Multilinear Algebra and Its Applications

Ana Cannas, Ozlem¨ Imamoglu, Alessandra Iozzi

March 2021

Contents

Introduction 1 Chapter 1. Review of Linear Algebra 7 1.1. Vector Spaces 7 1.2. Bases 9 1.3. The Einstein Convention 14 1.4. Linear Transformations 18 Chapter 2. Multilinear Forms 27 2.1. Linear Forms 27 2.2. Bilinear Forms 35 2.3. Multilinear Forms 41 Chapter 3. Inner Products 45 3.1. Definitions and First Properties 45 3.2.ReciprocalBasis 54 3.3. Relevance of Covariance and Contravariance 63 Chapter 4. Tensors 65 4.1. Towards General Tensors 65 4.2. Tensors of Type (p, q) 69 4.3. Tensor Product 71 Chapter 5. Applications 77 5.1. Inertia Tensor 77 5.2. Stress Tensor (Spannung) 89 5.3. Strain Tensor (Verzerrung) 98 5.4. Elasticity Tensor 102 5.5. Conductivity Tensor 104 Solutions to Exercises 107

Introduction

This text deals with physical or geometric entities, known as tensors, which can be thought of as a generalization of vectors. The quantitative description of tensors, i.e., their description in terms of numbers, changes when we change the frame of reference, a.k.a. the basis in linear algebra. Tensors are central in Engineering and Physics, because they provide the framework for formulating and solving problems in areas such as Me- chanics (inertia tensor, stress tensor, elasticity tensor, etc.), Electrodynamics (electrical conductivity and electrical resistivity tensors, electromagnetic tensor, magnetic suscep- tibility tensor, etc.), or General Relativity (stress–energy tensor, curvature tensor, etc.). Just like the main protagonists in Linear Algebra are vectors and linear maps, the main protagonists in Multilinear Algebra are tensors and multilinear maps. Tensors describe linear relations among objects in space, and are represented – once a basis is chosen – by multidimensional arrays of numbers:

T1 ...... Tn

T11n ...... T1nn 1. A tensor of order 1, Tj. T ...... T 111 1n1 ...... T11 T1n ...... T . T . . n1n nnn . . Tn11 ...... Tnn1

Tn1 ...... Tnn 3. A tensor of order 3, Tijk.

2. A tensor of order 2, Tij.

In the notation, the indices can be upper or lower. For tensors of order at least 2, some indices can be upper and some lower. The numbers in the arrays are called components of the tensor and give the representation of the tensor with respect to a given basis. 1 2 INTRODUCTION

Two natural questions arise: (1) Why do we need tensors? (2) What are the important features of tensors? (1) Scalars are not enough to describe directions, for which we need to resort to vectors. At the same time, vectors might not be enough, in that they lack the ability to “modify” vectors. Example 0.1. We denote by B the magnetic fluid density measured in V s/m2 and by H the magnetizing intensity measured in A/m.1 They are related by the· formula B = µH , where µ is the permeability of the medium in H/m. In free space, µ = µ0 = 4π 7 × 10− H/m is a scalar, so that the flux density and the magnetization are vectors that differ only by their magnitude. Other material however have properties that make these terms differ both in mag- nitude and direction. In such materials the scalar permeability is replaced by the tensor permeability µ and B = µ H . · Being vectors, B and H are tensors of order 1, and µ is a tensor of order 2. We will see that they are of different type, and in fact the order of H “cancels out” with the order of µ to give a tensor of order 1. 

(2) Physical laws do not change with different coordinate systems, hence tensors describing them must satisfy some invariance properties. While tensors remain intrinsi- cally invariant with respect to changes of bases, their components will vary according to two fundamental modes: covariance and contravariance, depending on whether the components change in a way parallel to the change of basis or in an opposite way. Here is an example of a familiar tensor from Linear Algebra, illustrating the effect of the change of basis. Example 0.2. We recall here the transformation property that vectors enjoy according to which they are an example of a contravariant tensor of first order. We use here freely notions and properties that will be recalled in the next chapter. Let = b1, b2, b3 and = ˜b1, ˜b2, ˜b3 be two basis of a vector space V . A vector v V canB be{ written} as B { } ∈ e 1 2 3 v = v b1 + v b2 + v b3 , or 1 2 3 v =v ˜ ˜b1 +˜v ˜b2 +˜v ˜b3 , where v1, v2, v3 (resp. v˜1, v˜2, v˜3) are the coordinate of v with respect to the basis B (resp. ). B 1The physical units here are: Volt V, second s, meter m, Ampere A, Henry H. e INTRODUCTION 3

Warning: Please keep the lower indices as lower indices and the upper ones as upper ones. You will see later that there is a reason for it!

We use the following notation:

v1 v˜1 2 2 (0.1) [v] = v and [v] e = v˜ , B v3 B v˜3     and we are interested in finding the relation between the coordinates of v in the two bases. ˜ The vectors bj, j =1, 2, 3, in the basis can be written as a linear combination of vectors in as follows: B B e ˜ 1 2 3 bj = Lj b1 + Lj b2 + Lj b3 , for some Li R. We consider the matrix of the change of basis from to , j ∈ B B 1 1 1 L1 L2 L3 e 2 2 2 L := L e = L1 L2 L3 BB  3 3 3 L1 L2 L3   whose jth-column consists of the coordinates of the vectors ˜bj with respect to the basis . The equalities B ˜ 1 2 3 b1 = L1b1 + L1b2 + L1b3 ˜ 1 2 3 b2 = L2b1 + L2b2 + L2b3  ˜ 1 2 3 b3 = L3b1 + L3b2 + L3b3  can simply be written as 

(0.2) ˜b1 ˜b2 ˜b3 = b1 b2 b3 L. (Check this symbolic equation using
the rules of matrix
multiplication.) Analogously, writing basis vectors in a row and vector coordinates in a column, we can write

v1 1 2 3 2 (0.3) v = v b1 + v b2 + v b3 = b1 b2 b3 v v3
  and v˜1 v˜1 1˜ 2˜ 3˜ 2 2 (0.4) v =v ˜ b1 +˜v b2 +˜v b3 = ˜b1 ˜b2 ˜b3 v˜ = b1 b2 b3 L v˜ , v˜3 v˜3

    4 INTRODUCTION where we used (0.2) in the last equality. Comparing the expression of v in (0.3) and in (0.4), we conclude that v˜1 v1 L v˜2 = v2 v˜3 v3     or equivalently v˜1 v1 2 1 2 v˜ = L− v v˜3 v3     We say that the components of a vector v are contravariant because they change by 1 L− when the basis changes by L; see Section 1.3.2. A vector v is hence a contravariant 1-tensor or tensor of order (1, 0).  Example 0.3 (A numerical example). Let 1 0 0 (0.5) = e , e , e = 0 , 1 , 0 E { 1 2 3}        0 0 1        be the standard basis or R3 and let   1 4 7 = ˜b , ˜b , ˜b = 2 , 5 , 8 B { 1 2 3}        3 6 0  e       1  be another basis of R3. The vector2 v = 1 has coordinates 1

  1 1 3 −1 [v] = 1 and [v] e = 3 . B 1 E  0      2 For a general basis , the notation [ ]B indicates the “operation” of taking the vector v and looking at its coordinatesB in the basis . However,· in order to “write down explicitly” a vector (that is three real numbers that we write in column),B one needs to give coordinates and the coordinates are usually given with respect to the standard basis. In this case there is the slightly confusing fact that v1 v1 . .  .  = v = [v]E =  .  . vn vn         INTRODUCTION 5

Since it is easy to check that ˜b =1 e +2 e +3 e 1 · 1 · 2 · 3 ˜b2 =4 e1 +5 e2 +6 e3 ,  · · · ˜b =7 e +8 e +0 e 3 · 1 · 2 · 3 the matrix of the change of coordinates from to is  E B 1 4 7 L = 2 5 8 . e 3 6 0 It is easy to check that   1 3 1 −1 1 3 = L− 1  0  1 or equivalently     1 3 1 −1 L 3 = 1 .  0  1     

CHAPTER 1

Review of Linear Algebra

1.1. Vector Spaces A vector space (or linear space) is a set of objects where addition and scaling are defined in a way that satisfies natural requirements for such operations, such as the properties listed in the definition below.

1.1.1. Vectors and Scalars.

In this text, we will only consider real vector spaces, a.k.a. vector spaces over R, where the scaling is by real numbers. Definition 1.1. A vector space V over R is a set V equipped with two operations: (1) Vector addition: V V V , (v,w) v + w, and (2) Multiplication by a× scalar:→R V V7→, (α, v) αv , × → 7→ satisfying the following properties: (1) (associativity) (u + v)+ w = u +(v + w) for every u,v,w V ; (2) (commutativity) u + v = v + u for every u, v V ; ∈ (3) (existence of the zero vector) There exists 0 ∈V such that v +0= v for every v V ; ∈ (4) (existence∈ of additive inverse) For every v V , there exists w V such that ∈ v ∈ v + wv =0. The vector wv is denoted by v. (5) α(βv)=(αβ)v for every α, β R and every− v R; (6) 1v = v for every v V ; ∈ ∈ (7) α(u + w)= αu + αv∈ for all α R and u, v V ; (8) (α + β)v = αv + βv for all α, β∈ R and v ∈V . ∈ ∈ An element of the vector space is called a vector and, mostly in the context of vector spaces, a real number is called a scalar. Example 1.2 (Prototypical example). The Euclidean space Rn, n = 1, 2, 3,... , is a vector space with componentwise addition and multiplication by scalars. Vectors in Rn are denoted by x1 . v =  .  , x  n 7  8 1. REVIEW OF LINEAR ALGEBRA with x1,...,xn R. Addition component-by-component translates geometrically to the parallelogram law∈ for vector addition, well-known in R2 and R3.  Examples 1.3 (Other examples). The operations of vector addition and scalar multi- plication are usually inferred from the context. (1) The set of real polynomials of degree n is a vector space, denoted by ≤ n n 1 V = R[x]n := anx + an 1x − + + a1x + a0 : aj R { − ··· ∈ } with the usual (degreewise) sum of polynomials and scalar multiplication. (2) The set of real matrices of size m n, × a11 ... a1m . . V = Mm n(R) := . . : aij R ×  . .  ∈   an1 ... anm      with componentwise addition and scalar multiplication.  (3) The space f : W R of all real-valued functions on a vector space W . (4) The space of{ solutions→ of} a homogeneous linear (ordinary or partial) differential equation. 

Exercise 1.4. Are the following vector spaces? 1 (1) The set V of all vectors in R3 perpendicular to the vector 2 . 3 (2) The set of invertible 2 2 matrices, that is   × a b V := : ad bc ==0 . c d − 6    (3) The set of polynomials of degree exactly n, that is n n 1 V := a0x + a1x − + + an 1x + an : aj R, an =0 . { ··· − ∈ 6 } (4) The set V of 2 4 matrices with last column zero, that is × a b c 0 V := : a, b, c, d, e, f R d e f 0 ∈    (5) The set of solutions f : R R of the equation f ′ =5, that is → V := f : R R : f(x)=5x + C, C R . { → ∈ } Definition 1.5. A function T : V W between real vector spaces V and W is a linear transformation if it satisfies the→ property T (αv + βw)= αT (v)+ βT (w) , for all α, β R and all v,w V . ∈ ∈ 1.2. BASES 9

Exercise 1.6. Show that the set of all linear transformations T : R2 R3 forms a vector space. → 1.1.2. Subspaces. Definition 1.7. A subset W of a vector space V that is itself a vector space is a subspace. By reviewing the properties in the definition of vector space, we see that a subset W V is a subspace exactly when the following conditions are verified: ⊆ (1) The 0 element is in W ; (2) W is closed under addition, that is v + w W for every v,w W ; (3) W is closed under multiplication by scalars∈, that is αv W for∈ every α R and every v W . ∈ ∈ ∈ Condition (1) in fact follows from (2) and (3) under the assumption that W = . Yet it is often an easy way to check that a subset is not a subspace. 6 ∅ Recall that a linear combination of vectors v ,...,v V is a vector of the form 1 n ∈ α1v1 + + αnvn for α1,...,αn R. With this notion, the above three conditions for a subspace··· are equivalent to the following∈ ones: (1)’ W is nonempty; (2)’ W is closed under linear combinations, that is αv + βw W for all α, β R and all v,w W . ∈ ∈ ∈ Definition 1.8. If T : V W is a linear transformation between real vector spaces V and W , then: → the kernel (or null space) of T is the set ker T := v V : T (v)=0 ; • the image (or range) of T is the set im T := T (v{): ∈v V . } • { ∈ } Exercise 1.9. Show that, for a linear transformation T : V W , the kernel ker T is a subspace of V and the image im T is a subspace of W . → 1.2. Bases The key to study and to compute in vector spaces is the concept of basis, which in turn relies on the fundamental notions of linear independence/dependence and of span. 1.2.1. Definition of Basis. Definition 1.10. The vectors b ,...,b V are linearly independent if α b + + 1 n ∈ 1 1 ··· αnbn =0 implies that α1 = = αn =0. In other words, if the only linear combination of these vectors that yields the··· zero vector is the trivial one. We then also say that the vector set b ,...,b is linearly independent. { 1 n} Example 1.11. The vectors 1 0 0 0 , 1 , 0 0 0 1       10 1. REVIEW OF LINEAR ALGEBRA are linearly independent in R3. In fact,

1 0 0 µ1 0 µ 0 + µ 1 + µ 0 =0 µ = 0 µ = µ = µ =0 . 1   2   3   ⇐⇒  2   ⇐⇒ 1 2 3 0 0 1 µ3 0            Example 1.12. The vectors 1 4 7 b1 = 2 , b2 = 5 , b3 = 8 3 6 0 are linearly independent inR3. In fact,    

µ1 +4µ2 +7µ3 =0

µ b + µ b + µ b =0 2µ1 +5µ2 +8µ30 µ = µ = µ =0 . 1 1 2 2 3 3 ⇐⇒  ⇐⇒··· ⇐⇒ 1 2 3 3µ1 +6µ2 =0 (If you are unsure how to fill in the dots look at Example 1.21.)   Example 1.13. The vectors 1 4 7 b1 = 2 , b2 = 5 , b3 = 8 3 6 9 are linearly dependent in R3, i.e., not linearly  independent. In fact,

µ1 +4µ2 +7µ3 =0 µ1 = µ2 µ1b1 + µ2b2 + µ3b3 =0 2µ1 +5µ2 +8µ30 ⇐⇒ ⇐⇒··· ⇐⇒ µ2 = 2µ3 , 3µ1 +6µ2 +9µ3 =0  − so  b 2b + b =0 1 − 2 3 and b , b , b are not linearly independent. For instance, we say that b = 2b b is a 1 2 3 1 2 − 3 non-trivial linear relation between the vectors b1, b2 and b3.  Definition 1.14. The vectors b ,...,b V span V , if every vector v V can be 1 n ∈ ∈ written as a linear combination v = α1b1 + + αnbn, for some α1,...,αn R. We then also say that the vector set b ,...,b ···spans V . ∈ { 1 n} Examples 1.15. 1 0 0 (1) The vectors 0 , 1 , 0 span R3. 0 0 1 1 0 0 1 (2) The vectors 0 , 1 , 0 , 1 also span R3. 0 0 1 1         1.2. BASES 11

1 0 (3) The vectors 0 , 1 span the xy-coordinate plane (i.e., the subspace given 0 0 by the equation z =0 ) in R3. 

Exercise 1.16. The set of all linear combinations of b1,...,bn V is denoted by span b ,...,b . Show that span b ,...,b is a subspace of V . ∈ { 1 n} { 1 n} Definition 1.17. The vectors b ,...,b V form a basis of V , if: 1 n ∈ (1) they are linearly independent and (2) they span V .

We then denote this basis as an ordered set := b1,...,bn , where we fix the order of the vectors. B { } Warning: In this text, we only consider bases for so-called finite-dimensional vector spaces, that is, vector spaces that admit bases consisting of a finite number of elements. Example 1.18. The vectors 1 0 0 e1 := 0 , e2 := 1 , e3 := 0 0 0 1 3     3   form a basis of R . This is called the standard basis of R and denoted := e1, e2, e3 . For Rn, the standard basis := e ,...,e is defined similarly. E { } E { 1 n} Example 1.19. The vectors in Example 1.12 span R3, while the vectors in Example 1.13 do not span R3. To see this, we recall the following facts about bases. 

1.2.2. Facts about Bases.

Let V be a vector space; as stated above, V is finite-dimensional. Then we have: (1) All bases of V have the same number of elements. This number is called the dimension of V and denoted dim V . (2) If = b1,...,bn is a basis of V , there is a unique way of writing any v V asB a linear{ combination} ∈ 1 n v = v b1 + ...v bn of elements in . The numbers v1,...,vn are the coordinates of v with respect to the basisB and we denote by B v1 [v] = . B  .  vn   the coordinate vector of v with respect to . B 12 1. REVIEW OF LINEAR ALGEBRA

(3) If we know that dim V = n, then: (a) More than n vectors in V must be linearly dependent; (b) Fewer than n vectors in V cannot span V ; (c) Any n linearly independent vectors span V ; (d) Any n vectors that span V must be linearly independent; (e) If k vectors span V , then k n and some subset of those k vectors must be a basis of V ; ≥ (f) If a set of m vectors is linearly independent, then m n and we can always complete that set to form a basis of V . ≤ 3 Example 1.20. The vectors b1, b2, b3 in Example 1.12 form a basis of R since they are linearly independent and they are exactly as many as the dimension of R3.  Example 1.21 (Gauss-Jordan elimination). We are going to compute here the coordi- 1 nates of v = 1 with respect to the basis := b1, b2, b3 from Example 1.12. The 1 B { }   v1 seeked coordinates [v] = . must satisfy the equation B  .  vn   1  4 7 1 v1 2 + v2 5 + v3 8 = 1 , 3 6 0 1 so to find them we have to solve the following  system  of linear equations: v1 +4v2 +7v3 =1 1 2 3 2v +5v +8v =1 3v1 +6v2 =1.

For that purpose, we may equivalently reduce the following augmented matrix 1 4 7 1 2 5 8 1 3 6 0 1 to echelon form, using the Gauss–Jordan elimination method. We are going to perform both calculations in parallel, which will also point out that they are indeed seemingly different incarnations of the same method. By multiplying the first equation/row by 2 (resp. 3) and subtracting it from the second (reps. third) equation/row we obtain 1 2 3 v +4v +7v =1 14 7 1 3v2 6v3 = 1 ! 0 3 6 1 .  − − −  − − −   6v2 21v3 = 2 0 6 21 2 − − −  − − −   1.2. BASES 13

By multiplying the second equation/row by 1 and by adding to the first (resp. third) − 3 equation/row the second equation/row multiplied by 4 (resp. 2) we obtain − 3 1 3 1 1 v v = 3 1 0 1 − − − − 3 v2 +2v3 = 1 ! 01 2 1 .  3  3   9v3 =0 0 0 9 0 −  −  The last equation/row shows that v3 = 0, hence by backward substitution we obtain the solution

1 1 1 v = 3 1 0 0 − − 3 v2 = 1 ! 0 1 0 1 .  3  3   v3 =0 0 0 1 0     Example 1.22. When V = Rn and = is the standard basis, the coordinate vector v Rn coincides with the vector itself!B InE this very special case, we have [v] = v.  ∈ E Exercise 1.23. Let V be the vector space consisting of all 2 2 matrices with trace zero, namely × a b V := : a,b,c,d R and a + d =0 . c d ∈    (1) Show that 1 0 0 1 0 0 := , , B 0 1 0 0 1 0  −       b1 b2 b3 is a basis of V . | {z } | {z } | {z } (2) Show that 1 0 0 1 0 1 := , − , B 0 1 1 0 1 0  −       e ˜b1 ˜b2 ˜b3 is another basis of V .| {z } | {z } | {z } (3) Compute the coordinates of 2 1 v = 7 2  −  with respect to and with respect to . B B e 14 1. REVIEW OF LINEAR ALGEBRA

1.3. The Einstein Convention

1.3.1. A Convenient Summation Convention.

We start by setting a notation that will turn out to be useful later on. Recall that if = b , b , b is a basis of a vector space V , any vector v V can be written as B { 1 2 3} ∈ 1 2 3 (1.1) v = v b1 + v b2 + v b3 for appropriate v1, v2, v3 R. ∈ Notation. From now on, expressions like the one in (1.1) will be written as ❤❤ ✭✭✭✭❤ 1 ✭❤✭❤2 ✭❤ 3 j (1.2) v = ✭v ✭b1✭+ v b2 +❤❤v ❤b❤3 = v bj . That is, from now on when an index appears twice – once as a subscript and once as a superscript – in a term, we know that it means that there is a summation over all possible values of that index. The summation symbol will not be displayed. k j On the other hand, indices that are not repeated in expressions like aijx y are free indices not subject to summation. Examples 1.24. For indices ranging over 1, 2, 3 , i.e. n =3: i k { } (1) The expression aijx y means i k 1 k 2 k 3 k aijx y = a1jx y + a2jx y + a3jx y , k k i k and could be called Rj (meaning that Rj and aijx y both depend on the indices j and k). (2) Likewise, k j k 1 k 2 k 3 k aijx y = ai1x y + ai2x y + ai3x y =: Qi . (3) Further i j 1 1 1 2 1 3 aijx y = a11x y + a12x y + a13x y 2 1 2 2 2 3 + a21x y + a22x y + a23x y 3 1 3 2 3 3 + a31x y + a32x y + a33x y =: P (4) An expression like i j ℓ ij A BkℓC =: Dk makes sense. Here the indices i, j, k are free (i.e. free to range in 1, 2,...,n ) and ℓ is a summation index. { } (5) On the other hand an expression like jk ℓ jk EijFℓ G = Hi does not make sense because the expression on the left has only two free indices, i and k, while j and ℓ are summation indices and neither of them can appear on the right hand side. 1.3. THE EINSTEIN CONVENTION 15

j Notation. Since v bj denotes a sum, we choose to denote the indices of the generic I term of a sum with capital letters. For example, we write v bI and the above expressions could have been written as (1) 3 i k I K 1 k 2 k 3 k aijx y = aIJ x y = a1jx y + a2jx y + a3jx y , I=1 X (2) 3 k j K J k 1 k 2 k 3 aijx y = aIJ x y = ai1x y + ai2x y + ai3x y . J=1 X (3) 3 3 i j I J aijx y = aIJ x y = J=1 I=1 XX1 1 1 2 1 3 = a11x y + a12x y + a13x y 2 1 2 2 2 3 + a21x y + a22x y + a23x y 3 1 3 2 3 3 + a31x y + a32x y + a33x y . 

