Linear Algebra Course 211

David Simms School of Mathematics

Trinity College Dublin

Original typesetting Brendan Griffin.

Alterations Geoff Bradley.

0–1

Table of Contents

Table of Contents  0–1
1  Vector Spaces  1–1
2  Linear Operators  2–1
2.1 The Definition  2–1
2.2 Basic Properties of Linear Operators  2–3
2.3 Examples  2–4
2.4 Properties Continued  2–8
2.5 Operator Algebra  2–10
2.6 Isomorphisms of L(M, N) with K^{m×n}  2–11
3  Changing Basis and Einstein Convention  3–1
4  Linear Forms and Duality  4–1
4.1 Linear Forms  4–1
4.2 Duality  4–3
4.3 Systems of Linear Equations  4–4
5 Tensors 5–1
5.1 The Definition  5–1
5.2 Contraction  5–3
5.3 Examples  5–4
5.4 Bases of Tensor Spaces  5–5
6 Vector Fields 6–1
6.1 The Definition  6–1
6.2 Velocity Vectors  6–3
6.3 Differentials  6–4
6.4 Transformation Law  6–7
7 Scalar Products 7–1
7.1 The Definition  7–1
7.2 Properties of Scalar Products  7–1
7.3 Raising and Lowering Indices  7–5
7.4 Orthogonality and Diagonal Matrix (including Sylvester’s Law of Inertia) 7–7
7.5 Special Spaces (including Jacobi’s Theorem)  7–12

0–2

8  Linear Operators 2  8–1
8.1 Adjoints and Isometries  8–1
8.2 Eigenvalues and Eigenvectors  8–4
8.3 Spectral Theorem and Applications  8–7
9 Skew-Symmetric Tensors and Wedge Product 9–1
9.1 Skew-Symmetric Tensors  9–1
9.2 Wedge Product  9–5
10 Classification of Linear Operators 10–1
10.1 Hamilton-Cayley and Primary Decomposition  10–1
10.2 Diagonalisable Operators  10–4
10.3 Conjugacy Classes  10–6
10.4 Jordan Forms  10–9
10.5 Determinants  10–18
11 Orientation 11–1
11.1 Orientation of Vector Spaces  11–1
11.2 Orientation of Coordinate Systems  11–6
12 Manifolds and (n)-dimensional Vector Analysis 12–1
12.1 Gradient  12–1
12.2 3-dimensional Vector Analysis  12–3
12.3 Results  12–4
12.4 Closed and Exact Forms  12–7
12.5 Contraction and Results (including Poincaré Lemma)  12–9
12.6 Surface in ℝ^{3}  12–14
12.7 Integration on a Manifold (Sketch)  12–16
12.8 Stokes Theorem and Applications  12–18

0–3

Chapter 1

Vector Spaces

Recall that in course 131 you studied the notion of a linear vector space. In that course the scalars were real numbers. We will study the more general case, where the set of scalars is any field K. For example ℚ, ℝ, ℂ, ℤ/(p).

Definition. Let K be a field. A set M is called a vector space over the field K (or a K-vector space) if

(i) an operation

M × M → M
(x, y) ↦ x + y

is given, called addition of vectors, which makes M into a commutative group;

(ii) an operation

K × M → M
(λ, x) ↦ λx

is given, called multiplication of a vector by a scalar, which satisfies:

(a) λ(x + y) = λx + λy,

(b) (λ + μ)x = λx + μx,

(c) λ(μx) = (λμ)x,

(d) 1x = x for all λ, μ ∈ K, x, y ∈ M, where 1 is the unit element of the field K.

1–1

The elements of M are then called the vectors, and the elements of K are called the scalars of the given K-vector space M.

Examples:

1. The set of 3-dimensional geometrical vectors (as in 131) is a real vector space (ℝ-vector space).

2. The set ℝ^{n} (as in 131) is a real vector space.

3. If K is any field then the following are K-vector spaces:

(a) K^{n} = {(α_{1}, …, α_{n}) : α_{1}, …, α_{n} ∈ K}, with vector addition:

(α_{1}, …, α_{n}) + (β_{1}, …, β_{n}) = (α_{1} + β_{1}, …, α_{n} + β_{n}),

and scalar multiplication:

λ(α_{1}, …, α_{n}) = (λα_{1}, …, λα_{n}).

(b) The set K^{m×n} of m × n matrices (m rows and n columns) with entries in K (m, n fixed integers ≥ 1), with vector addition:

( \begin{matrix}α_{11} & ⋯ & α_{1n} \\ ⋮ & & ⋮ \\ α_{m1} & ⋯ & α_{mn} \end{matrix} ) + ( \begin{matrix}β_{11} & ⋯ & β_{1n} \\ ⋮ & & ⋮ \\ β_{m1} & ⋯ & β_{mn} \end{matrix} )

= ( \begin{matrix}α_{11} + β_{11} & ⋯ & α_{1n} + β_{1n} \\ ⋮ & & ⋮ \\ α_{m1} + β_{m1} & ⋯ & α_{mn} + β_{mn} \end{matrix} ) ,

and scalar multiplication:

λ ( \begin{matrix}α_{11} & ⋯ & α_{1n} \\ ⋮ & & ⋮ \\ α_{m1} & ⋯ & α_{mn} \end{matrix} ) = ( \begin{matrix}λα_{11} & ⋯ & λα_{1n} \\ ⋮ & & ⋮ \\ λα_{m1} & ⋯ & λα_{mn} \end{matrix} ) .

(c) The set K^{X} of all maps from X to K (X a fixed non-empty set), with vector addition:

(f + g)(x) = f(x) + g(x),

and scalar multiplication:

(λf)(x) = λ(f(x))

for all x ∈ X, f, g ∈ K^{X}, λ ∈ K.

1–2 Definition. Let N ⊂ M, and let M be a K-vector space. Then N is called a K-vector subspace of M if N is non-empty, and

(i) x, y ∈ N ⇒ x + y ∈ N closed under addition; (ii) λ ∈ K, x ∈ N ⇒ λx ∈ N closed under scalar multiplication.

Thus N is itself a K-vector space. Examples:

1. {(α, β, γ) : 3α + β − 2γ = 0; α, β, γ ∈ ℝ} is a vector subspace of ℝ^{3}.

2. \{\underline{v} : \underline{v}.\underline{n} = 0\}, \underline{n} fixed, is a vector subspace of the space of 3-dimensional geometric vectors (see Figure 1.1).

Figure 1.1

3. The set C^{0}(ℝ) of continuous functions is a real vector subspace of the set ℝ^{ℝ} of all maps ℝ → ℝ.

4. Let V be an open subset of ℝ. We denote by C^{0}(V) the space of all continuous real valued functions on V, C^{r}(V) the space of all real valued functions on V having continuous rth derivative, C^{∞}(V) the space of all real valued functions on V having derivatives of all r.

Then

C^{∞}(V) ⊂ ⋯ ⊂ C^{r+1}(V) ⊂ C^{r}(V) ⊂ ⋯ ⊂ C^{0}(V) ⊂ ℝ^{V} is a sequence of real vector subspaces.

1–3 5. The space of solutions of the differential equation

\frac{d^{2}u}{dx^{2}} + w^{2}u = 0

is a real vector subspace of C^{∞}(ℝ).

Definition. Let u_{1}, …, u_{r} be vectors in a K-vector space M, and let α_{1}, …, α_{r} be scalars. Then the vector

α_{1}u_{1} + ⋯ + α_{r}u_{r}

is called a linear combination of u_{1}, …, u_{r}. We write

S(u_{1}, …, u_{r}) = {α_{1}u_{1} + ⋯ + α_{r}u_{r} : α_{1}, …, α_{r} ∈ K}

to denote the set of all linear combinations of u_{1}, …, u_{r}. S(u_{1}, …, u_{r}) is a K-vector subspace of M, and is called the subspace generated by u_{1}, …, u_{r}.

If S(u_{1}, …, u_{r}) = M, we say that u_{1}, …, u_{r} generate M (i.e. for each x ∈ M there exists α_{1}, …, α_{r} ∈ K such that x = α_{1}u_{1} + ⋯ + α_{r}u_{r}).

Examples:

1. The vectors (1, 2), (−1, 1) generate ℝ^{2} (see Figure 1.2), since

(α, β) = \frac{α + β}{3} (1, 2) + \frac{β − 2α}{3}(−1, 1).

2. The functions cos ωx, sin ωx generate the space of solutions of the differential equation:

\frac{d^{2}u}{dx^{2}} + w^{2}u = 0.

Figure 1.2

1–4 Definition. Let u_{1}, …, u_{r} be vectors in a K-vector space M. Then

(i) u_{1}, …, u_{r} are linearly dependent if there exist α_{1}, …, α_{r} ∈ K not all zero such that

α_{1}u_{1} + ⋯ + α_{r}u_{r} = 0;

(ii) u_{1}, …, u_{r} are linearly independent if

α_{1}u_{1} + ⋯ + α_{r}u_{r} = 0

implies that α_{1}, …, α_{r} are all zero.

Example: cos ωx, sin ωx (ω ≠ 0) are linearly independent functions in C^{∞}(ℝ).

Proof of This ▷ Let

α cos ωx + β sin ωx = 0; α, β ∈ ℝ

be the zero function. Put x = 0 : α = 0; put x = \frac{π}{2ω} : β = 0. ◁

1–5 Note. If u_{1}, …, u_{r} are linearly dependent, with

α_{1}u_{1} + α_{2}u_{2} + ⋯ + α_{r}u_{r} = 0,

and α_{1} (say) ≠ 0 then

u_{1} = −(α_{1}^{−1}α_{2}u_{2} + ⋯ + α_{1}^{−1}α_{r}u_{r}).

Thus u_{1}, …, u_{r} linearly dependent iff one of them is a linear combination of the others. Definition. A sequence of vectors u_{1}, …, u_{n} in a K-vector space M is called a basis for M if

(i) u_{1}, …, u_{n} are linearly independent; (ii) u_{1}, …, u_{n} generate M. Definition. If u_{1}, …, u_{n} is a basis for a vector space M then for each x ∈ M we have:

x = α^{1}u_{1} + ⋯ + α^{n}u_{n}

for a sequence of scalars:

(α^{1}, …, α^{n}),

which are called the coordinates of x with respect to the basis u_{1}, …, u_{n}.

The coordinates of x are uniquely determined once the basis is chosen because:

x = α^{1}u_{1} + ⋯ + α^{n}u_{n} = β^{1}u_{1} + ⋯ + β^{n}u_{n}

implies:

(α^{1} − β^{1})u_{1} + ⋯ + (α^{n} − α^{n})u_{n} = 0,

and hence

α^{1} − β^{1} = 0, …, α^{n} − β^{n} = 0,

by the linear independence of u_{1}, …, u_{n}. So α^{1} = β^{1}, …, α^{n} = β^{n}.

1–6 A choice of basis therefore gives a well-defined bijective map:

M → K^{n}
x ↦ coordinates of x,

called the coordinate map wrt the given basis.

The following theorem (our first) implies that any two bases for M must have the same number of elements.

Theorem 1.1. Let M be a K-vector space, u_{1}, …, u_{n} be linearly independent in M, and y_{1}, …, y_{n} generate M. Then n ≤ r.

Proof ▶

u_{1} = α_{1}y_{1} + ⋯ + α_{r}y_{r}

(say), since y_{1}, …, y_{r} generate M. α_{1}, …, α_{r} are not all zero, since u_{1} ≠ 0. Therefore α_{1} ≠ 0 (say). Therefore y_{1} is a linear combination of u_{1}, y_{2}, y_{3}, …, y_{r}. Therefore u_{1}, y_{2}, y_{3}, …, y_{r} generate M. Therefore

u_{2} = β_{1}u_{1} + β_{2}y_{2} + β_{3}y_{3} + ⋯ + β_{r}y_{r}

(say). β_{2}, …, β_{r} are not all zero, since u_{1}, u_{2} are linearly independent. Therefore β_{2} ≠ 0 (say). Therefore y_{2} is a linear combination of u_{1}, u_{2}, y_{3}, …, y_{r}. Therefore u_{1}, u_{2}, y_{3}, …, y_{r} generate M.

Continuing in this way, if n > r we get u_{1}, …, u_{r} generate M, and hence u_{n} is a linear combination of u_{1}, …, u_{r}, which contradicts the linear independence of u_{1}, …, u_{n}. Therefore n ≤ r. ◀

Note. If u_{1}, …, u_{n} and y_{1}, …, y_{r} are two bases for M then n = r.

Definition. A vector space M is called finite-dimensional if it has a finite basis. The number of elements in a basis is then called the dimension of M, denoted by dim M.

Examples:

1. The n vectors:

e_{1} = (1, 0, 0, …, 0), e_{2} = (0, 1, 0, …, 0), …, e_{n} = (0, 0, …, 0, 1)

form a basis for K^{n} as a vector-space, called the usual basis for K^{n}.

Proof of This ▷ We have

α_{1}e_{1} + ⋯ + α_{n}e_{n} = α_{1}(1, 0, …, 0) + ⋯ + α_{n}(0, …, 0, 1)
= (α_{1}, α_{2}, …, α_{n}).

Therefore

1–7 (a) e_{1}, …, e_{n} generate K^{n};

(b) α_{1}e_{1} + ⋯ + α_{n}e_{n} = 0 ⇒ ω_{1} = 0, …, ω_{n} = 0.

Therefore α_{1}, …, α_{n} are linearly independent. ◁

2. The mn matrices:

( \begin{matrix}1 & 0 & 0 & ⋯ & 0 \\ 0 & 0 & 0 & ⋯ & 0 \\ ⋮ & ⋮ & ⋮ & & ⋮ \\ 0 & 0 & 0 & ⋯ & 0 \end{matrix} ) , ( \begin{matrix}0 & 1 & 0 & ⋯ & 0 \\ 0 & 0 & 0 & ⋯ & 0 \\ ⋮ & ⋮ & ⋮ & & ⋮ \\ 0 & 0 & 0 & ⋯ & 0 \end{matrix} ) , …, ( \begin{matrix}0 & 0 & ⋯ & 0 & 0 \\ 0 & 0 & ⋯ & 0 & 0 \\ ⋮ & ⋮ & & ⋮ & ⋮ \\ 0 & 0 & ⋯ & 0 & 1 \end{matrix} )

form a basis for K^{m×n} as a K-vector space.

3. The functions cos ωx, sin ωx form a basis for the solutions of the equation

\frac{d^{2}u}{dx^{2}} + ω^{2}u = 0 (ω ≠ 0).

4. The functions

1, x, x^{2}, …, x^{n}

form a basis for the subspace of C^{∞}(ℝ) consisting of polynomial functions of degree ≤ n.

5. dim K^{n} = n; dim K^{m×n} = mn. We have:

dim ℂ^{m×n} = \begin{cases}mn & as a complex vector space; \\ 2mn & as a real vector space. \end{cases}

Given any linearly independent set of vectors we can add extra ones to form a basis. Given any generating set of vectors we can discard some to form a basis. More generally:

Theorem 1.2. Let M be a vector space with a finite generating set (or a vector subspace of such a space). Let Z be a generating set, and let X be a linearly independent subset of Z. Then M has a finite basis Y such that

X ⊂ Y ⊂ Z.

1–8 Proof ▶ Among all the linearly independent subsets of Z which contain X there is one at least

Y = {u_{1}, …, u_{n}},

with a maximal number of elements, n (say).

Now if z ∈ Z then z, u_{1}, …, u_{n} are linearly dependent. Therefore there exist scalars λ, α_{1}, …, α_{n} not all zero such that

λz + α_{1}u_{1} + ⋯ + α_{n}u_{n} = 0.

λ ≠ 0, since u_{1}, …, u_{n} are linearly independent. Therefore z is a linear combination of u_{1}, …, u_{n}.

But Z generates M. Therefore u_{1}, …, u_{n} generate M. Therefore u_{1}, …, u_{n} form a basis for M. ◀

1–9

Linear Operators 1

2.1 The Definition

Definition. Let M, N be K-vector spaces. A map

M →^{T} N

is called a linear operator (or linear map or linear function or linear transformation or linear homomorphism) if

(i) T (x + y) = T x + T y (group homomorphism); (ii) T αx = αT x for all x, y ∈ M, α ∈ K.

A linear operator is called a (linear) isomorphism if T is bijective. We say that M is isomorphic to N if there exists a linear isomorphism

M → N.

Note. Geometrically:

(i) means that T preserves parallelograms (see Figure 2.1); (ii) means that T preserves collinearity (see Figure 2.2).

