A Gentle Introduction to Tensors

Boaz Porat
Department of Electrical Engineering
Technion – Israel Institute of Technology
boaz@ee.technion.ac.il

May 27, 2014

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Opening Remarks

This document was written for the benefits of Engineering students, Electrical Engineering students in particular, who are curious about physics and would like to know more about it, whether from sheer intellectual desire or because one's awareness that physics is the key to our understanding of the world around us. Of course, anybody who is interested and has some college background may find this material useful.

I chose tensors as a first topic for two reasons. First, tensors appear everywhere in physics, including classical mechanics, relativistic mechanics, electrodynamics, particle physics, and more. Second, tensor theory, at the most elementary level, requires only linear algebra and some calculus as prerequisites. Proceeding a small step further, tensor theory requires background in multivariate calculus. For a deeper understanding, knowledge of manifolds and some point-set topology is required.

Accordingly, we divide the material into three chapters. The first chapter discusses constant tensors and constant linear transformations. Tensors and transformations are inseparable. To put it succinctly, tensors are geometrical objects over vector spaces, whose coordinates obey certain laws of transformation under change of basis. Vectors are simple and well-known examples of tensors, but there is much more to tensor theory than vectors. The second chapter discusses tensor fields and curvilinear coordinates. It is this chapter that provides the foundations for tensor applications in physics. The third chapter extends tensor theory to spaces other than vector spaces, namely manifolds. This chapter is more advanced than the first two, but all necessary mathematics is included and no additional formal mathematical background is required beyond what is required for the second chapter.

I have used the coordinate approach to tensors, as opposed to the formal geometrical approach. Although this approach is a bit old fashioned, I still find it the easier to comprehend on first learning, especially if the learner is not a student of mathematics or physics.

All vector spaces discussed in this document are over the field $\mathbb{R}$ of real numbers. We will not mention this every time but assume it implicitly.

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Chapter 1: Constant Tensors and Constant Linear Transformations

1.1 Plane Vectors

Let us begin with the simplest possible setup: that of plane vectors. We think of a plane vector as an arrow having direction and length.

The length of a physical vector must have physical units; for example: distance is measured in meter, velocity in meter/second, force in Newton, electric field in Volt/meter, and so on. The length of a "mathematical vector" is a pure number. Length is absolute, but direction must be measured relative to some (possibly arbitrarily chosen) reference direction, and has units of radians (or, less conveniently, degrees). Direction is usually assumed positive in counterclockwise rotation from the reference direction.

Vectors, by definition, are free to move parallel to themselves anywhere in the plane and they remain invariant under such moves (such a move is called translation).

Vectors are abstract objects, but they may be manipulated numerically and algebraically by expressing them in bases. Recall that a basis in a plane is a pair of non-zero and non-collinear vectors $(\mathbf{e}_1, \mathbf{e}_2)$.

Let $\mathbf{x}$ be an arbitrary plane vector and let $(\mathbf{e}_1, \mathbf{e}_2)$ be some basis in the plane. Then $\mathbf{x}$ can be expressed in a unique manner as a linear combination of the basis vectors; that is,

$$\mathbf{x} = \mathbf{e}_1 x^1 + \mathbf{e}_2 x^2 \tag{1.1}$$

The two real numbers $(x^1, x^2)$ are called the coordinates of $\mathbf{x}$ in the basis $(\mathbf{e}_1, \mathbf{e}_2)$. The following are worth noting:

1. Vectors are set in bold font whereas coordinates are set in italic font.
2. The basis vectors are numbered by subscripts whereas the coordinates are numbered by superscripts. This distinction will be important later.
3. In products such as $\mathbf{e}_1 x^1$ we place the vector on the left and the scalar on the right. In most linear algebra books the two are reversed — the scalar is on the left of the vector. The reason for our convention will become clear later.

Recalling notations from vector-matrix algebra, we may express (1.1) as

$$\mathbf{x} = \begin{bmatrix} \mathbf{e}_1 & \mathbf{e}_2 \end{bmatrix} \begin{bmatrix} x^1 \\ x^2 \end{bmatrix} \tag{1.2}$$

For now we will use row vectors to store basis vectors and column vectors to store coordinates. Later we will abandon expressions such as (1.2) in favor of more compact and more general notations.

1.2 Transformation of Bases

Consider two bases $(\mathbf{e}_1, \mathbf{e}_2)$, which we will henceforth call the old basis, and $(\tilde{\mathbf{e}}_1, \tilde{\mathbf{e}}_2)$, which we will call the new basis.

Since $(\mathbf{e}_1, \mathbf{e}_2)$ is a basis, each of the vectors $(\tilde{\mathbf{e}}_1, \tilde{\mathbf{e}}_2)$ can be uniquely expressed as a linear combination of $(\mathbf{e}_1, \mathbf{e}_2)$:

$$\tilde{\mathbf{e}}_1 = \mathbf{e}_1 S^1_1 + \mathbf{e}_2 S^2_1$$

$$\tilde{\mathbf{e}}_2 = \mathbf{e}_1 S^1_2 + \mathbf{e}_2 S^2_2 \tag{1.3}$$

Equation (1.3) is the basis transformation formula from $(\mathbf{e}_1, \mathbf{e}_2)$ to $(\tilde{\mathbf{e}}_1, \tilde{\mathbf{e}}_2)$. The 4-parameter object $\{S^j_i,\ 1 \le i, j \le 2\}$ is called the direct transformation from the old basis to the new basis. We may also write (1.3) in vector-matrix notation:

$$\begin{bmatrix} \tilde{\mathbf{e}}_1 & \tilde{\mathbf{e}}_2 \end{bmatrix} = \begin{bmatrix} \mathbf{e}_1 & \mathbf{e}_2 \end{bmatrix} \begin{bmatrix} S^1_1 & S^1_2 \\ S^2_1 & S^2_2 \end{bmatrix} = \begin{bmatrix} \mathbf{e}_1 & \mathbf{e}_2 \end{bmatrix} S \tag{1.4}$$

The matrix $S$ is the direct transformation matrix from the old basis to the new basis. This matrix is uniquely defined by the two bases. Note that the rows of $S$ appear as superscripts and the columns appear as subscripts; remember this convention for later.

A special case occurs when the new basis is identical with the old basis. In this case, the transformation matrix becomes the identity matrix $I$, where $I^i_i = 1$ and $I^j_i = 0$ for $i \ne j$.

Since $(\tilde{\mathbf{e}}_1, \tilde{\mathbf{e}}_2)$ is a basis, each of the vectors $(\mathbf{e}_1, \mathbf{e}_2)$ may be expressed as a linear combination of $(\tilde{\mathbf{e}}_1, \tilde{\mathbf{e}}_2)$. Hence the transformation $S$ is perforce invertible and we can write

$$\begin{bmatrix} \mathbf{e}_1 & \mathbf{e}_2 \end{bmatrix} = \begin{bmatrix} \tilde{\mathbf{e}}_1 & \tilde{\mathbf{e}}_2 \end{bmatrix} \begin{bmatrix} S^1_1 & S^1_2 \\ S^2_1 & S^2_2 \end{bmatrix}^{-1} = \begin{bmatrix} \tilde{\mathbf{e}}_1 & \tilde{\mathbf{e}}_2 \end{bmatrix} \begin{bmatrix} T^1_1 & T^1_2 \\ T^2_1 & T^2_2 \end{bmatrix} = \begin{bmatrix} \tilde{\mathbf{e}}_1 & \tilde{\mathbf{e}}_2 \end{bmatrix} T \tag{1.5}$$

where $T = S^{-1}$ or, equivalently, $ST = TS = I$. The object $\{T^j_i,\ 1 \le i, j \le 2\}$ is the inverse transformation and $T$ is the inverse transformation matrix.

In summary, with each pair of bases there are associated two transformations. Once we agree which of the two bases is labeled old and which is labeled new, there is a unique direct transformation (from the old to the new) and a unique inverse transformation (from the new to the old). The two transformations are the inverses of each other.

1.3 Coordinate Transformation of Vectors

Equation (1.2) expresses a vector $\mathbf{x}$ in terms of coordinates relative to a given basis $(\mathbf{e}_1, \mathbf{e}_2)$. If a second basis $(\tilde{\mathbf{e}}_1, \tilde{\mathbf{e}}_2)$ is given, then $\mathbf{x}$ may be expressed relative to this basis using a similar formula

$$\mathbf{x} = \tilde{\mathbf{e}}_1 \tilde{x}^1 + \tilde{\mathbf{e}}_2 \tilde{x}^2 = \begin{bmatrix} \tilde{\mathbf{e}}_1 & \tilde{\mathbf{e}}_2 \end{bmatrix} \begin{bmatrix} \tilde{x}^1 \\ \tilde{x}^2 \end{bmatrix} \tag{1.6}$$

The coordinates $(\tilde{x}^1, \tilde{x}^2)$ differ from $(x^1, x^2)$, but the vector $\mathbf{x}$ is the same.

We now pose the following question: how are the coordinates $(\tilde{x}^1, \tilde{x}^2)$ related to $(x^1, x^2)$? To answer this question, recall the transformation formulas between the two bases and perform the following calculation:

$$\begin{bmatrix} \tilde{\mathbf{e}}_1 & \tilde{\mathbf{e}}_2 \end{bmatrix} \begin{bmatrix} \tilde{x}^1 \\ \tilde{x}^2 \end{bmatrix} = \begin{bmatrix} \mathbf{e}_1 & \mathbf{e}_2 \end{bmatrix} S \begin{bmatrix} \tilde{x}^1 \\ \tilde{x}^2 \end{bmatrix} = \mathbf{x} = \begin{bmatrix} \mathbf{e}_1 & \mathbf{e}_2 \end{bmatrix} \begin{bmatrix} x^1 \\ x^2 \end{bmatrix} \tag{1.7}$$

Since (1.7) must hold identically for an arbitrary vector $\mathbf{x}$, we are led to conclude that

$$S \begin{bmatrix} \tilde{x}^1 \\ \tilde{x}^2 \end{bmatrix} = \begin{bmatrix} x^1 \\ x^2 \end{bmatrix} \tag{1.8}$$

Or, equivalently,

$$\begin{bmatrix} \tilde{x}^1 \\ \tilde{x}^2 \end{bmatrix} = S^{-1} \begin{bmatrix} x^1 \\ x^2 \end{bmatrix} = T \begin{bmatrix} x^1 \\ x^2 \end{bmatrix} \tag{1.9}$$

We have thus arrived at a somewhat surprising conclusion: the coordinates of a vector when passing from an old basis to a new basis are transformed via the inverse of the transformation from the old basis to the new.

As an example, the direct transformation between two bases is

$$S = \begin{bmatrix} 1 & 0.5 \\ 0.25 & 1 \end{bmatrix}$$

The inverse transformation is

$$T = 0.875 \begin{bmatrix} 1 & -0.5 \\ -0.25 & 1 \end{bmatrix}$$

The result obtained in this section is important and should be memorized: When a basis is transformed using a direct transformation, the coordinates of an arbitrary vector are transformed using the inverse transformation. For this reason, vectors are said to be contravariant ("they vary in a contrary manner").

1.4 Generalization to Higher-Dimensional Vector Spaces

We assume that you have studied a course in linear algebra; therefore you are familiar with general (abstract) finite-dimensional vector spaces. An $n$-dimensional vector space possesses a set of $n$ linearly independent vectors, but no set of $n+1$ linearly independent vectors. A basis for an $n$-dimensional vector space $V$ is any ordered set of linearly independent vectors $(\mathbf{e}_1, \mathbf{e}_2, \ldots, \mathbf{e}_n)$. An arbitrary vector $\mathbf{x}$ in $V$ can be expressed as a linear combination of the basis vectors:

$$\mathbf{x} = \sum_{i=1}^n \mathbf{e}_i x^i \tag{1.10}$$

The real numbers in (1.10) are called linear coordinates. Note again our preferred convention of writing the vector on the left of the scalar. If a second basis $(\tilde{\mathbf{e}}_1, \tilde{\mathbf{e}}_2, \ldots, \tilde{\mathbf{e}}_n)$ is given, there exist unique transformations $S$ and $T$ such that

$$\tilde{\mathbf{e}}_i = \sum_{j=1}^n \mathbf{e}_j S^j_i, \qquad \mathbf{e}_i = \sum_{j=1}^n \tilde{\mathbf{e}}_j T^j_i, \qquad T = S^{-1} \tag{1.11}$$

The coordinates of $\mathbf{x}$ in the new basis are related to those in the old basis according to the transformation law

$$\tilde{x}^i = \sum_{j=1}^n T^i_j x^j \tag{1.12}$$

Equation (1.12) is derived in exactly the same way as (1.9). Thus, vectors in an $n$-dimensional space are contravariant.

