Notes on General Relativity

Maurıcio O. Calvao1

Institute of PhysicsFederal University of Rio de Janeiro

August 13, 2008

1Email : orca@if.ufrj.br.

2

Contents

Introduction 1

I Geometrical foundations 3

1 Tensor algebra 51.1 Vector (linear) spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.2 Dual spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101.3 Tensor product spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131.4 Relative tensors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161.5 Additional operations and results . . . . . . . . . . . . . . . . . . . . . . . . . . 20

1.5.1 Contraction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201.5.2 Symmetrization and anti-symmetrization . . . . . . . . . . . . . . . . . . 211.5.3 Quotient rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

2 Manifolds 252.1 Differentiable manifolds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 252.2 The tangent space and tensor fields . . . . . . . . . . . . . . . . . . . . . . . . . 262.3 Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

2.3.1 Derivation in general . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 282.4 Symmetries and Lie derivative . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

i

ii CONTENTS

List of Figures

2.1 Lie derivative. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

iii

iv LIST OF FIGURES

List of Tables

v

vi LIST OF TABLES

Introduction

General relativity, that is, Einstein’s gravitational theory, is in a particularly blossoming epoch.This is due mainly to new experiments and observations, either already carried out, or takingplace right now or being planned.

binary pulsarLIGO, GEO, Gravity Probe BLISA, STEP, GAIA, LAGEOS, etcGlobal positioning systemA relatividade geral, ou seja, a teoria de gravitacao de Einstein, esta passando por uma epoca

particularmente efervescente. Isso deve-se, principalmente, a novos experimentos e observacoes,quer ja realizados, quer em curso, quer em planejamento. Valem citar (i) as observacoes do pul-sar binario PSR1913+16, que conferiram o premio Nobel a Hulse e Taylor, em 1993, com a usualinterpretacao de deteccao, mesmo indireta, de ondas gravitacionais; (ii) a construcao, em curso,de uma serie de detetores de ondas gravitacionais de nova geracao [LIGO (Laser InterferometerGravitational wave Observatory), Schomberg,. . . ] ; (iii) o planejamento de detetores assom-brosamente ambiciosos, como o LISA (Laser Interferometer Space Antenna), um interferometroespacial, com braco de 5 milhoes de quilometros (' 3,3 UA); (iv) planejamento de experimen-tos a serem realizados em satelites para teste da universalidade da queda livre ou princıpio deequivalencia fraco [STEP (Satellite Teste of Equivalence Principle)], e da precessao geodesica ouarrasto do referencial inercial local ou efeito de Sitter ou acoplamento spin-gravitoeletricidade,assim como do efeito Lense-Thirring ou acoplamento spin-gravitomagnetismo [Gravity Probe B,LAGEOS (Laser Geodynamic Sattelite Experiment)].

These notes are based on a graduate level course of general relativity, which I delivered onseveral occasion in the Institue of Physics of the Federal University of Rio de Janeiro, duringthe period 1991-2008. Our intention is to present a reasonably honest and deep exposition,in an intermediate level (= Weinberg; Anderson; Carroll; Hobson; > Rindler; Shutz; foster &Nightingale; Hughston & Tod; d’Inverno; Hartle; < Wald; Straumann; Sachs & Wu; de Felice &Clarke). Honest means we shall try to make clear the validity and certainty of the argumentsand results, at least from our point of view; deep, in the intermediate level means that we willnot take profit of mathematically or physically advanced arguments (algebraic geometry, fiberbundles; quantum gravity) but we sure will use the potential of the so-called “classical tensorcalculus” in all its richness.

The presentation is divided into four parts, besides this Introduction. In Part I, “Geometricalfoundations”, we deal first with tensor algebra, and then with tensor analysis (manifolds andtheir geometries). In Part II, “Physical foundations”, we deal with the very important notions ofinstantaneous observers, observers, and frames of reference, then with some “non-gravitational”physics, particularly, continuum physics (dynamical and thermal aspects) and electromagnetism

1

2 LIST OF TABLES

in arbitrary curved spacetimes and frames of reference. In Part III, “Einstein’s general relativ-ity”, we start with a very critical analysis of the historico-heuristical basis of general relativity,explicitly the general covariance principle, the general relativity principle, the equivalence prin-ciole(s), Mach’s principle and the minimal coupling principle. We present then Einstein’s fieldequations and its possibly most famous solution, the Schwarzschild one, together with the so-called classical tests: gravitational spectral shift, light deflection, perihelion precession, and radarecho delay. In Part IV, the last one, entitled “Further developments”, we study gravitationalcollapse, black holes and the linear perturbations around Minkowski spacetime, which allow asimple study of gravitational waves.

Part I

Geometrical foundations

3

Chapter 1

Tensor algebra

1.1 Vector (linear) spaces

Of course, the precise and plain concept of vector is the one provided by linear algebra; inparticular, it is obvious that, rigorously, a vector: (i) is neither a “quantity characterized by amagnitude, and a direction”, (ii) nor a “numeric (real, complex, rational) triple (or n-tuple)”.In fact, in a plain (“raw”) vector space, there does not exist the notion of magnitude (norm)for a vector, and the numerical triple (n-tuple) merely represents the abstract vector (by meansof its components) in a given basis (taken for granted from the context)1. Granted this, weshould hasten to immediately recognize the very fundamental, and possibly most important,from the heuristic point of view, role of the primitive association of vectors with displacementsin the physical three-dimensional space; by the way, in general, new concepts and theories arisefrom practice, enshrouded in a series of superfluous superstructures, which are only evidencedby future logical development and purging.

In the specific case of displacements, it urges to model the physical space as an affine spacein the mathematical sense[1, 2, 6]. Here, we would just like to remind that only within thisstructure can we assign some sense to the notion that the physical space, at least in the contextof Euclidean geometry, does not have a privileged origin; the most primitive concept is that ofpoint, wherefrom the concept of displacement is built and, finally, as a consequence, the conceptof vector2. Phrasing it another way, in a plain vector space, there is no notion of point at all.

