9 ofG such that a yeH 9K where 87 GH H GK here H acts on Zy by ha g lhga for a E These data must satisfy the following properties iii In Z ygk 2 k 1Z yg for k E K iv Zid e 2 id and Z y g o Zyg 2 Zy o ygg when M o y is de ned Given A H K a map c5 Y H Y from a ltzquot space Y to a ltitmspace Y determines and is determined by H maps C Cm Zx H Z Ax for objects 90 E KgH such that Z Ayg o 0 21 0 Z yg for pairs 79 as in Proof Given the speci ed data set Yx G X Zx and let Y y be the homo topy class of the Gmap associated to the pair Z ygg property iii ensures that Y y is independent of the choice of 9 Conversely given Y let Zx C Yx EQUIVARIANT ORIENTATIONS AND THOM ISOMORPHISMS 7 be the Hspace over the orbit 5H and let Z y g Zx H Zy be the composite of YW and multiplication by 9 1 Similarly given C let on Yx H Y Ax be the Gmap associated to the pair Che and conversely given c5 let Cm be the restriction Zx H Z Ax of 1 3 COHERENT FAMILIES OF CONNECTED COVERS A standard tool in equivariant algebraic topology is to study G spaces by means of their diagrams of xed point spaces On the diagram level it is quite trivial to give a notion of an H connected cover The following de nition encodes that trivial starting point of our wor De nition 31 An jgspace is a continuous contravariant functor 0G H 24 and a map of jgspaces is a natural transformation Let jgW denote the cat egory of jgspaces For a Gspace X de ne the xed point jgspace PX by DX XH For a xed point x E XH let XHx denote the component of x in XH De ne the H connected cover of PX at x to be the sub j39Hspace ltIgtXx of the j39Hspace X such that ltIgtX Xx for J C H We can lift this essentially nonequivariant structure to the equivariant level by means of a construction due to Elmendorf 12 see also 16 V 3 and VI 6 We shall gradually make sense of and prove the following result in this section Theorem 32 Let X be a Gspace There is a 77Xspace X such that for x GH H X Xx is the Hconnected cover of X at at There is a natural map of 71Xspaces X H X where X is regarded as a constant 77Xspace For C H 5 Xx H X is the composite of a canonical weak equivalence Xx H Xx and the inclusion Xx H X Here we are thinking of 77Xspaces in terms of the data speci ed in Lemma 25 We recall the main properties of Elmendorf s construction Theorem 33 There is afunctoqu jgW H G W and a natural transformation a FRI H Id such that for an jgspace T each GH IJTH H TGH is a homotopy equivalence IfX is a GCW complex then XIJTG E DXTG For an jgspace T evaluation of a at G 5 gives a Gmap 8Ge IJTG H TGe When T PX so that TGe X 8Ge 8GJ IJQXJ H X Thus 8Ge IJltIgtX H X is a weak Gequivalence for any G space X and 8Ge is a G homotopy equivalence if X is a G CW complex De nition 34 The Hconnected cover of X at x E XH is the Hspace Xx IJltIgtX Thus we have homotopy equivalences 8HJ Xx H Xx for J C H Applying 11 to the inclusion of j39Hspaces ltIgtXx C DX and composing with 8Ge IJltIgtX H X we obtain an Hmap Em Xx H X such that 5 is the composite of the homotopy equivalence 8HJ and the inclusion X x H X 8 JP MAY It remains to discuss the functoriality and naturality of this construction which is the crux of the matter We recall that the functor IJ jgW H Glt2 is given by a categorical twosided bar construction qu BTi39g339g Here 3b 0G H lt2 is the covariant functor that sends the object GH of 0G to the space GH The construction is suitably functorial in all three of its variables An alternative description of T may make the functoriality clearer De ne a small topological Gcategory EXT G as follows The object Gspace of EXT G is the disjoint union of the G spaces TGH gtlt GH where G acts on the orbit factors A morphism a 750 H if6 is a Gmap a GH H GH such that ac c and at t where the subscript and superscript s indicate the evaluation of covariant and contravariant functors There is an evident topology and Gaction on the set of morphisms such that the source7 target identity and composition functions are continuous Gmaps Up to canonical homeomorphism of Gspaces qu 13mm G For a homomorphism u G H G a Gfunctor EXT G H ltTCG induces a Gmap IJT H IJT where G acts on the targets by pullback along 4 similarly a Gnatural transformation induces a Ghomotopy Let aw x H y be a morphism in 71X Let a GH H GK be given by aeH 9K and let Cg H H K be the conjugacy injection that sends h to g lhg By Lemma 11 if we change a in its homotopy class then we replace 9 by cg for some 0 in the identity component of GgH Therefore7 although Cg depends on the choice ofg in its coset it does not depend on the choice of a in its homotopy class The homomorphism Cg determines a functor quotH H quotK that sends HJ to Kg IJg and we also have the Hmap HJ H Kg IJg that sends hJ to g lhg g ng Using the functoriality of the twosided bar construction there results an H map Xla wl791Xgt Xy The properties speci ed in iii and iv of Lemma 25 are sati ed Here iii is not obvious since a homotopy is required7 but it is easy to check that the two maps speci ed there are obtained by passage to classifying spaces of categories from naturally equivalent functors and are therefore homotopic Intuitively7 this transport along paths ensures that our Hconnected covers are related by the evident commutative diagrams to inclusions of components of xed point spaces The naturality of the construction with respect to Gmaps X H Y is checked similarly 4 ORIENTABILITY OF SPHERICAL GFIBRATIONS Nonequivariantly there is only one sensible de nition of an orientation of a vector bundle7 but this is a calculational fact that does not extend to the equivariant setting The point is that 22 2 mom 2 woltPLltngtgt 2 0ltTopltngtgt 2 own for all n 2 1 including 71 00 Nothing like this holds equivariantly There are at least eight different reasonable orientation theories for Gvector bundles cor responding to the linear piecewise linear topological and homotopical categories EQUIVARIANT ORIENTATIONS AND THOM ISOMORPHISMS 9 and their stable variants Similarly7 there are six orientation theories for PL G bundles7 four for topological Gbundles7 and two for spherical G brations We shall focus on the stable spherical G bration case7 but the modi cations for the other cases are easily imagined A general framework is given in We begin in this section with the simpler notion of o entability Even this depends on the type of Gbundle or G bration we consider By a G bration7 we understand a map that satis es the Gcovering homotopy property GCHP De nitions 41 Let j39gGlt2quotG be the category of Gspaces 5 X H GH with sections 039 GH H X and sectionpreserving maps of Gspaces over orbits For an Hrepresentation V let SV be the onepoint compacti cation of V We have a G bration G X SV H GH with section given by the points at in nity De ne the category 9 of n sphere G brations to be the full subcategory of j39gGlt2quotG whose objects are the G brations with section that are ber G homotopy equivalent to some G X H V ii A homotopy between maps in 9 is a sectionpreserving homotopy compare Remark 22 The homotopy category h is a groupoid over the category h 39g iii De ne the stable homotopy category sh of n sphere G brations over orbits to have the same objects as h and stable homotopy classes of maps Then sh is also a groupoid over h 39g and we have a canonical map i h H sh of groupoids over h 39g To control the colimits implicit in iii7 let U be the direct sum of countably many copies of each irreducible orthogonal representation of G since any representation of H C G extends to a representation of G on a possibly larger vector space U is also the sum of countably many copies of each irreducible representation of H For a G representation W G X SW GH gtlt SW over GH7 and we have the berwise smash products X A W of spherical G brations of dimension n and such trivial G brations The set of stable maps X H Y is the colimit over W C U of the set of maps of spherical G brations X A W H Y A W Restricting objects and morphisms appropriately7 we obtain analogous de ni tions for vector bundles or better7 their berwise onepoint compacti cations7 piecewise linear bundles7 and topological bundles We need a lemma before we can explain what it means for a spherical G bration p E H B to be orientable Lemma 42 An nsphere G bmtion p E H B determines a functm pquot 773 H h over h 39g A map D E qL LP ATE f H of nsphe7 e G bmtions determines a natuml isomomhism j 1 H pquot o f of functm s 71A H h over h 39g If DthHgtE AXITgtB 10 JP MAY is a homotopy between maps of spherical G brations f0d0 and fhdl then ff NHL 0 f3 I Proof This is an exercise in pulling back spherical G brations along Gmaps x G H H B and using the Gcovering homotopy property D De nition 43 A spherical G bration p E H B is orientable the stable sense if the composite of pquot 77B H h and i h H sh has the property that ipao1 ipoo for every pair of morphisms aw and oz40 such that a 0 Intuitively over a given a the stable homotopy class of the map of bers over orbits induced by a path between orbits is independent of the choice of path In the language of Construction 18 orientability requires ip to factor through the universal faithful groupoid associated to 77B We must distinguish between orientability in the stable sense and orientability in the unstable sense since it is possible to have ipao1 ipoo but paw pa o The following lemma is easily veri ed no matter how we de ne orientability Lemma 44 Each G X SV is an orientable spherical G bration Nonequivariantly when de ning orientations of bundles we implicitly compare bers to R with its standard two orientations This amounts to choosing a skeleton of the category of n dimensional vector spaces Equivariantly we must orient the G X SV and use their orientations as references and we must start by xing a skeleton of sh We have already discussed how to do this in Construction 18 De nition 45 Let Sph n denote the discrete groupoid over h 39g associated to sh and let p sh H Sph n be the canonical equivalence of categories Explicitly for each H C G choose one homomorphism f H H 0n in each conjugacy class and let Vf R with H acting through Choose one SV in each stable homotopy class of such Hlinear nspheres The objects of Sph n are the resulting nsphere G brations G X SV For each nsphere G bration X over GH we have an isomorphism A X H G X SV in Sh n and these chosen isomorphisms determine p De nition 46 Let p E H B be an nsphere G bration De ne pit to be the composite of 19 71B H sh and p sh H Sph n We continue to write pit for its restriction to a skeleton sk7139B of 77B Now recall Notations 110 The following immediate observation gives a con ceptual characterization of orientability in terms of the relationship between the fundamental groupoid and the component groupoid of B Lemma 47 The nsphere G bration p E H B is orientable and only pit shrrB H Spit factors through the associated discrete groupoid 770B 5 COHERENT FAMILIES OF THOM CLASSES AND THOM ISOMORPHISMS Let E be a commutative ring Gspectrum in the classical homotopical sense we have a unit map S H E and a product E A E H E satisfying the usual unity associativity and commutativity diagrams in the stable homotopy category of Gspectra of 15 see also 16 X111 5 We are interested in the ROGgraded cohomology theory E G represented by E Evaluated on Gspaces we understand EQUIVARIANT ORIENTATIONS AND THOM ISOMORPHISMS 11 the unreduced theory7 writing for the reduced theory on based Gspaces We shall make use of the precise treatment of ROGgrading given in 167 XIII 12 We may regard E as an Hring spectrum for any H C and we write E for the theory on H spaces represented by E For an H space Y and Grepresentation p EpcG X Y E ElfJAY We begin with a generalization of the notion of a cohomology class of a Gspace De nition 51 Let be a groupoid over h 39g let q H Sph n be a map of groupoids over h 39g and let Y H Glt2hjlg be a ltiffspace For an object as E KgH write qx G gtltH SW and describe Y as in Lemma 25 in terms of a system of Hspaces An E cohomology class 1 indexed on q of the ltzquot smzce Y consists of an element 1x E EEEZx for each object as E KgH The 1x are required to be compatible under Testn39ction in the sense that Zgvy We where y x Hgt y is a