Differential Geometry. Peter Petersen

Transcription

1 Differential Geometry Peter Petersen

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3 CHAPTER Preliminaries.. Vectors-Matrices Gien a basis e, f for a two dimensional ector space we expand ectors sing matrix mltiplication e e + f f e f apple e f and the matrix representation [L] for a linear map/transformation L can be fond from L (e) L (f) e f [L] e f apple L e e L e f L f e L f f Next we relate matrix mltiplication and the dot prodct in R 3. We think of ectors as being colmns or 3 matrices. Keeping that in mind and sing transposition of matrices we immediately obtain: X t Y X Y, X t X Y X X X Y apple t X X X Y X Y X apple t X X X Y X Y X Y, Y X Y Y 3 t X X X Y X Z X Y Z X Y Z 4 Y X Y Y Y Z 5 Z X Z Y Z Z These formlas can be sed to calclate the coefficients of a ector with respect to a general basis. Recall first that if E,E is an orthonormal basis for R, then X (X E ) E +(X E ) E E E E E t X So the coefficients for X are simply the dot prodcts with the basis ectors. More generally we hae Theorem... Let U.V be a basis for R, then X U V U V t U V U V t X U V U V t U V apple U X V X 3

4 .. VECTOR CALCULUS 4 Proof. First write X U V apple X X The goal is to find a formla for the coefficients X,X in terms of the dot prodcts X U, X V. To that end we notice apple U X U V t X V X U V t U V apple X Showing directly that apple X t apple U X X U V U V V X and conseqently X X U V U V t U V apple U X V X There is a similar formla in R 3 which is a bit longer. In practice we shall only need it in the case where the third basis ector is perpendiclar to the first two. Also note that if U, V are orthonormal then U V t U V apple 0 0 and we recoer the standard formla for the expansion of a ector in an orthonormal basis. Finally we mention the triple prodct formla... Chain Rle. d (V c) h V x det X Y Z X (Y Z) X t (Y Z).. Vector Calcls V y V z i apple dc... Directional Deriaties. If h is a fnction on R 3 and X (P, Q, R) then D X h P h x + Qh y + R h z (rh) X and for a ector field V we get [rh] t [X] D X V h h x h V x V y h y V z h z i [X] i [X].

5 .. VECTOR CALCULUS 5 We can also calclate directional deriaties by selecting any cre sch that ċ (0) X. Along the cre the chain rle says: d (V c) h i apple V V V dc x y z DċV Ths D X V d (V c) (0)

6 CHAPTER Cre Theory.. Cres Discss, parametrized cres, integral cres of a ector field, orthogonal cres, integral cres to second order system, implicitly gien cres (leel sets, intersections of leels sets (se cross prodct as ector field), as orthogonal cres. Geometry of cres is a way of measring that does not take in to accont how agiencreisparametrizedoreenplacedineclideanspace. Ontheotherhand it is ery hard to define geometric qantities withot referring to a parametrization. Ths the goal will first be to see if some sort of canonical parametrization might exist and secondly to also show that or measrements can be defined sing whateer parametrization the cre comes with. Imagine traeling in a car or flying an airplane. Yo can keep track of time and or the odometer and for each ale of the time and/or odometer reading do a GPS measrement of where yo are. This will gie yo a cre that is parametrized by time or distance traeled. The goal of cre theory is to decided what frther measrements are needed to retrace the precise path traeled. Clearly one mst also measre how one trns and that becomes the important thing to describe mathematically.. Arclength and Linear Motion The arclength is the distance traeled along the cre. This means that change in arclength measres how fast the cre moes. Specifically if there is no change in arclength, then the cre is stationary, i.e., yo stopped. More precisely, if the distance traeled is denoted by s, thenrelatiechangetothegeneralparameteris the speed d Ths s is the antideriatie of speed and is only defined p to an additie constant. This means that we shold define the length of a cre on [a, b] as follows ˆ b L ( ) b a s (b) s (a) a This is easily shown to be independent of the parameter t as long as the reparametrization is in the same direction. One also easily checks that a cre on [a, b] is stationary if and only if its speed anishes on [a, b]. Wesallysppresstheinteral and instead simply write L ( ). To see that arclength really is related to or sal concept of distance we hae Theorem... The straight line is the shortest cre between any two points in Eclidean space. 6

7 .3. CURVATURE 7 Proof. We shall gie two almost identical proofs. Withot loss of generality we assme that we hae a cre c (t) :[a, b]! R k where c (a) 0, and c (b) p. We wish to show that L (c) p. To that end we select a nit ector field X which is also a gradient field X rf. Two natral choices are possible: For the first p simply let f (x) x p, for the second f (x) x. In the first case the gradient is simply a parallel field and defined eerywhere, in the second case we obtain the radial field which is not defined at the origin. When sing the second field we need to restrict the domain of the cre to [a 0,b] sch that c (a 0 )0bt c (t) 6 0for t>a 0. This is clearly possible as the set of points where c (t) 0is a closed sbset of [a, b], so a 0 is jst the maximm ale where c anishes. This allows s to perform the following calclation sing Cachy-Schwarz, the chain rle, and the fndamental theorem of calcls. When we are in the second case the integrals are possibly improper at t a 0, bt clearly trn ot to be perfectly well defined since the integrand has a continos limit as t approaches a 0 L (c) ˆ b a 0 ċ ˆ b ċ rf a 0 ˆ b ċ rf a 0 ˆ b d (f c) a 0 ˆ b d (f c) a 0 f (c (b)) f (c (a 0 )) f (x) f (0) f (x) p We can een go backwar and check what happens when L (c) p. It appears that we mst hae eqality in two places where we had ineqality. Ths we hae d(f c) 0 eerywhere and ċ is proportional to rf eerywhere. This implies that c is a possibly singlar reparametrization of the straight line from 0 to p. Might also jst try to se the constant nit ector in the direction of p. Next discss linear motion and characterize it as cres whose tangents are all parallel to each other. Use both parameteric soltion and eqation soltion (perpendiclar to n ectors).3. Cratre Arclength measres how far the cre is from being stationary. Next we can consider the nit tangent ector. If it is stationary the the cre is a straight line so the degree to which the nit tangent is stationary is a measre of how fast it change and how far the cre is from being a line. If the nit tangent is also reglar then it to can be reparametrized by arclength. We let be the arclength parameter.