A vector (or linear) space T (over K) is a triple {T, +, ·,K}, where T is a (non-empty) set ofelements called vectors, +, · are two composition laws:

+ : T × T → T (addition)· : K × T → T (multiplication by a scalar)

and K := (K,⊕,¯) is a field, whose elements of the base set, K, are called, in this context,scalars (for instance, the real, or complex numbers). Furthermore, such compositions mustsatisfy, for any u,v,w ∈ T and a, b ∈ K, the following other axioms3:

1In a gross analogy, I, Maurıcio Ortiz Calvao, am “represented”, in Brazil, by a given identity card with acertain general registration number, whereas, in the United States, I have another card with a distinct number;am I, ipso facto, two distinct persons?

2It pays to ponder upon three concepts of vector traditionally introduced in the elementary physics literature:free, sliding and bound; think about the notion of equipolence in Euclidean geometry[10].

3Unfortunately, by a notational abuse, it is usual to denote both operations · and ¯ simply by juxtaposition,

5

6 CHAPTER 1. TENSOR ALGEBRA

1. v + w = w + v (commutativity of addition)

2. u + (v + w) = (u + v) + w (associativity of addition)

3. ∃ 0 |v + 0 = v (existence of a neutral element for addition)

4. ∃ −v ∈ T |v + (−v) = 0, ∀ v (existence of inverse elements for addition)

5. a · (v + w) = a · v + a ·w

6. (a⊕ b) · v = a · v + b · v

7. a · (b · v) = (a¯ b) · v

8. 1·v = v.

Axioms (1) to (4) render (T, +) an Abelian (or commutative) group. Axioms (2) and (7), withthe notational abuse mentioned in footnote 3, allow the elimination of parentheses in certainexpressions; e.g., u+v+w=(u+v)+w and abv = (ab)v.

Exercise 1.1 Prove the following immediate consequences of the axioms:

1. the neutral element, 0, for addition is unique.

2. for all v ∈ T , 0v = 0.

3. the inverse elements for addition are unique.

4. if a ∈ K,v ∈ T , and av = 0, then either a = 0 or v = 0.

Exercise 1.2 Let T := Rd :=

d times︷ ︸︸ ︷R× · · · ×R, where R is the set of real numbers. Define

(u1, . . . , ud) + (v1, . . . , vd) = (u1 + v1, . . . , ud + vd),

a · (v1, . . . , vd) = (av1, . . . , avd),∀a ∈ R.

Prove that (Rd, +, ·) is, then, a vector space (over the real field).

Exercise 1.3 Let R+ be the set of positive real numbers. Define the “sum” of two elementes ofR+ as the product in the usual sense (p + q := pq), and multiplication by a scalar of R as

· : R×R+

(r, p) 7→ r · p := pr.

With such operations, show that (R+, +, ·) is a vector space over R.

and the operations + and ⊕ by the same symbol +. You have been warned!

1.1. VECTOR (LINEAR) SPACES 7

If U and V are two vector spaces over the same field of scalars, then we build a new vectorspace, called the direct sum of U and V and denoted by U+V, in the following manner: the newset of vectors is U × V , the new addition and multiplication by a scalar are defined by (with anevident notational abuse)

(u,v) + (u′,v′) := (u + u′,v + v′)

a · (u,v) := (a · u, a · v).

Exercise 1.4 Show that, in Exercise 1.2 above, (Rd, +, ·) is the direct sum of R with itself.

Following [4], it is simplest to start with an example of matrix equation:

u = Av.

Here v is a column matrix (“vector”), of order, say, N × 1; A is a matrix of order M × N ;and u is thus a column matrix (“vector”), of order M × 1. This matrix equation tells us thateach individual element of u is determined from the individual elements of v via A. To writedown explicitly such an expression, we introduce a notation for the elements (“components”) ofu and v, as well as for the elements of A: let us agree that va stands for the a-th element of v(a = 1, 2, . . . , N), uα the α-th element of u (α = 1, 2, . . . , M), and Aα

a the element of the α-throw and a-th column of A. The matrix equation above is, thus, equivalent to the M equations

uα =N∑

a=1

Aαav

a.

The range convention arises from the observation that it is not necessary to spell out, in eachoccurrence of such a set of equations, that M equations are involved and that the validity of eachone of them is stated. This can be perceived from the presence of the index α in each memberof the equation: for α is a free index, as opposed to a, which is subject to the summation sign.On the other hand, the summation convention follows from the observation that, whenever asummation occurs in such an expression, it is a summation over an index (a in that case) whichoccurs precisely twice in the expression to be summed. Thus, a summation occurs when thereis a repeated index; and when an index is repeated, a summation is almost always implied.Under these circumnstances, the summation sign

∑Na=1 does not play any useful role, since the

summation may be recognized by the repetition of an index; the sign can, therefore, be omitted.Thus the “component” or “element” equation above is written, when the domain and sum-

mation conventions hold, in the simple form

uα = Aαav

a.

The presence of the repeated index a in the right hand side implies a summation over its alloweddomain of values (1, 2, . . . , N) by virute of the summation convention; whereas the presence ofa free index α, in both sides of the equation, implies an equality for each value 1, 2, . . . , M thatit may assume, by virtue of the domain convention.

In general, the range and summation conventions work in the following way. If, in an equationinvolving indexed quantities, there exist free (non-repeated) indices, then the equation holds for

8 CHAPTER 1. TENSOR ALGEBRA

all values in the domains of all free indices, such domains having been declared before: that is therange convention. In an expression involving indexed quantities, wherever an index is repeated,a summation over all possible values of the doamin of that index is implied, the domain againhaving been declared previously: that is the summation convention.

The way these conventions work is relatively straightforward. One or two rules–often em-ployed for an interactive verification of the correction of a calculation–should be mentioned. Thenumber of free indices in the two members of an equation must be the same; and, naturally,each different free index in an expression must be represented by a different letter (symbol). Re-peated indices in an expression may only occur in pairs. The replacement of a letter representingan index by another letter is allowed, provided all occurrences of the letter are altered at thesame time and in the same way, and provided it is understood that the new letter has the samerange of values as the one it replaces. The most convenient pratice to adopt, when indices withdifferent ranges are involved in the same calculation (expression), is to reserve a small section ofa particular alphabet to stand for the indices of a given range. Thereby, in the case discussedabove, we might take a, b, c to range and sum from 1 to N , and α, β, γ to range and sum from1 to M ; then uβ = Aβ

cvc would mean exactly the same as uα = Aα

ava.

