4
$\begingroup$

I've been learning about tensor analysis, and things have been going well so far, but I'm a bit stuck when it comes to the idea of a cobasis (by which I mean the reciprocal basis; not sure which term is more common). Basis vectors are simple enough, it's just a carry over from vector spaces in linear algebra. Cobasis vectors, on the other hand, don't seem well motivated to me. I understand that they're dual to the basis vectors, but why exactly do we need them? When should they be used over the normal basis vectors?

I'm familiar with the treatment of vectors as elements of a tangent space of a manifold and dual vectors as linear operators on vectors, if that's the sort of explanation that motivates the cobasis most effectively, although I don't have a good intuition with that sort of thing quite yet.

Improve this question
$\endgroup$

1 Answer 1

Reset to default
4
$\begingroup$

I think maybe you are just wondering why we need covectors at all. Once you accept that we need covectors, it makes sense to have cobases for your covectors just like you have bases for your vectors.

Ok so why do we need covectors. I will give one application---taking a derivative. Often in differential geometry you want to take a derivative. Suppose for example you have a function defined on your manifold. What kind of object should the derivative be? Well you should give it a tangent vector to the manifold and it gives you a number saying how quickly the function changes when you move at the velocity given by that tangent vector. This relationship between tangent vector and rate of change of the function value should be linear. Therefore it must be that the derivative of our function is a covector field.

Improve this answer
$\endgroup$
3
  • $\begingroup$ You're exactly right, understanding the need for covectors is all I need! Forgive me if this is a silly question, but just to make sure I understand: in your example, would the derivative be a directional derivative in the direction of the tangent vector? That is, would the covector be the gradient of the function, which we can then multiply by the tangent vector to get the number we're looking for? $\endgroup$
    – EtaZetaTheta
    Commented May 14, 2014 at 2:36
  • 2
    $\begingroup$ @EtaZetaTheta yes. A covector is a function that takes in vectors and spits out numbers. The gradient is most naturally viewed as (the linear extension of) a function which takes in unit vectors and spits out the rate of change of the function in that direction, which makes it a covector. $\endgroup$
    – Robert Mastragostino
    Commented May 14, 2014 at 3:03
  • $\begingroup$ Perfect, thank you! I've always read that the gradient of a function is sort of the prototypical covariant tensor, and your example helped connect that with the role of a covector as a thing that eats vectors and spits out numbers, so that was especially helpful. $\endgroup$
    – EtaZetaTheta
    Commented May 14, 2014 at 3:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.