morphism of b and g is an element of G such that a yeH 9K compare Lemma 25ii Here mgr E wzw H E ltzgtltzltxgtgt is the composite of restriction Ef www H EEltygtltZltygtgt along Cg H H K and the map E2ltygtltzltygtgt H EEltzgtltZltxgtgt induced by theNHmap Z yg Zx H Zy and the inverse of the stable Hequivalence f SW H SW9 such that g determines the stable G equivalence 17 G X SW H G gtltK SW9 compare Remark 22 The simultaneous functoriality in the grading and the space that we have used is explained in 167 XIII 12 Essentially this is just an exercise in the use of the suspension isomorphism in ROGgraded cohomology We shall apply this de nition with 773 taking q to be the functor p 773 H Sph n associated to an nsphere G bration p The relevant WBspace is the Thom WBspace Tp given by the following de nition De nition 52 Let p E H B be an nsphere G bration De ne the based Thom Gspace Tp to be the quotient space EO39B For example7 if we start with a Gvector bundle 5 then its Thom space is obtained from the berwise onepoint compacti cation of 5 by identifying all of the points at in nity We have the map of WBspaces 5 B H B of Theorem 32 De ne the Thom WBspace Tp by letting Tpx Tf9ac7 x 6 BH be the Thom Hspace of the pullback 1330 of p along 5 Bx H B The point of the de nition is that the Hspace 330 is Hconnected and7 as we now recall orientation theory for n sphere G brations over Gconnected base spaces is well understood We rst de ne orientations of spherical G brations over orbits then de ne orientations of spherical G brations over Gconnected base spaces7 and nally give our new de nition of orientations of general spherical G brations 12 JP MAY De nition 53 The Thom Gspace of G X SV H GH is G AH SV and E G AH 8V 2 EEW 2 EM An E orientation or Thom class 4 of 5 is an element 4 E EgG AH 8V that maps under this isomorphism to a unit of the ring Eg pt De nition 54 Let p E H B be an nsphere G bration over a Gconnected base space B For any x 6 BG p 1x is a based Gspace of the homotopy type of SV for some n dimensional representation V of G and V is independent of the choice of 30 Moreover for all x GH H B the pullback ofp along x is ber Ghomotopy equivalent to G X SV An E on39entation or Thom class 4 ofp is an element 4 E Tp that pulls back to an orientation along each orbit inclusion ac De nition 55 Let p E H B be an n sphere G bration An EEorientation or Thom class ofp is an E cohomology class 4 indexed on p 713 H Sph n of the Thom WBspace Tp such that7 for each x 6 BH ux E EgzTx is an orientation of the pullback 1330 ofp along a Bx H B We say that p is E orientable if it has an EEorientation Here for x 6 BH Vx is the ber Hrepresentation at at so that SW is stably Ghomotopy equivalent to p 1x the equivalence being xed by the speci cation of p Observe that the equivalence xes a stable H map 56 SV 2 197130 H The following observation should help clarify the force of the compatibility condition required of our orientations on H connected covers Lemma 57 Let p be an nsphe7 e G bmtion The following diagmm commutes for a momhism aw x H y in 773 with aeH 9K 7 WV EXltygtltTltpgtltygtgtHEXltygtltSVltygtgt e EEHM TltPgtltlagtwlgtg pagtwgtg 09 EElt gtltT ltpgtltxgtgtHW EEltmgtltSVltzgtgt e E529 Proof The map pa 01 g is de ned exactly as was Tpa 01 g in De nition 51 and the left square is a naturality diagram The right square commutes by a direct unravelling of de nitions D Remark 58 If the horizontal arrows are isomophisms7 then the left vertical arrow is determined by the right vertical arrow and the compatibility reduces to a question of compatible units in the rings comprising the Mackey functor E0 with E0 Eg pt As we shall see in the next section this is exactly what happens when p is orientable and we specialize to ordinary cohomology with Burnside ring coef cients Compatible Thom isomorphisms follow immediately from 15 X 5 where a generalization of the following theorem is proven EQUIVARIANT ORIENTATIONS AND THOM ISOMORPHISMS 13 Theorem 59 Let p ENH B be an nsphere G bration over a Gconnected base space B and let a E EETp be a Thom class Then cupping with a de nes a Thom isomorphism e 91 E B a EgVTp for all p E ROG Again we refer to 167 XIII 12 for precision about the grading Theorem 510 Let p E H B be an nsphere G bration and let be a Thom class ofp Then the Mac as 6 BH give rise to Thom isomorphisms mm 2 Emma H ELVlt gtltT ltpgtltxgtgt where the H space SW is stably equivalent to p 1x Moreover the following diagrams are commutative for aw x H y where aeH 9K andp E ROK Era ltygtgtME2Vltygtlt ltpgtltygtgt B 3agtwlgtgl LTWMQMM EzltBltxgtgt rm Tom Here the vertical arrows are as speci ed in De nition 51 6 ORIENTATIONS IN ORDINARY EQUIVARIANT COHOMOLOGY We have formalized the intuitive geometrical notion of orientability in De nition 43 and have expressed this notion categorically in Lemma 4 It is natural to hope that this notion coincides with the notion of orientability with respect to a suitable cohomology theory Nonequivariantly7 the relevant theory is integral cohomology The real reason this works is that orientability is a stable notion and Z coincides with the zeroth stable homotopy group of spheres Equivariantly the analogue of Z is the Burnside ring 14G7 which is the zeroth equivariant stable homotopy group of spheres As was rst explained by Bredon 3 ordinary equivariant cohomology theories are indexed on coef cient systems7 namely contravariant functors M h 39g H Ab where Ab denotes the category of abelian groups We have the Burnside ring coef cient system A such that AGH As was proven in 14 the ordinary cohomology theory indexed on M extends to an ROGgraded theory if and only if the coef cient system M extends to a Mackey functor See 157 V 9 or 167 IX 4 for a discussion of Mackey functors in the context of compact Lie groups The Burnside ring coef cient system does so extend7 hence we have the ordinary ROGgraded cohomology theory H HA It is represented by an EilenbergMac Lane Gspectrum HA 167 XIII 4 and HA is a commutative ring Gspectrum We abbreviate HAEorientability to Aorientability and HEXA to We proceed to relate orientability to Aorientability7 beginning with the case of G brations over Gconnected base spaces Theorem 61 Letp E H B be an nsphere G bration where B is Gconnected Let x 6 BG let V be the ber Grepresentation at x and consider the map i SV 2 p 1x C Tp The following statements are equivalent 14 JP MAY i p is orientable ii p is Aorientable iii iquot H gw v E AG is an isomorphism Proof By GCW approximation we may assume without loss of generality that B is a G CW complex with a single G xed base vertex 30 Let 3 1 be the q skeleton of B let Eq p 1Bq and let pq be the restriction of p to Eq Let Cg TpqTpq 1 Observe that SV 2 Tp0 If c GHC gtlt Dq H 3 1 is the characteristic map of a qcell of B then the pullback of p along 0 is trivial and is thus equivalent to G HC gtlt Dq gtlt SV Moreover the equivalence is determined by a choice of path connecting x to Ce 0 These equivalences determine an equivalence between the wedge over all q cells c of the G spaces GHCLr A Sq A SV and the quotient Gspace Cg Consider cohomology in degrees V i7 where i is an integer We have HEWHm A Sq A 8V 2 him This is zero unless i 2 q and it is AH when i q by the dimension axiom We conclude by long exact sequences and lim1 exact sequences that HV iTPq EV iTP 0 for i 2 1 and there is an exact sequence 0 a HgTpigtHgSVHg101 Here 5 may be viewed as a map AG H HAHC of 14Gmodules7 where the product runs over the lcells c A lcell c is speci ed by a loop at x in BHC The component of 5 in AHC can be interpreted geometrically as the difference between the identity map of SV and the stable Hequivalence of SV obtained by the action of this loop on SV The three statements of the theorem are each equivalent to the assertion that 5 0 D Observe the relevance of our de nition of orientability in the stable sense The conclusion would fail if we de ned orientability in the unstable sense Before generalizing this result we recall a standard fact about conjugation homomorphisms between Burnside rings Let a GH H GK be given by aeH 9K and consider Cg H H K Since clc AK H AK is the identity for k E K by inspection of the the standard inclusion of AK into a product of copies of Z eg 15 V 2 we see that cg AK H AH is independent of the choice of g in its coset 9K It is also independent of the choice of a in its homotopy class by Lemma 11 We write cg ca Theorem 62 Let p be an nsphere G bration The following statements are equivalent i p is orientable ii Each Mac is orientable iii Each Mac is Aorientable iv p is Aorientable Moreover an HAorientation a ofp is speci ed by a collection ofunits 1x E AH for points x 6 BH of the discrete groupoid 7103 that satisfy the compatibility condition cauy 1x for a map 7 x H y of 7103 with 87 a Equivalently a is speci ed by an automorphism of the fanctor p shrrB H Sph n over h 39g EQUIVARIANT ORIENTATIONS AND THOM ISOMORPHISMS 15 Proof Since the notion of orientability of p depends only on the behavior of the pullbacks of p along paths and paths lie in connected components the equivalence of i and ii is immediate from the properties of H connected covers given in Theorem 32 The equivalence of ii and iii is part of the previous theorem and it is trivial that iv implies iii by consideration of pullbacks Thus assume iii and consider the diagram of Lemma 57 with E HA Its horizontal arrows are isomorphisms by the previous theorem and Remark 58 applies to give the speci ed description of an Aorientation in terms of units of Burnside rings In particular we may take 1x to be the identity element for all x and this shows that p is Aorientable Finally the group of automorphisms of an object G X SV of Sph n is canonically isomorphic to the group of stable Hequivalences of SV and thus to a copy of the group of units of the Burnside ring For our functor pif sk7139B H Sph n the compatibility condition on units required of an Aorientation can be interpreted as the naturality condition required of an automorphism of functors 1 Let AOrp denote the set of Aorientations of an orientable nsphere G bration p Corollary 63 By multiplication of units or equivalently by composition of au tomorphisms of the functor pif skrrB H 53910an over h 39g AOrp acquires a stiuctui e of commutative group Nonequivariantly there are both topological and cohomological notions of an orientation and t ese notions coincide Equivariantly we have explained a coho mological notion of an orientation There is also a topological notion de ned in However these two notions do not coincide To explain this we sketch the de nition given in Working in the category of groupoids over h 39g consider maps into Sph n In 5 we construct and characterize a particular map p it H Sph n such that it is faithful over h 39g and any map from a faithful groupoid over h 39g into Sph n factors up to isomorphism through at least one map into 6 this is a weak universal property of p which intuitively is a kind of universal orientation Fix an orientable n sphere G brationp E H B for the rest of the section The functor p skvrB H Sph n factors through the discrete groupoid 7703 and we now agree to write pif for the resulting functor de ned on 710 The topological notion of an orientation is a pair C 7 consisting of a functor C 770B H it over h 39g together with a natural isomorphism n p H p o Let 07 denote the set of such orientations of p Precomposing with automorphisms of pif for xed C we obtain a free right action of AOrp on Orp Call the orbit set OrpA it can be identi ed with the set of those functors C 7103 H it that can be part of an orientation Cm Let Fp be the set of all functors C 710 B H it over h 39g such that p and p oC agree on objects In general not all such functors are components of orientations and we have an inclusion a OrpA H Let AX h 39g H Ab be the contravariant functor that sends GH to the group of units of AH and continue to write AX for its composite with a H h 39g for any groupoid over h 39g By analyzing the obstruction to the construction of n such that C n is an orientation one arrives at the following proposition We omit the proof as it is not very