Note that the rows of $S$ appear as superscripts and the columns appear as subscripts. This convention is important and should be kept in mind.

We remark that orthonormality of the bases is nowhere required or even mentioned. All that is needed here are the concepts of linear dependence/independence, finite dimension, basis, and transformation of bases.

1.5 Interlude: Implicit Summation Notation

Looking at the equations derived in the preceding section, we observe the frequent use of the summation symbol $\sum_{j=1}^n$. When it comes to tensors, the equations will contain more and more summations and will become cumbersome and potentially confusing. Albert Einstein has noted that, when the range of summation is understood from the context, the appearance of the summation symbol is redundant. Einstein thus proposed to eliminate the summation symbol and the summation range, and assume that the reader will mentally add them. This has come to be known as Einstein's notation, or implicit summation notation.

Thus, equations (1.10), (1.11), (1.12) can be written as

$$\mathbf{x} = \mathbf{e}_j x^j, \quad \tilde{\mathbf{e}}_i = \mathbf{e}_j S^j_i, \quad \mathbf{e}_i = \tilde{\mathbf{e}}_j T^j_i, \quad \tilde{x}^i = T^i_j x^j, \quad x^i = S^i_j \tilde{x}^j \tag{1.13}$$

Einstein's notation rules can be summarized as follows:

1. Whenever the same index appears twice in an expression, once as a superscript and once as a subscript, summation over the range of that index is implied.
2. The range of summation is to be understood from the context; in case of doubt, the summation symbol should appear explicitly.
3. It is illegal for a summation index to appear more than twice in an expression, once as a superscript and once as a subscript.
4. The index used in implicit summation is arbitrary and may be replaced by any other index that is not already in use. Such indices are called dummy or bound.
5. Indices other than dummy indices may appear any number of times and are free, in the same sense as common algebraic variables. A free index may be replaced by any other index that is not already in use, provided that replacement is consistent throughout the equation.
6. Attempt to place the symbol carrying the summation index as subscript on the left of the symbol carrying the summation index as superscript. For example, write $a^i_j x^j$ or $x^j a^i_j$, but avoid $x^j a^i_j$ in ambiguous order.

Einstein's notation takes some getting used to, but then it becomes natural and convenient. As an example, let us use it for multiplying two matrices. Let $a^i_j$ and $b^k_m$ stand for the square matrices $A$ and $B$, and recall that superscripts denote row indices and subscripts denote column indices. Then $c^i_k = a^i_j b^j_k$ stands for the product $C = AB$, and $d^i_j = b^i_k a^k_j$ stands for the product $D = BA$.

To conclude this interlude, we introduce the Kronecker delta symbol, defined as

$$\delta^i_j = \begin{cases} 1, & i = j \\ 0, & i \ne j \end{cases} \tag{1.14}$$

The Kronecker delta is useful in many circumstances. In connection with Einstein's notation it may be used as a coordinate selector:

$$\delta^i_j x^j = x^i \tag{1.15}$$

1.6 Covectors

Let $V$ be an $n$-dimensional space and $(\mathbf{e}_1, \mathbf{e}_2, \ldots, \mathbf{e}_n)$ a basis for $V$. As explained in Appendix A, to $V$ there corresponds a dual space $V^*$ and to $(\mathbf{e}_1, \ldots, \mathbf{e}_n)$ there corresponds a dual basis $(\mathbf{f}^1, \mathbf{f}^2, \ldots, \mathbf{f}^n)$. The members of $V^*$ are called dual vectors or covectors. A covector $\mathbf{y}$ can be expressed as a linear combination of the basis members $\mathbf{y} = y_i \mathbf{f}^i$; note the use of implicit summation and that superscripts and subscripts are used in an opposite manner to their use in vectors.

Let $S$ be the change-of-basis transformation from the basis $(\mathbf{e}_1, \ldots, \mathbf{e}_n)$ to the basis $(\tilde{\mathbf{e}}_1, \ldots, \tilde{\mathbf{e}}_n)$. What is the corresponding transformation from the dual basis $(\mathbf{f}^1, \ldots, \mathbf{f}^n)$ to the dual basis $(\tilde{\mathbf{f}}^1, \ldots, \tilde{\mathbf{f}}^n)$? To answer this question, recall the definition of the members of a dual basis as "coordinate selectors" to find

$$\tilde{\mathbf{f}}^i(\tilde{\mathbf{x}}) = \tilde{x}^i = T^i_j x^j = T^i_j \mathbf{f}^j(\mathbf{x}) \tag{1.16}$$

Since this equality holds for all $\mathbf{x} \in V$, necessarily

$$\tilde{\mathbf{f}}^i = T^i_j \mathbf{f}^j \tag{1.17}$$

We conclude that the members of the dual basis are transformed by change of basis using the inverse transformation $T$. It follows that the coordinates of covectors are transformed by change of basis using the direct transformation $S$:

$$\tilde{y}_i = y_j S^j_i \tag{1.18}$$

So, in summary, covectors behave opposite to the behavior of vectors under change of basis. Vector bases are transformed using $S$ and vector coordinates are transformed using $T$. Covector bases are transformed using $T$ and covector coordinates are transformed using $S$. Consequently, covectors are said to be covariant whereas vectors, as we recall, are contravariant.

Let us exemplify covectors by introducing functions of vectors. Consider a function $f(\mathbf{x})$ which assigns to every vector a scalar value $y = f(\mathbf{x})$. The gradient of $f(\mathbf{x})$ consists of the two partial derivatives $(\partial f/\partial x^1, \partial f/\partial x^2)$. Let us examine the behavior of the gradient under a change of basis. Using the chain rule for partial derivatives we obtain

$$\frac{\partial f}{\partial \tilde{x}^i} = \frac{\partial f}{\partial x^j} \frac{\partial x^j}{\partial \tilde{x}^i} \tag{1.20}$$

But from (1.13) we know that $x^j = S^j_i \tilde{x}^i$, so $\partial x^j / \partial \tilde{x}^i = S^j_i$. Introducing the notation

$$\nabla_k f = \frac{\partial f}{\partial x^k}, \qquad \tilde{\nabla}_k f = \frac{\partial f}{\partial \tilde{x}^k} \tag{1.22}$$

we can combine these to find

$$\tilde{\nabla}_i f = (\nabla_j f) S^j_i \tag{1.23}$$

It is now evident that $\nabla_i f$ cannot be regarded as a vector, but as a covector, because it is transformed under a change of basis using the direct transformation, rather than the inverse transformation. The gradient operator $\nabla_i$ is a covariant operator.

1.7 Linear Operators on Vector Spaces

A linear operator on an $n$-dimensional vector space $V$ is a function $f: V \to V$ which is additive and homogeneous:

$$f(\mathbf{x} + \mathbf{y}) = f(\mathbf{x}) + f(\mathbf{y}), \qquad f(a\mathbf{x}) = af(\mathbf{x}) \tag{1.24}$$

A linear operator acts on the coordinates of a vector in a linear way; if $\mathbf{y} = f(\mathbf{x})$, then

$$y^i = f^i_j x^j \tag{1.25}$$

Remark: Although the notation $f^i_j$ resembles $S^j_i$, $T^j_i$ used for the direct and inverse basis transformations, there is a subtle but important difference. The objects $S^j_i$, $T^j_i$ depend on two bases (the old and the new). By contrast, we interpret $f^i_j$ as a basis-independent geometric object, whose numerical representation depends on a single chosen basis.

Let us explore the transformation law for $f^i_j$ when changing from a basis $\mathbf{e}_i$ to a basis $\tilde{\mathbf{e}}_i$. From the contravariance of $x^i$ and $y^i$:

$$\tilde{y}^k = T^k_i y^i = T^k_i f^i_j x^j = T^k_i f^i_j S^j_m \tilde{x}^m \tag{1.26}$$

Hence we conclude that

$$\tilde{f}^k_m = T^k_i f^i_j S^j_m \tag{1.27}$$

Expression (1.27) is the transformation law for linear operators. As we see, the transformation involves both the direct and the inverse transformations. Therefore a linear operator is contravariant in one index and covariant in the second index. The transformation (1.27) can also be expressed in matrix form as $\tilde{F} = TFS = S^{-1}FS$, which is the similarity transformation of linear algebra.

1.8 Tensors

Vectors, covectors, and linear operators are all special cases of tensors. We will not attempt to define tensors in abstract terms, but settle for a coordinate-based definition, as follows.

A tensor of type (or valency) $(r, s)$ over an $n$-dimensional vector space is an object consisting of $n^{r+s}$ coordinates, denoted by the generic symbol $a^{i_1 \ldots i_r}_{j_1 \ldots j_s}$, and obeying the following change-of-basis transformation law:

$$\tilde{a}^{i_1 \ldots i_r}_{j_1 \ldots j_s} = T^{i_1}_{k_1} \cdots T^{i_r}_{k_r} \, a^{k_1 \ldots k_r}_{m_1 \ldots m_s} \, S^{m_1}_{j_1} \cdots S^{m_s}_{j_s} \tag{1.28}$$

Let us spend some time discussing this equation. According to Einstein's summation notation, summation must be performed $r+s$ times, using the indices $k_1, \ldots, k_r, m_1, \ldots, m_s$. The order of summation does not matter. Expanding fully:

$$\tilde{a}^{i_1 \ldots i_r}_{j_1 \ldots j_s} = \sum_{k_1=1}^n \cdots \sum_{k_r=1}^n \sum_{m_1=1}^n \cdots \sum_{m_s=1}^n T^{i_1}_{k_1} \cdots T^{i_r}_{k_r} \, a^{k_1 \ldots k_r}_{m_1 \ldots m_s} \, S^{m_1}_{j_1} \cdots S^{m_s}_{j_s} \tag{1.29}$$

The coordinates $i_1, \ldots, i_r$ are the contravariant coordinates and the coordinates $j_1, \ldots, j_s$ are the covariant coordinates. You should remember that:

• Contravariant coordinates appear as superscripts and are transformed using $T$
• Covariant coordinates appear as subscripts and are transformed using $S$

Like vectors, tensors are abstract objects. What we see in (1.28) is only a law of transformation of the coordinates of the tensor, while the tensor itself is invariant.

Looking back at what we presented in the preceding sections:

• A contravariant vector is a $(1,0)$-tensor
• A covariant vector (covector) is a $(0,1)$-tensor
• A linear operator on a vector space is a $(1,1)$-tensor
• A scalar is a $(0,0)$-tensor

1.9 Operations on Tensors

1.9.1 Addition

Two tensors of the same type can be added term-by-term:

$$c^{i_1 \ldots i_r}_{j_1 \ldots j_s} = a^{i_1 \ldots i_r}_{j_1 \ldots j_s} + b^{i_1 \ldots i_r}_{j_1 \ldots j_s} \tag{1.30}$$

We can write tensor addition symbolically as $\mathbf{c} = \mathbf{a} + \mathbf{b}$. Tensor addition is commutative. Furthermore, the change-of-basis transformation law holds for $\mathbf{c}$, hence $\mathbf{c}$ is indeed a tensor.

Remarks:

1. Tensors of different ranks cannot be added.
2. The tensor $\mathbf{0}$ can be defined as a tensor of any rank whose coordinates are all $0$.

