In Chapter TBD of Introduction to Tensor Calculus we introduced the concept of a Riemannian space as an analytical structure consisting of a metric tensor field $Z_{ij}$ imposed upon $\mathbb{R} ^{n}$. Thus, the switch from Euclidean to Riemannian spaces represented a reduction in the number of essential elements as we dispensed with the notion of a physical space along with geometric vectors, and no longer treated the concepts of length, area, and volume as primary notions. However, nearly all of the remaining elements of the tensor framework survived, starting with the concepts of coordinate changes and tensors. The availability of the metric tensor enabled index juggling, as well as the introduction of the Christoffel symbol and therefore the operation of the covariant derivative. In other words, the tensor framework remained fully intact and, as a matter of fact, became more general thanks to the severing of the connection to Euclidean spaces.

The subject of Differential Forms takes the reduction approach a decisive step further by eliminating the metric tensor from consideration. All that is left is the unadorned arithmetic space $\mathbb{R} ^{n}$.

What is the purpose of such an extreme reduction? The basic answer is generality. The fewer elements a framework has and the fewer assumptions it makes, the more general it is. Thus, a Riemannian space is a generalization of a Euclidean space, and a stripped down $\mathbb{R} ^{n}$ is a generalization of a Riemannian space. As one goes from a particular framework to a more general one, something is inevitably lost and some things are inevitably gained. For example, in going from Euclidean spaces to Riemannian spaces, we lost the absolute geometric reference that anchored and guided our investigations. On the other hand, we gained not only the ability to analyze analytically richer spaces but also a framework for questioning and reevaluating the very assumptions underpinning Euclidean geometry. This experience gives us the strong impetus to look for ways of generalizing the concept of a Riemannian space in the hope of gaining deeper insights into the nature of things.

In the broader pursuit of generality, one particular landmark will be of special interest to us: the Fundamental Theorem of Calculus. Ultimately, the Fundamental Theorem of Calculus is an arithmetic fact as it is essentially a statement that addition is opposite of subtraction. It is, of course, true that the integral on the left can be interpreted as the signed area of the domain "under" the graph of $f\left( x\right)$. However, this interpretation is after-the-fact. For example, if the integral is analyzed by a change of variables -- a technique often referred to as integration by substitution -- the shapes in the associated geometric interpretation must change accordingly.

Let us contrast it with the divergence theorem discussed in the previous Section. Both integrands $\nabla_{\alpha}T^{\alpha}$ and $n_{\alpha}T^{\alpha}$ are invariant under coordinate changes and represent geometric characteristics of the surface $S$, its boundary $L$, and the tensor field $T^{\alpha}$. Although the metric tensor does not appear explicitly in the statement of the divergence theorem it strongly permeates all of its elements. It is found in the covariant derivative $\nabla_{\alpha}$, in the components $n_{\alpha}$ of the geodesic normal, and -- perhaps most strongly -- in the fact that both integrals represent the limit of sums that involve measures of areas and lengths.

In other words, the elements of the Riemannian framework are essential to the formulation of the divergence theorem. It is of interest then to find an alternative multi-dimensional analogue of the Fundamental Theorem of Calculus that mimics its arithmetic nature. Differential Forms deliver that analogue in the form of the generalized Stokes theorem.

The subject of Differential Forms can be told without ever mentioning the concept of a tensor. But then again, the same can be said of Tensor Calculus itself. And much like in the case of Tensor Calculus, the tensor framework provides an essential underpinning for the subject of Differential Forms.

A change of coordinates is defined to be a map from $\mathbb{R} ^{n}$ to $\mathbb{R} ^{n}$ captured by a set of $n$ functions The inverse set is denoted by The change of coordinates is characterized by the Jacobians and which, as we are well aware by now, represent matrix inverses of each other. With the Jacobians in hand, we can give the same definition of a tensor as we did in a Riemannian space. However, due to the absence of the metric tensor, our ability to construct new tensors is far more limited. Nevertheless, we must mention the crucial fact that the permutation systems $e^{i_{1}\cdots i_{n}}$ and $e_{i_{1}\cdots i_{n}}$ are relative tensors of respective weights $-1$ and $1$, and that all delta systems $\delta_{j_{1}\cdots j_{m}} ^{i_{1}\cdots i_{m}}$ are absolute tensors. Furthermore, note that the Levi-Civita symbols $\varepsilon^{i_{1}\cdots i_{n}}$ and $\varepsilon _{i_{1}\cdots i_{n}}$ are excluded from consideration.

