{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

LectureNotes13U - Math 497C Curves and Surfaces Fall 2004...

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Math 497C Nov 11, 2004 1 Curves and Surfaces Fall 2004, PSU Lecture Notes 13 2.9 The Covariant Derivative, Lie Bracket, and Rie- mann Curvature Tensor of R n Let A R n , p A , and W be a tangent vector of A at p , i.e., suppose there exists a curve γ : ( , ) A with γ (0) = p and γ (0) = W . Then if f : A R is a function we define the (directional) derivative of f with respect to W at p as W p f := ( f γ ) (0) = df p ( W ) . Similarly, if V is a vectorfield along A , i.e., a mapping V : A R n , p V p , we define the covariant derivative of V with respect to W at p as W p V := ( V γ ) (0) = dV p ( W ) . Note that if f and V are C 1 , then by definition they may be extended to an open neighborhood of A . So df p and dV p , and consequently W p f and W p V are well defined. In particular, they do not depend on the choice of the curve γ or the extensions of f and V . Exercise 1. Let E i be the standard basis of R n , i.e., E 1 := (1 , 0 , . . . , 0), E 2 := (0 , 1 , 0 , . . . , 0) , . . . , E n := (0 , . . . , 0 , 1). Show that for any functions f : R n R and vectorfield V : R n R n ( E i ) p f = D i f ( p ) and ( E i ) p V = D i V ( p ) ( Hint: Let u i : ( , ) R n be given by u i ( t ) := p + tE i , and observe that ( E i ) p f = ( f u i ) (0), ( E i ) p V = ( V u i ) (0)). 1 Last revised: November 29, 2004 1
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
The operation is also known as the standard Levi-Civita connection of R n . If W is a tangent vectorfield of A , i.e., a mapping W : A R n such that W p is a tangent vector of A for all p A , then we set Wf ( p ) := W p f and ( W V ) p := W p V.
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern