By definition, a system is skew-symmetric if any of its elements related by a swap of two indices have opposite values. We will consider skew-symmetric systems of various orders. For second-order systems, such as $A_{ij}$, the skew-symmetric property reads which agrees with the familiar like-named property of matrices. For example, in four dimensions, $A_{ij}$ would be represented by a skew-symmetric matrix, such as The conspicuous zeros on the diagonal, which correspond to $i=j$, are an important characteristic of skew-symmetric systems.

To illustrate higher-order skew-symmetric systems, consider a fourth-order system $A_{ijkl}$. For the time being, we are not concerned with the tensor property of $A_{ijkl}$. Thus, the placement of the indices as subscripts is arbitrary and we could just as easily consider a fourth-order system $A^{ijkl}$ with four superscripts. We could even consider a mix of subscripts and superscripts, but that would unnecessarily complicate the notation.

For reasons that will become apparent shortly, we must require that the dimension $n$ of space is greater than or equal to the order of the system, i.e. $n\geq4$ in the case of $A_{ijkl}$. For the sake of the present discussion, assume that $n=6$.

The system $A_{ijkl}$ is skew-symmetric if, once again, any of its elements related by a swap of two indices have opposite values. This requirement can be reduced down to three identities involving swaps of consecutive indices, i.e. Indeed, there is no need to document the swaps for all possible pairs of indices since any swap can be accomplished by swaps of consecutive indices. For example, the swap can be accomplished by three swaps of consecutive indices, i.e.

Critical to the internal consistency of the definition is the fact that any given permutation can only be represented either by an odd number of swaps or by an even number of swaps. After all, if there was a way to achieve the permutation $ijkl\rightarrow kjil$ by an even number swaps, then the value of $A_{kjil}$ would be simultaneously $-A_{ijkl}$ and $+A_{ijkl}$ depending on whether we arrive at $kjil$ by an odd or even number of swaps. Permutations that can be obtained by an odd number of swaps from the identity permutation are called odd, while those that require an even number of swaps are called even. This property of permutations is called parity.

The skew-symmetric condition imposes a severe constraint on the values of $A_{ijkl}$. First, all elements whose indices are not all distinct must equal $0$. For example, consider the element $A_{1614}$ for which the first and the third indices are equal. By the skew-symmetric condition, we must have since swapping the $1^{\text{st}}$ and the $3^{\text{rd}}$ values in $1,6,1,4$ yields the same combination $1,6,1,4$. Therefore, Thus, in order for an element of $A_{ijkl}$ to be nonzero, its indices must have distinct values.

Second, any two elements whose indices are related by a permutation are either equal or opposite of each other. Indeed, consider the elements $A_{1256}$ and $A_{2561}$. Since the combinations $1256$ and $2561$ are related by an odd number of swaps, e.g. $1256\rightarrow2156\rightarrow2516\rightarrow2561$, we must have Thus, a fourth-order skew-symmetric system in an $n$-dimensional space can have at most degrees of freedom, i.e. independently assigned values. More generally, a skew-symmetric system of order $m$ in an $n$-dimensional space can have at most degrees of freedom. In particular, and crucially for the upcoming discussion, when the order of a skew-symmetric system matches the dimension of the space, the former can only have a single degree of freedom. For example, if $n=4$ then a system $A_{ijkl}$ has one degree of freedom represented by the element All of the remaining elements of $A_{ijkl}$ are either $0$ or $\pm A_{1234}$.

Finally, when the order of a skew-symmetric system exceeds the dimension of the space, all of its elements must vanish since the indices cannot all have distinct values.

The result of a double contraction between a skew-symmetric and a symmetric system is zero, i.e. a system in which all elements are $0$. For example, if $A_{ijkl}$ is a skew-symmetric system and $U^{ij}$ is a symmetric system, i.e. then

We will demonstrate the above identity in a way that will make it clear that the statement holds for more complicated indicial signatures as well. Let and consider the combination in which the indices on $U^{ij}$ are swapped, i.e. We will show that this new combination simultaneously equals $X_{kl}$ and $-X_{kl}$ which, of course, implies that $X_{kl}=0$.