Naturally, tensors can still be added, multiplied, and validly contracted to produce new tensors. As a result, we are still able to generate invariants if a proper combination of index flavors is available. For example, a contravariant tensor $U^{i}$ can be combined with a covariant tensor $V_{i}$ to produce the invariant On the other hand, for a pair contravariant tensors $U^{i}$ and $V^{i}$, we choose not to introduce a mechanism for constructing an invariant since we have decided to exclude the covariant tensor $Z_{ij}$ from consideration.

For the same reason, we will not consider the Christoffel symbol $\Gamma _{ij}^{k}$ which is an essential element in the construction of the covariant derivative. However, crucially, there exists an alternative means of constructing a differential operator that preserves the tensor property. The key is the operation skew-symmetrization.

Denote the partial derivative $\partial/\partial Z^{i}$ by the symbol $\partial_{i}$, i.e. As we observed in Chapter TBD of Introduction to Tensor Calculus, although the object is not a tensor for a tensor $T_{j}$, the commutator is a tensor.

There are several ways to see this, but the essential key to this insight is the fact that the non-tensor part of $\partial T_{j}/\partial Z^{i}$ is symmetric in $j$ and $i$. Indeed, under a change of coordinates, $\partial T_{j}/\partial Z^{i}$ transforms according to the rule as we showed in Chapter TBD of the Introduction to Tensor Calculus. Notice that the non-tensor term $T_{j}J_{j^{\prime}i^{\prime}}^{j}$ is proportional to the second-order Jacobian $J_{i^{\prime}j^{\prime}}^{j}$ which is symmetric in $j^{\prime}$ and $i^{\prime}$. Thanks to this property, we have and, due to the cancellation of the nontensor terms, we obtain that which proves the tensor property of the commutator

Alternatively, note that $\partial_{i}T_{j}$ can be expressed in terms of the covariant derivative $\nabla_{i}T_{j}$ by the equation where the nontensor part $\Gamma_{ij}^{k}T_{k}$ is proportional to the Christoffel tensor $\Gamma_{ij}^{k}$ which is once again symmetric in $i$ and $j$. As a result, we have which proves the tensor property of the commutator on the left by equating it to a tensor expression.

This observation will lead to the concept of the exterior derivative.

Thus, Differential Forms is a study of skew-symmetric systems. A differential form is a synonym for a skew-symmetric system. The term form is used because mathematicians often prefer to speak in terms of invariants rather than tensors or systems. For example, instead of saying "consider a system $A_{ijk}$", a mathematician might prefer to say "consider a form $A\left( U,V\right) =A_{ijk}U^{i}V^{j}W^{j}$". The term differential is used because the forms often include "differentials" $dS^{i}$, as in $A_{ijk}dS^{i}dS^{j}dS^{k}$.

The concept of skew-symmetric systems was initially discussed in Chapter TBD of Introduction to Tensor Calculus. We will now return to that topic and add a few details. Our discussion will target the $\mathbb{R} ^{n}$ for an arbitrary dimension $n$. However, for the sake of working with more transparent expression we will illustrate the concepts in and, when necessary, restate the result for a general $n$.

By definition, a system is skew-symmetric if any of its elements related by a switch of two indices have opposite values. Crucially, the result of a double contraction between a skew-symmetric and a symmetric system is zero. For example, if $A_{ijkl}$ is a skew-symmetric system and $U^{ij}$ is a symmetric system, i.e. then In fact, this property can be taken to be converted into a definition of a skew-symmetric, i.e. a system is called skew-symmetric if it produces zero when double-contracted with any symmetric system.

The permutation systems $e_{ijklm}$ and $e^{ijklm}$ are defined as the unique skew-symmetric systems that have as many indices as the dimension of the space such that In other words, As we have already mentioned, the permutation systems $e^{i_{1}\cdots i_{n}}$ and $e_{i_{1}\cdots i_{n}}$ are relative tensors of respective weights $-1$ and $1$.

