Copyright © 2003 jsd
We define differential forms to have the following properties:
 (1) 
for arbitrary scalarvalued functions f, g, et cetera. So we are using the set {[dx_{i}]} as a basis.
There exist pointy vectors, which are relatively familiar to most people. They can be represented by an arrow with a tip and a tail. In the language of linear algebra, these are column vectors.  There exist oneforms, which are less familiar to most people. They can be represented by contourlines and/or fishscales. In the language of linear algebra, these are row vectors. 
As we shall see, pointy vectors and oneforms have quite a few properties in common, but there are also some crucial differences, so be careful. Item 12 discusses one of the differences you need to watch out for.
df(x_{1}, x_{2}, ⋯) = 
 [dx_{i}] 
 ⎪ ⎪ ⎪ ⎪ 
 (2) 
where in the ith term of the sum, the partial derivative holds constant all the arguments to f() except for the x_{i} argument. The notation for this is clumsy, but the idea is important. The partial derivative is really a directional derivative in a direction specified by holding constant an entire set of variables except for one … so it is crucial to know the entire set, not just the one variable that is nominally being differentiated with respect to. For details on this, including ways to visualize what it means, see reference 1.
An example is shown in figure 2. The intensity of the shading depicts the height of the function F := sin(x_{1})sin(x_{2}) while the contourlines depict the exterior derivative dF.
 (3) 
which is convenient. It simplifies the notation.
Technically speaking, [dx_{1}] exists by fiat, according to item 2, while dx_{1} is something you can calculate according to equation 2. On a daytoday basis you don’t care about the distinction, but it would have been cheating to assume they are equal. We needed to keep them conceptually distinct just long enough to prove they are numerically equal.
Suppose we want to visualize the gradient of some landscape. If you visualize the gradient as a pointy vector, it points uphill. In many cases, though, you are better off visualizing the gradient as a oneform, corresponding to contour lines that run across the slope.
You can judge the magnitude of the 1form according to how closely packed the contour lines are. Closelypacked contours represent a largemagnitude 1form. To say the same thing the other way, the spacing between contours is inversely related to the magnitude of the oneform.
Contour lines have the wonderful property that they behave properly under a change of coordinates: if you take a landscape such as the one in figure 2 and stretch it horizontally (keeping the altitudes the same) as shown in figure 3, the slopes become less. The contour lines on the corresponding topographic map spread out by the same stretch factor, as they should, to represent the lesser slope. In contrast, if you try to represent the gradient by pointy vectors, the representation is completely broken by a change in coordinates. As you stretch the map, position vectors and displacement vectors get longer, but gradient vectors have to shorter, to represent the lesser slope. Therefore we need different representations. We represent positions and displacements using pointy vectors, but we represent gradients using 1forms.
To say the same thing the other way, representing a gradient using a pointy vector would be a bad idea; such vectors would not behave properly. They would not be “attached” to the landscape the way contour lines are.
Of course, pointy vectors are needed also; they are appropriate for representing the location of one point relative to another in this landscape. These location vectors do stretch as they should when we stretch the map. An example of this is shown in red in figure 3.
It is important to clearly distinguish the two types of vector:
Type of vector:  pointy vector  oneform  
Another name:  contravariant vector  covariant vector  
Represented by:  column vector  row vector  
Dirac notation:  ket: ⋯⟩  bra: ⟨⋯  
Example:  displacement  gradient  
Stretching the map:  increases distance  decreases steepness 
Consider the contrast:
In some spaces, we have a metric. That is, we have a dot product. That allows us to determine the length of a vector, and to determine the angle between two vectors. In such a space, we have a geometry (not just a topology). Ordinary Cartesian (x,y,z) space is a familiar example.  There are other spaces where we do not have a metric. We do not have a dot product. We do not have any notion of angle, and not much notion of length or distance. We have a topology, but not a geometry. Thermodynamic statespace is an important example. We can measure the distance (in units of S) between contours of constant S, but we cannot compare that to any distance in any other direction. 
In such a space, we can use the metric to transpose a vector. Transposing a row vector creates the corresponding column vector and vice versa. That gives us a onetoone correspondence: For any pointy vector you can find a corresponding oneform and vice versa.  In a nonmetric space, there is not any way of converting 1forms to pointy vectors or vice versa. There is not any way of finding a 1form that uniquely “corresponds” to a given pointy vector or vice versa. 
Technically, the row vectors always live in their own space, while the column vectors always live in another space. 
Given a metric, the two spaces are isomorphic, and people usually don’t bother to distinguish them.  Without a metric, the two spaces remain distinct. It makes sense to visualize dE as a oneform, i.e. as contours of constant E ... but it does not make sense to visualize dE as any kind of pointy vector. 
A 1form has a direction, but we cannot measure the angle between two such directions. You can say that we have a topology but not a geometry. This sounds like a terrible limitation, but it is actually the right thing for thermodynamics, because typically you have no way of knowing whether dS is “perpendicular” to dV or not, and it causes all sorts of trouble if you use a mathematical formalism that assumes you can measure angles when you can’t.
Among other things, this means that we do not require the coordinates x_{1}, x_{2}, ⋯ to be mutually perpendicular. Since there is no notion of distance or angle, so we could not make them perpendicular even if we wanted to. See figure 4.
dx_{i} ∧ dx_{j} = −dx_{j} ∧ dx_{i} (4) 
for all (i, j).
Nongrady force fields are common in the real world. See reference 2 for more about how to visualize such things.
A conspicuously ungrady form w is shown in figure 5. You can imagine that this represents the 1form w := PdV (aka “work”) in a slightlyidealized heat engine. The direction of the 1form (i.e. the uphill direction) is everywhere counterclockwise. This w is a perfectly fine 1form, but you cannot write w = dW because w cannot be the slope of any potential W. The concept of slope is locally welldefined, and you can integrate the slope along a particular path from A to B, but you cannot use this integral to define a potential difference W(B) − W(A) because the result depends very much on which path you choose. This is like Escher’s famous “Waterfall” shown in figure 6.
To repeat: You are free to write w = PdV. That is a perfectly fine 1form, welldefined at every point in the state space. In contrast, it is not OK to write w = dW or PdV = dW, because that cannot be welldefined throughout state space, except perhaps in trivial cases. (You might be able to define something like that on a onedimensional subspace, along a particular path through the system, but then you would need to decorate “W” with all sorts of subscripts to indicate exactly which subspace you are talking about.)
A more subtle example of an ungrady form is discussed in item 23 below.
d(A ∧ B) = dA ∧ B + (−1)^{k} A ∧ dB (5) 
where A has grade=k.
dd = 0 (6) 
This important result can be expressed in words: “the boundary of any boundary is zero”.
Before we explain why this is so, we should emphasize that dd is not the most general secondderivative operator. Rather, it is the antisymmetric part of the second derivative, in accordance with equation 5. So what we are saying is that the antisymmetric part of the second derivative vanishes.
The antisymmetric piece of the second derivative necessarily vanishes, because of the mathematicallyguaranteed symmetry of mixed partial derivatives:

