Topics in Geometry & Topology Synthetic Geometry
Topics in Geometry & Topology Synthetic Geometry MATH 4080
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01 ARBITRARY CURVES 1 01 Arbitrary Curves We have shown that all regular curves can be parametrized by arclength but to explicitly do this for all curves is impossible due to the nature of the functions involved We will need to modify the F renet formulas from the previous section to account for a nonconstant speed Let at be an arbitrary curve with u lot dsdt Theoretically we can parametrize Oz by arclength and get a unit speed curve 07st Even though we cannot be explicitly solve for 64 we can use it to de ne the curvature and torsion of 04 Also the tangent vector to 04 is in the same direction as the unit tangent vector Ts and Tst otlotl We can de ne these curve invariants in terms of our unit speed parametrization and then try to nd an easier way to compute them De nition 01 Let 04 be an arbitrary curve and 64 its unit speed parametrization 1 The unit tangent of at is de ned to be Tt 2 The curvature of at is de ned to be Mt 3 If a gt 0 then the principal normal of at is de ned to be Nt 4 If a gt 0 then the binormal of at is de ned to be Bt Est 5 If a gt 0 then the torsion of of at is de ned to be rt st In general a and E are different functions de ned on different intervals However they give exactly the same description of the turning of the common route of Oz and E Lemma 01 Frenet Formulas for Arbitrary Curves Fora regular curve 04 with speed 1 lsalt and curvature a gt 0 T HlN N HlT l THE B irzN PROOF We will take it one at a time using our de nitions The unit tangent Tt Tst by de nition Using the Chain Rule dT dT ds E EM 3 stNstzt by the F renet formulas ntNt1t by de nition HlN T t The second and third formulas follow in the same manner I This tells us what they are but it does not tell us how to nd them yet Let s see if we can nd 0 and a in terms of the F renet frame Lemma 02 o VT and a lT rel2N PROOF The rst is not hard The second is again a straightforward calculation 1 Mt d TlttgtulttgtT lttgt lT VHlT lT rel2N I Note that this last equation separates acceleration a into two components 7 a tan gential component lT and a normal component rel2N These are used in physics and are called the tangential and normal components of acceleration Now back to a computation of the Frenet apparatus for an arbitrary curve We know that T i M B T X N N B X T How is this going to help us From the preVious lemma we can show 0 X 04 VT X llT rel2N IIIT X T Hl3T X N rel3B Thus lo X 04 rel3 since lBl 1 and we have shown that 0 X a B 1 lo X a l lo X a l lo X a l H T 2 Now how do we nd the torsion We will use the third derivative 0 llT l HVZN lHT VT HVZN QHVVN HlZN lHT lHlN FLV2 2amp11N HZ27HlT THE 11 7 H2V3T HI2 I DV2 2HVIN HTlSB B 0 HTlS 01 ARBITRARY CURVES So7 04 X a OU OU HV3HTl3 FLZVGT 04 X a am 04 X a O T T We have just proven the following theorem Theorem 01 For any regular curve 04 the following formulas hold 0 X a o X 04 3 o X 0W1 r O 3 4 04 X a am T O X an z 5 B Example 01 Consider the curve at 3t 7 25373252325 253 The necessary derivatives are ot 317t22t1t2 a t 67t1t oquott 6717071 What we need for our computations are M 3 17t222t21t2 3 1t2 wquot 6V5 oon 1871t272t1t2 WWW 18 1t2 oXoo 216 o X 04W 18 1t2 W lt3 lt1t2gtgt3 1 31t22 7 o X 04 04 7 216 T W W i 1 7 3 1t22 T i 17t22t1t2 6W ow B 7 ozon 7 71t272t1t2 04 X 04W lt1t2 N BxT 7222172520 12 42 44 Surfaces of Revolution Many of the interesting examples that we have seen arise from revolving a curve about an axis This surface of revolution has a patch ofthe form xa7 v 9a7 hu cos 1 hu sin 12 The curves on the surface formed slicing the surface with a plane perpendicular to the axis of rotation are called parallels denoted 7139 The curves which look like the original curve formed by slicing the surface with a plane containing the axis of rotation are called meridians denoted Ia Also7 recall that 9a is the distance along the axis of revolution and hu is the radius of the parallel We have Xu 97hCosvhsinv xv Ofmini hcosv xu x xv hh7 79 cos 1 igh sin 12 H7 79 cos v 79 sin 12 U g2 h2 qu 9 7 71 cos U h sin 1 KW 07 771 sinv7 h cos 12 XM 07 ihcosv ihsinv Therefore E 92 he l W 7 Mg g2 be F 0 m 0 G h n hg g2 hZ39 Thus7 the Gaussian curvature is ggnh 7 hg hg2 h22 If the original curve at 9257 ht0 has the property that g t 7 0 for all 257 then 9 is strictly monotonic Thus it is one to one and has an inverse function 9 1 which is also differentiable This means that we can reparametrize the curve 4 De ne f h o 9 1 and BM a 09 1u 99 1u7 h9 1u70 u7fu70 In this case7 the computations now are very simple We will still write au a7 Ma 0 and the Gaussian curvature is then h h 1 112 We should be able to compute the normal curvatures for the median curves and the parallel curves for a surface of revolution To do this7 we will compute the eigenvalues of the shape operator because the median and parallel curves are xu and xv7 respectively 44 SURFACES OF REVOLUTION 43 5Xu quU XulUllyxulUZly XulUsl i gh 7 hgH h COSUh 7 hgH h sinvh 7 hggt 7 h2g232 7 h2g232 7 h2g232 hgn 7 hng 5Xv VXUU leU1l7 X11 Uzl7 X11 Wail 7 0 g sinv g cosv 7 7 7 h2 9277 hZ g2 g x hwy2 112 Thus the