1.3.2. Change of Basis.

Let and be two bases of a vector space V and let B B 1 1 L1 ... Ln e . . (1.3) L := L e = . . BB   Ln ... Ln  1 n   be the matrix of the change of basis from the “old” basis to the “new” basis . Recall that the entries of the j-th column of L are the coordinatesB of the new basisB ˜ vector bj with respect to the old basis . e B Mnemonic: Upper indices go up to down, i.e., they are row indices. Lower indices go left to right, i.e., they are column indices. With the Einstein convention we can write ˜ i (1.4) bj = Ljbi , or, equivalently,

˜b1 ... ˜bn = b1 ... bn L ,

16 1. REVIEW OF LINEAR ALGEBRA where we use some convenient formal notation: The multiplication is to be performed with the usual rules for vectors and matrices, though, in this case, the entries of the row vectors ˜b1 ... ˜bn and b1 ... bn are not real numbers but vectors themselves. 1 If Λ= L− denotes
the matrix of the
change of basis from to , then, using the same formal notation as above, we have B B e b1 ... bn = ˜b1 ... ˜bn Λ . Equivalently, this can be written in compact
form using
the Einstein notation as i ˜ bj =Λjbi . Analogously, the corresponding relations for the vector coordinates are 1 1 1 1 1 1 1 1 v L1 ...... Ln v˜ v˜ Λ1 ...... Λn v ......  .   . .   .   .   . .   .  i i i . i i i . v = L1 ...... Ln . and v˜ = Λ1 ...... Λn .  .   . .   .   .   . .   .   .   . .   .   .   . .   .   .   . .     .   . .     vn Ln ...... Ln v˜n  v˜n Λn ...... Λn vn    1 n      1 n   and these can be written with  the Einstein convention  respec tively as   

i i j i i j (1.5) v = Ljv˜ and v˜ =Λjv , or, in matrix notation, 1 [v] = L e [v] e and [v] e =(L e )− [v] = L e[v] . B BB B B BB B BB B Important: Note how the coordinate vectors change in a way opposite to the basis change. Hence, we say that the coordinate vectors are contravariant3 because they 1 change by L− when the basis changes by L. Example 1.25. We consider the following two bases of R2 1 2 = , B 0 1      b1 b2 (1.6) 3 1 = |{z}, |{z}− B 1 1   −   e ˜b1 ˜b2 and we look for the matrix of the change|{z} of| basis.{z } Namely we look for a matrix L such that 3 1 1 2 = ˜b ˜b = b b L = L. 1 −1 1 2 1 2 0 1  −   

3In Latin contra means “contrary” or “against”. 1.3. THE EINSTEIN CONVENTION 17

There are two alternative ways of finding L: (1) With matrix inversion: Recall that 1 a b − d b (1.7) = 1 , c d D c− a   −  a b where D = det is the determinant. Thus c d   1 1 2 − 3 1 1 2 3 1 1 1 L = = = . 0 1 1 −1 0− 1 1 −1 1 1    −    −   −  (2) With Gauss-Jordan elimination:

1 2 3 1 ! 1 0 1 1 0 1 1 −1 0 1 1 1  −   −  

1.3.3. The Kronecker Delta Symbol. Notation. i The Kronecker delta symbol δj is defined as 1 if i = j (1.8) δi := j 0 if i = j . ( 6 Examples 1.26. If L is a matrix, the (i, j)-entry of L is the coefficient in the i-th row i and j-th column, and is denoted by Lj. (1) The n n identity matrix × 1 ... 0 . . . I = . .. . 0 ... 1   i   has (i, j)-entry equal to δj. (2) Let L and M be two square matrices. The (i, j)-th entry of the product 1 1 1 1 1 M1 ...... Mn L1 ... Lj ... Ln . .  . .   . . . i i . . . ML = M1 ...... Mn . . .  . .   . . .  . .     n n  n n n M ...... M  L1 ... Lj ... Ln  1 n    equals the dot product of the i-th row of M and j-th column of L, 1 Lj M i ... M i · . = M iL1 + + M i Ln , 1 n  .  1 j ··· n j Ln
 j    18 1. REVIEW OF LINEAR ALGEBRA

or, using the Einstein convention, i k MkLj . Notice that, since in general ML = LM, it follows that 6 M i Lk = Li M k = M kLi . k j 6 k j j k 1 However, in the special case where M =Λ= L− , we have ΛL = LΛ= I and here we can write i k i i k ΛkLj = δj = LkΛj . 

Remark 1.27. Using the Kronecker delta symbol we can check that the notations in (1.5) are all consistent. In fact, from (1.2) we should have i i˜ (1.9) v bi = v =v ˜ bi , and, in fact, using (1.5), i˜ i j k k j j v˜ bi =Λjv Li bk = δj v bk = v bj , i k k 1 where we used that ΛjLi = δj since Λ= L− . Two words of warning: j k The two expressions v bj and v bk are identical, as the indices j and k are • dummy indices, that is, can be replaced by other symbols throughout, without changing the meaning of the expression (as long as the symbols do not collide with other symbols already used in the expression). When multiplying two different expressions in Einstein notation, you should • be careful to distinguish by different letters different summation indices. For i i j ˜ j example, if v˜ = Λjv and bi = Li bj, in order to perform the multiplication i v˜ ˜bi we have to make sure to replace one of the dummy indices in the two ˜ k i˜ i j k expressions. So, for example, we can write bi = Li bk, so that v˜ bi =Λjv Li bk.

1.4. Linear Transformations

1.4.1. Linear Transformations as (1, 1)-Tensors.

Recall that a linear transformation from V to itself, T : V V , is a function (or map or transformation) that satisfies the property → T (αv + βw)= αT (v)+ βT (w) , for all α, β R and all v,w V . Once we choose a basis = b1,...bn of V , the transformation∈ T is represented∈ by a matrix A, called theBmatrix{ of the} linear transformation with respect to that basis. The columns of A are the coordinate vectors of T (b ),...,T (b ) with respect to . Then that matrix A gives the effect of 1 n B 1.4. LINEAR TRANSFORMATIONS 19

T on coordinate vectors as follows: If T (v) is the value of the transformation T on the vector v, with respect to a basis we have that B (1.10) [v] [T (v)] = A[v] . B 7−→ B B If is another basis, we have also B (1.11) [v] e [T (v)] e = A[v] e , e B 7−→ B B where now A is the matrix of the transformation T with respect to the basis . e B We want to find now the relation between A and A. Let L := L e be the matrix of BB the change ofe basis from to . Then, for any v V , e B B ∈ 1 e (1.12) [v] e = L− [v] . e B B In particular the above equation holds for the vector T (v), that is 1 (1.13) [T (v)] e = L− [T (v)] . B B Using (1.12), (1.11), (1.13) and (1.10) in this order, we have

1 1 1 AL− [v] = A[v] e = [T (v)] e = L− [T (v)] = L− A[v] B B B B B 1 1 for every vector v V . If follows that AL− = L− A or equivalently e ∈ e 1 (1.14) A = L− AL , e which in Einstein notation reads e i i k m Aj =ΛkAmLj . We say that the linear transformation T is a tensor of type (1, 1). e Example 1.28. Let V = R2 and let and be the bases in Example 1.25. The matrices corresponding to the change ofB coordinatesB are e 1 1 1 1 1 1 1 1 2 2 L := L = and L− = = , e 1 1 2 −1− 1 1 1 BB  −  − −   2 − 2  where in the last equality we used the formula for the inverse of a matrix in (1.7). Let T : R2 R2 be the linear transformation that in the basis takes the form → B 1 3 A = . 2 4   Then according to (1.14) the matrix A of the linear transformation T with respect to the basis is B 1 1 e 1 2 2 1 3 1 1 5 2 A = L− AL = = . e 1 1 2 4 1 1 1− 0  2 − 2     −  −  e  20 1. REVIEW OF LINEAR ALGEBRA

Example 1.29. We now look for the standard matrix of T , that is, the matrix M that represents T with respect to the standard basis of R2, which we denote by 1 0 := , . E 0 1      e1 e2 We want to apply again the formula (1.14)|{z} |{z} and hence we first need to find the matrix S := L of the change of basis from to . Recall that the columns of S are the coordinatesBE of b with respect to the basisE ,B that is j E 1 2 S = . 0 1   According to (1.14), 1 A = S− MS, from which, using again (1.7), we obtain

1 1 2 1 3 1 2 1 2 1 1 5 1 M = SAS− = = = . 0 1 2 4 0− 1 0 1 2 0 2 0             

Example 1.30. Let T : R3 R3 be the orthogonal projection onto the plane of equation → P 2x + y z =0 . − This means that the transformation T is characterized by the fact that – it does not change vectors in the plane , and – it sends vectors perpendicular to to theP zero vector in . P P We want to find the standard matrix for T . Idea: First compute the matrix of T with respect to a basis of R3 well adapted to the problem, then use (1.14) after having found the matrix L Bof the change of basis. BE To this purpose, we choose two linearly independent vectors in the plane and a third vector perpendicular to . For instance, we set P P 1 0 2 := 0 , 1 , 1 , B        2 1 1  − b1 b2  b3  where the coordinates of b1 and b2 satisfy|{z} | the{z} equation| {z } of the plane, while the coordinates of b3 are the coefficients of the equation describing . Let be the standard basis of R3. P E Since

T (b1)= b1, T (b2)= b2 and T (b3)=0 , 1.4. LINEAR TRANSFORMATIONS 21 the matrix of T with respect to is simply B 1 0 0 (1.15) A = 0 1 0 , 0 0 0   where we recall that the j-th column is the coordinate vector [T (bj)] of the vector B T (bj) with respect to the basis . The matrix of the change ofB basis from to is E B 10 2 L = 01 1 , 2 1 1 − hence, by Gauss–Jordan elimination,   1 1 1 3 3 3 1 1 −5 1 L− = 3 6 6 . −1 1 1  3 6 6  −  Therefore 1 1 1 3 3 3 1 1 −5 1 M = LAL− = = 3 6 6 . ··· −1 1 5  3 6 6   

Example 1.31. Let V := R[x] be the vector space of polynomials of degree 2, and 2 ≤ let T : R[x]2 R[x]2 be the linear transformation given by differentiating a polynomial and then multiplying→ the derivative by x,

T (p(x)) := xp′(x) , so that T (a + bx + cx2)= x(b +2cx)= bx +2cx2. Let := 1, x, x2 and := x, x 1, x2 1 B { } B { − − } be two bases of R[x] . Since 2 e T (1)=0=0 1+0 x +0 x2 · · · T (x)= x =0 1+1 x +0 x2 · · · T (x2)=2x2 =0 1+0 x +2 x2 · · · and T (x)= x =1 x +0 (x 1)+0 (x2 1) · · − · − T (x 1) = x =1 x +0 (x 1)+0 (x2 1) − · · − · − T (x2 1)=2x2 =2 x 2 (x 1)+2 (x2 1) , − · − · − · − 22 1. REVIEW OF LINEAR ALGEBRA then 0 0 0 11 2 A = 0 1 0 and A = 0 0 2 . 0 0 2 00− 2 

  e  1 One can check that indeed AL = LA or, equivalently A = L− AL, where 0 1 1 L =e 11− − 0 e 00 1  is the matrix of the change of basis.   

1.4.2. Conjugate Matrices.

The above calculations can be summarized by the commutativity of the following diagram. Here, the vertical arrows correspond to the operation of change of basis from B to (recall that the coordinate vectors are contravariant tensors, that is, they transform B 1 as [v] e = L− [v] ) and the horizontal arrows correspond to the operation of applying B thee transformationB T with respect to the two different basis: [v] ✤ A / [T (v)] ❴B ❴ B L−1 L−1   / [v] e ✤ / [T (v)] e . B Ae B Saying that the diagram is commutative is saying that if one starts from the upper left hand corner, reaching the lower right hand corner following either one of the two paths has exactly the same effect. In other words, changing coordinates first then applying the transformation T yields exactly the same affect as applying first the transformation T 1 1 and then the change of coordinates, that is, L− A = AL− or, equivalently, 1 A = L− AL . e In this case we say that A and A are conjugate matrices. This means that A and A represent the same transformation withe respect to different bases. e e Definition 1.32. We say that two matrices A and A are conjugate if there exists and 1 invertible matrix L such that A = L− AL. e Example 1.33. The three matrices from Example 1.28 and Example 1.29 e 1 3 5 1 5 2 A = M = and A = 2 4 2 0 1− 0     −  are all conjugate. Indeed, we have e 1 1 1 A = L− AL , A = S− MS and A = R− MR,

e e 1.4. LINEAR TRANSFORMATIONS 23 where L and S may be found in those examples and where R := SL.  We now review some facts about conjugate matrices. Recall that the characteristic polynomial of a square matrix A is the polynomial p (λ) := det(A λI) . A − 1 Let us assume that A and A are conjugate matrices, that is A = L− AL for some invertible matrix L. Then e 1 1 e p e(λ) = det(A λI) = det(L− AL λL− IL) A − − 1 (1.16) = det(L− (A λI)L) ✘ − ✘e✘1 ✘✘ = ✘(det✘ L− ) det(A λI)✘(det✘ L) = p (λ) , − A which means that any two conjugate matrices have the same characteristic polynomial. Recall that the eigenvalues of a matrix A are the roots of its characteristic poly- nomial and we here usually allow complex roots. Then, by the so-called fundamental theorem of Algebra, each n n matrix has n (real or complex) eigenvalues counted with multiplicities as polynomial× roots. Recall also the definitions of determinant and trace of a square matrix. By analysing the characteristic polynomial, we see that (1) the determinant of a matrix is equal to the product of its eigenvalues (multi- plied with multiplicities), and (2) the trace of a matrix is equal to the sum of its eigenvalues (added with multi- plicities). From (1.16) if follows that, if the matrices A and A are conjugate, then: A and A have the same size; • the eigenvalues of A (as well as their multiplicities)e are the same as those of A; • det A =e det A; • tr A = tr A; e • A is invertiblee if and only if A is invertible. • e 1 3 1 2 Example 1.34. The matrices A = e and A = are not conjugate. In fact, 2 4 ′ 2 4     A is invertible, as det A = 2 =0, while det A′ =0, so that A′ is not invertible.  − 6

1.4.3. Eigenbases.

The possibility of choosing different bases is very important and often simplifies the calculations. Example 1.30 is such an example, where we choose an appropriate basis according to the specific problem. Other times, a basis can be chosen according to the symmetries and, completely at the opposite side, sometime there is just not a basis that is a preferred one. In the context of a linear transformation T : V V , a basis that is particularly convenient, when it exists, is an eigenbasis for that linear→ transformation. 24 1. REVIEW OF LINEAR ALGEBRA

Recall that an eigenvector of a linear transformation T : V V is a vector v =0 such that T (v) is a multiple of v, say T (v)= λv and, in that case,→ the scaling number6 λ is called an eigenvalue of T . An eigenbasis is a basis of V consisting of eigenvectors of a linear transformation T : V V . The point of having an eigenbasis is that, with respect to this eigenbasis, the linear transformation→ is representable by a diagonal matrix, D. Hence, the initial matrix representative A is actually conjugate to a diagonal matrix D. A linear transformation T : V V for which an eigenbasis exists is then called diagonalizable.4 Given→ a linear transformation T : V V , in order to find an eigenbasis of T , we first represent T by some matrix A (with→ respect to some chosen basis of V ), and then perform the following steps: (1) We find the eigenvalues by determining the roots of the characteristic polyno- mial of A (often allowing complex roots). (2) For each eigenvalue λ, we find the corresponding eigenvectors by looking for the nonzero vectors in the so-called eigenspace E := ker(A λI) . λ − When considering complex eigenvalues, the eigenspaces are determined as sub- spaces of the complex vector space Cn. However, in this text, we concentrate on real cases. (3) We determine whether there exists an eigenbasis. We will illustrate this in the following examples. Example 1.35. Let T : R2 R2 be the linear transformation given by the matrix 3 4 → A = with respect to the standard basis of R2. 4 −3 − −  (1) The eigenvalues are the roots of the characteristic polynomial pλ(A). Since 3 λ 4 pA(λ) = det(A λI) = det − − − 4 3 λ  − − −  = (3 λ)( 3 λ) 16 = λ2 25=(λ 5)(λ + 5) , − − − − − − hence λ = 5 are the eigenvalues of A. ± (2) If λ is an eigenvalue of A, the eigenspace corresponding to λ is given by Eλ = ker(A λI). Note that − v E Av = λv . ∈ λ ⇐⇒

4In general, when an eigenbasis does not exist, it is still possible to find a basis, with respect to which the linear transformation is as simple as possible, i.e., as close as possible to being diagonal. Such a best matrix representative of T : V V is called a Jordan canonical form and is, of course, conjugate to the first matrix representative→A. In this text, we will not address such more general canonical forms. 1.4. LINEAR TRANSFORMATIONS 25

With our choice of A and with the resulting eigenvalues, we have 2 4 2 E5 = ker(A 5I) = ker − − = span − 4 8 1 − −  −  8 4 1 E 5 = ker(A +5I) = ker − = span . − 4 2 2 −    (3) The following is an eigenbasis for this linear transformation: 2 1 = ˜b = , ˜b = B 1 1 2 2  −    and e T (˜b )=5˜b =5 ˜b +0 ˜b 1 1 · 1 · 2 T (˜b )= 5˜b =0 ˜b 5 ˜b , 2 − 2 · 1 − · 2 5 0 so that A˜ = . 0 5  −  b2

b1

x +2y =0

Notice that the eigenspace E5 consists of vectors on the line x +2y = 0 and these vectors get scaled by the transformation T by a factor of 5. On the other hand, the eigenspace E 5 consists of vectors perpendicular to the line x +2y =0 and these vectors get− flipped by the transformation T and then also scaled by a factor of 5. Hence T is just the reflection across the line x+2y =0 followed by multiplication by 5.  Example 1.36. Now let T : R2 R2 be the linear transformation given by the matrix 1 2 → A = with respect to the standard basis of R2. 4 3   (1) The eigenvalues are the roots of the characteristic polynomial: 1 λ 2 pA(λ) = det(A λI) = det − − 4 3 λ  −  = (1 λ)(3 λ) 2 4= λ2 4λ 5=(λ + 1)(λ 5) , − − − · − − − hence λ = 1 and λ =5 are the eigenvalues of A. − 26 1. REVIEW OF LINEAR ALGEBRA

(2) If λ is an eigenvalue of A, the eigenspace corresponding to λ is given by Eλ = ker(A λI). In this case we have − 2 2 1 E 1 = ker(A + I) = ker = span − 4 4 1   −  4 2 1 E5 = ker(A 5I) = ker − = span . − 4 2 2  −    (3) The following is an eigenbasis for this linear transformation: 1 1 = ˜b = , ˜b = . B 1 1 2 2  −    ˜ e ˜ ˜ ˜ ˜ 1 0 We have T (b1)= b1 and T (b2)=5b2, hence A = − . − 0 5   

Example 1.37. Now let T : R2 R2 be the linear transformation given by the matrix 5 3 → A = with respect to the standard basis of R2. 3 −1  −  (1) The eigenvalues are the roots of the characteristic polynomial: 5 λ 3 pA(λ) = det(A λI) = det − − − 3 1 λ  − −  = (5 λ)( 1 λ)+9= λ2 4λ +4=(λ 2)2 , − − − − − hence λ =2 is the only eigenvalue of A. (2) The eigenspace corresponding to λ =2 is 3 3 1 E2 = ker(A 2I) = ker − = span . − 3 3 1  −    (3) Since we cannot find two linearly independent eigenvectors (in order to form a basis of R2), we conclude that in this case there is no eigenbasis for this linear transformation. 

Summarizing, in Examples 1.28 and 1.29, we looked at how the matrix of a trans- formation changes with respect to two different bases that we were given. In Exam- ple 1.30, we looked for a particular basis appropriate to the transformation at hand. In Example 1.35, we looked for an eigenbasis with respect to the given transformation. Example 1.30 in this respect fits into the same framework as Example 1.35, but the orthogonal projection has a zero eigenvalue (see (1.15)). Example 1.36 illustrates how eigenvectors, in general, need not be orthogonal. In Example 1.37 we see that sometimes an eigenbasis does not exist. CHAPTER 2

Multilinear Forms

2.1. Linear Forms Linear forms on a vector space V are defined as linear real-valued functions on V . We will see that linear forms behave very much like vectors, only that they are elements not of V , but of a different, yet related, vector space. Whereas we represent regular vectors (from V ) by column vectors once a basis is fixed, we will represent linear forms (on V ) by row vectors. Then the value of a linear form on a specific vector is simply given by by the matrix product with the row vector (linear form) on the left and the column vector (actual vector) on the right.

2.1.1. Definition and Examples. Definition 2.1. Let V be a vector space. A linear form on V is a map α : V R such that for every a, b R and for every v,w V → ∈ ∈ α(av + bw)= aα(v)+ bα(w) . Alternative terminologies for “linear form” are tensor of type (0, 1), 1-form, linear functional and covector. Exercise 2.2. If V = R3 , which of the following are linear forms? (1) α(x, y, z) := xy + z; (2) α(x, y, z) := x + y + z +1; (3) α(x, y, z) := πx 7 z. − 2 Exercise 2.3. If V is the infinite dimensional vector space of continuous functions f : R R, which of the following are linear forms? → (1) α(f) := f(7) f(0); 33 x− (2) α(f) := 0 e f(x)dx; (3) α(f) := ef(4). R Example 2.4. [Coordinate forms] This is a most important example of linear form. Let i = b1,...,bn be a basis of V and let v = v bi V be a generic vector. Define βBi : V{ R by } ∈ → (2.1) βi(v) := vi ,

27 28 2. MULTILINEAR FORMS that is βi will extract the i-th coordinate of a vector with respect to the basis . The linear form βi is called coordinate form. Notice that B i i (2.2) β (bj)= δj , since the i-th coordinate of the basis vector bj with respect to the basis is equal to 1 if i = j and 0 otherwise. B  Example 2.5. Let V = R3 and let be its standard basis. The three coordinate forms are defined by E x x x β1 y := x , β2 y := y , β3 y := z . z z z        1 1 Example 2.6. Let V = R2 and let := , . We want to describe the B 1 1   −   b1 b2 1 2 elements of ∗ := β , β , in other words we want to find B { } |{z} | {z } β1(v) and β2(v) for a generic vector v V . To this purpose we∈ need to find [v] . Recall that if denotes the standard basis of R2 and L := L the matrix of the changeB of coordinateE from to , then BE E B 1 1 1 v [v] = L− [v] = L− 2 . B E v   Since 1 1 L = 1 1  −  and hence

1 1 1 1 L− = , 2 1 1  −  then 1 (v1 + v2) [v] = 2 . B 1 (v1 v2)  2 −  Thus, according to the definition (2.1), we deduce that β1(v)= 1 (v1 + v2) and β2(v)= 1 (v1 v2) . 2 2 −  2.1. LINEAR FORMS 29

2.1.2. Dual Space and Dual Basis.

We define

V ∗ := all linear forms α : V R , { → } and call this the dual (or dual space) of V .

Exercise 2.7. Check that V ∗ is a vector space whose null vector is the linear form identically equal to zero. Remark 2.8. Just like any function, two linear forms on V are equal if and only if their values are the same when applied to each vector in V . However, because of the defining properties of linear forms, to determine whether two linear forms are equal, it is enough to check that they are equal on each element of a basis of V . In fact, let α,α′ V ∗, let = b ,...,b be a basis of V and suppose that we know that ∈ B { 1 n} α(bj)= α′(bj) for all 1 j n. We verify that this implies that they are the same when applied to ≤ ≤ j each vector v V . In fact let v = v bj its representation with respect to the basis . Then we have∈ B j j j j α(v)= α(v bj)= v α(bj)= v α′(bj)= α′(v bj)= α′(v) . 

1 n Proposition 2.9. Let = b1,...,bn be a basis of V and β ,...,β the correspond- B { 1 } n ing coordinate forms. Then ∗ := β ,...,β is a basis of V ∗. As a consequence B { } dim V = dim V ∗ .

Proof. According to Definition 1.17, we need to check that the linear forms in ∗ B (1) are linearly independent and (2) span V ∗. (1) We need to check that the only linear combination of β1,...,βn that yields the zero i linear form is the trivial linear combination. Let ciβ =0 be a linear combination of the i β . Then for every basis vector bj, with j =1,...,n, i i i 0=(ciβ )(bj)= ci(β (bj)) = ciδj = cj , thus showing the linear independence.

