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# INTRO DIFFRNTL GEOM Math 162

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This 54 page Class Notes was uploaded by Adam Crona on Saturday September 12, 2015. The Class Notes belongs to Math 162 at University of California - Irvine taught by N. Donaldson in Fall. Since its upload, it has received 12 views. For similar materials see /class/201862/math-162-university-of-california-irvine in Mathematics (M) at University of California - Irvine.

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Math 162B Notes Neil Donaldson March 10 2009 1 Exterior calculus 11 Wedge products and nforms Recall the notion of a l form X on lRquot if X1 Xn are co ordinates on lRquot then X 311 m dxi is a l form where m an lRquot gt lR are smooth functions We introduce an operation A on l forms which satisfies the properties Cle39 A 13939 7 13939 A 13939 Cle39 A Cle39 0 and study the algebra generated by l forms under A De nition 11 A multieindex of length k is a list of numbers I i1 ik 1 g i1 ik g 71 when in lRquot We write ClXI ClXi1 A A ClXik A multi index is increasing if p lt q i i7 lt ii A kfarm X on lRquot is an object of the form 0c Z 111 de All multiindices I of length k where each 0C1 lRquot gt lR is a smooth function We also call X a farm afdegree k At the moment the symbol A is just a symbol for denoting k forms It will shortly become an operation It is clear that at the expense of a minus sign every multi index can be put in increasing order This is the standard way of writing k forms The definition also includes 0 forms with no multi indices a 0 form is just a function on lRquot Example k R2 R3 R4 0 functionf f 1 a1 dxl 112 dxz 0C1ClX10 2ClX20C3ClX3 a1 dxl 114 ClX4 2 dxlAdxz ldxlAdx2 3dx2Adx3 ldxlAdX2 6dX3AdX4 3 None 7 dxl A dxz A ClX3 71 dxl A dxz A dxs 74 dxz A dxs A dx4 4 None None 5 dxl A dxz A ClX3 A ClX4 Generally there are components for a k form on lRquot Note that there are no k forms on lRquot for k gt n The set of k forms at p E U C lRquot form a vector space this should be clear if you view k forms as alternating k linear maps lf X1 xn are co ordinates on U then the set of k forms dxi1 A A dxik i1 lt lt ik forms abasis of this vector space which thus has dimension More concretely if 0 an is a basis of 1 forms at p then the set 0a A 0c i lt is a basis of the set of 2 forms at p and 111 A Atkik i1 g g ik a basis of the set of k forms at p De nition 12 Suppose that k l g n The wedge product of a k form 0c and an l form 8 on Rquot is the k l form X A 8 and is formed in the obvious way The obvious way here can get complicated for large degree forms as the following formula shows if X Eng ClXI and Eb ClX are k and l forms respectively then D A Z txl IdXIAClX All mul indices 1 of lengths kl resp actively where ClXI A ClX dx1 A A dxik A dle A A dle Thankfully in this course we will rarely consider higher than 3 forms Examples 1 0c 2dx1 7 3x1 dxz and 8 1 7 xi dxl x2 dxz are 1 forms on W Then 0c A 8 2 dxl 7 3x1 dxz A 1 7 x dxl xz dxz 21 7 xi dxl A dxl sz dxl A dxz 7 3x11 7 x dxz A dxl 7 3xz dxz A dxz Zn 7 3x1 3x1xg dxl A dxz 2 0c dxl dez x1 ClX3 and 8 3X3 dxl A dxz 7 dxz A ClX3 are 1 and 2 forms on R3 respec tively Here we have 0c A 8 3X1X3 7 1 dxl A dxz A dX3 Proposition 13 If 0c 8 arefarms then 0c A 3 71deg deg 3 A 0c Note that a function f is a 0 form and that f A X fix ocf X A f so that the proposition still holds 12 The exterior derivative We are used to thinking of d of a function of a co ordinate function as dxl say or of a more general function on an open set H d f E 3 dx d is a more general operator on forms of any degree De nition 14 Given a k form 0c 21 0C ClXI on lRquot the exterior derivative of 0c is the k 1 form doc 2 dog A de I where dog is d of a function Example In 1R3 let 0c xlxg dxl 7 x2 dX3 Then doc dx1x A dxl 7 dx2 A dX3 xg dxl 2pr dxz A dxl 7 dxz A dX3 72x1x2 dxl A dxz 7 dxz A dX3 Proposition 15 For allfarms 0c 8 we have 11 doc dIB 2 dtx A doc A 8 71deg tx A d 1 d 3 d doc 0 The final result is often written as d2 0 Example Letfx1 x2 xz on W Then df 2x1x2 dxl 3 dxz and so d d f d2x1x2 dxl x dxz 2dx1x2 A dxl dx A dxz 2x2 dxl A dxl 2x1 dxz A dxl 2x1 dxl A dxz 0 Co ordinate invariance in R2 One of the main advantages that comes with thinking about forms is that they have an inbuilt co ordinate invariance otherwise said when you change co ordinates forms automatically change in the correct way Here is an example in R2 y Figure 1 Polar co ordinates With respect to the standard co ordinates Xy any 1 form may be written 0c d dx b dy The same 1 form may also be written 0c A d7 B de in polar co ordinates Similarly any 2 form is a multiple of dx A dy and simultaneously a multiple of Cl A d0 These must correspond somehow Indeed Xrcos0 i dxcos9d77rsin0d9 yrsin6 dysin9drrcos9d0 gadlede racos6drbsin6d6 Similarly X y y 7d 7d 7d WHWy x x 7X2 y2 dy Hence dxA dy cos9 rcostrA d6 7 rsine sin d9 A d7 rdrAdQ Recall this from change of variables in integration if f Xy gr 9 then fxydxdygr9rdrd6 The change of variables from integration is thus already built into forms We will come back to forms and integration earlier though we ve already seen its first steps l forms are integrated along curves 2 forms will be what we integrate over surfaces 3 forms over 3 dimensional regions etc 13 Forms as multilinear maps A k form X at a point p E lRquot is a multi linear alternating map alpzprnx gtlt TpRquot7R k times from k copies of the tangent space at p to lR The following formula can be taken either as a definition of a k form or as the definition of the determinant in terms of forms depending on your preference If X a1 A A oak is a k form where each X is a l form and v1 vk E Tlequot are vectors then 17 1 quot39 0610M mWWI a2v1 x2vk 1114171 W MW Alternating means that if you swap the positions of any two of the vi the result changes sign This is equivalent to swapping two rows in the determinant Example Let 8 dxl A dxz X3 dxz A ClX3 be aZ form on R3 and let u 371 7 373 390 x2372 Now dx1u dx1v dxzu dxzv Buv xz1 x3 De nition 16 In E2 the 2 form dx A dy is the standard areafarm Indeed if u v are two vectors then dxl A dx2uv x2 dxz A dX3uv x2 dx A dyuv ulvz 7 vluz 01 3902 1 2 is the signed area of the parallelogram spanned by u v Similarly dx A dy A dz is the standard volume form on E3 These terms will become clearer when we study integration 14 An aside on vector calculus The standard vector calculus operations of div grad and curl in lE3 are closely related to d For example the curl of a vector field v Dali txzj D gk is i k 1 A A VXV 3quot By 31 TltBy Bz 1 Bz 3x 3x By k39 0C1 0C2 0C3 while the exterior derivative of the l form 0c m1 dx 112 dy m3 dz is 7 3112 3111 3111 3113 3113 3112 doci dxAdyi a 7 dxAdz dyAdz Comparing coefficients gives part of the following table start on any line and compare what d does to the form with what the corresponding vector calculus operation does to the object on the right hand side Traditional vector elds Forms functionf lt gt f d gradiv l form a1 dxoc2 dytx3 dz lt gt a1i0 2j0 3k 1 m1 Vx 2form 31 dyAdzI32dzAdxf33 dxAdy lt gt 31i32j33k d diviVn 3 form 7 dx A dy A dz lt gt function 39y The single d operator is grad div and curl all in one The differential form notation has two distinct advantages over traditional vector calculus it works in all co ordinate systems and all dimensions The forms result d2 0 translates to the 2 theorems VgtltVf0 VVgtltv0 With a little calculation it can be seen that the wedge product of l forms translates to taking the cross product of vectors while the wedge product of a l form and a 2 form corresponds to taking the dot product Complicated formulae from vector calculus can be easily proved this way eg let f be a function and X a l form Then donc detxdtfo deocfdtx VvaVfgtltvfVgtltv Similarly if X 8 are l forms we have docA docA itxAd VugtltvVgtltuv7uVgtltv 2 Moving frames and the structure equations 21 Maps lRm a E3 De nition 21 A mavingfmme for a smooth map x U C lRm gt lE3 written x1u1um xu1um xzu1 um X3u1 is a triple of maps e U gt lE3 i 123 such that e1pe2pe3p is an oriented orthonormal basis oflE3 for each p u1um E U 937 U E3 Figure 2 A moving frame A moving frame is usually chosen in a way that is suited to the map x Examples 1 m 1 x U gt lE3 is a smooth curve If x is biregular we can choose e1e2e3 T N B to be the Frenet frame 2 m 3 If x is the cylindrical polar co ordinate map 739 cos 4 x 74z gt gt rsinzp 2 then it would be sensible to choose the moving frame 3x GTE 4 1 3x 7 Sin 4 3x 0 e 5134 e r c034 ezg 1 so that the moving frame s axes point in the directions of change with respect to the three variables1 3 m 2 x is a parameterized surface One direction at least is specified fix an orientation on the surface and choose a moving frame with eg U the unit normal X1 De nition 22 If x X2 U gt lE3 is a smooth function the we write dx for its exterior deriva X3 tive ClX1 dx Cle ClX3 dx is a matrix of lfarms or an lE3 valued l form since it maps tangent vectors in U to vectors in E3 Example Recalling the cylindrical co ordinate map we have coszpdrirsinzpdzp dx sinzpdrrcoszpdzp xdrx dzpxzdz eydrre d ezdz dz 22 Connection forms and the structure equations In this section we consider a map x U C lRm gt E3 and a moving frame e1 e2 e3 Instead of simply thinking about how the map x changes ie about dx we split the problem into two describe how x moves with respect to the moving frame and describe how the frame itself moves If the frame is chosen sensibly with respect to the map then the answer to the first problem should be simple and we transfer the difficulty to thinking about how the frame moves While this may seem to increase the complexity it in fact improves matters even allowing the application of group theory to the problem First we define l forms on U which encode how x changes with respect to the moving frame 9k dx ek k 123 Proposition 23 dx 21 lek Proof Since at each point p E U e1 e2 e3 form a basis of E3 it is clear that dx 21dx ekek 31 lek Examples 1 lfx U gt lE3 is a parameterized curve xt then dx x dt lfwe take e1e2e3 T N B to be the Frenet frame then 91 lx l dt while 92 93 0 2 In the cylindrical co ordinate example 91 dxe1 d7 92 rdzj 93 ClZ 1The fact that these directions are orthogonal is a special property of cylindrical coordinates which is not true for general co ordinate sys ems 2Indeed this process and its generalizations to higher dimensions and more specialized situations is one of the main ways that group theory finds applications in Physics etve1dt 91vdt 92930 8 Moreover the Frenet Serret equations imply that wlz 7m dt U13 0 U23 for dt In this context Theorem 26 is exactly the Frenet Serret equations multiplied through by the l form dt Inspired by the link to the Frenet equations for a curve we now turn to the analogues for the map x Theorem 27 6 and to satisfy the first structure equations d6 w A 6 0 This is equivalent to the three equations d91w12A92w13A93 0 d027w12A61w23A63 0 deg 7w13A01 iwngez 0 Proof Since d2 0 we have 0 dzx ddx dE9 dEA9Ed0 EwA6 do Writing a d9 w A 9 and multiplying out we see that 21 eigij 0 for eachj 123 The linear independence of the e forces all 9 coefficients 17 to be zero hence d9 w A 9 0 l Theorem 28 The connection form to satisfies the second structure equations dw w A w 0 This is equivalent to the three equations dwlz 7 U13 A U23 0 dwis w12 A wzs 0 ClUng 7 U12 A U13 0 Proof 0 ddE dEw dEAwEdw EwAwEdw I The first and second structure equations are easier to remember if we use the fact that wij iw for then they read i dwij wik A wk 0 ijk distinct It is straightforward to see the forms for our cylindrical co ordinate example