Two points should be emphasized regarding the way those conventions are used in thesenotes. First, we arrange things in such a way that the pair of repeated indices implying asummation will occur (almost always) with an index in the upper position and the other inthe lower position. This is already apparent from the way we chose to write down the matrixequation above, when something like uα = Aαava might be expected. This point is related tothe importance of distinguishing between a vector space and its dual (column vectors versus rowvectors), which will be dealt with later on.

The second point worth being mentioned is that an expression such as (xc) is often used torepresent (x1, x2, . . . , xN). Additionally, the value of a function, say f , of N variables, say xc,will be denoted by f(xc). In this situation, the index c is subject neither to the summation northe the range conventions. In this particular context, (xc) should be generally thought of as theset of coordinates of a “point” in a given “space”.

Let T be a vector space. A finite set of vectors, say {v1, . . . ,vr}, is said to be linearlydependent if there are scalars a1, . . . , ar, not all of them zero, such that aivi = 0 (here, it isunderstood, from the range and summation conventions, that we mean

∑ri=1 aivi = 0.). An

infinite set is linearly dependent if some finite subset is linearly dependent. A set of vectors islinearly independent if it is not linearly dependent.

A (finite) sum of the form aivi, where vi ∈ T and ai ∈ K, is called a linear combination ofv1, . . . ,vr.

As a simple consequence, we notice that two vectors are linearly dependent if one of themis a multiple of the other; we cannot state that each one is a multiple of the other, since one ofthem might be 0. If a set S includes 0, then it is linearly dependent whatever the other elements(if any).

Exercise 1.5 Prove these two statements.

The maximum number of linearly independent vectors in a vector space T is called thedimension of T and is denoted by dimT. Naturally, there might not be a finite maximum,in which case we write dimT = ∞; this means that, for every positive n, there is a linearlyindependent set of T having n elements.

1.1. VECTOR (LINEAR) SPACES 9

A basis of T is a linearly independent subset S of T such that every vector is a linearcombination of elements of S.4

Exercise 1.6 Prove that, if S is a basis, then a linear combination that expresses v ∈ T interms of elements of S is unique, except for the order of the summands.

If S is a basis of T, then, for each v ∈ T , the unique scalars that occur as coefficients inthe linear combination of elements of S that expresses v are called the components of v in (orwith respect to) the basis S. We consider that a component of v is assigned to each element ofS; however only a finite number of components is nonvanishing.5

Exercise 1.7 Prove that all bases have the same number of elements, the dimension of T.

Let {e1, . . . , eN} (or, simply, {eα}) be a basis of an N -dimensional vector space T, such thatany v ∈ T may be written as v = vαeα, for convenient scalars vα. The scalars vα are thecomponents of v with respect to the basis {eα}.

We now want to see how the components of a vector transform when a new basis is introduced.Let {eα′} be the new basis of T, and vα′ the components of v with respect to this new basis.Then

v = vα′eα′ . (1.1)

The new basis vectors can, as any vectors, be expressed as a linear combination of the old ones:

eα′ = Xβα′ eβ, (1.2)

and, inversely, the old ones as a linear combination of the new ones:

eγ = Xα′γ eα′ . (1.3)

(Though we are using the same kernel letter X, the N2 numbers Xβα′ are different of the N2

numbers Xα′β , the position of the primes indicating the difference.) Replacing, now, eα′ , from

(1.2), in 1.3, there comeseγ = Xα′

γ Xβα′ eβ. (1.4)

Due to the linear independence of {eβ}, we have then

Xα′γ Xβ

α′ = δβγ , (1.5)

where δβγ is the Kronecker delta, defined by

δβγ :=

{0, se β 6= γ,1, se β = γ.

(1.6)

4We mention, without proof, that a basis always exists. This is obvious if dimT is finite, but, otherwise,demands transfinite induction, or the axiom of choice, or Zorn’s lemma.

5In plain vector spaces, only linear combinations of a finite number of terms are defined, since no meaning canbe attributed to limits or convergence. Vector spaces where a notion of limit is defined and that satisfy certainadditional (pun intended) relations are called topological vector spaces. When such an additional structure derivesfrom a definite positive inner product, the space is called a Hilbert space.

10 CHAPTER 1. TENSOR ALGEBRA

(Notice that we cannot assert that δββ = 1, since β is present both as a superindex and as a

subindex and, according to the summation convention, a sum is implicit; in fact, δββ = N , the

dimension of T.) Analogously, we can show that

Xβα′X

γ′β = δγ′

α′ (= δγα). (1.7)

Exercise 1.8 Derive this.

The replacement of eα′ , from equation (1.2), in (1.1), furnishes

v = vα′Xβα′eβ, (1.8)

and, due to the linear independence of eβ,

vβ = Xβα′ v

α′ . (1.9)

Consequently,Xγ′

α vα = Xγ′α Xα

β′ vβ′ = δγ′

β′vβ′ = vβ′ . (1.10)

Summing up, if the primed and unprimed bases are related by

eα′ = Xβα′eβ, eα = Xβ′

α eβ′ , (1.11)

then the components of a given vector v are related by

vα′ = Xα′β vβ, vα = Xα

β′ vβ′ , (1.12)

and

Xαβ′ X

β′γ = δα

γ , Xα′β Xβ

γ′ = δα′γ′ (1.13)

hold

1.2 Dual spaces

Though it is usually suggested that it may be useful to visualize vectors of a given vector spaceas a set of arrows starting from a given origin, in a certain sense, this image may be somewhattroublesome, since several sets of objects with no similarity to arrows constitute vector spacesunder adequate definitions of addition and multiplication by a scalar. Among such of them, wehave functions in a given set; would you imagine one of them as an arrow?

Let us restrict our discussion to real functions defined on a real vector space of vector setT . Mathematically, such a function f is represented by f : T → R, indicating that it applies avector of T to a real number. We may endow the set of all these functions with the structure ofa vector space, by defining:

1. the sum f + g of two function f and g as

(f + g)(v) = f(v) + g(v), for all v ∈ T ;

1.2. DUAL SPACES 11

2. the product af of the scalar a by the function f as

(af)(v) = a(f(v)), for all v ∈ T ;

3. the function zero 0 as0(v) = 0, for all v ∈ T

(where, in the left, 0 stands for a function, whereas, in the right, it is the real numberzero);

4. the inverse function −f of function f as

(−f)(v) = −(f(v)), for all v ∈ T.