illuminating The essential ingredients are the cited weak universal property of p and the fact that Sph n is a uniquely divisible groupoid over h 39g 16 JP MAY Proposition 64 There is an exact sequence of pointed sets gt OrpAigtFpigtH17mBAX gt Thus H1770B AX measures the difference between topological and cohomo logical orientations if B is a bijection7 the notions are equivalent 7 CONCLUDING REMARKS Whenever one has cohomological orientations of a class of Gvector bundles that are suf ciently natural in G one will have cohomological orientations in the sense that we have de ned Since this paper was written around the deadline for submissions to this volume7 I have not had time to check details of the following two examples7 but they are surely correct Here it makes sense to use the variant of the theory appropriate to unstable Gvector bundles rather than to stable G brations Clearly orientations in the former sense give rise to orientations in the latter sense The methods of 2 should give the following result compare Example 7 1 Complex G vector bundles admit canonical KUgorientations Real Gvector bundles with Spin structures and dimension divisible by eight admit canonical KO Gorientations Tautological orientations should give the following result Example 72 Complex G vector bundles admit canonical M U orientations Real Gvector bundles admit canonical M OEorientations At the most structured extreme7 as in the nonequivariant case7 we have the following observation Example 73 A spherical G bration is Eorientable if and only if its pullbacks to Hconnected covers are stably ber homotopy trivial with suitably compatible trivializations To obtain a Poincare duality theorem along the present lines one would have to prove an Atiyah duality theorem for the H connected covers of smooth compact Gmanifolds M That is if M embeds in V with normal bundle 1 one might hope that the Hspaces TV and are Vdual for x 6 MH Although is in nite dimensional one has complete homotopical control on its xed point spaces7 which are homotopy equivalent to smooth manifolds I have not explored this question REFERENCES 1 MF Atiyah Bott periodicity and the index of elliptic operators Quart J Math 191968 113140 2 MF Atiah R Bott and A Shapiro Clifford modules Topology 31964 Supplement 1 338 3 G Bredon Equivariant cohomology theories Lecture Notes in Mathematics Vol 34 Springer Verlag 1967 4 RE Conner and EE Floyd Differentiable periodic maps Ergebnisse der Math N S 33 Academic Press 1 6 5 SR Costenoble JP May and S Waner Equivariant orientation theory Undistributed preprint 1989 Costenoble and S Waner Equivariant orientations and Gbordism theory Paci c J Math 1401989 6384 EQUIVARIANT ORIENTATIONS AND THOM ISOMORPHISMS 17 7 SR Costenoble and S Waner The equivariant Thom isomorphism Paci c J Math 1521992 21 39 S Costenoble and S Waner Equivariant Poincare duality Michigan Math J 391992 325351 9 SR Costenoble and S Waner The equivariant Spivallt normal bundle and equivariant surgery Michigan Math J 391992 415424 SR Costenoble and S Waner The equivariant Spivallt normal bundle and equivariant surgery for compact Lie groups Preprint 1996 11 T tom Dieck Transformation groups Studies in Mathematics Vol 8 Walter de Gruyter 10 1987 12 A Elmendorf Systems of xed point sets Trans Amer Math Soc 2771983 275284 13 A Grothendieck Revetements etale et groupe fondemental Lecture Notes in Mathematics Vol 224 SpringerVerlag 1971 14 LG Lewis JP May and JE McClure Ordinary ROGgraded cohomology Bull Amer Math Soc 4 1981 1287130 15 LG Lewis JP May and M Steinberger with contributions by J E McClure Equivariant stable homotopy theory Lecture Notes in Mathematics Vol 1213 SpringerVerlag 1986 16 J P May et al Equivariant homotopy and cohomology theory CBMS Regional Conference Series in Mathematics Number 91 American Mathematical Society 1996 17 JP May G spaces and fundamental groupoids Appendix to An equivariant Novikov con jecture by J Rosenberg and S Weinberger Journal of Ktheory 4 1990 503 18 M Rothenberg and J Sondow Nonlinear smooth representations of compact Lie groups Paci c J Math 841979 427 444 MA544 LECTURE NOTES7 FALL 2009 ANTONIO SA BARRETO 1 INTRODUCTION These are my lecture notes for MA 544 fall 2009 They were actually typed in the fall 2005 and have not been changed even though it obviously contains several typos But I guess they are helpful Here is hw 1 for the course Give me a list of typos There are no required text books 1 will use a bit of each of the following references which have been placed on reserve in the library I o A Friedman Foundations of Modern Analysis Dover Edition 1982 o W Rudin Principles of Mathematical Analysis 3rd Ed McGraw Hill 1986 o W Rudin Real and Complex Analysis 3rd Ed McGraw Hill 1986 o HIL Royden Real Analysis PrenticeHall 1988 o A Torchinsky Real Variables AddisonWesley 1988 2 PRELIMINARIES PROPERTIES OF R COMPACT SETS CONTINUOUS FUNCTIONS ETC We review some properties of the n dimensional Euclidean space R and its topology De nition 21 We denote R the set of real numbers The Euclidean space R is de ned as R z 11 zj E R As we all know R is a normed vector space One norm of an element 1 11 I In E R is M z 13 The distance between two points zy E R is 4179 HI yll We then de ne open and closed subsets De nition 22 A subset X C R is open iffor every I E X there exists an e gt 0 such that Baa y e R HyezH lt e c X A subset F C R is closed 1R F is open Exercise 21 If Um a E A is a family of open subsets of R show that U0 Ua is open Similarly show that if 1f Fm a E A is a family of closed subsets of R show that at Fa is closed The family of all open sets of R is described as a topology on R and R equipped with this family of open sets is called a topological space A subset U C R can be equipped with what is called the relative topology We say that a subset O C U is open with respect to the relative topology if 0 X N U where X C R is open 2 SA BARRETO De nition 23 A family Um a E A of open subsets of R is said to be an open cover of a subset K C R if K c U Ua aeA An open cover Um a E A of K is said to have a nite subcover if there exists Ua1Ua2 UaN with a1aN E A such that n K c U Ha j1 Theorem 21 Let K C R The following properties ofK are equivalent 1 K is bounded and closed 2 Every open cover of K has a nite subcover 5 Every sequence of points ofK has a subsequence that converges to a point of K Proof 1 gt 2 Since K is closed R K is open Since K is bounded then there exist ab E R a lt b lt 00 such that K C Q with Q ab gtlt ab gtlt gtlt ab If Um a E A is an open cover of K then Rn K U Um a E A is an open cover of Thus we need to prove that an arbitrary open cover of Q has a nite subcover Suppose by contradiction that this is not true ie there exists an open cover U04 of Q which does not have a nite subcover that covers Let us introduce some important notation Let U C R we de ne the diameter of U as diamU supHI Zxy E U Then d diamQ b 7 an Let c b 7 a2 The intervals ac and bc determine 2 cubes Qj with diamQj d2 At least one of these sets let s say Q1 cannot be covered by a nite subfamily of U0 Then we divide Q1 in 2 cubes of diameter d4 and repeat the process We obtain a family of cubes B1 j E N such that d d1amBj a B1 3 B2 3 B3 Bj cannot be covered by any nite subfamily of U0 j 12 We then need the following Lemma 21 Let Bj a1jb1jgtlt a2jb2j gtlt anjbnj B1 3 B2 3 B3 then B y 0 j1 Proof Let us suppose rst that n 1 Denote al v aj bl v bj and E ah j E N be the set of lower endpoints of the intervals Let x sup E If mn E N then an S amn S bmn S bm Therefore x S bm for every m Since it is obvious that x 2 am we have x E Ij for all j E N In the general case for each h E 1n there exists xk E 1akjbkj Then x1xn E 00 j1 Bj B Let xquot E Bj Since U04 covers Bj j E N there exists 10 such that x E U040 Since UDO is open there exists 6 gt 0 such that x E Bx6 y E R Z i lt 6 C U040 On the other hand since MA544 LECTURE NOTES FALL 2009 3 xquot E Bj j 12 and diamBj d2j we can pickj large enough such that Bj C Uaol Therefore Bj is covered by the set UDO and this is a contradiction This ends the proof of the rst implication 2 gt 1 Suppose every open cover of K has a nite subcover First we show that K is bounded Let 6 gt 0 andfor each x E K let Bx6 y E R 1 lt 6 The family Bx6 x E K is an open cover ofK and therefore has a nite subcover Bxk6 1 S h S N Hence K C Ug1Bxk6 It is then easy to see that K is bounded Now we want to show that K is closed Let p E R K For each 4 E K let Bq rq be the ball centered in q with radius rq lt Hp7qH4l This gives an open cover ofK Since K is compact there exists on M qN such that K C U lBqjrqj W Now let Bprqj be the ball centered at p with radius rqj It follows from the de nition of rq that Bprq Bqrq 0 Therefore if Bprqj V then V is open p E V and V N W 0 So in particular V N K 0 This proves the implication 2 gt 3 Let yj j E N be sequence contained in Kl For 6 gt 0 the family Bx6 x E K is an open cover of Kl Hence there exists xk 1 S h S N and a subsequence yjm C such that Hij 7ka lt 6 Repeating this process we deduce that has a Cauchy subsequence yjm Therefore it converges Since K is closed its limit belongs to Kl This proves the implication 3 gt 1 IfK is not bounded one can construct a sequence of elements ofK that does not converge 5 also clearly implies that K is closed This ends the proof of the Theorem There is one result that we will need later and it is important to review it Theorem 22 Let Km 1 E A be a family of compacts subset of R such that for any nite subcol lection K0 1 g j g N m K0 y 0 Then may K0 0 Proof Suppose by contradiction that aeA K0 0 Fix an element K01 of the family By assumption K041 QweA D ad K0 0 This is equivalent to saying that if GD R Kw Then K01 C UWEAX aa1Ga Since G0 is open and K01 is compact there exist 12 M aN such that K01 C G0 But this implies that K0 ll which is a contradiction D Now we discuss continuity De nition 24 Let X C R We say that a function f z X 7 R is continuous at a point x0 6 X if for every 6 gt 0 there exists a 6 gt 0 6 depending on e and x0 such that 21 for all x E X with 7 xOH lt 6 gt 7 fx0l lt e We say that f is continuous at a subest X C R f is continuous at every point of Xl Exercise 22 Show that a function f z X 7 R is continuous everywhere if and only if for every open subset V C R f 1V is an relatively open subset of Xl Notice that 21 in the de nition of continuity involves two inequalities that is f is continuous at x0 if and only if Hx 7 xOH lt 6 gt fx0 7 e lt lt fx0 6 This motivates the following de nition De nition 25 Let X C R A function f z X 7 R is upper semicontinuous at x0 6 X for any 6 gt 0 there exists 6 gt 0 such that 22 x E X 7x011 lt 6 implies lt fx0 5 4 SA BARRETO Similarly a function f z X 7 R is lower semicontinuous at 10 E X for any 6 gt 0 there exists 6 gt 0 such that 23 16X HI7IOH lt6 implies fzo7eltfzl A function is upper or lower semicontinous in X if it is upper or lower semicontinous at every 10 E X Theorem 23 Let X C R A function f z X 7 R is lower semicontinuous on X if and only if f 1hoo is a relative open subset of X for all h E R Similarly f is upper semicontinuous if and only if f 17oo k is a relative open subset ofX for all h E R Proof Suppose f is lower semicontinuous in X Let h E R and let 10 E f 1hoo Then fzo gt kl Let 6 fzo 7 h and let 6 gt 0 be such that 22 holds Then for Ioll lt 5 gt fzo 7 e kl So I z 7 roll lt 6 C f 1hool Conversely suppose that for every h f 1hoo is open for every h E R In particular for e gt 0 and 10 E X the set f 1fzo 7 600 9 10 is open Hence there exists 6 gt 0 such that for roll lt 6 z E f 1fzo 7 600 Therefore gt fzo 7 e D The following is a very important result which is a direct consequence of Theorem 23 Theorem 24 Let fun a E A be a collection of lower semicontinuous functions Then sup is lower semicontinuous aeA Similarly if fun a E A be a collection of upper semicontinuous functions Then inf is upper semicontinuous aeA Proof One just needs to show that if 161 SUI 16041 then f 1k700 Uf1k7007 as a if fltzgt7gggfaltzgt then f1ltlt7ookgtgt7Ufglltlt7oo7kgtl Let us consider the rst statement If f supaEA fa and z E f 1hoo then and gt kl This implies that gt h for some 1 Otherwise S kl