Mth 348 Fll 017 Lecture 6: Isometry Disclimer. As we hve textook, this lecture note is for guidnce nd sulement only. It should not e relied on when rering for exms. In this lecture we nish the reliminry discussion of the Guss m nd introduce the ide of isometry. We lso rove tht there cnnot e lnr m for shericl region. The exmles in this note re dierent from exmles in the textook. Plese red the textook crefully nd try your hnds on the exercises. During this lese don't hesitte to contct me if you hve ny questions. Tle of contents Lecture 6: Isometry...................................... 1 1. Proerties of the Guss M................................. Isometry............................................. 3. Geodesics on the Plne, Cylinder, nd Shere..................... 3 3.1. On the lne....................................... 3 3.. On the cylinder..................................... 3 3.3. On the shere...................................... 4 1
Dierentil Geometry of Curves & Surfces 1. Proerties of the Guss M Let G: S 7! S e the Guss m of the surfce S. Then y denition T S?N(). On the other hnd, since S is shere, there holds T G() S?G() = N(). Consequently the two surfces T S nd T G() S re rllel, nd re in fct the sme lne (when viewed s velocity sce. Consequently, we cn use the sis for T S, f u ; v g s the sis for T G() S. In fct it is enecl to think of D G s liner trnsformtion from T S to itself. Let S e rmetrized y. Then S cn e rmetrized s G which is exctly the function N(u; v) = u v k u v k : (1) In other words, it is nturl to rmetrize S y N. If we use fn u ; N v g s sis for T G() S, the mtrix reresenttion for D G is the identity mtrix 1 0 0 1. To see this, we note tht i. S is rmetrized y ; ii. S~ := G(S) S is rmetrized y N(u; v); iii. N 1 G = N 1 N = I the identity ming. iv. Consequently the derivtive is lso identity. If we use f u ; v g s sis for T G() S, things ecome interesting. Considering the curve (t)=(t;0), we see tht the corresonding curve on S is N(t;0). Consequently we hve nd similrly D G( u ) = N u () D G( v ) = N v : (3) From these we cn solve the mtrix reresenttion of D G. The entries of these mtrix re the most imortnt quntities in clssicl dierentil geometry. We will discuss much more out this lter in the course.. Isometry f: S 7! S~ is n isometry when it is ijective, nd reserves rc length. More seciclly, for ny curve (t) in S, the rc length of from () to () equls the rc length of the curve f((t)), on S~, from f(()) to f(()). One cn show tht if f is n isometry, then f lso reserves re, tht is, re of W S equls the re of f(w ) S~ for ny W. One cn lso show tht if f is n isometry, then t every S, D f reserves ngle. More seciclly, if v 1 = D f(u 1 ); v = D f(u ), then \(v 1 ; v ) = \(u 1 ; u ).
Mth 348 Fll 017 3. Geodesics on the Plne, Cylinder, nd Shere 3.1. On the lne The rolem. Let A; B e two oints in R n. There re innitely mny curves tht connect the two oints. Find the one with shortest rc length. Mthemticl formultion. Among ll curves (t) stisfying () = A; () = B, nd the one with miniml L := k_(t)k dt: (4) The solution. We notice tht i. There holds L > ii. For the curve L (t) = t y + t z, we hve L = k_ L (t)k dt = _(t) dt = k() ()k = kb Ak: (5) 1 ( ) Therefore the solution is _ L (t) which is stright line. (z y) dt = kb Ak: (6) Exercise 1. Prove tht L (t) is stright line nd nd its rc length rmetriztion. 3.. On the cylinder The rolem. Let A; B e two oints in R 3 lying on the cylinder with the se circle centered t the originl nd with rdius 1. Find the shortest th long the cyclinder surfce connecting the two oints. Mthemticl formultion. Among ll curves (t) = (cos t; sin t; z(t)) stisfying () = A = (x A ; y A ; z A ); () = B = (x B ; y B ; z B ), nd the one with miniml L := k_(t)k dt: (7) The solution. (Otionl) We hve L = 1 + z_(t) dt (8) with z(t) stisfying z() = z A ; z() = z B nd lso cos = x A ; cos = y A ; sin = x B ; sin = y B. 3
Dierentil Geometry of Curves & Surfces Since z_(t) is smooth function, we hve L = lim k!1 k Xk 1 1 + z_(t i ) where t i = + i ( ). Now notice tht the function f(x) := 1 + k x is convex, y Jensen's inequlity we hve v 1 Xk 1! u 1 + z_(t k i ) > 1 + 1 Xk 1 t z_(t k i ) : (10) As we hve lim k!1 k Xk 1 (9) z_(t i ) = z_(t) dt = z A z B (11) r za z B L > ( ) 1 + = ( ) + (z A z B ) > ( 0 0 ) + (z A z B ) : (1) where 0 0 < nd stises lso cos 0 = x A ; cos 0 = x B ; sin 0 = y A ; sin 0 = y B. On the other hnd, the rc length of the curve cos t; sin t; z A + t 0 (z B z A ) (13) 0 0 is exctly ( 0 0 ) + (z B z A ). Exercise. Visulize this shortest th. Wht would it look like if we tten the cylinder? 3.3. On the shere The rolem. Let A; B e two oints on unit shere centering t the origin. Find the shortest th on the shere connecting them. Mthemticl formultion. Among ll curves with () = A; () = B, x (t) + y (t) + z (t) = 1, nd the one minimizing the integrl L := k_(t)k dt: (14) The solution. (Otionl) Our gol is to show tht the minimizing curve is the gret rc connecting A; B. For ritrry (t) on the shere connecting A; B, we dene new curve: (t) = (0; r(t); z(t)) (15) 4
Mth 348 Fll 017 where r(t) := (x(t) + y(t) ) 1/. We notice tht (t) connects y; z nd covers the gret rc connecting y; z. Therefore the rc length of (t) is no less thn /. For X(t) we clculte L = r_(t) + z_(t) dt = 6 r (x(t) x_(t) + y(t) y_(t)) x(t) + y(t) + z_(t) x_(t) + y_(t) + z_(t) dt = L: (16) Technicl Aside Cuchy-Schwrtz. The crucil ste, the one with 6, is due to the so-clled Cuchy-Schwrtz inequlity for vectors: To rove it, notice tht ju vj 6 kuk kvk: (17) (u t v) (u t v) = ku t vk > 0 (18) for ll t. Exnding the left hnd side we see tht holds for ll t, nd the conclusion follows. kvk t (u v) t + kuk > 0 (19) Exercise 3. Finish the roof. (Hint: When is qudrtic eqution t A t + B = 0 hving t most one rel solution?) Remrk 1. From the ove we see tht for ech cse (lne, cyclinder, shere), new ide/technique is needed, worse still, we hve to somehow know the nswer efore we strt much the sme s the sitution in clssicl geometry. The reson for this is tht we hve used too little clculus. In the following weeks, we will ly more clculus to geometry rolems, nd eventully develo comlete theory for the rolem nding shortest th on surfce. In this theory, the discovery of such th will e reduced to the solution of single set of ODEs whose derivtion is mechnicl. dt No more d hoc ides, no more clever tricks. 5