Exercise 1.9 Prove that, under these operations, the set of functions f constitutes a vectorspace. What is its dimension?

The set of all real functions over the vector space T is too big for our purposes: we will thusrestrict ourselves to those functions which are linear; that is, functions satisfying

f(au + bv) = af(u) + bf(v), (1.14)

for all a, b ∈ R and all u,v ∈ T . Real linear functions over a real vector space are generallycalled linear functionals or forms. It is easy to verify that the sum of two linear functionals isalso a linear functional, and that the product by a scalar provides a linear functional as well.These observations guarantee that the set of all linear functionals over a vector space T is alsoa vector space. This space is called the dual space of T and is denoted by T∗.

Exercise 1.10 Prove the last statements.

Since the linear functionals or forms are vectors as well, we will henceforth use boldface forthem. Thus, if v ∈ T and f ∈ T ∗, then f(v) ∈ R, that is, it is a scalar, despited the boldface.

We now have two kinds of vectors, those in T and those in T ∗. To distinguish them, those inT will be called contravariant or primal vectors, whereas those in T ∗ will be called covariant ordual vectors. As an additional distinctive feature, the basis vectors of T∗ will carry superindicesand the components of vectors in T ∗ will carry subindices. Thus, if {eα} is a basis of T∗, theng ∈ T ∗ has a unique expression g = gαe

α, in terms of components. Indeed, the reason for thechoice of the expressions contravariant and covariant will be elucidated presently.

The use of the lowercase letter α in the implicit sum above suggests, according to our rangeconvention, that the range of the summation is from 1 to N , the dimension of T, i.e., that T∗has the same dimension as T, so that these spaces are isomorphic.6 This, indeed, is the case,as we shall prove now, by showing that a given basis {eα} of T induces, in a natural manner, adual basis {eα} in T∗ possessing N elements which satisfy eα(eβ) = δα

β .We start by defining eα as a function which takes each vector v ∈ T into the real number

which is its α-th component vα with respect to {eα}, i.e., eα(v) = vα, for all v ∈ T . This givesus N real functions which clearly satisfy eα(eβ) = δα

β ; it remains to show that they are linearand that they constitute a basis of T∗.

6However, if dimT is infinite, then it can be shown that dimT∗ > dimT, provided the usual meaning is givento the inequality between different orders of (infinite) cardinality.

12 CHAPTER 1. TENSOR ALGEBRA

Exercise 1.11 Check that the functions eα are, indeed, linear.

In order to show that they constitute a basis, we proceed in the following way.For any g ∈ T ∗, we may define N real numbers gα by g(eα) =: gα. Thus, for any v ∈ T ,

g(v) = g(vαeα) = vαg(eα) (pela linearidade de g)

= vαgα = gαeα(v).

Then, for any g ∈ T ∗, we have g = gαeα, showing that {eα} spans T ∗ and the issue of the

linear independence of {eα} is the only one that remains. This can be settled by noting that arelationship xαe

α = 0, where xα ∈ R and 0 is the functional zero, implies

0 = xαeα(eβ) = xαδα

β = xβ, for all β.

Thereby, we see that, given a basis {eα} of T , the components gα of g ∈ T ∗ with respect to thedual basis {eα} are given by gα = g(eα).

A change of basis (1.11) in T induces a change of dual basis. Let us denote the dual of theprimed basis {eα′} by eα′ , so that, by definition, eα′(eβ′) = δα′

β′ , and eα′ = Y α′β eβ, for some Y α′

β .Then,

δα′β′ = eα′(eβ′) = Y α′

γ eγ(Xµβ′eµ)

= Y α′γ Xµ

β′eγ(eµ) (by linearity of eγ)

= Y α′γ Xµ

β′δγµ = Y α′

γ Xγβ′ .

which means Y α′γ = Xα′

γ .

Exercise 1.12 Prove this last statement.

Thereby, through a change of basis of T given by (1.11), the dual bases of T ∗ transform as

eα′ = Xα′β eβ, eα = Xα

β′ eβ′ . (1.15)

It is immediately shown that the components of g ∈ T ∗, with respect to the dual bases, transformas

gα′ = Xβα′gβ, gα = Xβ′

α gβ′ . (1.16)

Exercise 1.13 Prove this.

Then, the same matrix [Xα′β ] and its inverse [Xα

β′ ] are involved, but their roles with respect tothe basis vectors and the components are changed and this finally gives some justification forthe terminology contravariant and covariant.

Given T and one of its bases {eα}, we have just seen how to build its dual space T∗ with dualbasis {eα} satisfying eα(eβ) = δα

β . We may apply this process again to get the dual space T∗∗ of

T∗, with dual basis {fα}, say, satisfying fα(eβ) = δβα, and the vectors h ∈ T ∗∗ may be expressed

in terms of components as h = hαfα. Under a change of basis of T, the components of vectorsin T transform according to vα′ = Xα′

βvβ. This induces a change of dual basis of T∗, under

which the components of the vectors in T ∗ transform according to gα′ = Xβα′gβ. On its turn, this

1.3. TENSOR PRODUCT SPACES 13

induces a change of basis of T∗∗, under which it is readily seen that the components of vectorsof T ∗∗ transform according to hα′ = Xα′

β hβ (because the inverse of the inverse of a matrix isthe original matrix itself). That is, the components of vectors in T ∗∗ transform exactly in thesame manner as the components of vectors in T . This means that, if we establish a one-to-onecorrespondence between vectors of T and of T ∗∗, making vαeα in T correspond to vαfα in T ∗∗,where {fα} is the dual of the dual of {eα}, then this correspondence will be independent of basis.

Exercise 1.14 Convince yourself of that!

A one-to-one basis-independent linear correspondence between two vector spaces is called anatural isomorphism and, of course, naturally isomorphic vector spaces are generally identified,by identifying the corresponding vectors. Consequently, we will identify T∗∗ and T.

1.3 Tensor product spaces

Given a vector space T, we saw how to create a new vector space, viz. its dual T∗, but the processis over here (when we identify T∗∗ with T). However, it is possible to generate a new vectorspace from two vector spaces, by forming what is called a tensor product. As a preliminary tothis, we need to define bilinear functionals in a pair of vector spaces.