Thanks to the symmetry of $U^{ij}$, we have On the other hand, due to the skew-symmetry of $A_{ijkl}$ -- in particular, the fact that $A_{ijkl}=-A_{jikl}$ -- we have Note that in the second step in this string of identities, we simply exchanged the letters $i$ and $j$. Thus, indeed, and, therefore, as we set out to show.

When both systems are second-order, this statement can be interpreted in the language of matrices. Namely, the product of a skew-symmetric and symmetric matrices has zero trace. For example, consider the product and observe that the trace of the resulting matrix is $-5-24+29=0$.

Let us now turn our attention to skew-symmetric systems whose order matches the dimension of the space. We will illustrate such systems with a third-order system $A_{ijk}$ in a three-dimensional space. Since the nonzero elements of $A_{ijk}$ correspond to the permutations of the numbers $1$, $2$, and $3$, out of a total of $27$ elements, there are only $6$ potentially nonzero ones: Denoting the sole degree of freedom shared by these elements by $\alpha$, we have and

When $\alpha=1$, the corresponding system is called a permutation system and is denoted by $e_{ijk}$. We will also consider the superscripted permutation system $e^{ijk}$ defined in the exact same way. Let us summarize the values of $e_{ijk}$ and $e^{ijk}$ in the language that generalizes to higher and lower dimensions: Any third-order skew-symmetric system $A_{ijk}$ in three dimensions is a scalar multiple of $e_{ijk}$, i.e.

We must now say a few words about the placement of the indices in the permutation systems $e_{ijk}$ and $e^{ijk}$ since, in Tensor Calculus, indices must "earn" their placements by accurately indicating how the system transforms under a change of coordinates. While the permutation systems are independent of coordinates, we can simply assume, in the context of a coordinate analysis, that $e_{ijk}$ and $e^{ijk}$ have the values summarized above at all points of a Euclidean space. Then we may legitimately ask whether $e_{ijk}$ and $e^{ijk}$ are tensors. In other words, suppose that, in the primed coordinates $Z^{i^{\prime}}$, $e_{i^{\prime}j^{\prime}k^{\prime}}$ and $e^{i^{\prime}j^{\prime}k^{\prime}}$ are defined in the exact same way as $e_{ijk}$ and $e^{ijk}$. Then, are the unprimed and primed systems related by the identities and The answer is no, but you may be surprised just how close to actually holding these identities are. In the next Chapter, we will introduce simple modifications of the permutation systems known as the Levi-Civita symbols which are, in fact, (nearly) tensors.

While the better part of this Chapter is devoted to the determinant, we would be remiss at this stage not to point out the inescapable connection between the permutation systems and the determinant. For example, for a $3\times3$ matrix $M$ with entries $M_{ij}$, the determinant is given by the formula where the summation takes place over all possible permutations $\pi$ of the numbers $1,2,3$. For example, the permutation $3,2,1$ corresponds to $\pi$ such that The sign of the permutation, denoted by $\operatorname{sign}\left( \pi\right)$ is defined to be $1$ if $\pi$ is even and $-1$ if $\pi$ is odd. When fully unpacked, the above formula reads From this expansion, it becomes apparent how the summation-based formula for the determinant essentially handpicks the $6$ contributing terms.

However, thanks to our present experience with the tensor notation, we immediately observe that the same sum can be captured far more compactly by the indicial expression Of course, this compactness comes at the expense of computational efficiency since the above contraction contains not $6$ but $27$ terms, all of which are zero except for the $6$ that matter. Nevertheless, the trade-off is well worth it. In fact, it seems almost as if the permutation systems were specifically formulated for the purpose of expressing the determinant.

Also note that for a set of three first-order systems $U^{i}$, $V^{i}$, and $W^{i}$, the combination corresponds to the determinant of the matrix whose columns (or rows) consist of $U^{i}$, $V^{i}$, and $W^{i}$, i.e. As we observed in Chapter 3, if we think of $U^{i}$, $V^{i}$, and $W^{i}$ as the components of the vectors $\mathbf{U}$, $\mathbf{V}$, and $\mathbf{W}$, then the sign of tells us whether the orientation of $\mathbf{U}$, $\mathbf{V}$, and $\mathbf{W}$ is the same as that of the covariant basis $\mathbf{Z}_{i}$.

We will leave the topic of determinants for now and will return to it later in this Chapter where we will discuss it in far greater detail.