1.9.2 Multiplication by a Scalar

Each of the coordinates of a tensor can be multiplied by a given scalar to yield a new tensor of the same type:

$$c^{i_1 \ldots i_r}_{j_1 \ldots j_s} = x \, a^{i_1 \ldots i_r}_{j_1 \ldots j_s} \tag{1.31}$$

We can write tensor multiplication by a scalar symbolically as $\mathbf{c} = x\mathbf{a}$. It is easy to check that

$$x(\mathbf{a} + \mathbf{b}) = x\mathbf{a} + x\mathbf{b}, \quad (x+y)\mathbf{a} = x\mathbf{a} + y\mathbf{a}, \quad (xy)\mathbf{a} = x(y\mathbf{a}) = y(x\mathbf{a}) \tag{1.32}$$

1.9.3 The Tensor Product

Let $\mathbf{a}$ be an $(r,s)$-tensor and $\mathbf{b}$ a $(p,q)$-tensor. We write the coordinates of the first tensor as $a^{i_1 \ldots i_r}_{j_1 \ldots j_s}$ and those of the second tensor as $b^{i_{r+1} \ldots i_{r+p}}_{j_{s+1} \ldots j_{s+q}}$. Note that all indices are distinct within and across tensors. The tensor product $\mathbf{c} = \mathbf{a} \otimes \mathbf{b}$ is defined as the $(r+p, s+q)$-tensor having the coordinates

$$c^{i_1 \ldots i_r \, i_{r+1} \ldots i_{r+p}}_{j_1 \ldots j_s \, j_{s+1} \ldots j_{s+q}} = a^{i_1 \ldots i_r}_{j_1 \ldots j_s} \, b^{i_{r+1} \ldots i_{r+p}}_{j_{s+1} \ldots j_{s+q}} \tag{1.33}$$

The tensor product is not commutative; that is, $\mathbf{a} \otimes \mathbf{b} \ne \mathbf{b} \otimes \mathbf{a}$.

The tensor product is bilinear:

$$(x\mathbf{a} + y\mathbf{b}) \otimes \mathbf{c} = x(\mathbf{a} \otimes \mathbf{c}) + y(\mathbf{b} \otimes \mathbf{c}), \qquad \mathbf{c} \otimes (x\mathbf{a} + y\mathbf{b}) = x(\mathbf{c} \otimes \mathbf{a}) + y(\mathbf{c} \otimes \mathbf{b}) \tag{1.34}$$

We can extend the tensor product to any finite number of tensors. The tensor product is multilinear:

$$\mathbf{a}^{(1)} \otimes \cdots \otimes [x\mathbf{a}^{(k)} + y\mathbf{b}] \otimes \cdots \otimes \mathbf{a}^{(m)} = x[\mathbf{a}^{(1)} \otimes \cdots \otimes \mathbf{a}^{(k)} \otimes \cdots \otimes \mathbf{a}^{(m)}] + y[\mathbf{a}^{(1)} \otimes \cdots \otimes \mathbf{b} \otimes \cdots \otimes \mathbf{a}^{(m)}] \tag{1.35}$$

1.9.4 Contraction

Let $\mathbf{a}$ be an $(r,s)$-tensor. Choose any contravariant index, say the $i$-th position, and any covariant index, say the $j$-th position, and rename both by the same new symbol $k$. Then

$$b^{g_1 \ldots g_{i-1} g_{i+1} \ldots g_r}_{h_1 \ldots h_{j-1} h_{j+1} \ldots h_s} = a^{g_1 \ldots g_{i-1} \, k \, g_{i+1} \ldots g_r}_{h_1 \ldots h_{j-1} \, k \, h_{j+1} \ldots h_s} \tag{1.36}$$

Note how the index $k$ disappears through the implied summation and the resulting tensor has type $(r-1, s-1)$. The operation (1.36) is called contraction. For a general $(r,s)$-tensor there are $rs$ possible contractions, one for each pair of contravariant and covariant indices.

In view of the simplest case where $a^g_h$ is a $(1,1)$ tensor: $a^g_g$ is the trace of $a$, and this trace is invariant under change of basis:

$$T^m_g \, a^g_h \, S^h_m = a^g_h \delta^h_g = a^g_g \tag{1.37}$$

Often, implied summation can be viewed as outer product followed by contraction. Consider the scalar product of a covector and a vector:

$$c = a_i b^i \tag{1.39}$$

The result (1.39) is a scalar, called the scalar product of $a_j$ and $b^i$. It follows that scalar product is invariant under change of basis.

1.10 Tensor Spaces as Vector Spaces; New Notations

Consider the set $\mathcal{T}(r,s)$ of all $(r,s)$-tensors, including the zero tensor. Equipped with the addition and scalar multiplication operations, this set becomes a vector space of dimension $n^{r+s}$. Let $(\mathbf{e}_1, \ldots, \mathbf{e}_n)$ be a basis for $V$ and $(\mathbf{f}^1, \ldots, \mathbf{f}^n)$ a basis for $V^*$. Then all tensor products

$$\mathbf{e}_{i_1} \otimes \cdots \otimes \mathbf{e}_{i_r} \otimes \mathbf{f}^{j_1} \otimes \cdots \otimes \mathbf{f}^{j_s}$$

constitute a basis for $\mathcal{T}(r,s)$. We can now express an $(r,s)$-tensor in a way similar to vectors:

$$\mathbf{a} = a^{i_1 \ldots i_r}_{j_1 \ldots j_s} \, \mathbf{e}_{i_1} \cdots \mathbf{e}_{i_r} \mathbf{f}^{j_1} \cdots \mathbf{f}^{j_s} \tag{1.40}$$

Expression (1.40) is a formal, unambiguous way to write a tensor (with implied summation over all indices).

Note: In general, the ordering of contravariant and covariant basis vectors in the tensor product matters (since $\otimes$ is not commutative). There are $\binom{r+s}{r}$ different spaces $\mathcal{T}(r,s)$ corresponding to different orderings.

1.11 Inner Products, The Gram Matrix, and Metric Tensors

The material in this section is extremely important, albeit not difficult. Be sure to understand it fully and review several times if necessary.

1.11.1 Inner Products

The defining axioms of vector space include no operations that act on vectors and produce scalars. Here we introduce one such function: the inner product. Let $\mathbf{u}$ and $\mathbf{v}$ be vectors in a vector space $V$ and denote by $\mathbf{u} \cdot \mathbf{v}$ a function acting on $\mathbf{u}$ and $\mathbf{v}$ and producing a scalar $a = \mathbf{u} \cdot \mathbf{v}$, such that the following properties hold:

1. Bilinearity:

   $$(\alpha \mathbf{u}_1 + \beta \mathbf{u}_2) \cdot \mathbf{v} = \alpha(\mathbf{u}_1 \cdot \mathbf{v}) + \beta(\mathbf{u}_2 \cdot \mathbf{v})$$
   $$\mathbf{u} \cdot (\alpha \mathbf{v}_1 + \beta \mathbf{v}_2) = \alpha(\mathbf{u} \cdot \mathbf{v}_1) + \beta(\mathbf{u} \cdot \mathbf{v}_2)$$
   for all $\mathbf{u}, \mathbf{u}_1, \mathbf{u}_2, \mathbf{v}, \mathbf{v}_1, \mathbf{v}_2 \in V$ and $\alpha, \beta \in \mathbb{R}$.

2. Symmetry:

   $$\mathbf{u} \cdot \mathbf{v} = \mathbf{v} \cdot \mathbf{u}$$
   for all $\mathbf{u}, \mathbf{v} \in V$.

3. Nondegeneracy:

   $$\mathbf{u} \cdot \mathbf{x} = 0 \text{ for all } \mathbf{u} \in V \implies \mathbf{x} = 0$$

A vector space equipped with an inner product is called an inner product space. We will only consider symmetric nondegenerate inner products.

When we substitute $\mathbf{u} = \mathbf{v}$ in the inner product, the resulting scalar-valued function $\mathbf{u} \cdot \mathbf{u}$ is called the quadratic form induced by the inner product. A quadratic form satisfying $\mathbf{u} \cdot \mathbf{u} > 0$ for all $\mathbf{u} \ne \mathbf{0}$ is called positive. An inner product space whose associated quadratic form is positive is called Euclidean.

An inner product space admits the concept of orthogonality. Vectors $\mathbf{u}$ and $\mathbf{v}$ are orthogonal if $\mathbf{u} \cdot \mathbf{v} = 0$. The notation $\mathbf{u} \perp \mathbf{v}$ is used to signify orthogonality. If the space is Euclidean, the length (or Euclidean norm) of $\mathbf{u}$ is $\|\mathbf{u}\| = \sqrt{\mathbf{u} \cdot \mathbf{u}}$.

1.11.2 The Gram Matrix

Let us express the inner product in some basis $(\mathbf{e}_1, \ldots, \mathbf{e}_n)$. Let $\mathbf{u} = \mathbf{e}_i u^i$ and $\mathbf{v} = \mathbf{e}_i v^i$. Then, using the bilinearity of the inner product:

$$\mathbf{u} \cdot \mathbf{v} = (\mathbf{e}_i \cdot \mathbf{e}_j) u^i v^j \tag{1.41}$$

The entity $\{\mathbf{e}_i \cdot \mathbf{e}_j,\ 1 \le i, j \le n\}$ is called the Gram matrix $G$ of the basis. By symmetry of the inner product, $G$ is symmetric.

Theorem 1. The Gram matrix $G$ is nonsingular.

Proof. Assume there exists a vector $\mathbf{x}$ such that $(\mathbf{e}_i \cdot \mathbf{e}_j) x^j = \mathbf{e}_i \cdot \mathbf{x} = 0$. Since this holds for all $i$, $\mathbf{x}$ is orthogonal to every member of the basis, hence to every vector in the space. By nondegeneracy, $\mathbf{x} = \mathbf{0}$. Since the only vector in the null space of $G$ is the zero vector, $G$ is nonsingular. $\square$

We now examine the behavior of $G$ under change of basis. Consider a new basis $(\tilde{\mathbf{e}}_1, \ldots, \tilde{\mathbf{e}}_n)$ related to the old basis through the transformation $S$. Then

$$\tilde{\mathbf{e}}_i \cdot \tilde{\mathbf{e}}_j = (\mathbf{e}_k S^k_i) \cdot (\mathbf{e}_m S^m_j) = (\mathbf{e}_k \cdot \mathbf{e}_m) S^k_i S^m_j \tag{1.42}$$

Or, in matrix form,

$$\tilde{G} = S^\top G S \tag{1.43}$$

Matrices $G$ and $\tilde{G}$ related by (1.43) with nonsingular $S$ are called congruent. The celebrated Sylvester law of inertia asserts:

Theorem 2 (Sylvester). Every real symmetric matrix $G$ is congruent to a diagonal matrix $\Lambda$ whose entries have values $+1$, $-1$, or $0$. The matrix $\Lambda$ is unique for all matrices congruent to $G$ (up to ordering of diagonal entries).

If $n$ is the dimension of $G$, then $n = n_+ + n_- + n_0$, according to the numbers of $+1$, $-1$, and $0$ along the diagonal of $\Lambda$. The triplet $(n_+, n_-, n_0)$ is called the signature of $G$. Since $G$ is nonsingular in our case, $n_0 = 0$ and $n = n_+ + n_-$. If the space is Euclidean, $n_- = n_0 = 0$ and $n = n_+$.

1.11.3 The Metric Tensor

Examination of (1.42) reveals that $(\mathbf{e}_k \cdot \mathbf{e}_m)$ is transformed like a $(0,2)$-tensor under change of basis. Defining $g_{ij} = \mathbf{e}_i \cdot \mathbf{e}_j$, we have

$$\mathbf{u} \cdot \mathbf{v} = g_{ij} u^i v^j \tag{1.44}$$

The $(0,2)$-tensor $g_{ij}$ is called the metric tensor of the inner product space. Like all tensors, it is a geometric object, invariant under change-of-basis transformations. By Sylvester's theorem, there exists a basis which makes the metric diagonal and reveals the signature of the space.

Since $G$ is nonsingular, it possesses an inverse $G^{-1}$. The entries of $G^{-1}$ may be viewed as the coordinates of a $(2,0)$-tensor, called the dual metric tensor, and usually denoted $g^{ij}$. It follows immediately that

$$g^{jk} g_{ki} = \delta^j_i \tag{1.45}$$

Theorem 3. For every finite-dimensional inner product space there exists a unique symmetric nonsingular $(0,2)$-tensor $g_{ij}$ such that $\mathbf{u} \cdot \mathbf{v} = g_{ij} u^i v^j$ for any pair of vectors $\mathbf{u}$ and $\mathbf{v}$. Conversely, if $g_{ij}$ is a symmetric nonsingular $(0,2)$-tensor on a finite-dimensional vector space, then an inner product $\mathbf{u} \cdot \mathbf{v}$ is uniquely defined such that $\mathbf{u} \cdot \mathbf{v} = g_{ij} u^i v^j$ for any pair of vectors $\mathbf{u}$ and $\mathbf{v}$.

If the vector space is Euclidean and the basis is orthonormal, then $g_{ij} = \delta_{ij}$ and the inner product is simply $\mathbf{u} \cdot \mathbf{v} = \sum_{i=1}^n u^i v^i$.