off diagonal entries of the matrix for Sp are zero and the values that we found here are the eigenvalues of Sp and hence the principal curvatures h 7 hNg g k 7 9 2 W k hW Example 43 The Torus The torus is a surface of revolution We take a circle of radius r centered R units from the y axis and rotate it about the y axis Our patch is xu v R rcosu cosv R rcosu sinv rsinu Looking closely we can see that gu r sinu and hu R rcosu This means that the Gaussian curvature is given by gu rcosu hu irsinu g u irsinu h u ircosu h 7 h K g g 2 29 hg h 2 r cos ur2 sin2 u 72 cos2 u R r cos ur2 cos u r2 sin2 74 os u K MR r cos u Now for 7 lt u lt cosu gt 0 so K gt 0 on the outer half of the torus If u g then K 0 and this is the top and the bottom of the torus Then for g lt u lt 3quot cosu lt 0 so K lt 0 on the inner half of the torus The maximum value for the Gaussian curvature occurs at u 0 and is 1 K 0 MR r The minimum value will occur at u 7139 and is K7r 7 Figure 41 Tractrix curve The sphere of radius R has constant Gaussian curvature of K 1132 Are there other surfaces that have constant Gaussian curuature There must be7 though we would think that ifa surface has constant positive curvature7 it must be part of a sphere If the curvature is 07 we would expect the surface to be part of a plane ls it possible to have a surface with constant negative curvature Example 44 The Pseudosphere Describe the curve characterized by a weight being dragged on the end of a xed straight length and the other end moves along a xed straight line Think of the curve as starting at the point 07 c The tractrix is the curve characterized by the condition that the length of the segment of the tangent line to the curve from the curve to the m axis is constant We insist that the curve be tangent to the y axis at the point 07 37 so this makes this constant distance equal to c This is also called the pursuit curve for it could be the path of a jet ghter trying to intercept a second airplane moving along the z axis Write au uhu and note that h u lt 0 Computing ou Lbu we see that the tangent line to the curve at the point u7 is pt au tou u t7 hu tit14 Then7 when t 0 we are at au and when t ihh the line intersects the z axis The length of this line segment has to be c so cwwwwmue wwm www 2 I ll141 Ii2 2 h Solve for h 2 i 3 5 am Now7 we can nd k7r7 kM and K 44 SURFACES OF REVOLUTION Since gu u our formulas become h 7h 1 h 22 h kt k 1 W fix1 H2 We have 71 and h above7 so 32h k7 3 2 027712 1 hz Czihz if h c 627712 7 1 hxlt Czihz hc K7 32 Thus7 the pseudosphere has constant negative curvature What is h We can nd h simply enough by solving the separable differential equation h W 710 c h 7 for h to get C u V 02 7 h 7 carctanh 0 W If we make the substitution h csechw will actually simplify this expression to u 7V02 7 h2 carctanh m 2 2 7 W 6mm COS w 0 cos w71 cosh w 7ctanh w carctanhtanh w u cw 7 ctanhw 46 Figure 42 Pseudosphere Thus7 we can parametrize the pursuit curve as aw Cw 7 tanh w7 sech w and put a patch on the surface as x0171 Cw 7 tanhw7 sechw cos v7 sech w7 sin 1 Actually7 the pseudosphere is determined by the condition that it have constant curva ture and certain initial conditions If we start with a surface of revolution M have constant Gaussian curvature K 71 Then we know that we can parametrize M with the patch xu7 v 9a7 hu cos 1 hu sin 12 with g u2h u2 1 This would mean that the curve is a unit speed curve Then differentiating we have 99 71 0 or 99 ih h So the Gaussian curvature will reduce to the formula ggnh 7 hg My h 2 7 ggnh 7 hHgZ T h 7hhh 7 h17 hZh h 7 The condition that K 71 then reduces this to a simple linear differential equation to solve h h The solution to this differential equation is hu A5 l B57 44 SURFACES OF REVOLUTION 47 Plugging this into the condition that we have a unit speed curve as the pro le curve and solving for gu will give gu VliAetiBe tdt uo Choosing different values for A and B give us different surfaces of constant curvature Our example above is A 1 and B O 4 2 OTHER CURVATURES 37 42 Other Curvatures De nition 42 The Gaussian curvature of a surface M at p E M is de ned to be Kltpgt dam The mean curvature of a surface M atp E M is de ned in terms of the trace Hp traceSp It appears that K and H are de ned solely in terms of the shape operator They are but we shall see that we can calculate them by calculus alone We have seen that the matrix of the shape operator with respect to a basis of principal vectors is given by V10 0 J 0 209 and the determinant and trace are then k1 kg and k1 k2 Since the determinant and trace do not change in a change of basis we have that Kk1k2 and H Now recall that we had to make a choice in the sign of U If we had chosen 7U as the normal instead of U then ku changes sign Since the Gaussian curvature is the product of two such changes K does not change sign by switching the choice of U However H does change sign under a change in unit normal Suppose that Kp gt 0 Since K klkg then both k1 and k2 have the same sign Since k1 max ku and k2 min ku then all ku have the same sign for all u So if ku gt 0 for all u then M bends toward U in every direction If ku lt 0 for all u then M bends away from U in every direction How can we