2–1 Figure 2.1 Figure 2.2

Examples:

1. If

A = (α_{j}^{i}) = ( \begin{matrix}α_{1}^{1} & … & α_{n}^{1} \\ ⋮ & & ⋮ \\ α_{1}^{m} & … & α_{n}^{m} \end{matrix} ) ∈ K^{m×n},

we denote by

K^{n} →^{A} K^{m}

2–2 the linear operator given by matrix multiplication by A acting on elements of K^{n} written as n × 1 columns. Since

A(x + y) = Ax + Ay,
Aαx = αAx

for matrix multiplication, it follows that A is a linear operator. E.g.

A = ( \begin{matrix}3 & 7 & 2 \\ −2 & 5 & 1 \end{matrix} ) ∈ ℝ^{2×3}

Now:

ℝ^{3} → ℝ^{2} : ( \begin{matrix}α \\ β \\ γ \end{matrix} ) ↦ ( \begin{matrix}3α + 7β + 2γ \\ −2α + 5β + γ \end{matrix} ) .

2. Take

\frac{d}{dt} : C^{∞}(ℝ) → C^{∞}(ℝ).

Now:

\frac{d}{dt}[x(t) + y(t)] = \frac{d}{dt}x(t) + \frac{d}{dt}y(t),
\frac{d}{dt}cx(t) = c\frac{d}{dt}x(t)

for all c ∈ ℝ. Therefore \frac{d}{dt} is a linear operator.

3. The Laplacian

Δ = \frac{∂}{∂x^{2}} + \frac{∂}{∂y^{2}} + \frac{∂}{∂z^{2}} : C^{∞}(ℝ^{3}) → C^{∞}(ℝ^{3})

is a linear operator.

2.2 Basic Properties of Linear Operators

1. If M →^{T} N is a linear operator and u_{1}, …, u_{r} ∈ M; α_{1}, …, α_{r} ∈ K

then

T (α_{1}u_{1} + ⋯ + α_{r}u_{r}) = α_{1}T u_{1} + ⋯ + α_{r}T u_{r},

2–3 i.e.

T ∑_{i=1}^{r} α_{i}u_{i} = ∑_{i=1}^{r} α_{i}u_{i},

i.e. T preserves linear combinations, i.e. T can be moved across summations and scalars.

2. If M →^{S,T} N are linear operators, if u_{1}, …, u_{m} generate M, and if Su_{i} = T u_{i} (i = 1, …, m) then S = T.

Proof of This ▷ Let x ∈ M. Then x = ∑_{i=1}^{m} α_{i}u_{i} (say). Therefore

Sx = S ∑_{i=1}^{m} α_{i}u_{i} = ∑_{i=1}^{m} α_{i}Su_{i} = ∑_{i=1}^{m} α_{i}T u_{i} = T ∑_{i=1}^{m} α_{i}u_{i} = T x.

◁

Thus two linear operators which agree on a generating set must be equal.

3. Let u_{1}, …, u_{n} be a basis for M, and w_{1}, …, w_{n} be arbitrary vectors in N. Then we can define a linear operator

M →^{T} N

by

T (α_{1}u_{1} + ⋯ + α_{n}u_{n}) = α_{1}w_{1} + ⋯ + α_{n}u_{n}.

Thus T is the unique linear operator such that

T u_{i} = w_{i} (i = 1, …, m).

We say that T is defined by T u_{i} = w_{i}, and extended to M by linearity.

Definition. Let M →^{T} N be a linear operator. Then

ker T = {x ∈ M : T x = 0}

is a vector subspace of M, called the kernel of M, and

im T = {T x : x ∈ M}

is a vector subspace of N, called the image of T. The dimension of im T is called the rank of T,

rank T = dim im T.

2–4

2.3 Examples

1. Consider the matrix operator

K^{n} →^{A} K^{m},

where A ∈ K^{m×n},

A = ( \begin{matrix}α_{11} & α_{12} & … & α_{1n} \\ ⋮ & & & ⋮ \\ α_{m1} & α_{m2} & … & α_{mn} \end{matrix} )

(say).

ker T = {x = (x_{1}, …, x_{n}) : Ax = 0}

is the space of solutions of

( \begin{matrix}α_{11} & … & α_{1n} \\ ⋮ & & ⋮ \\ α_{m1} & … & α_{mn} \end{matrix} ) ( x_{1} ; ⋮ ; x_{n} ) = ( 0 ⋮ 0 ) ,

i.e. The space of solutions of the m homogeneous linear equations in n unknowns, whose coefficients are the rows of A:

α_{11}x_{1} + α_{12}x_{2} + ⋯ + α_{1n}x_{n} = 0
⋮
α_{i1}x_{1} + α_{i2}x_{2} + ⋯ + α_{in}x_{n} = 0
⋮
α_{m1}x_{1} + α_{m2}x_{2} + ⋯ + α_{mn}x_{n} = 0

Number of equations = m = number of rows of A = dim K^{m}. Number of unknowns = n = number of columns of A = dim K^{n}.

We see that (x_{1}, x_{2}, …, x_{n}) ∈ ker A iff the dot product:

(α_{i1}, α_{i2}, …, α_{in}).(x_{1}, …, x_{n}) (i = 1, …, m)

with each row of A is zero. Therefore

ker A = (row A)^{⊥},

where row A is the vector subspace of K^{n} generated by the m rows of A (see Figure 2.3).

Now row A is unchanged by the following elementary row operations:

2–5 (i) multiplying a row by a non-zero echelon;

(ii) interchanging rows;

(iii) adding to one row a scalar multiple of another row.

So ker A is also unchanged by these operations.

To obtain a basis for row A, and from this a basis for ker A, carry out elementary row operations in order to bring the matrix to row echelon form (i.e. so that each row begins with more zeros than the previous row).

Example: Let

A = ( \begin{matrix}2 & 1 & −1 & 3 \\ −1 & 1 & 2 & 1 \\ 4 & 0 & −1 & 2 \end{matrix} ) : ℝ^{4} → ℝ^{3}.

Now

A ∼ ( \begin{matrix}2 & 1 & −1 & 3 \\ 0 & 3 & 3 & 5 \\ 0 & −2 & 1 & −4 \end{matrix} ) 2 row 2 + row 1 row 3 − 2 row 1

∼ ( \begin{matrix}2 & 1 & −1 & 3 \\ 0 & 3 & 3 & 5 \\ 0 & 0 & 9 & −2 \end{matrix} ) 3 row 3 + 2 row 2.

Since the new rows are in row echelon form they are linearly independent. Therefore row A is 3-dimensional, with basis (2, 1, −1, 3), (0, 3, 3, 5), (0, 0, 9, −2). Therefore

(α, β, γ, δ) ∈ ker A ⇔ 2α + β − γ + 3δ = 0
3β + 3γ + 5δ = 0
9γ − 2δ = 0
⇔ γ = \frac{2}{9}δ
3β = −3γ − 5δ = −\frac{2}{3}δ − 5δ = −\frac{17}{3}δ
2α = −β + γ − 3δ = \frac{17}{9}δ + \frac{2}{9}δ − 3δ = −\frac{8}{9}δ
⇔ (α, β, γ, δ) = (−\frac{4}{9}δ, −\frac{17}{9}δ, \frac{2}{9}δ, δ) = \frac{δ}{9}(−4, −17, 2, 9)

Therefore ker A is 1-dimensional, with basis (−4, −17, 2, 9).

2–6 If

A = ( \begin{matrix}α_{11} & … & α_{1j} & … & α_{1n} \\ α_{21} & … & α_{2j} & … & α_{2n} \\ ⋮ & & ⋮ & & ⋮ \\ α_{m1} & … & α_{mj} & … & α_{mn} \end{matrix} ) ∈ K^{m×n}

then

Ae_{j} = ( \begin{matrix}α_{11} & … & α_{1j} & … & α_{1n} \\ α_{21} & … & α_{2j} & … & α_{2n} \\ ⋮ & & ⋮ & & ⋮ \\ α_{m1} & … & α_{mj} & … & α_{mn} \end{matrix} ) ( \begin{matrix}0 \\ · \\ 1 \\ · \\ 0 \end{matrix} ) ← j^{th} slot

= ( α_{1j} ; ⋮ ; α_{mj} ) = j^{th} column of A.

Therefore

im A = {Ax : x ∈ K^{n}}

= {A(α_{1}e_{1} + ⋯ + α_{n}e_{n}) : α_{1}, …, α_{n} ∈ K} = {α_{1}Ae_{1} + ⋯ + α_{n}Ae_{n} : α_{1}, …, α_{n} ∈ K} = S(Ae_{1}, …, Ae_{n}) = column space of A = col A,

where col A is the vector subspace of K^{m} generated by the n columns of A. To find a basis for im A = col A we carry out elementary column operations on A.

Example: If

A = ( \begin{matrix}2 & 1 & −1 & 3 \\ −1 & 1 & 2 & 1 \\ 4 & 0 & −1 & 2 \end{matrix} )

2–7

then

A ∼ ( \begin{matrix}2 & 0 & 0 & 0 \\ −1 & 3 & 3 & 5 \\ 4 & −4 & 2 & −8 \end{matrix} ) 2 col 2 − col 1 2 col 3 + col 1 2 col 4 − 3 col 1

∼ ( \begin{matrix}2 & 0 & 0 & 0 \\ −1 & 3 & 0 & 0 \\ 4 & −4 & 6 & −4 \end{matrix} ) col 3 − 2 col 2 3 col 4 − 5 col 2

∼ ( \begin{matrix}2 & 0 & 0 & 0 \\ −1 & 3 & 0 & 0 \\ 4 & −4 & 6 & 0 \end{matrix} ) .

Therefore im A = col A has basis (2, −1, 4), (0, 3, −4), (0, 0, 6). Therefore rank A = dim im A = 3.

2. Let

D = \frac{d}{dt} : C^{∞}(ℝ) → C^{∞}(ℝ) (Dx(t) = \frac{d}{dt}x(t)).

(i) Let λ ∈ ℝ and D − λ be the operator

(D − λ) = \frac{d}{dt}x(t) − λx(t).

Then

x ∈ ker(D − λ) ⇔ (D − λ)x = 0 ⇔ \frac{dx}{dt} = λx ⇔ x(t) = ce^{λt}.

Therefore ker(D − λ) is 1-dimensional, with basis e^{λt}. (ii) To determine ker(D − λ)^{k} we must solve:

(D − λ)^{k}x = 0.

Put x(t) = e^{λt}y(t). Then

(D − λ)x = Dx(t) − λx(t)
= λe^{λt}y(t) + e^{λt}Dy(t) − λe^{λt}y(t)
= e^{λt}Dy(t).

Therefore

(D − λ)^{2}x = e^{λt}D^{2}y(t)
⋮
(D − λ)^{k}x = e^{λt}D^{k}y(t).

2–8

Therefore

(D − λ)^{k}x = 0 ⇔ e^{λt}D^{k}y(t) = 0
⇔ D^{k}y(t) = 0
⇔ y(t) = c_{0} + c_{1}t + c_{2}t^{2} + ⋯ + c_{k−1}t^{k−1}
⇔ x(t) = (c_{0} + c_{1}t + ⋯ + c_{k−1}t^{k−1})e^{λt}.

Therefore ker(D−λ)^{k} is k-dimensional, with basis e^{λt}, te^{λt}, t^{2}e^{λt}, …, t^{k−1}e^{λt}.

2.4 Properties Continued

Theorem 2.1. Let M →^{T} N be a linear operator, where M is finite dimensional. Let u_{1}, …, u_{k} be a basis for ker T, and let T w_{1}, …, T w_{r} be a basis for im T. Then

u_{1}, …, u_{k}, w_{1}, …, w_{r}

is a basis for M.

Proof ▶ We have two things to show:

(i) Linear independence: Let

∑ α_{i}u_{i} + ∑ β_{j}w_{j} = 0

Apply T:

0 + ∑ β_{j}T w_{j} = 0.

Therefore β_{j} = 0 for all j. Therefore α_{i} = 0 for all i.

Therefore u_{1}, …, u_{k}, w_{1}, …, w_{r} are linearly independent.

(ii) Generate: Let x ∈ M. Then

T x = ∑ β_{j}T w_{j} (say).

Therefore

T x = T ∑ β_{j}w_{j}.

Therefore

T [x − ∑ β_{j}w_{j}] = 0.

2–9 Therefore

x − ∑ β_{j}w_{j} ∈ ker T.

Therefore

x − ∑ β_{j}w_{j} = ∑ α_{i}u_{i} (say).

Therefore

x = ∑ α_{i}u_{i} + ∑ β_{j}w_{j}.

Therefore u_{1}, …, u_{k}, w_{1}, …, w_{r} generate M. ◀ Corollary 2.1. dim ker T + dim im T = dim M. Corollary 2.2. If dim M = dim N then

T is injective ⇔ ker T = {0} ⇔ dim im T = dim N ⇔ T is surjective.

2.5 Operator Algebra

If M, N are K-vector spaces, we denote by

L(M, N)

the set of all linear operators M → N, and we denote by

L(M)

the set of all linear operators M → M. Theorem 2.2. We have:

(i) L(M, N) is a K-vector space, with

(S + T)x = Sx + T x,
(αT)x = α(T x)

for all S, T ∈ L(M, N), x ∈ M, α ∈ K. (ii) Composition of operators gives a multiplication

\begin{matrix}L(L, M) × L(M, N) & → & L(L, N) \\ (T, S) & ↦ & ST \end{matrix} \} L →^{T} M →^{S} N,

with

(ST)x = S(T x) for all x ∈ L,

which satisfies

2–10 (a) (RS)T = R(ST),

(b) R(S + T) = RS + RT,

(c) (R + S)T = RT + ST,

(d) (αS)T = α(ST) = S(αT), provided each is well-defined.

Proof ▶ Straight forward verification. ◀ Corollary 2.3. L(M) is

(i) a K-vector space: S + T, αS; (ii) a ring: S + T, ST; (iii) (αS)T = α(ST) = S(αT) : αS, ST, i.e. L(M) is a K-algebra.

2.6 Isomorphisms of L(M, N) with K^{m×n}

Definition. Let u_{1}, …, u_{n} be a basis for M, and let w_{1}, …, w_{m} be a basis for N. Let M →^{T} N. Put Then we have:

T u_{1} = α_{1}^{1}w_{1} + α_{1}^{2}w_{2} + ⋯ + α_{1}^{i}w_{i} + ⋯ + α_{1}^{m}w_{m},
⋮
T u_{j} = α_{j}^{1}w_{1} + α_{j}^{2}w_{2} + ⋯ + α_{j}^{i}w_{i} + ⋯ + α_{j}^{m}w_{m},
⋮
T u_{n} = α_{n}^{1}w_{1} + α_{n}^{2}w_{2} + ⋯ + α_{n}^{i}w_{i} + ⋯ + α_{n}^{m}w_{m},

(say) where:

A = (α_{j}^{i}) = ( \begin{matrix}α_{1}^{1} & α_{2}^{1} & … & α_{j}^{1} & … & α_{n}^{1} \\ ⋮ & & & ⋮ & & ⋮ \\ α_{1}^{i} & … & … & α_{j}^{i} & … & α_{n}^{i} \\ ⋮ & & & ⋮ & & ⋮ \\ α_{1}^{m} & … & … & α_{j}^{m} & … & α_{n}^{m} \end{matrix} ) ∈ K^{m×n}.

2–11 Note. The coordinates of T u_{j} form the j^{th} column of A - NOTE THE TRANSPOSE! We call A the matrix of T wrt the bases u_{1}, …, u_{n} for M and w_{1}, …, w_{m} for N,

T u_{j} = ∑_{i=1}^{m} α_{j}^{i}ω_{i}.

Theorem 2.3. L(M, N) → K^{m×n} is a linear isomorphism where T → matrix of T w.r.t. basis u_{1}, ⋯, u_{n}; ω_{1}, cdots, ω_{m}.

Proof ▶ Let T have matrix A = (α_{j}^{i}), and let S have matrix B = (β_{j}^{i}). Then

(T + S)u_{j} = T u_{j} + Su_{j} = ∑_{i=1}^{m} α_{j}^{i}w_{i} + ∑_{i=1}^{m} β_{j}^{i}w_{i} = ∑_{i=1}^{m}(α_{j}^{i} + β_{j}^{i})w_{i}.

Therefore T + S has matrix (α_{j}^{i} + β_{j}^{i}) = A + B. Also

(λT)u_{j} = λ(T u_{j}) = λ ∑_{i=1}^{m} α_{j}^{i}w_{i} = ∑_{i=1}^{m} λα_{j}^{i}w_{i}.