1.11.4 Example: The Minkowski Space

The Minkowski space is a 4-dimensional inner product vector space possessing an orthogonal basis $(\mathbf{e}_0, \mathbf{e}_1, \mathbf{e}_2, \mathbf{e}_3)$ and a metric tensor whose coordinates in this orthogonal basis are

$$g_{ij} = \begin{cases} -1, & i = j = 0 \\ 1, & i = j = 1, 2, 3 \\ 0, & i \ne j \end{cases} \tag{1.46}$$

The metric of this space has signature $n_+ = 3, n_- = 1$. Some authors use the negative of (1.46), giving $n_+ = 1, n_- = 3$.

The Minkowski space is clearly non-Euclidean; indeed, this space underlies relativity theory, so it is the space in which our universe exists! In relativity theory, it is common to number the dimensions starting at $0$. The index $0$ is associated with $ct$ (time multiplied by the speed of light), and the remaining indices are associated with the usual space coordinates $x, y, z$.

Let $\mathbf{x}$ be a vector in the Minkowski space. Then

$$\mathbf{x} \cdot \mathbf{x} = -(x^0)^2 + \sum_{i=1}^3 (x^i)^2 \tag{1.47}$$

Clearly, $\mathbf{x} \cdot \mathbf{x}$ is not always nonnegative. The following terminology is in use:

$$\mathbf{x} \cdot \mathbf{x} \begin{cases} < 0 & \text{timelike} \\ = 0 & \text{lightlike} \\ > 0 & \text{spacelike} \end{cases}$$

1.12 Lowering and Raising of Tensors

Let $a^{i_1 \ldots i_r}_{j_1 \ldots j_s}$ be the coordinates of an $(r,s)$-tensor $\mathbf{a}$ in some basis and $g_{ij}$ be the metric tensor in this basis. Form the tensor product $g_{pq} a^{i_1 \ldots i_r}_{j_1 \ldots j_s}$ (type $(r, s+2)$). Now choose one of the contravariant coordinates of $\mathbf{a}$, say $i_k$; replace $i_k$ by $q$ and perform contraction with respect to $q$. The result is a tensor of type $(r-1, s+1)$:

$$b^{i_1 \ldots i_{k-1} i_{k+1} \ldots i_r}_{p \, j_1 \ldots j_s} = g_{pq} \, a^{i_1 \ldots i_{k-1} \, q \, i_{k+1} \ldots i_r}_{j_1 \ldots j_s} \tag{1.48}$$

This operation is called lowering. Lowering decreases the contravariant valency by 1 and increases the covariant valency by 1.

Raising is the dual of lowering. We start with the dual metric tensor $g^{pq}$ and choose an index $j_k$, replacing $j_k$ by $q$ and performing contraction:

$$c^{p \, i_1 \ldots i_r}_{j_1 \ldots j_{k-1} j_{k+1} \ldots j_s} = g^{pq} \, a^{i_1 \ldots i_r}_{j_1 \ldots j_{k-1} \, q \, j_{k+1} \ldots j_s} \tag{1.49}$$

Raising increases the contravariant valency by 1 and decreases the covariant valency by 1.

A common use of lowering and raising is in moving between vectors and covectors. If $v^i$ is a vector in some basis, we define its corresponding covector $v_i$ through the relationships

$$v_i = g_{ik} v^k, \qquad v^i = g^{ik} v_k \tag{1.50}$$

These relationships establish a natural isomorphism between the given vector space $V$ and its dual space of covectors $V^*$.

1.13 The Levi-Civita Symbol and Related Topics

1.13.1 Permutations and Parity

A permutation of the integers $(1, 2, \ldots, n)$ is a rearrangement of this set in a different order, say $(i_1, i_2, \ldots, i_n)$. There are $n!$ different permutations of $n$ integers.

The parity of a permutation is determined by the number of transpositions needed to restore it to the natural order. If the number is even, the parity is even; if the number is odd, the parity is odd.

1.13.2 The Levi-Civita Symbol

The Levi-Civita symbol $\varepsilon_{i_1 i_2 \ldots i_n}$ is a function of $n$ indices, each taking values from $1$ to $n$. It is defined as follows:

$$\varepsilon_{i_1 i_2 \ldots i_n} = \begin{cases} 1, & i_1 i_2 \ldots i_n \text{ is an even permutation of } 12\ldots n \\ -1, & i_1 i_2 \ldots i_n \text{ is an odd permutation of } 12\ldots n \\ 0, & i_1 i_2 \ldots i_n \text{ is not a permutation of } 12\ldots n \end{cases} \tag{1.51}$$

Using the Levi-Civita symbol, the determinant of an $n \times n$ matrix $A$ can be expressed as

$$\det A = \varepsilon_{i_1 i_2 \ldots i_n} A^{i_1}_1 A^{i_2}_2 \cdots A^{i_n}_n \tag{1.52}$$

with implied summation over all indices.

1.13.3 The Volume Tensor/Pseudotensor

Although the Levi-Civita symbol is written in tensor notation, it is not a tensor. If it were a $(0,n)$-tensor, its transformation law would have to be

$$\tilde{\varepsilon}_{j_1 j_2 \ldots j_n} = \varepsilon_{i_1 i_2 \ldots i_n} S^{i_1}_{j_1} S^{i_2}_{j_2} \cdots S^{i_n}_{j_n} \tag{1.53}$$

However, comparing the right side of (1.53) with (1.52), we see that it equals $(\det S)\, \varepsilon_{j_1 j_2 \ldots j_n}$. Hence

$$\tilde{\varepsilon}_{j_1 j_2 \ldots j_n} = (\det S)\, \varepsilon_{j_1 j_2 \ldots j_n} \tag{1.54}$$

Let us try correcting the problem by a small modification. From (1.43), the determinant of the Gram matrix undergoes the following change:

$$\det \tilde{G} = (\det S)^2 (\det G) \tag{1.55}$$

The sign of $\det G$ is invariant under change of basis (since $(\det S)^2 > 0$). We thus define a new object:

$$\omega_{i_1 i_2 \ldots i_n} = \frac{1}{\sqrt{|\det G|}} \varepsilon_{i_1 i_2 \ldots i_n} \tag{1.56}$$

Combining (1.53)–(1.56), we see that

$$\tilde{\omega}_{j_1 j_2 \ldots j_n} = \pm\, \omega_{i_1 i_2 \ldots i_n} S^{i_1}_{j_1} S^{i_2}_{j_2} \cdots S^{i_n}_{j_n} \tag{1.57}$$

where the sign is that of $\det S$. The conclusion is that $\omega_{i_1 \ldots i_n}$ is "almost" a tensor, except for possible sign change. As long as all transformations $S$ in the context have positive determinant, $\omega_{i_1 \ldots i_n}$ is a tensor. In the general case, we refer to it as a pseudotensor. We call $\omega_{i_1 \ldots i_n}$ the volume pseudotensor.

1.14 Symmetry and Antisymmetry

We say that $a_{j_1 j_2 \ldots j_s}$ is symmetric with respect to a pair of indices $p$ and $q$ if

$$a_{j_1 \ldots p \ldots q \ldots j_s} = a_{j_1 \ldots q \ldots p \ldots j_s} \tag{1.58}$$

We say that $a_{j_1 j_2 \ldots j_s}$ is antisymmetric with respect to a pair of indices $p$ and $q$ if

$$a_{j_1 \ldots p \ldots q \ldots j_s} = -a_{j_1 \ldots q \ldots p \ldots j_s} \tag{1.59}$$

A tensor is called completely symmetric if it exhibits symmetry under all possible transpositions; it is called completely antisymmetric if it exhibits antisymmetry under all possible transpositions. The Levi-Civita symbol provides an example of a completely antisymmetric symbol.

For a completely antisymmetric tensor, $a_{j_1 \ldots p \ldots p \ldots j_s} = 0$ — i.e., a completely antisymmetric tensor may have nonzero coordinates only when all indices are different.

The symmetric part of $a_{j_1 \ldots j_s}$ with respect to a pair of adjacent indices $p, q$ is defined by

$$a_{j_1 \ldots (pq) \ldots j_s} = \tfrac{1}{2}(a_{j_1 \ldots p \ldots q \ldots j_s} + a_{j_1 \ldots q \ldots p \ldots j_s}) \tag{1.60}$$

The antisymmetric part is defined by

$$a_{j_1 \ldots [pq] \ldots j_s} = \tfrac{1}{2}(a_{j_1 \ldots p \ldots q \ldots j_s} - a_{j_1 \ldots q \ldots p \ldots j_s}) \tag{1.61}$$

The tensor $a_{j_1 \ldots p \ldots q \ldots j_s}$ is the sum of its symmetric and antisymmetric parts:

$$a_{j_1 \ldots p \ldots q \ldots j_s} = a_{j_1 \ldots (p|...|q) \ldots j_s} + a_{j_1 \ldots [p|...|q] \ldots j_s} \tag{1.62}$$

The complete symmetrization of $a_{j_1 j_2 \ldots j_s}$, denoted $a_{(j_1 j_2 \ldots j_s)}$, is defined as the sum of all $s!$ permutations of indices divided by $s!$. The complete antisymmetrization, denoted $a_{[j_1 j_2 \ldots j_s]}$, is the alternating sum (even permutations added, odd permutations subtracted) divided by $s!$. Antisymmetrizations are of great importance in relativity theory.

1.15 Summary

In this chapter we introduced tensors in the simplest setting—that of common vector spaces. Using this approach enabled us to stay within the realm of linear algebra, with very little need for calculus. Even in this simple framework, there is a need for understanding dual vectors (or covectors) and dual bases. Once this difficulty is overcome, the road is clear for tensors of any rank.

We learned that a general tensor possesses a number of covariant coordinates and a number of contravariant coordinates. The former transform under the direct change-of-basis matrix $S$ and the latter transform under its inverse $T$.

The elementary operations on tensors include common vector operations (addition and multiplication by a scalar) as well as operations that are unique to tensors. Among the latter, the most important is the tensor product. The second is contraction. By combining tensor products and contractions we can form almost any algebraic tensor operation of interest.

Vector spaces may be equipped with an inner product. We considered inner products that are symmetric and nondegenerate, but not necessarily positive. An inner product space permits the definition of the important Gram matrix of a basis. The Gram matrix leads naturally to the metric tensor and to the operations of raising and lowering.

We define vector spaces to be flat, because all geometric objects are fixed, although their coordinates vary depending on the basis. Applications in physics require more complex objects, in particular ones which involve functions and calculus. The remaining chapters deal with tensors in more general settings.

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Chapter 2: Tensor Fields and Tensor Calculus

2.1 Tensor Fields

In this chapter we will consider tensors that vary from point to point in space. We therefore change our viewpoint on the underlying vector space $V$. Rather than an abstract space, we will think of $V$ as a real physical space, which can be the usual 3-dimensional Euclidean Newtonian space or the 4-dimensional Minkowski space.

To $V$ we attach a fixed origin and a reference basis $(\mathbf{e}_1, \mathbf{e}_2, \ldots, \mathbf{e}_n)$. Each point in $V$ has a radius vector $\mathbf{r}$ with respect to the origin. The coordinates of $\mathbf{r}$ can be expressed as linear coordinates $(x^1, x^2, \ldots, x^n)$, but they can also be expressed in other ways — for example, cylindrical and spherical coordinates are special cases of curvilinear coordinates.

To each point in the space $V$ we will assign a tensor $\mathbf{a}(\mathbf{r})$. Thus, $a^{i_1 \ldots i_r}_{j_1 \ldots j_s}(\mathbf{r})$ denotes the coordinates of a space-dependent tensor with respect to the reference basis. Such an object is called a tensor field over $V$.

The simplest way to think of $a^{i_1 \ldots i_r}_{j_1 \ldots j_s}(\mathbf{r})$ is as a collection of $n^{r+s}$ functions of $\mathbf{r}$. However, at every fixed $\mathbf{r}$, the values of the functions must obey the change-of-basis transformation laws defined in Chapter 1.