compute the curvature Let V and w be linearly independent vectors in TpM Since they are linearly independent and M is a surface they form a basis for TpM Thus SV av bw and Sw cV dw The matrix for S with respect to the basis V w is a b c d 39 From this we see that detS ad 7 be K and traceS a d 2H We will need the following calculations SV x Sw V aVbw x cvdw aeV x V adV x w bewx V edwx w 0ad7bcwaO Kwa SVgtltWVgtltSWaVbWgtltWVgtltCVdW awadwa traceSV x w 2Hwa 38 Lemma 43 Lagrange Identity For vectors u V w and x vxwuxx Vuwx7vxwu Applying this lemma to the formulas above we get SV x Sw V x w KV x w V x w 50 VSW W 7 SW WSW V KV VW W 7 V WW V solving for K we get 50 VSW W 7 SW WSW V VVww 7 Vwwv K Sv xwvxSwvxw 2Hvxwvxw 5V VW W 7 SM WWV VVSW W 7 VWSW V 2HVVWW 7 VWWV Dividing we obtain 5V VW W 7 SM WW V V VSW W 7 V WSW V 2V VW W 7 V WW V Now let s look at how the principal curvatures are determined by the Gaussian and mean curvatures We know that K klkg and 2H k1 k2 We want to nd ki in terms of K and H One interesting way to see this is that k1 and k2 are eigenvalues for the H shape operator S In terms of the principal directions the matrix for S is j Z The eigenvalues are the roots of the characteristic polynomial for this matrix The characteristic polynomial for this matrix is pm m2 7 a dm ad 7 bc 2 7 trace Sm det S m2 7 2Hz K The roots are k1 H VH2 7K and k2 H7 VH2 7K What is the geometry of the Gaussian curvature Let s go back and look at the Gauss map The derivative of the Gauss map G M 7 2 is the negative of the shape operator GV 7SV Then for the basis xmxv we have 04x7 x cum 7 75m x 75 K xu x x Geometrically lGun x Gxvl and lxu x xvl represent areas 7 in nitessimal areas on the image of the Gauss map on 52 and M respectively Our above formula says that the ratio of these two in nitessimal areas is Another way to say this is that for any small open neighborhood V containing p E M then Area C V K l l l V1an Area V This says that the magnitude of the Gaussian curvature measures the way in which the unit normal expands or contracts area A surface M is said to be flat if Kp 0 for every p E M and it is said to be minimal if Hp 0 for every p E M 322 The Gauss Map Let F M a N be a mapping between two surfaces We may de ne its derivative to be a certain linear transformation on the tangent planes FT TpM a TppN We do this as follows Let V E TpM be represented by a curve 04 I a M with 040 p and oO V De ne 1 FM Ema H Let Mt Fat This 3 is simply a curve in N and so all we have done is to assign the tangent vector to 04 at 0 the tangent vector to Fa at 0 Again the derivative mapping is the best linear approximation to F near the point p A linear transformation is determined by what does to a basis so we need to consider what Fi does to xu and xv Example 35 Let F R2 a R2 be a map of the plane to itself Then in terms of coordi nates we may write F as FWU u v79uw Let s compute We only need to take the composite of the u parameter curve au 14120 with F Fuv0 fuv0gu 120 and differentiate with respect to u We get fu gu Doing the similar calculation for the v parameter curve will give us that FXu fmgu and FX39u 10717971 Thus with respect to the basis xm x the matrix for Fi is We v the Jacobian matrix for F This indicates that the derivative mapping F for surfaces is an extension to maps of surfaces of the Jacobian Now we have that the shape operator is a linear transformation of the tangent plane 239e S TpM a TpM ls there a mapping G on M so that Gi i5 The answer turns out to be yes The mapping whose derivative is i5 is called the Gauss map The shape operator is a linear algebraic approach to answering geometric questions The Gauss map is a geometric alternative to the shape operator The Gauss map is a mapping G M a 52 de ned by Cp Up where Up is the unit normal to M at p Since Up is a unit vector in R3 it represents a point on the unit sphere 2 The induced derivative map is Gr TpM a TGpSZ and by de nition 04v ltGltalttgtgtito ltUltalttgtgtito VonOW VVU ism 32 where 040 p and oO V There is one more thing to discuss here GT V E TGpSZ while we saw that SV E TPM How can GT 75 Recall our geometry and our linear algebra The tangent plane TPM is a vector space and therefore it passes through the origin We may Visualize it sitting in a different location in R37 but it does contain the origin The tangent plane TPM is a plane through the origin perpendicular to Up For any point q 6 2 the tangent plane TqSZ is perpendicular to q itself7 thought of as a vector in R3 Because Cp Up7 the plane TGpSZ is a plane through the origin perpendicular to Up This makes TAM TGpSZ Thus7 it makes sense to have GT 7 75 Chapter 3 Surfaces 31 De nitions De nition 31 If S is a subset of R37 then S is a regular surface if7 for each point p E S there is an open set V C R37 an open set U E R2 and a function X U a V O S satisfying 1 X is differentiable7 226 writing Xu7 v 1 17L12 2u127 f3uv7 the functions fi have partial derivatives of all orders to X is oneto one7 onto7 continuous and has a continuous inverse X 1 V O S a U This means that X 1 is the restriction to V of a continous function W a R27 for some open set W that contains V O S 3 The Jacobian matriX ltwgt ltwgt JXuv w uu a 3 gm 0M has