8 .3. CURVATURE 8 The relatie change between the arclength parameters for the nit tangent and the cre is by definition the cratre apple d and with a general parametrization we see that it is related to the part of the acceleration that is orthogonal to the nit tangent ector. Note this ratio is nonnegatie as increases with s. apple d d dt d ( ) 4 ( ) ( T) T Frther note that when the nit tangent ector is reglar it too has a nit tangent ector denoted N satisfying dt d N This ector is in fact perpendiclar to T as 0 d T d T dt d It is called the nit normal to the cre. In terms of the arclength parameter for or eqialently and Note that apple dt applen dt N (T N) we obtain T dn ( T) T T ( T) T +( N) N ( T) T + apple N

9 .4. GENERAL FRAMES 9.4. General Frames We shall now consider the general problem of taking deriaties of a basis U (t),v (t) that depen on t, and iewed as a choice of basis at c (t). Gien U (t), anatralchoiceforv (t) wold be the nit ector orthogonal to U (t). Also we shall sally se U (t) ċ (t) or U (t) T (t). The goal is to identify the matrix [D] that appears in d U V d U d V U V [D] We know a complicated formla which simplifies to [D] U V t U V U V t d U d V Theorem.4.. Let U (t),v (t) be an orthonormal frame that depen on a parameter t, then d U V U V apple 0, 0 or U d V V d U d U V, d V U Proof. We se that apple t 0 U V U V 0 and the deriatie of this apple 0 0 d 0 0 apple Showing that U Or formla for [D] then becomes d V t t U V + U V d U d V d U U d U V apple U d d V U d V V + V d d U U d 0 V U V d U d V V, U U [D] U V t d U d V apple U d U U d V V d U V d apple V 0 0 U d V d V V

10 .6. SPACE CURVES 0 Occasionally we need one more deriatie d U V U V apple d d d U d U V, d V d U V..5. Planar Cres, Signed cratre. Arclength and signed cratre determine the cre. characterization of circlar cres. rotation index with tangent/normal circlar image. conex cres. oals. Isoperimetric..6. Space Cres Serret-Frenet. Obsere that there are no relations between cratre and torsion. Generalized helices and characterizations of planar and spherical cres.

11 CHAPTER 3 Local Srface Theory 3.. Srfaces We define a parametrized srface as a fnction (, ) :U R! R 3 where and are linearly independent. For parametrized srfaces we generally do not worry abot self intersections or other topological pathologies. This is jst as with cres and allows s a great deal of flexibility. When we need to worry abot these isses, or rather we wish to aoid them, then we resort to the more restrictie class of srfaces that come from the next two general constrctions. AspecialcaseistheMongepatch. Implicitly gien srfaces as leel sets. AsrfaceM R 3 is then a sbset which is locally represented as a graph oer two coordinates. Note that a parametrized srface might not be a srface in this sense if it intersects itself or otherwise gets arbitrarily close to itself. The tangent space T p M span,, and normal space N p M (T p M)? Proposition 3... Both tangent and normal spaces are sbspaces that do not depend on a choice of parametrization. Proof. This wold seem intitiely clear, jst as with cres, where the tangent line does not depend on parametrizations. For cres it boils down to the simple fact that elocities for different parametrizations are proportional and hence define the same tangent lines. With srfaces something similar happens, bt it is abitmoreinoled. Spposewehaetwodifferentparametrizationsofthesame srface: (s, t) (, ) This tells s that the parameters are fnctions of each other (s, t), (s, t) s s (, ),t t (, ) The chain rle then gies s s s + t t span s, t similarly span s, t

12 3.. SURFACES and in the other direction s, t span, This shows that at a fixed point p on a srface the tangent space does not depend on how the srface is parametrized. The normal space is then also well defined. with Note that the chain rle shows in matrix notation that apple s s s t t s t apple s t s t apple s s t apple It is often sefl to find coordinates sited to a particlar sitation. Howeer nlike for cres it isn t possible to some parametrize a srface sch that the coordinate cres are nit speed and orthogonal to each other. Bt there is one general constrction we can do. Theorem 3... Assme that we hae linearly independent tangent ector fiel X, Y defined on a srface M. Then it is possible to find a parametrization (, ) in a neighborhood of any point sch that is proportional to X and is proportional to Y. Proof. The ector fiel hae integral cres forming a net on the srface. Apparently the goal is to reparametrize the cres in this net in some fashion. The difficlty lies in ensring that the leels where is constant correspond to the - cres, and ice ersa. We proceed as with a classical constrction of Cartesian coordinates. Select a point p and let the -axis be the integral cre for X throgh p, similarly set the -axis be the integral cre for Y throgh p. Both of these cres retain the parametrizations that make them integral cres for X and Y. Ths p will natrally correspond to (, ) (0, 0). We now wish to assign (, ) coordinates to a point q near p. There are also niqe integral cres for X and Y throgh q. These will be or way of projecting onto the chosen axes and will in this way yield the desired coordinates. Specifically (q) is the parameter where the integral cre for Y throgh q intersects the -axis, and similarly with (q). In general integral cres can intersect axes in seeral places or might not intersect them at all. Howeer, a continity argment offers some jstification when we consider that the axes themseles are the proper integral cres for the qs thatlie on these axes and so q sfficiently close to both axes shold hae a well defined set of coordinates. We also note that as the projection happens along integral cres we hae ensred that coordinate cres are simply reparametrizations of integral cres. To completely jstify this constrction we need to know qite a bit abot the existence, niqeness and smoothness of soltions to differential eqations and the inerse fnction theorem also comes in handy. Exercise: A generalized cylinder is determined by a reglar cre and a ector that is not tangent to the cre. Constrct a natral parametrization and show that it gies a parametrized srface. Show that we can reconstrct the cone so that s s t t t t