A delta system $\delta_{j_{1}\cdots j_{m}}^{i_{1}\cdots i_{m}}$ of any order has the following definition. delta_{j_{1}cdots j_{m}}^{i_{1}cdots i_{m}}=left{ begin{tabular} {ll} $\begin{array} {c} \phantom{+} 1\text{, }\\ \\ \phantom{+} \end{array}$ & $\begin{array} {l} \text{if the superscripts and the subscripts are identical sets}\\ \text{of distinct numbers related by an \textit{even} permutation}\\ \phantom{+} \end{array}$ $\begin{array} {c} -1\text{,}\\ \\ \phantom{+} \end{array}$ & $\begin{array} {l} \text{if the superscripts and the subscripts are identical sets}\\ \text{of distinct numbers related by an \textit{odd} permutation}\\ \phantom{+} \end{array}$ $\begin{array} {c} \phantom{+} 0\text{,}\\ \end{array}$ & $\begin{array} {c} \text{for all other combinations of indices.}\\ \end{array}$ end{tabular} right. end{equation} All delta systems are tensors. The Kronecker delta $\delta_{j}^{i}$ satisfies the above definition and may thus be considered to be a delta system. The complete delta system $\delta_{rstuv}^{ijklm}$ is the tensor product of the two permutation systems, i.e. A delta system of any order can be expressed in terms of the Kronecker delta. For example, and Note that these relationships can be captured elegantly with the help of the determinant, i.e. and

A contraction of a delta system yields a lower-order delta system. For example, in $\mathbb{R} ^{5}$, In particular, It is clear how to generalize these relationships to $\mathbb{R} ^{n}$. The contraction of a complete delta system produces a factor of $1$, the next contraction produces a factor of $2$, the next one a factor of $3$, and so on. Finally, the contraction $\delta_{i}^{i}$ of the Kronecker delta $\delta_{j}^{i}$ produces a factor of $n$. Thus, and, in particular, Also, for $m\leq n$, we have

It is left as an exercise to show that a higher-order delta system "absorbs" the lower-order delta system in a contraction according to the formula where the factorial factor corresponds to the lower order.

Any skew-symmetric system $A_{ijklm}$ or $A^{ijklm}$ of order that matches the dimension of the space has a single degree of freedom which may be taken to be the value $A$ of the element $A_{12345}$ or $A^{12345}$. Then we have

A skew-symmetric system of order $m$ that less than $n$ has a greater number of degrees of freedom. We will now show that any such system can be expressed as a linear combination of the elements of the permutation system $e_{ijklm}$ and the coefficients of the linear combination can be treated as the degrees of freedoms of the skew-symmetric form.

For example, let us consider a system $A_{ijk}$ of order $m=3$. Then we can always find system $B^{lm}$ of order $2$ such that Of course, such a system $B^{lm}$ is not unique since it can be altered by any system symmetric in any two coefficients. However, if we stipulate that $A^{lm}$is itself skew-symmetric, then it becomes unique. In fact, the elements of $A^{lm}$ can be retrieved by contraction with $e^{ijklm}$. Indeed, Thus This calculation shows that we can move freely between the skew-symmetric systems $A_{ijk}$ and $B^{lm}$. In fact, there appears to be a perfect duality between $A_{ijk}$ and $B^{lm}$ and we can think of $B^{lm}$ as the degrees of freedom of $A_{ijk}$ and, conversely, of $A_{ijk}$ as the degrees of freedom of $B^{lm}$. In fact, the connection between $A_{ijk}$ and $B^{lm}$ is so strong that we ought to use the same symbol $A$ to denote both systems. While we are at it, let us balance the factorial factors between the two reciprocal equations. Thus, for a given skew-symmetric system $A_{ijk}$ in $\mathbb{R} ^{5}$, the dual skew-symmetric system $A^{lm}$ is given by while the reciprocal relationship reads Note that if $A_{ijk}$ is a relative covariant tensor of weight $m$, then the dual system $A^{kl}$ is a relative contravariant tensor of weight $m+1$.

It is clear how the relationship between a skew-symmetric system and its dual generalizes to $\mathbb{R} ^{n}$. For a system $A_{i_{1}\cdots i_{m}}$ of order $m$, the dual system $A^{j_{1}\cdots j_{n-m}}$ of order $n-m$ is given by while the reciprocal relationship reads If $A_{i_{1}\cdots i_{m}}$ is a relative covariant tensor of weight $m$, then the dual system $A^{j_{1}\cdots j_{n-m}}$ is a relative contravariant tensor of weight $m+1$.

A system of order $m$ has degrees of freedom. Thus, a system and its dual have the same number of degrees of freedom. In particular, systems of orders $0$ and $n$ have a single degree of freedom, while systems of orders $1$ and $n-1$ have $n$ degrees of freedom.