(2) To check that ∗ spans V we need to verify that any α V ∗ is a linear combination of β1,...,βn, thatB is that we can find α R such that ∈ i ∈ i (2.3) α = αiβ

To find such αi we apply both sides of (2.3) to the j-th basis vector bi, and we obtain i i (2.4) α(bj)= αiβ (bj)= αiδj = αj , 30 2. MULTILINEAR FORMS which identifies the coefficients in (2.3). i By hypothesis α is a linear form and, since V ∗ is a vector space, also α(bi)β is a linear form. Moreover, we have just verified that these two linear form coincide on the basis vectors. By Remark 2.8 the two linear forms are the same and, hence, we have written α as a linear combination of the coordinate forms. This completes the proof that the coordinate forms form a basis of the dual. 

The basis ∗ of V ∗ is called the basis of V ∗ dual to . We emphasize that the B B components (or coordinates) of a linear form α with respect to ∗ are exactly the values of α on the elements of , as we found in the above proof: B B αi = α(bi) . We build with these the coordinate-vector of α as a row-vector:

[α] ∗ := α1 ... αn . B

Example 2.10. Let V = R[x]2 be the vector space of
polynomials of degree 2, let α : V R be the linear form given by ≤ → (2.5) α(p) := p(2) p′(2) − and let be the basis 1, x, x2 of V . In this example, we want to: B { } (1) find the components of α with respect to ∗; 1 2 3 B (2) describe the basis ∗ = β , β , β ; B { } (1) Since α = α(b )= α(1) = 1 0=1 1 1 − α = α(b )= α(x)=2 1=1 2 2 − α = α(b )= α(x2)=4 4=0 , 3 3 − then

(2.6) [α] ∗ = 1 1 0 . B
(2) The generic element p(x) R[x]2 written as combination of basis elements 1, x and x2 is ∈ p(x)= a + bx + cx2 . 1 2 3 Hence ∗ = β , β , β , is given by B { } β1(a + bx + cx2)= a (2.7) β2(a + bx + cx2)= b β3(a + bx + cx2)= c .  2.1. LINEAR FORMS 31

Remark 2.11. Note that we have to be careful when referring to a “dual basis” of V ∗, as for every basis of V there is going to be a basis ∗ of V ∗ dual to the basis . In the next section weB are going to see how a dual basis transformB s with a change of basis.B

2.1.3. Covariance of Linear Forms.

We want to examine how a linear form α : V R behaves with respect to a change a basis in V . To this purpose, let → = b ,...,b and := ˜b ,..., ˜b B { 1 n} B { 1 n} be two bases of V and let e 1 n 1 n ∗ := β ,...,β and ∗ := β˜ ,..., β˜ B { } B { } be the corresponding dual bases. Let e [α] ∗ = α1 ... αn and [α] e∗ = α1 ... αn B B be the coordinate vectors of α with
respect to ∗ and ∗, that is
B Be e α(b )= α and α(˜b )= α . i i i ei Let L := L e be the matrix of the change of basis in (1.3) BB e ˜ i bj = Ljbi . Then ˜ i i i i (2.8) αj = α(bj)= α(Ljbi)= Ljα(bi)= Ljαi = αiLj , so that e i (2.9) αj = αiLj .

e Exercise 2.12. Verify that (2.9) is equivalent to saying that

(2.10) [α] e∗ = [α] ∗ L. B B i Note that we have exchanged the order of αi and Lj in the last equation in (2.8) to respect the order in which the matrix multiplication in (2.10) has to be performed. This i was possible because both αi and Lj are real numbers.

We say that a linear form α is covariant because its components change by L when the basis changes by L.5 A linear form α is hence a covariant tensor or a tensor of type (0, 1).

5In Latin, the prefix co means “joint”. 32 2. MULTILINEAR FORMS

Example 2.13. We continue with Example 2.10. We consider the bases as in Exam- ple 1.31, that is := 1, x, x2 and := x, x 1, x2 1 B { } B { − − } and the linear form α : V R as in (2.5). We will: → e (1) find the components of α with respect to ∗; 1 2 3 B (2) describe the basis ∗ = β , β , β ; B { } (3) find the components of α with respect to ∗; 1 2 3 B (4) describe the basis ∗ = β˜ , β˜ , β˜ ; B { } 1 (5) find the matrix of change of basis L := L ee and compute Λ= L− ; (6) check the covariancee of α; BB (7) check the contravariance of ∗. B (1) This is done in (2.6). (2) This is done in (2.7). (3) We proceed as in (2.6). Namely, α = α(˜b )= α(x)=2 1=1 1 1 − α = α(˜b )= α(x 1)=1 1=0 2 2 − − e 2 α3 = α(˜b3)= α(x 1)=3 4= 1 , e − − − so that e [α] e∗ = 1 0 1 . B − (4) Since β˜i(v)=˜vi, to proceed as in (2.7) we first
need to write the generic polynomial p(x)= a + bx + cx2 as a linear combination of elements in , namely we need to find B a,˜ ˜b and c˜ such that p(x)= a + bx + cx2 =˜ax + ˜b(x 1)+˜c(x2 e 1) . − − By multiplying and collecting the terms, we obtain that ˜b c˜ = a a˜ = a + b + c − − a˜ + ˜b = b that is ˜b = a c   − − c˜ = c c˜ = c .

Hence   p(x)= a + bx + cx2 =(a + b + c)x +( a c)(x 1) + c(x2 1) , − − − − so that it follows that β˜1(p(x)) = a + b + c β˜2(p(x)) = a c − − β˜3(p(x)) = c , 2.1. LINEAR FORMS 33

(5) The matrix of the change of basis is given by 0 1 1 − − L := L e = 1 1 0 , BB 0 0 1    since for example ˜b3 can be written as a linear combination with respect to as ˜b3 = 2 B x 1= 1b1 +0b2 +1b3, and these coordinates 1, 0, 1 build the third column of L. − − 1 − To compute Λ= L− we can use the Gauss–Jordan elimination process 0 1 1 1 0 0 1 0 0 1 1 1 1− 1− 0 0 1 0 ! ... ! 0 1 0 1 0 1 0 0 1 0 0 1 0 0 1 −0 0− 1  Therefore, we have    1 1 1 Λ= 1 0 1 . −0 0− 1    (6) The linear form α is indeed covariant, since 0 1 1 − − α1 α2 α3 L = 1 1 0 1 1 0 = 1 0 1 = α1 α2 α3 . 0 0 1  −



  (7) The dual basis ∗ is contravariant, since e e e B β˜1 β1 β˜2 =Λ β2 ,     β˜3 β3     as it can be verified by evaluating both sides on an arbitrary vector p(x)= a + bx + cx2: β1(p) 1 1 1 a a + b + c β˜1(p) Λ β2(p) = 1 0 1 b = a c = β˜2(p) .   − −     − −    β3(p) 0 0 1 c c β˜3(p)           

2.1.4. Contravariance of Dual Bases.

In fact, statement (7) in Example 2.10 holds in general, namely: Proposition 2.14. Dual bases are contravariant. Proof. We will check that when bases and are related by B B i ˜bj = L bi j e 34 2. MULTILINEAR FORMS

V real vector space V ∗ = linear forms α : V R with dim V = n dual{ vector space to V→ } 1 n = b ,...,b ∗ = β ,...,β B { 1 n} B { } basis of V dual basis of V ∗ w.r.t. B 1 n := ˜b ,..., ˜b ∗ = β˜ ,..., β˜ B { 1 n} B { } another basis of V dual basis of V ∗ w.r.t. ∗ 1 B Le := L e =matrix of the change Λ=e L− =matrix of the change of basisBB from to of basis from to e B B B B ˜ i ˜i i j Then we have bj = Ljbi or Then we have β =Λjβ or e e β˜1 β1 . 1 . ˜b ... ˜b = b1 ... bn L . = L− . 1 n     β˜n βn

covariance of a basis contravariance  of the dual basis

If v is any vector in V , If α is any linear form in V ∗, i i j j then v = v bi =v ˜ ˜bi, where then α = αjβ = αjβ˜ , where

i i j 1 i ∗ v˜ =Λjv i.e. [v] e = L− [v] or αj = Ljαi i.e. [α] e∗ = [α] L or B B B B v˜1 v1 e . 1 . e . = L− . α˜1 ... α˜n = α1 ... αn L     v˜n vn

contravariance  of the coordinate vectors covariance of linear forms components vectors are (1, 0)-tensors linear forms are (0, 1)-tensors → → Table 1. Summary regarding duality

the corresponding dual bases ∗ and ∗ of V ∗ are related by B B ˜j j i (2.11) β e=Λi β . j i ˜ It is enough to check that the Λi β are dual to the bj. In fact, since ΛL = I, then k ℓ ˜ k ℓ i k i ℓ k i ℓ k i k j ˜ (Λℓ β )(bj)=(Λℓ β )(Ljbi)=Λℓ Ljβ (bi)=Λℓ Ljδi =Λi Lj = δj = β (bj) . 

In Table 1, you can find a summary of the properties that bases and dual bases, coordinate vectors and components of linear forms satisfy with respect to a change of basis and hence whether they are covariant or contravariant. Moreover, Table 2 summarizes the characteristics of covariance and contravariance. 2.2. BILINEAR FORMS 35

covariance contravariace of a tensor of a tensor is denoted by lower indices upper indices coordinate-vectors are indicated as row vectors column vectors the tensor transforms w.r.t. a change of 1 basis from to by multiplication with L on the right L− on the left (for later use)B B a tensor of type (ep, q) has covariant order q contravariant order p Table 2. Covariance vs. contravariance

2.2. Bilinear Forms 2.2.1. Definition and Examples. Definition 2.15. A bilinear form on V is a function ϕ : V V R that is linear in each variable, that is × → ϕ(u,λv + µw)= λϕ(u, v)+ µϕ(u,w) ϕ(λv + µw,u)= λϕ(v,u)+ µϕ(w,u) , for every λ,µ R and for every u,v,w V . ∈ ∈ Examples 2.16. Let V = Rn. (1) If v,w Rn, the dot product (or scalar product) defined as ∈ ϕ(v,w) := v · w = v w cos θ , | || | where θ is the angle between v and w is a bilinear form. (2) Let n =3. Choose a vector u R3 and for any two vectors v,w R3, denote by v w their cross product.∈ The scalar triple product ∈ × u (2.12) ϕu(v,w) := u · (v w) = det v × w u   is a bilinear form in v and w, where v denotes the matrix with rows u, v and w w. The quantity ϕu(v,w) calculates the signed volume of the parallelepiped spanned by u,v,w: the sign of ϕu(v,w) depends on the orientation of the triple u,v,w. Since the cross product is defined only in R3, in contrast with the scalar product, the scalar triple product cannot be defined in Rn with n =3 (though there is a formula for an n dimensional parallelediped involving some6 “general- ization” of it).  36 2. MULTILINEAR FORMS

Exercise 2.17. Verify the equality in (2.12) using the Leibniz formula for the determi- nant of a 3 3 matrix. Recall that × a11 a12 a13 det a a a =a a a a a a + a a a  21 22 23 11 22 33 − 11 23 32 12 23 31 a31 a32 a33   a a a + a a a a a a − 12 21 33 13 21 32 − 13 22 31

= sign(σ)a1σ(1)a2σ(2)a3σ(3) , σ S3 X∈ where σ =(σ(1), σ(2), σ(3)) S := permutations of 3 elements ∈ 3 { } = (1, 2, 3), (1, 3, 2), (2, 3, 1), (2, 1, 3), (3, 1, 2), (3, 2, 1) , { } and the corresponding signs flip each time two elements get swapped: sign(1, 2, 3)=1 , sign(1, 3, 2) = 1 , sign(3, 1, 2)=1 , − sign(3, 2, 1) = 1 , sign(2, 3, 1)=1 , sign(2, 1, 3) = 1 . − − An even permutation is a permutation σ with sign(σ)=1; an odd permutation is a permutation σ with sign(σ)= 1. − Examples 2.18. Let V = R[x]2.

(1) Let p, q R[x]2. The function ϕ(p, q) := p(π)q(33) is a bilinear form. (2) Likewise,∈ 1 ϕ(p, q) := p′(0)q(4) 5p′(3)q′′( ) − 2 is a bilinear form. 

Exercise 2.19. Are the following functions bilinear forms? u (1) V = R2 and ϕ(u, v) := det ; v R 1  (2) V = [x]2 and ϕ(p, q) := 0 p(x)q(x)dx; R 1 (3) V = M2 2( ), the space of real 2 2 matrices, and ϕ(L, M) := L1 tr M, 1× R × where L1 it the (1,1)-entry of L and tr M is the trace of M; (4) V = R3 and ϕ(v,w) := v w; (5) V = R2 and ϕ(v,w) is the× area of the parallelogram spanned by v and w. (6) V = Mn n(R), the space of real n n matrices with n> 1, and ϕ(L, M) := tr L det M× , where tr L is the trace× of L and det M is the determinant of M. Remark 2.20. We need to be careful about the following possible confusion. A bilinear form on V is a function on V V that is linear in each variable separately. But V V is also a vector space and one might× wonder whether a bilinear form on V is also a linear× 2.2. BILINEAR FORMS 37 form on the vector space V V . But this is not the case. For example consider the case in which V = R, so that×V V = R2 and let ϕ : R R R be a function: × × → (1) If ϕ(x, y) := 2x y, then ϕ is not a bilinear form on R, but is a linear form on (x, y) R2; − (2) If ϕ(x,∈ y) := 2xy, then ϕ is a bilinear form on R (hence linear in x R and linear in y R), but it is not a linear form on R2, as it is not linear∈ in (x, y) R2. ∈ ∈ So a bilinear form is not a form that it is “twice as linear” as a linear form, but a form that is defined on the product of twice the vector space.  Exercise 2.21. Verify the above assertions in Remark 2.20 to make sure you understand the difference.

2.2.2. Tensor Product of Two Linear Forms on V .

Let α, β V ∗ be two linear forms, α, β : V R, and define ϕ : V V R, by ∈ → × → ϕ(v,w) := α(v)β(w) . Then ϕ is bilinear, is called the tensor product of α and β and is denoted by ϕ = α β . ⊗ Note 2.22. In general α β = β α, as there could be vectors v and w such that α(v)β(w) = β(v)α(w). ⊗ 6 ⊗ 6 Example . R 4 2.23 Let V = [x]2, let α(p)= p(2) p′(2) and β(p)= 3 p(x)dx be two linear forms. Then − 4 R (α β)(p, q)=(p(2) p′(2)) q(x)dx ⊗ − Z3 is a bilinear form.  Remark 2.24. The bilinear form ϕ : R R R defined by the formula ϕ(x, y) := 2xy is the tensor product of two linear forms× on→R, for instance, ϕ(x, y)=(α α)(x, y) ⊗ where α : R R is the linear form given by α(x) := √2x. On the other→ hand, not every bilinear form is simply the tensor product of two linear forms. As we will see below, the first such examples are found for bilinear forms on vector spaces of dimension at least 2. 

2.2.3. A Basis for Bilinear Forms.

Let Bil(V V, R) := all bilinear forms ϕ : V V R . × { × → } 38 2. MULTILINEAR FORMS

Exercise 2.25. Check that Bil(V V, R) is a vector space with the zero element equal to the bilinear form identically equal× to zero. Hint: It is enough to check that if ϕ, ψ Bil(V V, R), and λ,µ R, then λϕ + µψ Bil(V V, R). Why? (Recall Example∈ 1.3(3) on× page 8 and Exercise∈ 2.7 on page 29.)∈ × Assuming Exercise 2.25, we are going to find a basis of Bil(V V, R) and determine × 1 n its dimension. Let = b1,...,bn be a basis of V and let ∗ = β ,...,β be the B { i }i B { } dual basis of V ∗ (that is β (bj)= δj). Proposition 2.26. The bilinear forms βi βj, i, j =1,...,n form a basis of Bil(V V, R). As a consequence, dim Bil(V V, R⊗)= n2. × × Notation. We denote

Bil(V V, R)= V ∗ V ∗ × ⊗ and call this vector space the tensor product of V ∗ and V ∗. A justification for this notation will appear in §4.3.2. Remark 2.27. Just as it is for linear forms, to verify that two bilinear forms on V are the same it is enough to verify that they are the same on every pair of elements of a basis of V . In fact, let ϕ, ψ be two bilinear forms, let = b1,...,bn be a basis of V , and assume that B { } ϕ(bi, bj)= ψ(bi, bj) i j for all 1 i, j, n. Let v = v bi,w = w bj V be arbitrary vectors. We now verify that ϕ(v,w≤ )= ≤ψ(v,w). Because of the linearity∈ in each variable, we have i j i j i j i j ϕ(v,w)= ϕ(v bi,w bj)= v w ϕ(bi, bj)= v w ψ(bi, bj)= ψ(v bi,w bj)= ψ(v,w) . 

Proof of Proposition 2.26. The proof will be similar to the one of Proposi- tion 2.9 for linear forms. We first check that the set of bilinear forms βi βj, i, j = 1,...,n consists of linearly independent vectors, then that it spans Bil({ V ⊗ V, R). For the} linear independence we need to check that the only linear combination× of the βi βj that gives the zero bilinear form is the trivial linear combination. Let c βi ⊗βj =0 be a linear combination of the βi βj. Then for all pairs of basis vectors ij ⊗ ⊗ (bk, bℓ), with k,ℓ =1,...,n, we have 0= c βi βj(b , b )= c δi δj = c , ij ⊗ k ℓ ij k ℓ kℓ thus showing the linear independence. To check that span βi βj, i, j = 1,...,n = Bil(V V, R), we need to check that if ϕ Bil(V V, R{), there⊗ exists B R such} that × ∈ × ij ∈ ϕ = B βi βj . ij ⊗ Because of (2.2) on page 28, we obtain i j i j ϕ(bk, bℓ)= Bijβ (bk)β (bℓ)= Bijδkδℓ = Bkℓ , 2.2. BILINEAR FORMS 39 for every pair (bk, bℓ) V V . Hence, we set Bkℓ := ϕ(bk, bℓ). Now both ϕ and i j ∈ × ϕ(bk, bℓ)β β are bilinear forms and they coincide on . Because of the above Remark 2.27,⊗ the two bilinear forms coincide. B × B 

3 3 Example 2.28. We continue with the study of the scalar triple product ϕu : R R u1 × → R, that was defined in Example 2.16 for a fixed given vector u = u2 . We now want u3 3 to find the components Bij of ϕu with respect to the standard basis ofR . Recall the cross product in R3 is defined on the elements of the standard basis by 0 if i = j e e := e if (i, j, k) is a cyclic permutation of (1, 2, 3) i × j  k  ek if (i, j, k) is a noncyclic permutation of (1, 2, 3) , − that is  e e = e e e = e 1 × 2 3 2 × 1 − 3 e2 e3 = e1 and e3 e2 = e1  ×  × − cyclic   e3 e1 = e2 noncyclic e1 e3 = e2 × × − · k Since u ek = u , then  0 if i = j B = ϕ (e , e )= u · (e e )= uk if (i, j, k) is a cyclic permutation of (1, 2, 3) ij u i j i × j   uk if (i, j, k) is a noncyclic permutation of (1, 2, 3) − Thus  B = u3 = B 12 − 21 B = u2 = B 31 − 13 B = u1 = B 23 − 32 B11 = B22 = B33 =0 (that is, the diagonal components are zero) , which can be written as a matrix 0 u3 u2 B = u3 0 −u1 . −u2 u1 0  −   The components Bij of B are the components of this bilinear form with respect to the basis βi βj (i, j =1,...,n), where βi(e )= δi . Hence, we can write ⊗ k k ϕ = B βi βj = u β2 β3 β3 β2 u ij ⊗ 1 ⊗ − ⊗ +u β3 β1 β1 β3 + u β1 β2 β2 β1 . 2 ⊗ − ⊗
3 ⊗ − ⊗

 40 2. MULTILINEAR FORMS

2.2.4. Covariance of Bilinear Forms.

We have seen that, once we choose a basis = b1,...,bn of V , we automatically 1 n B {i j } have a basis ∗ = β ,...,β of V ∗ and a basis β β , i, j =1,...,n of V ∗ V ∗. This implies,B that any{ bilinear form} ϕ : V V R{can⊗ be represented by its} components⊗ × → (2.13) Bij = ϕ(bi, bj) , in the sense that ϕ = B βi βj . ij ⊗ Moreover, these components can be arranged in a matrix6

B11 ... B1n . . B :=  . .  B ... B  n1 nn called the matrix of the bilinear form ϕ with respect to the chosen basis . The natural question of course is: how does the matrix B change when we choose a differentB basis of V ? ˜ ˜ So, let us choose a different basis := b1,..., bn and corresponding bases ∗ = 1 n i j B { } B β˜ ,..., β˜ of V ∗ and β˜ β˜ , i, j =1,...,n of V ∗ V ∗, with respect to which ϕ { } { ⊗ } ⊗ ˜ ˜ will be represented by a matrix B, whosee entries are Bij = ϕ(bi, bj). e To see the relation between B and B, due to the change of basis from to , we B B start with the matrix of the changee of basis L := L e e, according to which BB ˜ e i e (2.14) bj = Ljbi . Then ˜ ˜ k ℓ k ℓ k ℓ Bij = ϕ(bi, bj)= ϕ(Li bk, Ljbℓ)= Li Ljϕ(bk, bℓ)= Li LjBkℓ , where the first and the last equality follow from (2.13), the second from (2.14) (after having renamede the dummy indices to avoid conflicts) and the remaining one from the bilinearity of σ. We conclude that

k ℓ Bij = Li LjBkℓ . Exercise . 2.29 Show that the formulae of the transformation of the component of a bilinear form in terms of the matrices of the change of coordinates is (2.15) B = tLBL , where tL denotes the transpose of the matrix L. e 6Note that, contrary to the matrix that gives the change of coordinates between two basis of the vector space, here we have only lower indices. This is not by chance and reflects the type of tensor a bilinear form is. 2.3. MULTILINEAR FORMS 41

We hence say that a bilinear form ϕ is a covariant 2-tensor or a tensor of type (0, 2).

