# Question about Differential Forms as Size of Projections

1. Feb 12, 2013

### mindarson

Hello, I have a somewhat conceptual question about differential forms. I have been studying differential forms off and on for some time now and things are starting to come together for me. However, there is an irritating gap in my understanding.

Regarding the geometric significance or visualization, etc. of differential forms, it is my understanding that the value of the form (the real number output) can be interpreted as the oriented size of the projection of the vector(s) on which the form acts - onto an axis.

What is not clear to me is exactly which axis the vector is projected onto before the projection size is 'calculated'. What space is this axis in? What is the role of this axis? Which coordinate or variable or quantity is 'tracked' on this axis?

Since the differential form is a generalization of the gradient or derivative, it seems to make sense to me that the projection would be onto an axis in the space of tangent vectors, but I cannot convince myself thoroughly of this.

Can anyone offer guidance in clarifying these ideas? If so, thanks so much!

2. Feb 13, 2013

### mathwonk

you might check out the nice geometric book by david bachmann on diff forms.

3. Feb 13, 2013

### mindarson

Mathwonk! I have read most of your posts here on differential forms, and they have been quite helpful to me. This one in particular:

'a differential form on a single vector space V is a function of k variables which is alternating and linear in each variable separately. If k = the dimension of the vector space, then essentially the only such animal is the determinant. Thus they are as uintuitive as determinants, no more no less. The intuition for a determinant to me is as a measure of oriented volume.

If k is less than the dimension of V, and the dimension is n > k, then there are exactly the binomial coefficient "n choose k" linearly independent such functions, formed as follows. Take k vectors from V and place them in order in a k by n matrix. Then choose k of the n columns, and take the determinant of the resulting square k by k subdeterminant.

There are exactly "n choose k" ways to do this. Moreover one can form linear combinations of these basic functions and get thus a vector space of such functions of dimension "n choose k". Thus there is an "n choose k" dimensional vector space of such k dimensional oriented volume measures on an n dimensional vector space, dependeing on which set of k axes one projects ones k dimensional parallepiped onto before taking the determinant.

Of course in the case of length, there is a formula known as pythgoras theorem that lest one calculate the usual length of a segment from the lengths of its projections noto the avrious axes, and this generalizes to the case of area of parallelograms in 3 space. So perhaps there is also a way of compouting k dimensional volume of a k block from the various projections onto different choices of sets of k axes in n space.'

which you wrote waaaay back in 2007. You are addressing the heart of my confusion right here! Unfortunately, I have not been able to convince myself of which axes and in which space these projections happen.

As to the Bachman book, I have it and have read through the first few chapters. Honestly, I found it less than enlightening. It is written in the right spirit for me (long on intuition, relatively low on formalism), but it is just too ... disorganized and jumbled, and sometimes honestly I find it hard to make heads or tails of his notations.

With regard to my specific confusion, Bachman does have a figure (in Chapter 2, the section entitled '1-forms' - page 19 of the first edition, I believe) in which he seems to be depicting a curve in the xy plane. At a point on that curve, he has set up another set of axes, the (dx,dy)-axes. I take the curve to be the manifold and the (dx,dy)-axes to be defining (or indicating?) the space of tangent vectors to the manifold at that particular point.

Bachman then goes on to say (same section):

'There is still another geometric interpretation of 1-forms. Let’s first look at the simple example ω(<dx, dy>) = dx. This 1-form simply returns the first coordinate of whatever vector you feed into it. This is also a projection; it’s the projection of the input vector onto the dx-axis. This immediately gives us a new interpretation of the action of a general 1-form, ω = a dx + b dy. Evaluating a 1-form on a vector is the same as projecting onto each coordinate axis, scaling each by some constant, and adding the results.'

So this indicates pretty clearly that the projection is onto the coordinate axis in the tangent space. This was my suspicion. However, I still do not understand how this relates to calculus as understood from multivariable calculus, i.e. chop the whole thing into little manageable bits, compute each bit, then add them all up. Specifically with regard to the spaces and their bases.

There is the space which the curve lives in, whose basis vectors, I presume, are i hat, j hat, k hat.

Then there is the tangent space to the curve (manifold) at each point, whose basis would be ∂x and ∂y? Correct?

And finally, there is the space of 1-forms dual to the tangent space (space of functions that act on the tangent vectors), whose basis vectors are the basis 1-forms dx, dy, dz. Correct?

It's my understanding that the 'dual' relationship is such that the basis 1-forms work as follows:

dxi(∂j) = ∂xi/∂xj = δij

(My indices may be a bit off here; I am pretty new to index notation. And sometimes I am using x and y, while sometimes I am indexing the coordinates with i.)

So my understanding of the above is that the basis 1-form dxi acts on the tangent space basis vectors ∂j, giving unity if they are associated with the same coordinate, but zero if they are associated with different coordinates. So in some sense the 1-form 'picks out' only those directions in which the vector has a 'shadow' or a non-zero projection.

If all this is more or less correct (more or less doubtful), then I am still working hard to try to make the connection between these projections of the input vector (which lives in a different space altogether) onto the tangent space axes - and the processes of differentiation and integration as I know them from old-fashioned vector calculus.

I feel like I have a fuzzy understanding of this. As we move along the curve (manifold), at each point we feed the vector that is tangent to the curve into a 1-form (which is living in the dual space). That 1-form projects the original curve onto the coordinate axes in the tangent space and returns the size of this projection. The integral is just the adding up of all these projection sizes. Is this correct (though I know it is informal)?