ectures 1: The Second Fundamental Form Disclaimer. As we have a textbook, this lecture note is for guidance sulement only. It should not be relied on when rearing for exams. In this lecture we introduce the second fundamental form, its roerties, alications. The required textbook sections are 7.17.3. I try my best to make the examles in this note dierent from examles in the textbook. Please read the textbook carefully try your hs on the exercises. During this lease don't hesitate to contact me if you have any questions. Table of contents ectures 1: The Second Fundamental Form.................. 1 1. The second fundamental form................................. The Weingarten ma..................................... 3. Examles............................................. 5 4. Alications of the second fundamental form...................... 7 1
Dierential Geometry of Curves & Surfaces 1. The second fundamental form Definition 1. (The second fundamental form) et S be a surface 0 S. et be a surface atch of S covering 0 : 0 (u 0 ; v 0 ). Then the second fundamental form of S at 0, denoted hh;ii 0 ;S (with ; S omitted when no confusion may arise), is a bilinear form on T 0 S dened through hhv; wii 0 ;S v 1 w 1 + M (v 1 w + v w 1 ) + N v w : (1) where v v 1 u (u 0 ; v 0 ) + v v (u 0 ; v 0 ) w w 1 u (u 0 ; v 0 ) + w v (u 0 ; v 0 ), (u 0 ; v 0 ) : uu (u 0 ; v 0 ) N(u 0 ; v 0 ) u N u ; () M(u 0 ; v 0 ) : uv (u 0 ; v 0 ) N(u 0 ; v 0 ) u N v v N u ; (3) N(u 0 ; v 0 ) : vv (u 0 ; v 0 ) N(u 0 ; v 0 ) v N v : (4) Remark. An alternative notation is (u 0 ; v 0 ) du + M(u 0 ; v 0 ) du dv + N(u 0 ; v 0 ) dv. Remark 3. et (s) (u(s); v(s)) where s is the arc length arameter. If (s 0 ), we clearly have n (u_) + M u_ v_ + N (v_) hh_; _ ii (s0 );S (5) We can further rove the following general formula for w T S n ( 0 ; w 0 ) hhw 0; w 0 ii 0 ;S : (6) hw 0 ; w 0 i 0 ;S As a consequence, when (t) is not arametrized by arc length, we have n ( 0 ; _(t 0 )) hh_; _ ii (t 0 );S h_; _ i (t0 );S (u_) + M u_ v_ + N (v_) E (u_) + F u_ v_ + G (v_) : (7). The Weingarten ma Definition 4. (Definition 7..1 in the textbook) We dene the Weingarten ma where G is the Gauss ma. W 0 ;S : D 0 G (8) Note the minus sign here. Examle 5. We try to calculate W 0 ;S( u ) W 0 ;S( v ) for the following surface atches. It is clear that W 0 ;S( u ) N u ; W 0 ;S( v ) N v : (9) a) S is the lane (u; v) (u; v; 3 u + v).
In this case we have which give N(u; v) G((u; v)) We see that W( u ) W( v ) 0. b) S is the cylinder (u; v) (cos u; sin u; v). In this case we have We have u (1; 0; 3); v (0; 1; ) (10) k k 1 ( 3; ; 1): (11) 14 u ( sin u; cos u; 0); v (0; 0; 1) (1) N(u; v) (cos u; sin u; 0): (13) k k N u ( sin u; cos u; 0) u ; N v (0; 0; 0): (14) Consequently we have W( u ) u ; W( v ) 0: (15) c) S is the unit shere (u; v) u; v; 1 u v. We have u Consequently 1; 0; u 1 u v ; v 0; 1; v 1 u v (16) N(u; v) u; v; 1 u v (u; v): (17) W( u ) N u ; W( v ) N v : (18) d) S is the hyerbolic araboloid (u; v) (u; v; u v) with 0 (0; 0; 0). We have Now we calculate u (1; 0; v); v (0; 1; u) (19) v u 1 N(u; v) ; ; 1 + u + v 1 + u + v 1 + u + v W( u ) N u W( v ) N v! u v (1 + u + v ) ; 1 + v 3/ (1 + u + v ) ; u 3/ (1 + u + v ) 3/ : (0) (1)! 1 + u (1 + u + v ) ; u v 3/ (1 + u + v ) ; v 3/ (1 + u + v ) 3/ : () 3
Dierential Geometry of Curves & Surfaces We see that u v W( u ) (1 + u + v ) 1 + v 3/ u + (1 + u + v ) 3/ v (3) W( v ) 1 + u (1 + u + v ) u v 3/ u (1 + u + v ) v: (4) 3/ We have seen that W( u ) N u, W( v ) N v. As W is linear, for a; b R we have W(a u + b v ) a N u b N v : (5) Therefore to underst W we need to underst N u ; N v. The crucial observation is the following. N u ; N v?n ) N u a 11 u + a 1 v ; N v a 1 u + a v. Theorem 6. We have a11 a 1 a 1 a F G M M N (6) where E du + F du dv + G dv is the rst fundamental form of S at 0, ; M; N are dened in ( 46). Proof. We notice