Therefore I E Ua f1h Reciprocally if z E Ua f1hoo then gt h for some 1 In this case gt h and hence I E f 1hool D Next we study the notion of oscillation of a function De nition 26 Let X C R and let f z X 7 R be a bounded function The oscillation of f over X is de ned to be 010 X suplf1 7 fyl 17y 6 X Exercise 23 Let X C R and let f z X 7 R be a bounded function on X C R show that wf7X 13 fI gggf l Next we localize to a point the notion of oscillation of a function De nition 27 Let X C R be an open subset let f z X 7 R be a bounded function and let 10 E a by Let Bzo6 y E R 7 roll lt 6 We de ne the oscillation of f at 10 as were 7 gmmm MA544 LECTURE NOTES FALL 2009 5 Notice that if 61 lt 62 then wfBzo61 S wfBzo62 so w6 wfBzo6 is a bounded nondecreasing function Hence the limit exists Theorem 25 Let X C R be open and let f z X 7 R be a bounded function and let 10 E X Then f is continuous at 10 and only if wfzo 0 Proof If f is continuous at 10 then for every 6 gt 0 there exists 6 gt 0 such that 7 roll lt 6 implies fzo 7 e lt lt fzo 6 This says that supfzz E Bzo6 lt fzo e and inffzz E Bzo6 gt fzo 7 6 Thus wfBzo6 lt 26 Hence wf Io 0 Conversely suppose that wfzo 0 Then ggwwm 6 7 0 That is for e gt 0 there exists 60 gt 0 such that for 6 lt 60 wfBzo6 lt 6 But wfBzo6 suplfz 7 fylzy E Bzo6 In particular for 7 roll lt 5 We have 7 fzol lt e D Theorem 26 Let X C R be open and let f z X 7 R be bounded and let 10 E X The function wfz is upper semicontinuous on X That is xed 10 E X then for all e gt 0 there exists 6 gt 0 such that z E X and lz 7 10 lt 6 implies that wfz lt wfzo 6 Proof Since wfzo lim5n0wf Bzo6 it follows that for any 6 gt 0 there exists 6 gt 0 such that wf7B1075 lt 010510 6 On the other hand recalling that if61 lt 62 then wf Bzo 61 S wf Bzo 62 wf z lt wf Bzo 6 z E Bzo6 This proves the theorem D We deduce from Theorem 26 and Theorem 23 that Corollary 21 Let X C R be open and let f z X 7 R be bounded Then for every 1 gt 0 the set ED z 6 ab wfz 2 a is closed Theorem 27 Let X C R be open and let f z X 7 R be bounded Let D be the set of points of ab where f is not continuous For 6 gt 0 let E5 z 6 ab wfz 2 6 Then D U135 UEln 6gt0 nEN Proof We know that f is not continuous at 10 if and only ifwf 10 gt 0 But it is easy to see that mm gt0 7 Unzww 2a Dtgt0 Since E5 C E7 for 6 gt 7 D UneN Eln D The structure of the set D is important and deserves a name De nition 28 A set F C R is an E if 00 F U Fn F7 closed 711 5 SA BARRETO AsetGClR isaG5 if G a G7 Gn open 711 So theorem 2 implies that D is an F01 3 SETS OF MEASURE ZERO AND CANTOR SETS We want to make sense of a way to characterize when a subset X C R is small For now we will restrict ourselves to one dimension De nition 31 A subset X C R has content zero and write cX 0 if for an arbitrary e gt 0 there exist a nite collection of open intervals 1112111IN such that N XC U 1 and 11 12M 1m lt e j1 Here Ij denotes the length of the interval Iji We say that X has Lebesgue measure equal to zero and denote mX 0 if for every 6 gt 0 there is a countable family of open intervals 1 j E N such that 00 oo XC U1 and 2m j1 j1 In particular cX 0 then mX 0 Exercise 31 Show that one can take the intervals to be closed in the de nition 31 Theorem 31 Sets of Lebesgue measure zero have the following properties 1 IfmX 0 andY C X then mY 0 2 IfX is compact and mX 0 then cX 0 5 fY Xj and mXj 0 for all j E N then mY 0 We leave the proof as an exercise Next we give examples of some 77weird77 sets which will be a rich source of examples and counter examples during the course and therefore it is useful to review themi Let C be the subset of 01 de ned in the following way Let r 6 01 Let E0 01i Let E1 be the set obtained by removing the open interval of length r centered at That is 1 1 E10 lt17 rgtiuiilt1rgt71i So E1 consists of two intervals of length 1 7 r Therefore the sum of the lengths of the intervals that makeup E1 is 1 7 r These can be thought of two copies of 01scaled by the factor 1 7 r Then repeat the process taking into account the scaling factor So let E2 be the closed set obtained from E1 by removing the open interval centered at the middle of each interval of E1 of length 1 7 rr and so on So E2 will consist of four intervals of length i1 7 r In the general case if n gt 1 the set En is obtained from En1 by removing 2 1 open intervals of length 7 rn71 ri So En consists of 2 intervals of length 7 Notice that the sum of the lengths of the intervals that make up the set En is 1 7 r MA544 LECTURE NOTES FALL 2009 7 So we have E03E13E23M3En The Cantor set is de ned by Theorem 32 C is compact and its content is equal to zero Proof C is the intersection of closed sets so it is closed Since it is contained in 01 it must be compacti Given 6 gt 0 pick N such that 1 7 7 N lt 62 Since EN 3 E one can cover E by nitely many intervals with sum of their lengths less than 6 D Now we will construct a similar set but with measure not equal to zero As before let F0 01i Let F1 be the set obtained by removing the middle open interval of length Ti That is 140507 TU1T711 So the sum of the lengths of the intervals that makeup E1 is 1 7 7 Now instead of removing the interval of length 1 7 7 7 from each interval we remove lessi Lets say we remove 1 7 7 7 2 from Fli So let F2 be the closed set obtained from F1 by removing the middle open interval of length 1 7 7 7 2 of each interval of Fli So F2 consists of four intervals of length l2 a1 7 7 1 7 7 2 eachi Let F3 be the set obtained from F2 by removing the middle open interval of length l2quot3 from each interval of F2 So F3 consists of 8 intervals of length 2 31 7 7 1 7 7 21 7 T3 In general if we let ln denote the length of each interval of F then Fn1 is the set obtained by removing the middle open interval of length lnrn1 from Fni Lemma 31 For n 2 1 Fn consists of 2 intervals of length n 31 Zn 2 H17r1i Proof By de nition F1 0 1 7 T U 7 1 So the claim holds for n 1i Supposes 31 holds for it holds and we want to show it holds for n 1 Since we have 2n1 intervals of the same size we work with the one which contains 0iTake the interval 0 ln ln01 and remove its middle open interval of length lnin1i So we get two intervals of the same length and the one which contains zero is ln 0 1 7 TWA 0 ln1739rn1i But ln1 l 1 7 TWA This ends the proof of the lemma D As above the modi ed Cantor set is de ned by K 61F Unlike the rst Cantor set C the sums of the lengths of intervals of Fn do not go to zero as n 7gt 00 In this case the sum of the intervals that makeup Fn is Mn 2nLn H 1 7T1 j1 3 SA BARRETO But we know that 14139 0 2 x H Theorem 33 The Cantor set K is compact has empty inten39m but it does not have measure zero Proof K is closed and contained in 01 Therefore it is compact Since Fn consists of 2 disjoint intervals of length In 2 H1l 7 Tj It is easy to see that K has empty interiori Otherwise K and hence Fn n 12M would contain an interval I of size 6 Since the intervals that makeup Fn are disjoint I would have to be contained entirely inside one of its intervalsi Just pick n so that 1 lt 6 If K had measure zero then for any 6 gt 0 there would be a family In n E N of open intervals such that K C Ur In and 221 lInl lt 6 Since K is compact this collection of intervals can be assumed to be nite so K would have content zeroi Then N KC 11 Mg um UIN Elm lt 5 j1 Pick 6 lt H1l 7 T1 Let U I1 U12 U UINi We claim that there exists m E N such that U D Fm and hence U D for allj 2 mi Suppose this is not true That is for all m E N Gm Fm U 3e 0 Since U is open Gm is closed and therefore compact and Gj D Gj1i By theorem 2 Gm gym U y 0A mEN That is absurd So U D Fm for some mi Therefore 6 would have to be bigger than or equal to the sum of the lengths of the intervals of Fmi That is e 2 1111 7 Tj gt 1111 7 T1 That is not possible in view of the choice of e D 4 THE RIEMANN INTEGRAL IN R In this section we will discuss the Riemann integrali For simplicity we restrict ourselves to R but all the results proved here can be easily extended to higher dimensions I believe most students in this class have studied this before Our goal is to do a brief review of the methods and to emphasize their limitations A partition 73 of an interval ab a b lt 00 is a nite collection of points 1011 C ab with a Io zj S 1 and b zni We say that a partition 731 is a re nement of if C 731 Let f ab 7 R be a bounded function and let 73 1112 be a partition of abi We de ne the upper and lower sums with respect to a xed partition 73 n Ult737fgt7ZMjltzj17zjgt7 M17 sup fltzgt j1 1gt 71 7 L73fa ij 1 7 zj mj inf j1 1106 1 Obviously M737 a S U037 a MA544 LECTURE NOTES FALL 2009 9 Proposition 41 Let 73 and 731 be partitions of a z Suppose 73 C 731 ie 731 is a re nement of 73 Let f a z 7 R be bounded and let M supab andm in ll Then mba S 14737133 14731716 S U7317f S U737f S M0271 Proof Since Mj S M and mj 2 m7 U73f S M02 7 a and 7211 7 a S L03 We use induction to prove the other inequalities Suppose that 731 contains one more point than 73 Let 73173 and suppose xquot E xhxjurlll Let M sup M sup 17W i wwil Tn inf mjg inf 17W lTgt171l Recall that Mj suphwidd x and mj inf 7 71fxl It is then clear that Mj1 S My M12 S My mil 2 my mm 2 mi Then U031 f U037 f Mj11 11 Mj2rj1 1 MjIj1 11 S 0 M7317 f M73716 mj11 11 mj2rj1 1 mjIj1 11 2 0 lf73173 consists of h points7 one just needs to repeat this argument h times D De nition 41 The upper and lower integrals of f in a7 12 are respectively 7b b f dx ingU If7 f dx supL7 77 a i P De nition 42 We say that f is integrable and denote f E R if Zfdxabfdxdefabfdx It is very important to characterize the class of functions Which are integrable in the sense of Riemannl The concept of oscillation of a function over a set is an important one7 not just for the theory of integration Theorem 41 Let f a7 12 7 R be a bounded function The following are equivalent 1 f e R 2 For every 6 gt 0 there exist partitions 73 and Q of a z such that Uf73 7 Lf7 Q lt 6 5 For every 6 gt 0 there exists a partition 73 of a z such that Uf73 7 L007 lt 6 4 For every 6 gt 0 there exists a partition 73 x07 M xn of a z such that ZWWH 11 lt 67 W wf7 lxjvxj1l 171 Proof The implications 1 ltgt 2 follow almost directly from the de ntion It is obvious that 3 gt 2 just take 73 Ql If 2 holds7 let 731 73 U Ql From Proposition 417 LO Q S 1105731 S Uf7731 S Uf773 Then Uf731 7 Lf731 lt e The implications 4 ltgt 3 follow from exercise 23 D 10 SA BARRETO Corollary 41 ff a7b 7 R is continuous then f E R Proof Since a7b is compact7 and f is continuous7 f is bounded The compactness of ab also implies that f is uniformly continuous So for any 6 gt 07 we can pick 6 gt 0 such that for any I y 6 a7 b with z 7m lt 6 was 7 fyl lt eltb7agt Let 73 10 be a partition of a7 b such that lzj 7 zj1l lt 6 Hence wjf wf7 zjzj1 lt eb 7 a Then 201195 11 lt 6b i a ZltIj1 11 6 j1 j1 In View of Theorem 417 f E R D The goal of this section is to prove Theorem 42 Let f ab 7 R be a bounded function and let D C a7b be the set of points where f is not continuous Then f E R if and only if mD 0 In View of Theorem 317 Corollary 21 and Theorem 27 Theorem 42 follows directly from Theorem 43 Let f a7 b 7 R be a bounded function Then f E R if and only iffor every 6 gt 0 the set E z 6 a7 b wf7 I 2 6 has content equal to zero lncleecl7 suppose Theorem 43 has been proved Suppose rst that mD 0 Since D UneN Eln7 we have mE1n 07 for all n E N Since E C E77 for 6 gt 7 it follows from Theorem 31 that mE5 0 for all 6 gt 0 Since E5 is compact7 by Corollary 217 Theorem 31 guarantees that cE5 0 Conversely7 if cE5 0 for all 6 gt 07 then in particular cE1n mE1n 0 for all n E N Then Theorem 31 implies that mD 0 Now we prove Theorem 43 Proof The implication f E R gt cE5 07 for all 6 gt 07 is easier and we prove it rst Suppose that f E R and x 6 gt 0 We want to show that cE5 0 Let 6 gt 0 Since f E R7 we know that there exists a partition 73 10771n of ab such that ijzj1 7 zj lt 66 j1 Let j zjzj1 and let A j l S j S n7 and 1 E5 Since for every I E E N 1k 6 S wfz S wfIk it follows that 5 Z lel 5 Emu 11 S 2011411 11 S ZWltI11 11 lt 65 jeA jeA jeA j1 Hence 2 HA lt e jEA The intervals j cover E5 it follows that cE5 0 Now we prove the converse We begin with the following Lemma 41 Suppose