satisfy both structure equations It is less easy to see that the wij are in fact determined by the structure equations and the 0 In this example we have 61 dr 62 rdzp 63 dz d91 c103 0 c192 drA d4 The first structure equations then give us d61 0 iwleydlpinAdZ c192 drAdzpw12Adriw23A dz d630w13Adrw23Ardzp A little thought leads us through the following o U13 is a combination of dzp and dz by the first equation and of dr and dzp by the third Since dr d4 dz are linearly independent at each point we must have U13 a multiple of dzp only Write U13 ra dzp for some function a o The first and third equations now say that U12 hdzp adz 0123 adr cdzp for some functions I c o Plugging all this into the second equation yields drAdzp bdszdradzAdriadrAdzicdszdz 1 bdrAdzp2adrAdzcdszdz 0 o The linear independence of the above three 2 forms at each point guarantees that all coefficients are zero so that a c 0 and h 71 thus recovering wij The point of the above exercise is to see that the connection l forms of a moving frame are gen erally determined directly by the forms 0 In practical examples surfaces later for instance the as are synonymous with the induced metric of the surface This method of imposing a metric choice of first fundamental form and calculating the connection l forms from it is critical in physical appli cations we shall do a little of this later It will be seen that the Gauss curvature of a surface can be calculated directly from the connection l forms The generalization of this to higher dimensions is the method by which the Riemann curvature tensor of the Levi Civita connection of a metric is calcu lated Indeed the relationship between a surface manifold metric connection and Gauss Riemann curvature is precisely what Physicists are talking about when they say that Spacetime is curved The structure equations are important in the same way that the Frenet Serret equations are im portant for curves they tell you everything there is to know about a moving frame As such it is a standard method in differential geometry to use the method of the moving frame reducing geometric problems to differential equations The trick of course is to choose your frame so that the equations are not too difficult The following Theorem may be regarded as the analogue of the Fundamental Theorem of biregular spacecurves Theorem 29 Let the domain U be simply connected Given forms wjk satisfying the second structure equa tions and given aframe e1pe2pe3p at a point p E U there exists a unique movingframe e1e2e3 on U which agrees with the given frame at p and has the wjk as connection forms Furthermore if we are also given forms 6k satisfying the first structure equations and one specifies xp then there is a unique map x U a E3 with xp as specified andfor which 6k dx ek The structure equations in other dimensions The first and second structure equations are equations relating 2 forms Since these vanish when m 1 the structure equations tell us nothing about curves In the case of maps x U C IV gt lEZ any moving frame has only two directions e1e2 hence there is only one connection l form U12 The first structure equations then reduce to d61w12A620 deziquQl When m 2 we have 91 92 forming a basis of l forms at each point hence U12 L191 1192 for some functions u b Plugging this into the first structure equations yields d91 91A920 d62b91A92 Since d01 c192 are also multiples of 91 A 02 it is clear that ub are uniquely determined by the first structure equations and thus so is U12 The second structure equations simply read dwlz 0 In higher dimensions the first structure equation stays exactly as in 23 while the second be comes only slightly more complicated with 24 being replaced with Clwly 2 wk A wk 0 k rj Understanding this one equation in a given geometric context is the key to understanding that ge ometry Group theory and differential geometry One of the main places group theory appears in geom etry is in the study of moving frames We have seen that knowing a moving frame together with the l forms 9 is equivalent to knowing the map x Thus it is often desirable to study the frame itself as a single object but where does it live In our examples we are think about the frame E e1 e2 e3 as taking values in the set of 1 X 3 matrices whose three entries form an orthonormal basis of E3 If instead we think of the e as column vectors with respect to some indeed any fixed basis of E3 then E may be viewed as taking values in the Special Orthogonal group SO3 As such the study of maps into lE3 is often reduced to the study of maps into SO3 This idea can be generalized and many different groups can be considered as the universe of choice for a problem 23 Surfaces and moving frames De nition 210 Let x U gt lE3 be an oriented local surface An uduptivefrume to the surface is any moving frame st e3 U Note that there are an enormous number of adaptive frames since we are still free to rotate the other two basis vectors about U In an adaptive frame 93 0 and thus dx e191 e202 The first and second structure equations now become the structure equations for u surface d61w12A92 0 deziquQl 0 U13 A 91 0123 A 92 0 The Symmetry equation dwlz 7 U13 A 0123 0 The Gauss equation dwis w12 A wzs 0 ClUng 7 U12 A U13 0 First structure equations The Codazzi equations Furthermore we have 93 91 Figure 3 An adaptive frame on the sphere Proposition 211 In an adaptive frame I 04 9 II 791w13 920123 Proof 1 dx dx elel eze22 9 e 11 7 dx dU 791e1 ezez dE3 791e1 9262 1w13 920123 7910J13 920123 Example Consider the sphere X2 y2 z2 1 parameterized by X sin 9 cos 4 x y sianinzp 6 E 071 416 0271 2 cos 9 together with the moving frame cos 9 cos 4 7 sin 4 sin 9 cos 4 e1 costinzp e2 coszp e3U sinesinzp 7 sin 9 0 cos 9 Here dx e1 d9 sin 962 dzp so that 91 d0 92 sin 9 214 It is easy to see that 0112 i cos9 d4 0113 d9 0123 sinedzp from which you can check that the structure equations are satisfied Moreover using the 0 tag we see from the proposition that I dQZ sin 92 dzpz I 7d62 sin 92 dzpz which are exactly the expressions obtained in 162A This frame is plotted in Figure 3 12 Since x is a local surface we have that the differential dx is a 1 1 linear map at each point ie dxl Tpll 7gt tangent vectors to x at p is a bijective linear map for each p It follows that 01 92 form a basis of the space of l forms at each point This suggests the following Lemma 212 There exist unique functions a h C st U13 a61 1792 U23 1791 C92 Proof That w13a01 1762 0123 1391 C92 are linear combinations of 9102 is automatic The sym metry equation then implies that 0w13A91w23A027bh61A62 The above gives us that ll 7a0 7 210192 7 C9 Proposition 213 The Gauss and mean curvatures are given by Kac7hz H7ac Proof First we construct a dual basis to 01 02 The map 2 gt gt dxz is invertible at each point so let v1v2 be the vector fields on U defined such that dxv39 e for i 12 Writing dx e191 e292 implies that 1 i j 639 v39 3 1M 1 0 iy j v1v2 are orthonormal with respect to I Now we solve for the Gauss and mean curvatures with respect to the basis v1v2 at each point The matrices of I ll with respect to this basis are the identity and lt 1 5 respectively We find the principal curvatures eigenvalues of ll with respect to l by solving 0detltltZ 1 726 aACAb2A2aCAaC7b2 Recalling that A 7 k1 A 7 k2 A2 7 ZHA K gives the required expressions for K and H Since our exterior calculus doesn t mention co ordinates at all it is clear from this proposition that K H are independent of any choice of co ordinates on a surface With our expressions for U13 0123 we have U13 A U23 a61 b92b91 C92 QC 7 17591 A 92 from which follows Theorem 214 The Gauss equation is equivalent to ClUJ12 K91 A 92 The theorem often gives a faster method of calculating the Gauss curvature of a surface than using the linear algebra method from 162A In particular you should observe that we need only calculate 1 forms that are related to the tangent part of the moving frame 01 92 decompose dx in terms of the tangent vectors e1 e2 while U12 describes how that tangent vector fields e1 e2 change with respect to each other The unit normal 33 doesn t need to be considered or calculated Here are a couple of Examples 1 With the unit sphere dwlz sinQdQA d4 91A92 K1 2 The unit cylinder parameterized by co ordinates 4 2 has the moving frame 7 sin 4 0 e1 coszp e2 0 e3 e1 X 32 don t need 0 1 from which we see that U12 0 so that dwlz and thus K are both zero coshucoszp xuzp coshusinzp 3 Consider the catenoid u A moving frame here is 1 sinh u cos 4 7 sin 4 1 7 cos 4 317 sinhusinzp e2 coszp 337 7sinzp cosh u cosh u 1 0 s1nh u Again we needn t have calculated 3 Thus U12 e1 dez 7 tanhudzp But here dx cosh ue1 du 32 dzp so that 91 cosh u du and 92 cosh u dzp The Gauss equation says that dwlz 7sech2uduAdzp 7sech4u91A92 x K 7 4 cosh u As an example of how exterior calculus is often used we re prove a theorem from 162A Theorem 215 If every paint an a surface is umbilic then the surface is part of a round sphere Proof We have ll Al where A might be a function It follows that with respect to an adapted frame we have a C 7A and h 0 Indeed U13 a91 and 0123 a02 Taking exterior derivatives and appealing to the Codazzi equations gives us 0 dw13w12Aw23 daA61ad61aw12A02 daAQl 7aw12A62aw12A62 daAQl Similarly da A 92 0 Since 01 92 form a basis at each point we have da 0 and so a is constant Now define c x 7 eg and calculate 1 dc 91e1 ezez 7 5w13e1 wzgez 0 x thus lies on the sphere of radius a 1 and center c 3 Isometry Gauss Theorem Egregium and Riemannian Geometry In this section we consider isometries both local and global and their effects of curvature We fin ish by discussing Riemannian Geometry how the imposition of an abstract first fundamental form induces curvature 31 Invariance under Euclidean motions There are two types of isometry to consider when it comes to surfaces De nition 31 Two surfaces x are globally isometric if they differ by an isometry of E3 That is Q Ax b where A is a constant orthogonal linear transformation3 and b is a constant vector Two surfaces x 2 are locally isometric if their first fundamental forms are identical l I In the first case the isometry is direct if det A 1 and indirect if det A 71 When the isometry is direct we often say that x 2 are related by a rigid motion or Euclidean motion In the second case when l I it is usual to say that x 2 are simply isometric Theorem 32 Let x U a E3 and 2 U a E3 be two surfaces related by a direct isometry 2uv Axuv b Then the fundamental forms of the two surfaces are equal and thus so are the two measures of curvature If the isometry is indirect then the Gauss curvature is unchanged while the mean curvature changes sign Proof Clearly d A dx Since A preserves dot products we have identical first fundamental forms Matrices of positive determinant preserve orientation and those of negative determinant reverse it so we have 0 AU direct isometry iAU indirect isometry Thus dU iA dU It follows that ll ill with a minus sign iff the isometry is indirect The converse is also true though a little harder to prove Two surfaces with equal fundamental forms can for most purposes be considered equivalent Lemma 33 Let x U a E3 be a local surface with first fundamental form I Let 91 62 be lforms on U such that l 6 9 Then there exists a unique adaptive frame such that 6 dx ei Proof Let 111112 be the dual vector fields to 0192 Then e dxv39 i 12 and eg e1 gtlt e2 defines the frame I Theorem 34 Bonnet Suppose that x 2 U a E3 are local surfaces with identical first and second funda mental forms Then x 2 di er by a Euclidean motion 33 gtlt 3 orthogonal matrix with respect to any fixed basis of E3 Proof First we show that there exist moving