Therefore λT has matrix (λα_{j}^{i}) = λA. ◀

Corollary 2.4. dim L(M, N) = dim M. dim N.

Theorem 2.4. If L →^{T} M has matrix A = (α_{j}^{i}) wrt basis v_{1}, …, v_{p}, u_{1}, …, u_{n},

and M →^{S} N has matrix B = (β_{j}^{i}) wrt basis u_{1}, …, u_{n}, w_{1}, …, w_{m} then

L →^{ST} N has basis

BA = (∑_{k=1}^{n} β_{k}^{i}α_{j}^{k}) = (γ_{j}^{i})

(say), wrt basis v_{1}, …, v_{p}, w_{1}, …, w_{m}.

Proof ▶

(ST)v_{j} = S(T v_{j}) = S (∑_{k=1}^{n} α_{j}^{k}u_{k}) = ∑_{k=1}^{n} α_{j}^{k}Su_{k}

= ∑_{k=1}^{n} α_{j}^{k} ∑_{i=1}^{m} β_{k}^{i}w_{i} = ∑_{i=1}^{m} (∑_{k=1}^{n} β_{k}^{i}α_{j}^{k}) w_{i} = ∑_{i=1}^{m} γ_{j}^{i}w_{i}.

◀

2–12 Corollary 2.5. If dim M = n then each choice of basis u_{1}, …, u_{n} of M defines an isomorphism of K-algebras:

L(M) → K^{m} : T ↦ matrix of T wrt u_{1}, …, u_{n}.

Note. If M →^{T} M has matrix A = (α_{j}^{i}) wrt basis u_{1}, …, u_{n} then

(i) T u_{j} = ∑_{i=1}^{n} α_{j}^{i}u_{i}, by definition; (ii) the elements of the j^{th} column of A are the coordinates of T u_{j}; (iii) λ_{0}1 + λ_{1}T + λ_{2}T^{2} + ⋯ + λ_{r}T^{r} has matrix α_{0}I + α_{1}A + ⋯ + α_{r}A^{r}; (iv) T^{−1} has matrix A^{−1}, since we have an algebra isomorphism. Theorem 2.5. Let M →^{T} N have matrix A = (α_{j}^{i}) wrt bases u_{1}, …, u_{n} for M and w_{1}, …, w_{m} for N. Let x have coordinates

X = (ξ^{i}) = ( ξ^{1} ; ⋮ ; ξ^{n} )

wrt u_{1}, …, u_{n}. Then T x has coordinates

AX = (∑_{i=1}^{m} α_{j}^{i}ξ^{j})

wrt w_{1}, …, w_{m}. Proof ▶

T x = T (∑_{j=1}^{n} ξ^{j}u_{j}) = ∑_{j=1}^{n} ξ^{i}T u_{j} = ∑_{j=1}^{n} ξ^{j} ∑_{i=1}^{m} α_{j}^{i}w_{i} = ∑_{i=1}^{m} (∑_{j=1}^{n} α_{j}^{i}ξ^{j}) w_{i},

as required. ◀ Note. We have thus a commutative diagram:

\begin{matrix}M & ⟶^{T} & N \\ ↓ & & ↓ \\ K^{n} & ⟶^{A} & K^{m} \end{matrix} : x  ↦^{T}  T x ↓  ↓ x coord. ↦^{A} T x coord

2–13

Changing Basis and Einstein Convention

Definition. If u_{1}, …, u_{n}^{old} and w_{1}, …, w_{n}^{new} are two bases for M then we have:

u_{1} = p_{1}^{1}w_{1} + p_{1}^{2}w_{2} + ⋯ + p_{1}^{n}w_{n}
⋮
u_{j} = p_{j}^{1}w_{1} + p_{j}^{2}w_{2} + ⋯ + p_{j}^{n}w_{n}
⋮
u_{n} = p_{n}^{1}w_{1} + p_{n}^{2}w_{2} + ⋯ + p_{n}^{n}w_{n}

(say). Put

P = (p_{j}^{i}) = ( \begin{matrix}p_{1}^{1} & … & p_{j}^{1} & … & p_{n}^{1} \\ p_{1}^{2} & … & p_{j}^{2} & … & p_{n}^{2} \\ ⋮ & & ⋮ & & ⋮ \\ p_{1}^{n} & … & p_{j}^{2} & … & p_{n}^{n} \end{matrix} )

Note. The new coordinates of the old basis vector u_{j} form the j^{th} column of P - NOTE THE TRANSPOSE! We call P the transition matrix from the (old) basis u_{1}, …, u_{n} to the (new) basis w_{1}, …, w_{n}:

u_{j} = ∑_{i=1}^{n} p_{j}^{i}w_{i}.

Theorem 3.1. If x has old coordinates

X = (ξ^{i}) = ( ξ^{1} ; ⋮ ; ξ^{n} )

3–1 then x has new coordinates

P X = ∑_{j=1}^{n}(p_{j}^{i}ξ^{j}) = (η^{i})

(say).

Proof ▶

x = ∑_{j=1}^{n} ξ^{j}u_{j} = ∑_{j=1}^{n} ξ^{j} ∑_{i=1}^{n} p_{j}^{i}w_{i} = ∑_{i=1}^{n} (∑_{j=1}^{n} p_{j}^{i}ξ^{j}) w_{i} = ∑_{i=1}^{n} η^{i}w_{i}.

◀

We shall often use the Einstein summation convention (s.c.) when dealing with basis and coordinates in a fixed n-dimensional vector space M. Repeated indices (one up, one down) are summed from 1 to n (contraction of repeated indices). Non-repeated indices may take each value 1 to n.

Example:

• α^{i} denotes

( α^{1} ; ⋮ ; α^{n} ) (column matrix; upper index labels the row).

• α_{i} denotes

(α_{1}, …, α_{n}) (row matrix; lower index labels the column).

• α_{j}^{i} denotes

( \begin{matrix}α_{1}^{1} & … & α_{n}^{1} \\ ⋮ & & ⋮ \\ α_{1}^{n} & … & α_{n}^{n} \end{matrix} ) (square matrix).

• u_{i} denotes u_{1}, …, u_{n} (basis).

• α^{i}u_{i} denotes α^{1}u_{1} + ⋯ + α^{n}u_{n}.

• α^{i}β_{i} denotes α^{1}β_{1} + ⋯ + α^{n}β_{n} (dot product).

3–2

• α_{k}^{i}β_{j}^{k} denotes AB (matrix product). Also

T u_{j} = α_{j}^{i}u_{i} (α_{j}^{i} matrix of operator T)

and

u_{j} = p_{j}^{i}w_{i} (p_{j}^{i} transition matrix from u_{i} to w_{i}).

If x has components ξ^{i} wrt u_{i} then T x has components α_{j}^{i}ξ^{i} wrt u_{i}. If x has components ξ^{j} wrt u_{i} then x has components p_{j}^{i}ξ^{j} wrt w_{i}.

• δ_{j}^{i} denotes the unit matrix

I = ( \begin{matrix}1 & 0 & … & 0 \\ 0 & 1 & … & 0 \\ ⋮ & & ... & ⋮ \\ 0 & … & 0 & 1 \end{matrix} ) .

• If Q = P^{−1} then (q_{j}^{i}) denotes Q (inverse matrix) and q_{k}^{i}p_{j}^{k} = δ_{j}^{i} = p_{k}^{i}q_{j}^{k}.

Theorem 3.2. Let M →^{T} N have matrix A wrt basis u_{1}, …, u_{n}. Let P be the transition matrix to (new) basis w_{1}, …, w_{n}. Then T has (new) matrix

P AP^{−1}

wrt w_{1}, …, w_{n}. Proof ▶ Let P = (p_{j}^{i}), A = (α_{j}^{i}), P^{−1} = Q = (q_{j}^{i}). Then

T u_{j} = α_{j}^{i}u_{i}; u_{j} = p_{j}^{i}w_{i}; w_{j} = q_{j}^{i}u_{i}.

Therefore

T w_{j} = T q_{j}^{l}u_{l} = q_{j}^{l}T u_{l} = q_{j}^{l}α_{l}^{k}u_{k} = q_{j}^{l}α_{l}^{k}p_{k}^{i}w_{i} = \underbrace{p_{k}^{i}α_{l}^{k}q_{j}^{l}}_{P AP^{−1}} w_{i},

as required. ◀

3–3

Linear Forms and Duality

4.1 Linear Forms

Definition. Fix M a K-vector space. A scalar valued linear function

f : M → K

is called a linear form on M.

If f is a linear form on M, and x is a vector in M, we write

⟨f, x⟩

to denote the value of f on x. This notation has the advantage of treating f and x in a symmetrised way:

(i) ⟨f, x + y⟩ = ⟨f, x⟩ + ⟨f, y⟩, (ii) ⟨f + g, x⟩ = ⟨f, x⟩ + ⟨g, x⟩, (iii) ⟨αf, x⟩ = α⟨f, x⟩ = ⟨f, αx⟩, (iv) ⟨∑_{i=1}^{r} α_{i}f^{i}, ∑_{j=1}^{s} β^{j}x_{j}⟩ = ∑_{i=1}^{r} ∑_{j=1}^{s} α_{i}β^{j}⟨f^{i}, x_{j}⟩.

If M is finite dimensional, with basis u_{1}, …, u_{n}, then each x ∈ M can be written uniquely as

x = α^{1}u_{1} + ⋯ + α^{n}u_{n} = ∑_{i=1}^{n} α^{i}u_{i} = α^{i}u_{i}.

We write

⟨u^{i}, x⟩ = α^{i}

4–1 to denote the i^{th} coordinate of x wrt basis u_{1}, …, u_{n}. We have:

⟨u^{i}, x + y⟩ = ⟨u^{i}, x⟩ + ⟨u^{i}, y⟩,
⟨u^{i}, αx⟩ = α⟨u^{i}, x⟩.

Thus u^{i} is a linear form on M, called the i^{th} coordinate function wrt basis u_{1}, …, u_{n}. We have:

1. ⟨u^{i}, u_{j}⟩ = \{ \begin{matrix}1 & if i = j \\ 0 & if i ≠ j \end{matrix} \} = δ_{j}^{i} (Kronecker delta);

2. x = ∑_{i=1}^{n}⟨u^{i}, x⟩u_{i} for all x ∈ M;

3. ⟨α_{1}u^{1} + ⋯ + α_{n}u^{n}, β^{1}u_{1} + ⋯ + β^{n}u_{n}⟩ = α_{1}β^{1} + ⋯ + α_{n}β^{n} = α_{i}β^{i} (dot product).

Theorem 4.1. If u_{1}, …, u_{n} is a basis for M then the coordinate functions u^{1}, …, u^{n} form a basis for the space M^{∗} of linear forms on M (called the dual space of M), called the dual basis, and

f = ∑_{i=1}^{n}⟨f, u_{i}⟩u^{i} for each f ∈ M^{∗}.

Proof ▶ We have to show that u^{1}, …, u^{n} generate M, and are linearly independent.

(i) Generate: Let f ∈ M^{∗}; ⟨f, u_{j}⟩ = β_{j} (say). Then

⟨∑_{i=1}^{n} β_{i}u^{i}, u_{j}⟩ = ∑_{i=1}^{n} β_{i}⟨u^{i}, u_{j}⟩ = ∑_{i=1}^{n} β_{i}δ_{j}^{i} = β_{j} = ⟨f, u_{j}⟩.

Therefore ∑_{i=1}^{n} β_{i}u^{i} and f are linear forms on M which agree on the

basis vectors u_{1}, …, u_{n}. Therefore

f = ∑_{i=1}^{n} β_{i}u^{i} = ∑_{i=1}^{n}⟨f, u_{i}⟩u^{i}.

(ii) Linear independence: Let ∑_{i=1}^{n} β_{i}u^{i} = 0. Then

⟨∑_{i=1}^{n} β_{i}u^{i}, u_{j}⟩ = 0

4–2 for all j = 1, …, n. Therefore

∑_{i=1}^{n} β_{i}δ_{j}^{i} = 0

for all j = 1, …, n. Therefore β_{j} = 0 for all j = 1, …, n. Therefore

u^{1}, …, u^{n} are linearly independent. ◀

Corollary 4.1. dim M^{∗} = dim M.

Note. We denote by x, y, z the coordinate function on K^{3} wrt basis e_{1}, e_{2}, e_{3}, and we denote by x^{1}, …, x^{n} the coordinate function on K^{n} wrt basis e_{1}, …, e_{n}. These coordinates are called the usual coordinates.

4.2 Duality

Let M be finite dimensional, with dual space M^{∗}. If x ∈ M and f ∈ M^{∗} then

(i) f is a linear form on M whose value on x is ⟨f, x⟩;

(ii) we identify x with the linear form on M^{∗} whose value on f is ⟨f, x⟩:

f = ⟨f, ·⟩,
x = ⟨·, x⟩.

What we are doing is identifying M with the dual of M^{∗}, by means of the linear isomorphism:

M → M^{∗∗}
x ↦ ⟨·, x⟩.

This is a linear map, and is bijective because:

(i) dim M^{∗∗} = dim M^{∗} = dim M,

(ii) ⟨·, x⟩ = 0 ⇒ ⟨u^{i}, x⟩ = 0 for all x ⇒ x = 0. So the map is injective (kernel = {0}), and hence by (i) surjective.

If u_{1}, …, u_{n} is a basis for M, and u^{1}, …, u^{n} the dual basis for M^{∗} then

⟨u^{i}, u_{j}⟩ = δ_{j}^{i}

shows that u_{1}, …, u_{n} is the basis dual to u^{1}, …, u^{n}.

The identification of vectors x ∈ M as linear forms on M^{∗} is called duality. A basis u^{1}, …, u^{n} for M^{∗} is called a linear coordinate system on M, and consists of coordinate functions wrt its dual basis u_{1}, …, u_{n}.

4–3

4.3 Systems of Linear Equations

Definition. If f^{1}, …, f^{k} are linear forms on M then we consider the vector subspace of M on which

f^{1} = 0, …, f^{k} = 0 (∗).

Any vector in this subspace is called a solution of the equations (∗). Thus x ∈ M is a solution iff

⟨f^{1}, x⟩ = 0, …, ⟨f^{k}, x⟩ = 0.

The set of solutions is called the solution space of the system of k homogeneous equations (∗). The dimension of the space S(f^{1}, …, f^{k}) generated by f^{1}, …, f^{k} is called the rank (number of linearly independent equations) of the system of equations.

In particular, if u^{1}, …, u^{n} is a linear coordinate system on M then we can write the equations as:

f^{1} ≡ β_{1}^{1}u^{1} + ⋯ + β_{n}^{1}u^{n} = 0
⋮
f^{k} ≡ β_{1}^{k}u^{1} + ⋯ + β_{n}^{k}u^{n} = 0

The coordinate map M^{∗} → K^{n} maps

f^{1} ↦ (β_{1}^{1}, …, β_{n}^{1})
⋮
f^{k} ↦ (β_{1}^{k}, …, β_{n}^{k}).

Thus it maps S(f^{1}, …, f^{k}) isomorphically onto the row space of B = (β_{j}^{i}). Therefore

rank of system = dimension of row space of B = dim row B.

Example: The equations

3x − 4y + 2z = 0,
2x + 7y + 3z = 0,

where x, y, z are the usual coordinates on ℝ^{3}, have

rank = dim row ( \begin{matrix}3 & −4 & 2 \\ 2 & 7 & 3 \end{matrix} ) = 2.

4–4 Theorem 4.2. A system of k homogeneous linear equations of rank r on an n-dimensional vector space M has a solution space of dimension n − r.

Proof ▶ Let

f^{1} = 0, …, f^{k} = 0

be the system of equations. Let u^{1}, …, u^{r} be a basis for S(f^{1}, …, f^{k}). Ex-

tend to a basis u^{1}, …, u^{r}_{equations}, u^{r+1}, …, u^{n} for M^{∗}. Let u_{1}, …, u_{r}, u_{r+1}, …, u_{n}_{solutions} be

the dual basis of M. Then

x = α^{1}u_{1} + ⋯ + α^{r}u_{r} + α^{r+1}u_{r+1} + ⋯ + α^{n}u_{n} ∈ solution space
⇔ α^{1} = ⟨u^{1}, x⟩ = 0, …, α^{r} = ⟨u^{r}, x⟩ = 0
⇔ x = α^{r+1}u_{r+1} + ⋯ + α^{n}u_{n}.