thatwf7 I lt 6 for all z E 174 Then there exists a partition 73 1011771N of p7 q such that 41 Mj 7mj lt 6 j 1727m7N7 Mj sup mj inf Tw a1 1 gt 7il MA544 LECTURE NOTES FALL 2009 11 Proof Extend for z lt p and for z gt L For each I E q there exists an open interval LE 9 I such that supxet 7 infxer lt 6 where E is the closure of LB This gives an open cover of 10 q and hence it has a nite subcover 1I N Now take a partition 73 of 10 4 such that an interval of 73 is contained in the closure of one of the intervals LE Then 41 holds Suppose cE5 0 for all 6 We want to show that f E R Let 6 gt 0 Since cE5 0 there exist open intervals 11IN such that E C I and that llll lINl lt 6 We begin with a partition 73 of a 12 consisting of intervals that fall in two categories 731 and 732 de ned as follows J E 731 if J C for some j and J E 732 if J N E 0 Let J be an interval in 732 Since wfz lt 6 for all z E J Lemma 41 shows that there exists a partition 73 of J such that sup 7 infgcgj lt 6 Doing this for every interval in 732 we get a re nement 73 of 73 such that each interval of 73 falls in two classes de ned as above which we denote by 731 and 735 Hence U05 110573 2W1 mJWl 2W1 mJWL M1 supfr7 mj supf Jep JePg J J Since f is bounded S M and hence M 7m S 2M for any J E 73 For J 673 M 7m lt 6 Hence U027 7Lf73 2 mm 2 am lt 2Mb7 a6 JePl JePz Since 6 gt 0 is arbitrary this shows that f E R D 41 Limitations of the Riemann Integral Theorem 42 shows that the class of functions integrable in the sense of Riemann is small The srt limitations is that one can have two functions which are equal outside a set of measure zero with one integrable and the other not integrable Consider the following function 1 if z E Q XQW 0 if r g Q Exercise 41 Show that wow I 1 Therefore XQ is discontinuous everywhere and hence not Riemann integrable On the other hand Fz 0 is Riemann integrable and Fz XQI I g So they are equal outside a set of measure zero one is integrable and the other one is not Proposition 42 Let X0 and XK be the characteristic functions of the Cantor sets de ned above Then the set of discontinuities of X0 and XK are C and K respectively Proof Let us take the case of X0 The other one is identical Since C is closed R C is open Since X0 0 if I g C X0 is continuous in R C Since C has empty interior for every point of C there is a sequence of points of R C converging to it Hence the set of discontinuities of X0 is exactly equal to Corollary 42 X0 6 R but XK g R Another limitation of the Riemann integral is in its relation with limit operations Before we discuss these examples we will review some concepts about convergence of functions in the next section 12 SA BARRETO 5 METRIC SPACES Several or perhaps all topics discussed in this section are not new for most students in MA 544 However these concepts are an important part of the course and worth reviewing De nition 51 A metric space is a pair M d consisting of a set M and a function d M X M 8 000 satisfying the following properties 1 dzy dyz for all zy E M 2 dzy dy 2 2 dz 2 for all zy2 E M triangle inequality 5 dzy 0 if and only y In what follows we denote a metric space either by M or M d we want to emphasize its metric Example 1 M R n 21 and dzy 11 7 y12 In 7 yn2 Exercise 51 Verify that R d is a metric space Example 2 Let C01 M denote the set of real valued functions de ned in 01 Let 159 sup WI 9IlA ace01 Properties 1 and 3 are obvious One just needs to verify property 2 Let fgh E C01 Since by the standard triangle inequality WI h1lSlfI 9Illy1 hIl7 we immediately see that df7h S 4059 d97hA De nition 52 Let X be a vector space over R or mJ A norm on X is a function NXgt0oo such that Nz 0 ltgt z 0 Nz lAlNz A E R or C NIySNINy One often uses the notation Nz Exercise 52 Let X N be a normed vector space Show that Who MI 7 y is a metric on X MA544 LECTURE NOTES FALL 2009 13 51 Sequences and their convergence A sequence of elements of a metric space M is a countable subset xj j E N of elements xj E M De nition 53 A sequence converges to x E M for every 6 gt 0 there exists N E N which depends on e such that for allj gt N dxjx lt e In this case we say that lim xjx or xjax jam An important concept is that of a Cauchy sequence De nition 54 A sequence is Cauchy if for every 6 gt 0 there exists N E N which depends on e such that for all jh gt N dxjxk lt 6 Theorem 51 Every convergent sequence is Cauchy Proof This is an easy consequence of the triangle inequality Let xj E M With xj A x For 6 gt 0 pick N such that dxjx lt 62 for all j gt N Then for j k gt N dxjxk S dxjx dxkx lt 6 Therefore xj is Cauchy D It is not the case that all Cauchy sequences in a metric space M d converge to an element x E M Example 51 Let M Q denote the set of rational numbers With the metric dpq lpi ql Q is not complete To see that pick a sequence qj E Q Which converges to qj is Cauchy as it converges in R but it does not converge in De nition 55 A metric space M d is complete for every Cauchy sequence C M there exists x E M such that converges to x 52 Completion of metric spaces One wonders if given a metric space there is a complete metric space that contains it De nition 56 Let M d and be two metric spaces A map i M a M is an isometry if di17iy dr7 y Notice thati is injective In this case we call i an embedding ofM into fi is onto we say that M and M are isomorphic Theorem 52 Let M d be a metric space Then there is a metric space with the following properties 1 is complete 2 There is an embedding i M a M 5 is dense in 4 If X p is another metric space satisfying 12 and 5 then and X p are isomorphic Proof Will be in the homework assignment D 14 SA BARRETO 6 SPACES OF FUNCTIONS The most important examples of metric spaces in MA 544 are certain spaces of functions We begin With the following fundamental examples7 Which are the motivation for the introduction of the Lebesgue measure Exercise 61 Let M CO7 1 be the space of continuous real valued functions in 01 Let Hflloo sup WIN xe01 1 117 llfllp lfzlde 7 p 2 1 This is the Riemann integral 0 Show that Hpr l S p S 00 are norms Theorem 61 The metric space C0717dp7 d1f7 g 7 ng is incomplete ifl S p lt 00 and is complete if p 00 Proof It is easier to prove that C0717dp7 l S p lt 007 is incomplete Let 1 if z e 0 g MI 17Iif167 She 0 if IE1 Let 1 if mpg x 0 if zeg Then 0 if IE0 fz7fnz 17zi if z 7 0 if IE1 Hence 1 n 1 2 7 17 7 7 7 p lt 7 0 mz 9Il dz lt1 2 z 2 dz dz n So dpfnf A 071 S p lt 007 as n A 00 But f g C071 Now we prove that the metric space C0717dm is complete Let be a Cauchy sequence in C0717dm Then for any 6 gt 0 there exists N E N such that 6 1 7 fmzHDo lt e for all mm 6 N Fix 1 E 01 Then is a Cauchy sequence in R and therefore it converges Let d f fltzgt hm fnltzgtz naoo In view of 61 fn converges to f uniformly on 01 Since fn is continuous and 01 is compact7 f E CO7 This ends the proof of the Theorem D MA544 LECTURE NOTES FALL 2009 15 We know from Theorem 52 that C0 1 dp 1 S p lt 00 can be completed The question is what is its completion This is a very important question in analysis Let fg E RGO 1 ie f and g are Riemann integrable in 01 We will say that f and g are equivalent if there exists a set F with mF 0 such that f1917 V I 610711FA This is an equivalence relation Theorem 62 Let Lp01R denote the space of equivalence classes offunctions which are Riemann integrable and that for each f E Lp01R let 1pr Ollfzlpdac1 Then Lp01R is a normed vector space but it is not complete Proof We will take p 1 but the general case is the same Since the set of discontinuities of is contained in the set of discontinuities of f and f is Riemann integrable then also is Riemann integrable Recall that by assumption if f E R then f is bounded So there is no convergence problem with the integral It is clear that does not depend on the choice of the representative of the class of The properties of the norm are clear with the exception of its nondegeneracy We need to show that if f E R and 0 then f E 0 This follows from Lemma 61 Ifgz 2 0 for all z in 01 M and folgzdz 0 then 91 0 for all I such that g is continuous at 1 Proof If 910 0 and g is continuous at 10 then there exists 6 gt 0 such that if z 7 10 lt 6 then 91 gt gzo2 This implies that folgzdz gt 69zo2 gt 0 D If 0 then 0 for all z in the set of points where is continuous Since f is Riemann integrable it follows that f 0 except on a set of measure zero Thus f E 0 To prove that L101R is incomplete we recall the construction of the Cantor set of nonzero measure Let fn XFW be the characteristic function of the set Fn de ned in Lecture 1 fn is obviously Riemann integrable Moreover fn 7 fn1 XFF1I By the construction Fn Fn1 consists of 2 disjoint intervals of length 2 H11 7 Tj7 1 Therefore 1 n f 7 mm He 7 NW W1 0 F1 If N gt 1 we write N N lfn 7 fnNl S Zlfnj fwja S Zlfnj fnjell j1 j1 Hence 71n 17f N llfn fnNll1 S 2 S j1 15 SA BARRETO Hence fn is a Cauchy sequence in L1017Ri But it obviously converges to the characteristic function of the Cantor set K Which is not Riemann integrable D Course Notes for MA 460 Version 3 Jim McClure 1 De nitions and Basic Facts The goal of this course is to help you become expert in geometry so that you can teach it with con dence and pleasure We begin more or less where you left off in high school and then advance rather quickly to a higher level Here s the big picture The goal of geometry is to know the truth about geometrical gures just as the goal of astronomy is to know the truth about planets and stars and the goal of chemistry is to know the truth about atoms molecules and chemical reactions To achieve this goal we must have a way of knowing which statements are true and which statements are false In sciences like astronomy and chemistry the way to do this is to test each statement against careful observations But this method is not foolproof because no observation can be perfectly accurate For example Newton7s laws in physics were tested by thousands of observations over almost 200 years but the small uncertainties in these observations concealed the fact that there is a more accurate and fundamental theory than Newton s namely Einstein s theory of relativity Mathematics is the only science in which there is a way to achieve complete certainty with no possibility of errors due to observation The way mathematicians do this is to begin with a few very simple statements called Basic Facts or Axioms that can be accepted as true on the basis of intuition These are then used to prove everything else This means that as long as the Basic Facts are correct everything else is guaranteed to be completely correct Before we can state the Basic Facts and begin to prove things it7s helpful to have some special words such as congruent or parallel that refer to geometric gures Thus we begin with a section of De nitions 11 De nitions In order to achieve complete certainty in our proofs we must avoid any uncertainty or ambiguity in the way we use geometric terms We need to be sure that we all mean the same thing by these terms and that we re always using them in the same way The way to accomplish this is to begin with precise de nitions I assume that you already know the meanings of some simple terms like point7 line7 line segment7 and ray Angle An angle is the gure formed by two rays which begin at the same point this point is called the vertex of the angle Straight angle A straight angle is an angle in which the two rays point in opposite directions on the same line Here is a picture You know that we measure angles in degrees But what is a degree L Degree A degree is the 180 th part of a straight angle Note we will not consider angles bigger than 180 Right Angle A 90 angle is called a right angle Perpendicular Two lines are called perpendicular if they form a right angle Triangle A triangle consists ofthree points called vertices and the three line segments which connect them called sides The vertices are not allowed to be collinear that is7 they are not allowed to all lie on the same line Congruent Triangles Two triangles AABC and ADEF are congruent written AABC g ADEF if all three corresponding angles and all three corresponding sides are equal Here is a picture C F Note when we say that AABC is congruent to ADEF