2.2 The Gradient Operator in Linear Coordinates

We have already met the gradient operator $\nabla_k$ in Section 1.6, applied to a scalar field. We wish to generalize the gradient operator to tensor fields. As long as we restrict ourselves to linear coordinates, this is not difficult. Let $a^{i_1 \ldots i_r}_{j_1 \ldots j_s}(\mathbf{r})$ be a tensor field. Upon expressing the radius vector $\mathbf{r}$ in terms of the reference basis (i.e., $\mathbf{r} = \mathbf{e}_i x^i$), the tensor $a^{i_1 \ldots i_r}_{j_1 \ldots j_s}(\mathbf{r})$ becomes a function of the coordinates $x^i$. We may now differentiate the tensor with respect to a particular coordinate $x^p$:

$$\nabla_p \, a^{i_1 \ldots i_r}_{j_1 \ldots j_s}(\mathbf{r}) = a^{i_1 \ldots i_r}_{j_1 \ldots j_s;\,p}(\mathbf{r}) = \frac{\partial a^{i_1 \cdots i_r}_{j_1 \ldots j_s}(\mathbf{r})}{\partial x^p} \tag{2.1}$$

There are several novelties in equation (2.1). First, each component of the tensor is differentiated separately. Second, the result depends on the choice of basis. Third, the resulting object is a tensor field of type $(r, s+1)$ — this must be proved. Fourth, the new covariant component appears last, separated by a semicolon.

Theorem 4. $\nabla_p$, as defined in (2.1) and (2.2), is a tensor.

Proof. The change-of-basis transformation of $a^{i_1 \ldots i_r}_{j_1 \ldots j_s}(\mathbf{r})$ gives

$$\tilde{a}^{i_1 \ldots i_r}_{j_1 \ldots j_s}(\mathbf{r}) = T^{i_1}_{k_1} \cdots T^{i_r}_{k_r} \, a^{k_1 \ldots k_r}_{m_1 \ldots m_s}(\mathbf{r}) \, S^{m_1}_{j_1} \cdots S^{m_s}_{j_s} \tag{2.3}$$

Differentiating with respect to $\tilde{x}^p$ and noting that $S$ and $T$ are constant:

$$\frac{\partial \tilde{a}^{i_1 \ldots i_r}_{j_1 \ldots j_s}}{\partial \tilde{x}^p} = T^{i_1}_{k_1} \cdots T^{i_r}_{k_r} \, \frac{\partial a^{k_1 \ldots k_r}_{m_1 \ldots m_s}}{\partial \tilde{x}^p} \, S^{m_1}_{j_1} \cdots S^{m_s}_{j_s} \tag{2.4}$$

Using the chain rule: $\partial a / \partial \tilde{x}^p = (\partial a / \partial x^q)(\partial x^q / \partial \tilde{x}^p) = (\partial a / \partial x^q) S^q_p$. Substituting:

$$\tilde{a}^{i_1 \ldots i_r}_{j_1 \ldots j_s;\,p}(\mathbf{r}) = T^{i_1}_{k_1} \cdots T^{i_r}_{k_r} \, a^{k_1 \ldots k_r}_{m_1 \ldots m_s;\,q}(\mathbf{r}) \, S^{m_1}_{j_1} \cdots S^{m_s}_{j_s} \, S^q_p \tag{2.7}$$

This is precisely the change-of-basis formula for an $(r, s+1)$-tensor, as claimed. $\square$

2.3 Curvilinear Coordinates

Let us assume that we are given $n$ functions of the coordinates of the reference basis, to be denoted by $y^i(x^1, \ldots, x^n)$, $1 \le i \le n$. These functions are assumed to be continuous, to possess continuous partial derivatives, and to be invertible. Such functions are called curvilinear coordinates.

Consider the partial derivatives of the radius vector $\mathbf{r}$ with respect to the curvilinear coordinates:

$$\mathbf{E}_i = \frac{\partial \mathbf{r}}{\partial y^i} = \mathbf{e}_j \frac{\partial x^j}{\partial y^i} \tag{2.8}$$

The $n^2$ partial derivatives $\partial x^j / \partial y^i$ form the Jacobian matrix. Let us denote

$$S^j_i = \frac{\partial x^j}{\partial y^i}, \qquad T^m_k = \frac{\partial y^m}{\partial x^k} \tag{2.9}$$

Then

$$T^m_j S^j_i = \delta^m_i, \qquad S^i_m T^m_k = \delta^i_k \tag{2.10}$$

The vectors $(\mathbf{E}_1, \ldots, \mathbf{E}_n)$ are called the tangent vectors of the curvilinear coordinates at the point $\mathbf{r}$. The vector space spanned by the tangent vectors is called the tangent space at the point $\mathbf{r}$.

The differential vector $d\mathbf{r}$ can be conveniently expressed in terms of the local basis as $d\mathbf{r} = \mathbf{E}_i \, dy^i$. Note that $d\mathbf{r}$ is also called the line element.

Equation (2.8) can be written as

$$\mathbf{E}_i = \mathbf{e}_j S^j_i, \qquad \mathbf{e}_k = \mathbf{E}_m T^m_k \tag{2.11}$$

This has the same form as an ordinary change-of-basis transformation (cf. (1.13)). We should keep in mind, however, that (2.11) is local at each $\mathbf{r}$ whereas (1.13) is global on the space.

2.4 The Affine Connections

When each of the tangent vectors $\mathbf{E}_i$ is differentiated with respect to each curvilinear component $y^j$, we obtain $n^2$ new vectors $\partial \mathbf{E}_i / \partial y^j$. Each such vector may be expressed in terms of the local basis $(\mathbf{E}_1, \ldots, \mathbf{E}_n)$:

$$\frac{\partial \mathbf{E}_i}{\partial y^j} = \mathbf{E}_k \Gamma^k_{ij} \tag{2.12}$$

where implicit summation over $k$ is understood. The $n^3$ coefficients $\Gamma^k_{ij}$ in (2.12) are called the affine connections or the Christoffel symbols of the second kind. Although the notation $\Gamma^k_{ij}$ may imply that the affine connections constitute a tensor, this is in fact not the case.

We use a comma to denote partial derivative: $f_{,p} = \partial f / \partial y^p$. So (2.12) can be written as

$$\mathbf{E}_{i,j} = \mathbf{E}_k \Gamma^k_{ij} \tag{2.13}$$

2.4.1 Formulas for the Affine Connections

Several explicit formulas for the affine connections will be useful later.

First formula: By differentiating (2.11) and substituting in (2.12):

$$\Gamma^p_{ij} = T^p_k S^k_{i,j} = T^p_k x^k_{,ij} \tag{2.17}$$

Therefore the affine connections are symmetric in their lower indices: $\Gamma^k_{ij} = \Gamma^k_{ji}$.

Second formula: By differentiating (2.10):

$$\Gamma^p_{ij} = -T^p_{k,j} S^k_i \tag{2.19}$$

Third formula (in terms of the metric tensor $g_{ij} = \mathbf{E}_i \cdot \mathbf{E}_j$):

$$\Gamma^k_{ij} = \tfrac{1}{2} g^{km}(g_{mi,j} + g_{mj,i} - g_{ij,m}) \tag{2.20}$$

Proof. Direct calculation using $g_{mi,j} = (\mathbf{E}_m \cdot \mathbf{E}_i)_{,j}$ and the definition (2.13) yields, after substitution and using the symmetry of the affine connection and the metric tensor:

$$\tfrac{1}{2} g^{km}(g_{pi} \Gamma^p_{mj} + g_{mp} \Gamma^p_{ij} + g_{pj} \Gamma^p_{mi} + g_{mp} \Gamma^p_{ji} - g_{pj} \Gamma^p_{im} - g_{ip} \Gamma^p_{jm}) = g_{km} g^{mp} \Gamma^p_{ij} = \delta^k_p \Gamma^p_{ij} = \Gamma^k_{ij}$$

$\square$

2.4.2 Example: Polar Coordinates

We illustrate the derivation of the affine connections for a two-dimensional vector space with Cartesian coordinates $(x, y)$ and curvilinear polar coordinates $(r, \varphi)$, related by

$$x = r\cos\varphi, \qquad y = r\sin\varphi$$

The Jacobian matrix and its inverse are

$$S = \begin{bmatrix} \cos\varphi & -r\sin\varphi \\ \sin\varphi & r\cos\varphi \end{bmatrix}, \qquad T = \begin{bmatrix} \cos\varphi & \sin\varphi \\ -r^{-1}\sin\varphi & r^{-1}\cos\varphi \end{bmatrix}$$

The resulting affine connections are:

$$\Gamma^r_{rr} = 0, \quad \Gamma^r_{r\varphi} = \Gamma^r_{\varphi r} = 0, \quad \Gamma^r_{\varphi\varphi} = -r$$

$$\Gamma^\varphi_{rr} = 0, \quad \Gamma^\varphi_{r\varphi} = \Gamma^\varphi_{\varphi r} = r^{-1}, \quad \Gamma^\varphi_{\varphi\varphi} = 0$$

2.5 Differentiation of Tensor Fields in Curvilinear Coordinates

We now turn our attention to the problem of differentiating tensor fields in curvilinear coordinates. We use uppercase letters for tensor coordinates in curvilinear coordinates (relative to the tangent basis $\mathbf{E}_i$) and lowercase for those in the reference basis $\mathbf{e}_i$.

We require that

$$A^{i_1 \ldots i_r}_{j_1 \ldots j_s;\,p} = T^{i_1}_{k_1} \cdots T^{i_r}_{k_r} \, a^{k_1 \ldots k_r}_{m_1 \ldots m_s;\,q} \, S^{m_1}_{j_1} \cdots S^{m_s}_{j_s} \, S^q_p \tag{2.24}$$

Deriving an expression for $A^{i_1 \ldots i_r}_{j_1 \ldots j_s;\,p}$ starting with the special case of a $(1,1)$-tensor, and using (2.15) and (2.18) to substitute for derivatives of $S$ and $T$, we arrive at:

$$A^i_{j;\,p} = A^i_{j,\,p} + \Gamma^i_{hp} A^h_j - A^i_h \Gamma^h_{jp} \tag{2.30}$$

Expression (2.30) is called the covariant derivative of the $(1,1)$-tensor $A^i_j$ in curvilinear coordinates. We now have two kinds of derivatives: the conventional derivative (denoted by a comma subscript), and the covariant derivative (denoted by a semicolon subscript), defined only for tensors.

The covariant derivative of a general tensor in curvilinear coordinates is:

$$A^{i_1 \ldots i_r}_{j_1 \ldots j_s;\,p} = A^{i_1 \ldots i_r}_{j_1 \ldots j_s,\,p} + \sum_{u=1}^r \Gamma^{i_u}_{h_u p} A^{i_1 \ldots i_{u-1} h_u i_{u+1} \ldots i_r}_{j_1 \ldots j_s} - \sum_{u=1}^s A^{i_1 \ldots i_r}_{j_1 \ldots j_{u-1} h_u j_{u+1} \ldots j_s} \Gamma^{h_u}_{j_u p} \tag{2.31}$$

The special cases for vectors and covectors:

$$A^i_{;\,p} = A^i_{,\,p} + \Gamma^i_{hp} A^h, \qquad A_{j;\,p} = A_{j,\,p} - A_h \Gamma^h_{jp} \tag{2.32}$$

Theorem 5. The covariant derivative of $g_{ij}$ is identically zero; that is, $g_{ij;\,p} = 0$.

Proof. It follows as a special case of (2.31) that $g_{ij;\,p} = g_{ij,\,p} - g_{kj}\Gamma^k_{ip} - g_{ik}\Gamma^k_{jp}$. Substituting the affine connections expression (2.20) and carrying out all cancellations yields $0$. $\square$

Let $T^{ij}$ be a symmetric $(2,0)$-tensor. The covariant divergence of $T^{ij}$ is defined as

$$\nabla_i T^{ij} = T^{ij}_{;\,i} = T^{ij}_{,\,i} + \Gamma^i_{ki} T^{kj} + \Gamma^j_{ik} T^{ik} \tag{2.34}$$

The covariant divergence is important in physics applications because it is typically associated with conservation laws.

