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This is a naive question; apologies in advance. For a point $p \in M$ on a smooth manifold $M$, a differential form can be viewed as a map $$T_p M\times \cdots \times T_p M \to \mathbb{R} \;.$$ What puzzles me about this object is that it is not "differential." Yes, I know, the tangent space $T_p M$ is differential in that it is tangent. But if $M=\mathbb{R}^n$, then I lose the intuitive sense of tangency, and just end up with a map from $k$ vectors to $\mathbb{R}$ with certain properties (the map is multilinear, and alternating).

I seek a way to view differential forms intuitively that somehow emphasizes their differential aspects. Help would be much appreciated---Thanks!

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Try to read a little bit about De Rham complexes. These differential forms works like $dx$ but for other dimensions. – Sigur Aug 19 '12 at 1:50
That's not what you get when $M = \mathbb{R}^n$; the map is still allowed to vary with $p$. Anyway, a differential $n$-form is a thing you can integrate infinitesimal $n$-dimensional paralleletopes against (multilinear just means it behaves nicely under concatenation of paralleletopes and alternating means it is zero if any two of the vectors defining the paralleletope agree). I think this perspective makes the link to e.g. Stokes' theorem a little clearer. – Qiaochu Yuan Aug 19 '12 at 1:51
Among physicists of certain generations, a favorite source of intuitive understanding of 1-forms, 2-forms and on is the book Gravitation by Misner, Thorne and Wheeler, published about 1973. The physics is woefully obsolete thanks to NASA and NSF, but the math is eternal... – DarenW Aug 19 '12 at 4:32
Googlers may also benefit from this answer: – isomorphismes Aug 7 '13 at 4:40
up vote 11 down vote accepted

In what one might call naive calculus, for each coordinate $x_i$, the differential $dx_i$ denotes a small (infinitesimal, even) change in $x_i$, so a covector $\sum_i a_i dx_i$ is an infinitesimal change.

On a manifold, coordinates are only local, not global, so we should also imagine that this covector sits at a particular point of $M$. If we want to have a covector varying smoothly at every point, this is a differential one-form.

If $f$ is a function, then the total differential of $f$ is the quantity $$df = \sum_i \dfrac{\partial f}{\partial x_i} dx_i, $$ which records how $f$ is changing, at each point.

A tangent vector at a point is a quantity $v = \sum_i a_i \dfrac{\partial}{\partial x_i}$; you should think of this as a vector pointing infinitesimally, based at whatever point we have in mind. You can measure the change of $f$ in the direction $v$ by pairing $df$ with $v$.

Summary: tangent vectors are infinitesimal directions based at a point, while covectors are measures of infinitesimal change. You can see how much of the change is occurring in a particular direction by pairing the covector with the vector.

Now higher degree differential forms are wedge products of $1$-forms. You can think of these as measuring infinitesimal pieces of oriented $p$-dimensional volumes. (Think about how the (oriented) volume of an oriented $p$-dimensional parallelapiped spanned by vectors $v_1,\ldots,v_p$ depends only on $v_1\wedge\cdots\wedge v_p$.)

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"covectors are measures of infinitesimal change": This is very helpful! – Joseph O'Rourke Aug 19 '12 at 16:43

There are probably several equivalent ways to define differential forms.

One natural starting point is as a vector bundle. I.e. just the same way $TM\rightarrow M$ in each point $p\in M$ gives the vector space $T_pM$ of tangent vectors at $p$, we can define $T^*_pM$ as the vector space of cotangent vectors at $p$. At least this makes it clear that $T^*M\rightarrow M$ is a vector bundle. A form is then just a section (i.e. vector field) on $T^*M$: i.e. a map $M\rightarrow T^*M$ which maps $p\in M$ to a vector in $T^*_pM$.

One obvious way to define $T^*M$ is as the dual vector bundle of $TM$: i.e. $T^*_pM$ is the dual vector space of $T_pM$. This is just a way of restating that it provides a mapping $TM\rightarrow\mathbb{R}$, but at least I find it more intuitive to think of them as vectors (dual to the tangent vectors) than linear maps from tangent vectors to $\mathbb{R}$.

Another way of defining $T^*_pM$ is as $\frak{m}_p/\frak{m}_p^2$ where $\frak{m}_p$ contains differentiable functions defined in a neighbourhood of $p$ which are zero in $p$: if $f,g$ are functions that are both zero at $p$, $f\sim g$ in $\frak{m}_p/\frak{m}_p^2$ if $f-g=\sum a_ib_i$ where $a_i,b_i$ are functions that are zero at $p$. I.e. the cotangent vectors (or differentials) capture the first order variation of functions. This definition perhaps captures more clearly that cotangent vectors represent the direction in which functions increase.

Yet another way is to look at integration, and see how the differentials used there transform e.g. under change of variables. This only provides the algebraic rules that differential forms need to satisfy, not a definition of what they are.

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Your emphasis on covectors is quite illuminating--Thanks! – Joseph O'Rourke Aug 19 '12 at 16:42

The integral version of an $n$-form is a gadget that "measures" (directed) $n$-dimensional shapes. For example, the flux of a vector field is a gadget that, when given a (directed) surface, produces a number: the flux of the vector field through the surface.

The differential version of such a gadget at a point, assuming it is "differentiable", would have to be able to produce a number given the differential version of an $n$-surface at that point -- i.e. when given a tangent $n$-vector. To finish the connection to the specific definition you gave, a (pure) tangent $n$-vector can be described by $n$ tangent vectors; e.g. as the $n$-vector described by the paralleltope Qiaochu mentions.

However, I frequently feel that one shouldn't think about the connection of differential forms to differential calculus in terms of infinitesimal geometry. Instead, you should let the properties of the derivative operator give you an intuition about 'infinitesimal algebra', of which differential forms are an example. (e.g. properties like the rule $d(fg) = f \, dg + g \, df$ when $f$ and $g$ are scalar fields)

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Thank you for mentioning the connection to flux, which is quite helpful. – Joseph O'Rourke Aug 19 '12 at 16:38

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