2.3. Multilinear Forms

2.3.1. Definition, Basis and Covariance.

We saw in §2.1.3 that linear forms are covariant 1-tensors – or tensors of type (0, 1) – and in §2.2.4 that bilinear forms are covariant 2-tensors – or tensors of type (0, 2). Analogously to what was done until now, one can define trilinear forms on V , that is functions T : V V V R that are linear with respect to each of the three arguments. The space× of× trilinear→ forms on V is denoted

V ∗ V ∗ V ∗ , ⊗ ⊗ has basis βj βj βk, i, j, k =1,...,n { ⊗ ⊗ } and, hence, has dimension n3. The tensor product is defined as above. Since the components of a trilinear form T : V ⊗ V V R satisfy the following transformation with respect to a change of basis × × → ℓ p q Tijk = Li Lj LkTℓpq , a trilinear form is a covariant 3-tensor or a tensor of type (0, 3). Of course, there is nothing speciale about k =1, 2 or 3: Definition 2.30. A k-linear form or multilinear form of order k on V is a function f : V V R from k-copies of V into R, that is linear in each of its arguments. ×···× → A k-linear form is a covariant k-tensor (or a covariant tensor of order k or a tensor of type (0,k)). The vectors space of k-linear forms on V , denoted

V ∗ V ∗ , ⊗···⊗ k factors has basis | {z } i1 i2 ik β β β , i1,...,ik =1,...,n ⊗ ⊗···⊗k and, hence, dim(V ∗ V ∗)= n . ⊗···⊗

2.3.2. Examples of Multilinear Forms.

Example 2.31. We once more address the scalar triple product,7 discussed in Exam- ples 2.16 and 2.28. This time we want to find the components Bij of ϕu with respect

7The scalar triple product is called Spatprodukt in German. 42 2. MULTILINEAR FORMS to the (nonstandard) basis 0 1 0 := 1 , 0 , 0 . B ( 0 1 1 ) e       ˜b1 ˜b2 ˜b3 The matrix of the change of coordinates|{z} from|{z} the|{z} standard basis to is B 0 1 0 L = 1 0 0 , e 0 1 1 so that   0 1 0 0 u3 u2 0 1 0 B = 1 0 1 u3 0 −u1 1 0 0 0 0 1 −u2 u1 0  0 1 1 − e  tL   B   L  3 2 2 1 3 1 | 0{z 1 0 } | u {z u }u| {z }0 u u u = 1 0 1 0 u1− u3 −u1 = u3 u1 −0 u2 . 0 0 1  u1 u−2 0   −u1 u2 −0  − −  tL   BL    It is easy to| check{z that} |B is antisymmetric{z just} like B is, and to check that the compo- nents of B are correct by using the formula for ϕ. In fact e ˜ ˜ 1 3 B12 = ϕ(b1, b2)= u · (e2 (e1 + e3)) = u u e × − B = ϕ(˜b , ˜b )= u · ((e ) e )= u1 e13 1 3 2 × 3 B = ϕ(˜b , ˜b )= u · ((e + e ) e )= u2 e23 2 3 1 3 × 3 − B = ϕ(˜b , b )= u · (e e )=0 e11 1 1 2 × 2 B = ϕ(˜b , b )= u · ((e + e ) (e + e ))=0 e22 2 2 1 3 × 1 3 B = ϕ(˜b , b )= u · (e e )=0 e33 3 3 3 × 3  e Example 2.32. If, in the definition of the scalar triple product, instead of fixing a vector a R, we let the vector vary, we have a function ϕ : R3 R3 R3 R, defined by ∈ × × → u ϕ(u,v,w) := u · (v w) = det v . × w One can verify that such function is trilinear, that is linea r in each of the three variables separately. 2.3. MULTILINEAR FORMS 43

The components Tijk of this trilinear form are simply given by the sign of the corre- sponding permutation: ϕ = sign(i, j, k)βi βj βk = β1 β2 β3 β1 β3 β2 + β3 β1 β2 ⊗ ⊗ ⊗ ⊗ − ⊗ ⊗ ⊗ ⊗ β3 β2 β1 + β2 β3 β1 β2 β1 β3 , − ⊗ ⊗ ⊗ ⊗ − ⊗ ⊗ where the sign of the permutation is given by +1 if (i, j, k)=(1, 2, 3), (2, 3, 1) or (3, 1, 2) (even permutations of (1, 2, 3))  sign(i, j, k) :=  1 if (i, j, k)=(1, 3, 2), (2, 1, 3) or (3, 2, 1) −  (odd permutations of (1, 2, 3)) 0 otherwise.    

Example 2.33. In general, the determinant defines an n-linear form in Rn by

v1 Rn Rn R . ϕ : ... , ϕ(v1,...,vn) := det . , × × −→   n factors vn   where we compute| the determinant{z } of the square matrix with rows (equivalently, columns) given by the n vectors. Multilinearity is a fundamental property of the determinant. In this case, the components of this multilinear form are also given by the permutation signs: ϕ = sign(i ,...,i )βi1 ... βin , 1 n ⊗ ⊗ where

+1 if (i1,...,in) is an even permutation of (1,...,n)

sign(i1,...,in) := 1 if (i ,...,i ) is an odd permutation of (1,...,n) − 1 n 0 otherwise.

A permutation of (1, 2,...,n ) is called an even permutation, if it is obtained from (1, 2,...,n) by an even number of two-element swaps; otherwise it is called an odd permutation. 

2.3.3. Tensor Product for Multilinear Forms.

Let T : V V R and U : V V R ×···× → ×···× → k times ℓ times | {z } | {z } 44 2. MULTILINEAR FORMS be, respectively, a k-linear and an ℓ-linear form. Then the tensor product of T and U is the function T U : V V R ⊗ ×···× → k+ℓ times defined by | {z } T U(v ,...,v ) := T (v ,...,v )U(v ,...,v ). ⊗ 1 k+ℓ 1 k k+1 k+ℓ This is a (k + ℓ)-linear form. Equivalently, this is saying that the tensor product of a tensor of type (0,k) and a tensor of type (0,ℓ) is a tensor of type (0,k + ℓ). Later we will see how this product extends to more general tensors. CHAPTER 3

Inner Products

3.1. Definitions and First Properties Inner products are a special case of bilinear forms. They add an important structure to a vector space, as for example they allow to compute the length of a vector and they provide a canonical identification between the vector space V and its dual V ∗.

3.1.1. Inner Products and their Related Notions.

Definition 3.1. An inner product g : V V R on a real vector space V is a bilinear form on V that is × → (1) symmetric, that is g(v,w)= g(w, v) for all v,w V and (2) positive definite, that is g(v, v) 0 for all v V∈, and g(v)=0 if and only if v =0. ≥ ∈

Exercise 3.2. Let V = R3. Verify that the dot product ϕ(v,w) := v · w, defined as v · w = viwi , v1 w1 where v = v2 and w = w2 is an inner product. This is called the standard inner v3 w3 product.     Exercise 3.3. Determine whether the following bilinear forms ϕ : Rn Rn R are inner products, by verifying whether they are symmetric and positive definite× → (the formulas are troughout defined for all v,w Rn): ∈ (1) ϕ(v,w) := v · w; (2) ϕ(v,w) := −v · w +2v1w2; (3) ϕ(v,w) := v1w1; (4) ϕ(v,w) := v · w 2v1w1; (5) ϕ(v,w) := v · w −+2v1w1; (6) ϕ(v,w) := v · 3w.

Exercise 3.4. Let V := R[x]2 be the vector space of polynomials of degree 2. Determine whether the following bilinear forms are inner products, by verifying whether≤ they are symmetric and positive definite: 45 46 3. INNER PRODUCTS

1 (1) ϕ(p, q) := 0 p(x)q(x)dx; (2) ϕ(p, q) := 1 p (x)q (x)dx; R0 ′ ′ (3) ϕ(p, q) := π exp(x)q(x)dx; R3 (4) ϕ(p, q) := p(1)q(1) + p(2)q(2); (5) ϕ(p, q) := pR(1)q(1) + p(2)q(2) + p(3)q(3). (6) ϕ(p, q) := p(1)q(2) + p(2)q(3) + p(3)q(1).

Definition 3.5. Let g : V V R be an inner product on V . × → (1) The norm v of a vector v V is defined as k k ∈ v := g(v, v) . k k (2) A vector v V is unit vector pif v =1; (3) Two vectors∈v,w V are orthogonalk k (that is, perpendicular denoted v w), if g(v,w)=0; ∈ ⊥ (4) Two vectors v,w V are orthonormal if they are orthogonal and v = w =1; ∈ k k k k (5) A basis of V is an orthonormal basis if b1,...,bn are pairwise orthonormal vectors,B that is

1 if i = j (3.1) g(b , b )= δ := i j ij 0 if i = j , ( 6 for all i, j = 1 ...,n. The condition for i = j implies that an orthonormal basis consists of unit vectors, while the one for i = j implies that it consists of pairwise orthogonal vectors. 6

Example 3.6. (1) Let V = Rn and g the standard inner product. The standard basis = B e1,...,en is an orthonormal basis with respect to the standard inner product. { R} 1 (2) Let V = [x]2 and let g(p, q) := 1 p(x)q(x)dx. Check that the basis − = p ,pR ,p , B { 1 2 3} where

p (x) := 1 , p (x) := 3 x, p (x) := 5 (3x2 1) , 1 √2 2 2 3 8 − q q is an orthonormal basis with respect to the inner product g. Up to scaling, p1,p2,p3 are the so-called first three Legendre polynomials.  3.1. DEFINITIONS AND FIRST PROPERTIES 47

An inner product g on a vector space V induces a metric8 on V , where the distance between vectors v,w V is given by ∈ d(v,w) := v w . k − k

3.1.2. Symmetric Matrices and Quadratic Forms.

t Recall that a matrix S Mn n(R) is symmetric if S = S, that is if ∈ × . a b .. ∗ .  a c ..  S = ∗ .  .  b c ..   ∗   . .  .. ..   ∗  Moreover, if S is symmetric, then   (1) S is positive definite if tvSv > 0 for all v Rn 0 ; (2) S is negative definite if tvSv < 0 for all v∈ Rn\{ 0} ; (3) S is positive semidefinite if tvSv 0 for all∈ v \{Rn;} (4) S is positive semidefinite if tvSv ≥ 0 for all v ∈ Rn; (5) S is indefinite if vtSv takes both≤ positive and∈ negative values for different v Rn. ∈ Definition 3.7. A quadratic form Q : Rn R is a homogeneous quadratic polyno- mial in n variables: → Q(x1,...,xn)= Q xixj , where Q R . ij ij ∈ n To any symmetric matrix S corresponds a quadratic form QS : R R defined by → v1 t 1 n . i j (3.2) QS (v)= vSv = v ... v S  .  = v v Sij . n v Einstein notation       matrix notation | {z } Note that Q is not linear in v. | {z } Let S be a symmetric matrix and QS be the corresponding quadratic form. The notion of positive definiteness, etc. for S can be translated into corresponding properties for QS, namely: (1) Q is positive definite if Q(v) > 0 for all v Rn 0 ; (2) Q is negative definite if Q(v) < 0 for all v∈ Rn\{ 0} ; ∈ \{ } 8An inner product induces a norm and a norm induces a metric on a vector space. However, the converses do not hold. 48 3. INNER PRODUCTS

(3) Q is positive semidefinite if Q(v) 0 for all v Rn; (4) Q is negative semidefinite if Q(v)≥ 0 for all v∈ Rn; (5) Q is indefinite if Q(v) takes both positive≤ and negative∈ values. Example 3.8. We consider R2 with the standard basis and the quadratic form9 v1 E Q(v) := v1v1 v2v2, where v = . The symmetric matrix corresponding to Q is − v2   1 0 v1 0 S := . If v = , then Q(v)=1 > 0, if v = , then Q(v) = 1 < 0, 0 1 0 v2 −  −      but any vector for which v1 = v2 has the property that Q(v)=0.  To find out the type of a symmetric matrix S (or, equivalently of a quadratic form QS) it is enough to look at the eigenvalues of S, namely:

(1) S and QS are positive definite when all eigenvalues of S are positive: (2) S and QS are negative definite when all eigenvalues of S are negative; (3) S and QS are positive semidefinite when all eigenvalues of S are non-negative; (4) S and QS are negative semidefinite when all eigenvalues of S are non-positive; (5) S and QS are indefinite when S has both positive and negative eigenvalues. The reason this makes sense is the same reason for which we need to restrict our attention to symmetric matrices and lies in the so-called Spectral Theorem: Theorem 3.9. [Spectral Theorem] Any n n symmetric matrix S has the following properties: × (1) it has only real eigenvalues; (2) it is diagonalizable; (3) it admits an orthonormal eigenbasis, that is, a basis b ,...,b of Rn such { 1 n} that the bj are orthonormal and are eigenvectors of S.

3.1.3. Inner Products vs. Symmetric Positive Definite Matrices.

Let = b1,...,bn be a basis of V and g an inner product. The components of g withB respect{ to are} B (3.3) gij := g(bi, bj) .

Let G be the matrix with entries gij

g11 ... g1n . . . (3.4) G =  . .. .  . g ... g  n1 nn We claim that G is symmetric and positive definite. In fact:

9Here, we avoid the usual notation for squares, because of the possible confusion with upper indices. 3.1. DEFINITIONS AND FIRST PROPERTIES 49

(1) Since g is symmetric, then for 1 i, j n, ≤ ≤ (3.5) g = g(b , b )= g(b , b )= g = G is a symmetric matrix; ij i j j i ji ⇒ (2) Since g is positive definite, then G is positive definite as a symmetric matrix. i j In fact, let v = v bi,w = w bj V be two vectors. Then, using the bilinearity of g, the definition (3.3) and Einstein∈ notation, we have: i j i j i j g(v,w)= g(v bi,w bj)= v w g(bi, bj) = v w gij

gij or, in matrix notation, | {z } w1 g(v,w)= t[v] G[w] = v1 ... vn G . . B B  .  wn       Conversely, if S is a symmetric positive definite matrix and = b1,...,bn is a basis of V , then the assignment B { } V V R , (v,w) t[v] S[w] × −→ 7−→ B B defines a map that is seen to be bilinear, symmetric and positive definite, hence an inner product.

3.1.4. Orthonormal Bases.

Suppose that there is a basis = b1,...,bn of V consisting of orthonormal vectors with respect to g, so that B { }

gij = δij ; cf. Definitions 3.5(5) and (3.3). In other words the symmetric matrix corresponding to the inner product g in the basis consisting of orthonormal vectors is the identity matrix. Moreover i j i j i i g(v,w)= v w gij = v w δij = v w , so that, if v = w, v 2 = g(v, v)= vivi = v1v1 + + vnvn . k k ··· We thus deduce that: Fact 3.10. Any inner product g can be expressed in the standard form g(v,w)= viwi , v1 w1 . . as long as [v] = . and [w] = . are the coordinates of v and w with respect B   B   vn wn to an orthonormal basis for g.   B 50 3. INNER PRODUCTS

Example 3.11. Let g be an inner product of R3 with respect to which 1 1 1 := 0 , 1 , 1 B ( 0 0 1 ) e       ˜b1 ˜b2 ˜b3 is an orthonormal basis. We want to|{z express} |{z}g |with{z} respect to the standard basis of R3 E 1 0 0 := 0 , 1 , 0 . E ( 0 0 1 )

e1 e2 e3 The matrices of the change of basis| are{z} |{z} |{z} 1 1 1 1 1 0 1 − L := L e = 0 1 1 and Λ= L− = 0 1 1 . BE 0 0 1 0 0− 1      Since g is a bilinear form, we saw in (2.15) that its matrices with respect to the bases and are related by the formula B E G = tLGL . Since the basis is orthonormal with respect to g, the associated matrix G is the identity e matrix, so thatB e t t e G = ΛGΛ= ΛΛ (3.6) 1 00 1 1 0 1 1 0 = e1 1 0 0− 1 1 = 1− 2 1 . −0 1 1 0 0− 1  −0 1− 2  − − It follows that, with respect to the  standard basis, gis given by  1 1 0 w1 g(v,w)= v1 v2 v3 1− 2 1 w2 −0 1− 2  w3 (3.7)
− = v1w1  v1w2 v2w1+2v2w2 − − v2w3 v3w2 +2v3w3 . − − 

Exercise 3.12. Verify the formula (3.7) for the inner product g in the coordinates of the basis by applying the matrix of the change of coordinate directly on the coordinates vectors [vB] . E e 3.1. DEFINITIONS AND FIRST PROPERTIES 51

Remark 3.13. The norm and the value of the inner product of vectors depend only on the choice of g, but not on the choice of basis: different coordinate expressions yield the same result: t t g(v,w)= [v] G[v] = [v] eG[v] e B B B B Example 3.14. We verify the assertion of the previous remark with the inner product e in Example 3.11. Let v,w R3 such that ∈ v1 3 v˜1 v1 1 2 2 1 2 [v] = v = 2 and [v] e = v˜ = L− v = 1 E v3 1 B v˜3 v3 1           and w1 1 w˜1 w1 1 2 2 1 2 − [w] = w = 2 and [w] e = w˜ = L− w = 1 . E w3 3 B w˜3 w3 −3            Then with respect to the basis we have that B g(v,w)=1 ( 1)+1 ( 1)+1 3=1 , e· − · − · and also with respect to the basis E g(v,w)=3 1 3 2 2 1+2 2 2 2 3 1 2+2 1 3=1 . · − · − · · · − · − · · · 

Exercise 3.15. Verify that v = √3 and w = √11, when computed with respect to either basis. k k k k

i Let = b1,...,bn be an orthonormal basis and let v = v bi be a vector in V . Then theB coordinates{ vi }of v can be obtained from the inner product g and the elements of the basis. In fact, we have

i i i j g(v, bj)= g(v bi, bj)= v g(bi, bj)= v δij = v , that is, the coordinates of a vector with respect to an orthonormal basis are the inner product of the vector with the basis vectors. This is particularly nice, so that we have to make sure that we remember how to construct an orthonormal basis from a given arbitrary basis. Recall. The Gram-Schmidt orthogonalization process is a recursive process that allows us to obtain an orthonormal basis starting from an arbitrary one. Let = B b1,...,bn be an arbitrary basis, let g : V V R be an inner product and the {corresponding} norm. × → k·k We start by recalling that the orthogonal projection of a vector V onto bk is defined as ∈ 52 3. INNER PRODUCTS

g(v, bk) (3.8) proj v = bk . bk v g(bk, bk)

bk

projbk v

In fact, projbk v is clearly parallel to bk and the following exercises shows that the component v proj v is orthogonal to b . − bk k Exercise . 3.16 With projbk v defined as in (3.8), check that we have v proj v b , − bk ⊥ k where the orthogonality is meant with respect to the inner product g.

Given the basis = b1,...,bn of V , we will find an orthonormal basis. We start by defining B { } 1 u := b . 1 b 1 k 1k Next, observe that g(b2,u1)u1 is the projection of the vector b2 in the direction of u1. It follows that

b⊥ := b g(b ,u )u 2 2 − 2 1 1 is a vector orthogonal to u1, but not necessarily of unit norm. Hence we set 1 u2 := b2⊥ . b⊥ k 2 k Likewise g(b3,u1)u1 +(b3,u2)u2 is the projection of b3 on the plane generated by u1 and u2, so that

b⊥ := b g(b ,u )u g(b ,u )u 3 3 − 3 1 1 − 3 2 2 is orthogonal both to u1 and to u2. Set 1 u := b⊥ . 3 b 3 k 3⊥k Continuing until we have exhausted all elements of the basis , we obtain an orthonormal basis u ,...,u . B { 1 n} b3 b3 b3

u3 b2 b2 u2 u2

b1 u1 u1 u1 3.1. DEFINITIONS AND FIRST PROPERTIES 53

Example 3.17. Let V be the subspace of R4 spanned by 1 2 1 1 2 1 b = b = b = . 1  1 2 0 3 1 −  1 0 0 −            (One can check that b1, b2, b3 are linearly independent and hence form a basis of V .) We look for an orthonormal basis of V with respect to the standard inner product , . Since h· ·i b = (11 +12 +( 1)2 +( 1)2)1/2 =2 , k 1k − − we find 1 u := b . 1 2 1 Moreover, 1 1 1 b2,u1 = (1+1)=2 = b⊥ := b2 b2,u1 u1 =   , h i 2 ⇒ 2 −h i 1 1     so that b⊥ =2 and k 2 k 1 1 1 u2 :=   . 2 1 1   Finally,   b ,u = 1 (1+1 1) = 1 and b ,u = 1 (1+1+1)= 3 h 3 1i 2 − 2 h 3 2i 2 2 imply that 0 0 b3⊥ := b3 b3,u1 u1 b3,u2 u2 =  1  . −h i −h i 2  1  − 2    √2 Since b⊥ = , we have k 3 k 2 0 √2 0 u3 :=   . 2 1  1 −     54 3. INNER PRODUCTS

3.2. Reciprocal Basis

3.2.1. Definition and Examples.

Let g : V V R be an inner product and = b1,...,bn any basis of V . From g and we can× define→ another basis of V , denotedB by{ } B g = b1,...,bn B { } and satisfying

i i (3.9) g(b , bj)= δj . The basis g is called the reciprocal basis of V with respect to g and . Note that,B strictly speaking, we are very imprecise here. Indeed, whileB it is certainly possible to define a set of n = dim V vectors as in (3.9), we should justify the fact that we call it a basis. This will be done in Claim 3.21. Remark 3.18. In general, g = and in fact, because of Definition 3.5(5), B 6 B = g is an orthonormal basis. B B ⇐⇒ B Example 3.19. Let g be the inner product in (3.7) in Example 3.11 and let the standard basis of R3. We want to find the reciprocal basis g, that is we want toE find g := e1, e2, e3 such that E E { } i i g(e , ej)= δj . If G is the matrix of the inner product in (3.6), using the matrix notation and considering j e as a row vector and ei as a column vector for i, j =1, 2, 3, we have

–– tei –– G e| = δi .  j j   | Letting i and j vary from 1 to 3, we obtain   –– te1 –– 1 1 0 1 0 0 t 2 − | | | –– e –– 1 2 1 e1 e2 e3 = 0 1 0 , –– te3 –– −0 1− 2    0 0 1 − | | | from which we conclude that      t 1 1 –– e1 –– − 1 1 0 − t 2 | | | − –– e –– = e1 e2 e3 1 2 1 –– te3 ––   −0 1− 2  | | | −     3 2 1 3 2 1 | | | = e1 e2 e3 2 2 1 = 2 2 1 .   1 1 1 1 1 1 | | |       3.2. RECIPROCAL BASIS 55

Therefore, 3 2 1 (3.10) e1 = 2 , e2 = 2 , e3 = 1 . 1 1 1

  1     Observe that in order to compute G− we used the Gauss–Jordan elimination method 1 1 0 1 0 0 1 1 0 1 0 0 1− 2 1 0 1 0 ! 0− 1 1 1 1 0 −0 1− 2 0 0 1 0 1− 2 0 0 1 − −   1 0 1 2 1 0  ! 0 1 −1 1 1 0 00− 1 1 1 1 1 0 0 3 2 1  ! 0 1 0 2 2 1 0 0 1 1 1 1   

Exercise 3.20. We put ourselves in the situation of Examples 3.19 and 3.11. Namely, let g be an inner product on R3, let = e , e , e be the standard basis and let E { 1 2 3} 1 1 1 := 0 , 1 , 1 B ( 0 0 1 ) e       ˜b1 ˜b2 ˜b3 be an orthonormal basis with respect|{z to}g.|{z} |{z}

(1) Compute [˜b1] e, [˜b2] e and [˜b3] e. B B B (2) Compute the matrix G e associated to g with respect to the basis , and the matrix G associated toB g with respect to the basis ,. B E E (3) We denote by g = e1, e2, e3 and g = ˜b1, ˜b2, ˜b3 the reciprocale bases E { } B { } respectively of and . By Remark 3.13, E B e g(˜bi, ˜b )= δi and g(ei, e )= δi j e j j j are independent of the choice of the basis. It follows that: i ˜i ˜ t˜i ˜ (a) δj = g(b , bj) = [ b ] eG e[bj] e; i ˜i ˜ t˜i B B ˜ B (b) δj = g(b , bj) = [ b ] G [bj] ; i i t i E E E (c) δj = g(e , ej) = [ e ] eG e[ej] e; i i t i B B B (d) δj = g(e , ej) = [ e ] G [ej] . E E E Using (1), (2) and the appropriate formula among (a), (b), (c) and (d), compute i i i i [˜b ] e, [˜b ] , [e ] and [e ] e. For some of these, you will probably want to apply theB sameE techniqueE as inB Example 3.19. 56 3. INNER PRODUCTS

3.2.2. Properties of Reciprocal Bases. Claim 3.21. Given a vector space V with a basis and an inner product g : V V R, a reciprocal basis exists and is unique. B × → As we pointed out right after the definition of reciprocal basis, what this claim really says is that there is a set of vectors b1,...,bn in V that satisfy (3.9), that form a basis and that this basis is unique. { }

1 n Proof. Let = b1,...,bn be the given basis. Any other basis b ,...,b is related to by theB relation{ } { } B i ij (3.11) b = M bj for some invertible matrix M. We want to show that there exists a unique matrix M i such that, when (3.11) is plugged into g(b , bj), we have i i (3.12) g(b , bj)= δj . From (3.11) and (3.12) we obtain

i i ik ik ik δj = g(b , bj)= g(M bk, bj)= M g(bk, bj)= M gkj , which, in matrix notation becomes I = MG, where G is the matrix of g with respect to whose entries are gij as in (3.4). Since G B 1 is invertible because it is positive definite, then M = G− exists and is unique.  Remark 3.22. Note that in the course of the proof we have found that, since M = L g , then B B 1 G =(L g )− = L g . B B BB ij 1 We denote with g the entries of M = G− . From the above discussion, it follows that with this notation

ik i (3.13) g gkj = δj as well as

i ij (3.14) b = g bj , or10

1 n 1 (3.15) b ... b = b1 ... bn G− .