Therefore u_{r+1}, …, u_{n} is a basis for the solution space. Therefore solution space has dimension n − r. ◀

Theorem 4.3. Let B ∈ K^{k×n}, where K is a field. Then

dim row B = dim col B (= rank B).

Proof ▶ Consider the k homogeneous linear equations on K^{n} with coefficients B = (β_{j}^{i}):

β_{1}^{1}x^{1} + ⋯ + β_{n}^{1}x^{n} = 0
⋮
β_{1}^{k}x^{1} + ⋯ + β_{n}^{k}x^{n} = 0.

Now

n − dim row B = n − rank of equations

= dimension of solution space = dim ker B = n − dim im B = n − dim col B.

Therefore dim col B = dim row B. ◀

4–5

Tensors

5.1 The Definition

Definition. Let M be a finite dimensional vector space over a field K, let M^{∗} be the dual space, and let dim M = n. A tensor over M is a function of the form

T : M_{1} × M_{2} × ⋯ × M_{k} → K,

where each M_{i} = M or M^{∗} (i = 1, …, k), and which is linear in each variable (multilinear).

Two tensors S, T are said to be of the same type if they are defined on the same set M_{1} × ⋯ × M_{k}.

Example: A tensor of type

T : M × M^{∗} × M → K

is a scalar valued function T (x, f, y) of three variables (x a vector, f a linear form, y a vector) such that

T (αx + βy, f, z) = αT (x, f, z) + βT (y, f, z) linear in 1^{st} variable,
T (x, αf + βg, z) = αT (x, f, z) + βT (x, g, z) linear in 2^{nd} variable,
T (x, f, αy + βz) = αT (x, f, z) + βT (x, f, z) linear in 3^{rd} variable.

If u_{i} is a basis for M, and u^{i} is the dual basis for M^{∗} then the array of n^{3} scalars

α_{i}^{j}_{k} = T (u_{i}, u^{j}, u_{k})

5–1 are called the components of T.

If x, f, y have components ξ^{i}, η_{j}, ρ^{k} respectively then

T (x, f, y) = T (ξ^{i}u_{i}, η_{j}u^{j}, ρ^{k}u_{k}) = ξ^{i}η_{j}ρ^{k}T (u_{i}, u^{j}, u_{k}) = ξ^{i}η_{j}ρ^{k}α_{i}^{j}_{k}

(using summation notation), i.e. the components of T contracted by the components of x, f, y.

The set of all tensors over M of a given type form a K-vector space if we define

(S + T)(x_{1}, …, x_{k}) = S(x_{1}, …, x_{k}) + T (x_{1}, …, x_{k}),
(λT)(x_{1}, …, x_{k}) = λ(T (x_{1}, …, x_{k})).

The vector space of all tensors of type

M × M^{∗} × M → K

(say) has dimension n^{3}, since T ↦ T (u_{i}, u^{j}, u_{k}) (components of T) maps it isomorphically onto K^{n^{3}}.

Definition. If S : M_{1} × ⋯ × M_{k} → K and T : M_{k+1} × ⋯ × M_{l} → K are tensors over M then we define their tensor product S ⊗ T to be the tensor:

S ⊗ T : M_{1} × ⋯ × M_{k} × M_{k+1} × ⋯ × M_{l} → K,

where

S ⊗ T (x_{1}, …, x_{l}) = S(x_{1}, …, x_{k})T (x_{k+1}, …, x_{l}).

Example: If S has components α_{i}^{j}_{k}, and T has components β^{rs} then S ⊗ T has components α_{i}^{j}_{k}β^{rs}, because

S ⊗ T (u_{i}, u^{j}, u_{k}, u^{r}, u^{s}) = S(u_{i}, u^{j}, u_{k})T (u^{r}, u^{s}).

Tensors satisfy algebraic laws such as:

(i) R ⊗ (S + T) = R ⊗ S + R ⊗ T,

(ii) (λR) ⊗ S = λ(R ⊗ S) = R ⊗ (λS),

(iii) (R ⊗ S) ⊗ T = R ⊗ (S ⊗ T).

5–2 But

S ⊗ T ≠ T ⊗ S

in general. To prove those we look at components wrt a basis, and note that

α^{i}_{jk}(β^{r}_{s} + γ^{r}_{s}) = α^{i}_{jk}β^{r}_{s} + α^{i}_{jk}γ^{r}_{s},

for example, but

α^{i}β^{j} ≠ β^{j}α^{i}

in general.

5.2 Contraction

Definition. Let T : M_{1} × ⋯ × M_{r} × ⋯ × M_{s} × ⋯ × M_{k} → K be a tensor, with

M_{r} = M^{∗}, M_{s} = M

(say). Then we can contract the r^{th} index of T with the s^{th} index to get a new tensor

S : M_{1} × ⋯ × M_{r}^{omit} × ⋯ × M_{s}^{omit} × ⋯ × M_{k} → K

defined by

S(x_{1}, x_{2}, …, x_{k−2}) = T (x_{1}, …, u^{i}_{r^{th} slot}, …, u_{i}_{s^{th} slot}, …, x_{k−2}),

where u_{i} is a basis for M.

To show that S is well-defined we need: Theorem 5.1. The definition of contraction is independent of the choice of basis.

Proof ▶ Put

R(f, x) = T (x_{1}, x_{2}, …, f, …, x, …, x_{k−2}).

Then if u_{i}, w_{i} are bases:

R(w^{i}, w_{i}) = R(p_{k}^{i}u^{k}, q_{i}^{l}u_{l}) = p_{k}^{i}q_{i}^{l}R(u^{k}, u_{l}) = δ_{k}^{l}R(u^{k}, u_{l}) = R(u^{k}, u_{k}),

as required. ◀ Example: If T has components α^{i}_{jk}^{lm} wrt basis u_{i} then contraction of the 2^{nd} and 4^{th} indices gives a tensor with components

β^{i}_{k}^{m} = T (u^{i}, u_{j}, u_{k}, u^{j}, u^{m}) = α^{i}_{jk}^{jm}.

Thus when we contract we eliminate one upper (contravariant) index and one lower (covariant) index.

5–3

5.3 Examples

A vector x ∈ M is a tensor:

x : M^{∗} → K

with components α^{i} = ⟨u^{i}, x⟩ (one contravariant index).

A linear form f ∈ M^{∗} is a tensor:

f : M → K

with components α_{i} = ⟨f, u_{i}⟩ (one covariant index).

A tensor with two covariant indices:

T : M × M → K,

with T (u_{i}, u_{j}) = α_{ij}, is called a bilinear form or scalar product.

Example: The dot product

K^{n} × K^{n} → K
((α^{1}, …, α^{n}), (β^{1}, …, β^{n})) ↦ α^{1}β^{1} + ⋯ + α^{n}β^{n}

is a bilinear form on K^{n}.

If M →^{T} M is a linear operator, we shall identify it with the tensor:

T : M^{∗} × M → K

by

T (f, x) = ⟨f, T x⟩.

This tensor has components

α^{i}_{j} = T (u^{i}, u_{j}) = ⟨u^{i}, T u_{j}⟩ = matrix of linear operator T

(one contravariant index, one covariant index).

Note (The Transformation Law). Let p_{j}^{i} be the transition matrix from basis u_{i} to basis w_{i}, with inverse matrix q_{i}^{j}. Let T be a tensor M ×M^{∗}×M → K (say). Then

\overbrace{T (w_{i}, w^{j}, w_{k})}^{new comps.} = T (q_{i}^{r}u_{r}, p_{s}^{j}u^{s}, q_{k}^{t}u_{t}) = q_{i}^{r}p_{s}^{j}q_{k}^{t} \overbrace{T (u_{r}, u^{s}, u_{t})}^{old comps.},

i.e. Upper indices contract with p, lower indices contract with q.

5–4

5.4 Bases of Tensor Spaces

Let M × M^{∗} × M → K (∗) (say) be a tensor with components α_{i}^{j}_{k} wrt basis u_{i}. Then the tensor:

α_{i}^{j}_{k} u^{i} ⊗ u_{j} ⊗ u^{k} (∗∗)

is of the same type as T, and has components

α_{i}^{j}_{k}u^{i} ⊗ u_{j} ⊗ u^{k}[u_{r}, u^{s}, u_{t}] = α_{i}^{j}_{k}⟨u^{i}, u_{r}⟩⟨u^{s}, u_{j}⟩⟨u^{k}, u_{t}⟩
= α_{i}^{j}_{k}δ_{r}^{i}δ_{j}^{s}δ_{t}^{k}
= α_{r}^{s}_{t}.

Therefore (∗∗) has the same components as T. Therefore

T = α_{i}^{j}_{k} u^{i} ⊗ u_{j} ⊗ u^{k}.

Therefore u^{i} ⊗ u_{j} ⊗ u^{k} is a basis for the n^{3}-dimensional space of all tensors of type (∗).

5–5

Chapter 6 Vector Fields

6.1 The Definition

Let V be an open subset of ℝ^{n}. Let x^{1}, …, x^{n} be the usual coordinate functions on ℝ^{n}. Let V →^{f} ℝ. If a = (a_{1}, …, a_{n}) ∈ V then we define the partial derivative of f wrt i^{th} variable at a:

\frac{∂f}{∂x^{i}} (a) = lim_{t→0} \frac{f(a_{1}, …, a_{i} + t, …, a_{n}) − f(a_{1}, …, a_{i}, …, a_{n})}{t}

= lim_{t→0} \frac{f(a + te_{i}) − f(a)}{t}

= \frac{d}{dt}f(a + te_{i})|_{t=0}

(see Figure 6.1). If it exists for each a ∈ V then we have:

\frac{∂f}{∂x^{i}} : V → ℝ.

6–1

F igure : 6.1

Note that

\frac{∂x^{i}}{∂x^{j}} = δ_{j}^{i}.

If all repeated partial derivatives of all orders:

\frac{∂^{r}f}{∂x^{i_{1}} ⋯ ∂x^{i_{r}}} = \frac{∂}{∂x^{i_{1}}} … \frac{∂}{∂x^{i_{r}}} f : V → ℝ

exist we call f C^{∞}. We denote by C^{∞}(V) the space of all C^{∞} functions V → ℝ. C^{∞}(V) is an ℝ-algebra:

(i) (f + g)(x) = f(x) + g(x), (ii) (fg)(x) = f(x)g(x), (iii) (αf)(x) = α(f(x)). Each sequence α^{1}, …, α^{n} of elements of C^{∞}(V) defines a linear operator

v = α^{1} \frac{∂}{∂x^{1}} + ⋯ + α^{n} \frac{∂}{∂x^{n}}

6–2 on C^{∞}(V), where

(vf)(x) = α^{1}(x)\frac{∂f}{∂x^{1}} (x) + ⋯ + α^{n}(x)\frac{∂f}{∂x^{n}} (x).

Such an operator

v : C^{∞}(V) → C^{∞}(V)

is called a (contravariant) vector field on V.

Now for each fixed a we denote by

\frac{∂}{∂x_{a}^{i}}

the operator given by:

\frac{∂}{∂x_{a}^{i}} f = \frac{∂f}{∂x^{i}} (a).

Thus \frac{∂}{∂x_{a}^{i}} acts on any function f which is defined and C^{1} on an open set containing a. We define the linear combination ∑_{i=1}^{n} α^{i} \frac{∂}{∂x_{a}^{i}} by

(α^{1} \frac{∂}{∂x_{a}^{i}} + ⋯ + α^{n} \frac{∂}{∂x_{a}^{n}} ) f = α^{1} \frac{∂f}{∂x^{1}} (a) + ⋯ + α^{n} \frac{∂f}{∂x^{n}} (a).

The set of linear combinations

\{α^{1} \frac{∂}{∂x_{a}^{1}} + ⋯ + α^{n} \frac{∂}{∂x_{a}^{n}} : α^{1}, …, α^{n} ∈ ℝ\}

is called the tangent space to ℝ^{n} at a, denoted T_{a}ℝ^{n}. Thus T_{a}ℝ^{n} is a real n-dimensional vector space, with basis

\frac{∂}{∂x_{a}^{1}} , …, \frac{∂}{∂x_{a}^{n}} .

The operators \frac{∂}{∂x_{a}^{i}} are linearly independent, since

α^{1} \frac{∂}{∂x_{a}^{1}} + ⋯ + α^{n} \frac{∂}{∂x_{a}^{n}} = 0 ⇒ (α^{1} \frac{∂}{∂x_{a}^{1}} + ⋯ + α^{n} \frac{∂}{∂x_{a}^{n}} ) x^{i} = 0 ⇒ α^{i} = 0,

since \frac{∂x^{i}}{∂x^{j}} (a) = δ_{j}^{i}.

6–3

If v = α^{1} \frac{∂}{∂x^{1}} + ⋯ + α^{n} \frac{∂}{∂x^{n}} (α^{i} ∈ C^{∞}(V)) is a vector field on V then we have (see Figure 6.2), for each x ∈ V a tangent vector

v_{x} = α^{1}(x)\frac{∂}{∂x_{x}^{1}} + ⋯ + α^{n}(x)\frac{∂}{∂x_{x}^{n}} ∈ T_{x}ℝ^{n}.

Figure 6.2

We call v_{x} the value of v at x, and note that

v_{x}f = (α^{1}(x)\frac{∂}{∂x_{x}^{1}} + ⋯ + α^{n}(x)\frac{∂}{∂x_{x}^{n}} )f

= α^{1}(x)\frac{∂f}{∂x^{1}} (x) + ⋯ + α^{n}(x)\frac{∂f}{∂x^{n}} (x)

= (vf)(x)

for all x ∈ V. Thus v is determined by its values {v_{x} : x ∈ V }, and vice versa. Thus a contravariant vector field is a function on V

x ↦ v_{x},

which maps to each point x ∈ V a tangent vector v_{x} ∈ T_{x}ℝ^{n}.

6–4

6.2 Velocity Vectors

β(t)

Figure 6.3

Let β(t) = (β^{1}(t), …, β^{n}(t)) be a sequence of real valued C^{∞} functions defined on an open subset of ℝ. Thus β = (β^{1}, …, β^{n}) is a curve in ℝ^{n} (see Figure 6.3). If f is a C^{∞} real-valued function on an open set in ℝ^{n} containing β(t) then the rate of change of f along the curve β at parameter t is

\frac{d}{dt}f(β(t)) = \frac{d}{dt}f(β^{1}(t), …, β^{n}(t))

= \frac{∂f}{∂x^{1}} (β(t))\frac{d}{dt}β^{1}(t) + ⋯ + \frac{∂f}{∂x^{n}} (β(t))\frac{d}{dt}β^{n}(t) (by the chain rule)

= [\frac{d}{dt}β^{1}(t)\frac{∂}{∂x_{β(t)}^{1}} + ⋯ + \frac{d}{dt}β^{n}(t)\frac{∂}{∂x_{β(t)}^{n}} ] f

= β^{.}(t)f,

where

β^{.}(t) = \frac{d}{dt}β^{1}(t)\frac{∂}{∂x_{β(t)}^{1}} + ⋯ + \frac{d}{dt}β^{n}(t)\frac{∂}{∂x_{β(t)}^{n}} ∈ T_{β(t)}ℝ^{n}

is called the velocity vector of β at t.

We note that if β(t) has coordinates

β^{i}(t) = x^{i}(β(t))

then β^{.}(t) has components

\frac{d}{dt}β^{i}(t) = \frac{d}{dt}x^{i}(β(t))
= rate of change of x^{i} along β at t wrt basis\frac{∂}{∂x_{β(t)}^{1}} , …, \frac{∂}{∂x_{β(t)}^{n}} .

6–5 In particular, if α = (α^{1}, …, α^{n}) ∈ ℝ^{n} and a = (a^{1}, …, a^{n}) ∈ ℝ^{n} then the straight line through a (see Figure 6.4) in the direction of α:

(a^{1} + tα^{1}, …, a^{n} + tα^{n})

α

Figure 6.4

has velocity vector at t = 0:

α^{1} \frac{∂}{∂x_{a}^{1}} + ⋯ + α^{n} \frac{∂}{∂x_{a}^{n}} ∈ T_{a}ℝ^{n}.

Thus each tangent vector is a velocity vector.

6.3 Differentials

Definition. If a ∈ ℝ^{n}, and f is a C^{∞} function on an open neighbourhood of a then the differential of f at a, denoted

df_{a},

is the linear form on T_{a}ℝ^{n} defined by

⟨df_{a}, β^{.}(t)⟩ = \frac{d}{dt}f(β(t)) = β^{.}(t)f

β̇(t)
= β(t)

Figure 6.5

for any velocity vector β^{.}(t), such that β(t) = a.