rank 2 This means that for each 1471 6 U7 the function JXuv R2 a R3 thought of as a linear mapping7 is one to one or equivalently7 ker JXu7 v This condition on the function7 X7 is called regularity The function X U a S is called a coordinate chart Coordinate charts allow us to put local coordinates on a surface It makes precise the idea that in a neighborhood of a point7 a surface is two dimensional A chart is in some respects the at cartographic map that describes the surface near that point EXarnple 31 1 Let uV E R3 with v 7 0 Let Hu denote the plane in R3 that is normal to V and contains the vector u Let u 141142143 and V 121122123 and suppose that 123 7 0 We can create a coordinate chart for Hu by X R2 a R3 with m s lt7 3 W ug US 21 22 CHAPTER 3 SURFACES Note that Xu1 742 u and that XT s 7 u V 0 Thus X maps R2 to Hum It is clear that X is differentiable Let fv1v2v3 121122 Then fXTS r s This is continuous so X is oneto one and onto Hum The Jacobian is 1 0 JX 0 1 771 1 2 13 13 This has rank 2 Thus any plane in R3 is a regular surface to Let S2 1z2z3 E R3 l 3 3 1 denote the unit sphere in R3 Let D u v E R2 l 742 l v2 lt 1 denote the open unit disk in R2 De ne z D a S2 by Xuv ltuv 17 U2 7 U2 This chart only covers the upper hemisphere of S2 The reason for this is to make it a reasonable map and to keep the one to one nature of X This X is differentiable and has inverse X 1u1u2u3 u1u2 This is a regular chart as well 03 Let f R2 a R be a smooth function Then the graph of f is the set u v u and is a surface in R3 F Let 9 ab a R be a smooth function with gt gt 0 for all t 6 ab Rotate the graph ofg about the z aXis De ne a coordinate chart z a b x 0 27139 a R3 by zu v ugu cosvgu sinv Let s consider this in a slightly different manner We have our mapping X U a R3 given by Xuv zuvyuvzu If we X a value of v v0 and let u vary we have a function of one real variable 7 a curve in R3 Xu U0 is called the u pammeter curve Similarly Xing u uo and letting v vary we get the v pammeter curve Xu0 12 Of course both of these curves pass through the point duo 120 in R3 The tangent vectors for these u parameter and v parameter curves are found by taking the partials of the component functions with respect to u and 1 respectively Q m Xui aulaulau X UT 81178117811 39 Evaluating these at the point 140120 gives us two tangent vectors at the point Xu0 120 If we are to have a real surface then it should appear to be two dimensional at this point and we would hope that the two tangent vectors point in two different non opposite directions 239e they are linearly independent De nition 32 A mapping X U a R3 from an open set U is regular if Xu x XU eXists and is non zero for all points in U Our coordinate patch is a oneto one regular mapping X U a R3 A surface in R3 is a subset S C R3 such that each point of S has a neighborhood in S contained in the image of some coordinate patch 31 DEFINITIONS 23 Let s call these patches 1 1 etc Each of the patches on a surface is oneto one by de nition then the respective inverse functions exist and are continuous Then 1 U a R3 and jfl R3 a R2 If the point p is contained in the image of j and 1 then we can de ne a function jfl o 1 U a R2 is a map from R2 to R2 This map is smooth if all of the partial derivatives for all of the component functions exist and are continuous A surface is smooth if 171 o 1 is smooth for any pair of patches on the surface We will be dealing mostly with smooth surfaces If they are not smooth it will be only at a nite number of points eg the cone We are going to look for global patches ie a single patch that covers all but a nite number of points on the surface A function f S a R from a surface S is smooth if the composition f o gt U a R is smooth for each patch 1 on S A curve on a surface is simply a mapping from an interval in the real line to the surface S 04 I a S The surface is said to be path connected if for any two point pq E S there is a curve 04 01 a S with 040 p and 041 q We will be dealing with path connected surfaces unless otherwise stated A curve is smooth if jfl o 04 I a R2 is smooth for all patches a5 Lemma 31 LetM be a surface Ifa I gt C M is a curue in R3 which is contained in the image of a patch j on M then there are unique smooth functions ut ut I gt R so that To complete our process here we need to talk about the situation of a function between two surfaces Suppose that gtu 12 de nes a patch on the surface M and 110 5 de nes a patch on N Suppose that F M a N sends the point p E M in the image of the patch 1 to the point q E N in the image of the patch 1 We will say that f M a N is smooth if the composition nil o f o gt U a Uw is smooth as a map of open subsets of R2 That is all of the usual partial derivatives must exist and be continuous 311 Examples We have seen a few examples already Example 32 1 Any plane in R3 is a regular surface 2 Let f R2 a R be a smooth function Then the graph of f is the set u u fu and is a surface in R3 To see this let gtuu uufuu Then lt1 10 8f3u 10 fu and m 018f8u 01fv Is this patch regular All we need to do is to compute u gtlt U39 L J a u gtlt u 1 0 7107177101171 7g 0 0 1 saw This patch is called the Mange patch 24 CHAPTER 3 SURFACES 3 Spherical coordinates