10Note that the following, like previously remarked, is a purely symbolic expression that has the only advantage of encoding the n expressions in (3.15). 3.2. RECIPROCAL BASIS 57

1 (you need to understand why G− has to be multiplied on the right). Note that this is consistent with the findings in §1.3.2. We can now compute g(bi, bj)

i j (3.14) ik jℓ ik jℓ g(b , b ) = g(g bk,g bℓ)= g g g(bk, bℓ)

ik jℓ (3.5) ik jℓ (3.13) ik j ij = g g gkℓ = g g gℓk = g δk = g , where we used in the second equality the bilinearity of g. Thus, similarly to (3.1), we have (3.16) gij = g(bi, bj) .

Exercise 3.23. In the setting of Exercise 3.20, verify (3.15) in the particular cases of and g and of and g, that is verify that E E 1 2 B 3 B 1 (1) e e e = e1 e2 e3 G− , and ˜1 ˜ e˜ e˜ ˜ ˜ E1 (2) b b2 b3 = b1 b2 b3 G−e .

B Recall in fact that in
(3.15), because
of the way it was obtained, G is the matrix of g with respect to the basis . B Given that we just proved that reciprocal bases are unique, we can talk about the reciprocal basis (of a fixed vector space V associated to a basis and an inner product). Claim 3.24. The reciprocal basis is contravariant.

Proof. Let and be two bases of V and L := L e be the corresponding matrix B B 1 BB of the change of basis, with Λ= L− . Recall that this means that e ˜ j bi = Li bj . We have to check that if g = b1,...,bn is a reciprocal basis for , then the basis B { } B ˜b1,..., ˜bn defined by { } ˜i i k (3.17) b =Λkb is a reciprocal basis for . Then the assertion will be proven, since ˜b1,..., ˜bn is contravariant by construction.B { } To check that ˜b1,...,e˜bn is the reciprocal basis, we need with check that with the { } choice of ˜bi as in (3.17), the property (3.1) of the reciprocal basis is verified, namely that ˜i ˜ i g(b , bj)= δj . But in fact, ˜i ˜ (3.17) i k ℓ i ℓ k (3.12) i ℓ k i k i g(b , bj) = g(Λkb , Ljbℓ)=ΛkLjg(b , bℓ) = ΛkLjδℓ =ΛkLj = δj , where the second equality comes from the bilinearity of g, the third from the property 1 (3.9) defining reciprocal basis and the last from the fact that Λ= L− .  58 3. INNER PRODUCTS

Suppose now that V is a vector space with a basis and that g is the reciprocal basis of V with respect to and to a fixed inner productB g : V VB R. Then there are two ways of writing a vectorB v V , namely × → ∈ i j v = v bi = vjb .

with respect to with respect to g B B Recall that the (ordinary) coordinates|{z} of v with|{z} respect to are contravariant (see Example 0.2). B Claim 3.25. Vector coordinates with respect to the reciprocal basis are covariant. Proof. This will follow from the fact that the reciprocal basis is contravariant and the idea of the proof is the same as in Claim 3.24. Namely, let , be two bases of V , L := L e the matrix of the change of basis and 1 Bg B g BB j Λ = L− . Let and be the corresponding reciprocal bases and v = vjb a vector with respect toB g.e B B It is enough to checke that the numbers j v˜i := Li vj are the coordinates of v with respect to g, because in fact these coordinates are covariant by definition. But in fact, using thisB and (3.17), we obtain ˜i j i k je i k j v˜ib =(Li vj)(Λkb )= Li Λk vjb = vjb = v

j δk | {z } 

Definition 3.26. The coordinates vi of a vector v V with respect to the reciprocal basis g are called the covariant coordinates of v.∈ B 3.2.3. Change of Basis from a Basis to its Reciprocal Basis g. B B We want to look now at the direct relationship between the covariant and the con- travariant coordinates of a vector v. Recall that we can write i j v bi = v = vjb .

with respect to with respect to g B B from which we obtain |{z} |{z} i j i j i j (v gij)b = v (gijb )= v bi = v = vjb , and hence, by comparing the coefficients of bj, i t (3.18) vj = v gij or [v] g = G[v] ,. B B Likewise, from i j ji ji v bi = v = vjb = vj(g bi)=(vjg )bi , 3.2. RECIPROCAL BASIS 59 follows that

i ji 1t (3.19) v = vjg or [v] = G− [v] g . B B Example 3.27. Let = e , e , e be the standard basis of R3 and let E { 1 2 3} 1 1 0 G = 1− 2 1 −0 1− 2  −   be the matrix of g with respect to . In (3.10) Example 3.19, we saw that E 3 2 1 g = b1 = 2 , b2 = 2 , b3 = 1 E ( 1 1 1 )       4 is the reciprocal basis. We find the covariant coordinates of v = 5 with respect to 6 g using (3.18), namely   E 1 1 0 4 − [v] g = G[v] = 1 2 1 5 = 1 0 7 . E E −0 1− 2  6 − −
The following computation double checks this result:  3 2 1 4 i vib =( 1) 2 +0 2 +7 1 = 5 . − 1 1 1 6         

Example 3.28. Let V := R[x]1 be the vector space of polynomials of degree 1 (that is, “linear” polynomials of the form a + bx). Let g : V V R be defined by≤ × → 1 g(p, q) := p(x)q(x)dx , Z0 and let := 1, x be a basis of V . Determine: B { } (1) the matrix G; 1 (2) the matrix G− ; (3) the reciprocal basis g; (4) the contravariant coordinatesB of p(x)=6x (that is the coordinates of p(x) with respect to ); (5) the covariant coordinatesB of p(x)=6x (that is the coordinates of p(x) with respect to g). B 60 3. INNER PRODUCTS

g 1 n = b1,...,bn = b ,...,b B {basis} B reciprocal{ basis}

i i these bases are related by g(b , bj)= δj

a vector v has now v = vib v = v bi two sets of coordinates i i contravariant covariant coordinates coordinates

ij i j the matrices of g are gij = g(bi, bj) g = g(b , b )

these matrices are inverse gikg = δi of each other, that is, kj j

the basis and the b = g bj bi = gijb reciprocal basis satisfy i ij j

covariant and contravariant vi = gijv v = g vj coordinates are related by j i ij

Table 3. Summary of covariance and contravariance of vector coordinates

(1) The matrix G has entries gij = g(bi, bi), that is 1 1 2 g11 = g(b1, b1)= (b1) dx = dx =1 Z0 Z0 1 1 1 g = g(b , b )= b b dx = x = 12 1 2 1 2 2 Z0 Z0 1 1 g = g(b , b )= b b dx = 21 2 1 2 1 2 Z0 1 1 1 g = g(b , b )= (b )2dx = x2dx = , 22 2 2 2 3 Z0 Z0 3.2. RECIPROCAL BASIS 61 so that 1 1 2 G = 1 1 . 2 3 !

(2) Since det G =1 1 1 1 = 1 , then by using (1.7), we get · 3 − 2 · 2 12 1 4 6 G− = ,. 6− 12 −  (3) Using (3.15), we obtain that

1 2 1 4 6 b b = 1 x G− = 1 x − = 4 6x 6+12x , 6 12 − − −  so that g =
4 6x,
6+12x .

B { − − } 0 (4) p(x)=6x =0 1+6 x, so that p(x) has contravariant coordinates [p(x)] = . · · B 6   (5) From (3.18) it follows that if v = p(x), then v v1 1 1 0 3 1 = G = 2 = . v v2 1 1 6 2  2    2 3      We can check this result: v b1 + v b2 =3 (4 6x)+2 ( 6+12x)=6x . 1 2 · − · − 

3.2.4. Isomorphisms Between a Vector Space and its Dual.

We saw already in Proposition 2.9, that if V is a vector space and V ∗ is its dual, then dim V = dim V ∗. In particular, this means that V and V ∗ can be identified, once we choose a basis of V and a basis ∗ of V ∗. In fact, the basis ∗ of V ∗ is given B B B once we choose the basis of V , as the dual basis of V ∗ with respect to . Then there is the following correspondence:B B

v V ! α V ∗ , ∈ ∈ exactly when v and α have the same coordinates, respectively with respect to and B ∗. However, this correspondence depends on the choice of the basis and hence not Bcanonical. B If however V is endowed with an inner product, then there is a canonical identifi- cation of V with V ∗ that is, an identification that does not depend on the basis of V . In fact, let g : V V R be an inner product and let v V . Then B × → ∈ g(v, ): V R · −→ w g(v,w) 7−→ 62 3. INNER PRODUCTS is a linear form and hence we have the following canonical identification given by the metric V V ∗ (3.20) ←→ v v∗ := g(v, ) . ←→ · Note that the isomorphism11 sends the zero vector to the linear form identically equal to zero, since g(v, ) 0 if and only if v =0 by positive definiteness of g. So far, we have· two≡ bases of the vector space V , namely the basis and the g B reciprocal basis and we have also the dual basis of the dual vector space V ∗. It turns out that, underB the isomorphism (3.20), the reciprocal basis of V and the dual basis of V ∗ correspond to each other. This follows from the fact that, under the isomorphism (3.20) an element of the reciprocal basis bi corresponds to the linear form g(bi, ) · bi g(bi, ) ←→ · and the linear form g(bi, ): V R has the property that · → i i g(b , bj)= δj . We conclude that g(bi, ) βi , · ≡ so under the canonical identification between V and V ∗ the reciprocal basis of V corre- sponds to the dual basis of V ∗.

3.2.5. Geometric Interpretation.

Let g : V V R be an inner product and = b1,...,bn a basis of V with × g → i B { } reciprocal basis . Let v = vib V be a vector written in terms of its covariant coordinates (thatB is, the coordinates∈ with respect to the reciprocal basis). Then i i g(v, bk)= g(vib , bk)= vi g(b , bk) = vk ,

i δk so that (3.8) becomes | {z } vk projbk v = bk . g(bk, bk)

If we assume that the elements of the basis = b1,...,bn are unit vectors, then (3.8) further simplifies to give B { }

(3.21) projbk v = vkbk . This equation shows the following: Fact 3.29. The covariant coordinates of v give the orthogonal projection of v onto b1,...,bn.

11Recall that an isomomorphism T : V W is an invertible linear transformation. → 3.3. RELEVANCE OF COVARIANCE AND CONTRAVARIANCE 63

Likewise, the following holds basically by definition: Fact 3.30. The contravariant coordinates of v give the “parallel” projection of v onto

b1,...,bn.

| | v2

{z v2

v {z

b2

} }

| b1 {z } 1 | v {z } v1

3.3. Relevance of Covariance and Contravariance Why do we need or care for covariant and contravariant components?

3.3.1. Physical Relevance.

Consider the following physical problem: Calculate the work performed by a force F on a particle to move the particle by a small displacement dx, in the Euclidean plane. The work performed should be independent of the choice of the coordinate system (i.e choice of basis) used. For the work to remain independent of choice of basis we will see that, if the components of the displacement change contravariantly, then the components of the force should change covariantly. To see this let = b1, b2 be a basis of the Euclidean plane. Suppose the force B { } 1 2 F =(F1, F2) is exerted on a particle that moves with a displacement dx =(dx , dx ). Then the work done is given by 1 2 dW = F1dx + F2dx . ˜ ˜ Suppose we are given another coordinate system := b1, b2 and let F =(F˜1, F˜2) and dx =(dx˜1,dx˜2). Then B { } 1 2 dW = F˜1dx˜ + F˜2dx˜e. Now assume that the coordinates of dx change contravariantly; i i j dx˜ =Λjdx , or, equivalently, i i j dx = Ljdx˜ , 64 3. INNER PRODUCTS

1 i where Λ= L− and L =(L ) is the change of basis matrix from to . j B B 1 2 e dW =F1dx + F2dx 1 1 1 2 2 1 2 2 =F1[L1dx˜ + L2dx˜ ]+ F2[L1dx˜ + L2dx˜ ] 1 2 1 1 2 2 =(F1L1 + F2L1)dx˜ +(F1L2 + F2L2)dx˜ Since the work performed is independent of basis chosen we also have 1 2 dW = F˜1dx˜ + F˜2dx˜ . This gives that ˜ 1 2 F1 = F1L1 + F2L1 ˜ 1 2 F2 = F1L2 + F2L2 . ˜ j Hence the coordinates of F transform covariantly; Fi = Li Fj. Using matrices this can be written as (F˜1, F˜2)=(F1, F2)L.

3.3.2. Distinction Vanishes when Restricting to Orthonormal Bases.

In the presence of an inner product g : V V R and if we restrict ourselves to orthonormal bases, the distinction between covariance× → and contravariance vanishes! In fact, it all becomes a matter of transposition: writing vectors as columns or as rows. Why is that? First, as we saw, the reciprocal of an orthonormal basis is equal to itself, so covariant and contravariant coordinates are equal in this special case. Moreover, when we change from one orthonormal basis to another by a change of basis matrix L, the inverse change is given simply by the transpose of L. Here is a proof. Let = b ,...,b and = ˜b ,..., ˜b be two orthonormal bases of V and let B { 1 n} B { 1 n} L := L e be the corresponding matrix of the change of basis. This means, as always, that BB e ˜ j bi = Li bj . Since and are orthonormal, we have g(b , b )= δ = g(˜b , ˜b ). Therefore, we have B B i j ij i j k ℓ k ℓ k ℓ k k δij = g(˜bi, ˜bj)= g(L bk, L bℓ)= L L g(bk, bℓ)= L L δkℓ == L L , e i j i j i j i j t 1 t showing that L L = I, that is, L− = L. Such a matrix L is called an orthogonal matrix. We conclude that, in this special case, we see no distinction between covariant and contravariant behaviour, it is simply a matter of transposition. CHAPTER 4

Tensors

4.1. Towards General Tensors Let V be a vector space. Up to now, we saw several objects related to V , which we called tensors. We summarize them in Table 4. From these, we infer the definition of a tensor of type (0, q) for all q N, but we cannot say the same for a tensor of type (p, 0) for all p N, even less of∈ general type (p, q). The next discussion will lead us to tensors of type∈(p, 0), and in the meantime we will discuss an important isomorphism.

4.1.1. Canonical Isomorphism between V and (V ∗)∗.

We saw in §3.2.4 that any vector space is isomorphic to its dual, though in general the isomorphism is not canonical, that is, it depends on the choice of a basis. We also saw that, if there is an inner product on V , then there is a canonical isomorphism. The point of this section is to show that, even without an inner product, there is always a canonical isomorphism between V and its bidual (V ∗)∗, that is the dual of its dual. To see this, let us observe first of all that

(4.1) dim V = dim(V ∗)∗ .

Tensor Components Behavior under a change of basis Type vector vi contravariant tensor (1, 0) in V linear form α covariant tensor (0, 1) V R j linear transformation→ tensor of mixed character: Ai (1, 1) V V j contravariant and covariant bilinear→ form B covariant 2-tensor (0, 2) V V R ij k-linear× → form F covariant k-tensor (0, k) V V R i1i2...ik ×···× → Table 4. Covariance and contravariance of earlier tensors

65 66 4. TENSORS

In fact, for any vector space W , we saw in Proposition 2.9 that dim W = dim W ∗. If we apply this equality both to W = V and to W = V ∗, we obtain

dim V = dim V ∗ and dim V ∗ = dim(V ∗)∗ , from which (4.1) follows immediately. We deduce (for instance, using §3.2.4) that V and (V ∗)∗ are isomorphic, and we only have to see that there is a canonical isomorphism. To this end, observe that a vector v V gives rise to a linear form on V ∗ defined by ∈ ϕ : V ∗ R v −→ α α(v) . 7−→ Then we can define a linear map as follows:

Φ: V (V ∗)∗ (4.2) −→ v ϕ . 7−→ v Since, for any linear map T : V W between vector spaces, we have the dimension formula (known as the rank-nullity→ theorem in Linear Algebra): dim V = dim im (T ) + dim ker(T ) , it will be enough to show that ker Φ = 0 , because then { } dim V = dim im (Φ) = dim(V ∗)∗ , and we can conclude that im (Φ) = (V ∗)∗, hence Φ is an isomorphism. Notice that we have not chosen a basis to define the isomorphism Φ. To see that ker Φ = 0 , observe that this kernel consists of all vectors v V such { } ∈ that α(v)=0 for all α V ∗. We want to see that the only vector v V for which ∈ ∈ this happens is the zero vector. In fact, if v V is nonzero and = b1,...,bn is i ∈ B { j } any basis of V , then we can write v = v bi, where at least one coordinate, say v , is not j zero. In that case, if ∗ = β ,...,β is the dual basis, we have β (v)= v =0, thus B { 1 n} j 6 we have found an element in V ∗ not vanishing on this v. We record this fact as follows:

Fact 4.1. Let V be a vector space and V ∗ its dual. The dual (V ∗)∗ of V ∗ is canonically isomorphic to V . The canonical isomorphism Φ: V (V ∗)∗ takes v V to the linear → ∈ form on the dual ϕ : V ∗ R, ϕ (α) := α(v). v → v

4.1.2. (2, 0)-Tensors.

Recall that the dual of a vector space V is the vector space

V ∗ := linear forms α : V R = (0, 1)-tensors . { → } { } Taking now the dual of the vector space V ∗, we obtain

(V ∗)∗ := linear forms ϕ : V ∗ R . { → } 4.1. TOWARDS GENERAL TENSORS 67

Using the canonical isomorphism Φ: V (V ∗)∗ and the fact that coordinate vectors are contravariant, we conclude that →

linear forms ϕ : V ∗ R =(V ∗)∗ = V = (1, 0)-tensors . { → } ∼ { } So, changing the vector space from V to its dual V ∗ seems to have had the effect of converting covariant tensors of type (0, 1) into contravariant ones of type (1, 0). We are going to apply the above principle to convert covariant tensors of type (0, 2) into contravariant ones of type (2, 0). Recall that (0, 2)-tensors = bilinear maps ϕ : V V R { } { × → } and consider now

bilinear maps σ : V ∗ V ∗ R . { × → } Anticipating the contravariant character of such bilinear maps (to be proven in §4.1.4), we advance the following definition:

Definition 4.2. A tensor of type (2, 0) is a bilinear form on V ∗, that is, a bilinear function σ : V ∗ V ∗ R. × → Then we have

(2, 0)-tensors = bilinear maps σ : V ∗ V ∗ R { } { × → } and we denote this set Bil(V ∗ V ∗, R). × Exercise 4.3. Check that Bil(V ∗ V ∗, R) is a vector space. Just like in the case of Bil(V V, R) (cf. Exercise 2.25),× it is enough to show that the zero map is in × Bil(V ∗ V ∗, R) and that if σ, τ Bil(V ∗ V ∗, R) and c,d R, then the linear × ∈ × ∈ combination cσ + dτ is also in Bil(V ∗ V ∗, R). ×

4.1.3. Tensor Product of Two Linear Forms on V ∗.

If v,w V are two vectors (i.e., are two (1, 0)-tensors), we define ∈ σ : V ∗ V ∗ R v,w × → by

σv,w(α, β) := α(v)β(w) , for any two linear forms α, β V ∗. Then σv,w is indeed bilinear, i.e., linear in each variable, α and β, so is indeed∈ a (2, 0)-tensor. We denote σ =: v w v,w ⊗ and call this the tensor product of v and w. 68 4. TENSORS

Note 4.4. In general, we have v w = w v , ⊗ 6 ⊗ as there can be linear forms α, β such that α(v)β(w) = α(w)β(v). 6 Similar to what we saw in §2.2.3, we find a basis for the space of (2, 0)-tensors by considering the (2, 0)-tensors defined by b b , where = b ,...,b is a basis of V . i ⊗ j B { 1 n} Proposition 4.5. The elements bj bj, i, j =1,...,n form a basis of Bil(V ∗ V ∗, R). 2 ⊗ × Thus dim Bil(V ∗ V ∗, R)= n . × The proof of this proposition is analogous to the one of Proposition 2.26 and we will not write it here. However, we state the crucial remark for the proof, analogous to Remark 2.27. Remark 4.6. As for linear forms and bilinear forms on V , in order to verify that two bilinear forms on V ∗ are the same, it is enough to verify that they are the same on every pair of elements of a basis of V ∗. In fact, let σ, τ be two bilinear forms on V ∗, let 1 n γ ,...,γ be a basis of V ∗ and let us assume that { } σ(γi,γj)= τ(γi,γj) i j for all 1 i, j, n. Let α = αiγ and β = βjγ be arbitrary elements of V ∗. We now verify that≤ σ(α,≤ β)= τ(α, β). Because of the linearity in each variable, we have i j i j i j i j σ(α, β)= σ(αiγ , βjγ )= αiβjσ(γ ,γ )= αiβjτ(γ ,γ )= τ(αiγ , βjγ )= τ(α, β) . 

