# Elementary Differential Forms Question

1. Apr 19, 2012

### mindarson

Let me preface by saying I am a physics major. So I am coming at differential forms from the perspective of physics, i.e. work, flows, em fields, etc.

My question is this. My understanding is that a basic 1-form dx, dy, or dz takes a vector v = (v1,v2,v3) and gives back the corresponding component of that vector, i.e. the size of that vector's projection onto the corresponding coordinate axis. The complete 1-form, then, such as phi = Adx + Bdy + Cdz, is the sum of multiples of these projection sizes, where the coefficients are functions of x,y,z.

My confusion is regarding which vector space these projections "occur" in: R^3, the tangent space or the dual space? In any case, how is it possible to project onto the x, y, or z axes within the tangent or dual spaces? Wouldn't the "axes" in the dual space be the "dx-axis", etc.?

Or am I just confusing myself unnecessarily by assuming that the 1-form and the projection (or the input vector and the projection) have to be in the same space?

2. Apr 19, 2012

### Dickfore

3. Apr 19, 2012

### wisvuze

the "projections" occur as projections onto tangent space bases. that is, {e1,..., en} situated at a point p

4. Apr 20, 2012

### lavinia

1 forms map tangent vectors into the base field - either the real numbers or the complex numbers.

For instance dx,dy,and dz pick out the x,y, and z components of tangent vectors to a 3 dimensional coordinate system.

In Euclidean space this can get confusing because the tangent space to Euclidean space can be naturally identified with Euclidean space itself and often the distinction is overlooked. Try working with 1 forms on a surface. It will be less confusing.

5. Apr 20, 2012

### Matterwave

One should be careful calling this kind of operation a "projection". A real projection requires a metric, and one forms naturally act on vectors without need for metric.

The easiest way to see that the one forms dx, dy, and dz don't orthogonally project vectors is to consider the case where the basis vectors are not orthonormal. Consider, if they are skewed, your components for your vectors are no longer the result of orthogonal projections, but dx, dy, and dz always pick out components.

6. Feb 13, 2013

### mindarson

First, much-belated thank-you's for taking the time to reply!

This has definitely been a source of confusion for me. I knew that the dx, dy, dz picked out the corresponding components, but I did not know of what vector! The original vector or the tangent vector? The tangent vector definitely makes more sense, since the 1-form is analogous to the gradient/derivative, i.e. you add them all up over the whole manifold in order to get the total.

Thanks again for the insight!