we mean that the vertices match up in that order that is A matches with D B matches with E and 0 matches with Similar Triangles Two triangles AABC and ADEF are similar written AABC w ADEF if all three corresponding angles are equal Note a common mistake is to say that the de nition of similarity tells you that the corresponding sides are proportional This is not part of the de nition it is Basic Fact BF 4 in Section 12 Here is a picture Parallel Lines Two lines are parallel if they do not intersect Corresponding angles If two lines m and n are crossed by a third line the third line is called a transversal then the following pairs of angles are called corresponding angles7 1 and 5 2 and 6 3 and 7 4 and 8 Midpoint of a line segment The midpoint of a segment AB is the point M on the segment for which MA MB Angle bisector The bisector of an angle is the line that goes through the vertex of the angle and splits the angle into two equal parts Quadrilateral A quadrilateral consists of four points A7 B7 C7 and D called vertices and the line segments AB7 BC7 CD and AD called sides Parallelogram A quadrilateral is a parallelogram if the opposite sides are parallel Note a common mistake is to say that the de nition of parallelogram tells you that the opposite sides are equal this is not part of the de nition7 it is Theorem 10 Rectangle A quadrilateral is a rectangle if it has four right angles Square A quadrilateral is a square if it has four equal sides and four right angles Comment on the de nitions of similar triangles and parallelogram As l7ve mentioned7 the de nition of similarity doesn t include the fact that corresponding sides of similar triangles are proportional7 and the de nition of parallelogram doesn7t include the statement that opposite sides are equal This means that when you want to say that similar triangles have equal angles7 you use the de nition of similarity7 but when you want to say that similar triangles have proportional sides7 you have to use Basic Fact 4 which is stated in Section 12 And when you want to say that opposite sides of a parallelogram are parallel7 you use the de nition of parallelogram7 but when you want to say that opposite sides of a parallelogram are equal7 you have to use Theorem 10 Since this makes your life more complicated7 I should explain why we do it this way As l7ve said7 our method for achieving complete certainty in geometry is to begin with Basic Facts that we can accept on the basis of intuition and then to prove everything else using only those But for this method to work7 we have to be careful to know exactly what were assuming In particular7 we must be careful to distinguish between Basic Facts which are statements about reality that we re assuming and De nitions which are agreements about the meanings of words The fact that when two triangles have equal angles they also have proportional sides is a rather surprising statement about reality7 it isn t a statement about the meanings of the words If we included this fact in the de nition of the word similar7 we would be making a hidden assumption about reality Here s an analogy that can make this clearer If we want to prove the statement that there is water on Mars7 we can t do it by including this statement in the de nition of the word Mars or of the word water We can only prove it by making observations7 for example by sending a spacecraft to Mars and nding water there This way of using de nitions is also what we do in ordinary life For example the dictionary de nition of presiden is the highest executive of cer in a modern republic 7 the de nition does not include the statement that Washington was the rst president of the United States although this is a true fact 12 Basic Facts Before we can begin to prove things we have to have as a starting point a list of intuitively clear facts that we accept without proof Here is a list of basic facts that you know from high school which will be the starting points for our proofs When we get to Euclid we will see that most of these basic facts can themselves be proved by starting from even simpler facts Euclid gives proofs of BF 1 BF 2 BF 3 BF 4 and the second part of BF 5 An important point In geometry its more satisfying to prove things than to base them on intuition Because of this we want to keep the list of Basic Facts as short as possible We shouldn7t add a statement to the list of Basic Facts if there s a way to prove it from the other Basic Facts BF 1 SSS if two triangles have three pairs of corresponding sides equal then the tri angles are congruent BF 2 SAS if two triangles have two pairs of corresponding sides and the included angles equal then the triangles are congruent Note it is possible for two triangles to have two pairs of corresponding sides and a pair of ntmincluded corresponding angles equal and still not be congruentican you draw an example of this BF 3 ASA if two triangles have two pairs of corresponding angles and the included side equal then the triangles are congruent Note as you know from high school the AAS criterion for congruence is also valid but we don t include it in the list of Basic Facts because we can prove itisee the end of Section 22 BF 4 If two triangles are similar then their corresponding sides are proportional that is if AABC is similar to ADEF then AB 7 AC 7 BC DE DF EF Note BF 4 is a one way street liit does not say that if the corresponding sides are proportional then the triangles are similar Eventually we will prove that ifthe corresponding sides are proportional then the triangles are similar BF 5 If two parallel lines I and m are crossed by a transversal then all corresponding angles are equal lftwo lines I and m are crossed by a transversal and at least one pair of corresponding angles are equal then the lines are parallel Note BF 5 is a two way street you can use it in either direction Here are some simpler Basic Facts which are included for completeness we will need these in order to give reasons for every step in our proofs Euclid gives proofs of BF 10 BF 11 BF 12 BF 13 BF 14 and part of BF 8 BF 6 The whole is the sum of its parts this applies to lengths angles areas and arcs BF 7 Through two given points there is one and only one line This means two things First it is possible to draw a line through two points Second if two lines have two or more points in common they must really be the same line BF 8 On a ray there is exactly one point at a given distance from the endpoint This means two things First it is possible to nd a point on the ray at a given distance from the endpoint Second if two points on the ray have the same distance from the endpoint they must really be the same point BF 9 It is possible to extend a line segment to an in nite line BF 10 It is possible to nd the midpoint of a line segment BF 11 It is possible to draw the bisector of an angle BF 12 Given a line I and a point P which may be either on I or not on I it is possible to draw a line through P which is perpendicular to Z BF 13 Given a line I and a point P not on I it is possible to draw a line through P which is parallel to Z BF 14 If two lines are each parallel to a third line then they are parallel to each other BF 15 The area of a rectangle is the base times the height 2 Some familiar theorems from highschool geome try Now we can start proving things The facts we state from now on will be called Theorems not Basic Facts Both Basic Facts and Theorems are true statements about realityithe difference between them is that a Basic Fact is something we accept without proof based on intuition while a Theorem is something we can prove by using the Basic Facts Basic Facts are also called Axioms A very important point In proving theorems we are allowed to use only three ingredients de nitions Basic Facts and theorems that have already been proved We have to be careful not to use Theorems that have not yet been proved because we have to avoid circular arguments we aren7t allowed to use one theorem to prove a second and then use the second to prove the rst 21 Angles formed by intersecting and parallel lines Before continuing we need a few more de nitions Adjacent and Vertical Angles When two lines cross four angles are formed The pairs of angles that share a side are called adjacent and the pairs of angles that do not share a side are called vertical In the picture the adjacent pairs are 1 and 2 1 and 4 2 and 3 3 and 4 The vertical pairs are 1 and 3 2 and 4 Interior and exterior angles If two lines m and n are crossed by a transversal then angles 1 2 7 and 8 are called exterior angles and angles 3 4 5 and 6 are called interior angles Alternate interior angles In the picture above7 the following pairs of angles are called alternate interior pairs 7 3 and 6 4 and 5 Theorem 1 When two lines cross a adjacent angles add up to 180 and b vatlcal angles are equal Proof Every possible picture which illustrates the theorem will look like Figure 17 so it is enough if we show in Figure 1 that 1 and 2 add up to 180 and that 1 and 3 are equal Now 11 2 ABC by BF 67 and M ABC 180 by the de nition of degree Combining and gives 6 11 2 180 which is what we were to prove for part a An exactly similar argument shows that MM 2 3 180 and we conclude that 1 3 by and algebra This is what we were to prove for part QED Figure 1 Note we always end a proof by writing QED This is an abbreviation for the Latin phrase Quod erat dernonstrandurn77 which means this is what was to be proved 7 Comment on the use of pictures in proofs We are allowed to use a picture in a proof as long as we use a typical picture7 that is7 one which has only the features guaranteed by the hypothesis7 and no special features An argument which is valid for such a picture will automatically be valid for every other picture which satis es the hypothesis Examples of the sort of thing you should be careful about if the theorem is to be valid for every rectangle7 your proof shouldn t be based on a picture of a square if the theorem is to be valid for every triangle7 your proof shouldnt be based on a picture of an isosceles triangle Our next theorern collects some useful variations on BF 5 Theorem 2 Suppose that Z and m are two lines crossed by a transversal a fl and m are parallel then both pairs of alternate interior angles are equal If at least one pair of alternate interior angles are equal then I and m are parallel b fl and m are parallel then each pair of interior angles on the same side of the transversal adds up to 180 If at least one pair of interior angles on the same side of the transversal adds up to 180 then I and m are parallel c fl and m are parallel then each pair of castcrior angles on the same side of the transversal adds up to 180 If at least one pair of castcrior angles on the same side of the transversal adds up to 180 then I and m are parallel Proof Every possible picture that illustrates the Theorem will look like Figure 2 Figure 2 Part a contains two statements and we have to prove both For the rst statement we are given that Z and m are parallel and we have to prove that the two pairs of alternate interior angles are equal We know that Q 11 15 by BF 5 We also know that M 11 14 by Theorem 1b Combining and tells us that 4 5 An exactly similar argument gives the equality of the other alternate interior pair For the second statement in part a we are given that a pair of alternate interior angles are equal lt7s enough to give the proof when this pair is 4 and 5 because the proof for the other pair is exactly similar So we are given 9Q 14 15 Theorem 1b tells us that M 14 11 Combining and tells us that 1 5 Since this is a pair of corresponding angles BF 5 tells us that Z and m are parallel For the rst statement of part b we are given that the lines I and m are parallel and by BF 5 this tells us that Q 11 15 Theorem 1a tells us that M 41 13 1800 10 Combining and tells us that 3 5 180 An exactly similar argument tells us that the other two interior angles also add up to 180 For the second statement of part b7 we are given that a pair of interior angles on the same side of the transversal adds up to 180 lt7s enough to give the proof when this pair is 3 and 5 since the argument is