6–6 Thus

(i) ⟨df_{β(t)}, β^{.}(t)⟩ = rate of change of f along β at t (see Figure 6.5), (ii) ⟨df_{a}, v⟩ = vf (for all v ∈ T_{a}ℝ^{n}) = rate of change of f along v. Theorem 6.1. dx_{a}^{i}, …, dx_{a}^{n} is the basis of T_{a}ℝ^{n}^{∗} dual to the basis \frac{∂}{∂x_{a}^{1}} , …, \frac{∂}{∂x_{a}^{n}} for T_{a}ℝ^{n}. Proof ▶

⟨dx_{a}^{i}, \frac{∂}{∂x_{a}^{j}} ⟩ = \frac{∂x^{i}}{∂x^{j}} (a) = δ_{j}^{i},

as required. ◀ Definition. If V is open in ℝ^{n} then a covariant vector field ω on V is a function on V:

ω : x ↦ ω_{x} ∈ T_{x}ℝ^{n}^{∗}.

The covariant vector fields on V can be added:

(ω + η)_{x} = ω_{x} + η_{x},

and multiplied by elements of C^{∞}(V):

(fω)_{x} = f(x)ω_{x}.

Each covariant vector field ω on V can be written uniquely as

ω_{x} = β_{1}(x)dx_{x}^{1} + ⋯ + β_{n}(x)dx_{x}^{n}.

Thus

ω = β_{1}dx^{1} + ⋯ + β_{n}dx^{n}

(we confine ourselves to β_{i} ∈ C^{∞}(V)).

If f ∈ C^{∞}(V) then the covariant vector field

df : x ↦ df_{x}

is called the differential of f. Thus we have:

• contravariant vector fields:

v = α^{1} \frac{∂}{∂x^{1}} + ⋯ + α^{n} \frac{∂}{∂x^{n}} , α^{i} ∈ C^{∞}(V);

6–7 • covariant vector fields:

ω = β_{1}dx^{1} + ⋯ + β_{n}dx^{n}, β ∈ C^{∞}(V);

and more general tensor fields, e.g.

S = α_{i}^{j}_{k} dx^{i} ⊗ \frac{∂}{∂x^{j}} ⊗ dx^{k}, α_{i}^{j}_{k} ∈ C^{∞}(V),

a function on V whose value at x is

S_{x} = α_{i}^{j}_{k}(x)dx_{x}^{i} ⊗ \frac{∂}{∂x_{x}^{j}} ⊗ dx_{x}^{k},

a tensor over T_{x}ℝ^{n}.

We can add, multiply and contract tensor fields pointwise (carrying out the operation at each point x ∈ V). For example:

(i) (R + S)_{x} = R_{x} + S_{x},

(ii) (R ⊗ S)_{x} = R_{x} ⊗ S_{x},

(iii) (contracted S)_{x} = contracted (S_{x}),

(iv) (fS)_{x} = f(x)S_{x} f ∈ C^{∞}(V).

Contracting the covariant vector field ω = β_{1}dx^{1} + ⋯ + β_{n}dx^{n} with the contravariant vector field v = α^{1} \frac{∂}{∂x^{1}} + ⋯ + α^{n} \frac{∂}{∂x^{n}} gives the scalar field

⟨ω, v⟩ = β_{1}α^{1} + ⋯ + β_{n}α^{n}.

In particular, if f ∈ C^{∞}(V) has differential df then the scalar field

⟨df, v⟩ = vf

is the rate of change of f along v.

If ω = β_{1}dx^{1} + ⋯ + β_{n}dx^{n} then

β_{i} = i^{th} component of ω = ⟨ω, \frac{∂}{∂x^{i}} ⟩ .

In particular:

i^{th} compoment of df = ⟨df, \frac{∂}{∂x^{i}} ⟩ = \frac{∂f}{∂x^{i}} .

Therefore

df = \frac{∂f}{∂x^{1}} dx^{1} + ⋯ + \frac{∂f}{∂x^{n}} dx^{n} Chain Rule,
rate of change of f = \frac{∂f}{∂x^{1}} .rate of change of x^{1} + ⋯ + \frac{∂f}{∂x^{n}} .rate of change of x^{n}.

6–8

6.4 Transformation Law

A sequence

y = (y^{1}, …, y^{n}) (y^{i} ∈ C^{∞}(V))

is called a (C^{∞}) coordinate system on V if

V → W
x ↦ y(x) = (y^{1}(x), …, y^{n}(x))

maps V homeomorphically onto an open set W in ℝ^{n}, and if

x^{i} = F^{i}(y^{1}, …, y^{n}),

where F^{i} ∈ C^{∞}(W).

π θ)

θ)

θ θ

−π

Figure 6.6

Example: (r, θ) is a C^{∞} coordinate system on {(x, y) : y or x > 0} (see Figure 6.6), where r = \sqrt{x^{2} + y^{2}}, θ unique solution of x = r cos θ, y = r sin θ (−π < θ < π).

If a ∈ V, and β is the parametrised curve – the curve along which all y^{j} (j ≠ i) are constant, and y^{i} varies by t – such that

y(β(t)) = y(a) + te_{i}

6–9

i

Figure 6.7

(see Figure 6.7) then the velocity vector of β at t = 0 is denoted:

\frac{∂}{∂y_{a}^{i}}

Thus if f is C^{∞} in a neighbourhood of a then \frac{∂f}{∂y^{i}} (a) = \frac{∂}{∂y_{a}^{i}} f = \frac{d}{dt}f(β(t))|_{t=0} = rate of change of f along the curve β. If we write f as a function of y^{1}, …, y^{n}:

f = F (y^{1}, …, y^{n})

(say), then

\frac{∂f}{∂y^{i}} (a) = \frac{d}{dt}f(β(t))|_{t=0} = \frac{d}{dt}F (y(β(t)))|_{t=0} = \frac{d}{dt}F (y(a) + te_{i})|_{t=0} = \frac{∂F}{∂x^{i}} (y(a)),

i.e. to calculate \frac{∂f}{∂y^{i}} (a) write f as a function F of y^{1}, …, y^{n}, and calculate \frac{∂F}{∂x^{i}} (partial derivative of F wrt i^{th} slot):

\frac{∂f}{∂y^{i}} = \frac{∂F}{∂x^{i}} (y^{1}, …, y^{n}).

6–10

Now if β is any parametrised curve at a, with β(t) = a (see Figure 6.8), then

⟨df_{a}, β(t)^{.}⟩ = \frac{d}{dt}f(β(t))

= \frac{d}{dt}F (y^{1}(β(t)), …, y^{n}(β(t)))

= ∑_{i=1}^{n} \frac{∂F}{∂x^{i}} (y^{1}(β(t)), …, y^{n}(β(t)))\frac{d}{dt}y^{i}(β(t))

= ∑_{i=1}^{n} \frac{∂f}{∂y^{i}} (β(t))⟨dy_{a}^{i}, β^{.}(t)⟩

β̇(t)

β(t)

Figure 6.8

Therefore

df_{a} = ∑_{i=1}^{n} \frac{∂f}{∂y^{i}} (a)dy_{a}^{i}.

Therefore

df = ∑_{i=1}^{n} \frac{∂f}{∂y^{i}} dy^{i}.

The operators

\frac{∂}{∂y_{a}^{1}} , …, \frac{∂}{∂y_{a}^{n}}

are linearly independent, since \frac{∂}{∂y_{a}^{i}} y^{i} = δ_{j}^{i}. Therefore these operators form a basis for T_{a}ℝ^{n}, with dual basis

dy_{a}^{1}, …, dy_{a}^{n},

6–11 since ⟨dy_{a}^{i}, \frac{∂}{∂y_{a}^{i}} ⟩ = \frac{∂y^{i}}{∂y^{j}} (a) = δ_{j}^{i}.

If z^{1}, …, z^{n} is a C^{∞} coordinate system on W then on V ∩ W:

dz^{i} = ∑_{i=1}^{n} \frac{∂z^{i}}{∂y^{j}} dy^{j}.

Therefore \frac{∂z^{i}}{∂y^{j}} is the transition matrix from basis \frac{∂}{∂y^{i}} to basis \frac{∂}{∂z^{i}}. Therefore

\frac{∂}{∂y^{j}} = ∑_{i=1}^{n} \frac{∂z^{i}}{∂y^{j}} \frac{∂}{∂z^{i}}

on V ∩ W.

If (say) g = g_{ij}dy^{i} ⊗ dy^{j} is a tensor field on V, with component g_{ij} wrt coordinates y^{i}, then

g = g_{ij} (\frac{∂y^{i}}{∂z^{k}} dz^{k}) ⊗ (\frac{∂y^{j}}{∂z^{l}} dz^{l}) = \frac{∂y^{i}}{∂z^{k}} \frac{∂y^{j}}{∂z^{l}} g_{ij}dz^{k} ⊗ dz^{l},

using s.c., and therefore g has component

\frac{∂y^{i}}{∂z^{k}} \frac{∂y^{j}}{∂z^{l}} g_{ij}

wrt coordinates z^{i}. Example: On ℝ^{n}:

(i) usual coordinates x, y;

(ii) polar coordinates r, θ.

x = r cos θ, y = r sin θ.

So

dx = \frac{∂x}{∂r}dr + \frac{∂x}{∂θ}dθ = cos θ dr − r sin θ dθ,
dy = \frac{∂y}{∂r} dr + \frac{∂y}{∂θ} dθ = sin θ dr + r cos θ dθ.

The matrix

( \begin{matrix}cos θ & −r sin θ \\ sin θ & r cos θ \end{matrix} )

is the transition matrix from r, θ to x, y:

\frac{∂}{∂r} = \frac{∂x}{∂r} \frac{∂}{∂x} + \frac{∂y}{∂r} \frac{∂}{∂y} = cos θ\frac{∂}{∂x} + sin θ\frac{∂}{∂y} ,
\frac{∂}{∂θ} = \frac{∂x}{∂θ} \frac{∂}{∂x} + \frac{∂y}{∂θ} \frac{∂}{∂y} = −r sin θ\frac{∂}{∂x} + r cos θ\frac{∂}{∂y} .

6–12

Scalar Products

7.1 The Definition

Definition. A tensor of type M × M → K is called a scalar product or (bilinear form) (i.e. two lower indices).

Example: The dot product K^{n} × K^{n} → K. Writing X, Y as n × 1 columns:

((α_{1}, …, α_{n}), (β_{1}, …, β_{n})) ↦ α_{1}β_{1} + ⋯ + α_{n}β_{n}
(X, Y) ↦ X^{t}Y.

7.2 Properties of Scalar Products

1. If (·|·) is a scalar product on M with components G = (g_{ij}) wrt basis u_{i}, if x has components X = (ϕ^{i}) and y has components Y = (ν^{i}) (g_{ij} = (u_{i}|u_{j}) and (·|·) = g_{ij}u^{i} ⊗ u^{j}) then

(x|y) = (ϕ^{i}u_{i}|ν^{j}u_{j})

= ϕ^{i}ν^{j}(u_{i}|u_{j}) = g_{ij}ϕ^{i}ν^{j}

= ( ϕ^{1} … ϕ^{n} ) ( \begin{matrix}g_{11} & … & g_{1n} \\ ⋮ & & ⋮ \\ g_{n1} & … & g_{nn} \end{matrix} )( ν^{1} ; ⋮ ; ν^{n} )

= X^{t}GY.

Note. The dot product has matrix I wrt e_{i}, since e_{i}.e_{j} = δ_{j}^{i}.

7–1 2. If P = (p_{j}^{i}) is the transition matrix to new basis w_{i} then new matrix of

(·|·) is Q^{t}GQ, where Q = P^{−1}.

Proof of This ▷ As a tensor with two lower indices, new components of (·|·) are:

q_{i}^{k}q_{j}^{l}g_{kl} = q_{i}^{k}g_{kl}q_{j}^{l} = Q^{t}GQ.

Check:

(P X)^{t}Q^{t}GQ(Y) = X^{t}P^{t}Q^{t}GQY = X^{t}GY.

◁

3. (·|·) is called symmetric if

(x|y) = (y|x)

for all x, y. This is equivalent to G being a symmetric matrix G^{t} = G:

g_{ij} = (u_{i}|u_{j}) = (u_{j}|u_{i}) = g_{ji}.

A symmetric scalar product defines an associated quadratic form

F : M → K

by

F (x) = (x|x)

= X^{t}GX

= ( ξ^{1} … ξ^{n} ) ( \begin{matrix}g_{11} & … & g_{1n} \\ ⋮ & & ⋮ \\ g_{n1} & … & g_{nn} \end{matrix} )( ξ^{1} ; ⋮ ; ξ^{n} )

= g_{ij}ξ^{i}ξ^{j},

i.e.

F = ( u^{1} … u^{n} ) ( \begin{matrix}g_{11} & … & g_{1n} \\ ⋮ & & ⋮ \\ g_{n1} & … & g_{nn} \end{matrix} )( u^{1} ; ⋮ ; u^{n} ) = g_{ij}u^{i}u^{j}.

u^{i}u^{j} is a product of linear forms, and is a function:

(u^{i}u^{j})(x) = u^{i}(x)u^{j}(x).

7–2 Example: If x, y, z are coordinate functions on M then

F = ( x y z ) ( \begin{matrix}3 & 2 & 3 \\ 2 & −7 & −1 \\ 3 & −1 & 2 \end{matrix} )( \begin{matrix}x \\ y \\ z \end{matrix} )
= 3x^{2} − 7y^{2} + 2z^{2} + 4xy + 6xz − 2yz.

(Thus quadratic form ≡ homogeneous 2^{nd} degree polynomial).

The quadratic form F determines the symmetric scalar product (·|·) uniquely because:

(x + y|x + y) = (x|x) + (x|y) + (y|x) + (y|y),
2(x|y) = F (x + y) − F (x) − F (y) ( if 1 + 1 ≠ 0),

and g_{ij} = (u_{i}|u_{j}) are called the components of F wrt u_{i}. Definition. (·|·) is called non-singular if

(x|y) = 0 for all y ∈ M ⇒ x = 0,

i.e.

X^{t}GY = 0 for all Y ∈ K^{n} ⇒ X = 0,

i.e.

X^{t}G = 0 ⇒ X = 0,

i.e.

det G ≠ 0.

Definition. A tensor field (·|·) with two lower indices on an open set V ⊂ ℝ^{n}:

(·|·) = g_{ij}dy^{i} ⊗ dy^{j}

(say), y^{i} coordinates on V, is called a metric tensor if

(·|·)_{x}

is a symmetric non-singular scalar product on T_{x}ℝ^{n} for each x ∈ V, i.e.

g_{ij} = g_{ji} and det g_{ij} nowhere zero.

The associated field ds^{2} of quadratic forms:

ds^{2} = g_{ij}dy^{i}dy^{j}

is called the line-element associated with the metric tensor.

7–3 Example: On ℝ^{n} the usual metric tensor

dx ⊗ dx + dy ⊗ dy,

with line element ds^{2} = (dx)^{2} + (dy)^{2}, has components

( \begin{matrix}1 & 0 \\ 0 & 1 \end{matrix} )

wrt coordinates x, y.

If

v = v^{1} \frac{∂}{∂x} + v^{2} \frac{∂}{∂y} , w = w^{1} \frac{∂}{∂x} + w^{2} \frac{∂}{∂y}

then

(v|w) = ( v^{1} v^{2} ) ( \begin{matrix}1 & 0 \\ 0 & 1 \end{matrix} )( \begin{matrix}w^{1} \\ w^{2} \end{matrix} ) = v^{1}w^{1} + v^{2}w^{2} (dot product)
ds^{2}[v] = (v|v) = (v^{1})^{2} + (v^{2})^{2} = ∥v∥^{2} (Euclidean norm).

If r, θ are polar coordinates:

x = r cos θ, y = r sin θ,

then

dx = cos θ dr − r sin θ dθ,
dy = sin θ dr + r cos θ dθ

and

ds^{2} = (dx)^{2} + (dy)^{2}
= (cos θ dr − r sin θ dθ)^{2} + (sin θ dr + r cos θ dθ)^{2}
= (dr)^{2} + r^{2}(dθ)^{2}

has components

( \begin{matrix}1 & 0 \\ 0 & r^{2} \end{matrix} )

wrt coordinates r, θ.