 f ≡ 

 f (7) 
This is true for all f, assuming f is sufficiently differentiable, so that the indicated derivatives exist.
Figure 7 and Figure 8 show what’s going on. We will use these figures to discuss finite steps (as indicated by Δ) instead of infinitesimals, but the same ideas apply in the limit of very small steps. In particular, Δxy means to take a step toward increasing x along a contour of constant y. Similarly Δyx means to take a step toward increasing y along a contour of constant x.
In accordance with the usual operatorprecedence rules, the interpretation of the LHS of equation 7 is:

 f means  ⎛ ⎜ ⎜ ⎝ 
 ⎛ ⎜ ⎜ ⎝ 
 (f)  ⎞ ⎟ ⎟ ⎠  ⎞ ⎟ ⎟ ⎠  (8) 
That is, we work from right to left, first taking a step toward increasing y along a contour of constant x, then taking a step toward increasing x along a contour of constant y. For example, in the figure, this would correspond to proceeding clockwise from (0,0) via (0,1) to (1,1).
Meanwhile, the RHS of equation 7 tells us to proceed counterclockwise in the figure, from (0,0) via (1,0) to (1,1). The point is that we get to same point either way. That is, the clockwise trip we just took, together with the counterclockwise trip, form a “closed” figure.
This result is nontrivial. Although the boundary of a boundary is zero, the boundary of “something else” is not necessarily zero. For example:
 (9) 
Forms that are closed, including figure 2 and figure 9, have the property that the “contour” lines in one region mesh nicely with the lines in adjacent regions. In a nonclosed form such as figure 5, the meshing fails somewhere. (Commonly it fails everywhere.)
Beware that this notion of “closed oneform” is not equivalent to the notion of “closed set” (containing its limit points) nor to the notion of “closed manifold” (compact without boundary). See reference 3 and reference 4.
∫ 
 dF = F(B) − F(A) (10) 
The meaning is simple: the integral measures the number of contours that you cross in going from point A to point B. For a grady 1form, this number is independent of the path you take along the way from A to B.
This integral is, of course, a linear operator.
B = f_{i}(x) dx_{i} (11) 
We are using the Einstein summation convention, i.e. implied summation over repeated indices, such as index i in this equation.
As explained in section 2, the integral of this is:
 (12) 
To understand how we integrate a oneform B along the curve C, start by breaking the curve into small segments and integrating each segment separately:
 (13) 
and if f is a sufficiently smooth function and if C is a sufficiently smooth curve, and if the points {θ1, θ2, ⋯} are sufficiently close together, then we can treat f as being locally constant and pull it out front of the integrals:
 (14) 
Now we have grady forms inside the integral, so we can integrate them immediately using equation 10. We get
 (15) 
where we have described the point C(θ) using an expansion in terms of the basis vectors:
C(θ) = C_{i}(θ) x_{i} (16) 
Equation 15 is beginning to look like a familiar Riemann integral. In fact it is just
 (17) 
In equation 17, do not think of the integrand as a dot product, even though it involves the same sumofproducts you would use for evaluating f · ∂C/∂θ. We do not have a dot product. The operation here is a contraction. A contraction involves a oneform acting on a pointy vector. In this case the oneform is f and the pointy vector is ∂C/∂θ. In equation 15, you can visualize [C_{i}(θ2) − C_{i}(θ1)] as a pointy vector with its tip at C(θ2) and its tail at C(θ1).
We can carry out the contraction of a oneform with a pointy vector. We cannot carry out the dot product of two oneforms, nor the dot product of two pointy vectors. Think of oneforms as 1×D matrices (one row and D columns) and pointy vectors as D×1 matrices.
As an example, consider integrating the oneform
f := 
 dx_{1} + 
 dx_{2} (18) 
where r := √(x_{1}^{2} + x_{2}^{2}). This oneform is depicted, with fair accuracy, in figure 5. We wish to integrate it along a curve C which is a circular path of radius R, centered on the origin, so that along C:
 (19) 
Plugging in to equation 17 we find
 (20) 
For a discussion of how differential forms apply to thermodynamics, see reference 5.
(beware: at some points this assumes the existence of a dot product.)
Copyright © 2003 jsd