frames for which the connection l forms and forms 9 are identical for both maps We use these frames to define an orthogonal matrix A use the structure equations to show that A is constant and then calculate that 2 7 Ax is constant With respect to a choice of adaptive frame E e1e2e3 we have I 9 9 By Lemma 33 there exists a unique moving frame E 1 z 3 of gt2 with 9 9 i 12 Writing the identical second fundamental forms in terms of coefficients a b C E f we have 11 7a 7 213992 7 e g 719 7 2179192 7 c9 11 Evaluating these on the pairs 111111 111112 112112 respectively yields 9 L1 b E C It follows that 213 U13 and 223 U23 Now consider the first structure equations for both surfaces since 9 9 we have d910cA920 d9270cA91 for X U12 and 2112 respectively It follows that E z 2112 7 U12 satisfies 9 A 91 0 E A 92 Hence 8 0 We thus have that the moving frames E E have identical connection l forms of 0 Define a possibly non constant orthogonal matrix A by E AE Ae1Ae2Ae3 Then dE dAE AdE dAE AEw dAE Ea It follows that dA 0 and A is constant Finally note that d 9121 92 2 A91e1 92632 Adx dAx Thus gt2 Ax b for some constant vector b and orthogonal A note that detA 1 since both E E are oriented frames I We could have appealed to Theorem 29 to see that gt2 is determined from of w and 9 9 by initial conditions 32 Gauss Theorem Egregium We have already used differential forms and moving frames for a new method to compute the Gauss curvature We can improve on this even further so that you need not even compute the moving frame 1 Find 1 2 Write l 9 9 Do this by inspection or by finding an adaptive frame 3 Solve the first structure equations for U12 4 Use the Gauss equation to find K This process is in most cases much easier than trying to solve for the eigenvalues of ll with I Not only that but you needn t calculate the moving frame if you can find 91 92 by inspection A Lemma below shows that it is always possible to write a first fundamental form as 9 92 16 u cos 4 Examples 1 A surface of revolution x sin 4 1o wowaMMW b 91 92 W cmquot c d61 0 70112 A 92 6192 1 d A dip wlz A 61 Hence f u i d wiz 1fu2 47 d Here we have f u f u d 7 d Ad 7 9 A9 12 1fu23 4 fu1f u23 1 239 7 f so thatKi 7W X2 2 Letxxy y thenl 4x2dx2 1 4y2 dyz Choosing 91 Zxdx and 92 ii 4y2dy 2 we have d01 0 deg By the structure equations U12 0 and so K 0 Since the first fundamental form encodes what angle and length mean for inhabitants of the sur face the local geometry of isometric surfaces is the same for an ant living on said surfaces By an ant inhabiting a surface we mean that the ant has no notion of what outside the surface means or normal to the surface all his experience comes intrinsically from the surface The following theorem says that the Gauss curvature of a surface may in principle be detected by an ant Theorem 35 Gauss39 Theorem Egregium Isometric surfaces have the same Gauss curvature Proof If l I then by Lemma 33 we may choose the same 9102 for each surface By the above method we get the same K in each case I De nition 36 A surface with K E 0 is called at Example A cone can be opened out and laid at as part of a plane hence K 0 The same is true for a cylinder Algebraically the cone below is sliced along the line 4 0 and laid at as in Figure 4 Z C05 47 2 cos 7 azsinzp gt gt1 2 39 j12 Z 2 sin W4 Conversely to the fact that the Gauss curvature is intrinsic detectible by our hypothetical ant the mean curvature is not intrinsic For example even if our ant decides that his home is at he has no local way to tell whether he is crawling on a plane a cone a cylinder or some other at surface Figure 4 Unwrapping a cone A corollary of the Theorem says that no part of a sphere is isometric to part of a plane This has applications to map making for it implies that the perfect map is impossible any map of part of the Earth must distort distances in some way The Theorem implies that I gives us K without needing to know lI Since the usual formula K 57quot involves lI there must exist a formula for K involving only E P G and their derivatives This formula is very complicated and is given below Suppose that X1X2 are co ordinates on x The Gauss curvature can be written in terms of the Christa el symbols Let the first fundamental form be written I E d 2Fdx1 dxz G dxg gm d g12 g21dx1 dxz gzz dxg where gm gm and define gif to be the ij th entry of the inverse matrix 71 If we define the Christoffel symbols by v 1 2 v agm39 3g ag39k z 7 7 1m J mk 7 7 J39kizmgllg Back 3x megt then the Gauss curvature is given by K 7 7L LE 7 Lr r r3 7 rh r r 7 r rz gt gm 3X1 3X2 2 2 2 2 22 Working with curvatures in terms of Christoffel symbols is very popular in Physics Good luck if that s where you re going 33 Riemannian geometry The idea of Riemann geometry is to consider a domain LI and specify an abstract first fundamental form or metric without it necessarily having arisen from a map x LI gt IEquot By the above procedure it makes sense to talk about and calculate the Gauss curvature 0ftlie metric In this situation there is no mean curvature or second fundamental form being no unit normal vector to U there is simply no notion of either H or H To do this in the abstract without any notion of a moving frame or a map x we need a Lemma Lemma 37 Any first fundamental form curl be written us I 9 I egfar some lfarms 9 Proof Suppose that I o dx2 2b dx dy cdy2 written in co ordinates xy Then I lt dx dygtz WW lt suffices to check that c 7 bZo gt 0 This is equivalent to ac 7 b2 gt 0 which asserts the positivity of the determinant of the matrix of I in these co ordinates However first fundamental forms are positive definite so the eigenvalues of this matrix are both positive and thus so is the determinant I There are in fact an infinity of possible choices of 0 the above is just an example A basic example of Riemannian geometry is hyperbolic space De nition 38 Twodimertsiorml hyperbolic space is the upper half plane equipped with the metric 1ltdxgt2ltdygt2 yz Theorem 39 Hyperbolic space has K 71 Proof Take the obvious choice 61 y l dx 62 y l dy and follow the recipe Thus if U12 o dx b dy then 0 d61 U12 A 62 y 2 oy 1dx A dy 0 do2 7 0112 A 91 by ldxAdy so that U12 7y 1 dx Then dwlz y zdy A dx 761 A 92 Thus K 71 Hyperbolic space could be considered as the negative analogue of the sphere Why cut off at y 0 Consider a curve zt 0 1 7 t where t runs from 0 to 1 The length of the curveis t t t lZZClT1ClTCl f 0 0 y o 17T 7ln17t7gtoo ast7gt1 The X axis is thus infinitely far away from all points in hyperbolic space We will think about hyper bolic space in the next section on geodesics Finding metrics from a prescribed curvature Since the curvature is determined entirely from the metric we can ask what metrics give a particular curvature This is useful in practical examples because we often want to design a metric that will have particular curvature properties For example Find all the metrics of the form I fr2 dr2 r2 d92 on lR2 which have constant Gauss curvature Here we have 91 d7 92 rdQ It is not hard to see that U12 7 d0 and that i f 7 K wmy We want this to be constant hence integrate f 7107 1 7C7K72 f 73 T where C is constant Solving gives f 72 7 i1 2 c172 72 d6 1 V i 39 163 An example with a singularity Consider the metric I 1 7 2 d72 1392d02 on lR2 with the origin removed Here we choose 61 1 7 2 dr and 62 rde Then writing 0112 g dr I de we have 0 d61 w12A62 ardrAdQ 0 d9 7 ule 91 1 bx1 r2dmd9 Thus U12 71 7 2 12 d0 from which we have 774 1 7 22 Hence K 7 1 72 2 While it looks like the metric has a singularity at r 0 in fact the curvature has limit 71 there In this example we can Visualize the metric as being the first fundamental form of the surface 2 f r 9 ln 7 which certainly has a singularity at r 0 Thus singularities in surfaces do not necessarily correspond to singularities in the Gauss curvature dwlz 1272r31 r z 32dr A de 7 7 91 A 92 An example with a singularity on a curve Consider the region of the plane outside the unit circle equipped with the first fundamental form I 1 7 r1 c172 72 c192 For large 739 this is close to the standard Euclidean metric but as we get close to r 1 the first term blows up Here 91 V 1 7 7 1 dr and 92 rdQ Writing U12 g dr I de and applying the structure equations 0 d61w12A62mdrAd6 0 d627w12A61 1 bx1 771drAd6 so that U12 71 7 FHA2 d0 Thus 1 dw12 17r 1 327 dede Zemez 1 27771 20 so that K 270171 For large 7 the curvature is close to zero but as r gt 1 the curvature goes to 00 This example cannot be written as the first fundamental form of a surface 2 f r 9 A very different curvature is obtained if you start instead with the metric I 1 77 1 1d72 rzdez on r gt 1 This time we can see thatK 7 Black holes and Relativity The previous example can be viewed as a simpler though far from identical version of the following example which is very much for the Physicists Spacetime The following example is a metric on 4 dimensional spacetime written R For us this is identical to lR4 but equipped with a first fundamental form I which has 3 orthogonal spacelike directions g and one timelike direction A tangent vector v E Tle3391 is spacelike lvv gt 0 timelike lvv lt 0 lightlike lvv 0 l is like a standard first fundamental form in all ways except that it is not positive definite indeed l lt 0 etc The standard at metric afspacetime is I dxz dyz dz2 7 czdtz where C is the speed of light4 If you think about a tangent vector i a a a a v 7 ma vya 111 mg then the speed of a particle moving with tangent vector v is the infinitessimal spacial change divided by the time change ie A 032 v 21 speed lvtl Evaluating gives us lvv speed2 7 C239Ut2 from which we see that spacelike gt C v is timelike ltgt speed is lt C lightlike C One of the principal ideas of relativity is that physical objects can only travel at speeds less than that of light ie they must travel only in timelike directions 4Alternativelyl drZ 1392 d92 1392 sin2 9 d z 7 L2 dt2 in spherical polar coordinates 21 The effect of mass on spacetime The above standard spacetime metric is the background metric without the presence of any mass or as a limiting case for deep space where gravity is negligible It is the metric of spacetime used in special relativity In general relativity the effect of mass energy is considered It affects spacetime by changing the metric Here is a model of how this might happen Consider a rotationally symmetric object of mass rn and fix a co ordinate system r 9 4t of spacetime centered at the star spherical polar co ordinates together with time The Schwarzschild metric is defined outside the object by 71 1 lt1 7 23 dr2 72 c192 r2sin26dzp2 7 c2 lt1 7 23m dtz c r c r where G is the gravitational constant and c the speed of light If we remove the mass the the metric is simply the standard at metric Thusfar we have only considered the Gauss curvature for surfaces or 2 dimensional domains The Gauss curvature is the 2 dimensional avatar of the more general Riemann curvature tensor Although we have not intro duced this object its entries5 may be computed in an analogous way to how we calculate the Gauss curvature in 2 dimensions ie from the l forms 9 such that l 9 9 9 7 9 we can use the structure equations to find the 6 connection l forms 01 1 g i lt g 4 the curvature tensor being the components of the matrix valued 2 form6 Q dw w A w This is the essence of the notion that massenergy curves spacetime While we re not going to calculate the full curvature of the Schwarzschild metric we can calculate certain sectional curvature the Gauss curvature of a 2 dimensional subspace of Tle3391 In particular we can calculate the sectional curvature of the space lt gt g Tle3391 which is equivalent to calcu lating