De ne Xu v R cos 11 cos 11131 sin 11 cos 11131 sin 11 Then xu 7R sin 11 cos 11131 cos 11 cos 110 X1 7R cos u sin 117131 sinu sin 11131 cos 11 Xu x X1 R2 cos 11 cos2 111312 sin 11 cos2 111312 sin 11 cos 11 KM x Xv R2 cosv 4 Let 1 11 a R be a smooth function with ft gt 0 for all t 6 11 Rotate the graph of 1 about the z axis De ne a coordinate chart X 11 x 0 27r a R3 by KW U U7 u 6080 u sinv If the curve C in the zy plane is parameterized by au gu and is revolved about the z axis the patch is de ned by Xu v gu hu cosv hu sinv 43 CALCULATING CURVATURE 39 43 Calculating Curvature The formulas that we generated for K and H were using general vectors V and w Let us specialize to the case xu and xv7 when we have a patch X for the surface M The following notation is traditional EXuXu ZSXuxu FXuX1 mSXuXv GXvxv nSXvXv Then7 replacing V and w by Xu and X1 we have 7 nimz TEEin iGlEn72Fm 7 2EF7G2 This is almost what we need to compute the curvature independent of the shape oper ator7 except for 17 m7 and n We can handle this situation by using Lemma 41 Corollary4 lUXWmUXM andnUXW Example 42 Let s revisit the hyperboloid of one sheet m2 yZ 22 E L We need a patch for this surface This is a surface of revolution7 by taking a hyperbola in the yz plane and rotating it about the z axis How can we parametrize this surface Since it is a surface of revolution and the curve can be parametrized by au 01 cosh u csinhu7 then the patch for the surface is xu7 v a cosh u cos U71 cosh u sin 1 c sinh So Xu a sinhu cosv7 b sinhu sinv7 c coshu X1 7a coshu sinv7 b coshu cosv7 0 Then xu x XU ibc cosh2u cosv7 iaccosh2u sinv7 ab sinhu coshu7 and the unit normal is U xu x x xu x x lxu x x l W where W xu x x b202 cosh4u cos2v 1202 cosh4u sin2v 1sz sinh2 cosh2u Now we need to compute E F G l m and n E xu xu a2 sinh2u cos2v 2 sinh2u sin2v 02 cosh2u F xu X1 7a2 sinhu cosv coshu sinv 2 sinhu sinv coshu cosv G X1 X1 a2 cosh2u sin2v 2 cosh2 cos2v Now EC 7 F2 19202 cosh4u cos2v 1202 cosh4u sin2v azbz sinh2u cosh2 W2 Compute the second partials XW a coshu cosv b coshu sinv csinhu X7W 7a sinhu sinv bsinhu cosv 0 KW 7a coshu cosv 7b coshu sinv 0 l KW U 7 7abc cosh3u csinhuab sinhu cos2v coshu ab sinhu sin2v coshu W 7 abc coshu W m KM U 0 n X39u39u U abc cosh3u cos2v abc cosh3 sin2v abc cosh3u The Gaussian curvature is K 7 Zn 7 m2 7 1213202 cosh4u 7 EC 7 F2 7 W4 which may be rewritten as K71 W2 b b m 6 cosh2 cos2v cosh2u sin2v 0 sinhZ 43 CALCULATING CURVATURE Using our patch7 we can see that W2 2 bcm2acy2ab22 2 abc cosh2u 7 a 2 b b2 c 02 z2 y z 2 7 222 iabc b 4C 4 K7 64 NN m 2 2 azbzc2 jg 7 The mean curvature is given by H 2 i 1abccoshuz2 y2 22 7 a2 b2 7 32 7 2 2 azbzcz 217 2y2227a2b2762 2 Qabc 3 14 7 1 abc coshu7a2 7 b2 a2 cosh2u cos2v 7 2 cosh2u cos2v 2 cosh2u 02 coshz 7 19202 cosh2u cos2v 1202 cosh2u 7 1202 cosh2u cos2v azbz cosh2u 7 azbz Chapter 1 Curves 11 De nitions and Examples De nition 11 A curve is a continuous mapping 04 I 7 R where I is an interval in the real line The interval could be closed ab open ab half open ab an open ray a 00 or a closed ray 700 b The function 04 is called a parameterization for the curve The curve is called smooth if 04 has derivatives of all orders It is called differentiable of order k if all derivatives of order k exist and are continuous The curve is called regular if 71 7g 0 for all t 6 ab We will be mostly interested in curves in R3 Then we can represent the curve in terms of its coordinate functions at z1tz2t z3t and so 51 z1tz2t Note that the notation in our text and hence throughout the remainder of these notes ttogt The image of Oz in R is called the trace of the curve and this is the geometric object We often identify a curve with its trace but realize that there may be many curves for a given trace to3 dt 1042 dt Oz 041 at a1ta2ta3t and oto ii t tto tto Example 11 We know that two points determine a unique line and in the plane we know how to nd a unique representation of this line using the slopeintercept equation for a line The slope is a unique number that is associated to a line in the plane We have no such number in three space What we do have joining two points P and Q is the following Let p denote the vector terminating at P and q denote the vector terminating at Q Then the vector joining P to Q is the vector q 7 p The line joining P and Q must run along this vector 239e it has the same direction as the vector q 7 p We can use a parameter t to tell us how far along the direction of q 7 p we are and put this together to get that the parametrized equation for a line in R3 is W p tq 719 Note that when t 0 you are at the point P and when t 1 you are at the point Q 2 CHAPTER 1 CURVES 0 as we have de ned it is called the velocity vector of a Note that for our line at p tq 7 p the velocity vector is our direction vector ot q 7 p We may see this written as at p tv By going through the usual de nitions of the derivative we can show that M ia w t7 to 7 W732 7 which looks like our slope de nition of the derivative De nition 12 Given a smooth curve 04 I a R3 the tangent vector to 04 at at is given by da t t yo ao The speed of 04 at at is the norm lotl The unit tangent vector to 04 is MW wwww lf at a1ta2t 04305 is a curve in R3 then its