The delta systems are objects of utmost elegance and usefulness. The delta systems are highly structured, expressive, and are governed by a tight set of algebraic rules. Thus, they go a long way towards the algebraization of our subject. Furthermore, the delta systems are tensors that have an equal number of superscripts and subscripts. As such, they introduce a great deal of balance into tensorial calculations.

The complete delta system $\delta_{rst}^{ijk}$ is defined as the product of the permutation systems $e^{ijk}$ and $e_{rst}$, i.e. The term complete refers to the fact that it has as many superscripts and as many subscripts as the dimension of the space. The choice of the letter $\delta$ is not arbitrary. The Kronecker delta $\delta_{j}^{i}$ and the complete delta system $\delta_{rst}^{ijk}$ are part of the same family of related objects.

The delta system $\delta_{rst}^{ijk}$ has the greatest number of indices of any system we have encountered so far. Is this a d\'{ebauches} of indices or a choreographed dance? You be the judge. What is undeniable is that the delta system $\delta_{rst}^{ijk}$ possesses a great deal of structure that will enable us to work with it on an intuitive algebraic level.

Let us now explore its $3^{6}=729$ elements. Since the permutation systems consist of $-1$s, $0$s, and $1$s, so does the delta system $\delta_{rst} ^{ijk}$. Its values can be summarized in the following table. Thus, out of the $729$ elements, only $\left( 3!\right) ^{2}=36$ are not zero.

Naturally, the delta system $\delta_{rst}^{ijk}$ is skew-symmetric in both its superscripts and its subscripts. It follows that simultaneously subjecting its superscripts and subscripts to the same permutation yields in an equivalent symbol. For example, where the first two superscripts and the first two subscripts were switched. Similarly, where we moved the last superscript and the last subscript to the first position. More generally, subjecting its superscripts and subscripts to permutations of the same parity results in equivalent symbols. Finally, it is left as an exercise to demonstrate the frequently used identity which will also be used in our upcoming discussion of the determinant.

Of great utility is the following identity that expresses the delta system $\delta_{rst}^{ijk}$ in terms of the Kronecker deltas: It is left up to the reader to describe this identity in words. Undoubtedly, one can clearly see that the right side is related to the $3\times3$ determinant. Indeed, the above relationship can be captured by the elegant equation Elegant though it may be, the preceding expanded form is more transparent and useful. Note that this relationship can also be adopted as the definition of the delta system $\delta_{rst}^{ijk}$.

Explaining in complete detail why the identity is valid is left as an exercise. It is a good idea to consider a number of specific examples and to articulate the precise reason for each conclusion. To help you in this endeavor, let us consider three such examples that will shed some light on the "inner mechanics" of the expression on the right.

First, consider a typical nonzero element $\delta_{231}^{123}$ which should equal $1$ since $2,3,1$ is two switches away from $1,2,3$. We have Observe that the only nonzero term is $\delta_{1}^{1}\delta_{2}^{2}\delta _{3}^{3}$, which indeed equals $1$. For a second example, consider $\delta_{122}^{123}$ which should equal $0$ because $1,2,2$ is not a permutation. We have where all terms equal $0$ because it is impossible to have a nonzero term when the superscripts and the subscripts are not identical sets of numbers. Finally, consider the element $\delta_{113}^{113}$ which should also equal $0$, because while the superscripts and the subscripts represent identical sets of numbers, neither is a permutation. We have This time, the two nonzero terms -- first and last -- cancel each other.

Despite its bulkiness, the identity finds numerous practical applications. However, its immediate theoretical impact is the crucial implication that the delta system $\delta_{rst}^{ijk}$ is a tensor, as it is expressed by sums and products of the tensor Kronecker delta. Additionally, the same identity shows that $\delta_{rst}^{ijk}$ vanishes under the covariant derivative, i.e. In other words, the metrinilic property of the covariant derivative extends to the delta system $\delta_{rst}^{ijk}$.

Interestingly, not only can the delta system be expressed in terms of the Kronecker delta, but the latter can also be expressed in terms of the former. This relationship is captured by the identity which is one of many that will be discussed next.