the Gauss curvature of the metric 71 17mm dr27c2 17ml dtz czr czr on lRZ This can be seen to be K A homework question walks you through some of this By analogy with the previous example the sectional curvature of the r 0 plane is K 7 The Schwarzschild metric is only valid outside a mass thus at the surface of the Earth r 6 X 106 meters the metric is approximately I 1 10 9 dr2 r2 d62 r2 sin2 6 d42 7 1 7 10 9c2 dtz differing from the standard at metric by only a tiny tiny amount Even at the surface of the sun the difference from the standard metric is only 425 X 10 6 dr2 7 c2 dtz and the sectional curvature calculated above is only K 88 X 10 This very slight difference from at K 0 space is just detectable In a famous 1919 experiment of was Eddington and Dyson the light from stars very close to the sun was observed during a total eclipse and the bending of light was found to be within observational error of Einstein s prediction This combined with general relativity s prediction of the correct precession of Mercury s orbit helped convince people that Einstein had a good theory The process of light bending around massive objects is known as gravitational lensing The Schwarzschild metric was one of the first explicit non trivial solutions to Einstein s equations of general relativity to be found Analyzing the curvature of this metric allows Physicists to model 5In n dimensions it has nznZ 7 1 12 independent components 6In Zdimensions w 7212 06 so that Q 73 d3 K 701 1 91 A 92 22 what an object might experience as it fell towards a huge mass does it stretch does its volume change etc At the time the solution was found the physical significance of the seeming singularity at the Schwarzschild radius r R z 2 was not understood First of all for the metric to be valid at such a radius requires all the mass of the object to be concentrated in the region r lt R Since 27 m 1485 X 10 27 meters per kilogram the Schwarzschild radius is minuscule for non stellar masses for a human it s about 10 25 meters roughly one ten7billiontli the size of a proton while a Schwarzschild radius of 1 meter would have to contain 6733 X 1026 kg m 113 times the mass of the Earth An object who s mass is contained entirely within its Schwarzschild radius is a black hole Here are some of the predictions of the Schwarzschild black hole model 0 The region of spacetime given by r R is termed the event horizon As 739 gets very close to R from above all terms in the metric become dwarfed by the ClVZ term and the metric looks like it becomes singular The curvature does not however we ve already seen that the sectional curvature of the r t plane is finite at the event horizon 0 Suppose you are an observer sitting a long way from the black hole t is your measure of time Now throw a ticking clock towards the black hole An infinitessimal change dt in t corresponds via the metric to an infinitessimal change 1 7 Rr dt in the time on the clock As 739 7gt R the time change on the sacrificial clock approaches zero this is time dilation It takes an infinite amount of time as measured by the observer a long way away for the clock to fall in but according to the clock it takes only a finite amount of time o A 2 dimensional visualization of the shape of space around a black hole is available if you fix time and restrict to a constant angle 9 712 from the north pole The metric is then 1 7 71 ClVZ 72 dzpz This is the first fundamental form of the surface z 2Rx1 7 RV in polar co ordinates see Figure 5 1172 7 ma a Figure 5 Representation of a black hole at constant time 0 At the Schwarzschild radius r R radial and timelike directions switch sign in the metric This seems to make no sense until you remember that the metric describes spacetime from the point of view of an observer at rest The switching of the signs says that it is impossible to be at rest inside the event horizon there can be no stationary observers No matter what is powering your spacecraft it is impossible to resist the pull of the black hole even light cannot stay still 23 let alone escape from within Indeed if you pretend for a moment that classical mechanics applies the Schwarzschild radius is exactly the radius for which the escape velocity of a given mass is the speed of light 0 The center of the black hole at r 0 is a genuine singularity both of the metric and of the curvature At present there is no good physical understanding of what this might mean Since nothing can be observed with the event horizon of a black hole no one knows whether there really is a singularity of spacetime at the center There have recently been attempts eg Chap line s Dark Energy star model to meld quantum physics with relativity to explain what might happen inside the event horizon 4 Geodesics Parallel transport and covariant derivatives The concept of a geodesic is very important for applications It replaces the notion of straight line for surfaces and more complicated objects For example it is well known and we prove below that a straight line is the shortest path between two points in Euclidean space How would we go about finding the path of shortest length between two points on a surface 41 Geodesics in Euclidean space We want to find the shortest curve in lE3 joining two given points Let xt be unit speed such that xa A xb B Lete bea small number and y am gt lE3 be acurve such thatya yb 0 and x y 0 Define re X0 Yt Since y is orthogonal to the tangent direction of x the point r t is obtained from xt by a normal movement for each t Clearly r01 A and rb B so that for any choice of y we have a family of curves r connecting A and B see Figure 6 Figure 6 Family of curves r De nition 41 The curve xt has stationary length if the lengths e of the curves r t satisfy a 0 ae 0 for all choices of yt as defined above Theorem 42 The curve xt above has stationary length if it is a straight line Proof Throughout we ignore terms of order 62 or higher Since we are eventually evaluating ate 0 and at most are performing a single derivative with respect to e in the final analysis all such terms contribute zero to the final answer Wherever terms of order 62 or higher have been deleted we use the 2 sign Since the derivative of the curve r is given by r x ey its speed is Mt m V lx lz 26 5 62 WV 2 1 26 5quot 2 1 ex y where the final 2 uses the Taylor approximation Then 17 17 v tdt x ydt 0 ibiw dtibx dt t dt y i y 17 7 x ydt since x y 0 This expression vanishes for all y iff x is parallel to x However x has constant speed hence x x 0 Thus x 0 and so x is a straight line g Be 1 0 de The above is a calculus of variations argument Note that the proposition only shows a straight line to be of stationary length not of minimum length You can think about this as an infinite dimen sional calculus problem However of all the curves of unit speed between the points A B x is the only one that has stationary length The function Z smooth curves A w B gt lR is smooth on an open set so if there exists a minimum then said curve must have stationary length In fact the straight line is the minimum length curve joining two point We say that straight lines are geodesics in Euclidean space 42 Geodesics in surfaces We want to find a unit speed curve xzt which is the shortest curve between two points in a surface x Again we consider curves of the form ret XZt 6370 where ya y17 0 and y U 0 re is thus a curve in lE3 which is perturbed a small amount from xzt in the tangent direction to x at all points Note that re is no longer a curve in the surface when 6 51 0 De nition 43 xzt has stationary length if for all such y g 7 0 d6 0 xzt is a geodesic if x Oiit22xzt is normal to the surface everywhere Theorem 44 A unit speed curve lying in a surface has stationary length if it is a geodesic Proof The argument begins in the same way as in Theorem 42 recall throughout that x xzt etc HZ d l7 l7 7 7 t dt dt 36 0 d6 0 v x y b d b A axydt7Z x ydt x 177 b if b a 7 y x ydt7 x ydt where this time the first term vanishes due to ya yl1 0 xzt has stationary length iff this last term vanishes for all y But y U 0 hence x cannot have any component tangent to the surface x is thus normal I A little more is true the definition of geodesic forces it to have constant speed Indeed if xzt is a geodesic then lx lz 2x x 0 since x is normal to the surface and x is tangent It is just common practice to parameterize geodesics by unit speed Example On the sphere geodesics are great circles radius equal to that of the sphere These mini mize length if we are less than half way round The long great circle between two points is a saddle of the length function Here s how to calculate this naively Suppose that xzt is a geodesic Since x is a sphere we may orient things so that that x 7U where r is the radius of the sphere The geodesic condition then reads X f x wherefis some scalar function7 Take dot products of with x to getO x x lx lz Hence xzt has constant speed 390 Now take dot products of with x to obtain y2f XH x d 2 i 2 dtxx7 x 7 7v so that f is constant and indeed 2 v a 7 x 7 772x Finally the vector k x X x is constant k x X x 0 thus x lies in the intersection of the constant plane kL and the sphere the definition of a great circle 7fx is multiplication of the value off by x evaluation which makes no sense Contrary to 3 w m sanly a w We ve goms minimums s s L y w m mm s pmjeeuon Fxgum The great emle from Irvme CA to Irvme Scotland Wenow consider geodesics m terms of the structure equations Theorem 45 Z2 x u a 133 be a snjfzoz 07 which a mavivzgfmme ha xzl Iyivzgivz the smsz is a gma39zs c 9m satis es lhefo owivzg aquariums lt911u42z 92z 0 ag 7 magma 0 9112 922 chmesyenheachmeMappemcp Hg m2 cm yThus xmgthe Home M23 Geodest equauons nergy equahons s 1122 chosen 9 A unit spem39 awe Proof The energy equation is just the unit speed condition 12 2 1 Now a curve having x normal to the surface is equivalent to e1 x 0 e2 x Expanding the first as e x 7 e x yields 0 e1x e1x7 e1x e1 dxz 7 de1z dxz 7 gm wultz gtezltz gt e2 x 0 similarly expands to the second geodesic equation I Any parameterized curve xzt satisfying just the geodesic equations ie x normal to the sur face automatically has constant speed observe that the geodesic equations force game 622 7 o The energy equation simply normalizes this speed to be 1 In practice geodesic often simply refers to a non parameterized curve as a subset of the surface which if parameterized by arc length satis fies our definition Proposition 46 Given a point p on a surface and a unit tangent vector v to the surface at that point there exists a unique geodesic through p in the direction v equivalently xz0 p and x z0 v Proof This is a consequence of the usual theorem of existence of solutions to ODE s the geodesic equations are a pair of second order ODE s I Given two points on a nice surface there exists a geodesic joining those points Nice in this situation essentially means that you wouldn t want the geodesic to leave or touch the edge of the surface Certainly any two points on a handlehody a complete surface closed with no edge and no self intersections may be joined by a geodesic However you ll need good luck to succeed in computing many of these explicitly 43 The surface of revolution As an example we consider a standard surface of revolution f cos 4 xu 4 sin 4 u We already know that 91 xl f1 du 92 fdzp and U12 7 dzp For all curves xzt we have u a a 7 i i Z Bu 4 34 Thus 912 7 i1 au 922 7 we WW 7 iv 1 12 28 The geodesic equations and the energy equation thus become amt 7 223 0 0415 fuu 47 0 1 HEW f247 2 1 Observe that the lines 4 constant lines of longditude are geodesics the second equation is trivially satisfied while the 1st and the energy