exactly similar if it is the other pair So we are given i 3 5 180 Theorem 1a tells us M 41 3 180 Combining and we see that 1 5 and by BF 5 we conclude that Z and m are parallel The proof of part c is a homework problem QED A convenient shortcut When one part of an argument repeats an earlier part word for word7 with only the names of the points and lines changed7 you7re allowed and encouraged to skip the repetition and just say this part is similar 7 This happened in the proof of Theorem 1 and several times in the proof of Theorem 2 But the part that s skipped must be an exact repetition of an earlier partiif it merely resembles an earlier part you still have to give it in full 22 The sum of the angles of a triangle Theorem 3 The angles of a triangle add up to 180 Proof Refer to Figure 3 BF 13 allows us to draw a line m through 0 which is parallel to AB Now i 41 2 3 180 by BF 6 Theorem 2a tells us that M 1 1A and x x as 3 AB Combining x7 and as x as we see that A AB 2 180 7 which is what we were to show QED Figure 3 Note We dont need to consider the different ways the picture might look because the argument works exactly the same way for all possible pictures Theorem 4 If two triangles ABC and DEF have A 1D and AB 1E then also 0 1F Proof We know from Theorem 3 that 1A AB 0 180 Combining this with the given7 we have AD E 0 180 But using Theorem 3 again we have M D E 1F 180 By 9Q and algebra we conclude that AC 1F QED Comment on the AAS criterion for congruence Suppose that you have two triangles and you know that two pairs of corresponding angles and a nonlncluded pair of corre sponding sides are equal You cant apply BF 3 directly to this situation7 but Theorem 4 implies that all corresponding pairs of angles are equal7 and then BF 3 does apply So when you are in the AAS situation7 you may conclude that the triangles are congruent7 with the justi cation Theorem 4 and BF 377 For an exarnple7 see the proof of Theorem 5 23 Isosceles triangles Theorem 5 a If two sides of a triangle are equal then the opposite angles are equal b If two angles of a triangle are equal then the opposite sides are equal C Comment This theorem is a two way street it says that if you are given either one of the statements two sides are equal or two angles are equal then the other statement must also be true There is a convenient abbreviation for this kind of situation if we have a theorem that says if A is true then B is true and if B is true then A is true we can state it more brie y by saying A is true if and only lfB is true or even more brie y by A ltgt B For example we can restate Theorem 5 as Two sides of a triangle are equal ltgt the opposite angles are equal lt7s important to note that when we want to prove a theorem that has ltgt we have to give two proofs one for each direction Proof of Theorem 5 We begin with part a This means that in Figure 4 we are given that AC BC and we want to prove 1A AB Find the midpoint M of AB which we are allowed to do by BF 10 and connect it to C In triangles AMC and BMC we have AC BC given AM MB de nition of midpoint and MC MC Now by BF 1 we have AAMC E ABMC and from this we conclude that A AB de nition of congruent triangles Figure 4 For part b we are given that A AB and we want to prove that AC BC Draw the perpendicular line m from C to AB allowed by BF 12 and give the intersection of m and AB the name D see Figure 5 Then 1 2 since both are right angles by the de nition of perpendicular and 1A AB given Thus 3 4 by Theorem 4 Furthermore CD DC so AADC g ABDC by BF 3 and from this we conclude that AC BC de nition of congruent triangles QED Figure 5 Note In this proof M and D are actually the same point However because they are constructed by different recipes the information which is available for use is different in the two parts ofthe proof for part a we are allowed to use the fact that AM MB whereas for part b what we are allowed to use is that ADC and ABDC are right angles There is a standard name for the kind of triangle described in Theorem 5 14 Isosceles A triangle with two equal sides is called isosceles The word isosceles is a Greek word meaning equal sides Using this terminology7 Theorem 5 says that a triangle is isosceles ltgt it has two equal angles More about de nitions As l7ve mentioned7 a frequently asked question about de nitions is why they don t include more information For example7 why doesnt the de nition of isosceles say that an isosceles triangle has both two equal sides and two equal angles At rst sight7 this might seem to make Theorem 5 unnecessary But more careful thought shows otherwise Suppose we did de ne isosceles to mean two equal sides and two equal angles We would still want to know that whenever two sides are equal we are guaranteed that two angles will be equalithat is7 we would want to know that if a triangle has two equal sides then it is isosceles in the new sense And we would also want to know that if it has two equal angles then it is isosceles in the new sense In other words7 we would still want to know both parts of Theorem 57 and the proof of Theorem 5 wouldn t be any easier than before The moral of this discussion is that by making the de nition more complicated we wouldn t actually have made anything else simpler So we might as well at least make the de nition as simple as possible7 and this is what mathematicians usually do To put it another way7 Theorem 5 is a fact about reality7 so our only choice is to prove it or add it to the list of Basic Facts We aren t allowed to hide it inside a de nition 24 The area of a triangle Our goal in this section is to prove that the area of a triangle is one half of the base times the height We begin with a special case Theorem 6 In triangle ABC if B is a right angle then the area of the triangle is AB BC Proof Using BF 137 draw a line m through A parallel to BC and a line n through C parallel to AB see Figure 6 Give the intersection of m and n the name D Then 1 4 by Theorem 2a using the fact that AD and BC are parallel and 2 3 by Theorem 2a again this time using the fact that AB and CD are parallel Furthermore7 AC AC7 so by BF 3 we have AABC g ACDA Now the area of ABCD is equal to the sum of the areas of ABC and CDA by BF 67 and the areas of these two triangles are the same since they are congruent so we conclude that the area of ABCD is twice the area of ABC in other words7 area of ABC area of ABCD Next we want to show that ABCD is a rectangle We are given that B is a right angle7 and so by Theorem 2b we know that ABCD is also a right angle Using Theorem 2b 15 again7 this implies that AD is a right angle7 and this in turn implies by one more use of Theorem 2b that DAB is a right angle Now we have shown that ABCD is a rectangle de nition of rectangle and so we know by BF 15 that M area of ABCD AB BC Combining and gives us the formula which was to be proved QED Figure 6 Next we need some de nitions Distance from a point to a line The distance from a point P to a line rn is de ned to be the length of the line segment from P to rn which is perpendicular to rn p m Base and height of a triangle ln triangle ABC7 any side can be chosen as the base Once we have chosen the base7 the height is the distance from the remaining vertex to the line containing the base For example7 if we choose BC as the base7 then the height is the distance from A to gt BC Remember that distance 7 has just been de ned to mean the perpendicular distance The perpendicular from A to BC will be inside of the triangle if the base angles are both less than 90 but if one of them is bigger than 90 it will be outside the triangle7 as shown in the following picture 16 Theorem 7 The area of a triangle is one half of the base times the height Note For each triangle we are really getting three formulas for the area because there are three ways of choosing the base In the proof of Theorem 7 for the rst time we have to consider dz erent ways that the picture might look because the argument will be different Proof of Theorem 7 Its enough to give the proof when AB has been chosen as the base the other two possibilities are proved by exactly similar arguments Draw the line gt m that goes through 0 and is perpendicular to AB as we may by BF 12 and give the gt intersection of this line with AB the name D There are four cases to consider Case D is between A and B Case ii D is to the right of B or to the left of A Case iii D is the same point as A Case iv D is the same point as B C c d A o B A 00 Case Case ii D Case iii BD Case iv In all four cases CD is the height of the triangle by de nition of height and the angle formed by CD and line AB is a right angle by de nition of perpendicular In case we have area of ABC area of ADC area of BDC by BF 6 1 EAD DC BD DC by Theorem 6 1 AD BDDC by algebra 1 iAB DC by BF 6 and this is what was to be shown for this case In case ii7 the area of ADC is equal to the area of ABC plus the area of BDC by BF 6 and so area of ABC area of ADC 7 area of BDC 1 1 EAD DC 7 EBD DC by Theorem 6 1 5AD i BDDC 1 7AB DC 2 by algebra by BF 6 and algebra and this is what was to be shown for this case In case iii A must be a right angle7 so we can apply Theorem 6 Case iV is similar QED 25 The Pythagorean Theorem and the Hypotenuse Leg The orem Theorem 8 Pythagorean theorem In a right triangle the sum of the squares of the two legs is equal to the square of the hypotenuse 18 Proof In Figure 77 we are given that ACB is a right angle7 and we want to prove that a2 62 02 where a7 6 and c are the lengths of B07 AC and AB respectively We draw a perpendicular line I from C to AB which we are allowed to do by BF 12 and label the intersection of Z and AB by F Let d and 8 stand for the lengths of AF and BF note that dec by BF 6 Now 1 ACB because both are right angles and 1A 1A so by Theorem 4 and the de nition of similarity we see that AABC w AACF Using BF 4 we see that 06 bd and algebra tells us M 62 Cd An exactly similar argument shows that AABC w ACBF7 and BF 4 tells us that ca ae which by algebra gives 6 a2 08 Combining Os and 6 as we see that a262 cecdced c cc2 and this is what we wanted to show QED Figure 7 From the Pythagorean Theorem it is easy to prove the Hypotenuse Leg criterion for congruence Theorem 9 Hypotenuse Leg Theorem In triangles ABC and DEF if A and D are right angles and ifBC EF and AB DE then AABC E ADEF That is if two right triangles have the hypotenuse and a leg matching then they are congruent C A B D E Proof The Pythagorean Theorem says that A32 A02 302 and DE2 DF2 EF2 Combining these equations with the given we see that AC2 DFQ and since AC and DF are positive numbers this implies AC DF Now BF 2 tells us that AABC g ADEF QED 26 Parallelograms Theorem 10 If ABCD is a parallelogram then opposite sides of ABCD are equal Proof In Figure 8 we are given that ABCD is a parallelogram Connect A and C by a line segment Now AB is parallel to CD by de nition of parallelogram s0 11 14 Theorem 2a Also AD is parallel to BC de nition of parallelogram s0 M 12 13 Theorem 2a Furthermore AC AC and s0 AACD g ACAB BF 3 From this we conclude that AB CD and AD BC de nition of 2 QED 20 Figure 8 Theorem 11 If ABCD is a parallelogram then opposite angles of ABCD are equal Proof By Theorem 2b7 1A 1B 180 and 1B AC 180 Combining these two equations gives 1A AC The proof that B 1D is similar QED Theorem 12 If a quadrilateral has a pair of sides which are equal and parallel then it is a parallelogram The proof of Theorem 12 is a homework problem Theorem 13 A quadrilateral is a parallelogram ltgt the diagonals biscct each other that is ltgt the intersection of the two diagonals is the midpoint of each diagonal The proof of Theorem 13 is a homework problem 3 More about triangles 31 The line through the midpoints of two sides of a triangle We begin with a convenient piece of algebra Theorem 14 a Suppose that C is a point on the segment AB C is the midpoint of AB ltgt AB 2A0 b A line segment can have only one midpoint Proof For a7 we begin with the gt direction7 so we are given that C is the midpoint of AB Now AB A0 BC by BF 6 and also AC BC de nition of midpoint Combining these equations gives AB 2A0 In the lt direction7 we are given that AB 2A0 But