DIFFERENTIAL GEOMETRY: A First Course in Curves and Surfaces

APPENDIXReview of Linear Algebra and Calculus1. Linear Algebra ReviewRecall that the set {v 1 ,...,v k } of vectors in R n gives a basis for a subspace V of R n if andonly if every vector v ∈ V can be written uniquely as a linear combination v = c 1 v 1 + ···+ c k v k .In particular, v 1 ,...,v n will form a basis for R n if and only if the n × n matrix⎡⎤| | |⎢⎥A = ⎣ v 1 v 2 ··· v n ⎦| | |is invertible, and are said to be positively oriented if the determinant det A is positive. In particular,given two linearly independent vectors v, w ∈ R 3 , the set {v, w, v × w} always gives a positivelyoriented basis for R 3 .We say e 1 ,...,e k ∈ R n form an orthonormal set in R n if e i · e j =0for all i ≠ j and ‖e i ‖ =1for all i =1,...,k. Then we have the followingProposition 1.1. If {e 1 ,...,e n } is an orthonormal set of vectors in R n , then they form a basisfor R n and, given any v ∈ R n ∑,wehavev = n (v · e i )e i .i=1We sayann × n matrix A is orthogonal if A T A = I. Itiseasy to check that the column vectorsof A form an orthonormal basis for R n (and the same for the row vectors). Moreover, from thebasic formula Ax·y = x·A T y we deduce that if e 1 ,...,e k form an orthonormal set of vectors in R nand A is an orthogonal n × n matrix, then Ae 1 ,...,Ae k are likewise an orthonormal set of vectors.An important issue for differential geometry is to identify the isometries of R 3 (although thesame argument will work in any dimension). Recall that an isometry of R 3 is a function f : R 3 → R 3so that for any x, y ∈ R 3 ,wehave‖f(x) − f(y)‖ = ‖x − y‖. Wenowprove theTheorem 1.2. Any isometry f of R 3 can be written in the form f(x) =Ax + c for someorthogonal 3 × 3 matrix A and some vector c ∈ R 3 .Proof. Let f(0) =c, and replace f with the function f − c. Ittoo is an isometry (why?) andfixes the origin. Then ‖f(x)‖ = ‖f(x) − f(0)‖ = ‖x − 0‖ = ‖x‖, sothat f preserves lengths ofvectors. Using this fact, we prove that f(x) · f(y) =x · y for all x, y ∈ R 3 .Wehave‖f(x) − f(y)‖ 2 = ‖x − y‖ 2 =(x − y) · (x − y) =‖x‖ 2 − 2x · y + ‖y‖ 2 ;on the other hand, in a similar fashion,‖f(x) − f(y)‖ 2 = ‖f(x)‖ 2 − 2f(x) · f(y)+‖f(y)‖ 2 = ‖x‖ 2 − 2f(x) · f(y)+‖y‖ 2 .107