equation become the same first order ODE for u which by the usual theorem from ODE s has a solution The lines of latitude u constant are geodesics if only if f fuzp z 0 which is iff f1 0 the remaining equations allow one to solve for Thus lines of latitude are only geodesics at critical values of the radius Notice that since f fuu the second geodesic equation is equivalent to A fazp constant Substituting back into the energy equation we have 2 1 f5u 2 1 Hence lAl g f This shows that a geodesic starting at a given point which comes equipped with a value for A is confined to regions of the surface where f 2 A This means that geodesics can bounce off narrow necks of surfaces of revolution It s not advisable to try to solve these equations explicitly for many surfaces Even for the cone it is extremely messy The cone is one that we can manage If you let f u where a is constant then it can be seen that A2 t B 2 xl 2 t B uti i 4w tan 1 a C a2 1 a2 a A 1 g2 where A BC are constants Normalizing B 0 C and fixing u gt 0 we see that lAl and u increases as t moves in either direction 4 keeps increasing as t does but never quite makes it to angle V 1212quot If you take a shallow cone a 1 is the cone 22 X2 yz then the geodesics have no self intersections However once a lt i the geodesics will begin to meet round the far side of the cone Figure 8 illustrates a geodesic with a Note that this is clearly not the shortest path between the two points on the geodesic at the bottom of the picture indeed as a decreases and the cone becomes more sharp the number of distinct geodesic paths between two points increases 44 Riemannian geometry Since the geodesic equations depend only on quantities derived from the first fundamental form I we can apply them to abstract first fundamental forms as we did in the section on Riemannian geometry Notice first that isometric surfaces xy have the same geodesics in the sense that if xzt is a geodesic then so is yzt Geodesics are extremely important in Physics For example light is always assumed to travel along the shortest path between two points If your geometry is that of refraction between two mediums for which the speed of light is different you will obtain Snell s law If your geometry is 29 2 W V gg a aza y El V Figure 8 Geodesic on a cone some curved spacetime such as that of the Schwarzschild metric then light is seen to curve Indeed the concept of motion along a geodesic is the relativistic replacement of Newton s first law instead of a body moving at constant speed along a straight line if unaffected by external forces a body will move along a geodesic in spacetime This can be difficult to visualize when you have curved spacetime because the notion of speed is different in spacetime and normal space indeed a curve with zero acceleration in curved spacetime will when translated into 3 dimensional space not have constant speed In at spacetime however geodesics are straight lines and these translate to straight lines in our 3 dimensional reference frame Indeed if a curve has zero acceleration in at spacetime and constant speed ikz then its speed in Euclidean space will be 4 C2 7 k2 11 where 39Ut is the nec essarily constant t component of the tangent vector to the curve in spacetime First we consider the upper half plane model of hyperbolic space The geodesic and energy equa tIOl lS become 36 xy V 7 T 7 I Notice that the first geodesic equation is equivalent to A 3 constant Taking A 0 gives us X 0 i y i1 i y Ceit Hence vertical straight lines are geodesics Otherwise first rearrange the energy equation yZ yZ 1 W W szT Now observe that so that dy 2 1 dy ZiliAzyz 1Iltagt Azyz lta Azyz Solving this we have dxAih17Azy2xic 30 where C is a constant Squaring and rearranging yields X 7 C2 y2 A72 thus circles centered on the X axis are the other geodesics see Figure 9 Figure 9 Geodesics leaving a point in the hyperbolic upper half plane Geometry in this picture is a little strange For example if we wanted to talk about a triangle the only sensible definition is that its edges be geodesics since geodesic is what we mean by straight in this geometry Figure 10 gives an example Figure 10 A geodesic triangle Geodesics are infinitely long in hyperbolic space Given two points there exists a unique geodesic joining them The situation is like that of lE2 except that Euclid s quotparallel postulatequot does hold Given a line geodesic and a point not on the line there exist many geodesics through the point not meeting the original line Euclid s postulate was that there is only one such parallel line Hyper bolic space shows that the postulate cannot be proved from the other axioms of Euclidean geometry If you know a little complex analysis specifically Mobius transforms then it can be shown that the geodesics for Poincare s disc model of hyperbolic space given by the metric 4 WUZ drz 72 c192 on the unit disc are arcs of circles which intersect the edge of the disc at right angles The Circle Limit artworks by MC Escher illustrate this geometry very nicely Figure 11 is Circle Limit If you look at a string of angels and demons connected head to toe you ll see that they re lined up along one of 31 gure 11 MC Max s 04me h a he hypenmhe meme he mdwxdual gures a have he same 1eng h and area As Covarianl derivatives and geodes39cs vedm elds A he hean of he 5 he xdea hat m a Vader space we have an mhexem mum of ummhg a Vader along a eme gust shde so ha pom m he same dhecmh When mu the correct thing to do in this situation De nition 47 Suppose that 7t xzt is a parameterized curve in a surface x U gt lE3 A vectorfield v along 39y is a smooth assignment vt dxXzt of a vector tangent to the surface at 7t here Xzt is a vector field X on U restricted to the parameterizing curve zt The assignment is smooth in the sense that the function t gt gt vt is infinitely differentiable De nition 48 Let v be a vector field along a curve 39y in a surface The covariant derivative of v is the vector field Div 7TT dxv 7TT dz dxXzt 71Tv where 7TT is orthogonal projection onto l the tangent space to the surface at each point We differentiate the vector field v with respect to the parameter t and then project back onto the tangent space Example If 7t is a curve in a surface then 39y t is a vector field along 39y The covariant derivative of 39y is then HT39y t often written Dw39y De nition 49 A vector field v along a curve 39y is parallel if its covariant derivative is zero everywhere By appealing to the above example we see that we have proved the following Theorem 410 A curve 7 in a surface is a geodesic if the vector field 7 is parallel along 7 Let vt dxXzt be a vector field along a curve 7t xzt If we choose an adaptive frame e1 e2 for the surface then there exist functions at ht such that VG tE1Ztbtezztr in which case the covariant derivative may be written D v nTgvo do blttgtwultzxtgtgtgtelltzlttgtgt Iat alttgtw21ltzxtgtgtgtezltzlttgtgt It follows that v is parallel along 39y iff the coefficients a h satisfy the differential equation a 7 0 7w12z a b 7 01122 0 b Note that w12z is simply a function of t and so we may define a function gt w12z dt It is then easily checked that the solution to the above differential equation with initial condition prescribes att to is at 7 cosgt isingt at0 1 Mt T singt cosgt ht0 We have thus proved the following theorem Theorem 41 Given a smooth curve 7 in a surface x and an initial vector v0 tangent to the surface at 7t0 there exists a unique parallel vector field v along 7 such that vt0 v0 De nition 412 The vector field v is above termed the parallel transport of v0 along 39y The concept of parallel transport is exactly what sliding a vector along a curve means inside a surface Indeed it is clear that lvt 2 a2 b2 71 7 11 is constant so that the length of a parallel transported vector is constant Moreover it can be seen see homework that the parallel transport of a vector along a geodesic has constant angle with the tangent field to that geodesic Examples 1 If we jump dimensions for a moment and think about the covariant derivative in E3 then D is simply differentiate then project onto the tangent space at each point But this tangent space is exactly lE3 itself so the projection is the identity map and D is just normal differentiation In this case let 39y be a curve in E3 and v0 a vector at 70 We may take our moving frame to be the standard basis e1e2e3 the connection 1 forms 07 1 g i g 3 of which are all zero Comparing with the differential equation 1 we see that the coefficients of the parallel transport v of v0 with respect to the standard basis are all constant Thus v v0 and parallel transport really just moves the vector v0 keeping it parallel and of the same length sin 9 cos 4 Now let x U 7gt lE3 be the unit sphere x9 4 sin 9 sin 4 and consider the parallel trans cos 9 N 0 sint 712cost port ofthevector v0 14 at the point39y0 along the curve 7t sint 7 712 sint 2 cost 7 712 7 sint 91 d0 92 sinedzp and U12 7 costzp Now zt 9tzpt t 712t so that z t Thus 01122 7 cost 712 sint Hence gt jot 01122 dt 1 7 cost Noting that 710 7 and 110 14 we apply equation to see that the coefficients of the parallel transport of v0 along 7t are cos t cost cos t sin t In terms of the moving frame on page 12 we have dx e1 d9 7 sin 962 dzp so that at 7 cos1 7 cos t isin1 7 cost bt 7sin1 7 cost 731cos1 7 cost so that the parallel transport of v0 along 39y is 7sintcost 7sint vt at 7sintsint bt cost 7 cost 0 Notice that the length of v is constant Figure 12 shows the curve and the transported vector field for 7712 lt t lt 712 It is not hard to show see homework that by parallel transporting around closed curves on general curved surfaces you can transform a tangent vector to any other Figure 13 shows the par allel transport of the red tangent vector 700T at the north pole down to the equator round the equator by an angle 10 and back to the north pole 700T has become the blue vector 7 cos 410 7 sin 10 0T Figure 12 A parallel transported vector field Figure 13 Transporting a vector field around a closed curve 46 Covariant derivatives more generally The concept of covariant derivative is much more general than that along a curve One may take any vector field on a surface differentiate it and project back onto the tangent space at each point However when you taking the covariant derivative of a general vector field the result is only a vector field once you specify a direction in which to differentiate Thus De nition 413 Let x C E3 be a parameterized surface and suppose that v dxY is a vector field on S ie Y is a vector field on U Let X be a vector field on U The covariant derivative of v with respect to X is the vector field DXv which may be written alternately DXv DX dxY nT dXv nT dvX nT dX dyx nTXYx As before we differentiate and then project onto the tangent plane at each point In keeping with our program of moving all calculations to the parameterization space we can define the covariant derivative of a vector field Y on U by X to be VXY where dxVXY DX dxY The operator D is often referred to as the LE39UZLCII39UZIM connection or just the connection of the surface x and V as the Levi Civita connection of the induced metric l dx dx on U The full curvature 35 tensor can be written in terms of V for any vector fields X Y Z on U we have that10 RXYZ VXVy 7 VyVX 7 VX1yZ KIXZY 7 lYZX where K is the Gauss curvature Why connection The covariant derivative tells you how to parallel transport tangent vectors along curves If we imagine two nearby points p a on a surface S joined by a geodesic then the covariant derivative defines an invertible linear map TpS 7gt TqS where a tangent vector at p is parallel trans ported