13 3.. THE ABSTRACT FRAMEWORK 3 the cre lie in the plane perpendiclar to the ector. What if the cre is gien by two eqations and yo also want the srface to be gien by an eqation? Exercise: A generalized cone is generated by a reglar cre and a point not on the cre. Constrct a natral parametrization and determine where it yiel a parametrized srface. Show that we can se a cre that lies on a sphere centered at the ertex of the cone. What if the cre is gien by two eqations and yo also want the srface to be gien by an eqation? Exercise: A rled srface is gien by a parametrization of the form (s, t) (s)+t It is eidently a srface that is a nion of lines (rlers). Gie conditions on, and the parameter t that garantee we get a parametrized srface. A special case occrs when is nit speed and 0. These are also called tangent deelopables. Exercise: A srface of reoltion is determined by a planar reglar cre and a line that is neer perpendiclar to the tangent ectors of the cre. The srface is generated by rotating the cre arond the line. Constrct a natral parametrization and show that it is a parametrized srface. What if the planar cre is gien by an eqation and yo also want the srface to be gien by an eqation? Exercise: Many classical srfaces are of the form (s) F (x, y, z) ax + by + cz + dx + ey + fz + g 0 Gie conditions on the coefficients sch that it is generates a srface (g 0takes special care). Under what conditions does it become a srface of reoltion arond the z-axis? Under what conditions does it become a cylinder or cone? Why are these elliptic when abc > 0 and hyperbolic when abc < 0? When abc 6 0 rewrite it in the form F (x, y, z) a (x x 0 ) + b (y y 0 ) + c (z z 0 ) + h The Abstract Framework As with cres, parametrized srfaces can hae intersections and other nasty complications that do not come p with the other three cases. Neertheless it is sally easier to deelop formlas for parametrized srfaces. For a parametrized srface (, ) we hae the elocities of the coordinate ector fiel, While these can be normalized to be nit ectors we can t garantee that they are orthogonal. Nor can we find parameters that make the coordinate fiel orthonormal. We shall see that there are geometric obstrctions to finding sch parametrizations. Before discssing general srfaces it might be instrctie to see what happens if (, ) is simply a reparametrization of the plane. Ths, form a basis at each point. Taking partial deriaties of these fiel gie s h h i i [ ], [ ]

14 3.. THE ABSTRACT FRAMEWORK 4 or in condensed form w h w w i [ w],w, The matrices [ w] tell s how the tangent fiel change with respect to themseles. Agoodexamplecomesfromconsideringpolarcoordinates (r, ) (rcos, r sin ) apple cos, apple r sin r sin r cos apple apple r r sin, 0, cos r r cos r sin so r r r h h rr r [ r] r i i apple r r r, apple r apple 0 r r 0 apple 0 r [ ] r 0 The key is that only Cartesian coordinates hae the property that its coordinate fiel are constant. When sing general coordinates we are natrally forced to find these qantities. To see why this is consider a cre c (t) (r (t) cos (t),r(t)sin (t)) in the plane. Its elocity is natrally gien by ċ ṙ r + If we wish to calclate its acceleration then we mst compte the deriaties of the coordinate fiel. This inoles the chain rle as well as the formlas jst deeloped c r r + +ṙ d r r + +ṙ dr r + d r + d r r + +ṙ +ṙ r r r r + +ṙ r r r r r r + + ṙ r! r + dr + r + d Note that r correspon to the centrifgal force that yo feel when forced to moe in a circle. The term ṙ + r is related to Kepler s second law nder a central force field. In this context this simply means that + ṙ r 0

15 3.. THE ABSTRACT FRAMEWORK 5 if the force and hence acceleration is radial. This in trn implies that r is constant as Kepler s law states. The general goal will be to deelop a similar set of ideas for srfaces and in addition to find other ways of calclating [ w] that depend on the geometry of the tangent fiel. Before generalizing we make another rather startling obseration. Taking one more deriatie we obtain w w w w [ w ] [ w ]+ [ w ][ w ]+ [ w ][ w ]+apple w w apple w w apple w w Switching the order of the deriaties shold not change the otcome, w w w [ w ][ w ]+apple w bt it does look different when we se w and w. Therefore we can conclde that apple apple [ ][ ]+ [ ][ ]+ or apple apple +[ ][ ] [ ][ ] 0. For polar coordinates this can be erified directly: apple apple apple apple r r r 0 r 0 apple apple apple apple r 0 r 0 0 [ r][ ] [ ][ r] 0 r r 0 r 0 0 r apple 0 r 0 This means that the two matrices of fnctions [ ], [ ] hae some nontriial relations between them that are not eident from the definition. For a srface x (, ) in R 3 we add to the tangent ectors the normal n (, ) in order to get a basis. While n does depend on the parametrizations we note that as it is normal to a plane in R 3 there are in fact only two choices ±n, jst as with planar cres. This means we shall consider frames frames w n h w w n w i n and deriaties of sch n [D w ]

16 3.. THE ABSTRACT FRAMEWORK 6 where w can be either or. The entries of D w are diided p into parts. The first depen only on tangential information, the first two rows and colmns, and correspon to the [ w] that we defined in the plane sing general coordinates. The second depen on normal information, the third row and colmn. Since n is a nit ector the 33 entry actally anishes: 0 n w n n w showing that n w lies in the tangent space and hence does not hae a normal component. As before we hae w w n w w n In particlar, or apple D [D ][D ]+ apple D apple D apple D [D ][D ]+ +[D ][D ] [D ][D ]0 As we shall see, other interesting featres emerge when we try to restrict attention to the tangential and normal parts of these matrices. Elie Cartan deeloped an approach that ses orthonormal bases, bt he clearly had to gie p on the idea of sing coordinate ector fiel. Ths he chose an orthonormal frame E,E,E 3 along part of the srface with the property that E 3 n is normal to the srface, and conseqently E,E form an orthonormal basis for the tangent space. The goal is again to take deriaties. For that prpose we can still se parameters E E E 3 E w w E w E 3 w E E E 3 [Dw ] The first obseration is that [D w ] is skew-symmetric since we sed an orthonormal basis: 3 t 0 0 E E E 3 E E E so 0 t E E E 3 E E E 3 w t E E E 3 E E E 3 w + t E E E 3 E E E 3 w E E E 3 [Dw ] t E E E 3 + E E E 3 t E E E 3 [Dw ] [D w ] t +[D w ]