Let T and U be two real finite-dimensional vector spaces. The Cartesian product T × U isthe set of all (ordered) pairs of the form (v,w), where v ∈ T and w ∈ U . A bilinear functionalf over T × U is a real function f : T × U → R, which is bilinear, that is, satisfies

f(au + bv,w) = af(u,w) + bf(v,w),

for all a, b ∈ R, u,v ∈ T and w ∈ U,

and

f(v, cw + ex) = cf(v,w) + ef(v,x),

for all c, e ∈ R, v ∈ T and w,x ∈ U.

With definitions of addition, multiplication by a scalar, the zero function and inverses anal-ogous to those given for linear functionals, it is immediate to show that the set of bilinearfunctionals over T × U is a vector space and, henceforth, we will use boldface for the bilinearfunctionals.

Exercise 1.15 Demonstrate the statement above.

We are now in a position to define the tensor product T⊗U of T and U as the vector spaceof all bilinear functionals over T ∗ × U∗. Notice that, in this definition, we use the base sets T ∗

and U∗ of the dual spaces, not the spaces T and U themselves.There naturally arises the question of the dimension of T ⊗ U. It is indeed NM , where N

and M are the dimensions of T and U, respectively; we prove this by showing that, from givenbases of T and U, we can define NM elements of T⊗ U, which constitute a basis for it.

Let {eα}, α = 1, . . . , N and {fa}, a = 1, . . . , M , be bases of T∗ and U∗, dual to the bases {eα}and {fa} of T and U, respectively (Note that we used two different alphabets for the suffixeswhich possess different ranges.). Let us define NM functions eαa : T ∗ × U∗ → R as

eαa(g,h) = gαha, (1.17)

14 CHAPTER 1. TENSOR ALGEBRA

where gα are the components of g ∈ T ∗ relatively to {eα} and ha those of h ∈ U∗ relatively to{fa}. In particular,

eαa(eβ, f b) = δβ

αδba. (1.18)

It is simple to show that eαa are bilinear and thus belong to T⊗U. To show that they constitutea basis we must prove that they span T⊗ U and that they are linearly independent.

Exercise 1.16 Following a reasoning similar to one in Section 1.2, show that {eαa} (i) spansT⊗ U, and (ii) is linearly independent.

With the preceding exercise, we proved that the dimension of T⊗ U is NM , the product ofthe dimensions of T and U, and that, in a natural manner, bases {eα} of T and {fa} of U inducea basis {eαa} of T ⊗ U, the components ταa of any τ∈ T ⊗ U, relatively to this basis, beinggiven, in terms of the dual bases, by ταa =τ (eα, fa).

Let us investigate now how the components ταa and the basis vectors eαa transform whennew bases are introduced in T and U. Let us suppose the bases of T and U transform accordingto

eα′ = Xβα′ eβ, fa′ = Y b

a′ fb. (1.19)

This induces a new basis {eα′a′} of T⊗ U, and, for any (g,h) ∈ T ∗ × U∗,

eα′a′(g,h) = gα′ha′ = Xβα′Y

ba′gβhb

= Xβα′Y

ba′eβb(g,h).

Therefore,

eα′a′ = Xβα′Y

ba′eβb. (1.20)

Analogously, for the components, we obtain

τα′a′ = Xα′β Y a′

b τβb. (1.21)

Exercise 1.17 Prove this.

A vector which is an element of the tensor product of two (or more) spaces (cf. below) iscalled a tensor. The tensor product, as defined above, is a product of spaces. It is possible todefine a tensor which is the tensor product g ⊗ h of individual vectors g ∈ T and h ∈ U, bydemanding that

g ⊗ h = gαhaeαa, (1.22)

where gα and ha are the components of g and h relatively to bases of T and U which induce abasis {eαa} of T ⊗ U. Although this definition is given by means of bases, it is, in fact, basisindependent.

Exercise 1.18 Prove this.

In particular we have

eα ⊗ fa = eαa. (1.23)

1.3. TENSOR PRODUCT SPACES 15

The tensor product g⊗ h belongs to T⊗U, but not all tensors of T⊗U have this form. Thosewhich do have are called decomposable or simple tensors.

Having established the basic idea of the tensor product of two vector spaces, we can nowextend it to three or more spaces. However, given three vector spaces T,U,V, we can form theirtensor product in several ways: (T ⊗ U) ⊗ V or T ⊗ (U ⊗ V). These two spaces clearly havethe same dimension and are, in fact, naturally isomorphic, in the sense that we can establisha bijective basis-independent linear correspondence between their elements, just as we havedone with T and T ∗∗. This is achieved by choosing bases {eα}, {fa} and {gA} in T, U e V,respectively (three ranges, therefore three “alphabets”), and letting ταaAeα⊗(fa⊗gA) correspondto (ταaAeα⊗fa)⊗gA, and showing, then, that such a correspondence is basis-independent. Due tothis natural isomorphism, we identify these spaces, and the notation T⊗U⊗V is not ambiguous.

Exercise 1.19 Prove the existence of the mentioned natural isomorphism.

An alternative way to define T⊗U⊗V is as the space of trilinear functionals over T ∗×U∗×V ∗.This leads to a space which is naturally isomorphic to those of the preceding paragraph, all ofthem being identified.

Exercise 1.20 Convince yourself of this.

There are other natural isomorphisms, for example between T⊗U and U⊗T, or between (T⊗U)∗

and T∗ ⊗ U∗, and, whenever they do exist, the vector spaces are identified.

Exercise 1.21 Convince yourself of this.

From now on, we will restrict ourselves to tensor produt spaces obtained by taking repeatedtensor products of a unique vector space T and its dual T∗. We introduce the following notation:

r times︷ ︸︸ ︷T⊗ T⊗ · · · ⊗ T =: Tr = T r (this last notation, by abuse),

s times︷ ︸︸ ︷T∗ ⊗ T∗ ⊗ · · · ⊗ T∗ =: Ts = Ts (this last notation, by abuse),

T r ⊗ Ts =: T rs .

