# Differential forms and visualizing them

## Main Question or Discussion Point

I made a post titled the same thing but it didnt seem to show up for some reason so if i am just reposting this over again i apologize.

I recently got the book A geometrical approach to differential forms by David Bachman. At the moment the biggest issue i am having is just visualizing what the form should look like/ be. The calculations for a one form and a vector are pretty much the same for a dot product so should the form be visualized as a vector? if so where is it and why does it have such an odd notation if it cant be visualized as a vector then how should we (I) picture it?

HallsofIvy
Homework Helper
I think you will find it helpful to think in terms of "linear functionals". A linear functional is a linear transformation that maps a vector into a real number. One thing that can be shown is that, given a vector space, V, the set of all linear functionals from V to the real numbers is itself a vector space where, if f and g are linear functionals, a, b numbers, af+ bg is defined as the linear functional that maps v to af(v)+ bg(v).

If V has finite dimension n, then the set of linear functions, V*, has dimension n also and there is a one to one correspondence. Choosing basis for V, through that correspondence, automatically assigns a basis to V*. Specifically, if $\{v_1, v_2, ..., v_n\}$ is a basis for v, then the set of linear functionals, $\{f_1, f_2, ..., f_n\}$ where $f_k$ is defined by "$f_k(v_k)= 1$, $f_k(v_i)= 0$ if $i\ne k$ and extended to all vectors by "linearity"- if $v= a_1v_1+ a_2v_2+ ...+ a_nv_n$ then the functional corresponding to v is $f_v= a_1f_{v_1}+ a_2f_{v_2}+ ... + a_nf_{v_n}$.

Now, a more formal definition of "dot product" would be $v\cdot u= f_v(u)$ where $f_v$ is the linear functional corresponding to v.

The purpose of all that is to say that we can think of differential forms in exactly the same way- they are "functionals that assign numbers to functions (vectors)":
The differential form $d\mu$ assigns, to each function, x, $\int f d\mu$.