17 3.3. THE FIRST FUNDAMENTAL FORM 7 In particlar, there will only be 3 entries to sort ot. This is a significant redction from what we had to deal with aboe. What is more, the entries can easily be fond by compting the dot prodcts E i E j w This is also in sharp contrast to what happens in the aboe sitation as we shall see. Taking one more deriatie will again yield a formla apple apple Dw Dw [D w ][D w ] [D w ][D w ] w w where both sides are skew symmetric. Gien the simplicity of sing orthonormal frames it is perhaps pzzling why one wold bother deeloping the more cmbersome approach that ses coordinate fiel. The answer lies, as with cres, in the nfortnate fact that it is often easier to find coordinate fiel than orthonormal bases that are easy to work with. Monge patches are prime examples. For specific examples and many theoretical deelopments, howeer, Cartan s approach has many adantages The First Fndamental Form Let (, ) :U! R 3 be a parametrized srface. At each point of this srface we get a basis (, ), (, ), n (, ) These ectors are again parametrized by,. The first two ectors are tangent to the srface and gie s an nnormalized ersion of the tangent ector for a cre, while the third is the normal and is natrally normalized jst as the normal ector is for a cre. One of the isses that make srface theory more difficlt than cre theory is that there is no canonical parametrization along the lines of the arclength parametrization for cres. The first fndamental form is the symmetric positie definite form that comes from the matrix [I] t apple apple g g g g For a cre the analogos term wold simply be the sqare of the speed t d d d d. This form dictates how one comptes dot prodcts of ectors tangent to the srface assming they are expanded according to the basis,

18 3.3. THE FIRST FUNDAMENTAL FORM 8 X X + X Y Y + Y apple X X apple Y Y I(X, Y ) X X apple g g g X X x X t Y X Y x apple X X g apple Y t Y t apple Y Y apple Y In particlar, we see that while the metric coefficients g ww depend on or parametrization. The dot prodct I(X, Y ) of two tangent ectors remains the same if we change parameters. Note that I stan for the bilinear form I(X, Y ) which does not depend on parametrizations, while [I] is the matrix representation for a fixed parametrization. Or first srprising obseration is that the normalization factor can be compted from [I]. Lemma We hae det[i]g g (g ) Proof. The proof is a bit more general. Fix two ectors m, n R 3. The qantity m n represents the area of the parallelogram with sides m and n. This area can also be calclated by the height base formla. If m is the base then we hae to find h m. The height can be calclated by the Pythagorean theorem if we know the projection onto m. The projection of n onto m is So we hae h + (n m) m m (n m) m m Isolating h and mltiplying my m yiel m n h m 0 m n n (n m) m m A m n m (n m) m m 4 (m m)(n n) (m n) Y

19 3.3. THE FIRST FUNDAMENTAL FORM 9 This is what we wanted to proe. The inerse [I] apple g g g g apple g can be sed to find the expansion of a tangent ector by compting its dots prodcts with the basis: X Proposition If X T p M, then g X + g X [I] and more generally for any Z R 3 Z g Z + g Z [I] t X + g + t Z +(Z n) n g g g X + g X g Z + g Z +(Z n) n Proof. We already sspect that this formla works for X T p M as we worked with it in R. Clearly a similar formla hol in R 3 as well. Note that the operation [I] t can be applied to any ector in R 3. It simply projects the ector to a ector in the tangent space. For a general ector Z R 3 we therefore hae to split it p in the tangential component and normal component Z X +(Z n) n, X Z (Z n) n and then apply or reslt to X. Defining the gradient of a fnction is another important se of the first fndamental form as well as its inerse. Let f (, ) be iewed as a fnction on the srface (, ). Or definition of the gradient shold definitely be so that it conforms with the chain rle for a cre c (t) ( (t),(t)). Ths on one hand we want d (f c) while the chain rle also dictates Ths d (f c) (rf) rf ċ (rf) (rf) [I] f (rf) f f apple d d f apple d d [I]

20 3.3. THE FIRST FUNDAMENTAL FORM 0 or rf apple (rf) (rf) f [I] f f f g + g f [I] t f + t f f g + g In particlar, we see that changing coordinates changes the gradient in sch a way that it isn t simply the ector corresponding to the partial deriaties! The other nice featre is that we now hae a concept of the gradient that gies a ector field independently of parametrizations. The defining eqation d (f c) rf ċ I(rf,ċ) gies an implicit definition of rf that makes sense withot reference to parametrizations of the srface. Exercise: If we hae a parametrization where apple 0 [I] 0 g then the coordinate fnction f (, ) has r. Exercise: Show that it is always possible to find an orthogonal parametrization, i.e., g anishes. Exercise: Show that if g g g 0 then we can reparametrize and separately, i.e., (s) and (t), in sch awaythatwehaecartesian coordinates: then Exercise: Show that if g ss g tt, g st 0 0 (, ) F ()+G () and conclde that we we are in the sitation of the preios exercise. Discss Cartesian, eqiareal, and isothermal parameterizations as sbstittes for arclength parameter.

21 3.4. THE GAUSS FORMULAS 3.4. The Gass Formlas With all of this in mind we are now going to compte the partial deriaties of or basis in both the and directions. Since these deriaties might not be tangential we get a formla that looks like w w [I] t + n n w w w w The goal here and in the next section is to show that the tangential part of this formla w w [I] t w w can be compted directly form the first fndamental form and withot knowledge of the second deriaties w w. Note that this is similar to what we did for a reparametrization of the plane. To accomplish this we need some more notation: w w w L ww w w w n w w The first line defines the Christoffel symbols of the first kind. The second line the second fndamental form II n (X, Y ) X X apple [II n Y ] Y X X apple apple L L Y L L Y The sperscript n refers to the choice of normal and is sally sppressed since there are only two choices for the normal ±n. This also tells s that nii n is independent of the normal. To frther simplify expressions we also need to do the appropriate mltiplication with g w4w5 to find the coefficients also called the Christoffel symbols of the second kind: apple w w w g w w w + g w w w, apple apple g w w w w [I] g g g This now gies s the tangential component as w w w w + [I] w w w w w w t w w t w w The second deriaties of (, ) can now be expressed as follows in terms of the Christoffel symbols of the second kind and the second fndamental form. These