4.1.4. Contravariance of (2, 0)-Tensors.

Let σ : V ∗ V ∗ R be a bilinear form on V ∗, that is, a so-called (2, 0)-tensor. We want to verify× that→ it behaves as we expect with respect to a change of basis. After 1 n choosing a basis = b ,...,b of V , we have the dual basis ∗ = β ,...,β of B { 1 n} B { } V ∗ and the basis bi bj : i, j =1,...,n of the space of (2, 0)-tensors. The (2, 0)-tensor{ ⊗σ is represented by its} components Sij = σ(βi, βj) , in the sense that σ = Sijb b , j ⊗ j and the components Sij can be arranged into a matrix12 S11 ... S1n . . . S =  . .. .  Sn1 ... Snn     12Once again, contrary to the change of basis matrix L, here we have only upper indices. This reflects the contravariance of the underlying tensor, σ. 4.2. TENSORS OF TYPE (p, q) 69 called the matrix of the (2, 0)-tensor σ with respect to the chosen basis of V . We look now at how the components of a (2, 0)-tensor change with a change of basis. ˜ ˜ 1 n Let = b1,...,bn and = b1,..., bn be two basis of V and let ∗ := β ,...,β B { 1 } n B { } B { } and ∗ := β˜ ,..., β˜ be the corresponding dual bases of V ∗. Let σ : V ∗ V ∗ R be aB(2, 0)-tensor{ with} componentse × → e Sij = σ(βi, βj) and Sij = σ(β˜i, β˜j) with respect to ∗ and ∗, respectively. Let L := L e be the matrix of the change of B B 1 e BB basis from to , and let Λ := L− . Then, as seen in (1.4) and (2.11), we have that B B e ˜b = Li b and β˜i =Λi βj . e j j i j It follows that ij ˜i ˜j i k j ℓ i j k ℓ i j kℓ S = σ(β , β )= σ(Λkβ , Λℓβ )=ΛkΛℓσ(β , β )=ΛkΛℓS , where the first and the last equalities follow from the definition of Sij and of Skℓ, e respectively, the second from the change of basis and the third from the bilinearity of σ. We conclude that e ij i j kℓ (4.3) S =ΛkΛℓS . Hence, the bilinear form σ is a contravariant 2-tensor. e Exercise 4.7. Verify that, in terms of matrices (4.3) translates into

t S =ΛSΛ . Compare with (2.15). e 4.2. Tensors of Type (p, q) In general, a (p, q)-tensor (with p, q = 0, 1, 2,...) is defined to be a real-valued function of p covectors and of q vectors, which is linear in each of its arguments: Definition 4.8. A tensor of type (p, q) or (p, q)-tensor is a multilinear form (or (p + q)-linear function)

T : V ∗ ...V ∗ V V R . × × ×···× −→ p q By convention, a tensor of| type{z(0, 0)} is a| real{z number,} a.k.a. scalar (a constant funtion of no arguments). The order of a tensor is the number of arguments that it takes: a tensor of type (p, q) has, thus, order p + q.13

13The order of a tensor is sometimes also called rank. However, rank of a tensor is often reserved for another notion closer to the notion of rank of a matrix and related to decomposability of tensors (see §4.3.1 and §4.3.2). 70 4. TENSORS

Earlier tensor Viewed as multilinear function Type vector V R ∗ (1, 0) v V β −→ β(v) linear∈ form V 7−→ R −→ (0, 1) α V ∗ w α(w) linear transformation∈ V V 7−→ R ∗ (1, 1) F : V V (β,w× ) −→ β(F (w)) bilinear→ form V V 7−→ R (0, 2) ϕ Bil(V V, R) (v,w× ) −→ ϕ(v,w) ∈k-linear× form V V 7−→ R (0, k) ϕ on V (v×···×,...,v ) −→ ϕ(v ,...,v ) 1 k 7−→ 1 k bilinear form on V ∗ V ∗ V ∗ R × −→ (2, 0) σ Bil(V ∗ V ∗, R) (α, β) σ(α, β) ∈ × 7−→ Table 5. Earlier tensors viewed within general definition

If all the arguments of a tensor are vectors, i.e. p = 0, the tensor is said to be (purely) covariant. If the arguments are all linear forms, i.e. q =0, the tensor is said to be (purely) contravariant. Otherwise, a (p, q)-tensor is of mixed character, p being its order of contravariance and q its order of covariance. Purely covariant tensors are what we earlier called multilinear forms. Purely contravariant tensors are sometimes called polyadics.14 Table 5 gives an overview of how the earlier examples of tensors fit in the above general definition. 1 n Let T be a (p, q)-tensor, = b ,...,b a basis of V and ∗ = β ,...,β the B { 1 n} B { } corresponding dual basis of V ∗. The components of T with respect to these bases are

i1,...,ip i1 ip Tj1,...,jq := T (β ,...,β , bj1 ,...,bjq ) .

1 n If, moreover, = ˜b1,..., ˜bn is another basis, ∗ = β˜ ,..., β˜ the corresponding B { } B { } 1 dual basis of V ∗ and L := L e the matrix of the change of basis with inverse Λ := L− , then the componentse of T withBB respect to these newe bases are

i1,...,ip i1 ip ℓ1 ℓq k1,...,kp Tj1,...,jq =Λk1 ... Λkp Lj1 ...Ljq Tℓ1,...,ℓq .

The above formula displayse the p-fold contravariant character and the q-fold covariant character of T .

14Whereas Latin roots are used for covariant tensors, like in bilinear form, Greek roots are used for contravariant tensors, like in dyadic. 4.3. TENSOR PRODUCT 71

The set of all tensors of type (p, q) on a vector space V with the natural operations of addition and scalar multiplication on tensors is itself a vector space denoted by p(V ) := all (p, q)-tensors on V . Tq { }

4.3. Tensor Product We saw already in §2.2.3 and §2.3.3 the tensor product of two multilinear forms. Since multilinear forms are covariant tensors, we said that this corresponds to the tensor product of two covariant tensors. We can now define the tensor product of two tensors in general. This will further lead us to the tensor product of vector spaces.

4.3.1. Tensor Product for Tensors. Definition 4.9. Let

T : V ∗ V ∗ V V R ×···× × ×···× −→ p q be a (p, q)-tensor and | {z } | {z }

U : V ∗ V ∗ V V R ×···× × ×···× −→ k ℓ a (k,ℓ) tensor. The tensor product T U of T and U is the (p + k, q + ℓ)-tensor | {z }⊗ | {z } T U : V ∗ V ∗ V V R ⊗ ×···× × ×···× −→ p+k q+ℓ defined by | {z } | {z } (T U)(α ,...,α , v ,...,v ) := ⊗ 1 p+k 1 q+ℓ T (α1,...,αp, v1,...,vq)U(αp+1,...,αp+k, vq+1,...,vq+ℓ) .

Although both T U and U T are tensors of the same type, in general we have ⊗ ⊗ T U = U T. ⊗ 6 ⊗ So we say that the tensor product is not commutative. On the other hand, the tensor product is associative, since we always have (S T ) U = S (U T ) . ⊗ ⊗ ⊗ ⊗ Analogously to how we proceeded in the case of (0, 2)-tensors, we compute the p dimension of the vector space q (V ). Let = b1,...,bn be a basis of V and ∗ := 1 n T B { } B β ,...,β the corresponding dual basis of V ∗. Just like we saw in Proposition 4.5 { } p in the case of (0, 2)-tensors, we form a basis of q (V ) by collecting all elements of the form T b b b βj1 βj2 βjq i1 ⊗ i2 ⊗···⊗ ip ⊗ ⊗ ⊗···⊗ 72 4. TENSORS where the indices i1,...,ip and j1,...,jq take all values between 1 and n. Since there are npnq = np+q elements in the above basis (corresponding to all possible choices of jℓ bik and β ), we deduce that dim p(V )= np+q . Tq If T is a (p, q)-tensor with components

i1,...,ip i1 ip Tj1,...,jq := T (β ,...,β , bj1 ,...,bjq ) , then we have, with a (p + q)-fold application of the Einstein convention, that

T = T i1,...,ip b b b βj1 βj2 βjq . j1,...,jq i1 ⊗ i2 ⊗···⊗ ip ⊗ ⊗ ⊗···⊗ A simple tensor (also called a tensor of rank 1 or pure tensor or decomposable tensor or elementary tensor) of type (p, q) is a tensor T that can be written as a tensor product of the form T = a b ... α β ... ⊗ ⊗ ⊗ ⊗ ⊗ p q where a, b, . . . V and α, β, . . . | V ∗{z. The} rank| of{z a tensor} T is then the minimum number of simple∈ tensors that sum∈ to T .15 By convention, the zero tensor has rank 0 and a nonzero tensor of order zero, i.e., a nonzero scalar, has rank 1. A nonzero tensor of order 1 always has rank 1. Already among tensors of order 2 (and when dim V 2) there are tensors of rank greater than 1. Example 4.11 provides such an instance. ≥

4.3.2. Tensor Product for Vector Spaces.

To complement the previous exposition and justify the notation V ∗ V ∗ for the vector space of all bilinear forms on V (cf. §2.2.3), we aim in this section⊗ to give an idea of what the tensor product for finite-dimensional vector spaces should mean and of how the tensor product for vector spaces relates to the tensor product for tensors. Let V and W be two vector spaces with dim V = n and dim W = m. Choose b ,...,b a basis of V and a ,...,a a basis of W .16 { 1 n} { 1 m} Definition 4.10. The tensor product of V and W is the (n m)-dimensional vector space V W with basis · ⊗ b a : 1 i n, 1 j m . { i ⊗ j ≤ ≤ ≤ ≤ } 15This notion of rank of a tensor extends the notion of rank of a matrix, as can be seen by considering tensors of order two and their corresponding matrices of components. 16There is a way of defining the tensor product for vector spaces without involving bases, but we will not do it here. That other, more abstract way shows elegantly that the tensor product of vector spaces does not depend on the choice of bases. 4.3. TENSOR PRODUCT 73

Elements of V W are naturally referred to as tensors; further below we address the connection to the⊗ previous notion of tensor. By definition, they are linear combinations of the b a . By viewing b V as a linear map (cf. Table 5) i ⊗ j i ∈ b : V ∗ R , β β(b ) , i −→ 7−→ i and similarly for a W as j ∈ a : W ∗ R , α α(a ) , j −→ 7−→ j we may regard the symbol b a as a bilinear map i ⊗ j V ∗ W ∗ R , (β,α) β(b )α(a ) . × −→ 7−→ i j The tensor product V W is endowed with a bilinear map from the cartesian product V W ⊗ × Ψ: V W V W × −→ ⊗ defined as follows. If v = vib V and w = wja W, i ∈ j ∈ then Ψ(v,w) =: v w is the element of V W with coordinates viwj with respect to the basis b a :⊗ 1 i n, 1 j m ⊗, so that the following holds: { i ⊗ j ≤ ≤ ≤ ≤ } v w = vib wja = viwj b a . ⊗ i ⊗ j i ⊗ j Notice that the ranges for the indices
are different:
1 i
n, 1 j m. The numbers viwj may be viewed as obtained by the so-called outer≤ product≤ of≤ the≤ coordinate vectors of v and w yielding an n m matrix: × v1 v1w1 ...v1wm t . 1 m . . [v] [w] = . w ... w = . . . B A     vn vnw1 ...vnwm
    An element of V W that is in the image of the map Ψ, that is, an element of V W that can be written⊗ as v w for some v V and w W is called a simple tensor⊗ (or tensor of rank 1 or pure⊗ tensor or decomposable∈ ∈ tensor or elementary tensor). Yet keep in mind, that the map Ψ is usually by far not surjective. In particular, one can show that, if v , v V are linearly independent and w ,w W are also 1 2 ∈ 1 2 ∈ linearly independent, then the sum v1 w1 + v2 w2 is not a pure tensor. Checking the first instance of this phenomenon is left⊗ as the next⊗ exercise. The proof in general goes along somewhat similar lines, but gets sophisticated.17 Exercise 4.11. Check that if both V and W are 2-dimensional with respective bases b , b and a , a , then b a + b a is not a pure tensor. { 1 2} { 1 2} 1 ⊗ 1 2 ⊗ 2 The following proposition gives some useful identifications.

17If you are interested in learning more, look up the Segre embedding from Algebraic Geometry. 74 4. TENSORS

Proposition 4.12. Let V and W be vector spaces with dim V = n and dim W = m and let

Lin(V, W ∗) := linear maps V W ∗ . { → } Then

Bil(V W, R) = Lin(V, W ∗) × ∼ ∼= Lin(W, V ∗) = V ∗ W ∗ ∼ ⊗ = (V W )∗ ∼ ⊗ = Lin(V W, R) . ⊗ Proof. Here is the idea behind this chain of identifications. Let f Bil(V W, R), that is, f : V W R is a bilinear function, in particular it takes two∈ vectors,× v V and w W ,× as input→ and gives a real number f(v,w) R as output. If, however,∈ we only∈ feed f one vector v V as input, then there is∈ a remaining spot waiting for a vector w W to produce a∈ real number. Since f is linear in V and in W , the map ∈ f(v, ): W R is a linear form, so f(v, ) W ∗, hence f gives us an element in · → · ∈ Lin(V, W ∗). There is then a linear map

Bil(V W, R) Lin(V, W ∗) × −→ f T , 7−→ f where

Tf (v)(w) := f(v,w) .

Conversely, any T Lin(V, W ∗) can be identified with a bilinear map fT Bil(V W, R) defined by ∈ ∈ ×

fT (v,w) := T (v)(w) .

Since fTf = f and TfT = T , we have proven the first identification in the proposition. Analogously, if the input is only a vector w W , then f( ,w): V R is a linear ∈ · → map and we now see that f Bil(V W, R) defines a linear map U Lin(W, V ∗). The ∈ × f ∈ same reasoning as in the previous paragraph, shows that Bil(V W, R) = Lin(W, V ∗). × ∼ To proceed with the identifications, observe that, because of our definition of V ∗ ⊗ W ∗, we have

Bil(V W, R) = V ∗ W ∗ , × ∼ ⊗ since these spaces both have basis βi αj : 1 i n, 1 j m , { ⊗ ≤ ≤ ≤ ≤ } 1 n where b1,...,bn is a basis of V with corresponding dual basis β ,...,β of V ∗, { } 1{ n } and a ,...,a is a basis of W with corresponding dual basis α ,...,α of W ∗. { 1 n} { } 4.3. TENSOR PRODUCT 75

i j Finally, an element D β α V ∗ W ∗ may be viewed as a linear map V W R, ij ⊗ ∈ ⊗ ⊗ → that is as an element of (V W )∗ by ⊗ V W R ⊗ −→ Ckℓb a D Ckℓ βi(b ) αj(a ) = D Ckℓ . k ⊗ ℓ 7−→ ij k ℓ ij δi j k δℓ | {z } | {z }  Because of the identification Bil(V W, R) = Lin(V W, R), we can say that × ∼ ⊗ the tensor product linearizes what was bilinear (or multilinear). There is no reason to restrict oneself to the tensor product of only two factors. One can equally define the tensor product V1 Vk, and obtain a vector space of dimension dim V dim V . Note that we⊗···⊗ do not need to use brackets, since the 1 ×···× k tensor product is associative: (V1 V2) V3 = V1 (V2 V3). We have ⊗ ⊗ ⊗ ⊗ p (V )= V V V ∗ V ∗ , Tq ⊗···⊗ ⊗ ⊗···⊗ p q p since both spaces have the same| basis.{z An} | element{z T of} q (V ) was first regarded according to Definition 4.8 as a multilinear map T

T : V ∗ ...V ∗ V V R . × × ×···× −→ p q 1 n Now, with respect to bases| {z= b1},...,b| n {zof V }and ∗ = β ,...,β of V ∗, the components of T are B { } B { }

i1,...,ip i1 ip Tj1,...,jq := T (β ,...,β , bj1 ,...,bjq ) , hence we may view T as

i1,...,ip j1 jq T b ... b β ... β V V V ∗ V ∗ . j1,...,jq i1 ⊗ ⊗ ip ⊗ ⊗ ⊗ ∈ ⊗···⊗ ⊗ ⊗···⊗ p q In particular, the pth tensor power of a vector| space{z is } | {z } p p V ⊗ := V V = (V ) . ⊗···⊗ T0 p Therefore, we may write | {z } 2 2 Bil(V ∗ V ∗, R)= V V = V ⊗ , Bil(V V, R)= V ∗ V ∗ =(V ∗)⊗ × ⊗ × ⊗ and p p q (V )= V ⊗ (V ∗)⊗ . Tq ⊗

CHAPTER 5

Applications

Mathematically-speaking, a tensor is a real-valued function of some number of vec- tors and some number of covectors (a.k.a. linear forms), which is linear in each of its arguments. On the other hand, tensors are most useful in connection with concrete phys- ical applications. In this chapter, we borrow notions and computations from Physics and Calculus to discuss important classical tensors.

5.1. Inertia Tensor

5.1.1. Physical Preliminaries.

We consider a rigid body M fixed at a point O and rotating about an axis through O with angular velocity ω. Denoting the time variable t and an angle variable θ around the axis of rotation, the angular velocity will be viewed as a vector18 with magnitude dθ ω = , k k dt

with direction given by the axis of rotation and with orientation given by the right-hand rule. The position vector of a point P in the body M relative to the origin O is

x := −→OP while the linear velocity of that point P is v := ω x . × The linear velocity v has, hence, magnitude v = ω x sin α , k k k k k k dθ =:r k dt k where α is the angle between ω and x|{z}, and| has{z direction} tangent at P to the circle of radius r perpendicular to the axis of rotation.

18Warning: The angular velocity is actually only what physicists call a pseudovector because it does not follow the usual contravariance of a vector in case of orientation flip. Luckily, this issue does not affect the inertia tensor, since the sign flip cancels out thanks to squaring. 77 78 5. APPLICATIONS

ω

α .O x P. r v

The kinetic energy of an infinitesimal region dM of M around P is 1 dE = v 2dm , 2k k where v 2 = v · v is the square of the norm of the linear velocity and dm is the mass of dM.k Thek total kinetic energy of M is 1 1 E = v 2dm = ω x 2dm . 2 k k 2 k × k ZM ZM Actually, depending on the type of rigid body, we might take here a sum instead of integral, or some other type of integral (line integral, surface integral, etc) such as: (1) If M is a solid in 3-dimensional space, then 1 E = ω x 2ρ dx1dx2dx3 , 2 k × P k P ZZZM x 2 where the norm squared ω P and the density ρP are functions of the point P with coordinates k(x1,× x2, xk3). (2) If M is a flat sheet, then 1 E = ω x 2ρ dx1dx2 , 2 k × P k P ZZM where the integrand only depends on two cartesian coordinates. 5.1. INERTIA TENSOR 79

(3) If M is a (curvy) surface in 3-dimensional space, then 1 E = ω x 2ρ dσ , 2 k × P k P ZZM where dσ is the infinitesimal element of the surface for a surface integral. (4) If M is a wire in 3-dimensional space, then 1 E = ω x 2ρ ds , 2 k × P k P ZM where ds is the infinitesimal element of length for a line integral. (5) If M is a finite set of N point masses mi with rigid relative positions, then

N 1 2 E = ω xi m . 2 k × k i i=1 X We will keep writing our formulas for the first case (with a volume integral); these should be adjusted for situations of the other types. In any case, we need to work out the quantity ω x 2 k × k for vectors ω and x in 3-dimensional space. To this purpose, we use the Lagrange identity19, according to which a · c a · d (5.1) (a b) · (c d) = det . × × b · c b · d   Applying (5.1) with a = c = ω and b = d = x, we obtain ω · ω ω · x ω x 2 =(ω x) · (ω x) = det = ω 2 x 2 ω · x 2 . k × k × × x · ω x · x k k k k −k k   Let now = e , e , e be an orthonormal20 basis of R3, so that B { 1 2 3} i i ω = ω ei and x = x ei . Then ω 2 = ω · ω = δ ωiωj = ω1ω1 + ω2ω2 + ω3ω3 k k ij x 2 = x · x = δ xkxℓ = x1x1 + x2x2 + x3x3 k k kℓ i k 1 1 2 2 3 3 ω · x = δikω x = ω x + ω x + ω x

19The Lagrange identity can be patiently proven in coordinates. 20 3 We could use any basis of R . Then, instead of the δij , the formulas would have involved the components of the metric tensor gij . However, computations with orthonormal bases are simpler; in particular, the inverse of an orthonormal basis change L is simply tL. Moreover, the inertia tensor is symmetric, hence admits an orthonormal eigenbasis. 80 5. APPLICATIONS so that ω x 2 = ω 2 x 2 ω · x 2 k × k k k k k −k k =(δ ωiωj)(δ xkxℓ) (δ ωixk)(δ ωjxℓ) ij kℓ − ik jl =(δ δ δ δ )ωiωjxkxℓ . ij kℓ − ik jℓ Therefore, the total kinetic energy is 1 E = (δ δ δ δ )ωiωj xkxℓdm 2 ij kℓ − ik jℓ ZZZM and it depends only on ω1,ω2,ω3 (since we have integrated over the x1, x2, x3).

5.1.2. Moments of Inertia. Definition 5.1. The inertia tensor is the covariant 2-tensor whose components with respect to an orthonormal basis are B I =(δ δ δ δ ) xkxℓdm . ij ij kℓ − ik jℓ ZZZM Then the kinetic energy of the rotating rigid body is 1 E = I ωiωj 2 ij which in matrix notation amounts to 1 1 E = ω · Iω = tωIω . 2 2 Remark 5.2. If, instead of an orthonormal basis, we had used any basis of R3, we would have gotten

I =(g g g g ) xkxℓdm . ij ij kℓ − ik jℓ ZZZM where gij are the components of the metric tensor. This formula also makes apparent the covariance and the symmetry of I inherited from the metric: Iij = Iji for all i and j. We will see that the inertia tensor is a convenient way to encode all moments of inertia of an object in one quantity and we return now to the case of an orthonormal basis. The first component of the inertia tensor is

k ℓ I11 =(δ11δkℓ δ1kδ1ℓ) x x dm . − M =0 =0 ZZZ unless unless k=ℓ k=ℓ=1 | {z } | {z } 5.1. INERTIA TENSOR 81

If k = ℓ =1, then δ11δ11 δ11δ11 =0, so that the non-vanishing terms have k = ℓ =1. In this way, one can check− that 6

2 2 3 3 I11 = (x x + x x ) dm ZZZM 1 1 3 3 I22 = (x x + x x ) dm ZZZM 1 1 2 2 I33 = (x x + x x ) dm ZZZM I = I = x2x3 dm 23 32 − ZZZM I = I = x1x3 dm 31 13 − ZZZM I = I = x1x2 dm , 12 21 − ZZZM so that with respect to an orthonormal basis , the inertia tensor is represented by the symmetric matrix B

I11 I12 I13 I = I I I .  21 22 23 I31 I32 I33  

The diagonal components I11,I22,I33 are the moments of inertia of the rigid body M with respect to the coordinate axes Ox1, Ox2, Ox3, respectively. The off-diagonal components I12,I23,I31 are the polar moments of inertia or the products of inertia of the rigid body M.

Example 5.3. We want to find the inertia tensor of a homogeneous rectangular plate with sides a and b and total mass m, assuming that the rotation preserves the center of mass O. We choose a coordinate system (corresponding to an orthonormal basis) with origin at the center of mass O, with x-axis parallel to the side of length a, y-axis parallel to the side of length b, z-axis perpendicular to the plate, and adjust our previous formulas to double integrals. Since the plate is assumed to be homogeneous, it has a constant mass density equal to

total mass m ρ = = . area ab 82 5. APPLICATIONS

e2

b . e1

a Then a b 2 2 2 2 I11 = (y + z ) ρ dy dx a b 2 2 m Ixx − − =0 Z Z ab b |{z} m 2 |{z} |{z} = a y2 dy ab b Z− 2 b m y3 2 m = = b2 . b 3 b 12  − 2 Similarly, m I = a2 , 22 12 Iyy and |{z}

a b 2 2 2 2 m 2 2 I33 = (x + y )ρdydx = (a + b ) a b 2 2 12 Izz Z− Z− turns out to be just|{z} the sum of I11 and I22. Furthermore, a b 2 2 I23 = I32 = y z ρ dydx =0 , − a b Z− 2 Z− 2 =0 and, similarly, I31 = I13 =0. Finally, we have|{z} a b a b 2 2 m 2 2 I21 = I21 = xyρdydx = x dx ydy . − a b −ab a b Z− 2 Z− 2 Z− 2 ! Z− 2 ! =0 =0

because the integral of an odd function | on a{z symmetric} | interval{z is 0 } | {z } 5.1. INERTIA TENSOR 83

We conclude that the inertia tensor is given by the matrix b2 0 0 m 0 a2 0 . 12  0 0 a2 + b2   

Exercise 5.4. Compute the inertia tensor of the same plate, but now with center of rotation O coinciding with a vertex of the rectangular plate.