2.6 The Second Covariant Derivative

The second covariant derivative of $A^i$, denoted $A^i_{;\,pq}$, is the first covariant derivative of $A^i_{;\,p}$. This is a $(1,1)$-tensor; therefore its first covariant derivative follows from (2.30):

$$A^i_{;\,pq} = (A^i_{;\,p})_{,q} + \Gamma^i_{hq} A^h_{;\,p} - A^i_{;\,h} \Gamma^h_{pq} \tag{2.35}$$

Expanding:

$$A^i_{;\,pq} = A^i_{,pq} + \Gamma^i_{hp,q} A^h + \Gamma^i_{hp} A^h_{,q} + \Gamma^i_{hq} A^h_{,p} + \Gamma^i_{hq} \Gamma^h_{kp} A^k - A^i_{;\,h} \Gamma^h_{pq} \tag{2.38}$$

2.7 The Riemann Curvature Tensor

In multivariate calculus, second derivatives possess the symmetry $X_{,pq} = X_{,qp}$ if the second derivatives exist and are continuous. Let us explore if this property holds for second covariant derivatives. Computing $A^i_{;\,pq} - A^i_{;\,qp}$ and canceling terms:

$$A^i_{;\,pq} - A^i_{;\,qp} = (\Gamma^i_{hp,q} + \Gamma^i_{kq}\Gamma^k_{hp} - \Gamma^i_{hq,p} - \Gamma^i_{kp}\Gamma^k_{hq}) A^h \tag{2.39}$$

Let us introduce a new symbol:

$$R^i{}_{hqp} = \Gamma^i_{hp,q} + \Gamma^i_{kq}\Gamma^k_{hp} - \Gamma^i_{hq,p} - \Gamma^i_{kp}\Gamma^k_{hq} \tag{2.40}$$

Equation (2.39) then becomes

$$A^i_{;\,pq} - A^i_{;\,qp} = R^i{}_{hqp} A^h \tag{2.41}$$

An immediate consequence is that the second covariant derivative is not symmetric in general.

Theorem 6. $R^i{}_{hqp}$ is a tensor.

Proof. The left side of (2.41) is a difference of tensors and is therefore a tensor. Denoting it temporarily as $\delta A^i_{pq}$, we have $\delta A^i_{pq} = R^i{}_{hqp} A^h$, where both $\delta A^i_{pq}$ and $A^h$ are tensors. Making a change-of-basis transformation and using the arbitrariness of $A^h$, one can show that $R^i{}_{hqp}$ must transform as a tensor. $\square$

The tensor $R^i{}_{hqp}$ is called the Riemann curvature tensor. It plays a central role in differential geometry and the theory of manifolds.

The first thing to note is that $R^i{}_{hqp}$ depends only on the affine connections $\Gamma^k_{ij}$, which in turn depend only on the metric $g_{ij}$. The Riemann tensor depends only on the metric. When the curvilinear coordinates are linear, the metric is constant, the affine connections vanish, and so does the Riemann tensor.

The Riemann tensor is antisymmetric in the indices $q$ and $p$:

$$R^i{}_{hqp} = -R^i{}_{hpq} \tag{2.42}$$

The Riemann tensor satisfies the First Bianchi identity:

$$R^i{}_{hqp} + R^i{}_{qph} + R^i{}_{phq} = 0 \tag{2.43}$$

Note that the lower indices of the three terms in (2.43) are cyclic permutations of one another.

To make further symmetries explicit, we introduce the purely covariant Riemann tensor, obtained by lowering the contravariant index:

$$R_{ihqp} = g_{ij} R^j{}_{hqp} \tag{2.44}$$

The covariant Riemann tensor possesses the symmetries

$$R_{ihqp} = -R_{ihpq} = -R_{hiqp} = R_{qpih} \tag{2.45}$$

2.8 Some Special Tensors

The space-geometry tensors of general relativity are, in a figure of speech, children of the Riemann tensor.

The Ricci tensor is the result of contracting the contravariant and last covariant indices of the Riemann tensor:

$$R_{ij} = R^k{}_{ikj} \tag{2.46}$$

The Ricci tensor is symmetric:

$$R_{ij} = R^k{}_{ikj} = g^{ku} R_{uikj} = g^{ku} R_{jkui} = g^{ku} R_{kjiu} = R^u{}_{jiu} = R_{ji} \tag{2.47}$$

The curvature scalar is the full contraction of the Ricci tensor:

$$R = g^{ij} R_{ij} \tag{2.48}$$

The Einstein tensor is defined by

$$E_{ij} = R_{ij} - \tfrac{1}{2} g_{ij} R \tag{2.49}$$

Einstein's tensor is symmetric, since both $R_{ij}$ and $g_{ij}$ are symmetric. Einstein's tensor can also be expressed in mixed and contravariant forms:

$$E^i_j = R^i_j - \tfrac{1}{2} \delta^i_j R, \qquad E^{ij} = R^{ij} - \tfrac{1}{2} g^{ij} R \tag{2.50}$$

2.9 Summary

In this section we extended tensor theory from constant tensors in constant bases to tensor fields — tensors that vary from point to point in space. The space itself is still a vector space with an inner product; therefore tensors can still be expressed relative to a fixed (reference) basis. However, they may also be expressed relative to bases that vary from point to point. We introduced the concept of curvilinear coordinates.

Calculus demands the ability to differentiate functions. We distinguished between two kinds of derivatives: the usual partial derivative; and the covariant derivative. The latter is a bona-fide tensor and therefore transforms properly under change of basis.

The definition of the covariant derivative relies on the extension of the concept of metric to local bases on curvilinear coordinates, and on the affine connections. The second covariant derivative is not symmetric in general, and an important consequence of this lack of commutativity is the ability to define the curvature tensor — the Riemann tensor, which leads to the Ricci tensor and to the Einstein tensor fundamental to general relativity.

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Chapter 3: Tensors on Manifolds

3.1 Introduction

The space $\mathbb{R}^n$ is one of mathematics' greatest success stories: it is at the same time the star of linear algebra, geometry, and analysis. To mention but one example of why we need something more general, consider the surface of a sphere. It follows from the Pythagorean theorem that the surface of a sphere is the set of all points of constant distance from the origin. But, although the surface of a sphere is intimately related to the Euclidean space in which it is embedded, it does not at all look like a Euclidean space of any dimension.

The surface of a sphere is an example of a smooth topological manifold. We can loosely define such an object as a set patched up from subsets, each of which is "like" a Euclidean space. "Like" in this description means that each patch can be mapped in a one-to-one way to a patch of a Euclidean space such that the map and its inverse are continuous.

This chapter extends tensor calculus from curvilinear coordinates in Euclidean spaces to tensor fields on manifolds.

3.2 Mathematical Background

3.2.1 Sets and Functions

We assume that you know the concept of an abstract set. We remind that $x \in A$ means that $x$ belongs to the set $A$, and $x \notin A$ means that $x$ does not belong to $A$.

If $A$ is a set and $B$ is another set such that for all $x \in B$ it holds that $x \in A$, then $B$ is a subset of $A$, denoted $B \subseteq A$. The notation $A \setminus B$ stands for the set of all $x \in A$ and $x \notin B$.

The union $A \cup B$ is the set of all $x$ such that $x \in A$ or $x \in B$ or both. The intersection $A \cap B$ is the set of all $x$ such that $x \in A$ and $x \in B$.

A function $f: X \to Y$ is called:

• injective (one-to-one) if $f(x_1) \ne f(x_2)$ unless $x_1 = x_2$
• surjective (onto) if for every $y \in Y$ there exists $x \in X$ such that $y = f(x)$
• bijective if it is both injective and surjective

If $f: X \to Y$ is bijective, its inverse $f^{-1}: Y \to X$ is defined such that $f^{-1}(f(x)) = x$ for all $x \in X$.

The image and inverse image satisfy:

$$f[A \cup B] = f[A] \cup f[B] \tag{3.2a}$$

$$f[A \cap B] \subseteq f[A] \cap f[B] \tag{3.2b}$$

$$f^{-1}[A \cup B] = f^{-1}[A] \cup f^{-1}[B] \tag{3.2c}$$

$$f^{-1}[A \cap B] = f^{-1}[A] \cap f^{-1}[B] \tag{3.2d}$$

If $f$ is injective, then (3.2b) changes to equality.

The composition $g \circ f: X \to Z$ of $f: X \to Y$ and $g: Y \to Z$ is defined by $z = g(f(x))$.

3.2.2 The Topological Structure of $\mathbb{R}^n$

The space $\mathbb{R}^n$ consists of all $n$-tuples of real numbers $(x^1, \ldots, x^n)$. A distance function is defined for all pairs of vectors:

$$|\mathbf{x} - \mathbf{y}| = \left(\sum_{i=1}^n (x^i - y^i)^2\right)^{1/2} \tag{3.4}$$

The distance function has three fundamental properties: (D1) it is zero iff $\mathbf{x} = \mathbf{y}$ and positive otherwise; (D2) it is symmetric; (D3) it satisfies the triangle inequality $|\mathbf{x} - \mathbf{y}| \le |\mathbf{x} - \mathbf{z}| + |\mathbf{z} - \mathbf{y}|$.

Let $\mathbf{x}_0$ be a point in $\mathbb{R}^n$ and $d$ a positive number. The set

$$B(\mathbf{x}_0, d) = \{\mathbf{y} : |\mathbf{x}_0 - \mathbf{y}| < d\} \tag{3.6}$$

is called an open ball centered at $\mathbf{x}_0$ and having radius $d$.

A subset $O$ of $\mathbb{R}^n$ is open if it is a union (finite, countable, or uncountable) of open balls. Open sets have three fundamental properties:

• (T1) The empty set $\emptyset$ and $\mathbb{R}^n$ are open sets.
• (T2) The union of any number of open sets is an open set.
• (T3) The intersection of two open sets is an open set.

3.2.3 General Topological Spaces

A topological space is a set $S$ equipped with a collection $\mathcal{T}$ of subsets of $S$, such that axioms T1, T2 and T3 are satisfied. The member sets of $\mathcal{T}$ are the open sets of the topology.

Two simple examples:

• Any set $S$ with $\mathcal{T} = \{\emptyset, S\}$: the indiscrete topology on $S$.
• Any set $S$ with $\mathcal{T}$ containing all subsets of $S$: the discrete topology on $S$.

A set $C$ is closed if its complement $S \setminus C$ is open.

If $x$ is a point and $O$ is an open set containing $x$, then $O$ is an open neighborhood of $x$. A neighborhood of $x$ is any set containing an open neighborhood of $x$.

3.2.4 More on $\mathbb{R}^n$

The usual topology on $\mathbb{R}^n$ has several important properties:

Hausdorff: If $\mathbf{x}_1$ and $\mathbf{x}_2$ are two different points, there exist open neighborhoods $O_1$ and $O_2$ of $\mathbf{x}_1$ and $\mathbf{x}_2$ such that $O_1 \cap O_2 = \emptyset$.

Separability: A subset $A$ of a topological space is dense if every open set contains a point of $A$. A topological space is separable if it contains a countable dense set. $\mathbb{R}^n$ is separable (the rationals are dense in $\mathbb{R}$).

Second Countability: A base for a topology is a collection $\mathcal{B}$ of open sets such that every open set is a union of members of $\mathcal{B}$. $\mathbb{R}^n$ has a countable base (open balls with rational centers and rational radii).

3.2.5 Continuity and Homeomorphisms

A function $f$ on a topological space $X$ to a topological space $Y$ is continuous at a point $x$ if, for any open neighborhood $V$ of $f(x)$, there is an open neighborhood $U$ of $x$ such that $f[U] \subseteq V$.

A function is continuous on $X$ if and only if, for any open set $V$ in $Y$, the inverse image $U = f^{-1}[V]$ is an open set in $X$.

Let $X$ and $Y$ be two topological spaces. If there exists a bijective function $f$ on $X$ onto $Y$ such that both $f$ and $f^{-1}$ are continuous, then the two spaces are homeomorphic and $f$ is called a homeomorphism. You may think of two spaces as homeomorphic if one can be obtained from the other by arbitrary stretching, squeezing, or bending, but no tearing or punching holes.

For functions $f: \mathbb{R}^n \to \mathbb{R}^m$: a function is of class $C^k$ if all its partial derivatives up to order $k$ exist and are continuous. A function is $C^\infty$ if it is of class $C^k$ for all $k$. A $C^\infty$ function is also called smooth.