In particular T = T 1 and T ∗ = T1.An element of T r is a (totally) contravariant tensor of rank r, an element of Ts is a (totally)

covariant tensor of rank s, whereas an element of T rs is a (mixed) tensor of rank (r, s). Notice

that this naming labels contravariant and covariant vectors as tensor of rank (1, 0) and (0, 1),respectively. Scalar can be included in the general scheme by considering them as tensors ofrank (0, 0).

A basis {eα} of T (of dimension N) induces a dual basis {eα} of T ∗ and these, together,induce a basis {eβ1···βs

α1···αr} of T r

s . Each tensor τ ∈ T rs has N r+s unique components unıvocas

relatively to the induced dual basis:

τ = τα1···αrβ1···βse

β1···βsα1···αr

. (1.24)

A change of basis of T induces a change of basis of T rs , under which the components transform

according to:

τα′1···α′rβ′1···β′s = Xα′1

µ1· · ·Xα′r

µrXν1

β′1· · ·Xνs

β′sτµ1···µr

ν1···νs . (1.25)

16 CHAPTER 1. TENSOR ALGEBRA

For instance, for a tensor τ ∈ T 12 ,

τα′β′γ′ = Xα′

ρ Xσβ′X

λγ′τ

ρσλ.

It is usual to define tensors as objects having components which transform accoding to theequations (1.25). Such a way of conceiving tensors is justified noticing that if to each basis ofT are associated N r+s real numbers, which, under a change of basis given by equations (1.11),transform as (1.25), then these numbers are the components of a tensor τ of rank (r, s), inaccordance with the way we defined such an object; simply make

τ = τα1···αrβ1···βse

β1···βsα1···αr

.

1.4 Relative tensors

We will show now that “there are more things in heaven and earth than are dreamt of in [our]philosophy”, that is, there are still more general (and equally useful) geometric objects thanthe tensors we defined above, whose components transform not simply with factors of the basistransformation matrix, but also with a factor depending on the determinant of such a matrix.To that end, let us remind something about determinants

Let there be given an N × N square matrix[Zαβ], where, as usual, we will suppose the

superindex α indicates a row and the subindex β indicates a column, that is,

[Zαβ] =

Z11 Z1

2 · · · Z1N

Z21 Z2

2 · · · Z2N

......

. . ....

ZN1 ZN

2 · · · ZNN

. (1.26)

The determinant of this matrix, det[Z], or simply Z, can be defined through the generalCramer’s rule or, in a more geometrical fashion, through the study of the notion of volume orextension in an N -dimensional space. Here we want you to convince yourself that it may alsobe written as:

det[Z] = εα1α2...αNZα1

1Zα2

2 · · ·ZαNN ,

= εα1α2...αN Z1α1Z

2α2 · · ·ZN

αN, (1.27)

where we introduced the so called Levi-Civita permutation symbols:

εα1α2...αN:= εα1α2...αN :=

1, if (α1, α2, . . . , αN) for even permutation of (1, 2, . . . , N);−1, if (α1, α2, . . . , αN) for odd permutation of (1, 2, . . . , N);

0, otherwise, i.e., if there are repeated indices.(1.28)

Observations:

1. Notice that we follow the convention that ε12...N = ε12...N = 1 and not that ε12...N =−ε12...N = 1, sometimes adopted by certain authors. Watch out!

1.4. RELATIVE TENSORS 17

2. Notice that, with equation (1.27), it is indeed obvious that, under an exchange of two rowsor columns, the determinant changes sign, due to the anti-symmetry of the Levi-Civitasymbols.

3. As a consequence of the former item, or right from equation (1.27) itself, the determinantof a matrix with two proportional rows or columns turns out to be vanishing.

4. To help accept the expression (1.27) for the determinant, remember (or prove now), frombasic vector calculus, the expression for (i) the vector product in terms of Cartesian com-ponents:

A×B = det

x y zAx Ay Az

Bx By Bz

= εijkAiBj ek (implicit sums in i, j, k).

or (ii) the mixed product, also in terms of Cartesian components:

A · (B×C) = det

Ax Ay Az

Bx By Bz

Cx Cy Cz

,

which also connects to the idea, mentioned above, of the determinant as a measure ofvolume (in this case, of a parallelepiped with sides A, B e C).

Taking into account the anti-symmetry of the Levi-Civita symbols, we can recast (1.27) as

εα1α2...αN Z = εβ1β2...βN Zα1β1Z

α2β2 . . . ZαN

βN, (1.29)

or

εα1α2...αNZ = εβ1β2...βN

Zβ1α1Z

β2α2 . . . ZβN

αN, (1.30)

forms which will be useful later on.

Exercise 1.22 Convince yourself of the validity of these formulae.

It would be interesting if we could have a final expression for the determinant, Z, isolated ina member of this mentioned expressioni as a function of the Levi-Civita symbols and the matrixelements. To that end, it is expedient to introduce new objects, which are the generalizedKronecker delta tensors, δα1α2···αr

β1β2···βr, defined by

δα1···αk···αr

β1···βk···βr:= det

δα1β1

· · · δα1βk

· · · δα1βr

.... . .

.... . .

...δαkβ1

· · · δαkβ2

· · · δαkβr

.... . .

.... . .

...δαrβ1

· · · δαrβk

· · · δαrβr

(1.31)

18 CHAPTER 1. TENSOR ALGEBRA

Obviously, the (usual) Kronecker delta is a particular case of that definition, corresponding tothe value r = 1. With r = 2 in (1.31), we see that

δαβµν = δα

µδβν − δα

ν δβµ .

In general, δα1α2···αr

β1β2···βris a sum of r! terms, each one of which is a product of r (usual) Kronecker

deltas. Since, as we have already seen, the (usual) Kronecker delta is a type (1,1) tensor, itfollows immediately that the generalized Kronecker delta is a tensor of rank (r, r). From its verydefinition, it is easy to show that: (i) the generalized Kronecker delta is anti-symmetric in all itsupper indices and all its lower indices; (ii) if r > N , where N is the dimension of the underlying(primal) space, then δα1α2···αr

β1β2···βr≡ 0.

Exercise 1.23 Convince yourself of these statements.