22 3.5. CALCULATING CHRISTOFFEL SYMBOLS are often called the Gass formlas: or or w w w w w + + L n + L n + L n w w + L w w n n 4 w w L w w w L w This means that we hae introdced notation for the first two colmns in [D w ]. We shall wait a bit to deal with the last colmn. As we shall see, and indeed already saw when considering polar coordinates in the plane, these formlas are important for defining accelerations of cres. They are howeer also important for giing a proper definition of the Hessian or second deriatie matrix of a fnction on a srface. This will be explored in an exercise later Calclating Christoffel Symbols Next we seek formlas for the Christoffel symbols that inole only the first fndamental form. This shows that they can be compted knowing only the first deriaties of (, ) despite the fact that they are defined sing the second deriaties! Proposition We hae that g g g g g g g g Proof. We select to proe only two of these as the proofs are all similar. We se the prodct rle jst as we did when compting deriaties of dot prodcts for the Frenet-Serret formlas. Note that we se the first calclation to finish off the second calclation. 3 5 g

23 3.5. CALCULATING CHRISTOFFEL SYMBOLS 3 g g g There is a nified formla for all of these eqations. While it nifies it also complicates and is less sefl for actal calclations: w w w gww + g w w g ww w w w The prodct rle for deriaties also tells s that g ww w ww w + www Note that this formla is now also a direct conseqence of or new formlas for the Christoffel symbols in terms of the deriaties of the metric coefficients. The proposition can also be sed to find the Christoffel symbols of the second kind. For example g + g g g + g g While this can t be made simpler as sch, it is possible to be a bit more efficient when calclations are done. Specifically we often do calclations in orthogonal coordinates, i.e., g 0. In sch coordinates g 0 g (g ) g (g ) g g g g g g

24 3.6. GENERALIZED AND ABSTRACT SURFACES 4 and g g g g g g g g g g g g ln g ln g ln g ln g We often hae more specific information. This cold be that the metric coefficients only depend on one of the parameters, or that g. In sch circmstances it is qite manageable to calclate the Christoffel symbols. What is more, it is always possible to find parametrizations where g and g 0 as we shall see Generalized and Abstract Srfaces It is possible to work with generalized srfaces into Eclidean spaces of arbitrary dimension: (, ) :U! R k for any k. What changes is that we no longer hae anormalectorn. In fact for k we cold jst let n be (0, 0, ) after letting R be the (x, y) coordinates in space. While for k 4 we get a whole family of normal ectors, not nlike what happened for space cres. What all of these srfaces do hae in common is that we can define the first fndamental form and with it also the Christoffel symbols of the first and second kind sing the formlas in terms of deriaties of g. This lea s to the possibility of an abstract definition of a srface that is independent of a particlar map into some coordinate space R k. One of the simplest examples of a generalized srface is the flat tors in R 4. It is parametrized by (, ) (cos, sin, cos, sin ) and its first fndamental form is I apple 0 0 jst as we hae for Cartesian coordinates in the plane. This is why it is called the flat tors. It is in fact not possible to hae a flat tors in R 3. An abstract parametrized srface consists of a domain U R and a first fndamental form I apple g g g g that defines inner prodcts of ectors X, Y with the same base point p U where I(X, Y ) X X apple apple g g Y g g Y X (X,X ),Y (Y,Y )

25 3.6. GENERALIZED AND ABSTRACT SURFACES 5 are the representations of the ectors sing the standard (, ) coordinates on U. Note the first fndamental form consists of three fnctions and so gies an inner prodct that aries from point to point. For this to gie s an inner prodct we also hae to make sre that it is positie definite: 0 < I(X, X) X X apple apple g g X g g X X X g +X X g + X X g Proposition I is positie definite if and only if g + g, and g g (g ) are positie. Proof. If I is positie definite, then we see that g and g are positie by letting X (, 0) and (0, ) resp. Next let X p g, ± p g to get Ths we hae showing that 0 < I(X, X) g g ± p g p g g ±g < p g p g g g > (g ). To check that I is positie definite we hae to se that it is symmetric. The characteristic polynomial is (g + g ) + g g (g ) The minimm of this pward pointing parabola is obtained at and has the ale (g + g ) g g (g ) 4 (g + g ) 4 (g g ) (g ) < 0 Ths there are two real roots. The spectral theorem for symmetric matrices cold also hae been inoked at this point to establish that the eigenales are both real. It is now easy to see that [I] is positie definite if its eigenales are positie. Two real nmbers hae to be positie if their sm and prodct are both positie. In this case the sm of the eigenales is the trace g + g while the prodct is the determinant g g (g ) so or assmptions garantee that the eigenales are positie. There is an interesting example of an abstract srface on the pper half plane defined by H {(, ) :>0}, where the metric coefficients are apple g g [I] apple 0 g g 0 One can show that for each point p H, there is a small neighborhood U containing p and (, ) :U! R 3 sch that t apple apple 0 0

26 3.6. GENERALIZED AND ABSTRACT SURFACES 6 In other wor we can locally represent the abstract srface as a srface in space. Howeer, a ery difficlt theorem of Hilbert shows that one cannot represent the entire srface in space, i.e., there is no fnction (, ) :H! R 3 defined on the entire domain sch that apple t apple 0 0 Janet-Brstin-Cartan showed that if the metric coefficients of an abstract srface are analytic, then one can always locally represent the abstract srface in R 3. Nash showed that any abstract srface can be represented by a map (, ) :U! R k on the entire domain, bt only at the expense of making k ery large. Based in part on Nash s work Greene and Gromo independently showed that one can always locally represent an abstract srface in R 5.