to a tangent vector at a This connects the two tangent spaces Indeed a smooth choice of connection is equivalent to a choice of covariant derivative The Levi Civita is just one of many connections albeit one with very nice properties One of the best ways of thinking about D is that it is the natural restriction of the operator d to the tangent space at each point Indeed if we think about the equivalent of the moving frame equation dE Ew a 3 X 3 matrix equation we can write De1e2 631632 7112 0 82 For this reason it is common for to think of splitting up d into pieces D 7 11V 11 0 39 where D ll can be thought of as the matrices of the connection D and the second fundamental form ll If you differentiate a vector field D tells you how much points in a tangent direction and ll how much in a normal direction This idea can be used to describe differentiating any even non tangent vector field In higher dimensions when you have a k dimensional surface in Equot the matrix ll of the second fundamental form is k X rt 7 k and the 0 in the lower right of 0 becomes a rt 7 k X rt 7 k skew symmetric matrix of l forms which describes how to differentiate normal vector fields De nition 414 A vector field v dxY is parallel iff DXv 0 for all vector fields X on U The existence of parallel vector fields is of great importance in applications In higher dimensions the notion of a parallel tangent frame is very common for example in the case of a surface in E3 we would want e1 e2 to be parallel vector fields The generalization of the following theorem to higher dimensions is extremely useful Proposition 415 There exists a parallel tangent frame i K 0 Proof Suppose that e1 e2 are parallel then Del 7TT de1 920121 0 hence 0112 vanishes and thus so does K Conversely let K 0 Then dwlz 0 and so at least locally 0112 d f for some function Now consider the new tangent frame A A 7 cosf 7 sinf el EZ T el EZ sinf cosf gt Then D l 7 sinfdf sinfwlze1 cosfdf 7 cosfw12e2 0 hence l is parallel z is parallel similarly and we have a parallel frame I 10Here X Y X o Y 7 Y o X is a vector field on 11 known as the Lie bracket of X Y see homework 36 47 Geodesic curvature We can think of the curvature of a curve in lE2 as measuring how much the curve deviates from a straight line Since a geodesic replaces the concept of straight line on a surface we can look for a measure of how much a general curve deviates from a geodesic De nition 416 Let 7t be a unit speed curve contained in an oriented surface Then 39y t is a unit length vector field along 39y Indeed the covariant derivative of 39y must be orthogonal both to 39y and to the unit normal U of the surface Hence D 7 KgU X 39y for some function Kg called the geodesic curvature of 7 Note that the sign of Kg depends on the orientation of the surface changing orientation switches the sign of U It is clear that 39y is a geodesic ltgt Kg is identically zero The above can be reformulated using cross products so that we never see the covariant derivative Indeed for any curve unit speed or otherwise the formula becomes 7 X 7 U W l3 i This is reminiscent of the formula for curvature of a spacecurve see homework for a proof The geodesic curvature of 39y should be Viewed as the curvature of the curve detectible to a dweller of the surface who knows nothing of Euclidean space and normal directions Indeed it is the curva ture of the curve that forms the best planar approximation to 39y at each point as the following theorem shows Kg Theorem 417 Let 7t be a unit speed curve in a surface x and let U U0 and T T0 39y 0 be the unit normal and tangent vectors to 7t at t 0 1 The geodesic curvature Kg0 of 39y at zero is the plane curvature at t 0 of the projection of 7 onto the tangent plane to x at 70 Le of the curve t 71370 7t 7 7tUU Moreover lKgl WW N The normal curvature 1910 of39yt at zero is the plane curvature at t 0 ofthe projection of39yt onto the space SpanT U Le ofthe curve t 7tTT 7t UU Moreover Kn 7 3 K2 K K3 Proof 1 7t 7 7tUU where is the scalar product Thus 39 39y 7 39y UU and 1 39y 7 7 UU Evaluating att 0 yields AV0 W0 Tr AW0 WNW 7quot01UU If we take the orientation of the tangent plane at t 0 given by the unit normal we have that IT U X T where I is rotate 90 degrees ie T U X T U form an oriented basis The plane curvature of the curve 39 at t 0 is thus WW WW 39IT 7 07quot01UU U X T Y 0 U X T T X 7quot0 39 U Kg0 Moreover since 7 is unit speed at t 0 we see that Kg0 WW MTV0 37 2 It is clear that 3970 T and 39Y 0 7 07 0U 7 0UU 39y 0UU Taking U to be the unit normal N for the curve 397 at t 0 we clearly have 3970 39y 0 UN so that 130 7 0U Now suppose that 39y cos zpel sin 1162 where e1 e2 are principal curvature directions on x Thus K0 7 0 U 7 cos zpw130 7 sin zpw250 However e1e2 are curvature directions thus w130 7k1 cos 4 and w250 7kg sin 4 so that 130 k1 cos2 4 k2 sin2 4 1910 is the normal curvature11 of 39y at t 0 3 At each point 39y HT39y 39y UU KgU X 39y KnU is an orthogonal decomposition hence Pythagoras gt K2 K K3 I Examples 1 On the surface of a sphere of radius r it can be seen that curves of constant geodesic curvature are all circles Kg 0 are great circles Kg a are circles of geodesic radius r tan 1r 1a 1 see the homework 2 Consider the curve 39y on the surface of the cylinder given by wrapping round a sine wave cos t 7t sint sin t We calculate all the curvatures 7 sint 0 7 0 cost I Wt 7W 7 X 7 71 cost 71 Since U cos t sin t 0 T is the usual outward pointing normal of the cylinder we have W Xv WU 7sth K WXVNl 7 3 7 1cos2t3239 7 7 3 7 1cos2t32 Kg It follows that the normal curvature is given by 2 2 2 27sin2t 1 K K 7K 4 n g 1 cos2 t3 1 cos2 t2 Since the surface is curving away from the unit normal the non zero principal curvature is 71 we must have Kn g 0 and so 1 K 7 n lcos2t 11We use Euler s formula from 162A We can check this in an alternative way using Euler s formula The angle 9 between 39y and the vertical is given by 0 cos9 7 l 0 7 COSt T i39li 1 T 1cos2t Since k1 0 and k2 71 we have Kn 7 sin2 9 71 7 cos2 9 71 cos2 t 1 as above 5 Integration We wish to develop integration in lRquot rather than lEquot so that the theory will be independent of Eu clidean structure Rather than integrate functions over open sets in lRquot we integrate n forms The advantage is that our story becomes co ordinate independent when changing variables n forms transform through scaling by the Iacobian of the transformation 51 Orientation De nition 51 An orientation of lRquot is a choice of which n forms are positive lf X1 Xn are the standard co ordinate functions on lRquot then the standard orientation is given by taking asserting that all positive multiples of dx1AAdxm are positive Co ordinates y1 yn are said to be oriented if dy1 A A dyn is positive From now on we will forget in the abstract that X1 xn are the standard co ordinate functions and just assume that these are oriented co ordinates on some open set U Now let to be an n form defined on U C lRquot Let X1 Xn be oriented co ordinates on U Then since the set of n forms on lRquot has dimension 1 at each point there exists a smooth function f U gt lR such that w fx1xdx1A Adxn De nition 52 The integral of to over U is defined as Uw Ufx1xldx1 dxn provided the integral on the right hand side exists Proposition 53 Let x1 xn be oriented coordinates on U C Rquot Let yl y U a R be di erentiahle functions Then dy1AAdyndetltgt dx1AAdxn J The determinant is the usual Iacohian matrix12 ie the determinant of the matrix whose i jeth entry is It l follows that the integral of an nform is independent of the choice of oriented coeordinates which we integrate with respect to 12This matrix is often written a Proof Recall that dyi 3 3 Since any two n forms differ only by a multiple and are multi linear maps we need only evaluate on a family on n linearly independent vectors Using the formula on page4we see that a a 7 By dy1 A Adyn EPWE 7 det 339 However this is clearly the same result as obtained by evaluating the right hand side of the formula in the proposition on the same n vectors Suppose thatw gy1yn dy1 A A dyn and thatfx1xn gy1 y lfy1yn are oriented co ordinates then the determinant above is positive From multi variable calculus we know that Agwhwmdypdy l mwn a det dx1 dxn The left hand side is the integral of 0 evaluated using the co ordinates yi while the right hand side since the determinant is positive without having to take absolute values is the same integral evaluated using the co ordinates Xi The integral jaw is thus independent of choice of oriented co ordinates I It is an obvious follow on from the proof that if you change the orientation of the co ordinates the sign of fu 0 changes The Proposition says that change of variables is built into this definition of an integral We simply have to be careful to change variables only to other oriented co ordinates Examples 1 Let U dx A dy U unit disk Then 1dd Awny ln polar co ordinates dx A dy 739 d7 A de and so 271 1dd AwiJOrrzpim 2 Let U WZHXA dy A dz and U is the positive octant xyz x gt 0y gt 02 gt 0 Then w w w w HV dx dy dz 1 u 0 0 0 52 Integration over surfaces We now know how to integrate 3 forms over open sets U C lR3 and l forms over segments of curves in lR3 We now want to integrate 2 forms over surfaces in lR3 We will see that a Euclidean structure is not necessary in order to integrate De nition 54 Letf U gt lR3 and g lR3 gt lR be smooth functions Then dg is a l form on lR3 The pullback of dg by f is the l form on U defined by fdg dg0f 40 Now let X Eiltj my dx A dxj be a general 2 form on lR3 The pull back of X by f is the 2 form 1N 2W 0ff 013939 A0 dxj 20117 of C109 of A C109 0f 39 39 iltj There is a little more exterior calculus going on here We have as is common brushed something about the exterior derivative under the carpet We ve said that if g U gt lRm is a function then for each p E U dgl7 Tpll gt lRm is a linear map It is more correct to say that dgl7 Tpll gt TANKquot le dgl7 maps tangent vectors to U at p to tangent vectors to lRm at gp The map g sends points in U to points in lRm thus dgl7 sends vectors based at p to vectors based at gp When n 1 it is standard practice to write Tle lR and treat dgv as a number rather than a tangent vector to lR Consider the following labelling of bases and attendant co ordinate systems required to describe a compound map f o g For example we choose a basis E1 Em of RV and co ordinates y1 ym lRm gt lR with respect to this basis eg y32E1 3E3 7 5E4 3 The identification Rm 2 Tlem is then Ej gt gt i By Functions U C R Lgt Rm Rquot d d Differentials Tpu ggt TgpIRm fgt TfgpIRn Basis vectors Ej Ek Co ordinate functions Xi yj Zk l forms dx dyj dzk a B a Tangent vectors 371 37 37k In what follows the tensor product notation dx denotes the rank one linear map Tpll gt TgplRm which sends 3 to 3 and all other basis tangent vectors to zero With respect to the bases E and 2k there exist functions fk lRm gt lR and g U gt lR such that 71 m f kaEk g Zngj k1 j1 But then m n a a I ma 3 dfzzaiffdyj g dg22 gfdxi j1k1 J k 39 39 J With respect to the bases 3 3 the matrices of the above linear maps are simply their Jacobians It follows by the chain rule that 1 afkag a 3 mm i1j1k1 3 axi afkog 1 11 339 dx aZkidfog x Applying this reasoning to the notion of a pull back we see that f dx dx 0 f may be writ ten dx 0 df Indeed if X is any form we may compactly write ftx 0c 0 df De nition 55 Let f U 7gt lR3 be an oriented local surface and let X be a 2 form on lR3 The integral of 0c over 2 z is given by ac a A uf The orientation of Z induces an orientation on U in the following way Suppose that uv are co ordinates on U The unit normal vector U to Z is a positive multiple of