In order to obtain other important properties of the generalized Kronecker delta, following [9],we now return to (1.31) and restrict ourselves to the case r ≤ N . We can expand (1.31) in termsof its elements of the last column, to obtain

δα1α2···αk···αr−1αr

β1β2···βk···βr−1βr= (−1)r+rδαr

βrδ

α1···αr−1

β1···βr−1+ (−1)r+r−1δ

αr−1

βrδ

α1···αr−2αr

β1···βr−2βr−1+ · · ·

· · ·+ (−1)r+kδαkβr

δα1···αk−1αk+1···αr−1αr

β1···βk−1βkβk+1···βr−1+ · · ·

· · ·+ (−1)r+2δα2βr

δα1α3···αr−1αr

β1β2β3···βr−1+ (−1)r+1δα1

βrδ

α2···αr−1αr

β1β2···βr−1. (1.32)

Contract now over the indices αr, βr, to derive

δα1α2···αk···αr−1αr

β1β2···βk···βr−1αr= Nδ

α1···αr−1

β1···βr−1− δ

α1···αr−2αr−1

β1···βr−2βr−1+ · · ·

· · ·+ (−1)r+kδα1···αk−1αk+1···αr−1αk

β1···βk−1βkβk+1···βr−1+ · · ·

· · ·+ (−1)r+2δα1α3···αr−1α2

β1β2β3···βr−1+ (−1)r+1δ

α2···αr−1α1

β1β2···βr−1. (1.33)

It is easy to check that, in the right hand side of this expression, we have, besides the first term,Nδ

α1···αr−1

β1···βr−1, r − 1 others; furthermore these other terms are all equal to −δ

α1···αr−1

β1···βr−1.

Exercise 1.24 Prove these two last statements.

Thus, we getδ

α1···αr−1αr

β1···βr−1αr= (N − r + 1)δ

α1···αr−1

β1···βr−1. (1.34)

We now contract, in this formula, αr−1 and βr−1, applying the formula to itself for the case whenr is reduced to r − 1, obtaining

δα1···αr−2αr−1αr

β1···βr−2αr−1αr= (N − r + 2)(N − r + 1)δ

α1···αr−2

β1···βr−2. (1.35)

Repeating this process r − s times, we get the following general identity

δα1···αsαs+1···αr

β1···βsαs+1···αr=

(N − s)!

(N − r)!δα1···αs

β1···βs. (1.36)

Exercise 1.25 Prove this.

1.4. RELATIVE TENSORS 19

An equivalent form of (1.36) is

δα1···αtαt+1···αr

α1···αtβt+1···βr=

(N − r + t)!

(N − r)!δ

αt+1···αr

βt+1···βr. (1.37)

Exercise 1.26 Prove this.

In particular, when t = r, we have

δα1···αrα1···αr

=N !

(N − r)!. (1.38)

Now we want to establish a fundamental relation or identity among εα1α2···αN , εβ1β2···βNand

δα1α2···αNβ1β2···βN

:

εα1α2···αN εβ1β2···βN= δα1α2···αN

β1β2···βN. (1.39)

To that end, consider the quantity

Aα1α2···αNβ1β2···βN

:= εβ1β2···βNεα1α2···αN − δα1α2···αN

β1β2···βN, (1.40)

which is obviously anti-symmetric in the subindices and the superindices. Therefore, the onlypossible nonvanishing components of Aα1α2···αN

β1β2···βNwill occur when (α1, α2, . . . , αN) and (β1, β2, . . . , βN)

are permutations (without repetition) of (1, 2, . . . , N). However, from (1.28), (1.40) and (1.31),we easily see that

A12···N12···N = 0.

Thus, we have just shown that

Aα1α2···αNβ1β2···βN

≡ 0,

which, due to (1.40), establishes (1.39).

Exercise 1.27 Show, from (1.39), that, generically,

1

j!εα1...αN−jγ1...γjεβ1...βN−jγ1...γj

= δα1...αN−j

β1...βN−j. (1.41)

Thereby, or from (1.39) itself, it comes, in particular,

εα1α2···αN εα1α2···αN= N !

Exercise 1.28 Prove that!

Finally, we can then have the expression we looked for:

Z := det[Zµν ] =

1

N !εα1α2···αN εβ1β2···βN

Zβ1α1Z

β2α2 · · ·ZβN

αN. (1.42)

Exercise 1.29 Prove it!

20 CHAPTER 1. TENSOR ALGEBRA

Let us consider now, as a particular case for the matrix, [Zαβ], dealt with above a matrix for

the change of bases, in a certain vector space:

eα′ = Xβα′eβ. (1.43)

In this situation, the expression for the determinant of [Xβα′ ] or its inverse [Xα′

β ], according to(1.29) or (1.30), shows that, if we postulate, as is all too natural, that, regardless of basis, thevalues of the (homonimous) components of the Levi-Civita symbols are the same (εα′1α′2···α′N =εα1α2···αN and εα′1α′2···α′N = εα1α2···αN

), then we conclude that the transformation laws for thecomponents of these symbols (invariant by postulation or definition) turn out to be:

εα′1α′2···α′N = X−1Xα′1β1

Xα′2β2· · ·Xα′N

βN

andεα′1α′2···α′N = XXβ1

α′1Xβ2

α′2· · ·XβN

α′Nεβ1β2···βN

.

Such laws are equal to those for tensors, except for the presence of a factor which is a powerof the determinant fo the matrix of basis change. This suggests the importance of dealingwith algebraic (geometric) objects whose components transform in a more general way. Hence,deviating a little from our geometrical or basis-independent presentation up to now, we will saythat a set of N r+s numbers Λα1···αr

β1···βs constitute the components of a relative tensor relativo ofrank (r, s) and weight w, if, under a basis change (1.43), these numbers (called the componentsof the relative tensor) transform according to

Λα′1···α′rβ′1···β′s = X−wXα′1

µ1· · ·Xα′r

µrXν1

β′1· · ·Xνs

β′sΛµ1···µr

ν1···νs . (1.44)

Notice the sign of the exponent of the determinant X. We can then notice that:

1. εα1···αN and εα1···αNconstitute the components of relative tensors of weight 1 and −1,

respectively.

2. the tensors we have dealt with so far are relative tensors of weight 0; sometimes they arecalled absolute tensors.