If

v = α^{1} \frac{∂}{∂r} + α^{2} \frac{∂}{∂θ}, w = β^{1} \frac{∂}{∂r} + β^{2} \frac{∂}{∂θ}

then

(v|w) = α^{1}β^{1} + r^{2}α^{2}β^{2},
∥v∥^{2} = (α^{1})^{2} + r^{2}(α^{2})^{2}.

7–4

7.3 Raising and Lowering Indices

Definition. Let M be a finite dimensional vector space with a fixed nonsingular symmetric scalar product (·|·). If x ∈ M is a vector (one upper index), we associate with it

x̃ ∈ M^{∗},

a linear form (one lower index) defined by:

⟨x̃, y⟩ = (x|y) for all y ∈ M.

We call the operation

M → M^{∗}
x ↦ x̃

lowering the index. Thus

x̃ ≡ (x|·) ≡ ‘take scalar product with x^{,}.

If x = α^{i}u_{i} has components α^{i} then x̃ has components

α_{j} = ⟨x̃, u_{j}⟩ = (x|u_{j}) = (α^{i}u_{i}|u_{j}) = α^{i}(u_{i}|u_{j}) = α^{i}g_{ij}.

Since (·|·) is non-singular, g_{ij} is invertible, with inverse g^{ij} (say), and we have

α^{j} = α_{i}g^{ij}.

Thus

M → M^{∗}
x ↦ x̃

is a linear isomorphism, with inverse

f_{∼} ← f

(say), called raising the index. So

x = α^{i}u_{i} = f_{∼},
x̃ = α_{i}u^{i} = f

and

(x|y) = (f_{∼}|y) = ⟨f, y⟩ = ⟨x̃, y⟩.

7–5 To lower: contract with g_{ij} (α_{j} = α^{i}g_{ij}).

To raise: contract with g^{ij} (α^{j} = α_{i}g^{ij}).

Let M →^{T} M be a linear operator and (·|·) be symmetric. The matrix of T is:

α^{i}_{j} = ⟨u^{i}, T u_{j}⟩,

one up, one down mixed components of T.

α_{ij} = (u_{i}|T u_{j}),

two down covariant components of T.

α_{ij} = (u_{i}|α^{k}_{j}u_{k}) = (u_{i}|u_{k})α^{k}_{j} = g_{ik}α^{k}_{j}

(lower by contraction with g_{ij}). Therefore

α^{i}_{j} = g^{ik}α_{kj}

(raise by contraction with g^{ij}).

If we take the covariant components α_{ij}, and raise the second index we get

α_{i}^{j} = α_{ik}g^{kj}.

α_{ij} are the components of the tensor B (two lower indices) defined by:

B(x, y) = (x|T y),

since

B(u_{i}, u_{j}) = (u_{i}|T u_{j}) = α_{ij}.

α_{j}^{i} are the components of an operator T^{∗} (one upper index, one lower index) defined by:

(T^{∗}x|y) = (x|T y),

since T^{∗} has components

γ_{ij} = (u_{i}|T^{∗}u_{j}) = (T^{∗}u_{j}|u_{i}) = (u_{j}|T u_{i}) = α_{ji},

and therefore T^{∗} has mixed components:

γ^{i}_{j} = g^{ik}γ_{kj} = α_{jk}g^{ki} = α_{j}^{i}.

T^{∗} is called the adjoint of operator T.

7–6

7.4 Orthogonality and Diagonal Matrix

Definition. If (·|·) is a scalar product on M and

(x|y) = 0,

we say that x is orthogonal to y wrt (·|·).

If N is a vector subspace of M, we write

N^{⊥} = {x ∈ M : (x|y) = 0 for all y ∈ N},

and call it the orthogonal complement of N wrt (·|·) (see Figure 7.1).

⊥

Figure 7.1

We denote by (·|·)_{N} the scalar product on N defined by

(x|y)_{N} = (x|y) for all x, y ∈ N,

and call it the restriction of (·|·) to N.

Definition. Let N_{1}, …, N_{k} be vector subspaces of a vector space M. Then we write

N_{1} + ⋯ + N_{k} = {x_{1} + ⋯ + x_{k} : x_{1} ∈ N_{1}, …, x_{k} ∈ N_{k}},

and call it the sum of N_{1}, …, N_{k}. Thus M = N_{1} + ⋯ + N_{k} iff each x ∈ M can be written as a sum

x = x_{1} + ⋯ + x_{k}, x_{i} ∈ N_{i}.

7–7 We call M a direct sum of N_{1}, …, N_{k}, and write

M = N_{1} ⊕ ⋯ ⊕ N_{k}

if for each x ∈ M there exists unique (x_{1}, …, x_{k}) (for example, see Figure 7.2) such that

x = x_{1} + ⋯ + x_{k} and x_{i} ∈ N_{i}.

2

1

Figure 7.2

Theorem 7.1. Let (·|·) be a scalar product on M. Let N be a finite-dimensional vector subspace such that (·|·)_{N} is non-singular. Then

M = N ⊕ N^{⊥}.

7–8 Proof ▶ Let x ∈ M (see Figure 7.3). Define f ∈ N^{∗} by

⟨f, y⟩ = (x|y)

for all y ∈ N.

Since (·|·)_{N} is non-singular we can raise the index of f, and get a unique vector z ∈ N such that

⟨f, y⟩ = (z|y)

for all y ∈ N, i.e.

(x|y) = (z|y)

for all y ∈ N, i.e.

(x − z|y) = 0

for all y ∈ N, i.e.

x − z ∈ N^{⊥},

Figure 7.3

i.e.

x = z_{∈N} + (x − z)_{∈N^{⊥}}

uniquely, as required. ◀

7–9 Lemma 7.1. Let (·|·) be a symmetric scalar product, not identically zero on a vector space M over a field K of characteristic ≠ 2. (i.e. 1 + 1 ≠ 0). Then there exists x ∈ M such that

(x|x) ≠ 0.

Proof ▶ Choose x, y ∈ M such that (x|y) ≠ 0. Then

(x + y|x + y) = (x|x) + (x|y) + (y|x) + (y|y).

Hence (x + y|x + y), (x|x), (y|y) are not all zero. Hence result. ◀

Theorem 7.2. Let (·|·) be a symmetric scalar product on a finite-dimensional vector space M. Then M has a basis of mutually orthogonal vectors:

(u_{i}|u_{j}) = 0 if i ≠ j,

i.e. the scalar product has a diagonal matrix

( \begin{matrix}α_{1} & 0 & … & 0 \\ 0 & α_{2} & … & 0 \\ ⋮ & & ... & ⋮ \\ 0 & … & 0 & α_{n} \end{matrix} ) ,

where α_{i} = (u_{i}|u_{i}).

Proof ▶ Theorem holds if (x|y) = 0 for all x, y ∈ M. So suppose (·|·) is not identically zero.

Now we use induction on dim M. Theorem holds if dim M = 1. So assume dim M = n > 1, and that the theorem holds for all spaces of dimension less than n.

Choose u_{1} ∈ M such that

(u_{1}|u_{1}) = α_{1} ≠ 0.

Let N be the subspace generated by u_{1}. (·|·)_{N} has 1 × 1 matrix (α_{1}), and therefore is non-singular. Therefore

M = N ⊕ N^{⊥}
dim : n = 1 + n − 1.

By the induction hypothesis N^{⊥} has basis

u_{2}, …, u_{n}

7–10 (say) of mutually orthogonal vectors. Therefore u_{1}, u_{2}, …, u_{n} is a basis for M of mutually orthogonal vectors, as required. ◀

If M is a complex vector space, we can put

w_{i} = \frac{u_{i}}{\sqrt{α_{i}}}

for each α_{i} > 0. Then (w_{i}|w_{i}) = 1 or 0, and rearranging we have a basis wrt which (·|·) has matrix

( \begin{matrix}1 1 ... 1 & 0 \\ 0 & 0 0 ... 0 \end{matrix} )

(r × r diagonal block top left), and the associated quadratic form is a sum of squares:

(w^{1})^{2} + ⋯ + (w^{r})^{2}.

If M is a real vector space, we can put

w_{i} = \begin{cases}u_{i}/\sqrt{α_{i}} & α_{i} > 0; \\ u_{i}/\sqrt{−α_{i}} & α_{i} < 0; \\ u_{i} & α_{i} = 0. \end{cases}

Then (w_{i}|w_{i}) = ±1 or 0, and rearranging we have a basis wrt which (·|·) has matrix

( \begin{matrix}1 ... 1 & 0 & 0 \\ 0 & \begin{aligned}−1 \\ & . & . & . \\ & & & & −1 \end{aligned} & 0 \\ 0 & 0 & 0 ... 0 \end{matrix} ) ,

7–11 and the associated quadratic form is a sum and difference of squares:

(w^{1})^{2} + ⋯ + (w^{r})^{2} − (w^{r+1})^{2} − ⋯ − (w^{r+s})^{2}.

Example: Let (·|·) be a scalar product on a 3-dimensional space M which has matrix

A = ( \begin{matrix}4 & 2 & 2 \\ 2 & 0 & −1 \\ 2 & −1 & −3 \end{matrix} )

wrt a basis with coordinate functions x, y, z.

To find new coordinate functions wrt which (·|·) has a diagonal matrix. Method: Take the associated quadratic form

F = 4x^{2} − 3z^{2} + 4xy + 4xz − 2yz,

and write it as a sum and difference of squares, by ‘completing squares’. We have:

F = 4(x^{2} + xy + xz) − 3z^{2} − 2yz
= 4(x + \frac{1}{2}y + \frac{1}{2}z)^{2} − y^{2} − z^{2} − 2yz − 3z^{2} − 2yz
= 4(x + \frac{1}{2}y + \frac{1}{2}z)^{2} − (y^{2} + 4yz + 4z^{2})
= 4(x + \frac{1}{2}y + \frac{1}{2}z)^{2} − (y + 2z)^{2} + 0z^{2}
= 4u^{2} − v^{2} + 0w.

Therefore (·|·) has diagonal matrix

D = ( \begin{matrix}4 & 0 & 0 \\ 0 & −1 & 0 \\ 0 & 0 & 0 \end{matrix} )

wrt to coordinate functions

u = x + \frac{1}{2}y + \frac{1}{2}z,
v = y + 2z,
w = z.

The transition matrix is

P = ( \begin{matrix}1 & \frac{1}{2} & \frac{1}{2} \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end{matrix} ) .

7–12 Check: P^{t}DP = A?

( \begin{matrix}1 & 0 & 0 \\ \frac{1}{2} & 1 & 0 \\ \frac{1}{2} & 2 & 1 \end{matrix} ) ( \begin{matrix}4 & 0 & 0 \\ 0 & −1 & 0 \\ 0 & 0 & 0 \end{matrix} ) ( \begin{matrix}1 & \frac{1}{2} & \frac{1}{2} \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end{matrix} ) = ( \begin{matrix}4 & 2 & 2 \\ 2 & 0 & −1 \\ 2 & −1 & −3 \end{matrix} ) .

For a symmetric scalar product on a real vector space the number of + signs, and the number of − signs, when the matrix is diagonalised, is independent of the coordinates chosen:

Theorem 7.3 (Sylvester’s Law of Inertia). Let u_{1}, …, u_{n} and w_{1}, …, w_{n} be bases for a real vector space, and let

F = (u^{1})^{2} + ⋯ + (u^{r})^{2} − (u^{r+1})^{2} − ⋯ − (u^{r+s})^{2}
= (w^{1})^{2} + ⋯ + (w^{t})^{2} − (w^{t+1})^{2} − ⋯ − (w^{t+k})^{2}

be a quadratic form diagonalised by each of the two bases. Then r = t and s = k.

Proof ▶ Suppose r ≠ t, r > t (say). The space of solutions of the n − r + t homogeneous linear equations

u^{r+1} = 0, …, u^{n} = 0, w^{1} = 0, …, w^{t} = 0

has dimension at least

n − (n − r + t) = r − t > 0.

Therefore there exists a non-zero solution x so

F (x) = (u^{1}(x))^{2} + ⋯ + (u^{r}(x))^{2} > 0
= −(w^{t+1}(x))^{2} − ⋯ − (w^{t+k}(x))^{2} ≤ 0,

which is clearly a contradiction. Therefore r = t, and similarly s = k. ◀

7.5 Special Spaces

Definition. A real vector space M with a symmetric scalar product (·|·) is called a Euclidean space if the associated quadratic form is positive definite, i.e.

F (x) = (x|x) > 0 for all x ≠ 0,

7–13 i.e. there exists basis u_{1}, …, u_{n} such that (·|·) has matrix

( \begin{matrix}1 & 0 & … & 0 \\ 0 & 1 & … & 0 \\ ⋮ & & ... & ⋮ \\ 0 & … & 0 & 1 \end{matrix} )

(all + signs).

F = (u^{1})^{2} + ⋯ + (u^{n})^{2},
(u_{i}|u_{j}) = δ_{j}^{i},

i.e. u_{1}, …, u_{n} is orthonormal.

We write

∥x∥ = \sqrt{(x|x)} (x ∈ M),

and call it the norm of x. We have

∥x + y∥ ≤ ∥x∥ + ∥y∥ (Triangle Inequality).

Thus M is a normed vector space, and therefore a metric space, and therefore a toplogical space.

The scalar product also satisfies:

|(x|y)| ≤ ∥x∥∥y∥ (Schwarz Inequality).

We define the angle θ between two non-zero vectors x, y by:

\frac{(x|y)}{∥x∥∥y∥} = cos θ (0 ≤ θ ≤ π)

(see Figure 7.4).

7–14

π

Figure 7.4

If M is an n-dimensional vector space with scalar product having an orthonormal basis (e.g. a complex vector space or a Euclidean vector space) then the transition matrix P from one orthonormal basis to another satisfies:

P^{t}IP_{new} = I_{old},

i.e.

P^{t}P = I

i.e. P is an orthogonal matrix, i.e.

( ^{⋯ i^{th} col of P ⋯} ) ( ⋮ ; j^{th} ; col ; of P ) = ( \begin{matrix}1 & 0 & … & 0 \\ 0 & 1 & … & 0 \\ ⋮ & & ... & ⋮ \\ 0 & … & 0 & 1 \end{matrix} ) ,

i.e.

(i^{th} col of P).(j^{th} col of P) = δ_{ij},

i.e. the columns of P form an orthonormal basis of K^{n}.

Also

P orthonormal ⇔ P^{t} = P^{−1}

⇔ P P^{t} = I

⇔ the rows of P form an orthonormal basis of K^{n}.

7–15 Definition. A real 4-dimensional vector space M with scalar product (·|·) of type + + +− is called a Minkowski space. A basis u_{1}, u_{2}, u_{3}, u_{4} is called a Lorentz basis if wrt u_{i} the scalar product has matrix

( \begin{matrix}1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & −1 \end{matrix} ) ,

i.e.

F = (u^{1})^{2} + (u^{2})^{2} + (u^{3})^{2} − (u^{4})^{2}.

The transition matrix P from one Lorentz basis to another satisfies:

P^{t} ( \begin{matrix}1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & −1 \end{matrix} ) P = ( \begin{matrix}1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & −1 \end{matrix} ) .

Such a matrix P is called a Lorentz matrix.

Example: On ℂ^{n} we define the hermitian dot product (x|y) of vectors

x = (α_{1}, …, α_{n}), y = (β_{1}, …, β_{n})

to be

(x|y) = α_{1}\overline{β_{1}} + ⋯ + α_{n}\overline{β_{n}}.

This has the property of being positive definite, since:

(x|x) = α_{1}\overline{α_{1}} + ⋯ + α_{n}\overline{α_{n}} = ∥α_{1}∥^{2} + ⋯ + ∥α_{n}∥^{2} > 0 if x ≠ 0.

More generally:

Definition. If M is a complex vector space then a hermitian scalar product (·|·) on M is a function

M × M → ℂ

such that

(i) (x + y|z) = (x|z) + (y|z),

7–16 (ii) (αx|z) = α(x|z), (iii) (x|y + z) = (x|y) + (x|z), (iv) (x|αy) = \overline{α}(x|y),

(v) \overline{(x|y)} = (y|x).

(i) and (ii) imply linear in the first variable, (iii) and (iv) imply conjugatelinear in the second variable, (v) implies conjugate-symmetric.

If, in addition,

(x|x) > 0

for all x ≠ 0 then we call (·|·) a positive definite hermitian scalar product. Definition. A complex vector space M with a positive definite hermitian scalar product (·|·) is called a Hilbert space. Note. For a finite dimensional complex space M with an hermitian form (·|·) we can prove (in exactly the same way as for a real space with symmetric scalar product):

1. There exists basis wrt which (·|·) has matrix

( \begin{matrix}1 ... 1 & 0 & 0 \\ 0 & \begin{aligned}−1 \\ & . & . & . \\ & & & & −1 \end{aligned} & 0 \\ 0 & 0 & 0 ... 0 \end{matrix} ) .