acceleration vector is given by 12041 12042 12043 7 0 t 7 lt dtz t7 dtz t7 dtz 39 Proposition 11 The curve 04 is a line if and only if a 0 PROOF lfat ptv is a line then ot V and hence Ot 0 since V is a constant If Ot 0 for all t then 120 it This means that each component of the vector 0 is constant 0 Lz3 7 do U dt and hence ot pi tvi and at p1 tvlp2 tU2p3 tug p tV Thus the equation can be parametrized as a line I What else can calculus tell us We know that distance is the integral of speed so this would imply that b rm a lot dt is the arclength of the curve 04 11 a R3 from aa to 041 11 DEFINITIONS AND EXAMPLES 3 Proposition 12 Schwarz7s Inequality The dot product obeys the inequality lvwl S lval PROOF First from properties of the dot product we know that v w lval cos0 where 9 is the angle between the vectors v and w Thus lv wl lvaH COMM lv wl lval since lcos0l 1 and this completes the proof I Proposition 13 Let 04 B a b 7gt R3 be two curves then the derivative satis es Leibnitz Rule with respect to the dot product in R3 at 316475 PM at Bt Mt WW PROOF Follow this through the components 0104 5 d 1 1 2 2 3 3 d da a a6 da1 1 1161 1042 2 Zd z la3 3 gd s W6 tawtw QWWB WW 3 dan id i ltdt a dt 3 dam 3 id i WWW 71 a a Theorem 11 In R3 a line is the curve of least arclength between two points PROOF We need to nd the distance between two points along the line and then show that it is shorter than the arclength of any other curve joining those two points Let P Q E R3 be two points The line between them can be parametrized by lt p tq 7 p Then Ht q 7 p and the speed ll tl la 7 pl is constant Therefore 1 1 La l tdt7 M7291 1617291 and the lenght of the vector or line segmentO from P to Q is the distance from P to Q Now let 04 be another curve joining P to Q Why should 04 be longer than t Well at some point 04 must point away from the vector from P to Q for if it did not it would 4 CHAPTER 1 CURVES be the line segment above To measure how much this is we can measure the angle that 04 makes with the vector in the direction of q 7 p We will do this by taking the dot product of 0 with this direction vector Let V q 7 plq 7 pl be the unit vector in the direction of q 7 p The total deviation of 04 from the line segment can be added up by integration Precisely we need to compute b ot th 1 to see why La is larger than L Note that at V ot V at V ot V because V is a constant Thus by the Fundamental Theorem of Calculus Abet Volt 17 QOzt Vdt 041 V 7 aa V qVpVqpV qifq7pqip q pl Now by Schwarz s Inequality b b lothdt lo llvldt b 7 la tldt La Thus by these two we have that L La Now ot V lothl only when cos0 1 or 9 0 Thus ot is parallel to q 7 p for all values oft In this case at is the line from P to Q Therefore the strict inequality holds unless the curve is the line I Other Curves There are a lot of curves that we have and have not studied The circle of radius r 04 t rcost rsint at 7 sinta17 cost a cos3t a sin3t 2a tant 2a cos2 a cost a sint bt a sint a lntant2 a cost The cycloid Oz The astroid Oz The Witch of Agnesi Oz A circular helix 04 t TractriX or pursuit curve 04 t Additionally we have the polar curves 7 the cardiod the limacon of Pascal the lem niscate the roses the Archimedean spiral We will look at many of these and more 52 GEODESIC EQUATIONS AND CLAIRAUT RELATION 55 52 Geodesic Equations and Clairaut Relation Now that we know what geodesics are how can we do the calculations Let s consider an orthogonal patch xu u ie F xu xv 0 Let a be a geodesic in the patch x Then we know that we can write 0t as 0t xut ut and it follows that 0 xuu xvu and 2 2 a qu xwu u xuu xwu 1 xwu Xv U Now from our previous work we can replace KW KM and XM by their representations in xmme We nd g 7 X u 2 u2 JuU 7 flu2 E G G X 1 2412 My 2 U lit2 QmuU nuZ So the rst two terms give us the tangential part of a For a to be a geodesic these terms must be zero so it is necessary and suf cient that the following geodesic equations are satis ed UH uZ uU 7 y 0 2E E 2E Geodesic Equations Eu 2 Cu Go 2 W t KW W 0 The geodesic equations are a system of 2nd order differential equations Theorem 51 Let p Xu0u0 be a point on a surface M with patch Xuu and let V E TPM Then there is a unique geodesic 039 777 H M with 00 p and aO V The proof of this is based on the theory of differential equations and the existence of a solution to a system with certain intial values Our geodesic equations satisfy the hypothesis of the theorem so such a solution exists Let s look at an example Example 53 The Unit Sphere 82 Take the standard patch for the unit sphere xu u cos u cos 1 sinu cos 1 sin 12 We have calculated previously that E cos2 1 F O and G 1 The geodesic equations for the unit sphere become H A 2 u i2tanuu u 0 u s1nucosuu 0 How in the world will we ever be able to solve such a system of 2nd order non linear differential equations In fact it is the case that we usually CANNOT solve the differential equations directly In the case of the sphere though we have a few tricks up our collective sleeves to bring to bear on the problem Assume that 0t xut7 vt is a unit speed geodesic Then since 0 u xu UX1 we have the unit speed relation H E712 112 1 which on the sphere is the condition coszv 12 12 1 Solve the rst