Between the Kronecker delta $\delta_{j}^{i}$ and the complete delta system $\delta_{rst}^{ijk}$, there exists an object with an intermediate number of indices known as the partial delta system $\delta_{rs}^{ij}$. In fact, when we generalize this discussion to $n$ dimensions, we will discover that between the Kronecker delta $\delta_{j}^{i}$ and the complete delta system $\delta_{j_{1}\cdots j_{n}}^{i_{1}\cdots i_{n}}$, there exists a whole family of partial delta systems $\delta_{j_{1}\cdots j_{k}} ^{i_{1}\cdots i_{k}}$, one for every $k$ between $1$ and $n$. The partial deltas are defined in a way that bridges the Kronecker delta and the complete delta system.

For the remainder of this Section, we will operate in three dimensions which will limit our discussion to the delta system $\delta_{rs}^{ij}$. However, the reader is invited to think about how all of the presented concepts generalize to $n$ dimensions. In describing the elements of $\delta_{rs}^{ij}$, we will choose our words so that the resulting definition also encompasses the Kronecker delta $\delta_{j}^{i}$ and the complete delta system $\delta _{rst}^{ijk}$ and generalizes to $n$ dimensions. We have: A few examples of the values of $\delta_{rs}^{ij}$ and the reasons for those values are given in the following table. Thus, out of a total of $81$ elements, only $6$ are not zero. Specifically, and

All delta systems, from the Kronecker delta to the complete delta system are connected by various identities. Generally speaking, a contraction of a higher-order delta system yields a lower-order delta system. Meanwhile, higher-order delta systems can be expressed in terms of the Kronecker delta by determinant-like combinations. Demonstrations of all identities are left as exercises.

Let us begin by considering the result of contracting higher-order delta systems. Starting with the complete system $\delta_{rst}^{ijk}$, we have This is one of the most frequently used identities in practical applications. In fact, $\delta_{rs}^{ij}$ almost always arises as the result of contracting $\delta_{rst}^{ijk}$. Furthermore, by the contraction property of tensors, the above identity immediately tells us that the partial delta system $\delta_{rs}^{ij}$ is a tensor, leading to the conclusion that all delta systems are tensors. Additionally, thanks to the above identity, we know that $\delta_{rs}^{ij}$ vanishes under the covariant derivative, i.e.

When $\delta_{rs}^{ij}$ is itself contracted, the result is a multiple of the Kronecker delta, i.e. To understand where the factor of $2$ comes from, note that When $i=r$ and both values equal, say, $1$, we have and it is clear that the same result would be obtained for any common value of $i$ and $r$.

In summary, the contraction of a higher-order delta system results in a lower-order delta system or a multiple thereof.

Now, let us talk about expressing higher-order delta systems in terms of the Kronecker delta. Recall that the complete delta system $\delta_{rst}^{ijk}$ is given in terms of the Kronecker delta by the formula Contracting on $k$ and $t$, we discover that This is one of the most commonly used identities, along with In fact, these identities are usually used in tandem: a contraction reduces $\delta_{rst}^{ijk}$ to $\delta_{rs}^{ij}$ which is followed by the representation of $\delta_{rs}^{ij}$ in terms of the Kronecker delta. Thus, for convenience, let us combine the two identities into one, i.e. Note that the expression on the right once again exhibits the familiar $2\times2$ determinant pattern and can therefore be captured elegantly, albeit impractically, in the following way:

This completes our discussion of permutation and delta systems in three dimensions and we will now turn our attention to lower and higher-dimensional cases.

The permutation systems, the delta systems, and all of the related identities naturally generalize to any number of dimensions. The two-dimensional case is particularly important since it plays a crucial role in the study of two-dimensional surfaces.

In two dimensions, all indices assume values $1$ and $2$. The permutation systems $e_{ij}$ and $e^{ij}$ are given by However, it may be even easier simply to list all of the elements. For $e_{ij}$, we have and for $e^{ij}$ we have In matrix terms, both $e_{ij}$ and $e^{ij}$ correspond to the matrix

The complete delta system $\delta_{rs}^{ij}$ is the product of the permutation systems, i.e. Note that, generally speaking, the symbol $\delta_{rs}^{ij}$ is ambiguous since it represents different systems in different dimensions. Thus, when working with these symbols, it is essential to indicate the relevant dimension.

