Studying vector calculus

1. Nov 13, 2005

Diophantus

I've recentlty been studying vector calculus at uni and now that we are almost up to the general Stokes's theorem and its many froms I am just beginning to understand it and actually enjoy it.

One thing that is really annoying me though is that I have heard one-forms being refered to as covectors. I am reasonably familiar with covectors in that I know that they are elements of a dual space and are constucted as a linear combination of the daul basis vectors ei* which satisfy <ei,ei*> = {kronecker delta} where ei is a standard basis vector.

Is a single one-form such as dx then an element of the dual basis?

I've only ever studied the dual space of R^n, which as I recall, is the space of linear functionals. I fail to see how a vector of the form (dx, dy, dz) could represent a linear functional.

Does it have something to do with work?

2. Nov 13, 2005

hypermorphism

They are dual vectors, but not to Rn.
Think of a smooth surface (a 2-dimensional manifold) in R3. At each point of the surface, there is a tangent plane. Attach these planes to each point, respectively. Each plane is a vector space called the tangent space to the manifold at that particular point. The dual space is called the cotangent space. Now, you're familiar with vector fields that assign a vector to each point of a set. A 1-form assigns a dual vector to each point of the manifold (that are elements of the cotangent space to the manifold at that particular point).
As a concrete example, the gradient operator is a 1-form you may already be familiar with. It attaches a covector to each point that can then be used (as a 1-tensor) on the tangent space to that point to get directional derivatives at that point.

Last edited: Nov 13, 2005
3. Nov 14, 2005

Diophantus

Ok, let me get this straight. Each point on the manifold has a tangent space associated with it and each tangent space has a cotangent space associated with it containing covectors. Am I right in saying then that if a manifold is defined and we fix the tangent space for an arbitary point, then there is only one one form associated with that point; moreover, this one form is a particular member of the cotangent space (or do we have to define a field aswell?)

Since the cotangent vectors are always normal to the tangent vectors, does this mean that if we are in R^n and the manifold is of dimension k, then the cotangent space is of dimension n-k?

Thanks a lot for your time.

4. Nov 14, 2005

George Jones

Staff Emeritus
Terminology varies: see the thread "One-forms" in the Linear & Abstract Algebra forum. I tried to sum up the situation in post #23, but you might find the other posts interesting.

Regards,
George

5. Nov 14, 2005

hypermorphism

In most mathematics texts, a 1-form is a function that attaches covectors to points (it is a covector field). There is only 1 covector associated with that point by the 1-form.
In other texts, these covectors are called 1-forms, and the function is just called a covector field/field of 1-forms. In those texts, your statement would be correct.
This is not true! The cotangent space is the dual space of the tangent space, hence it has the same dimension. Juxtaposing the gradient operator with a surface may have given the wrong idea. Note that the gradient operator acts on functions from R3 into R (this can be generalized). The gradient 1-form is a form on the tangent bundle (the collection of all tangent spaces, each of which is 3-dimensional) to the graph of f (the collection of points (x, f(x)), a 3-dimensional subset of R4). The cotangent space is also 3-dimensional. It is an easy exercise to show that the gradient vector, the dual of the gradient covector, is normal to the tangent spaces of level surfaces of f (submanifolds of the graph of f) at each point, but don't confuse this fact with the tangent and cotangent spaces of the graph of f.

Last edited: Nov 14, 2005
6. Nov 21, 2005

Thanks

Thank you.