we also have AB ACBC by BF 6 Combining these equations gives AC BC7 s0 0 is the midpoint of AB de nition of midpoint 21 For b7 suppose that C and C were both midpoints of AB Then AC AB and AC AB both by part a and so AC AC Now BF 8 tells us that C and C are the same point QED Theorem 15 In triangle ABC let D be the midpoint of AC and suppose that E is a point on BC with DE parallel to AB Then E is the midpoint of BC and DE ABi Proof See Figure 9 By BF 57 1 2 and 3 4 Also7 AC 0 Thus AABC w ADEC de nition of similarity7 and therefore AC 7 BC 7 AB DC EC DE by BF 4 Since we are given that D is the midpoint of A07 we can apply Theorem 14a to get AC 7 2 M Do i i 30 i i i Combining and we see that m 27 so that E is the midp01nt of BC by Theorem 14a Finally7 and also tell us that AB 72 DE so that DE AB QED Figure 9 Our next theorem is closely related to Theorem 15 but considerably trickier to prove 22 Theorem 16 In triangle ABC let D be the midpoint ofAC and let E be the midpoint of BC Then DE is parallel to AB and DE AB Notice that the picture which illustrates this theorem is the same as that for Theorem 15 but the information we are given about this picture is different Proof Use BF 13 to draw a line in through D which is parallel to AB Give the intersection of m and BC the name F Then DF is parallel to AB so we can apply Theorem 15 to conclude that F is the midpoint of BC But then F is really the same point as E by Theorem 14b Now BF 7 tells us that DE is the same line as DF so DE is parallel to AB by Finally since we now know that E satis es the hypothesis of Theorem 15 we can apply Theorem 15 to conclude DE AB QED The proof of Theorem 16 is the trickiest we have seen so far You should study this proof carefully because we will be using similar ideas in several other proofs and homework problems Note that Theorem 16 would be rather easy if we had a Basic Fact that said that triangles with two sides proportional and the included angle equal are similar But we dont include this fact among the Basic Facts because we can prove it using the Basic Facts we already have We will do so in the next section 32 The SAS and SSS criteria for similarity Theorem 17 SAS for similarity In triangles ABC and DEF if 0 1F and BC m then AABC w ADEF The proof is rather tricky The basic idea is to make a copy of ADEF on top of AABC and then show that the copy is similar to AABC But we wont know that the copy is actually congruent to ADEF until almost the end of the proof Proof See Figure 10 On ray 7A mark a point D with CD FD this is allowed by BF 8 Draw the line in through D which is parallel to AB allowed by BF 13 and give the intersection of m and BC the name E Our rst goal is to show that AD E C w AABC Since in is parallel to AB we can use BF 5 to see that 1 2 and 3 4 Also 0 0 so AD E C w AABC Our next goal is to see that AD E C E ADEF Because of the way D was con structed we know M D C DE 23 Next we use and BF 4 to get AC 7 BC D C E C Combining and 6 as we have 0 BC DF EC and comparing this with the given we obtain BC 7 BC E C EF which by algebra gives E C EF But we were given that AC 1F so we can apply BF 2 to get MM AD E C E ADEF Now tells us that 1 2 and tells us that 1 1D so we conclude that 2 1D Sirnilarly7 4 1E And we were given 0 1F so AABC w ADEF by de nition of similarity QED Figure 10 There is also a criterion for sirnilarity analogous to the SSS criterion for congruence its the other direction of BF 4 Theorem 18 SSS for similarity In triangles ABC andDEF 239 A73 AC 7 BO DE DF W then AABO w ADEF The proof is a homework problem 24 33 The sine of an angle In this section we introduce the sine of an angle and use it to give a new formula for the area of a triangle First we need an of cial de nition for the sine7 and before stating it we need to set a the stage Let ABC be any angle7 and choose a point D on BC By BF 12 we can gt gt draw a line m through D perpendicular to AB Give the intersection of m and AB the name E Notice that there are three cases7 depending on whether ABC is less than 90 7 equal to 90 7 or greater than 90 B E A BE A E B A ABC lt 90 ABC90 ABC gt 90 In all three cases the sine is given by the same formula i DE De nltlon of Slne The sine of ABC is Notice that for angles less than 90 this agrees with the formula you learned in high school opposite sin hypotenuse When applying the de nition of sine to an angle you have the freedom to choose any point D you want as long as D is on one of the sides of the angle All choices are guaranteed to give the same answer7 because if you choose D and your friend chooses DE E D the triangles DEB and D E B will be sirnilar7 and so E will be equal to by BF 4 See Figure 11 25 D D 0 0 B E E39 A Figure 11 There is a useful relationship between the sines of angles that add up to 180 Theorem 19 sin1ABC sin180 7 ABC Proof In Figure 127 FBC 180 7 ABC by BF 6 and algebra When we apply the de nition of sine to ABC we get DE ABC 7 s1n DB and when we apply the de nition of sine to FBC we get DE AFBC 7 s1n DB so sin1ABC sinzFBC QED 0407 A B E F Figure 12 Theorem 20 For any triangle ABC the area can be calculated by any of the following three formulas area of AABC AB AC sin A 26 area of AABC AB BC sin 1B area of AABC AC BC sin 0 that is the area is one half the product of two sides times the sine 0f the included angle Proof It is enough to give the proof of the rst formulaithe proofs of the other two are completely sirnilar There are three cases A lt 90 7 1A 90 7 and LA gt 90 see Figure 137 but in fact the proof is the same in all three cases Let us choose AB to be the base ofthe triangle7 and as usual use BF 12 to draw a line through 0 perpendicular gt gt to AB which intersects AB at a point E Then CE is the height of the triangle7 and so we have 1 90 area of AABC AB CE by Theorem 7 But the de nition of sine says CE 1A 7 s1n 0A and so M CE CA sin1A Combining and gives the formula we wanted to prove QED c C AE Figure 13 By comparing the three formulas in Theorem 207 we get an interesting relationship called the Law of Sines Theorem 21 Law of Sines In any triangle ABC sin1A 7 sinzB 7 sinzC BC AC AB 27 Proof Theorem 20 tells us that m AB ACsin1A AB Basinma since both sides of the equation are equal to the area of AABC Now and algebra give sin1A sinzB BC AC The proof of the other equality is completely similar QED 28 4 Concurrence Theorems De nition of concurrent lines Three lines are concurrent if they meet at a single point concurrent lines nonconcurrent lines Its very unusual for three lines to be concurrentiordinarily three lines will form a triangle But in this chapter we will see that certain kinds of lines associated with a triangle are forced to be concurrent 41 Concurrence of the perpendicular bisectors7 angle bisec tors7 and altitudes Our rst concurrence theorem77 concerns the perpendicular bisectors of the sides of a triangle De nition of perpendicular bisector The perpendicular bisector of a line segment is the line that goes through the midpoint and is perpendicular to the segment A common mistake Suppose you have a point A and a line segment BC7 and you put the following statement in a proof Draw the perpendicular bisector of BC through A77 This is wrong because the perpendicular bisector of BC may not go through A Its a useful exercise to see why a step like this is not justi ed by BF 10 and BF 12 BF 10 allows us to nd the midpoint M of BC7 and BF 12 allows us to draw a line m through M which is perpendicular to BC then m is the perpendicular bisector of BC On the other hand7 BF 12 also allows us to draw a line n through A which is perpendicular to line BC But m and n will not ordinarily be the same line7 so its not allowable to say Draw the perpendicular bisector to BC through A7 because ordinarily there is no such line 29 Theorem 22 For any triangle ABC the perpendicular bisectm s of AB AC and BC are concurrent It turns out the that obvious way of trying to prove Theorem 22 leads to a dead end Anyone s rst idea would be to draw the three perpendicular bisectors and see what happens But then there would be two cases the three perpendicular bisectors might meet in a point as in Figure 14 Figure 14 but since we don t yet know Theorem 22 we cant assume that the picture wouldn t look like Figure 15 30 Figure 15 and Figure 15 is a dead end there s nothing interesting we can say about it lnstead7 we do something trickier We draw two of the perpendicular bisectors7 then draw the line segment connecting their intersection to the third rnidpoint7 and use congruent triangles to show that this line segment is the third perpendicular bisector Here are the details Proof of Theorem 22 Find the rnidpoints M7 N7 and P of AB7 A07 and BC re spectively which we are allowed to do by BF 10 Now draw a line m through M perpendicular to AB and a line n through N perpendicular to AC allowed by BF 12 and give the intersection of m and n the name X Connect X to P There are three cases Case X is inside the triangle Case ii X is outside the triangle Case iii X is on BC 1711 give the proof of Case here and ask you to do the other two on the homework See Figure 16 Our strategy is to show that XP is the perpendicular bisector of BC First observe that AM MB de nition of rnidpoint7 MX MX7 and AAMX BMX by de nition of perpendicular So AAMX E ABMX by BF 27 and hence AX BX by the de nition of congruent triangles Sirnilarly7 AANX E ACNX7 so M AX OX 31 Combining and gives BX CX But also OP BP de nition of midpoint and PX PX so ACPX E ABPX by BF 1 and therefore 9F 11 2 de nition of congruent triangles On the other hand7 Theorem 1a says that MM 11 2 180 Combining 6 as and we see that 1 90 7 so PX is perpendicular to BC de nition of perpendicular Since we already knew that P is the midpoint of BC7 we conclude that T PX is the perpendicular bisector of BC de nition of perpendicular bisector Now X is on the perpendicular bisectors of AB and AC because of the way X was constructed and also on the perpendicular bisector of BC by That is7 all three perpendicular bisectors contain X7 and so they are concurrent de nition of concurrent QED A M B Figure 16 De nition of circumcenter The point where the three perpendicular bisectors of the sides of a triangle meet is called the circumcenter of the triangle 32 Note In the proof of Theorem 227 X turned out to be the circumcenter of AABC and we showed that XA XB XC This means that the circle with center X that goes through A also goes through B and 0 Thus X is the center of a circle that goes through the three vertices of the triangle This circle is called the circumscribed circle of AABC7 and because of this its center X is called the circumcenter We conclude this section with two more concurrence theorems Theorem 23 For any triangle ABC the bisectors of 1A 1B and 0 are concurrent The proof is a homework problem De nition of incenter The point where the three angle bisectors meet is called the incenter of the triangle The incenter turns out to be the center of a circle which is tangent to all three sides of the triangle this circle is called the inscribed circle of the triangle and this is the reason for the name incenter7 De nition of altitude An altitude of a triangle is a line that goes through a vertex of the triangle and is perpendicular to the opposite side Theorem 24 For any triangle ABC the three altitudes are concurrent The proof is a homework problem De nition of orthocenter The point where the three altitudes meet is called the or thocenter of the triangle The name orthocenter 7 is somewhat misleading since this point isn t the center of an interesting circle The name comes from the fact that orthogonal is another word for perpendicular 42 Concurrence of the medians De nition of median A median of a triangle is a line that goes through a vertex of the triangle and through the midpoint of the opposite side We will show that the medians of a triangle are always concurrent In order to prove this7 we need a preliminary fact which is interesting for its own sake Theorem 25 The point where two medians of a triangle intersect is 23 of the way from each of the two vertices t0 the opposite midpoint 33

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