27 CHAPTER 4 Cres on Srfaces 4.. Acceleration and Geodesics We ll now consider cres on a parametrized srface (, ) :U! R 3. The cre is parametrized in U as ( (t),(t)) and becomes a space cre c (t) ( (t),(t)) that lies on or parametrized srface. The elocity is ċ dc d d + d apple d d Next we calclate the acceleration as if it were a space cre, bt sing the elocity representation we jst gae. Recall that we can decompose any ector into normal and tangential components. For the acceleration this is c [I] t c +( c n) n The goal is to calclate each of these components in terms of d This will lead s to another srprising reslt. where Theorem 4... The acceleration can be calclated as 3 c n d + (ċ, ċ) 4 d + (ċ, ċ) 5 w (ċ, ċ) d + (ċ, ċ) + X w,w, II (ċ, ċ) w dw dw w w d d + (ċ, ċ) d, d + nii (ċ, ċ), apple w w w w and d, d. apple d d Proof. We start from the formla for the elocity and take deriaties. This clearly reqires s to be able to calclate deriaties of the tangent fiel,. Fortnately the Gass formlas tell s how that is done. This lea s to the acceleration as follows c d apple d d which after the chain rle " d d # + d d d + d 7 apple d d

28 4.. ACCELERATION AND GEODESICS 8 becomes c + d + d " d d # apple d d apple d d The Gass formlas help s with the last two terms w apple d d + w w n 4 w w w w L w apple d d apple d d +n L w L w apple d d w w L w 3 5 apple d d which after frther rearranging allows s to conclde c + + d d +n d " d d d d d # apple apple apple L n 4 L d L L + (ċ, ċ) d + (ċ, ċ) II (ċ, ċ) apple d d apple d d apple d d 3 5

29 4.. ACCELERATION AND GEODESICS 9 Alternately the whole calclation cold hae been done sing smmations c d c d + + d d + + +n d + d + d + d + X w,w, X w,w, w,w, d + (ċ, ċ) d d + X w,w, dw w w! dw dw w w X dw dw w w! dw dw L ww + d + d d! dw d + (ċ, ċ) + nii (ċ, ċ) Note that we hae shown Theorem 4... [Mesnier] The normal component of the acceleration satisfies ( c n) n c II nii (ċ, ċ) In particlar two cres with the same elocity at a point hae the same normal acceleration components. The tangential component is more complicated [I] t c c I d + (ċ, ċ) + d + (ċ, ċ) Bt it seems to be a more genine acceleration as it incldes second deriaties. It actally tells s what acceleration we feel on the srface. Note that the tangential acceleration only depen on the first fndamental form. We say that c is a geodesic on the srface if the tangential part of the acceleration anishes c I 0, or specifically d + (ċ, ċ) 0, d + (ċ, ċ) 0. This is eqialent to saying that c is normal to the srface or that c nii (ċ, ċ). Proposition A geodesic has constant speed.

30 4.. ACCELERATION AND GEODESICS 30 Proof. Let c (t) be a geodesic. We compte the deriatie of the sqare of the speed: d I(ċ, ċ) d (ċ ċ) c ċ II (ċ, ċ) n ċ 0 since n and ċ are perpendiclar. Ths c has constant speed. Note that we sed the second fndamental form to gie a simple proof of this reslt. It is desirable and indeed possible to gie a proof that only refers to the first fndamental form. The key lies in showing that we hae a prodct rle for I(ċ, ċ) that works jst inside the srface. Since I c I, ċ c ċ this is fairly obios. The goal wold be to do the calclation sing only the first fndamental form and that takes qite a bit more work. Next we address existence of geodesics. Theorem Gien a point p ( 0, 0 ) and a tangent ector V V ( 0, 0 )+V ( 0, 0 ) T p M there is a niqe geodesic c (t) ( (t),(t)) defined on some small interal t ( ", ") with the initial ales c (0) p, ċ (0) V. Proof. The existence and niqeness part is a ery general statement abot soltions to differential eqations. In this case we note that in the (, ) parameters we mst sole a system of second order eqations with the initial ales d d d d d d apple apple ( (0),(0)) ( 0, 0 ), ( (0), (0)) (V,V ). apple d d apple d d As long as is sfficiently smooth there is a niqe soltion to sch a system of eqations gien the initial ales. The domain ( ", ") on which sch a soltion exists is qite hard to determine. It ll depend on the domain of parameters U, the initial ales, and finally on. This theorem allows s to find all geodesics on spheres and in the plane withot calclation. In the plane straight lines c (t) p + t are clearly geodesics. And since these sole all possible initial problems there are no other geodesics. On S we claim that the great circles c (t) p cos ( t)+ sin ( t) p S, p 0

31 4.. ACCELERATION AND GEODESICS 3 are geodesics. Note that this is a cre on S, and that c (0) p, ċ (0). Next we see that the acceleration c (t) p cos ( t) sin ( t) c (t) compted in R 3 is normal to the sphere. Ths c I 0. This means that we hae also soled all initial ale problems on the sphere. Exercise: Let c (s) be a nit speed cre on a srface with normal n. Show that it is a geodesic if and only if [c 0,c 00,n]0 Exercise: Let c (s) be a nit speed cre on a srface with normal n. Define T as the sal tangent to the cre and S n T as the normal to the cre in the srface. Show that d 0 apple g apple n T S n T S n 4 apple g 0 g apple n g for fnctions apple g,apple n, g. They are called geodesic cratre, normal cratre, and geodesic torsion respectiely. Frther show that S and c I are proportional and that apple g I S, c I S dt, apple n II(ċ, ċ) n dt, g II(S, ċ) n ds. Exercise: Show that the geodesic cratre can be compted as apple g T p det [I] T Exercise: Define the Hessian of a fnction on a srface abstractly by Hessf (X, Y )I(D X rf,y ) Show that the entries in the matrix [Hessf] defined by Hessf (X, Y ) X X apple Y [Hessf] are gien as f w w + f f Frther relate these entries to the dot prodcts rf w w apple ww w w Y