1.5 Additional operations and results

1.5.1 Contraction

So far, we have three basic (closed) operations with tensors (relative or absolute): addition ofrelative tensors of the same rank and weight, multiplication of a tensor by a scalar, and formationof the tensor product. There is a fourth operation, which is most easily explained in terms ofcomponents. This operation is the contraction, which associates N r+s−2 numbers (components)Rα1···αp−1αp+1···αr

β1···βq−1βq+1···βs with N r+s numbers (components) Qα1···αrβ1···βs

, defined by

Rα1···αp−1αp+1···αrβ1···βq−1βq+1···βs := Qα1···αp−1γαp+1···αr

β1···βq−1γβq+1···βs. (1.45)

That is, rendering one subindex equal to a superindex and summing, as the summation conven-tion implies. Of course there are rs ways to acomplish that, each one of which leading to an (inprinciple) different contraction of the original set of numbers.

1.5. ADDITIONAL OPERATIONS AND RESULTS 21

The special meaning this operation has for (absolute or relative) tensors is that, if the originalnumbers are components of a relative tensor of rank (r, s) and weight w, then its contractionsare the components of a relative tensor of rank (r − 1, s− 1) and same weight, w.

Exercise 1.30 Prove this!

1.5.2 Symmetrization and anti-symmetrization

Given a matrix [Mαβ], we can always express it as a sum of two other matrices, [M(αβ)] and[M[αβ]], such that

Mαβ = M(αβ) + M[αβ], (1.46)

where

M(αβ) :=1

2(Mαβ + Mβα) (1.47)

M[αβ] :=1

2(Mαβ −Mβα). (1.48)

The matrix [M(αβ)] is called the symmetric part of [Mαβ] and the process shown in (1.47) iscalled symmetrization of [Mαβ], whereas the matrix [M[αβ]] is called the anti-symmetric (or skew-symmetric) part of [Mαβ] and the process indicated in (1.48) is called the anti-symmetrizationof [Mαβ]; such a terminology is justified by the fact that

M(αβ) = M(βα)

andM[αβ] = −M[βα].

Besides, if, indeed, Mαβ constitute the components of a relative tensor of rank (2, 0) and weightw, so will M(αβ) e M[αβ], in contradistinction to

(Mα

β −Mβα

)/2.

Exercise 1.31 Prove this!

In the more general case, the component M(α1···αr) of the so called (totally) symmetric part ofMα1···αr is obtained by summing all components obtained by all possible r! permutations of theindices (α1, . . . , αr) and dividing the result by r! ; i.e., in the case of three indices, we wouldhave:

M(αβγ) :=1

3!(Mαβγ + Mαγβ + Mβαγ + Mβγα + Mγαβ + Mγβα) .

Something analogous holds for the so called (totally) anti-symmetric part, but here the even per-mutaions of the indices (α1, . . . , αr) are summed, whereas the odd permutations are subtracted,that is:

M[αβγ] :=1

3!(Mαβγ −Mαγβ −Mβαγ + Mβγα + Mγαβ −Mγβα) .

Naturally, all this can be extended to “covariant indices”, always preserving the (relative) tensorcharacter of the resulting objects (the symmetric and anti-symmetric parts). On the other hand,the symmetrization or anti-symmetrization over indices in distinct “levels” do not yield (relative)tensors.

22 CHAPTER 1. TENSOR ALGEBRA

Exercise 1.32 Prove this!

We can relate the anti-symmetrization operation to the previously defined generalized Kro-necker deltas. In fact, as an example, given the tensor Tαβγ, consider the tensor δαβγ

ρσλ Tαβγ. From(1.32) (with r = 3), we have

δαβγρσλ Tαβγ = δαβ

ρσ Tαβλ − δαγρσ Tαλγ + δβγ

ρσ Tλβγ

= 3!T[ρσλ] . (1.49)

Exercise 1.33 Show this.

For any integer r ≤ N , it is indeed easily seen that

δα1···αr

β1···βrTα1···αr = r!T[β1···βr] . (1.50)

Exercise 1.34 Convince yourself of this.

1.5.3 Quotient rules

The quotient rules allow us to establish directly the (relative) tensor character of an object onceit is known that its product with an arbitrary (relative) tensor always yields a (relative) tensor.Let us argue with a concrete example, again in terms of components.

Let there be given, in a certain basis, a set of numbers Y αβγ, which, when multiplied by the

components T γµ of an arbitrary tensor, we know always furnishes a tensor Cαβ

µ, say; that is,

Cαβ

µ = Y αβγT

γµ (1.51)

is a tensor for any tensor T γµ. Then, the quotient rule, in this case, states that Y αβγ will

constitute the components of a tensor as well, of rank (1,2), as suggested by the structure of itsindices.

To show that, we use the characteristic law of transformation of tensor components. Weimagine tha, in a new basis, there still hodls the equation (1.51), as if by definition of the new

1.5. ADDITIONAL OPERATIONS AND RESULTS 23

components of the object Y, whose character we wish to discover. Then,

Cα′β′

µ′ = Y α′β′γ′T

γ′µ′

wwwwwÄ(for C and T are tensors)

Xα′α Xβ

β′Xµ′µ Cα

βµ = Y α′

β′γ′Xγ′γ Xµ′

µ T γµ

wwwwwÄreplacing (1.51)

Xα′α Xβ

β′Xµ′µ Y α

βγTγµ = Y α′

β′γ′Xγ′γ Xµ′

µ T γµ

wwwwwÄsince T is arbitrary

(Y α′

β′γ′Xγ′γ −Xα′

α Xββ′Y

αβγ

)Xµ′

µ = 0wwwwwÄ

multiplying by Xµµ′

Y α′β′γ′X

γ′γ = Xα′

α Xββ′Y

αβγwwwwwÄ

multiplying by Xγσ′

Y α′β′σ′ = Xα′

α Xββ′X

γσ′Y

αβγ,

which is exactly what we wished to demonstrate. The “quotient rule” expression itself is ex-plained by the manner in which Y α

βγ appears in (1.51).

Exercise 1.35 How would you adapt the statement of this rule for the case of relative tensors?

Exercise 1.36 If, for a symmetric but otherwise arbitrary tensor, with components Sαβ, theresult

Cα = Y αβγS

βγ

is always a contravariant vector, what can you deduce about the character of Y αβγ or of any of

its parts?

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[12] B. O’Neill. Semi-Riemannian geometry with applications to relativity. Academic Press,1983.

[13] B. Schutz. Geometrical methods of mathematical physics. Cambridge University Press,1980.

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