5.1.3. Moment of Inertia About any Axis.

We compute the moment of inertia of the body M about an axis through the point O and defined by the unit vector p.

p α P x . .O r

The moment of inertia of an infinitesimal region of M around P is dI = r2 dm = p x 2 dm , k × k r is the distance infinitesimal from P to the axis mass where the last equality follows|{z} from the fact|{z} that p x = p x sin α = r, since p is a unit vector. Hence, the total moment of inertiak × kof Mk withkk k respect to the axis given by p is

I := p x 2 dm 0 . p k × k ≥ ZZZM This is very similar to the total kinetic energy E: just replace ω by p and omit the factor 1 2 . By the earlier computations, we conclude that i j I = Iijp p , 84 5. APPLICATIONS where Iij is the inertia tensor. This formula shows that the total moment of inertia of the rigid body M with respect to an arbitrary axis passing through the point O is determined only by the inertia tensor of the rigid body. Example 5.5. For the rectangular plate in Example 5.3, we now want to compute the moment of inertia with respect to the diagonal of the plate.

b .O

a We choose the unit vector p = 1 (ae + be ) (the other possibility is the negative √a2+b2 1 2 of this vector, yielding the same result), so that a b p1 = , p2 = , p3 =0 √a2 + b2 √a2 + b2 and use the matrix for I found in Example 5.3. The moment of inertia is

i j Ip =Iijp p m 2 a 2 2 12 b 0 0 √a +b a b m 2 b = 2 2 2 2 0 0 a 0 √a +b √a +b 12 √a2+b2  0 0 m (a2 + b2)     12 0 m a2b2     = 6 a2+b2 . 

Exercise 5.6. Double-check the above result for the moment of inertia of the rectan- gular plate in Example 5.3 with respect to the diagonal of the plate, now by using the inertia tensor computed in Exercise 5.4 (with center of rotation O in a vertex belonging also to that diagonal). Exercise 5.7. Compute the moment of inertia of the rectangular plate in Example 5.3 with respect to an axis perpendicular to the plate and passing through its center of mass. Exercise 5.8. Compute the moment of inertia of the rectangular plate in Example 5.3 with respect to an axis perpendicular to the plate and passing through one vertex. 5.1. INERTIA TENSOR 85

5.1.4. Angular Momentum.

Let M be a body rotating with angular velocity ω about an axis through the point O. Let x = −→OP be the position vector of a point P and v = ω x the linear velocity of P . ×

ω α.O x P. v

Then the angular momentum of an infinitesimal region of M around P is

dL =(x v) dm , × so that the total angular momentum21 of M is

L = (x (ω x)) dm . × × ZZZM

21Just like the angular velocity, the angular momentum is not an honest vector, but only a pseu- dovector, since there is an issue with orientation. In this subsection, we should thus assume that we 3 work with an oriented orthonormal basis e1,e2,e3 of R , so that e1 e2 = e3 (and not e3). This amounts to assuming that the change of{ basis matrix} L from the standard× basis has det L−= 1 (and not 1). − 86 5. APPLICATIONS

We need to work out x (ω x) for vectors x and ω in 3-dimensional space. We apply the following identity22×for the× triple vector product: (5.2) x (ω x)=(x · x)ω (ω · x)x . × × − 3 Let = e1, e2, e3 be an orthonormal basis of R . Then, replacing the following equalitiesB { } i i j (5.3) ω = ω ei = δjω ei x i i k (5.4) = x ei = δkx ei k ℓ (5.5) x · x = δkℓx x j ℓ (5.6) ω · x = δjℓω x into (5.2), we obtain x (ω x)=(δ xkxℓ) δiωje (δ ωjxℓ) δi xke =(δi δ δi δ )ωjxkxℓe . × × kℓ j i − jℓ k i j kℓ − k jℓ i (5.5) (5.3) (5.6) (5.4) Therefore, the total| angular{z } | momentum{z } | {z is } | {z } i L = L ei , where the components Li are

Li =(δi δ δi δ ) ωj xkxℓ dm . j kℓ − k jℓ ZZZM Since we have always 1 if i = j δi = δ = δij = , j ij 0 if i = j ( 6 if we violate index balance for one instance, the above expression for Li can be written in terms of the inertia tensor Iij as i j L = Iijω , which corresponds to the matrix form L = Iω. We see that the angular momentum L is proportional (or parallel) to the angular velocity ω only when ω is an eigenvector of the inertia tensor I. Example 5.9. Suppose the rectangular plate in the previous examples is rotating about an axis through the center of mass O with angular velocity ω1 1 2 ω = e1 +2e2 +3e3 , or ω = 2 . ω3 3 We want to compute its angular momentum.    

22To prove this vector equality use coordinates, consider only the case in which ω is a standard basis vector and then use the linearity in ω. 5.1. INERTIA TENSOR 87

The inertia tensor is given by the matrix Iij found in Example 5.3: m 2 12 b 0 0 m 2 0 12 a 0 .  m 2 2  0 0 12 (a + b ) The total angular momentum has components given by 1 m 2 m b2 L 12 b 0 0 1 12 2 m 2 m 2 L = 0 12 a 0 2 = 6 a , L3  0 0 m (a2 + b2) 3  m 2 2  12 4 (a + b ) so that         m m m L = b2e + a2e + (a2 + b2)e . 12 1 6 2 4 3 

5.1.5. Principal Moments of Inertia.

Observe that the inertia tensor of a rigid body M is symmetric and recall the spectral theorem (Theorem 3.9). Then we know that an orthonormal eigenbasis e˜ , e˜ , e˜ { 1 2 3} exists for the inertia tensor. Let I1,I2,I3 be the corresponding eigenvalues. The matrix representing the inertia tensor with respect to this eigenbasis is

I1 0 0 0 I 0 .  2  0 0 I3 The orthonormal eigenbasis gives a preferred coordinate system in which to formulate a problem pertaining to rotation of this body. The axes of the eigenvectors are called the principal axes of inertia of the rigid body M. The eigenvalues Ii are called the principal moments of inertia. For instance, if a homogeneous body is symmetric with respect to the xy-plane, then the polar moments of inertia I23 = I32 and I13 = I31 vanish, thus the z-axis is necessarily a principal axis (because of the block-form of I). The principal moments of inertia are the moments of inertia with respect to the principal axes of inertia, hence they are non-negative I , ,I , ,I 0 . 1 2 3 ≥ A rigid body is called

(1) an asymmetric top if I1 = I2 = I3 = I1; (2) a symmetric top if exactly6 two6 eigenvalues6 are equal, say I = I = I : any 1 2 6 3 axis passing through the plane determined by e˜1 and e˜2 is then a principal axis of inertia; (3) a spherical top if I1 = I2 = I3: any axis passing through O is a principal axis of inertia. 88 5. APPLICATIONS

With respect to the eigenbasis e˜ , e˜ , e˜ the kinetic energy is { 1 2 3} 1 E = (I (˜ω1)2 + I (˜ω2)2 + I (˜ω3)2) , 2 1 2 3 i i where ω =ω ˜ e˜i, with ω˜ the components of the angular velocity with respect to the basis e˜1, e˜2, e˜3 . In particular, we see that the kinetic energy can be conserved, even if the angular{ velocity} ω changes, as long as the above combination of squares is preserved. This is related to the phenomenon of precession. The surface determined by the equation (with respect to the coordinates x, y, z)

2 2 2 (5.7) I1x + I2y + I3z =1 is called the ellipsoid of inertia. The symmetry axes of the ellipsoid coincide with the principal axes of inertia. Note that for a spherical top, the ellipsoid of inertia is actually a sphere. The ellipsoid of inertia gives the moment of inertia with respect to any axis as follows: Consider an axis given by the unit vector p and let q = cp be a vector of intersection of the axis with the ellipsoid of inertia, where c is ( ) the distance to O of the intersection of the axis with the ellipsoid of inertia. ±

q axis p O

The moment of inertia with respect to this axis is 1 1 I = I pipj = I qiqj = , ij c2 ij c2 where the last equality follows from the fact that, since q is on the ellipsoid, then i j Iijq q =1 by equation (5.7). Example 5.10. The principal axes of inertia for the rectangular plate in Example 5.3 are the axes parallel to the sides and the axis perpendicular to the plate. The corresponding principal moments of inertia are m m m I = b2 , I = a2 and I = (a2 + b2) . 11 12 22 12 33 12 If a = b, that is, if the rectangle is a square, we have a symmetric top.  5.2. STRESS TENSOR (SPANNUNG) 89

5.2. Stress Tensor (Spannung)

5.2.1. Physical Preliminaries.

Let us consider a rigid body M acted upon by external forces but in static equilibrium, and let us consider an infinitesimal region dM around a point P . There are two types of external forces: (1) The body forces, that is forces whose magnitude is proportional to the vol- ume/mass of the region. For instance, gravity, attractive force or the centrifugal force. (2) The surface forces, that is forces exerted on the surface of the element by the material surrounding it. These are forces whose magnitude is proportional to the area of the region in consideration. The surface force per unit area is called the stress. We will concentrate on homoge- neous stress, that is stress that does not depend on the location of the element in the body, but depends only on the orientation of the surface given by its tangent plane. Moreover, we assume that the body in consideration is in static equilibrium.

Remark 5.11. It was the concept of stress in mechanics that originally led to the invention of tensors

tenseur9 s9 ▲▲ sss ▲▲▲ sss ▲▲▲ sss ▲▲% stress tensor9 ❑❑ r9 ❑❑ rr ❑❑ rr ❑❑ rr ❑% rr tension

Choose an orthonormal basis e , e , e and the plane Π through P parallel to the { 1 2 3} e2e3 coordinate plane. The vector e1 is normal to this plane. Let ∆A1 be the area of the slice of the infinitesimal region around P cut by the plane and let ∆F be the force acting on that slice. We write ∆F in terms of its components 1 2 3 ∆F = ∆F e1 + ∆F ee + ∆F e3 and, since the stress is the surface force per unit area, we define ∆F j σ1j := lim , for j =1, 2, 3 . ∆A1 0 → ∆A1 Similarly, we can consider planes parallel to the other coordinate planes Π2 Π1 e3 P P Π3 e1 e2 P 90 5. APPLICATIONS and define ∆F j σij := lim . ∆Ai 0 → ∆Ai It turns out that the resulting nine numbers σij are the components of a contravariant 2-tensor called the stress tensor, as we will see in §5.2.5. The stress tensor encodes the mechanical stresses on an object. We now compute the stress across other slices through P , that is, across other planes with other normal vectors. Let Π be a plane passing through P , n a unit vector through P perpendicular to the plane Π, ∆s = Π dM the area of a small element of the plane Π containing P and ∆F the force acting∩ on that element.

n

P

∆s Π

Claim 5.12. The stress at P across the surface perpendicular to n is

∆F ij σ(n) := lim = σ (n · ei)ej . ∆s 0 → ∆s It follows from the claim that the stress σ is a vector-valued function that depends linearly on the normal n to the surface element.

Proof. Consider the tetrahedron OA1A2A3 bound by the triangular slice on the plane Π having area ∆s and three triangles on planes parallel to the coordinate planes

A3

e3 n

O e2 A2 e1

A1

Consider all external forces acting on this tetrahedron, which we regard as a volume element of the rigid body: 5.2. STRESS TENSOR (SPANNUNG) 91

(1) Body forces amounting to f · ∆v, where f is the force per unit of volume and ∆v is the volume of the tetrahedron. We actually do not know these forces, but we will see later that these are not relevant. (2) Surface forces amounting to the sum of the forces on each of the four sides of the tetrahedron. We want to assess each of the four surface contributions due to the surface forces. If ∆s is the area of the slice on the plane Π, the contribution of that slice is, by definition of stress equal to σ(n)∆s .

If ∆s1 is the area of the slice on the plane with normal e1, the contribution of that slice is − σ1j e ∆s , − j 1 and, similarly, the contributions of the other two slices are σ2j e ∆s and σ3je ∆s . − j 2 − j 3

A3 A3

n e1 −

O O A2 A2

A1 A1

A3 A3

e2 − O O A2 A2

e3 A1 A1 −

Note that the minus sign comes from the fact that we use everywhere outside pointing normals. So the total surface force is σ(n)∆s σ1je ∆s σ2je ∆s σ3je ∆s . − j 1 − j 2 − j 3 Since there is static equilibrium, the sum of all (body and surface) forces must be zero f∆v + σ(n)∆s σije ∆s =0 . − j i The term f∆v can be neglected when ∆s is small, as it contains terms of higher order (in fact, ∆v 0 faster than ∆s 0). We conclude that → → ij σ(n)∆s = σ ej∆si . 92 5. APPLICATIONS

It remains to relate ∆s to ∆s1, ∆s2, ∆s3. The side with area ∆si is the orthogonal projection of the side with area ∆s onto the plane with normal ei. The scaling factor for the area under projection is cos αi, where αi is the convex angle between the plane normal vectors

n

αi

ei

∆s i = cos α = cos α n e = n · e . ∆s i ik kk ik i Therefore ,

ij σ(n)∆s = σ ej(n · ei)∆s or, equivalently,

ij σ(n)= σ (n · ei)ej .

Note that, in the above formula, the quantities n · ei are the coordinates of n with respect to the orthonormal basis e , e , e , namely { 1 2 3} 1 2 3 n =(n · e1)e1 +(n · e2)e2 +(n · e3)e3 = n e1 + n e2 + n e3 . 

Remark 5.13. For homogeneous stress, the stress tensor σij does not depend on the point P . However, when we flip the orientation of the normal to the plane, the stress tensor changes sign. In other words, if σ(n) is the stress across a surface with normal n, then

σ( n)= σ(n) . − − The stress considers orientation as if the forces on each side of the surface have to balance each other in static equilibrium.

n σ( n) − σ(n) n −

 5.2. STRESS TENSOR (SPANNUNG) 93

5.2.2. Principal Stresses.

Claim 5.14. The stress tensor is a symmetric tensor, that is σij = σji.

Proof. Consider an infinitesimal cube of side ∆ℓ surrounding P and with faces parallel to the coordinate planes.

D′ C′ e3

B′ A′

e2 CD

e1 A B

The force acting on each of the six faces of the cube are: σ1j∆A e and σ1j∆A e , respectively for the front and the back faces, • 1 j − 1 j ABB′A′ and DCC′D′; 2j 2j σ ∆A e and σ ∆A e , respectively for the right and the left faces BCC′B′ • 2 j − 2 j and ADD′A′; σ3j∆A e and σ3j∆A e , respectively for the top and the bottom faces • 3 j − 3 j ABCD and A′B′C′D′, 2 where ∆A1 = ∆A2 = ∆A3 = ∆s = (∆ℓ) is the common face area. We compute now the torque µ, assuming the forces are applied at the center of each face, whose distance 1 to the center point P is 2 ∆ℓ. Recall that the torque is the tendency of a force to twist or rotate an object. It is given by the cross product of the distance vector and the force vector.

µ = ∆ℓ e σ1j∆s e + ∆ℓ e ( σ1j∆s e ) 2 1 × j − 2 1 × − j + ∆ℓ e σ2j∆s e + ∆ℓ e ( σ2j∆s e ) 2 2 × j − 2
2 × − j + ∆ℓ e σ3j∆s e + ∆ℓ e ( σ3j∆s e ) 2 2 × j − 2 3
× − j =∆ℓ∆s (e σije ) i × j
=∆ℓ∆s (σ23 σ32)e +(σ31 σ13)e +(σ12 σ21)e . − 1 − 2 − 3
Since the equilibrium is static, then µ =0, so that σij = σji.  94 5. APPLICATIONS

We can hence write σ11 σ12 σ13 σ = σ12 σ22 σ23 , σ13 σ23 σ33 where the diagonal entries σ11, σ22and σ33 are the normal components, that is the components of the forces perpendicular to the coordinate planes and the remaining entries σ12, σ13 and σ23 are the shear components, that is the components of the forces parallel to the coordinate planes. Since the stress tensor is symmetric, again by the spectral theorem (Theorem 3.9), it can be orthogonally diagonalized, that is σ1 0 0 σ = 0 σ2 0 ,  0 0 σ3 where now σ1, σ2 and σ3 are the principal stresses , that is the eigenvalues of σ. The eigenspaces of σ are the principal directions and the shear components disappear for the principal planes, i.e., the planes orthogonal to the principal directions.

5.2.3. Special Forms of the Stress Tensor.

We consider the stress tensor with respect to an orthonormal eigenbasis or another special basis, so that the corresponding matrix has a simpler form. We use the following terminology: Uniaxial stress for a stress tensor given by • σ 0 0 0 0 0 0 0 0 Example 5.15. This is the stress tensor in a long vertical rod loaded by hang- ing a weight on the end.  Plane stressed state or biaxial stress for a stress tensor given by • σ1 0 0 0 σ2 0  0 0 0 Example 5.16. This is the stress tensor in plate on which forces are applied as in the picture.

 5.2. STRESS TENSOR (SPANNUNG) 95

Pure shear for a stress tensor given by • σ 0 0 0 σ 0 (5.8) −0 σ 0 or σ 0 0 .  0 00 0 0 0     This is special case of the biaxial stress, in the case where σ1 = σ2. In (5.8) the first is the stress tensor written with respect to an eigenbasis,− while the second is the stress tensor written with respect to an orthonormal basis obtained by rotating an eigenbasis by 45◦ about the third axis. In fact 0 σ 0 √2 √2 0 σ 0 0 √2 √2 0 2 2 − 2 − 2 σ 0 0 = √2 √2 0 0 σ 0 √2 √2 0    2 2     2 2  0 0 0 −0 01 0 00 0 0 1         tL L where L is the matrix| of{z the change} of coordinates.| {z }

Shear deformation for a stress tensor given by • 0 σ12 σ13 σ12 0 σ23 σ13 σ23 0    (with respect to some orthonormal basis). It turns out that the stress tensor σ is equivalent to a shear deformation if and only if its trace is zero. Example 5.17. The stress tensor 2 4 0 4− 0 4 −0 4 2 − represents a shear deformation. In fact, one can check that √2 0 √2 2 4 0 √2 0 √2 0 0 2 2 2 2 2 − 0 10 4− 0 4 01− 0 = 0 0 4√2   −      √2 0 √2 0 4 2 √2 0 √2 2 4√2 0 2 2 − 2 2 − −        tL L  | {z } | {z } 96 5. APPLICATIONS

Hydrostatic pressure with stress tensor given by • p 0 0 −0 p 0 ,  0− 0 p − where p =0 is the pressure. Here all eigenvalues are equal to p. 6 − Example 5.18. Pressure of a fluid on a bubble.  Exercise 5.19. Any stress tensor can be written as the sum of a hydrostatic pressure and a shear deformation. Hint: look at the next section.

5.2.4. Invariants.

ij Let A be a 3 3 matrix with entries a . The characteristic polynomial pA(λ) of A is invariant under× conjugation (see §1.4.2), so its coefficients remain unchanged. a11 λ a12 a13 21− 22 23 pA(λ) = det(A λI) = det a a λ a −  a31 a32− a33 λ − = λ3 + tr Aλ2   − a11a22 + a11a33 + a22a33 a12a21 a23a32 a13a31 λ + det A. − − − − quadratic expression in the entries of A
Applying this to| the stress tensor σ = A,{z we obtain some stress invariants} , namely: 11 22 33 I1 := tr σ = σ + σ + σ I :=(σ12)2 +(σ23)2 +(σ13)2 σ11σ22 σ22σ33 σ33σ11 2 − − − I3 := det σ .

That means, that the above quantities I1,I2 and I3 are invariant when we change the orthonormal basis.23 The stress tensor can be expressed as the sum of 2 other stress tensors; The hydrostatic stress tensor • π 0 0 πδij = 0 π 0 , 0 0 π

11 22 33  where π := I1/3=(σ + σ + σ )/3. This relates to volume change.

23By contravariance, when we change basis via a matrix L, the matrix of the stress tensor changes from σ to σ = tΛσΛ, where Λ= L−1. But since we are restricting to orthonormal bases, we have that the change of basis matrix is orthogonal, i.e., Λ= tL, so this is in fact a conjugation: σ = LσL−1. e e 5.2. STRESS TENSOR (SPANNUNG) 97

The deviatoric stress tensor •

σ11 π σ12 σ13 sij := σij πδij = σ12− σ22 π σ23 . −  σ13 σ23− σ33 π −   This relates to shape change. Remark 5.20. The hydrostatic pressure is generally defined as the negative one third of the stress invariant I , i.e. p = π 1 − Remark 5.21. The stress deviator tensor is a shear deformation since tr(sij)= σ11 + σ22 + σ33 3π =0. − Clearly σij = sij + πδij and hence the stress tensor is a sum of a shear deformation and a hydrostatic pressure as claimed in Exercise 5.19.

5.2.5. Contravariance of the Stress Tensor.

Let = e , e , e and = e˜ , e˜ , e˜ be two orthonormal bases, and let B { 1 2 3} B { 1 2 3} (5.9) e˜ = Lje and e =Λje˜ , i e i j i i j 1 where L := L e is the matrix of the change of basis and Λ := L− is the inverse. Let n be a given unitBB vector and σ the stress across a surface perpendicular to n. Then σ can be expressed in two ways, respectively with respect to and to , B B (5.10) σ = σij(n · e )e and σ =σ ˜ij(n · e˜ )˜e , i j i j e and we want to relate σij to σ˜ij. We start with the first expression for S in (5.10) and rename the indices for later convenience: km n · km n · i j km i j n · σ = σ ( ek)em = σ ( Λke˜i)(Λme˜j)= σ ΛkΛm( e˜i)˜ej , where in the second equality we used (5.9), and in the third we used linearity. Comparing the last expression with the second expression in (5.10) we obtain

ij km i j σ˜ = σ ΛkΛm , thus showing that σ is a contravariant 2-tensor or a tensor of type (0, 2). Heuristically, we may think that the stress at each point takes one plane, thought of as the kernel of a linear form, and gives a vector, thus is a linear map V ∗ V or, → equivalently, a bilinear map V ∗ V ∗ R, i.e., a tensor of type (0, 2). × → 98 5. APPLICATIONS

5.3. Strain Tensor (Verzerrung)

5.3.1. Physical Preliminaries.

Consider a slight deformation of a body, where we compare the relative positions of two particles before and after the deformation:

P1 . Pe1 .

P .

Pe .

e displacement of P P1

P1 ∆u

∆x ∆˜x

P Pe displacement of P1

We have

∆˜x = ∆x + ∆u , where ∆x is the old relative position of P and P1, ∆˜x is their new relative position and ∆u is the displacement difference, which hence measures the deformation. Assume that we have a small homogeneous deformation, that is

∆u = f(∆x) , in other words, f is a small linear function independent of the point P . If we write the components of ∆u and ∆x with respect to an orthonormal basis e , e , e , the { 1 2 3} function f will be represented by a matrix with entries that we denote by fij,

j ∆ui = fij∆x . 5.3. STRAIN TENSOR (VERZERRUNG) 99

The matrix (fij) can be written as a sum of a symmetric and an antisymmetric matrix as follows:

fij = ǫij + ωij , where 1 ǫ = (f + f ) ij 2 ij ji is a symmetric matrix and is called the strain tensor or deformation tensor and 1 ω = (f f ) ij 2 ij − ji is an antisymmetric matrix called the rotation tensor. We will next try to understand where these names come from.