32 4.. UNPARAMETRIZED GEODESICS Unparametrized Geodesics It is often simpler to find the nparametrized form of the geodesics, i.e., in agienparametrizationtheyareeasiertofindasfnctions () or (). We start with a tricky characterization showing that one can characterize geodesics withot referring to the arclength parameter. The idea is that a reglar cre can be reparametrized to be a geodesic if and only if its tangential acceleration c I is tangent to the cre. Lemma 4... A reglar cre c (t) a geodesic if and only if ( (t),(t)) can be reparametrized as d d + (ċ, ċ) d d + (ċ, ċ). Proof. Let s correspond to a reparametrization of the cre. When switching from t to s we note that the left hand side becomes d d + (ċ, ċ) d d dc +, dc d d d s d + d s d + d s d d + d + d + 3 d d + with a similar formla for the right hand side. Here the first term d s d d dc,! dc dc, dc! dc, dc is the same on both sides, so we hae shown that the eqation is actally independent of parametrizations. In other wor if it hol for one parametrization it hol for all reparametrizations. If c is a geodesic then the formla clearly hol for the arclength parameter. Conersely if the eqation hol for some parameter then it also hol for the arclength parameter. Being parametrized by arclength gies s the eqation I ċ, c I d d apple g g g g " d + (ċ, ċ) d + (ċ, ċ) # 0 Ths we hae two eqations d g + g d d + (ċ, ċ) d d + (ċ, ċ) d + g + g d d + (ċ, ċ) d + (ċ, ċ) d 0, 0

33 4.. UNPARAMETRIZED GEODESICS 33 Since apple det g d d + g d d d d g + g ċ the only possible soltion is d + showing that c is a geodesic. g d + d d + g d d g + g (ċ, ċ) 0 d + (ċ, ċ), d Depending on or parametrization (, ) geodesics can be pictred in many ways. We ll stdy a few cases where geodesics take on some familiar shapes. Consider the sphere where great circles are described by ax + by + cz 0, x + y + z If we se the parametrization p +s +t (s, t, ), or in other wor x z s, y z t then these eqations simply become straight lines in (s, t) coordinates: as + bt + c 0 Or we cold se,, p and note that the eqations become a + c +ab + b + c c which are the eqations of ellipses whose axes go throgh the origin and are inscribed as well as tangent to the nit circle. This is how yo draw great circles on the sphere! The first fndamental form is gien by " # + [I] + Here is an intrinsic metric on the (, ) plane where we hae simply switched signs from aboe " # [I] Using the parameter independent approach to geodesics one can show that they trn ot to be hyperbolas whose axes go throgh the origin a c +ab + b c c This metric can also be reparametrized to hae its geodesics be straight lines. The later reparametrization is: s p + + t p + + and the geodesics gien by as + bt + c 0.

34 4.3. SHORTEST CURVES 34 We shall later explicitly find the geodesics on the pper half plane sing the eqations deeloped here. Exercise: Show that geodesics satisfy a second order eqation of the type d 3 d d d A + B + C d d d d + D and identify the fnctions A, B, C, D with the appropriate Christoffel symbols Shortest Cres The goal is to show that the shortest cres are geodesics, and conersely that sfficiently short geodesics are minimal in length. For the latter sing geodesic coordinates makes an argment that is similar to the Eclidean ersion sing a nit gradient field.

35 CHAPTER 5 The Second Fndamental Form 5.. Inariance Isses We offer a geometric approach to show that the second fndamental form is, like the first fndamental form, defined in sch a way that selecting a different parametrization will not affect it. The key obseration is that if we hae a srface M and a point p M, then the tangent space T p M is defined independently of or parametrizations. Correspondingly the normal space N p M (T p M)? of ectors in R 3 perpendiclar to the tangent space is also defined independently of parametrizations. Therefore, if we hae a ector Z in Eclidean space then its projection onto both the tangent space and the normal space are also independently defined. Consider a cre c (t) on the srface. We know that the elocity ċ and acceleration c can be calclated withot reference to parametrizations. This means that the projections of c onto the tangent space, c I, and onto the normal space, nii n (ċ, ċ), can also be compted withot reference to parametrizations. This shows that tangential and normal accelerations are well defined. This also takes care of nii n (X, Y ) if we se two important obserations. The first is called polarization, theideaisthatsymmetricbilinearformshaetheproperty: nii n (X, Y ) (niin (X + Y,X + Y ) nii n (X, X) nii n (Y,Y )) Ths it sffices to show that nii n (Z, Z) is well defined. Bt this follows from knowing that nii n (ċ, ċ) is inariant and that any tangent ector is the elocity of some cre. 5.. The Weingarten Map and Eqations There is a similar set of eqations for the entries in the second fndamental form that also lead s to the partial deriaties of n (, ). Together these are also known as the Weingarten eqations. Bt first we need to introdce the Weingarten map. It is related to the second fndamental form in the same way the Christoffel symbols of the second kind are related to the symbols of the first kind. Its matrix or the entries of its matrix are apple L L L w L L w g w L w + g w L w, [L] [I] [II], apple apple g g L L g g L L 35

36 5.. THE WEINGARTEN MAP AND EQUATIONS 36 When sing matrix langage we mst be carefl in or definitions as [I] [II] and [II] [I] are generally not the same. The abstract Weingarten map L will be a self-adjoint map with respect to the first fndamental form I(L (X),Y)I(X, L (Y )) bt this does not garantee that its matrix representation [L] is symmetric. This will only be the case if we are lcky enogh to hae sed an orthonormal basis. Proposition 5... We hae I(L (X),Y)II(X, Y ) In particlar L is self-adjoint as II is symmetric. Proof. We hae II (X, Y ) X X apple Y [II] Y, I(L (X),Y) X X apple [L] t Y [I] Y and by definition [L] [I] [II] so [L] t [II][I] as [I] and [II] are symmetric. This shows that I(L (X),Y)II(X, Y ). Next we find a new formla for L. Proposition 5... [Weingarten Eqations] L ww [II] n L n L n, w w n n t t n L L n L L Proof. The strategy is jst as with Christoffel symbols, bt works ot a bit more easily L ww were we sed that n is perpendiclar to w. n w w n w w n n w w w w n w w