that as u N v N 0, there holds similarly This leads to uu N ( u N) u u N u u N u (7) M v N u u N v ; N v N v : (8) E a 11 + F a 1 u (a 11 u + a 1 v ) u N u ; (9) F a 11 + G a 1 v (a 11 u + a 1 v ) v N u M: (30) Consequently a11 a 1 F G M : (31) Similarly we have a 1 a M F G N the conclusion follows. emma 7. et v; w T S. Then hhv; wii ;S hw ;S (v); wi ;S hv; W ;S (w)i ;S : (3) Proof. Since hh; ii ;S, hw ;S (); i ;S, h; W ;S ()i ;S are all bilinear, it suces to rove the following cases: v u ; w v ; v w u ; v w v ; v v ; w u. We rove the rst one leave the other three as exercises. 4
We calculate hh u ; v ii ;S M: (33) On the other h, W ;S ( u ) N u a 11 u + a 1 v where a11 : (34) a 1 F G M Consequently Note that we have used F G F G hw ;S ( u ); v i ;S 1 0 0 1 a 11 h u ; v i ;S + a 1 h v ; v i ;S a 11 F + a 1 G ( F G ) F G M ( 0 1 ) M: (35) M ) ( F G ) F G ( 0 1 ): (36) The roof that h u ; W ;S ( v )i ;S M is similar. 3. Examles Examle 8. Consider the unit shere u; v; 1 u v. We calculate which gives Therefore u 1; 0; N(u; v) u 1 u v ; v 0; 1; v 1 u v (37) k k u; v; 1 u v (u; v): (38) (u; v) u N u v 1 1 u v; (39) M(u; v) u N v u v 1 u v; (40) N(u; v) v N v u 1 1 u v: (41) Examle 9. Consider the unit shere in sherical coordinates (cos u cos v; cos u sin v; sin u). We calculate u ( sin u cos v; sin u sin v; cos u); v ( cos u sin v; cos u cos v; 0) (4) which gives N(u; v) (cos u cos v; cos u sin v; sin u): (43) 5
Dierential Geometry of Curves & Surfaces Therefore (u; v) ; (44) M(u; v) 0; (45) N(u; v) cos u: (46) Examle 10. Consider the surface atch (u; v) (u; v; u + v ). We have Thus we have u (1; 0; u); v (0; 1; v); (47) uu vv (0; 0; ); uv (0; 0; 0); (48) N So the second fundamental form is k k ( u; v; 1) : (49) 1 + 4 u + 4 v uu N ; (50) 1 + 4 u + 4 v M uv N 0; (51) N vv N : (5) 1 + 4 u + 4 v (du + dv ): (53) 1 + 4 u + 4 v Exercise 1. Does this mean at any oint S, the normal curvature n is a constant in every direction? Examle 11. Consider a ruled surface (u; v) (u) + v l(u) where l(u) is of unit length. We calculate This gives We further calculate N(u; v) Therefore if we set A k k. u _(u) + v l _ (u); v l(u): (54) k k _(u) l(u) + v l_ (u) l(u) _(u) l(u) + v l _ (u) l(u) : (55) uu (u) + v l (u); uv l _ (u); vv 0: (56) (u; v) uu N A + v l _(u) l(u) + v l _ (u) l(u) ; (57) M(u; v) uv N A l _ (_ l); (58) N(u; v) vv N 0: (59) Recalling our discussion on develoable surfaces, we see that a ruled surface is develoable if only if M 0. 6
4. Alications of the second fundamental form Proosition 1. et S be a surface whose second fundamental form is identically zero. Then S is art of a lane. Proof. et be a surface atch for S. Then by assumtion we have N u u N 0. As N is the unit normal, naturally N u N 0. Consequently N u 0 as f u ; v ; N g form a basis of R 3. Similarly N v 0. Thus N is a constant vector therefore is art of a lane. Proosition 13. et S be a suface whose second fundamental form at every S is a nonzero scalar multile of its rst fundamental form at. Then S is art of a shere. Exercise. Prove that if S is art of a shere, then its second fundamental form is a non-zero scalar multile of its rst fundamental form. Proof. et (u; v) be a surface atch for S. Then there holds (u; v) c(u; v) E(u; v); M(u; v) c(u; v) F(u; v); N(u; v) c(u; v) G(u; v) (60) for every (u; v). This leads to M F G M N As a consequence, we have 1 0 c(u; v) 0 1 : (61) N u + c(u; v) u 0; N v + c(u; v) v 0 (6) at every (u; v). Taking v; u derivatives of the two equations resectively, we have N uv + c v u + c uv 0 N vu + c u v + c vu ) c v u c u v : (63) As u ; v form a basis of T S, there must hold c v c u 0, that is c(u; v) c is a constant. Now (6) becomes is a constant. In other words, we have (N + c ) u (N + c ) v 0 ) N + c r 0 (64) + c N c r 0 (65) is a constant which means is art of the shere centered at c r 0 with radius jcj. 7