Remark 5.22. First we verify that a (small) antisymmetric 3 3 matrix represents a (small) rotation in 3-dimensional space. ×

Fact 5.23. Let V be a vector space with orthonormal basis = e1, e2, e3 , and let a B { } ω = b . The matrix Rω of the linear map V V defined by v ω v with respect c → 7→ × to the basis is B 0 c b − Rω = c 0 a .  b a −0  −   Indeed, we have

a x e1 e2 e3 ω v = b y = det a b c × c × z  x y z  bz cy  0  c b x = cx − az = c −0 a y . ay − bx  b a −0  z − −       0 ω ω 12 − 31 Note that the matrix Rω = (ωij) := ω12 0 ω23 corresponds to the cross −ω ω 0  31 − 23 ω   − 23 product with the vector ω = ω31 .  −ω  − 12   100 5. APPLICATIONS

5.3.2. The Antisymmetric Case: Rotation.

Suppose that the matrix (fij) was already antisymmetric, so that

ωij = fij and ǫij =0 . Note that 1 ω = (f f )=0. ii 2 ii − ii ω − 23 By the Fact 5.23, if ω = ω31 , then −ω  − 12  R ∆x = ω ∆x ω × and the relation i j (5.11) ∆u = fij∆x is equivalent to ∆u = ω ∆x , × so that ∆˜x = ∆x + ∆u = ∆x + ω ∆x . × When ω is small, this represents an infinitesimal rotation of an angle ω about the axis Oω. k k

r

∆x ∆˜x

ω α

O 5.3. STRAIN TENSOR (VERZERRUNG) 101

In fact, since ω ∆x is orthogonal to the plane determined by ω and by ∆x, it is tangent to the circle× with center along the axis Oω and radius determined by ∆x. Moreover, ∆u = ω ∆x = ω ∆x sin α , k k k × k k kk k r and hence, since the length of an arc of a circle of radius r corresponding to an angle θ | {z } is rθ, infinitesimally this represents a rotation by an angle ω . k k

5.3.3. The Symmetric Case: Strain.

The opposite extreme case is when the matrix fij was already symmetric, so that

ǫij = fij and ωij =0 .

We will see that it is ǫij that encodes the changes in the distances: in fact, ∆˜x 2 = ∆˜x · ∆˜x = (∆x + ∆u) · (∆x + ∆u) k k (5.12) = ∆x · ∆x + 2∆x · ∆u + ∆u · ∆u ∆x 2 +2ǫ ∆xi∆xj , ≃k k ij where in the last step we neglected the term ∆u 2 since it is small compared to ∆u when ∆u 0 and used (5.11). k k → Remark 5.24. Even when fij is not purely symmetric, only the symmetric part ǫij is relevant for the distortion of the distances. In fact, since ωij is antisymmetric, the term i j 2ωij∆x ∆x =0, so that ∆˜x 2 ∆x 2 +2f ∆xi∆xj = ∆x 2 +2ǫ ∆xi∆xj . k k ≃k k ij k k ij 

Recall that a metric tensor (or inner product) encodes the distances among points. It follows that a deformation changes the metric tensor. Let us denote by g the metric before the deformation and by g˜ the metric after the deformation. By definition, we have (5.13) ∆˜x 2 def=g ˜(∆˜x, ∆˜x)=˜g ∆˜xi∆˜xj =g ˜ (∆xi + ∆ui)(∆xj + ∆uj) k k ij ij and (5.14) ∆x 2 def= g(∆x, ∆x)= g ∆xi∆xj . k k ij For infinitesimal deformations (that is, if ∆u 0), (5.13) becomes ∼ ∆˜x 2 =g ˜ ∆xi∆xj . k k ij This, together with (5.14) and (5.12), leads to g˜ ∆xi∆xj g ∆xi∆xj +2ǫ ∆xi∆xj ij ≃ ij ij 102 5. APPLICATIONS and hence 1 ǫ (˜g g ) , ij ≃ 2 ij − ij that is, ǫij measures the change in the metric. By definition the strain tensor ǫij is symmetric

ǫ11 ǫ12 ǫ13 = ǫ ǫ ǫ , E  12 22 23 ǫ13 ǫ23 ǫ33 where the terms on the diagonal (ingreen) determine the elongation or the contraction of the body along the coordinate directions e1, e2, e3, and the terms above the diagonal (in orange) are the shear components of the strain tensor; that is ǫij is the movement of a line element parallel to Oej towards Oei. Since it is a symmetric tensor, it can be orthogonally diagonalized (cf. Theorem 3.9), so we can find an orthonormal basis with respect to which is given by E ǫ1 0 0 0 ǫ 0 ,  2  0 0 ǫ3 The eigenvalues of are the principal coefficients of the deformation and the eigenspaces are the principal directionsE of the deformation.

5.3.4. Special Forms of the Strain Tensor.

We use the following terminology: (1) Shear deformation when is traceless E tr = ǫ + ǫ + ǫ =0 . E 11 22 33 (2) Uniform compression when the principal coefficients of are equal (and nonzero) E k 0 0 0 k 0 0 0 k Exercise 5.25. Any strain tensor can be written as the sum of a uniform compression and a shear deformation.

5.4. Elasticity Tensor The stress tensor represents an external exertion on the material, while the strain tensor represents the material reaction to that exertion. In crystallography these are called field tensors because they represent imposed conditions, opposed to matter tensors, that represents material properties. 5.4. ELASTICITY TENSOR 103

Hooke’s law says that, for small deformations, stress is related to strain by a matter tensor called elasticity tensor or stiffness tensor E: ij ijkℓ σ = E ǫkl , while the tensor relating strain to stress is the compliance tensor S: ij ǫkℓ = Sijkℓσ . The elasticity tensor has order 4, and hence in 3-dimensional space it has 34 = 81 components. Luckily, symmetry reduces the number of independent components for Eijkℓ. (1) Minor symmetries: The symmetry of the stress tensor σij = σji implies that Eijkℓ = Ejikℓ for each k,ℓ ; it follows that for each k,ℓ fixed there are only 6 independent components Eijkℓ E11kℓ E12kℓ E13kℓ E12kℓ E22kℓ E23kℓ E13kℓ E23kℓ E33kℓ . Having taken this in consideration, the number of independent components decreases to 6 32 at the most. Moreover, the symmetry also of the strain tensor ×

ǫkℓ = ǫℓk implies that Eijkℓ = Eijℓk for each i, j . This means that for each i, j fixed there are also only 6 independent components Eijkℓ, so that Eijkℓ has at most 62 = 36 independent components. (2) Major symmetries: Since (under appropriate conditions) partial derivatives commute, if follows from the existence of a strain energy density functional U satisfying ∂2U = Eijkℓ ∂ǫij ∂ǫkℓ that Eijkℓ = Ekℓij , that means the matrix with rows labelled by (i, j) and columns labelled by (k,ℓ) is symmetric. Since, from (1), there are only 6 entries (i, j) for a fixed (k,ℓ), 104 5. APPLICATIONS

Eijkℓ can be written in a 6 6 matrix with rows labelled by (i, j) and columns labelled by (k,ℓ) ×

∗∗∗∗∗∗  ∗∗∗∗∗ ∗∗∗∗    ∗ ∗ ∗    ∗ ∗    ∗ so that Eijkℓ has in fact only 6+5+4+3+2+1=21 components.

5.5. Conductivity Tensor Consider a homogeneous continuous crystal. Its properties can be divided into two classes: Properties that do not depend on a direction, and are hence described by scalars. • Examples are density and heat capacity. Properties that depend on a direction, and are hence described by tensors. • Examples are elasticity, electrical conductivity and heat conductivity. We say that a crystal is anisotropic when it has such tensorial properties.

5.5.1. Electrical Conductivity.

Let E be the electric field and J the electrical current density. We assume that these are constant throughout the crystal. At each point of the crystal: (1) E gives the electric force (in Volts/m) that would be exerted on a positive test charge (of 1 Coulomb) placed at the point; (2) J (in Amperes/m2) gives the direction the charge carriers move and the rate of electric current across an infinitesimal surface perpendicular to that direction. The electrical current density J is a function of the electric field E, J = f(E) . Consider a small increment ∆J in J caused by a small increment ∆E in E, and write these increments in terms of their components with respect to a chosen orthonormal basis e , e , e . { 1 2 3} i i ∆J = ∆J ei and ∆E = ∆E ei . By Calculus, the increments are related by ∂f i ∆J i = ∆Ej + higher order terms in (∆Ej)2, (∆Ej)3,... ∂Ej 5.5. CONDUCTIVITY TENSOR 105

If the quantities ∆Ej are small, we can assume that ∂f i (5.15) ∆J i = ∆Ej ∂Ej ∂f i If we assume that ∂Ej is independent of the point of the crystal, ∂f i = κi R ∂Ej j ∈ we obtain the relation i i j ∆J = κj∆E or simply ∆J = κ ∆E, where κ if the electrical conductivity tensor. Thisisa (1, 1)-tensor and may depend24 on the initial value of E, that is the electrical conductivity may be different for small and large electric forces. If initially E =0 and κ0 is the corresponding electrical conductivity tensor, we obtain the relation J = κ0 E that is called the generalized Ohm law. This is always under the assumption that ∆E and ∆J are small and that the relation is linear. The electrical resistivity tensor is the inverse of κ: 1 ρ := κ− , that is, it is the (1, 1)-tensor such that j ℓ ℓ ρi κj = δi . The electrical conductivity measures the material’s ability to conduct an electrical cur- rent, while the electrical resistivity quantifies the ability of the material to oppose the flow of the electrical current. For an isotropic crystal, all directions are equivalent and these tensors are spherical, meaning 1 (5.16) κj = kδj and ρj = δj , i i i k i where k is a scalar, called the electrical conductivity of the crystal. Equation (5.16) can also be written as 1 k 0 0 k 0 0 1 0 k 0 and 0 k 0 .    1  0 0 k 0 0 k     24Typically, if the dependence between E and J is linear for any value, and not only for small ones, the tensor κ will not depend on the initial value of E. 106 5. APPLICATIONS

j In general, κi is neither symmetric nor antisymmetric (and actually symmetry does not even make sense for a (1, 1) tensor unless a metric is fixed, since it does require a canonical identification of V with V ∗).

5.5.2. Heat Conductivity.

Let T be the temperature and H the heat flux vector. For a homogeneous crystal and constant H and for a constant gradient of T , Fourier heat conduction law says that (5.17) H = K grad T. − At each point of the crystal: (1) grad T points in the direction of the highest ascent of the temperature and mea- sures the rate of increase of T in that direction. The minus sign in (5.17) comes from the fact that the heat flows in the direction of decreasing temperature. (2) H measure the amount of heat passing per unit area perpendicular to its direc- tion per unit time. Here, K is the heat conductivity tensor or thermal conductivity tensor. In terms of components with respect to a chosen orthonormal basis, we have Hi = Kij(grad T ) . − j Exercise 5.26. Verify that the gradient of a real function is a covariant 1-tensor. The heat conductivity tensor is a contravariant 2-tensor and experiments show that it is symmetric and hence can be orthogonally diagonalized. The heat resistivity tensor is its inverse: 1 r := K− , and hence is also symmetric. With respect to an orthonormal basis, K is represented by

K1 0 0 0 K 0 ,  2  0 0 K3 where the eigenvalues of K are called the principal coefficients of heat conductivity. The fact that heat flows always in the direction of decreasing temperature shows that the eigenvalues are positive

Ki > 0 . The eigenspaces of K are called the principal directions of heat conductivity. Solutions to Exercises

Exercise 1.4: (1) yes; (2) no, (3) no, (4) yes, (5) no. Exercise 1.23: (1) The vectors in span V , since any element of V is of the form B a b 1 0 0 1 0 0 = a + b + c . c a 0 1 0 0 1 0  −   −      Moreover, the vectors in are linearly independent since B 1 0 0 1 0 0 0 0 a + b + c = 0 1 0 0 1 0 0 0  −        if and only if a b 0 0 = , c a 0 0  −    that is, if and only if a = b = c =0. (2) We know that dim V =3, as is a basis of V and has three elements. Since B also has three elements, it is enough to check either that it spans V or that itB consists of linearly independent vectors. We will check this last condition. Indeed,e 1 0 0 1 0 1 0 0 a c b 0 0 a + b − + c = − = , 0 1 1 0 1 0 0 0 ⇐⇒ b + c a 0 0  −         −    that is, a =0 a =0 b + c =0 b =0  ⇐⇒  c b =0 c =0. − (3) Since   2 1 1 0 0 1 0 0 =2 +1 +7 , 7 2 0 1 0 0 1 0  −   −      we have 2 [v] = 1 . B 7

107 108 SOLUTIONS TO EXERCISES

To compute the coordinates of v with respect to we need to find a, b, c R such that B ∈ 2 1 1 0 0 1 0e 1 = a + b + c . 7 2 0 1 1− 0 1 0  −   −      Solving the corresponding system of linear equations as above yields 2 [v] e = 3 . B 4   Exercise 2.2: (1) no; (2) no; (3) yes.

Exercise 2.3: (1) yes; (2) yes; (3) no.

Exercise 2.7: We show that V ∗ is a subspace of the vector space of all real-valued functions on V (cf. Example 1.3(3) on page 8), by checking the three conditions: (1) The 0-function 0 associating the number zero to each vector in V is linear because 0+0=0 and k0=0 for every k R, so 0 V ∗; ∈ ∈ (2) V ∗ is closed under addition since, if α : V R and β : V R are linear, then α + β : V R defined by (α + β)(v)= α→(v)+ β(v) is also→ linear (in v V ); → ∈ (3) V ∗ is closed under multiplication by scalars since, if α : V R is linear and k R, then kα : V R defined by (kα)(v)= k (α(v)) is also→ linear. ∈ → Exercise 2.12: In fact: 1 1 L1 ... Ln . . [α] ∗ L = α1 ... αn . . B  . .  Ln ... Ln
 1 n i  i  = αiL1 ... αiLn = α1 ... αn
=[α] e∗
eB e Exercise 2.17: By the Leibniz formula, we have u u1 u2 u3 det v = det v1 v2 v3 w w1 w2 w3   = u1 v2w3 v3w2 + u2 v3w1 v1w3 + u3 v1w2 v2w1 − − − 1 2 3 3 2 u v w
v w

= u2 · v3w1 − v1w3 = u · (v w) . u3 v1w2 − v2w1 × −     SOLUTIONS TO EXERCISES 109

Exercise 2.19: (1) yes; (2) yes; (3) yes; (4) no, because v w is not a real number; (5) no, because it fails linearity (the area of the parallelogram× spanned by v and w is the same as that of the parallelogram spanned by v and w); (6) no, because because it fails linearity in the second argument (the determinant− of a matrix with n> 1 is not linear in that matrix).

Exercise 2.29: In fact, 1 n 1 1 L1 ... L1 B11 ... B1n L1 ... Ln t ...... LBL =  . .   . .   . .  L1 ... Ln B ... B Ln ... Ln  n n  n1 nn  1 n  1 n  i  i  L1 ... L1 B1iL1 ... B1iLn . . . . =  . .   . .  L1 ... Ln B Li ... B Li  n n  ni 1 ni n  k i   k i  L1BkiL1 ... L1BkiLn . . =  . .  Lk B Li ... Lk B Li  n ki 1 n ki n =B. 

e Exercise 3.3: (1) no, as ϕ is negative definite, that is ϕ(v, v) < 0 if v V , v =0; (2) no, as ϕ is not symmetric; ∈ 6 (3) no, as ϕ is not positive definite; (4) no, as ϕ is not positive definite; (5) yes; (6) yes.

Exercise 3.4: (1) Yes, in fact: 1 1 (a) 0 p(x)q(x)dx = 0 q(x)p(x)dx because p(x)q(x)= q(x)p(x); (b) 1(p(x))2dx 0 for all p R[x] because (p(x))2 0, and 1(p(x))2dx = R0 R 2 0 0 only when p≥(x)=0 for∈ all x [0, 1], that is only≥ if p 0. R 1 2 ∈ ≡ R (2) No, since 0 (p′(x)) dx = 0 implies that p′(x)=0 for all x [0, 1], but such p is not necessarily the zero polynomial. ∈ (3) Yes. R 2 2 (4) No. Is there p R[x]2, p =0 such that (p(1)) +(p(2)) =0? (5) Yes. Is there a∈ non-zero polynomial6 of degree 2 with 3 distinct zeros? (6) No, since this is not symmetric. 110 SOLUTIONS TO EXERCISES

Exercise 3.12: We write v˜1 w˜1 2 2 [v] e = v˜ and [w] e = w˜ . B v˜3 B w˜3     We know that g with respect to the basis has the standard form g(v,w)=˜viw˜i and B 1 we want to verify (3.7) using the matrix of the change of coordinates L− =Λ. If v1 e w1 [v] = v2 and [w] = w2 B v3 B w3 then we have that     v˜1 v1 v1 v2 v˜2 =Λ v2 = v2 − v3 v˜3 v3  v−3  and       w˜1 w1 w1 w2 w˜2 =Λ w2 = w2 − w3 w˜3 w3  w−3  It follows that       g(v,w)=˜viw˜i =(v1 v2)(w1 w2)+(v2 v3)(w2 w3)+ v3w3 − − − − = v1w1 v1w2 v2w1 +2v2w2 v2w3 w3v2 +2v3w3 . − − − −

Exercise 3.15: With respect to , we have B v = (12 +12 +12)1/2 = √3 k k e w = (( 1)2 +( 1)2 +32)1/2 = √11 k k − − and with respect to E v = (3 3 3 2 2 3+2 2 2 2 1 1 2+2 1 1)1/2 = √3 k k · − · − · · · − · − · · · w = (1 1 1 2 2 1+2 2 2 2 3 3 2+2 3 3)1/2 = √11 . k k · − · − · · · − · − · · ·

Exercise 3.16: Saying that the orthogonality is meant with respect to g, means that we have to show that g(v proj v, b )=0. In fact, − bk k g(v, bk) g(v, bk) ✘✘✘ ✘✘✘ g(v projbk v, bk)= g(v bk, bk)= g(v, bk) ✘✘ g(bk, bk)=0 − − g(bk, bk) − ✘g(✘bk, bk)

Exercise 3.20: SOLUTIONS TO EXERCISES 111

(1) The coordinate vectors of basis vectors with respect to that same basis are simply standard vectors, in this case: 1 0 0 [˜b1] e = 0 , [˜b2] e = 1 and [˜b3] e = 0 . B 0 B 0 B 1       1 1 0 − (2) As in Example 3.11, we have G e = I and G = 1 2 1 . B E −0 1− 2  − (3) In parts (a) and (b), note that, for an orthonormal basis , we have g = . In parts (c) and (d), we use the computations in ExampleB 3.19 andB the factB that [v] = L e [v] e. e e e E BE B 1 0 0 ˜1 ˜ ˜2 ˜ ˜3 ˜ (a) [b ] e = [b1] e = 0 , [b ] e = [b2] e = 1 and [b ] e = [b3] e = 0 . B B 0 B B 0 B B 1 1 1  1  1 2 3 (b) [˜b ] = [˜b1] = 0 , [˜b ] = [˜b2] = 1 and [˜b ] = [˜b3] = 1 . E E 0 E E 0 E E 1 1   0  0   1 2 3 (c) [e ] e = 1 , [e ] e = 1 and [e ] e = 0 . B 1 B 1 B 1 3 2 1  (d) [e1] = 2 , [e2] = 2 and [e3] = 1 . E 1 E 1 E 1       Exercise 3.23: (1) The assertion in the case of the bases and g follows from E E 3 2 1 1 1 |1 |2 |3 G− =(L g )− = e e e = 2 2 1 . E EE   1 1 1 | | | ˜j ˜ 1     (2) Since b = bj, we have G−e = L e eg = I and (3.15) is immediately verified. B BB Exercise 4.7: We first use upper indices for rows and lower indices for columns and Assume that A and B are square matrices of the same size. If A has (i, j)-entry (where i i i labels the row and j the column) Aj and B has (i, j)-entry Bj, then by the definition of matrix product the matrix C := AB has (i, j)-entry i k AkBj j t and the transpose of A has (i, j)-entry Ai , so C A has (i, j)-entry i j CℓAℓ 112 SOLUTIONS TO EXERCISES and ABtA has (i, j)-entry i k j AkBℓ Aℓ . k kℓ We obtain the formula (4.3) by replacing A by Λ and Bℓ by S .

Exercise 4.11: The coordinates of b1 a1 + b2 a2 with respect to the basis bi aj are δij. So the task amounts to showing⊗ that there⊗ is no solution to the equation ⊗ v1 1 0 w1 w2 = v2 0 1    
where the vi and wj are the coordinates of arbitrary vectors v V and w W w.r.t. the given bases. Indeed, the system ∈ ∈ v1w1 =1 1 2 v w =0 2 1 v w =0  v2w2 =1  2 has no solution, since the first two equations force that w be zero, yet this is impossible for the last equation.

Exercise 5.4: We choose a coordinate system with origin at the vertex O, with x-axis along the side of length a, y-axis along the side of length b and z-axis perpendicular to m the plate. We already know that the mass density is constant equal to ρ = ab . Then a b 2 2 I11 = (y + z ) ρ dy dx 0 0 m Ixx =0 Z Z ab |{z} m b |{z} = a y2 dy |{z} ab Z0 m y3 b m = = b2 . b 3 3  0 Similarly, m I = a2 , 22 3 Iyy and |{z} a b 2 2 m 2 2 I33 = (x + y )ρdydx = (a + b ) 0 0 3 Izz Z Z is again just the sum|{z} of I11 and I22. SOLUTIONS TO EXERCISES 113

Furthermore, a b I = I = y z ρ dydx =0 , 23 32 − Z0 Z0 =0 and, similarly, I31 = I13 =0. Finally, we have|{z} a b m x2 a y2 b mab I = I = xyρdydx = = . 21 21 − −ab 2 2 − 4 Z0 Z0  0  0 We conclude that the inertia tensor is given by the matrix 4b2 3ab 0 m − 2 12 3ab 4a 0 . −0 0 4(a2 + b2)   Exercise 5.6: We again choose the unit vector a √a2+b2 a b 1 p = = 2 2 b  √a2+b2  √a +b   0 0 defining the axis of rotation, but now use the inertia tensor computed in Exercise 5.4, 4b2 3ab 0 m − 2 12 3ab 4a 0 , −0 0 4(a2 + b2) thus obtaining   i j Ip =Iijp p 4b2 3ab 0 a m − 2 = 12(a2+b2) a b 0 3ab 4a 0 b −0 0 4(a2 + b2) 0
m a2b2     = 6 a2+b2 , necessarily the same result as in Example 5.5.

Exercise 5.7: We use the inertia tensor calculated in Example 5.3, where the origin of the coordinate system is at the center of mass, and choose p = e3. The moment of inertia is then i j Ip =Iijp p b2 0 0 0 m 2 = 12 0 0 1 0 a 0 0  0 0 a2 + b2 1
m 2  2    =I33 = 12 (a + b ) . 114 SOLUTIONS TO EXERCISES

Exercise 5.8: We use the inertia tensor calculated in Exercise 5.4, where the origin of the coordinate system is at a vertex of the plate, and choose p = e3. The moment of inertia is then i j Ip =Iijp p 4b2 3ab 0 0 m − 2 = 12 0 0 1 3ab 4a 0 0 −0 0 4(a2 + b2) 1
m 2 2    =I33 = 3 (a + b ) .