For an elementary example illustrating degrees of freedom, consider a second-order system $A_{ij}$ in $\mathbb{R} ^{3}$. Such a system is represented by a $3\times3$ skew-symmetric matrix, such as and has $n=3$ degrees of freedom. The dual system $A^{k}$ is given by It is left as an exercise to show that Thus, the $3$ degrees of freedom $A^{k}$ of $A_{ij}$ are arranged within the matrix representing $A_{ij}$ as follows:

Let us also consider the extreme example of a fifth-order skew-symmetric system $A_{ijklm}$ in $\mathbb{R} ^{5}$.Then its dual is a system $A$ of order $0$ given by the contraction The system $A_{ijklm}$ can be recovered from $A$ by the equation

For any system $T_{ijk}$, the combination is skew-symmetric. Owing to the identity we have It $T_{ijk}$ is skew-symmetric to begin with, then it equals its skew-symmetric part, i.e.

We can also think of the operation of skew-symmetrization in the following way. Treat $T_{ijk}$ as if it were already skew-symmetric and calculate its dual $T^{lm}$ by the formula Then, calculate the dual to $T^{lm}$, i.e. The result is the skew-symmetric part $\bar{T}_{ijk}$ of $T_{ijk}$, as demonstrated by the following chain of identities:

We pointed out at the top of the Chapter that while the partial derivative of a tensor $T_{j}$ is not itself a tensor, the commutator is one. When we investigated the underlying reasons for this phenomenon, we observe that the "nontensor" part of $\partial_{i}T_{j}$ is symmetric in $i$ and $j$ and is therefore eliminated by alternatization. Naturally, the same alternatization approach works for tensors of arbitrary order and suggests the idea of a new differential operator that, much like the covariant derivative, preserves the tensor property of its input.

The resulting differential operator is known as the exterior derivative. For a covariant tensor $T_{jkl}$ with a representative collection of indices, the exterior derivative, denoted by $d_{i}$, is defined by

Let us consider one interesting example in $\mathbb{R} ^{n}$ and investigate what happens when the exterior derivative is applied to a tensor of order $n-1$. One again, using $\mathbb{R} ^{5}$ for simplicity, consider Represent $T_{jklm}$ in terms of its dual system $T^{n}$, i.e. Then, we have By the product rule (and the fact that $\partial_{r}e_{stuvw}=0$), we have Observe that Therefore,

The generalized Stokes theorem applies to $m$-dimensional surfaces in $\mathbb{R} ^{n}$. To formalize what we mean by that, denote the elements of $\mathbb{R} ^{m}$ by $S^{\alpha}$ and consider within $\mathbb{R} ^{m}$ a finite domain $S$ with boundary $L$.

Denote the elements of $\mathbb{R} ^{n}$ by $Z^{i}$ and let the equations denote a mapping from $\mathbb{R} ^{m}$ to $\mathbb{R} ^{n}$. Then the image of $S$ under this mapping can be considered a surface patch in $\mathbb{R} ^{n}$ with the values $S^{\alpha}$ acting as surface coordinates.

Note that in the previous Chapter, we used the symbol $S$ and $\bar{S}$ to distinguish between the actual surface patch and its arithmetic domain. In the present context however, there is hardly a need to make that distinction since the statement of the generalized Stokes theorem does not make any references to any of the geometric characteristics, such as the volume element $\sqrt{S}$, of the "actual" surface $S$ but only to the arithmetic domain $\mathbb{R} ^{m}$.

The closed boundary $L$ is an $\left( m-1\right)$-dimensional surface. Let us refer it to coordinates $L^{\Phi}$. In other words, represent $L$ as the image of a mapping from $\mathbb{R} ^{m-1}$ to $\mathbb{R} ^{m}$ given by the equations Once again, we do not use the distinct symbols $L$ and $\bar{L}$ for the surface $\left( m-1\right)$-dimensional surface $L$ in $\mathbb{R} ^{m}$ and the arithmetic domain $L$ in $\mathbb{R} ^{m-1}$ since the Stokes theorem does not reference and of the geometric characteristics of the surface $L$.

Each mapping is characterized by the shift tensor. The shift tensor $Z_{\alpha}^{i}$ characterizing $S$ and the shift tensor $S_{\Phi}^{\alpha}$ characterizing $L$ are given by the familiar equations and The shift tensors may still be thought as representing the respective tangent spaces. However, no further geometric characteristic of the surfaces $S$ and $L$ are involved.