37 5.3. THE GAUSS CURVATURE AND MAP 37 For the second set of eqations we note L L [L] [I] [II] [I] Bt n is a nit ector field so n n w w n 0 showing that n w is a tangent ector. In particlar L L n n t n The Weingarten eqations can also be combined into one eqation n w L w L w L. w The Gass formlas and Weingarten eqations together tell s how the deriaties of or basis,,n relate back to the basis. They can be collected as follows: Corollary [Gass and Weingarten Formlas] w n n [D w ] n w w 4 w w L w L w The Gass Cratre and Map One of the interesting featres of the Weingarten map is that its trace H L + L and determinant K L L L L yield fnctions on the srface that are independent of the chosen parametrization. Clearly the entries themseles do depend on parametrizations. This means that if we hae two parametrizations arond a point p M, then the calclation of H and K at p will not depend on what parametrization we se! H is called the mean cratre and K the Gass cratre. We saw that for a fixed choice of normal the second fndamental form is defined independently of the parameters. This will clearly also be tre of the Weingarten map. The next theorem is therefore obios. Neertheless it is instrctie to offer alessinspiredproof. Theorem The mean and Gass cratres do not depend on the parametrizations. Proof. The key to proing this is to first realize that we know that the trace and determinant of a matrix do not depend on the basis that is sed to represent the matrix, second we need to see that the Weingarten map changes according to the change of basis rles when we change parametrizations. In these calclations we assme that the normal ector field is fixed rather than gien as a formla that depen on the tangent fiel. There are only two choices for the normal field ±n, L w L w n 3 5

38 5.3. THE GAUSS CURVATURE AND MAP 38 and II as well as L will also change sign if we change sign for n. Note that this sign change does affect H, bt not K! The Weingarten map is calclated by n n apple L L L L and similarly in (s, t) coordinates n s n t s t apple L s s L t s L s t L t t Changing parametrizations is done sing the chain rle which in matrix form looks like s n s t n t n n apple s s apple s s t t t t Ths n n apple s s t t n s n t s t apple L s s L t s apple s s L s t L t t t t apple L s s L t s L s t L t t showing that n n apple s s t t apple L s s L t s L s t L t t apple s s t t or apple L L L L apple s s t t apple L s s L t s L s t L t t apple s s t t This in trn gies s apple L det L L L det apple s s apple det s s t t t t apple L s s L t s L s t L t t apple L s det s L s t L t s L t t apple s t s t apple det s s! t t L s detapple s L s t L t s L t t

39 5.3. THE GAUSS CURVATURE AND MAP 39 and apple L tr L L L tr tr apple s s apple s s t t t t apple L s s L t s apple s s L s t L t t t t apple s s apple L s s L t s t t L s t L t t!! L s trapple s L s t L t s L t t Note that we hae in fact shown that the linear map L : T p M! T p M does not depend on the parametrizations we se. We can frther find a ery interesting formla for the Gass cratre Proposition [Gass] K n n Proof. Simply se the Weingarten eqations to calclate n n L L L L L L + L L n (L L L L ) K n Note that the denominator is already compted in terms of the first fndamental form g g (g ) The nmerator is actally ery similar in natre as it is simply the corresponding expression for the so called Gass map n (, ) :U! S () R for the srface, i.e., compted from the first fndamental form of n (, ). Thismapisoranalog of the tangent spherical image. Note that for the nit sphere the nit normal at n n is ±n. Ths n n represents the oriented area for the parallelogram whose sides are n, n.recallfromcretheorythatthetangentsphericalimage was also related to cratre in a similar way. Here the formlas are a bit more complicated as we se arbitrary parameters. One classically defines the third fndamental form III as the first fndamental form for n [III] n t n n apple n n n n n n n n n

40 5.3. THE GAUSS CURVATURE AND MAP 40 This certainly makes sense, bt n might not be a genine parametrization if the Gass cratre anishes. Note howeer that n is not jst the normal to the srface, bt also to the nit sphere at n n n n n n This is part of what we jst established. The three fndamental forms and two cratres are related by a ery interesting formla which also shows that the third fndamental form is almost redndant. Theorem We hae III HII + KI 0 Proof. We first redce this statement to the Cayley-Hamilton theorem for the linear map L. This relies on showing I(L (X),Y) II(X, Y ), I L (X),Y III(X, Y ) and then proing that any matrix satisfies: L (tr (L)) L +det(l) I 0 where I is the identity matrix. The last step can be done by a straightforward calclation. We already saw that I(L (X),Y)II(X, Y ) as that followed directly from [II] [I] [L]. Wesimilarlyhae [III] L L t L L [L] t [L] t [I] [L] [II][L] [I][L][L] [I][L] t [L] showing that I L (X),Y III(X, Y ). In a related ein we mention Gass amazing discoery that the Gass cratre can be compted knowing only the first fndamental form. Gien the definition of K this is certainly a big srprise. A different proof that ses or abstract framework will be gien in a later section. Here we se a more direct approach. Theorem [Theorema Egregim] The Gass cratre can be compted knowing only the first fndamental form. Proof. We start with the obseration that K detl det[i] det [I] g g (g ) det [II],

41 5.3. THE GAUSS CURVATURE AND MAP 4 So we concentrate on apple L det [II] det L " L L # det n n n n " g g (g ) det # Which then redces s to consider det " # Here each entry in the matrix is a triple prodct and hence a determinant of a 3 3 matrix h det w w i w w With that obseration and recalling that a matrix and its transpose hae the same determinant we can calclate the prodcts that appear in or determinant h i h det det h i t h det det h i t h det 6 det4 det4 i 3 g g g g det [I] + det i i 0 g g g g 3 5

42 5.4. PRINCIPAL CURVATURES 4 and similarly 6 det4 det4 g g 3 g g det [I] + det 4 g g g g Since we need to sbtract these qantities we are finally redced to check the difference Ths K det + " det 4 det [I] (det [I]) 0 g g g g 3 5 det 4 (det [I]) # g g g g Exercise: Compte the mean and Gass cratres of the generalized cones, cylinders, and tangent deelopables. It trns ot that these are essentially the only srfaces in space with anishing Gass cratre Principal Cratres The principal cratres at a point p on a srface are the eigenales of the Weingarten map associated to that point, and the principal directions are the corresponding eigenectors. The fact that L is self-adjoint with respect to the first 3 5