2. The number of + signs and the number of − signs are each uniquely determined by (·|·).

3. M is a Hilbert space iff all the signs are +.

7–17 Thus M is a Hilbert space iff M has an orthonormal basis. The transition matrix P from one orthonormal basis to another satisfies:

P^{t}I\overline{P}_{new} = I_{old},

i.e.

P^{t}\overline{P} = I.

Such a matrix is called a unitary matrix.

A Hilbert space M is a normed space, hence a metric space, hence a topological space if we define:

∥x∥ = \sqrt{(x|x)}.

To test how many +, − signs a quadratic form has we can use determinants:

Example:

F = ax^{2} + 2bxy + cy^{2} = a (x + \frac{b}{a}y)^{2} + \frac{ac − b^{2}}{a} y^{2}

on a 2-dimensional space, with coordinate functions x, y and matrix ( \begin{matrix}a & b \\ b & c \end{matrix} ).

Therefore

++ ⇔ a > 0, | \begin{matrix}a & b \\ b & c \end{matrix} | > 0,
−− ⇔ a < 0, | \begin{matrix}a & b \\ b & c \end{matrix} | > 0,
+−  ⇔  | \begin{matrix}a & b \\ b & c \end{matrix} | < 0.

More generally:

Theorem 7.4 (Jacobi’s Theorem). Let F be a quadratic form on a real vector space M, with symmetric matrix g_{ij} wrt basis u_{i}. Suppose each of the determinants

Δ_{i} = | \begin{matrix}g_{11} & … & g_{1i} \\ ⋮ & & ⋮ \\ g_{i1} & … & g_{ii} \end{matrix} |

7–18 is non-zero (i = 1, …, n). Then there exists a basis w_{i} such that F has matrix

( \frac{1}{Δ_{1}} ; \frac{Δ_{1}}{Δ_{2}} ; ... ; \frac{Δ_{n−1}}{Δ_{n}} ) ,

i.e.

F = \frac{1}{Δ_{1}} (w^{1})^{2} + \frac{Δ_{1}}{Δ_{2}} (w^{2})^{2} + ⋯ + \frac{Δ_{n−1}}{Δ_{n}} (w^{n})^{2}.

Thus

F is +ve definite ⇔ Δ_{1}, Δ_{2}, …, Δ_{n} all positive,
F is −ve definite ⇔ Δ_{1} < 0, Δ_{2} > 0, Δ_{3} < 0, ….

Proof ▶ F (x) = (x|x), where (·|·) is a symmetric scalar product, (u_{i}|u_{j}) = g_{ij}. Let

N_{i} = S(u_{1}, …, u_{i}).

(·|·)_{N_{i}} is non-singular, since Δ_{i} ≠ 0 for i = 1, …, n.

Now

{0} ⊂ N_{1} ⊂ N_{2} ⊂ ⋯ ⊂ N_{i−1} ⊂ N_{i} ⊂ ⋯ ⊂ N_{n} = M.

Therefore

N_{i} = N_{i−1} ⊕ (N_{i} ∩ N_{i−1}^{⊥})
dim : i = (i − 1) + 1.

Choose non-zero w_{i} ∈ N_{i} ∩ N_{i−1}^{⊥}. Then

w_{1}, …, w_{i−1}, w_{i}, …, w_{n}

are mutually orthogonal, and w_{i} is orthogonal to u_{1}, …, u_{i−1}. Therefore w_{i} is not orthogonal to u_{i}, since (·|·) is non-singular. Therefore we can choose w_{i} such that (u_{i}|w_{i}) = 1.

It remains to show that

(w_{i}|w_{i}) = \frac{Δ_{i−1}}{Δ_{i}} .

7–19 To do this we write

λ_{1}u_{1} + ⋯ + λ_{i−1}u_{i−1} + λ_{i}u_{i} = w_{i}.

Taking scalar product with w_{i}, u_{1}, u_{2}, …, u_{i} we get:

0 + ⋯ + 0 + λ_{i} = (w_{i}|w_{i})
λ_{1}g_{11} + ⋯ + λ_{i−1}g_{1,i−1} + λ_{i}g_{1i} = 0
λ_{1}g_{21} + ⋯ + λ_{i−1}g_{2,i−1} + λ_{i}g_{2i} = 0
⋮
λ_{1}g_{i−1,1} + ⋯ + λ_{i−1}g_{i−1,i−1} + λ_{i}g_{i−1,i} = 0
λ_{1}g_{i1} + ⋯ + λ_{i−1}g_{i,i−1} + λ_{i}g_{ii} = 1

Therefore

(w_{i}|w_{i}) = λ_{i} = \begin{matrix}| & \begin{matrix}g_{11} & … & g_{1,i−1} & 0 \\ ⋮ & & ⋮ & ⋮ \\ g_{i−1,1} & … & g_{i−1,i−1} & 0 \\ g_{i1} & … & g_{i,i−1} & 1 \end{matrix} & | \\ | & \begin{matrix}g_{11} & … & g_{1,i−1} & g_{1i} \\ ⋮ & & ⋮ & ⋮ \\ g_{i−1,1} & … & g_{i−1,i−1} & g_{i−1,i} \\ g_{i1} & … & g_{i,i−1} & g_{ii} \end{matrix} & | \end{matrix} = \frac{Δ_{i−1}}{Δ_{i}} ,

as requried. ◀

This has an application in Calculus:

Theorem 7.5 (Criteria for local maxima or minima). Let f be a scalar field on a manifold X such that df_{X} = 0, and let y^{i} be coordinates on X at a. Put

Δ_{i} = | \begin{matrix}\frac{∂^{2}f}{∂y^{1}^{2}} & … & \frac{∂^{2}f}{∂y^{1}∂y^{i}} \\ ⋮ & & ⋮ \\ \frac{∂^{2}f}{∂y^{i}∂y^{1}} & … & \frac{∂^{2}}{∂y^{i}^{2}} \end{matrix} | .

Then

1. If Δ_{i}(a) > 0 for i = 1, …, n then there exists open nbd V of a such that

f(x) > f(a) for all x ∈ V, x ≠ a,

i.e. a is a local minima of f;

7–20 2. If Δ_{1}(a) < 0, Δ_{2}(a) > 0, Δ_{3}(a) < 0, … then there exists open nbd V of

a such that

f(x) < f(a) for all x ∈ V, x ≠ a,

i.e. a is a local maxima of f. To make sure that ∥x∥ = \sqrt{(x|x)} is a norm on a Euclidean or a Hilbert

space we need to show that the triangle inequality holds. Theorem 7.6. Let M be a Euclidean or a Hilbert space. Then

(i) |(x|y)| ≤ ∥x∥∥y∥ Schwarz, (ii) ∥x + y∥ ≤ ∥x∥ + ∥y∥ Triangle. Proof ▶

(i) Let x, y ∈ M. Then

(x|y) = |(x|y)|e^{iθ}, (y|x) = |(x|y)|e^{−iθ}

(say). So for all λ ∈ ℝ we have:

0 ≤ (λe^{−iθ}x + y|λe^{−iθ}x + y)
= ∥x∥^{2}λ^{2} + λe^{−iθ}(x|y) + λe^{iθ}(y|x) + ∥y∥^{2}
= ∥x∥^{2} + 2λ|(x|y)| + ∥y∥^{2}.

Therefore

|(x|y)|^{2} ≤ ∥x∥^{2}∥y∥^{2} (‘b^{2} ≤ 4ac^{,}).

Therefore

|(x|y)| ≤ ∥x∥∥y∥.

(ii)

∥x + y∥^{2} = (x + y|x + y)
= ∥x∥^{2} + (x|y) + (y|x) + ∥y∥^{2}
≤ ∥x∥^{2} + 2∥x∥∥y∥ + ∥y∥^{2}
= (∥x∥ + ∥y∥)^{2}.

Therefore

∥x + y∥ ≤ ∥x∥ + ∥y∥.

◀

7–21

Linear Operators 2

8.1 Adjoints and Isometries

Let M be a finite dimensional vector space with a fixed non-singular symmetric or hermitian scalar product (·|·). Recall that if

M →^{T} M

is a linear operator then the adjoint of T is the operator

M →^{T^{∗}} M,

which satisfies

(x|T y) = (T^{∗}x|y)

for all x, y ∈ M.

If (·|·) has matrix G wrt basis u_{i} and T has matrix A then T^{∗} has matrix

A^{∗} = G^{−1}A^{t}G (\overline{G^{−1}A^{t}G} in hermitian case)

because

X^{t}GAY = X^{t}GAG^{−1}GY = [G^{−1}A^{t}GX]^{t}GY,

and similarly

X^{t}G\overline{AY} = X^{t}G\overline{A}G^{−1}G\overline{Y} = [\overline{G^{−1}A^{t}G}X]^{t}G\overline{Y} .

8–1 Examples:

1. M Euclidean, basis orthonormal:

A^{∗} = A^{t}.

2. M Hilbert, basis orthonormal:

A^{∗} = \overline{A}^{t}.

3. M Minkowski, basis Lorentz:

A^{∗} = ( \begin{matrix}1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & −1 \end{matrix} ) A^{t} ( \begin{matrix}1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & −1 \end{matrix} ) .

Definition. An operator M →^{T} M is called an isometry if

(T x|T y) = (x|y) for all x, y ∈ M,

i.e. T preserves (·|·), i.e.

(T^{∗}T x|y) = (x|y),

i.e.

T^{∗}T = 1,

i.e.

T^{∗} = T^{−1}.

Examples:

1. M Euclidean, basis orthonormal, A matrix of T:

T is an isometry ⇔ A^{t}A = I,

i.e. A is an orthogonal matrix.

2. M Hilbert, basis orthonormal, A matrix of T:

T is an isometry ⇔ \overline{A}^{t}A = I,

i.e. A is a unitary matrix.

8–2 3. M Minkowski, basis Lorentz, A matrix of T:

T is an isometry ⇔ GA^{t}GA = I ⇔ A^{t}GA = G,

i.e. A is a Lorentz matrix. Definition. An isometry of a Euclidean space is called an orthogonal transformation. An isometry of a Hilbert space is called a unitary transformation. An isometry of a Minkowski space is called a Lorentz transformation. Definition. An operator M →^{T} M is called self-adjoint if

T^{∗} = T,

i.e.

(T x|y) = (x|T y) for all x, y ∈ M,

i.e.

(u_{i}|T u_{j}) = (T u_{i}|u_{j}) = (u_{j}|T u_{i}),

i.e. covariant components of T are symmetric.

(In quantum mechanics physical quantities are always represented by selfadjoint operators). Examples:

1. M Euclidean, basis orthonormal, A matrix of T:

T is self-adjoint ⇔ A^{t} = A,

i.e. A is symmetric.

2. M Hilbert, basis orthonormal, A matrix of T:

T is self-adjoint ⇔ \overline{A}^{t} = A,

i.e. A is hermitian.

3. M Minkowski, basis Lorentz, A matrix of T: T is self-adjoint ⇔ GA^{t}G = A.

8–3 Summary: Let M →^{T} M have matrix A wrt orthonormal or Lorentz basis. Then:

Space:  Euclidean  Hilbert  Minkowski
Matrix of T^{∗}:  A^{t}  \overline{A}^{t}  GA^{t}G
T self-adjoint:  A^{t} = A  \overline{A}^{t} = A  GA^{t}G = A
A symmetric A Hermitian
T an isometry: A^{t}A = I \overline{A}^{t}A = I A^{t}GA = G
A orthogonal A unitary A Lorentz

8.2 Eigenvalues and Eigenvectors

Definition. A vector space N ⊂ M is called invariant under a linear oper-

ator M →^{T} M if

T (N) ⊂ N,

i.e. x ∈ N ⇒ T x ∈ N.

A non-zero vector in a 1-dimensional invariant subspace under T is called an eigenvector of T:

(i) x ∈ M is called an eigenvector of T, with eigenvalue λ if

(a) x ≠ 0,

(b) T x = λx, where λ is a scalar (see Figure 8.1);

λx

Figure 8.1

8–4 (ii) λ ∈ K is called an eigenvalue of T if there exists x ≠ 0 such that

T x = λx,

i.e.

(T − λ1)x = 0,

i.e.

ker(T − λ1) ≠ {0}.

ker(T − λ1) = {x ∈ M : T x = λx}

is called the λ-eigenspace of T. It is the vector subspace consisting of all eigenvectors of T having eigenvalue λ, together with the zero vector.

Definition. If M →^{T} M is a linear operator on a vector space of finite dimension n, with matrix A = (α_{j}^{i}) wrt basis u_{i}, then the polynomial of degree n with coefficients in K:

char T = det | \begin{matrix}α_{1}^{1} − X & α_{2}^{1} & … & α_{n}^{1} \\ α_{1}^{2} & α_{2}^{2} − X & & ⋮ \\ ⋮ & & ... & ⋮ \\ α_{1}^{n} & … & … & α_{n}^{n} − X \end{matrix} | = det(A − XI)

is called the characteristic polynomial of T.

char T is well-defined, independent of choice of basis u_{i}, since if B is the matrix of T wrt another basis then

B = P AP^{−1}.

Therefore

det(B − XI) = det(P AP^{−1} − XI)
= det P (A − XI)P^{−1}
= det P det(A − XI) det P^{−1}
= det(A − XI),

since det P det P^{−1} = det P P^{−1} = det I = 1.

8–5 Theorem 8.1. If M →^{T} M is a linear operator and dim M < ∞ and λ ∈ K then

λ is an eigenvalue of T ⇔ λ is a zero of char T.

Proof ▶ Let T have matrix A wrt basis u_{i}. Then

λ is an eigenvalue of T ⇔ there exists y ∈ M such that (T − λ1)y = 0
⇔ there exists Y ∈ K^{n} such that (A − λI)Y = 0
⇔ det(A − λI) = 0
⇔ λ is a zero of det(A − XI).

◀

Corollary 8.1. If T is a linear operator on a finite dimensional complex space then T has an eigenvalue, and therefore eigenvectors.

Theorem 8.2. Let M →^{T} M be a linear operator on a finite dimensional vector space M. Then T has a diagonal matrix

( \begin{matrix}λ_{1} \\ & . & . & . \\ & & & & λ_{n} \end{matrix} )

wrt a basis u_{1}, …, u_{n} iff u_{i} is an eigenvector of T, with eigenvalue λ_{i}, for i = 1, …, n.

Proof ▶

T u_{1} = λ_{1}u_{1} + 0u_{2} + ⋯ + 0u_{n}
T u_{2} = 0u_{1} + λ_{2}u_{2} + ⋯ + 0u_{n}
⋮
T u_{n} = 0u_{1} + 0u_{2} + ⋯ + λ_{n}u_{n},

hence result. ◀

Theorem 8.3. Let M →^{T} M ba a self-adjoint operator on a Hilbert space M. Then all the eigenvalues of T are real.

Proof ▶ Let T x = λx, x ≠ 0, λ ∈ ℂ. Then

λ(x|x) = (λx|x) = (T x|x) = (x|T x) = (x|λx) = \overline{λ}(x|x).

(x|x) ≠ 0. Therefore λ = \overline{λ}. Therefore λ is real. ◀

8–6 Corollary 8.2. Let A ba a hermitian matrix. Then ℂ^{n} →^{A} ℂ^{n} is a selfadjoint operator wrt hermitian dot product. Therefore all the roots of the equation

det(A − XI) = 0

are real.

Corollary 8.3. Let T be a self-adjoint operator on a finite dimensional Euclidean space. Then T has an eigenvector.

Proof ▶ Wrt an orthonormal basis T has a real symmetric matrix A:

\overline{A}^{t} = A^{t} = A.

Therefore A is hermitian. Therefore det(A − XI) = 0 has real roots. Therefore T has an eigenvalue. Therefore T has eigenvectors. ◀

Theorem 8.4. Let N be invariant under a linear operator M →^{T} M. Then N^{⊥} is invariant under T^{∗}.

Proof ▶ Let x ∈ N^{⊥}. Then for all y ∈ N we have:

(T^{∗}x|y) = (x|T y) = 0.

Therefore T^{∗}x ∈ N^{⊥}. ◀

Definition. M →^{T} M is a normal operator if

T^{∗}T = T T^{∗},

i.e. T commutes with T^{∗}.