3.3 Manifolds

3.3.1 Definition of a Manifold

A smooth topological manifold of dimension $n$ is a set $M$ satisfying the following axioms:

• (M1) $M$ is a topological space whose topology is Hausdorff and second countable.
• (M2) There is a fixed collection of open sets $\mathcal{O} = \{O_i, i \in I\}$ on $M$ that covers $M$; i.e., $\bigcup_{i \in I} O_i = M$.
• (M3) For each $O_i \in \mathcal{O}$ there is an injective function $\psi_i: O_i \to \mathbb{R}^n$ such that $\psi_i$ (with range restricted to $\psi_i[O_i]$) is a homeomorphism between $O_i$ and $\psi_i[O_i]$. The pair $(O_i, \psi_i)$ is called a chart and the collection of all charts is called an atlas.
• (M4) Two charts $(O_i, \psi_i)$ and $(O_j, \psi_j)$ are compatible if either $O_i \cap O_j = \emptyset$ or $O_i \cap O_j = U \ne \emptyset$ and the function $\psi_i \circ \psi_j^{-1}: \psi_j[U] \to \psi_i[U]$ is smooth. Every pair of charts in the atlas is compatible.
• (M5) The atlas is maximal: if $(O, \psi)$ is a chart that is compatible with every chart $(O_i, \psi_i)$ in the atlas, then $(O, \psi)$ is in the atlas.

Comments:

1. The Hausdorff and second countability requirements are technical and needed for deeper theoretical aspects.
2. It is the dimension $n$ of $\mathbb{R}^n$ in axiom M3 that makes us call the manifold $n$-dimensional.
3. A chart is also called a coordinate system. Given a chart $(O, \psi)$, the set $O$ is the chart set and the function $\psi$ is the chart function.
4. Axiom M5 makes the definition of a manifold unique by expanding the atlas to include all compatible charts.

Example: The surface of a sphere $S^2$ in $\mathbb{R}^3$ can be covered by two charts: $O_1 = \{(x,y,z) \in S^2 : z \ne 1\}$ and $O_2 = \{(x,y,z) \in S^2 : z \ne -1\}$. For $O_1$, the stereographic projection

$$u = \frac{2x}{1-z}, \qquad v = \frac{2y}{1-z}$$

maps $O_1$ continuously onto $\mathbb{R}^2$. The surface of a sphere is therefore a two-dimensional manifold.

3.3.2 Smooth Functions on Manifolds

We want to define a smooth function $f: M \to \mathbb{R}^m$ on a manifold $M$. Since $M$ is not numerical, we use charts. Let $\mathcal{F}$ be the set of all smooth functions $f: M \to \mathbb{R}$, with the requirement:

(MF1) For every $(O_i, \psi_i)$ in the atlas, the function $f \circ \psi_i^{-1}: \mathbb{R}^n \to \mathbb{R}^m$ is smooth.

Then $f$ is defined to be smooth.

A particular class of smooth functions is given by the coordinate functions of a chart $(O, \psi)$: the functions $\xi^k \circ \psi_i: O_i \to \mathbb{R}$, where $\xi^k(\mathbf{x}) = x^k$ selects the $k$-th coordinate.

3.3.3 Derivatives on Manifolds

We want to define derivatives of functions on manifolds. Let $p \in M$ be a fixed point, and define an operator $d_p: \mathcal{F} \to \mathbb{R}$ to be a derivative operator at $p$ if:

(MD1) Linearity:

$$d_p(\alpha f + \beta g) = \alpha d_p(f) + \beta d_p(g) \text{ for all } f, g \in \mathcal{F} \text{ and all real } \alpha, \beta \tag{3.9}$$

(MD2) Product rule:

$$d_p(fg) = f\, d_p(g) + g\, d_p(f) \text{ for all } f, g \in \mathcal{F} \tag{3.10}$$

Note that $f$, $g$, $d_p(f)$, and $d_p(g)$ in (3.9), (3.10) are all evaluated at $p$, so they are all real numbers.

Example: The derivative of a constant function $f(p) = c \ne 0$ is zero. Using (3.9), $d_p(f^2) = c \cdot d_p(f)$. Using (3.10), $d_p(f^2) = 2c \cdot d_p(f)$. These agree only if $d_p(f) = 0$.

3.3.4 Directional Derivatives Along Cartesian Coordinates

Let $(O_p, \psi_p)$ be a fixed chart such that $p \in O_p$. For $f \in \mathcal{F}$, the function $f \circ \psi_p^{-1}: \psi_p[O] \to \mathbb{R}$ is smooth on an open subset of $\mathbb{R}^n$. The conventional partial derivatives

$$\frac{\partial}{\partial x^k}(f \circ \psi_p^{-1})$$

are well defined. Evaluating at $\psi(p)$ for every $f \in \mathcal{F}$ gives an operator $\partial/\partial x^k|_p: \mathcal{F} \to \mathbb{R}$ that satisfies (3.9) and (3.10). Therefore $\partial/\partial x^k|_p$ is a derivative operator on $M$.

The operators $(\partial/\partial x^k|_p,\ 1 \le k \le n)$ are called the directional derivatives along the coordinates.

3.3.5 Tangent Vectors and Tangent Spaces

Given any two derivative operators $d_{p,1}$ and $d_{p,2}$, we can define their linear sum:

$$(\alpha_1 d_{p,1} + \alpha_2 d_{p,2})(f) = \alpha_1 d_{p,1}(f) + \alpha_2 d_{p,2}(f) \tag{3.11}$$

The collection of all derivative operators at a point $p$ is therefore a vector space. We denote this space by $\mathcal{D}_p$ and call it the tangent space of the manifold at $p$. The elements of $\mathcal{D}_p$ (the derivative operators) are called tangent vectors.

Theorem 7. The tangent space $\mathcal{D}_p$ is $n$-dimensional and the directional derivatives $(\partial/\partial x^k|_p,\ 1 \le k \le n)$ constitute a basis for the space.

This basis is called the coordinate basis for short.

3.4 Tensors and Tensor Fields on Manifolds

3.4.1 Coordinate Transformations

Suppose that a point $p \in M$ belongs to two charts $(O, \psi)$ and $(\tilde{O}, \tilde{\psi})$. The two different charts have different coordinate bases $(\partial/\partial x^1, \ldots, \partial/\partial x^n)$ and $(\partial/\partial \tilde{x}^1, \ldots, \partial/\partial \tilde{x}^n)$ at $p$. These bases are related by the chain rule:

$$\frac{\partial}{\partial \tilde{x}^i} = \frac{\partial}{\partial x^j} \frac{\partial x^j}{\partial \tilde{x}^i}, \qquad \frac{\partial}{\partial x^i} = \frac{\partial}{\partial \tilde{x}^j} \frac{\partial \tilde{x}^j}{\partial x^i} \tag{3.12}$$

If $v$ is a tangent vector at $p$, then $v = v^i \partial/\partial x^i = \tilde{v}^j \partial/\partial \tilde{x}^j$, from which we deduce the change-of-basis transformation rules:

$$v^i = \frac{\partial x^i}{\partial \tilde{x}^j} \tilde{v}^j, \qquad \tilde{v}^i = \frac{\partial \tilde{x}^i}{\partial x^j} v^j \tag{3.14}$$

Comparing with (1.11) and (1.12), we see that with $S^i_j = \partial x^i / \partial \tilde{x}^j$ and $T^i_j = \partial \tilde{x}^i / \partial x^j$, the transformation rules for vectors on manifolds become identical to the transformation rules for conventional vectors. The essential difference is that (1.11) and (1.12) apply globally, whereas (3.14) holds for a given point of the manifold. Consequently, the matrices $S^i_j$ and $T^i_j$ vary from point to point.

3.4.2 Cotangent Spaces

We have denoted by $\mathcal{D}_p$ the tangent space at $p$. We can assign to $\mathcal{D}_p$ a dual space $\mathcal{D}^*_p$, called the cotangent space at $p$. A common notation for the covectors comprising the dual basis is $(dx^1, dx^2, \ldots, dx^n)$. The change-of-basis transformation for dual bases is

$$d\tilde{x}^i = \frac{\partial \tilde{x}^i}{\partial x^j} dx^j, \qquad dx^i = \frac{\partial x^i}{\partial \tilde{x}^j} d\tilde{x}^j \tag{3.16}$$

and the change-of-basis rules for the coordinates of a covector $w$ are

$$w_i = \tilde{w}_j \frac{\partial \tilde{x}^j}{\partial x^i}, \qquad \tilde{w}_i = w_j \frac{\partial x^j}{\partial \tilde{x}^i} \tag{3.17}$$

3.4.3 Vector Fields

A vector field $d$ on a manifold is a collection $\{d_p : p \in M\}$ of derivative operators. For a fixed $f$, $\{d_p(f) : p \in M\}$ defines a function $d_f: M \to \mathbb{R}$. A vector field $d$ is smooth if $d_f$ is smooth on $M$ for every smooth function $f$ on $M$.

3.4.4 Tensors and Tensor Fields

We define an $(r,s)$-tensor field as a collection of $n^{r+s}$ smooth functions on $M$, denoted $a^{i_1 \ldots i_r}_{j_1 \ldots j_s}$, and obeying the transformation law

$$\tilde{a}^{i_1 \ldots i_r}_{j_1 \ldots j_s} = \frac{\partial \tilde{x}^{i_1}}{\partial x^{k_1}} \cdots \frac{\partial \tilde{x}^{i_r}}{\partial x^{k_r}} \, a^{k_1 \ldots k_r}_{m_1 \ldots m_s} \, \frac{\partial x^{m_1}}{\partial \tilde{x}^{j_1}} \cdots \frac{\partial x^{m_s}}{\partial \tilde{x}^{j_s}} \tag{3.20}$$

Compare this definition with (1.28): the two are essentially identical, except for the difference in interpretation. Whereas (1.28) defines constant tensors over a vector space, (3.20) defines a tensor field over a manifold.

We can use ordinary partial differentiation on tensor fields:

$$a^{i_1 \ldots i_r}_{j_1 \ldots j_s,\,p} = \frac{\partial a^{i_1 \ldots i_r}_{j_1 \ldots j_s}}{\partial x^p} \tag{3.21}$$

The result is not a tensor field, since it does not satisfy the transformation law (3.20). To obtain a derivative tensor field, we need the covariant derivative, as in Chapter 2.

3.4.5 The Metric Tensor

A metric tensor on a manifold $M$ is a smooth $(0,2)$-tensor field $g_{ij}$ satisfying:

• (MM1) $g_{ij}$ is symmetric.
• (MM2) $g_{ij}$ is nondegenerate.
• (MM3) The signature $\Lambda$ of $g_{ij}$ is constant on the entire manifold.

These axioms allow us to write $g_{ij} = S^k_i S^m_j \Lambda_{km}$, where $S$ is a nonsingular matrix and $\Lambda$ is diagonal with a constant pattern of $\pm 1$'s.

The inverse (dual) metric tensor $g^{ij}$ is given by $g^{ij} = T^i_k T^j_m \Lambda^{km}$, where $T = S^{-1}$. It satisfies

$$g^{jk} g_{ki} = g_{ik} g^{kj} = \delta^i_j \tag{3.24}$$

The metric tensor can be expressed in full form as

$$ds^2 = g_{ij} dx^i dx^j \tag{3.25}$$

The notation $ds^2$ is called the (square of the) line element.

The transformation law of the metric tensor is the same as any $(0,2)$-tensor:

$$\tilde{g}_{ij} = \frac{\partial x^p}{\partial \tilde{x}^i} \frac{\partial x^q}{\partial \tilde{x}^j} g_{pq} \tag{3.26}$$

3.4.6 Interlude: A Device

A simple device will enable us to save many pages of definitions and derivations. Let us conjure up an $n$-dimensional inner product space $V$, together with a basis $(\mathbf{e}_1, \ldots, \mathbf{e}_n)$ whose Gram matrix is $\mathbf{e}_k \cdot \mathbf{e}_m = \Lambda_{km}$. Now define a local basis at each point of the manifold:

$$\mathbf{E}_i = S^k_i \mathbf{e}_k \tag{3.28}$$

Then

$$\mathbf{E}_i \cdot \mathbf{E}_j = S^k_i S^m_j (\mathbf{e}_k \cdot \mathbf{e}_m) = S^k_i S^m_j \Lambda_{km} = g_{ij} \tag{3.29}$$

We thus find ourselves in exactly the same framework as in Chapter 2. The constancy of $\Lambda$ (axiom MM3) facilitates the use of a fixed inner product space with a fixed signature for the entire manifold. This artificial device ($V$ and the bases) is not part of the definition of a manifold — it is introduced only for technical usefulness.

3.4.7 (Almost) All Work Done

Once the device in the preceding subsection is understood, all the material in Chapter 2 from Section 2.4 onwards applies to tensors on manifolds, with no changes. In particular, affine connections, covariant derivatives, and the special tensors are defined and used as in Chapter 2. The coordinates $x^k$ on some chart $(O, \psi)$ take the place of the curvilinear coordinates $y^k$.