either fu gtlt f1 or f1 gtlt fu If the former then we say that uv are oriented co ordinates on U and take du A dv to be positive Changing the orientation of 2 changes the orientation on U Examples 1 Let 0c X3 dxl A dxz 7 x1 dxz A ClX3 and uv fuv uz 390 defined on U 01 gtlt 01 with orientation13 U V72 X f Then xlofuv xzofu2 x3ofv fdx1vduudv fdxz2udu fdX3 dv fdx1Adxz 72u2duAdv fdszdX3 ZuduAdv ftx 74u2v du A dz 1 1 2 txfoc 74u2vdudv77 f u 0 0 3 2 Let 0c X1X3 dxz A ClX3 be a 2 form of R3 U 01 X 0271 and ucoszp u Thus be the cone with orientation U W12 gtlt fv The composition x1 0 f here is just ucoszp Thus ftx uz coszpdusinzp Adu u3 coszzpdzp Adu and so 7 3 2 d d 7 7 2 1 3 2 d d 7 if 7 A 7 7 foc ucoszpzp u 00ucoszpuzp 53 Stokes theorem De nition 56 For the purposes of what follows a k dimensional submanifold Z of lR3 with bound ary 32 is14 k 1 An oriented curve with boundary the two ends and no self intersections 131e we take the orientation such that du A dv is positive 14We are being very loose with terminology here a submanifold is a much more precisely defined object 42 k 2 An oriented surface bounded by a single curve 32 k 3 A domain subset with no holes 2 C lR3 oriented as usual and bounded by a surface 32 We can even put an orientation on a 0 dimensional manifold a point simply by attaching a choice of i1 De nition 57 1 Let 2 be an oriented curve in R3 The induced orientation on 32 the two end points of the curve is to attach a 1 to starting end of the curve and a 1 to the terminal end Alternatively let u1 be an oriented co ordinate on the curve if aim points from the curve to wards the endpoint then that endpoint is positively oriented N Let 2 be an oriented surface in lR3 with boundary an oriented curve 32 The induced orientation on 32 is such that if u1u2 are oriented co ordinates on 2 with u1 g 0 and u1 0 on the boundary then uz is an oriented co ordinate for 32 U Now let 2 be an oriented domain in lR3 with boundary 32 The induced orientation on 32 is such that if u1 uz u are oriented co ordinates on 2 with u1 g 0 and u1 0 on the boundary then uz u3 are oriented co ordinates for 32 While this definition seems complicated it really exists just so that we could deal with induced orientations in a dimension independent way What is really going on is this on a surface 2 with co ordinates as described in part 1 then at the boundary curve 32 the tangent vector aim is pointing out of the surface orthogonally to the boundary Thus if u1 uz are oriented co ordinates on 2 then 3142 3741 M39 induced orientation is found using the right hand rule when you walk round the boundary with your head pointing in the direction of the unit normal then the surface is on your left Figure 14 shows the induced orientation on 32 when 2 is first a curve then a surface i must point along the curve in such a way that a a U is an adapted frame for 2 Le the Figure 14 Induced orientations on boundaries The induced orientation on a domain 2 with bounding surface 32 is for the normal to be outward pointing indeed this normal is aim according to the definition Theorem 58 Stokes Let 2 be a kedirnensional submanifold of Rquot with boundary 32 given the induced orientationi Let a be a k 7 1forrn de ned on a neighborhood of Z in R Then da 11 3 a 43 Praaffar a square in R Let X1 X2 be co ordinates on 2 scaled such that the boundary of Z is the set 32 X10 X11 Z 0 S X1 S 1 U 0X2 1X2 0 S X2 S Let X fX1X2 ClX1 gX1 X2 Cle be a general 1 form on 2 Taking ClX1 A Cle as positive on 2 the induced orientation on 32 is then counter clockwise We thus have 1 1 Aim 7 0 ltfltx1ogt7fltx11gtgtdx10 g1lxz7g0xzdxz However 1 1 doc ltgx17fxzgtdxmdxz7 gxrfxadndxz 3 3 0 0 1 1 1 1 gx1 ClX1 Cle 7 fx2 Cle ClX1 0 0 0 0 1 1 7 0 ltglt1x2gt7gltoxzgtgtdxz7 0 ltfltx11gt7fltx1ogtgtdx1 0C 3 I Example Let 2 be the triangular region Xy E lR2 0 g X g 10 g y g X Let X 7X dX 7 6Xy dy If we take the orientation such that dX A dy is positive then the induced orienta tion is to traverse 32 counter clockwise Now fa 0c is calculated by restricting 0c along each edge of the triangle On the first horizontal edge 0c 7X dX on the second vertical edge 0c 76y dy while on the third we have X y and so 0c 7X 7 6X2 dX Hence 1 Cl 176d 0 762d 777 777 71 i 7 7 i 3 27 A211 0 XX 0 yy 1X X X 2 2 Now doc 76y dX A dy so that 1X 1 2 d76dd73d71 20C 00 yyX 0 XX The full proof of Stokes theorem for a general region is beyond this course ourselves with observing that in lE3 it reduces to one of three fundamental theorems 15 so we ll content k 1 The fundamental theorem of calculus fC d f f b 7 f a Integration of a O form function over a point is equivalent to evaluation the i signs on the right hand side coming from the induced orientation on the endpoints of the curve k 2 Here X is a 1 form Using our identification of 1 forms and 2 forms with vectors as in Section 11 we recover what is often referred to as Stokes theorem in vector calculus classes VXAdSfAdr S C 1 1 n 15Proving for cubic is a o U of the above argument but for general submani folds we need the concept of partitions of unity in order to patch together integrals on overlapping cubic regions 44 The surface integral of the curl of a vector field is identified with the path integral of said vector field over the boundary curve Recall our identification of l forms with Euclidean vector fields 0C 0C1 ClX1 0C2 Cle 0C3 ClX3 W A alel txzez 0C3E3 Thus if C is a curve parameterized by 7t on a curve I then Ca I39ya Ia39y tdt Similarly identifying t a a a d 7 7 7 W9 Y Yl 3X1 YzaXZ Ysaxg r Yiei Yzez 7363 we see that 3 74C a gamomodt 74CA dr Thus 35C 11 is the line integral of the vector field A round the curve Now recall that doc em V X A the curl of A Suppose that S is parameterized by f U gt lE3 Then f doc doc 0 df We have da l Cle A ClX3 3932 ClX3 A ClX1 g ClX1 A Cle W V X A lel zez gEEg We thus have to identify the pull back f doc For this note that dx1 o is simply the el component of the differential df The pull backs can thus be seen to be f dxi df ei for each i 123 From this we have that f dxl A dxz is the e3 component of the cross product of fu gtlt f1 du A do and similarly It follows that 3 fda fuvaduA dvVgtltAfugtltfduAdv s Hence f5 doc fV X A fu gtlt fv du dv fV gtlt A dS where dS is the surface element of S k 3 In a similar way as for k 2 we get the divergence theorem VAdVAd V 5 Corollary 59 Green s theorem in the plane Let U be a domain in the plane with boundary curve C i The induced orientation on C is countereclockwise due to the standard orientation on LL Stokes theorem for the lforrn a Pxy dx Qxy dy is 3 ap y CdeQdyUlt7gt dxAdy Stokes theorem in fact holds for more general objects than our definition of submanifold For example a segment of a cylinder has two boundary circles so that 32 consists of 2 curves 45 6 Applications in Euclidean space 61 Integrating functions From multi variable calculus we have that the integral of a function f lE3 gt IR over a local surface x U gt IE3 is given by surfacef Ufxu39vlxu gtlt le dudv The lxu gtlt xvl du dv factor is the area of an infinitessimal parallelogram on the surface Taking f 1 gives us the area of a piece of a local surface In Euclidean space we can integrate a function over a surface rather than 2 f0rms over surfaces in R3 This is because the notion of distance in Euclidean space means that we also have a notion of area Proposition 61 Suppose we have a local surface in E3 referred to an adapted frame and let uv be oriented coordinates Then 61A62 lxu gtlt xvlduAdv Proof Consider dx gtAlt dx a vector of 2 forms where we simultaneously take wedge products of 1 forms and cross products of vectors On the one hand dxgtAlt dx xu du x1 dv gtAlt xu du x1 dv 2xu gtlt xv du A dz leu gtlt xyleg du A dv Here we use the fact that u v are oriented so that X X x1 is a positive multiple of the unit normal e3 However on the other hand ax dx 9131 Qzel 9 9121 Qzel 23391 A 92 giving the result The Proposition shows that the integral of a function over a surface is given by Ufxuv91 A 92 In particular if f E 1 we have the area of the surface For this reason the 2 form 91 A 92 is often called the area form for the surface x Examples 1 The sphere of radius a has 91 a d0 92 a sine Clip If 2 is the standard vertical co ordinate on IE3 we have a AZ Uzx9zp91A92 Uacos9a2sin9d9dzp 0 b Surface Area 1 a2 sinzocdoc d4 47Ia2 S U 2 In hyperbolic space the area form is W To find the area of the infinite region U shown in Figure 15 bounded by the geodesics X i1 X2 y2 1 we evaluate 1 co 1 91A92 lzdxdy dx7r u 71 My 71 17x2 Conversely the hyperbolic area between the geodesic X2 y2 1 and the X axis is infinite 46 71 0 1 3 Figure 15 A region of hyperbolic space with area 71 The same notation may be used to integrate functions over curves in Euclidean space If we write fy f we mean fs1f39ys ds where 7s039ys1 are the endpoints of 39y and l7 sl 1 39y is parameterized by arc length Why do we need the Euclidean structure for this Because without it we have no notion of length by which to parameterize and so we are forced to integrate only l forms 62 Minimal surfaces Now that we know how to integrate functions on surfaces and calculate surface area we can ap proach the problem of finding minimal surfaces ie those whose area is minimal for all surfaces with a given boundary The argument is similar to the stationary distance approach to geodesics Let x U gt lE3 be a surface and consider the family of surfaces x uv xuv efuvUuv nearby x Here f is a smooth function on U of compact support16 The idea is that we start with x and perturb it by a small amount in the normal direction at each point in such a way that the boundary is unchanged De nition 62 A local surface has stationary area if for all such families E OAreax 0 Theorem 63 A surface has stationary area if its mean curvature vanishes Proof Take the exterior derivative of xg dx dxedede Just like in the geodesic arguments we throw away all terms of order 62 and higher We first calculate the first fundamental forms of the surfaces xg le 2 dx dx2 de dx172efll 2 9 6 Zefw1391 w2392 16The only important point is that if f extends to the curve parameterizing the boundary of x then f would be zero there Z 91 efw132 92 efw232 We can therefore diagonalize l up to first order in e be choosing 9M 2 91 efwlg and 9 2 92 e f 0123 Hence ere A 92e 2 91 A 92 fw13 A 92 91A23 91 A 92 efa91 1792 A 92 91 A 1791 92 2 61 A 0 efiz C 1 izefHWl A 92 where we ve written U13 i191 1192 and 0123 1191 C92 It follows that d E 0Areax 72 UfH61 A 92 which vanishes for all functions f iff H E 0 I Thus minimal surfaces as defined in 162A are critical surfaces of the area functional It certainly follows that any surface which has minimal area for a given perimeter must have mean curvature zero although it is perfectly possible to have H 0 for a surface which is not area minimizing 63 Relations with Complex analysis Let lR2 C the complex plane with complex co ordinate z X iy and Z X 7 iy We can consider complex functions and complex differential forms Exterior derivatives of real and imaginary parts can be taken In particular we can define dzdxiydxidy dZdxiidy dz and d can be used as a basis for complex l forms instead of dx dy Defining two complex vector fields 3313 i 33133 quotaz z ax lay 39 ai z ax lay 39 it is easy to see that any complex function satisfies df dez l 5fd2 Proposition 64 f is halamarphic i gf Proof Let f u iv be the decomposition of f into real and imaginary parts Then 3f ltuiv iuivgt aux ivyivxuy Thus 3f 0 ltgt f satisfies the CauchyiRiemann equations The Cauchy Riemann equations are exactly the condition that a complex function f z is differ entiable we say that f is holomorphic