For the sake of simplicity assume that $m=3$. We do not need to assume a specific $n$ since it does not figure in the statement of the theorem. The theorem applies to a system $F_{jkl}$ of order $m-1$, so that the collection $\partial_{i}F_{jkl}$ of its partial derivative is of order $m$. Clearly, in order to be able to carry out the partial differentiation, the field $F_{jkl}$ must be defined not only at the points on the surface $S$ but in the wider ambient space $\mathbb{R} ^{n}$.

The generalized Stokes theorem reads Observe that both repeated integrals are arithmetic in nature and are meant to be interpreted in the sense of ordinary Calculus. Furthermore, the fields $\partial_{i}F_{jk}$ and $F_{jk}$ receive analogous treatments where each live index is engaged in a contraction with the shift tensor and then each of the live surface indices is contracted with the permutation system.

With the help of the two shorthand symbols and the generalized Stokes theorem reads

Notice that the theorem is stated in terms of the partial derivative $\partial_{i}F_{jk}$ rather than the exterior derivative $d_{i}F_{jk}$. This is possible owing to the presence of the skew-symmetric permutation symbol $e^{\alpha\beta\gamma}$ which makes the combination $e^{\alpha\beta\gamma }Z_{\alpha}^{i}Z_{\beta}^{j}Z_{\gamma}^{k}$ skew-symmetric in $i$, $j$, $k$, thus eliminating the need for alternating $\partial_{i}F_{jk}$.

For the first proof of the generalized Stokes' theorem, let us assume that the divergence theorem holds for a domain $S$ in a Riemannian space of any dimension. Although the statement of the Stokes theorem references only the arithmetic elements of $S$ does not mean that we cannot arbitrarily impose a Riemannian structure upon it for the sake of proving the theorem.

Introduce a Cartesian metric in the ambient space $\mathbb{R} ^{n}$ and let $Z_{ij}$ denote the associated metric tensor. Naturally, $Z_{ij}$ corresponds to the $n\times n$ identity matrix. Let $S_{\alpha\beta}$ be the induced metric tensor on $S$, i.e. and $\sqrt{S}$ be the corresponding volume element.

Let $F_{jk}\left( S\right)$ be the restriction of the field $F_{jk}$ to the surface $S$. Then $F_{jk}\left( S\right)$ is given by the composition and an application of the chain rule yields i.e. Thus, the integral becomes Next, observe that the partial derivative $\partial_{\alpha}$ can be applied to the entire product $e^{\alpha\beta\gamma}F_{jk}Z_{\beta}^{j}Z_{\gamma}^{k}$, i.e. Indeed, by the product rule The first term on the right vanishes since due to the fact that the permutation symbol $e^{\alpha\beta\gamma}$ has constant elements. The third and fourth terms vanish because the partial derivative of the shift tensor, e.g. is symmetric with respect to its surface indices and therefore vanishes in a double contraction with $e^{\alpha\beta\gamma}$. This is the crucial moment where the skew-symmetry of the permutation symbol is used. Thus, the identity is confirmed.

Therefore, we have Let us now take two steps towards "tensorizing" the integrand. First, replace the permutation system $e^{\alpha\beta\gamma}$ with its expression $\sqrt {S}\varepsilon^{\alpha\beta\gamma}$ in terms of the Levi-Civita symbol $\varepsilon^{\alpha\beta\gamma}$. We have By the product rule, Since we have Next, let us focus on the combination and convert the partial derivative to the covariant derivative. This will prepare the integrand for an application of the divergence theorem. Since the combination $\varepsilon^{\alpha\beta\gamma}F_{jk}Z_{\beta}^{j}Z_{\gamma}^{k}$ has a single live index $\alpha$, we have or Thus, the net effect of the two tensorization steps is captured by the identity and we have

Now, observe that the arithmetic integral on the right corresponds to the geometric integral of the quantity $\nabla_{\alpha}\left( \varepsilon ^{\alpha\beta\gamma}F_{jk}Z_{\beta}^{j}Z_{\gamma}^{k}\right)$ over $S$ subject to the induced metric tensor $S_{\alpha\beta}$, i.e. By the divergence theorem, the integral on the right can be converted to a geometric integral over the boundary $L$, i.e. Recall that the normal $n_{\alpha}$ is given by the equation Therefore, for the combination $n_{\alpha}\varepsilon^{\alpha\beta\gamma}$, we have which yields Convert the integral on the right to its arithmetic form, i.e. Finally, recall that $\sqrt{L}\varepsilon^{\Phi\Psi}=e^{\Phi\Psi}$, which yields