3.5 Curves and Parallel Transport

A smooth curve on a manifold $M$ is a smooth function $\gamma: \mathbb{R} \to M$. Smoothness is defined by requiring that $f \circ \gamma: \mathbb{R} \to \mathbb{R}$ be smooth for all $f \in \mathcal{F}$.

The tangent vector at a point $p(t)$ on the curve is defined as

$$\tau(f) = \frac{d(f \circ \gamma)}{dt} \tag{3.30}$$

The tangent vector can be expressed in terms of the coordinate basis:

$$\tau = \frac{dx^k}{dt} \frac{\partial}{\partial x^k} = \tau^k \frac{\partial}{\partial x^k} \tag{3.31}$$

where $\tau^k = dx^k/dt$ are the coordinates of the tangent vector.

Let $\gamma(t)$ be a curve with tangent vector $\tau^k(t)$, and let $v^i(t)$ be a vector field defined on all points of the curve. Then $v^i(t)$ is parallel transported on the curve if

$$\tau^k(t) \, v^i_{;\,k}(t) = 0 \text{ for all } t \text{ and all } i \tag{3.32}$$

This can be rewritten as a differential equation:

$$\frac{dv^i}{dt} + \Gamma^i_{km} \frac{dx^k}{dt} v^m = 0 \text{ for all } t \text{ and all } i \tag{3.37}$$

Given $v^i$ at a single point on the curve, (3.37) uniquely defines the parallel transported vector $v^i(t)$ on the entire curve. The conclusion from a geometric analysis: a parallel transported vector remains unchanged — parallel to itself — when viewed as an abstract geometrical object in the hypothetical reference space $V$.

3.6 Curvature

Let $\gamma(t)$ be a closed curve on a manifold ($\gamma(0) = \gamma(1)$). Let $v^k(0)$ be a fixed vector at $\gamma(0)$ and perform parallel transport on $\gamma$. Surprisingly, $v^k(1) \ne v^k(0)$ in general. This follows from the curvature of the manifold.

Example: Place an arrow at the North Pole of the earth, pointing south along the Greenwich meridian. Move south to the equator, east to longitude 90°, then north back to the North Pole. The arrow will have rotated 90° relative to its original direction — even though it was "always pointing south."

3.6.1 The Area of a Closed Curve

For a closed planar curve $\gamma(t)$ surrounding the origin, parameterized so that $\gamma(0) = \gamma(1)$, the enclosed area is

$$A = \int_0^1 \frac{dy(t)}{dt} x(t)\, dt = -\int_0^1 \frac{dx(t)}{dt} y(t)\, dt \tag{3.43}$$

3.6.2 Approximate Solution of the Parallel Transport Equations

Choose a point $x^k_0$ and construct a closed curve $x(t) = x_0 + \varepsilon\gamma(t)$, where $\gamma(t)$ is a fixed curve and $\varepsilon$ is a small scalar. Approximating the parallel transport equations (3.37) to order $\varepsilon^2$ and integrating, we find that there is no first-order effect:

$$v^i_1(1) - v^i_1(0) = -[\gamma^k(1) - \gamma^k(0)] \Gamma^i_{km}(x_0) v^m_0 = 0 \tag{3.52}$$

Proceeding to the second-order term and using the area $A^{k\ell}$ of the projection of the curve on the $k$-$\ell$ plane (cf. (3.43)), we arrive at:

$$v^i(1) - v^i(0) \approx \tfrac{1}{2} \varepsilon^2 R^i{}_{k\ell j} A^{k\ell} v^j_0 \tag{3.59}$$

This result establishes an interesting interpretation of the Riemann curvature tensor: when a vector is parallel transported along an infinitesimal closed curve in a manifold, there is a second-order nonzero difference between the vectors at the end and the beginning of the curve. This difference is proportional to the Riemann tensor at the central point of the loop and to the area of the loop.

Equivalently: when a given vector is parallel transported from a point $p$ to a point $q$ on a manifold, the resulting vector is not uniquely determined by $p$ and $q$, but depends on the chosen path between the points.

3.7 Geodesics and Line Length

3.7.1 Geodesics

A geodesic in a manifold is a curve having the property that its tangent vector is parallel transported along the curve. Substituting the tangent vector in place of the transported vector in (3.37) yields the geodesic equation:

$$\frac{d^2 x^i}{dt^2} + \Gamma^i_{km} \frac{dx^k}{dt} \frac{dx^m}{dt} = 0 \text{ for all } t \text{ and all } i \tag{3.60}$$

This is a set of $n$ second-order, nonlinear, coupled differential equations in the unknown functions $x^i(t)$. Note that the affine connections $\Gamma^i_{km}$ are functions of $x^i(t)$ and are not constant in general on a curved space.

Given $x^i(0)$ and $dx^i(0)/dt$ at a point $p$, the geodesic equation has a unique solution.

3.7.2 Length in Euclidean Spaces

In an $n$-dimensional Euclidean space, the length of a curve $\gamma(t)$ with Cartesian coordinates $x^i(t)$, $0 \le t \le t_f$, is

$$S = \int_0^{t_f} \left(\sum_{k=1}^n \left(\frac{dx^k}{dt}\right)^2\right)^{1/2} dt \tag{3.61}$$

It is well known that the curve of shortest length between two points is a straight line. A straight line satisfies the geodesic equation because the affine connections vanish on flat space and $d^2x^i/dt^2 = 0$. So the curve of shortest length satisfies the geodesic equation.

The natural parameterization $x^i(s)$ is defined by making the arc length $s$ the parameter:

$$s(t) = \int_0^t \left(\sum_{k=1}^n \left(\frac{dx^k}{du}\right)^2\right)^{1/2} du \tag{3.62}$$

3.7.3 Length in a Manifold

Suppose the metric is positive ($g_{ij} \ge 0$). We define the length of a curve by:

$$S = \int_0^{t_f} \left[g_{ij}(x^k(t)) \frac{dx^i(t)}{dt} \frac{dx^j(t)}{dt}\right]^{1/2} dt \tag{3.65}$$

The natural parameterization is defined analogously via the partial length

$$s(t) = \int_0^t \left[g_{ij}(x^k(u)) \frac{dx^i(u)}{du} \frac{dx^j(u)}{du}\right]^{1/2} du \tag{3.66}$$

The curve of minimum length satisfies the geodesic equation, provided the curve is parameterized in the natural parameterization. (This condition is necessary but not sufficient.) A proof is given in Appendix C.

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Appendix A: Dual Vector Spaces

Let $V$ be an $n$-dimensional vector space over $\mathbb{R}$. A function $f: V \to \mathbb{R}$ is called a functional. A functional is linear if

$$f(\mathbf{x}_1 + \mathbf{x}_2) = f(\mathbf{x}_1) + f(\mathbf{x}_2), \qquad f(a\mathbf{x}) = af(\mathbf{x}) \tag{A.1}$$

The set of all linear functionals on $V$ is a vector space, called the dual space of $V$ and denoted $V^*$. The elements of $V^*$ are called dual vectors or covectors.

We will denote $\langle \mathbf{y}, \mathbf{x} \rangle = \mathbf{y}(\mathbf{x})$ for a dual vector $\mathbf{y}$ and vector $\mathbf{x}$.

Let $(\mathbf{e}_1, \ldots, \mathbf{e}_n)$ be a basis for $V$. Define the functional $\mathbf{f}^i \in V^*$ by

$$\langle \mathbf{f}^i, \mathbf{x} \rangle = x^i \tag{A.2}$$

Thus $\mathbf{f}^i$ selects the $i$-th coordinate of $\mathbf{x}$ when expressed in the basis.

Theorem 8. The space $V^*$ is $n$-dimensional and the set $(\mathbf{f}^1, \ldots, \mathbf{f}^n)$ is a basis for $V^*$.

Proof. Independence: Let $g = \sum_{i=1}^n a_i \mathbf{f}^i$ be the zero functional. Then $\sum_{i=1}^n a_i x^i = 0$ for all $(x^1, \ldots, x^n)$, which implies $a_i = 0$ for all $i$.

Spanning: Let $g \in V^*$ and define $g_i = \langle g, \mathbf{e}_i \rangle$. For an arbitrary vector $\mathbf{x}$:

$$\langle g, \mathbf{x} \rangle = \left\langle g, \sum_{i=1}^n \mathbf{e}_i x^i \right\rangle = \sum_{i=1}^n \langle g, \mathbf{e}_i \rangle x^i = \sum_{i=1}^n g_i x^i = \sum_{i=1}^n g_i \langle \mathbf{f}^i, \mathbf{x} \rangle \tag{A.3}$$

Since this holds for all $\mathbf{x}$, $g = \sum_{i=1}^n g_i \mathbf{f}^i$. $\square$

The basis $(\mathbf{f}^1, \ldots, \mathbf{f}^n)$ is called the dual basis of $(\mathbf{e}_1, \ldots, \mathbf{e}_n)$. It is worthwhile noting that

$$\langle \mathbf{f}^i, \mathbf{e}_j \rangle = \delta^i_j \tag{A.4}$$

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Appendix B: Derivation of the Symmetries of the Covariant Riemann Tensor

We derive an expression for the covariant Riemann curvature tensor that makes its symmetries transparent. Define the lowered affine connection by

$$\Gamma_{ijk} = \tfrac{1}{2}(g_{ij,k} + g_{ik,j} - g_{jk,i}) \tag{B.1}$$

The following identity is verified by direct substitution of (B.1):

$$g_{ij,k} = \Gamma_{ijk} + \Gamma_{jik} \tag{B.2}$$

To find the covariant Riemann tensor, we compute each of the four terms in (2.40) and lower the contravariant index. After calculation:

$$R_{ihqp} = g_{iu} R^u{}_{hqp} = (\Gamma_{ihp,q} - \Gamma_{ihq,p}) + g^{ut}(\Gamma_{tip}\Gamma_{uhq} - \Gamma_{thp}\Gamma_{uiq}) \tag{B.8}$$

The antisymmetry $R_{hiqp} = -R_{ihqp}$ follows from (B.8). The symmetry

$$R_{ihqp} = R_{qpih} \tag{B.11}$$

can also be read from (B.8) and (B.9).

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Appendix C: Proof that the Curve of Minimum Length Satisfies the Geodesic Equation

Let $S$ be the length of the minimum-length curve $x(t)$ parameterized in natural parameterization. Consider a neighboring curve $x(t) + \varepsilon y(t)$, where $\varepsilon$ is small and $y(t)$ is smooth with $y(0) = y(t_f) = 0$.

Define $F(x(t), \dot{x}(t)) = g_{ij}(x(t)) \dot{x}^i(t) \dot{x}^j(t)$. Under natural parameterization, $F = 1$. The length of the neighboring curve is

$$\int_0^{t_f} [F + \varepsilon\, \Delta F + O(\varepsilon^2)]^{1/2}\, dt$$

After expanding $\Delta F$ using the product rule and the chain rule, and using the formula (2.20) for the affine connections, one obtains:

$$\Delta F = 2\frac{d}{dt}[g_{ij}(x(t))\dot{x}^i(t)y^j(t)] - 2[g_{ik}\ddot{x}^i(t) + g_{k\ell}\Gamma^\ell_{ij}\dot{x}^i(t)\dot{x}^j(t)]y^k(t) \tag{C.13}$$

Approximating the square root to first order and integrating, using $y^j(0) = y^j(t_f) = 0$:

$$\int_0^{t_f} [F + \varepsilon\,\Delta F]^{1/2}\, dt = S - 2\varepsilon \int_0^{t_f} [g_{ik}\ddot{x}^i(t) + g_{k\ell}\Gamma^\ell_{ij}\dot{x}^i(t)\dot{x}^j(t)] y^k(t)\, dt + O(\varepsilon^2) \tag{C.16}$$

For $S$ to be a minimum, the integral must vanish for all smooth $y^k(t)$. Therefore

$$g_{ik}\ddot{x}^i(t) + g_{k\ell}\Gamma^\ell_{ij}\dot{x}^i(t)\dot{x}^j(t) = 0 \tag{C.18}$$

Multiplying by $g^{mk}$ and contracting:

$$\frac{d^2 x^m}{dt^2} + \Gamma^m_{ij} \frac{dx^i}{dt}\frac{dx^j}{dt} = 0 \tag{C.21}$$

This is the geodesic equation. $\square$