2. Note that each f on M induces a covector field of. Then at each point p, the cotangent vector df and the tangent vect
Posted: Mon May 09, 2022 1:34 pm
2. Note that each f on M induces a covector field of. Then at each point p, the cotangent vector df and the tangent vector X would evaluate to some number. So df(x) is a function M + R. Show that df(x)= X(f), i.e., the two are the same. (Hint: just use definitions and calculate directly.) 3. If X,Y :V + V are vector fields, then note that X OY :V +V might not be a vector field. (Leibniz rule might fail.) However, show that X - Y - YoX is always a vector field. 4. On a related note, show that if A, B are skew-symmetric matrices, then AB - BA is still skew- symmetric. (Skew-symmetric matrices actually corresponds to certain vector fields on the manifold of orthogonal matrices. So this is no coincidence.)
a Exercise 1.7.2 (What is a derivative). The discussions in this problem holds for all manifolds M. But for simplicities sake, suppose M =Rfor this problem. Let V be the space of all analytic functions from M to R. Here analytic means f(1,9, 2) is a infinite polynomial series (its Taylor expansion) with variables x, y, 2. Approximately f(1,9, 2) = 20 +212 +day+ 232 +04:22 +agry+0622+27y2 + ..., and things should converge always. Then a dual vector v EV* is said to be a "derivation at P EM" if it satisfy the following Leibniz rule (or product rule): v(f9) = f(p) (g) +g(p)u(f). (Note the similarity with your traditional product rule ($9)' (*) = f(x)g' ()+g()f'(x).) Prove the following: 1. Constant functions in V must be sent to zero by all derivations at any point. PI 2. Let 2,4, 2 € V be the coordinate function. Suppose p= then for any derivation v at p, then we have v((2 - P.)f) = f(p)u(x), v((y – P2) f) = f(p)(y) and v((2 - P3)f) = f(p)+(2). P1 3. Letr, y, 2 E V be the coordinate function. Suppose p = P2 then for any derivation v at p, then we have v((r - P)"(y – P2)(2 – Ps)) = 0 for any non-negative integers a, b,c such that a+b+c>1. P1 4. Let 2, 4,2 € V be the coordinate function. Suppose p= P2, then for any derivation v at p, v(f) = (p)u(x) + (p)+(y) + (p)u(2). (Hint: use the Taylor expansion of f at p.) [u() 5. Any derivation v at p must be eractly the directional derivative operator V, where v = (y) (v(2)] P21 P3 P3 P3
a Exercise 1.7.2 (What is a derivative). The discussions in this problem holds for all manifolds M. But for simplicities sake, suppose M =Rfor this problem. Let V be the space of all analytic functions from M to R. Here analytic means f(1,9, 2) is a infinite polynomial series (its Taylor expansion) with variables x, y, 2. Approximately f(1,9, 2) = 20 +212 +day+ 232 +04:22 +agry+0622+27y2 + ..., and things should converge always. Then a dual vector v EV* is said to be a "derivation at P EM" if it satisfy the following Leibniz rule (or product rule): v(f9) = f(p) (g) +g(p)u(f). (Note the similarity with your traditional product rule ($9)' (*) = f(x)g' ()+g()f'(x).) Prove the following: 1. Constant functions in V must be sent to zero by all derivations at any point. PI 2. Let 2,4, 2 € V be the coordinate function. Suppose p= then for any derivation v at p, then we have v((2 - P.)f) = f(p)u(x), v((y – P2) f) = f(p)(y) and v((2 - P3)f) = f(p)+(2). P1 3. Letr, y, 2 E V be the coordinate function. Suppose p = P2 then for any derivation v at p, then we have v((r - P)"(y – P2)(2 – Ps)) = 0 for any non-negative integers a, b,c such that a+b+c>1. P1 4. Let 2, 4,2 € V be the coordinate function. Suppose p= P2, then for any derivation v at p, v(f) = (p)u(x) + (p)+(y) + (p)u(2). (Hint: use the Taylor expansion of f at p.) [u() 5. Any derivation v at p must be eractly the directional derivative operator V, where v = (y) (v(2)] P21 P3 P3 P3