geodesic equation as follows u 2tanv U U lnu 72lncosv C C to 7 cos2 1 C Now replace 1 in the unit speed equation by 2 cos 1 03 2 2 1 4coszvv cos 1 2 C U 1i 2 cos 1 cos2v702 U 2 cos 1 4 Now7 diVide u by 1 and we get a separable differential equation du c 11 cos m cos2 1 7 02 5 lntegrate this to nd 0 u dv cos vx cos 1 7 02 0 sec2 12 7 U 1 7 02 sec2 1 0 sec2 1 2 U 1 7 02 7 02 tan 1 w ctan v 7 substitute w 7 17w2 V1 762 d ltctanvgtd arcs1n w arcs1n V1762 Thus sinu7d tanv where Now7 use the Angle Sum formula for Sines to get sinu 7 d sin u cos d 7 sin 1 cos u 52 GEODESIC EQUATIONS AND CLAIRAUT RELATION 57 6 This gives us sinucosd7 sindcosu 7 tanv 0 sinu cosv cosucosv sinv icosd7s1nd 7 cosv cosv cosv Going back to our spherical coordinates x cosucosv y sinu cost and z sinv and considering only the numerator gives 7 sindz cos dy 7 z 0 Thus the geodesic equations imply that 0 lies on a plane through the origin 7 just what we expected Let us look at yet another piece of information afforded to us by the geodesic equations Let 1 denote the smaller angle between the vectors 0 and X1 at any point along the the unit speed curve a The makes the angle between 0 and Xu 7 1 Thus 7r 0 Xu 1 sin gt cos lt77 gtgt LLX 1X X uEucosv 2 HU HHMH xE v u but u cos v from the geodesic equations so sin gt cosv39 This is a special case of a result known as Claimut s relation Example 54 The Torus Take the torus with the parametrization Xu v R r cos u cos 1 R r cos u sin 127 sin We have previously computed E r2 F O and G R rcos 702 so the geodesic equations are u R T COSU sinu 12 0 7 r sin u 7 2 0 U R r cos u u U Then second equation is separable giving 1 4 R r cos u2 39 Again assume that a has unit speed replace 1 in the unit speed relation by the above to get 2 R r cos 74 CZ 1 u r Rrcosu2 58 Dividing 1 by u we get do crxR rcosu E R rcosu2 7 02 chRrcosu U Rrcosu27c2 We cannot integrate the right hand side of this equation using elementary functions This means that our Clairaut relation will have added importance du We say that an orthogonal patch xu7 u is a Clairaut parametrization in u if Ev 0 and GE 0 The patch is Clairaut in u if Eu 0 and GM 0 We have shown that the sphere is Clairaut in u and the torus is Clairaut in u In these cases the geodesic equations simplify to E 2 G 2 u 7u u 2Eu 2E1 0 u u u 0 Clairaut in u u uu 0 H a v a Clairaut in u 7 0 U 2G l 2GU The following theorem is stated for the u Clairaut parametrizations7 but everything also applies to the u Clairaut case as well Theorem 52 Let M be a surface with a u Clairaut patch Xu7 1 Then euery u parameter curue is a geodesic and a u parameter curue with u uo is a geodesic precisely when Guu0 0 Corollary 6 For a surface of reuolution hauing parametrization xu7 u gu7 hu cos y hu sin u7 any meridian is a geodesic and a parallel is a geodesic precisely when huo 0 16 16 Implications of Curvature and Torsion We have seen what happens if either the curvature or torsion is zero The next step is to see what happens if we move them up a little First we know that if the torsion is 0 and the curvature is constant then the curve forms an arc of a circle The next thing we should do is to see what happens if the torsion is constant Consider the curve 043 a cos asin bf where c xaz 1 b2 The unit tangent vector is T 7sin i cos Thus the third component is constant so T e3 Thus the angle 0 between T and eg satis es cos 0 We generalize this property by saying that a curve 04 with unit speed is a cylindrical helix if there is some constant vector u so that Tu cos 9 is constant along the curve The direction of the vector u is called the axis of the helix We can characterize all cylindrical helices with the following theorem Theorem 111 Suppose 04 has a gt 0 then 04 is a cylindrical helix if and only if rn is constant PROOF We may assume that 04 has unit speed For the rst direction assume that T u cosd is constant for some vector u Then 0 d T u T u N u L ds Thus we see that N is perpendicular to u since a gt 0 Since u is perpendicular to N we can write u in terms of the Frenet frame as uuTTuBB cos 0T 1 sin 0B But u is constant so u 0 cos 0T 1 sin 0B 0 acost 7 rsindN acosd 7 739 sin 0N Thus a cos0 7 r sind 0 and cot0 ra is constant since 9 is a constant Now suppose that ra is constant The cotangent function takes on all real values so there is some 9 so that cot0 ra De ne u cos 0T 1 sin 0B Thus u acos07 rsin0N 0 Thus u is constant and thus T u cos0 is constant is a cylindrical helix I We know even more Theorem 112 Fundamental Theorem for Curves in R3 Let ar a b 7 R be continuous functions with a gt 0 Then there exists a curve 04 a b 7gt R3 parametrized by arclength whose curvature and torsion are a and 739 respectively Furthermore any two such curues di er by a proper rigid motion that is if B is any such curue then there is a linear transformation A R3 7gt R3 which preserves arclength and a uector V E R3 so that 33 14043 V 16 IMPLICATIONS OF CURVATURE AND TORSION 17 This result is essentially a result about the existence and uniqueness of solutions of systems of differential equations The F renet Formulas provide the system of differential equations and the unique solution provides the one curve Let alttgt a1lttgt7a2lttgt7aslttgtgt7 