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Simple surfaces, unit normal, vectors, tangent planes

Assumptions. All functions are C∞. Notations. ∂~x ∂~x ~x1 = ∂u1 , ~x2 = ∂u1 spanning tangent vectors M = ~x (U) image of a simple TpM tangent at P Definitions. (1) A simple surface is a 1-1 C∞ function ~x : U → R3, where U ⊂ R2 is open and ~x1 × ~x2 6= ~0. (2) The tangent plane to a simple surface ~x : U → R3 at P = ~x (a, b) is the plane spanned by ~x1 and ~x2, i.e., the plane to ~x1 × ~x2. (3) The unit normal ~n (a, b) is

~x × ~x ~n = 1 2 . |~x1 × ~x2|

(4) X~ is a tangent vector at P ∈ ~x (U) if there exists a curve ~α : d~α ~ (−ε, ε) → ~x (U) such that ~α (0) = P and dt (0) = X. Results. Lemma 1.16. The set of tangent vectors at P is a vector subspace of R3. Proof. Let X~ and Y~ be tangent vectors at P and let r ∈ R. Then there exist curves

~α :(−ε, ε) → ~x (U) , β~ :(−ε, ε) → ~x (U)

~ d~α ~ dβ~ ~ such that ~α (0) = β (0) = P , dt (0) = X, and dt (0) = Y. The lemma follows from showing that: (a) rX~ is a tangent vector at P. (b) X~ + Y~ is a tangent vector at P. ε ε Proof of (a). Define ~η :(− r , r ) → ~x (U) by ~η (t) = ~α (rt) . Then by the chain rule, d~η d~α (0) = r (0) = rX.~ dt dt Proof of (b). Define functions α1 (t) , α2 (t) , β1 (t) , β2 (t) [from (−ε, ε) to R] by ¡ ¢ ¡ ¢ ~α (t) = ~x α1 (t) , α2 (t) , β~ (t) = ~x β1 (t) , β2 (t) .

1 ¡ ¢ ¡ ¢ If you like, you may write αi = ~x−1 i ◦ ~α and βi = ~x−1 i ◦ β~ for i = 1, 2, ³¡ ¢ ¡ ¢ ´ where ~x−1 = ~x−1 1 , ~x−1 2 . Note that since ¡ ¢ ~x (a, b) = P = ~α (0) = ~x α1 (0) , α2 (0) , ¡ ¢ ~x (a, b) = P = β~ (0) = ~x β1 (0) , β2 (0) and since ~x is 1-1, we have ¡ ¢ ¡ ¢ α1 (0) , α2 (0) = β1 (0) , β2 (0) = (a, b) .

We compute by the chain rule d~α dα1 dα2 X~ = (0) = ~x (a, b) (0) + ~x (a, b) (0) , dt 1 dt 2 dt dβ~ dβ1 dβ2 Y~ = (0) = ~x (a, b) (0) + ~x (a, b) (0) . dt 1 dt 2 dt Note we find a curve corresponding to X~ + Y.~ Define

~γ :(−δ, δ) → ~x (U) by ¡ ¢ ~γ (t) = ~x γ1 (t) , γ2 (t) , γ1 (t) = α1 (t) + β1 (t) − a, γ2 (t) = α2 (t) + β2 (t) − b ¡ ¢ where 0 < δ ≤ ε is chosen small enough so that α1 (t) + β1 (t) − a, α2 (t) + β2 (t) − b ∈ U for t ∈ (−δ, δ) . Note that ~γ (0) = ~x (a, b) . We compute by the chain rule

d~γ dγ1 dγ2 (0) = ~x1 (a, b) (0) + ~x2 (a, b) (0) dt µdt dt¶ µ ¶ dα1 dβ1 dα2 dβ2 = ~x (a, b) (0) + (0) + ~x (a, b) (0) + (0) 1 dt dt 2 dt dt dα1 dα2 = ~x (a, b) (0) + ~x (a, b) (0) 1 dt 2 dt dβ1 dβ2 + ~x (a, b) (0) + ~x (a, b) (0) 1 dt 2 dt = X~ + Y.~

2 This completes the proof of Lemma 1.16. dα1 1 dα2 2 Remark. Often the notation dt (0) = X and dt (0) = X is used, so that 1 2 X~ = X ~x1 + X ~x2. Proposition 1.18. The set of tangent vectors is the same as the tangent plane. Remark. Tangent vector is defined by (4) above, whereas tangent plane is defined by (2) above. Proof. Recall ~x : U → R3 is a simple surface. The tangent plane is the span of {~x1, ~x2}. Define the curves ~γ1 :(−ε, ε) → ~x (U) and ~γ2 :(−ε, ε) → ~x (U) by

¡ 1 2 ¢ ~γ1 (t) = ~x γ1 (t) , γ1 (t) , 1 γ1 (t) = a + t, 2 γ1 (t) = b

1 2 dγ1 dγ1 (so that dt (0) = 1 and dt (0) = 0) and ¡ 1 2 ¢ ~γ2 (t) = ~x γ2 (t) , γ2 (t) , 1 γ2 (t) = a, 2 γ2 (t) = b + t

1 2 dγ2 dγ2 (so that dt (0) = 0 and dt (0) = 1). Then d~γ dγ1 dγ2 1 (0) = ~x 1 (0) + ~x 1 (0) = ~x , dt 1 dt 2 dt 1 d~γ dγ1 dγ2 2 (0) = ~x 2 (0) + ~x 2 (0) = ~x . dt 1 dt 2 dt 2

Thus ~x1 and ~x2 are both tangent vectors. Since the set of tangent vectors is a vector space containing ~x1 and ~x2, we conclude that the span of {~x1, ~x2}, i.e., the tangent plane, is a subset of the set of tangent vectors. Conversely, given a tangent vector, we shall show that it is a linear combination of ~x1 and ~x2. Let Let X~ be a tangent vector at P. Then there exists a curve ~α : d~α ~ (−ε, ε) → ~x (U) such that ~α (0) = P and dt (0) = X. We have previously shown dα1 dα2 X~ = (0) ~x + (0) ~x . dt 1 dt 2 This completes the proof of Proposition 1.18.

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