This point is highlighted by the contraction property satisfied by $\delta_{rs}^{ij}$ in two dimensions. Note that the analogous identity in three dimensions read $\delta_{r}^{i}=2\delta_{rj}^{ij}$. On the other hand, the expression for $\delta_{rs}^{ij}$ in terms of the Kronecker delta reads which is identical to the three-dimensional case. It turns out that the form of this identity is universal for all dimensions.

We must begin by noting that the case $n\gt 3$ does not correspond to a geometric Euclidean space as we have defined it. Nevertheless, the analytical framework holds up perfectly for any $n$. Furthermore, in the future -- in Chapter 20, to be precise -- we will generalize the concept of Euclidean spaces to higher dimensions.

In an $n$-dimensional space, indices run the values between $1$ and $n$. The permutation systems $e_{i_{1}\cdots i_{n}}$ and $e^{i_{1}\cdots i_{n}}$ are defined as follows. Thus, each permutation system has $n^{n}$ elements of which only $n!$ are not zero.

The complete delta system $\delta_{j_{1}\cdots j_{n}}^{i_{1}\cdots i_{n}}$ is the product of the permutation systems, i.e. and has the values The complete delta system can be expressed in terms of the Kronecker deltas by the identity which proves that it is a tensor, and that it vanishes under the covariant derivative, i.e.

The partial delta systems $\delta_{j_{1}\cdots j_{k}}^{i_{1}\cdots i_{k}}$ for every $k$ between $1$ and $n$ can be defined by the very same language as in the two- and three-dimensional cases, i.e. It is, once again, important to point out that this language includes the Kronecker delta $\delta_{j}^{i}$ as well as the complete delta system $\delta_{j_{1}\cdots j_{n}}^{i_{1}\cdots i_{n}}$. For every $k$, the partial delta system $\delta_{j_{1}\cdots j_{k}}^{i_{1}\cdots i_{k}}$ can be expressed in terms of the Kronecker delta by which shows the great universality of this identity.

A contraction of any delta system results in a scalar multiple of the lower-order delta system. It is left as an exercise to show that and so on. In general, the multiple corresponding to the contraction of the $k^{\text{th}}$-order system $\delta_{j_{1}\cdots j_{k-1}j_{k}}^{i_{1}\cdots i_{k-1}i_{k}}$ is $n-\left( k-1\right)$, i.e.

For a specific example, let us consider the case $n=5$ and denote the complete delta system by the symbol $\delta_{rstuv}^{ijklm}$. Then we have the following relationships Thus, in particular, and, in general,

In Matrix Algebra, it is common to represent a square matrix $A$ as a sum of a symmetric matrix $S$ and a skew-symmetric matrix $K$, i.e. For example, for we have

Such a decomposition is unique and, as it is easy to show, we must have and The matrix $S$ is known as the symmetric part of $A$ and $K$ is known as the skew-symmetric part of $A$.

In indicial notation, the same identities read where and In the expression for $K_{ij}$, we are beginning to see the now-familiar skew-symmetric indicial pattern.

The concepts of symmetric and skew-symmetric parts generalize to higher-order systems (although it is no longer true that a system is a sum of its symmetric and skew-symmetric parts). For example, for a third-order system $A_{ijk}$, its skew-symmetric part $S_{ijk}$ is defined as and it is clear how to generalize this definition to higher-order systems. The operation that converts $A_{ijk}$ into its skew-symmetric part is known as alternatization and it is valid in any dimension that is equal or greater than the order of the system. Alternatization is relevant to our narrative because it can be used as an alternative approach to constructing invariant differential operators.

As we established, the partial derivative for a tensor $T_{i}$ is not a tensor in its own right. We have remedied this problem by replacing the partial derivative with the combination which is a tensor. Observe, however, that the combination which represents the non-tensor portion of the partial derivative is symmetric in $i$ and $k$. Therefore, in the alternatization of $\nabla_{k}T_{i}$, i.e. the terms containing the Christoffel symbol cancel each other. As a result, Thus, the alternatization is also a tensor. In other words, the tensor property can be achieved by alternatization -- once again, by virtue of cancelling the non-tensor contributions. This remains true for tensors of any order as well as for partial derivatives of any order. For example, the alternatization of the variant is a tensor.