Tlttgt T1lttgt7T2lttgt7Tslttgtgt7 Nlttgt N1lttgt7N2lttgt7 Nslttgtgt7 and Bt B1t B2t B3t The system of differential equations that has to be solved comes from ot Tt T t ntNt N t iwtTt 7tBt B t 77tNt If we insist that 040 000 T0 100 N0 010 and B0 001 then we have a unique solution For plane curves it is a little simpler Remember that we proved that 593 So de ne 0u A and and 53 cos0udusin0udugt Now 33 is a curve with the given curvature function s 48 45 A Formula for Gaussian Curvature The Gaussian curvature can tell us a lot about a surface We compute K using the unit normal U7 so that it would seem reasonable to think that the way in which we embed the surface in three space would affect the value of K while leaving the geometry of M un changed This would mean that the Gaussian curvature would not be a geometric invariant and7 therefore7 would not be as helpful in studying surfaces If we can nd a formula for K which does not depend on U7 we would then show that the value of K does not depend on how M is situated in space We will give a formula for K when depends only on E7 F7 and G These three quantities E7 F7 G are called the metric of the surface I will give the more general formula later7 but we will derive this for the case that F xu XU 0 Note that this means that the u and u parameter curves form perpendicular families of curves Theorem 42 Gauss7 Theorem Egregrium The Gaussian curuature depends only on the metric E F and G where 8 8 8 8 5E awn Xu and GM G X1 Xv Eu 8u 8u PROOF This is nothing more than nding the coefficients of a vector with respect to a particular basis Since we assumed that our patch is regular7 we know that Km Xv U forms a basis for R3 Now7 we need to eliminate U from our formula for Gaussian curvature and l XW U7 m X7W U7 n xW U Expand KW X7W7 and XM in terms of this basis KW Faxu P uxw W XW szxu 1 va mU 42 X u39u P vxu 1 5va nU Our job is to nd the T s While they are just the coefficients in the basis expansion7 they are traditionally known as Christo el symbols qu xuquXu Xu00 PZME likewise KW X39u FZuG so if we can compute KW Xu we can nd P3 45 A FORMULA FOR GAUSSIAN CURVATURE 49 o E xu xu so by taking the derivative with respect to u we get that Eu qu xu l xu xuu 2Xu xuu Therefore E E XW xu 2 and F3 o xu XU 0 so differentiating with respect to u gives 0qu X1Xu xuv orquXu7XuXW o E xu xu and differentiating with respect to 1 gives Ev 2xu XW So E11 7 Xu Xu u qu xv Thus X X E X X E P2 MUG 17 U39U U o GXUXUSOGu2XMXUSO pvampL Eg W G 2G 0 0 xu X1 so differentiating with respect to v and following the same technique as above will give us P31 X1111 Xu fan E E39 Finally XU XU C so XW X1 iv2 and X39Lm x39u G39u We end up with the following formulas E E Kw m mtw E G KW M M7 G G MF Mt mtw l W EM M m 71 EM EM In order to complete our proof it is necessary to look at certain mixed third partial derivatives We know that the mixed partials are equal regardless of the order in which we 50 take the partials Thus me qu or me 7 qu O This means that when we write me 7 qu in terms of Km Xv U all of the coefficients are zero We will concentrate on the X1 term for our result Other terms give other results Eu Eu Ev Ev 7 7 l U U qu39u 2Exu39u 2GX U U 1 l 1 Expand KW KM and Uv by their basis expansions 7 Eu 7 EuEv 7 EEG 71m 7 EuGu 7 Eu 7 Eva 7 Zn KW 2E v 4E2 4EG E X 4EG 2G v 4G2 G X mEu nEv E E G G x 7 7 x 7 7 x muU 1U 7 Eu EuEv EEG lm Eva Gu GuGu m2 Eli 4E2 4EG Elxul 4EG lt gtu 4G2 Glx Ev mGu 2G mu U Now the X1 coefficient of me 7 qu must be zero so 07 EuGu 7 Eu 7Eva 7m 7 7Eva7 Gu 7GuGu7m2 4EG 2G v 4G2 G 4EG 2G u 4G2 G zn7m2 7 EuGu 7 Eu 7 Eva Eva 7 GM 7 GuGu G 4EG 2G v 4G2 4EG 2G u 4G2 ln7m2 EuGu 7 1 Eu 7 Eva 7 Eva 7 1 GM 7 GuGu EG T4E2G E 2G v 4EG2 4E2G E 2G u 4EG2 EuGu 1ltEEgt Eva Eva 1ltGugt GHQ 11 u 4E2G E 4EG24E2G E 4EG2 We have to check that right hand side of the above equation is the same as the right hand side of Equation 41 That means we must compute the derivatives 1 8 Eu a GM 7 WWW 1 7 EvEG 12EvG 7 EG 7 GEM 7 GuEG 12EMG 7 EGugt EG EC 7 7 EM 7 EvEvG 7 EGv 7 GW 7 GuEuG7 EGu 2EG 4E2G2 2EG 4E2G2 GEuGu 7 QEGEM EEEGE GEEEE 7 QEGGW EGuGu 2EG2 7 EuGu 2GEvv 7 2Eva Eva Eva 2GGW 7 2GuGu GuGu 7 K 4E2G 4EG2 4EG2 4E2G 4EG2 4EG2 Thus we have developed the Gaussian curvature as a quantity involving only E G and I implicitly F 46 SOME EFFECTS OF CURVATURE Lemma 44 In general the Gaussian curuature is giuen by vi EG F F G F G 46 Some Effects of Curvature Recall that a point on a surface is an umbilic point if the principal curvatures a p are equal Theorem 43 A surface M consisting entirely of umbilic points is contained in either a plane or a sphere A surface in R3 is compact if it is closed and bounded Here bounded means that it is contained in a sphere of sufficiently large7 but finite7 radius Closed means that every sequence of points on the surface converges to a point on the surface Theorem 44 On euery compact surface M C R3 there is some pointp with Kp gt 0 Corollary 5 There are no compact surfaces in R3 with K g 0 In particular no minimal surface embedded in R3 is compact Theorem 45 Liebmann IfM is a compact surface of constant Gaussian curuature K then M is a sphere of radius 1K
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