Indeed we require hm fz fzo 7 zgtzo z 7 20 fZo to be independent of the direction in which we approach 20 It is easy to see that independence is equivalent to the Cauchy Riemann equations The proposition makes clear the statement that a holomorphic function is independent of 2 Complex l forms and 2 forms can be integrated be integrating real and imaginary parts respec tively One nice conclusion of Green s theorem in the plane comes form applying it to the l form 0c fzZ dz Then doc dedzdez3deAdz3deAdz Green s theorem then says that f d 3 d Ad Cfz z 2 U f z z In particular if f is holomorphic then the right hand side is 0 in which case we have Cauchy s theo rem that the integral of a holomorphic function around a closed curve is zero 7 The GaussBonnet theorem In this final section we think about the geometry of polygons specifically geodesic triangles on surfaces and prove using Stokes theorem a famous result that relates topological and differential geometric data on a surface 71 Polygons on surfaces First we think about how the angle made by a curve with an adapted vector field changes along the curve Let xzt be a unit speed curve in an oriented surface x with adaptive frame field e1 e2 e3 U In terms of the angle 0 the tangent vector x t dxz t makes with e1 we have xt cos lPte1 sin lPtez Lemma 71 1V0 Kg w12z l Proof Recall that the geodesic curvature is defined in terms of the covariant derivative D all x l ngg X x Calculating we have Ddix lPi sin lpel cos lpez w12z7 cos lpez sin lpez l IP 7 w12z7 sin lpel cos lPez IP 7 U12ZEg X X hence Kg 1V7 w12z De nition 72 Suppose that xzt is a unit speed curve which runs from t to to t1 The total rotation of the curve with respect to the frame field is given by t1 f w12z Kg dt 1W1 A we 0 In particular if xzt is a simple closed curve traversed counter clockwise then the above inte gral is 27139 We now restrict to polygons whose edges are geodesics De nition 73 A polygon on a surface is a piecewise differentiable curve 39y lying in a surface in such a way that 39y has no self intersections and the differentiable pieces of 39y are all geodesics Figure 16 shows an example If we orient the surface then 39y bounds a region of the surface S and has an induced orientation We label the geodesic curves of 39y as 71 39yn in order of the orientation and the internal angles A1 VA Figure 16 A polygon on a cylinder Theorem 74 Let 39y be a polygon on an oriented surface boundinga region S Then the internal angles satisfy em l l H Ai1 l727 SK Proof Moving all the way around the polygon clearly rotates the tangent vector field through 27139 radians We know that the total rotation of the vector field along each curve 39yi is fy U12 while the rotation at each vertex is through an angle 71 7 Ai Adding up we have 27139 2717Ai i 0112 11 11 7 Hi turning at ver ces turning over each edge n n7r 0112 7 ZAi 7 1 50 However using Stokes theorem w12dw12K91A62K 7 s s s which gives the result 72 The geometry of a geodesic triangle Before we move on we can restrict to the simplest case where 39y is a geodesic triangle De nition 75 A geodesic triangle in a surface is three points joined to each other by non intersecting geodesics The above theorem says that the angles in a geodesic triangle A satisfy A B C 7 7r K A Think about what this says for a minute The familiar fact that the angles in a triangle add up to 71 is false on any surface which is not at Example If A is a geodesic triangle with angles A B C and area S then 1 intheplaneABC 7t 2 ontheunitsphereABC 71S 3 in hyperbolic space A B C 7t 7 S Notice that in hyperbolic space the area S of any geodesic triangle satisfies S g 71 Equality is iff the three angles A B C are all zero This only happens if the geodesics meet on the X axis We are not saying that the total area of hyperbolic space is g 71 in fact it is infinite rather than the largest area you can fit inside a geodesic triangle is 7 Somewhat counter intuitively all three edges of this geodesic triangle have infinite length A similar but less geometrically interesting result comes from observing that the fact that all angles in a triangle must be less than 71 forces the maximum area bounded by a geodesic triangle on a sphere of radius 1 to be 27139 the triangle in question is not an interesting triangle rather it is a complete great circle with three marked pointsl There are many other things we could do in spherical or hyperbolic geometry There are Sine and Cosine rules in spaces of constant curvature Suppose we have a geodesic triangle in a space of constant Gauss curvature K whose angles A B C are opposite the sides of lengths a b C respectively Then COSCR cosm cosb sinmI sinbI cos C sinmI sinbI 7 sinCI sin A sin B sin C For the unit sphere we have K 1 and the formulae simplify For hyperbolic space with K 71 the expressions may be rewritten using cosia COSh and sinia isinha so that the sines and 51 cosines of lengths become hyperbolic Sines and cosines with appropriate sign changes To recover the cosine and sine rules in at space divide both sides by R and take the limit as K gt 0 The cosine rule allows us to recover the constant curvature Pythagoras theorem for any right angled triangle with hypotenuse of length c we have cosc cosa cosh This doesn t look like the Pythagoras theorem you re used to but if you take Taylor approximations up to order two in K and substitute c4 2 a2 h22 we recover 7 c2 2 a2 b2 7 gazbz You should expect the hypotenuse of a small right angled triangle on a sphere to have shorter hy potenuse than in the plane On a pseudosphere the hypotenuse should be longer The approximation is good if the lengths of a hc are small when compared to the curvature typically ah lt 1 i lKl or a h lt r for a sphere Indeed if a h ur then the ratio of the length of the above Pythagorean which is within 2 of 1 for a lt 1 and within 9 for u 712 This last is a geodesic triangle with edges going 1 4 of the way round the sphere larger than you d ever likely need in navigation prediction to the correct hypotenuse is cp c 73 The GaussBonnet theorem For the full version of this result we will need a stronger version of the theorem of the rotation of a vector field around a geodesic polygon Theorem 76 Let S be a region on an oriented surface bounded by a piecewise di erentiable closed curve 7 Suppose that 7 has n corners and that its internal angles are A1 A Then M An727IKKg 1 S r i where the last integral18 is understood to be the sum over all the di erentiable pieces of 39y The proof is identical to that of Theorem 74 except that the rotation along each curve 39y is as given in Definition 72 where the geodesic curvature is not necessarily zero Consider a surface S with a boundary 3S De nition 77 A dissection of S is a cutting up of S into polygons for us geodesic polygons The number of faces P of the dissection is the number of polygons plus the number of regions with one edge as part of the boundary Let E be the number of edges of the dissection number of edges of polygons plus resulting boundary segments and V the number of vertices of the dissection The Euler characteristic of S is defined to be XP7EV Figure 17 Two dissections of a disk The surface S could be simply a region of a plane Figure 17 shows two possible dissections of a disk Both have x 1 Proposition 78 The Euler characteristic is independent afdissectian and thus well7defined Proof Start by observing that if you subdivide a polygon into triangles then x is unaltered join up one vertex of a polygon with another and you get one extra edge one extra face and no extra vertices This means that we need only consider triangular dissections Triangles can now be sub divided in various ways again leaving x unchanged Thus given two dissections into triangles we can subdivide each until we get a common subdissection x is thus independent of the choice of dissection I There is something missing here A general dissection does not have the edges being geodesics so a complete proof would have to deal with this A region of the plane bounded by a simple closed curve no intersections has x 1 The sphere no boundary has x 2 as can be seen in many ways The torus has x 0 for example take as a dissection to orthogonal circles of curvature to get x P 7 E V 1 7 2 1 0 To get some examples of Euler characteristics we do a little surgery and consider how the Euler characteristic changes when you glue surfaces together De nition 79 Given two topological surfaces 21 22 their connected sum 2122 is given by cutting a small hole in each surface and pasting them together By topological surface we mean that nothing matters except the topology we may deform 21 22 and glue them together in any way we like and the topology of 2122 is unchanged Note that the connected sum of any surface with a sphere effectively leaves the surface un changed Theorem 710 x2122 x21 x22 7 2 Proof Dissect 2122 into triangles Remove one triangle on each surface to create a hole and stick these together Let P V E are the total number of faces vertices and edges of both original dissections added together and P V E the faces vertices and edges of the resulting dissection of 21Zz Then PP72 EE73 VV73Theresultfollows I 17You can similarly take Taylor approximations of the cosine and sine rules to obtain the standard rules as approxima tions 18Recall that the integral f7 Kg of a function over a curve is defined in Section 61 This allows us to easily calculate x for any handlebody formed by taking the connected sums of tori since moms 0 taking the connected sum with a torus decreases x by 2 The genus g of a handlebody is the number of holes in the surface It is clear from the theorem that x 21 7 g Now that we have many examples of surfaces and their Euler characteristics we proceed to the theorem Theorem 711 Gauss Bonnet Let 2 be u surface in E3 with di erentiuble boundary 32 Then K 2 Z 32Kg 7176 Proof Dissect 2 into triangles where all edges except those that are part of the boundary 32 are geodesics Let the internal triangles be labeled A1 AFL and let the triangular faces bordering the boundary be labeled A pn1 A 1 The boundary 32 is thus made up of n curved edges and there are n triangles bordering the boundary First consider the geodesic internal edges of which there are E 7 Each internal edge is com mon to two faces Counting each internal edge twice there are three edges per internal face P 7 n of them and two edges for each boundary face Thus 2E 7 n 3F 7 n 2n and so for such a dissection E 32quot The Euler characteristic may therefore be written 3P 1 xP7 VV7EPn Now for each triangle Aj internal and external Aj Bf Ci 7 7r fA K fBA Kg Summing over 7 7 j gives F K Kg 2AjBjCj 77 L39P 3 a j1 Adding up the angles around all the V 7 n internal vertices yields 27139 for each while for each of the n boundary vertices we get an angle of 71 Hence gtKA2Kg 27IV7n7tn77IP27IV7 Pn2 an I The theorem may be modified to include the case where the boundary 32 is piecewise differen tiable with internal angles at its corners 4 It can also be simplified to the case where the boundary 32 is a closed geodesic or is empty In the latter case we call 2 a handlebody eg a sphere torus pretzel etc Corollary 712 Let 2 be u handlebody in E3 Then K27Ix 3 The Gauss Bonnet theorem is particularly powerful because it relates the topological Euler char acteristic to the differential gauss curvature x is a single number which depends only on topology twisting squashing or stretching or otherwise altering a surface in any way that does not involve puncturing it leaves x unchanged For example the famous quip that a doughnut is topologically 54

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