Examples:

(i) T self-adjoint ⇒ T normal.

(ii) T an isometry ⇒ T normal.

Theorem 8.5. Let S, T be commuting linear operators M → M (ST = T S). Then each eigenspace of S is invariant under T.

Proof ▶

Sx = λx ⇒ S(T x) = T (Sx) = T (λx) = λ(T x),

i.e. x ∈ λ-eigenspace of S ⇒ T x ∈ λ-eigenspace of S. ◀

8–7

8.3 Spectral Theorem and Applications

Theorem 8.6 (Spectral theorem). Let M →^{T} M be either a self-adjoint operator on a finite dimensional Euclidean space or a normal operator on a finite dimensional Hilbert space. Then M has an orthonormal basis of eigenvectors of T.

Proof ▶ (By induction on dim M). True for dim M = 1. Let dim M = n, and assume the theorem holds for spaces of dimension ≤ n − 1.

Let λ be an eigenvalue of T, M_{λ} the λ-eigenspace. (·|·)_{M_{λ}} is non-singular, since (·|·) is positive definite. Therefore

M = M_{λ} ⊕ M_{λ}^{⊥}.

M_{λ} is T -invariant. Therefore M_{λ}^{⊥} is T^{∗}-invariant. T^{∗} commutes with T. Therefore M_{λ} is T^{∗}-invariant. Therefore M_{λ}^{⊥} is T -invariant.

Now

(T^{∗}x|y) = (x|T y)

for all x, y ∈ M_{λ}^{⊥}. Therefore (T^{∗})_{M_{λ}^{⊥}} is the adjoint of T_{M_{λ}^{⊥}}. Therefore

T self-adjoint ⇒ T_{M_{λ}^{⊥}} is self-adjoint,

and

T normal ⇒ T_{M_{λ}^{⊥}} is normal.

But dim M_{λ}^{⊥} ≤ n − 1. Therefore, by induction hypothesis M_{λ}^{⊥} has an orthonormal basis of eigenvectors of T. Therefore M = M_{λ} ⊕ M_{λ}^{⊥} has an orthonormal basis of eigenvectors of T. ◀

Applications:

1. Let A be a real symmetric n × n matrix. Then

(i) ℝ^{n} has an orthonormal basis of eigenvectors u_{1}, …, u_{n} of A, with eigenvalues λ_{1}, …, λ_{n} (say),

(ii) if P is the matrix having u_{1}, …, u_{n} as rows then P is an orthogonal matrix and

P AP^{−1} = ( \begin{matrix}λ_{1} & 0 & … & 0 \\ 0 & λ_{2} & … & 0 \\ ⋮ & & ... & ⋮ \\ 0 & … & 0 & λ_{n} \end{matrix} ) = Q^{t}AQ,

where Q = P^{−1}.

8–8 Proof ▶ ℝ^{n} is a Euclidean space wrt the dot product, e_{1}, …, e_{n} is an orthonormal basis. Operator ℝ →^{A} ℝ^{n} has symmetric matrix A wrt orthonormal basis e_{1}, …, e_{n}. Therefore A is self-adjoint. Therefore ℝ^{n} has an orthonormal basis u_{1}, …, u_{n}, with eigenvalues λ_{1}, …, λ_{n}.

Let P be the transition matrix from orthonormal e_{1}, …, e_{n} to orthonormal u_{1}, …, u_{n}, with inverse matrix Q. P is an orthogonal matrix, and therefore

Q = P^{−1} = P^{t}.

Q is the transition matrix from u_{i} to e_{i}. Therefore

u_{j} = q_{j}^{1}e_{1} + ⋯ + q_{j}^{n}e_{n} = (q_{j}^{1}, …, q_{j}^{n})
= j^{th} column of Q
= j^{th} row of P.

Matrix of operator A wrt basis u_{i} is:

P AP^{−1} = P AP^{t} = ( \begin{matrix}λ_{1} & 0 & … & 0 \\ 0 & λ_{2} & … & 0 \\ ⋮ & & ... & ⋮ \\ 0 & … & 0 & λ_{n} \end{matrix} ) .

◀

2. (Principal axes theorem) Let F be a quadratic form on a finite dimensional Euclidean space M. Then M has an orthonormal basis u_{1}, …, u_{n} which diagonalises F:

F = λ_{1}(u^{1})^{2} + ⋯ + λ_{n}(u^{n})^{2}.

Such a basis is called a set of principal axes for F.

Proof ▶ F (x) = B(x, x), where B is a symmetric bilinear form. Raising an index of B gives a self-adjoint operator T:

(x|T y) = B(x, y) = (T x|y).

Let u_{1}, …, u_{n} be an orthonormal basis of M of eigenvectors of T, with eigenvalues λ_{1}, …, λ_{n} (say). Then wrt u_{i} the quadratic form F has matrix:

B(u_{i}, u_{j}) = (u_{i}|T u_{j}) = (u_{i}|λ_{j}u_{j}) = λ_{j}δ_{j}^{i},

8–9 i.e.

( \begin{matrix}λ_{1} & 0 & … & 0 \\ 0 & λ_{2} & … & 0 \\ ⋮ & & ... & ⋮ \\ 0 & … & 0 & λ_{n} \end{matrix} ) ,

as required. ◀

Note. If F has matrix A = (α_{ij}), and (·|·) has matrix G = (g_{ij}) then T has matrix

(α^{i}_{j}) = (g^{ik}α_{kj}) = G^{−1}A.

Therefore λ_{1}, …, λ_{n} are the roots of

det(G^{−1}A − XI) = 0,

i.e.

det(A − XG) = 0.

3. Consider the surface

ax^{2} + by^{2} + cz^{2} + 2hxy + 2gxz + 2fyz = k (k > 0)

in ℝ^{3}. The LHS is a quadratic form with matrix

A = ( \begin{matrix}a & h & g \\ h & b & f \\ g & f & c \end{matrix} )

wrt usual coordinate functions x, y, z. By the principal axes theorem we can choose new orthonormal coordinates X, Y, Z such that equation becomes:

λ_{1}X^{2} + λ_{2}Y^{2} + λ_{3}Z^{2} = k,

where λ_{1}, λ_{2}, λ_{3} are eigenvalues of A.

The surface is:

an ellipsoid if λ_{1}, λ_{2}, λ_{3} are all > 0, i.e. if the quadratic form is positive definite, i.e.

a > 0, ab − h^{2} > 0, det A > 0 by Jacobi;

8–10 a hyperboloid of 1-sheet (see Figure 8.2) if the quadratic form is of type + + − (e.g. X^{2} + Y^{2} = Z^{2} + 1), i.e.

a > 0, ab − h^{2} > 0, det A < 0
or a > 0, ab − h^{2} < 0, det A < 0
or a < 0, ab − h^{2} < 0, det A < 0;

Figure 8.2

a hyperboloid of 2-sheets (see Figure 8.3) if the quadratic form is of type + − − (e.g. X^{2} + Y^{2} = Z^{2} − 1), i.e.

a > 0, ab − h^{2} < 0, det A > 0
or a < 0, ab − h^{2} < 0, det A > 0
or a < 0, ab − h^{2} > 0, det A > 0.

8–11 Figure 8.3

8–12

Skew-Symmetric Tensors and Wedge Product

9.1 Skew-Symmetric Tensors

Definition. A bijective map

σ : {1, 2, …, r} → {1, 2, …, r}

is called a permutation of degree r. The group S_{r} of all permutations of degree r is called the symmetric group of degree r. Thus S_{r} is a group of order r!.

Let T^{r}M denote the space of all tensors over M of type

M × M × ⋯ × M → K.

Thus T^{r}M consists of all tensors T with components

T (u_{i_{1}}, …, u_{i_{r}}) = α_{i_{1}...i_{r}} (r lower indices),
T = α_{i_{1}...i_{r}}u^{i_{1}} ⊗ ⋯ ⊗ u^{i_{r}}.

u^{i_{1}} ⊗ ⋯ ⊗ u^{i_{r}} is a basis for T^{r}M.

For each σ ∈ S_{r}, and each T ∈ T^{r}M we define σ.T ∈ T^{r}M by:

(σ.T)(x_{1}, …, x_{r}) = T (x_{σ(1)}, …, x_{σ(r)}).

If T has components α_{i_{1}...i_{r}} then σ.T has components β_{i_{1}...i_{r}}, where

β_{i_{1}...i_{r}} = (σ.T)(u_{i_{1}}, …, u_{i_{r}}) = T (u_{i_{σ(1)}}, …, u_{i_{σ(r)}}) = α_{i_{σ(1)}} … α_{i_{σ(r)}}.

9–1 Theorem 9.1. The group S_{r} acts on T^{r}M by linear transformations, i.e.

(i) σ.(αT + βS) = α(σ.T) + β(σ.S), (ii) σ.(τ.T) = (στ).T, (iii) 1.T = T for all α, β ∈ K, σ, τ ∈ S_{r}, S, T ∈ T^{r}M. Proof ▶ e.g. (ii)

[σ.(τ.T)](x_{1}, …, x_{r}) = (τ.T)[x_{σ(1)}, …, x_{σ(r)}]
= T (x_{σ(τ(1))}, …, x_{σ(τ(r))})
= [(στ).T](x_{1}, …, x_{r}).

Therefore σ.(τ.T) = (στ).T. ◀ Note. If σ ∈ S_{r}, we put

ϵ^{σ} = \{ \begin{matrix}+1 & if σ is an even permutation; \\ −1 & if σ is an odd permutation \end{matrix} \} = sign of σ.

We have:

(i) ϵ^{στ} = ϵ^{σ}ϵ^{τ}, (ii) ϵ^{1} = 1, (iii) ϵ^{σ^{−1}} = ϵ^{σ}. Definition. T ∈ T^{r}M is skew-symmetric if

σ.T = ϵ^{σ}T for all σ ∈ S_{r},

i.e.

T (x_{σ(1)}, …, x_{σ(r)}) = ϵ^{σ}T (x_{1}, …, x_{r})

for all σ ∈ S_{r}, x_{1}, …, x_{r} ∈ M, i.e. the components α_{i_{1}...i_{r}} of T satisfy: α_{i_{σ(1)}...i_{σ(r)}} = ϵ^{σ}α_{i_{1}...i_{r}}.

9–2 Example: T ∈ T^{3}M, with components α_{ijk} is skew-symmetric iff

α_{ijk} = −α_{jik} = α_{jki} = −α_{kji} = α_{kij} = −α_{ikj}.

It follows that if T is skew-symmetric, with components α_{i_{1}...i_{r}} (from now on assume K has characteristic zero, i.e. α ≠ 0 ⇒ α + α + ⋯ + α ≠ 0) then

1. α_{i_{1}...i_{r}} = 0 if i_{1}, …, i_{r} are not all distinct;

2. if we know α_{i_{1}...i_{r}} for all increasing sequences i_{1} < ⋯ < i_{r} then we know α_{i_{1}...i_{r}} for all sequences i_{1}, …, i_{r};

3. if T is skew-symmetric, with components α_{i_{1}...i_{r}} and S is skew-symmetric, with components β_{i_{1}...i_{r}}, and if α_{i_{1}...i_{r}} = β_{i_{1}...i_{r}} for all increasing sequences i_{1} < ⋯ < i_{r} then T = S.

Theorem 9.2. Let T ∈ T^{r}M. Then

∑_{σ∈S_{r}} ϵ^{σ}σ.T

is skew-symmetric. Proof ▶ Let τ ∈ S_{r}. Then

τ. (∑_{σ∈S_{r}} ϵ^{σ}σ.T ) = ϵ^{τ} ∑_{σ∈S_{r}} ϵ^{τσ}(τσ).T = ϵ^{τ} ∑_{σ∈S_{r}} σϵ^{σ}(σ.T),

as required. ◀ Definition. The linear operator

A : T^{r}M → T^{r}M

defined by

AT = \frac{1}{r!} ∑_{σ∈S_{r}} ϵ^{σ}σ.T

is called the skew-symmetriser.

9–3 Example: Let T ∈ T^{3}M have components α_{ijk}. Then

AT (x, y, z) =

\frac{1}{6}[T (x, y, z) − T (y, x, z) + T (y, z, x) − T (x, z, y) + T (z, x, y) − T (z, y, x)],

and AT has components

β_{ijk} = \frac{1}{6}(α_{ijk} − α_{ikj} + α_{jki} − α_{jik} + α_{kij} − α_{kji}).

Theorem 9.3. Let S ∈ T^{s}M, T ∈ T^{t}M. Then

(i) A[(AS) ⊗ T] = A[S ⊗ T] = A[S ⊗ AT], (ii) A(S ⊗ T) = (−1)^{st}A(T ⊗ S). Proof ▶

(i) We first note that if τ ∈ S_{s} then

[(τ.S) ⊗ T](x_{1}, …, x_{s}, x_{s+1}, …, x_{s+t}) = (τ.S)(x_{1}, …, x_{s})T (x_{s+1}, …, x_{s+t})
= S(x_{τ(1)}, …, x_{τ(s)})T (x_{s+1}, …, x_{s+t})
= S(x_{τ′(1)}, …, x_{τ′(s)})T (x_{τ′(s+1)}, … x_{τ′(s+t)})
= [τ′.(S ⊗ T)](x_{1}, …, x_{s}, x_{s+1}, …, x_{s+t}),

where

τ′ = ( \begin{matrix}1 & … & s & s + 1 & … & s + t \\ τ(1) & … & τ(s) & s + 1 & … & s + t \end{matrix} ) .

Thus

(τ.S) ⊗ T = τ′.(S ⊗ T)

and ϵ^{τ′} = ϵ^{τ}. Now

A[(AS) ⊗ T] = \frac{1}{(s + t)!} ∑_{σ∈S_{s+t}} ϵ^{σ}σ. [(\frac{1}{s!} ∑_{τ∈S_{s}} ϵ^{τ}τ.S) ⊗ T ]

= \frac{1}{s!} ∑_{τ∈S_{s}} \frac{1}{(s + t)!} ∑_{σ∈S_{s+t}} ϵ^{στ′}(στ′).(S ⊗ T)

= \frac{1}{s!} ∑_{τ∈S_{s}} A(S ⊗ T)

= A(S ⊗ T).

9–4 (ii) Let

τ = ( \begin{matrix}1 & … & s & s + 1 & … & s + t \\ t + 1 & … & t + s & 1 & … & t \end{matrix} )

so that ϵ^{τ} = (−1)^{st}. Then

[τ.(S ⊗ T)](x_{1}, …, x_{s}, x_{s+1}, …, x_{s+t}) = S ⊗ T [x_{t+1}, …, x_{t+s}, x_{1}, …, x_{t}]
= T (x_{1}, …, x_{t})S(x_{t+1}, … x_{t+s})
= (T ⊗ S)[x_{1}, …, x_{t+s}].

Therefore

τ.(S ⊗ T) = T ⊗ S.

Therefore

A(S ⊗ T) = \frac{1}{(s + t)!} ∑_{σ∈S_{s+t}} ϵ^{στ}στ.(T ⊗ S)

= ϵ^{τ} \frac{1}{(s + t)!} ∑_{σ∈S_{s+t}} ϵ^{σ}σ.(T ⊗ S)

= (−1)^{st}A(T ⊗ S),

as required. ◀

9.2 Wedge Product

Definition. If S ∈ T^{s}M and T ∈ T^{t}M, we define their wedge product (also called exterior product) by

S ∧ T = \frac{1}{s!t!} ∑_{σ∈S_{s+t}} ϵ^{σ}σ(S ⊗ T) = \frac{(s + t)!}{s!t!}A(S ⊗ T).

Example: Let S, T ∈ M^{∗} have components α_{i}, β_{i} wrt u_{i}. Then

S ∧ T = S ⊗ T − T ⊗ S.

Therefore

S ∧ T [x, y] = S(x)T (y) − T (x)S(y),

and S ∧ T has components γ_{ij} = S ∧ T [u_{i}, u_{j}] = S(u_{i})T (u_{j}) − T (u_{i})S(u_{j}) = α_{i}β_{j} − β_{i}α_{j}.

9–5 Theorem 9.4. The wedge product has the following properties:

1. (R + S) ∧ T = R ∧ T + S ∧ T,

2. R ∧ (S + T) = R ∧ S + R ∧ T,

3. (λR) ∧ S = λ(R ∧ S) = R ∧ (λS),

4. R ∧ (S ∧ T) = (R ∧ S) ∧ T,

5. S ∧ T = (−1)^{st}T ∧ S,

6. R_{1} ∧ ⋯ ∧ R_{k} = \frac{(r_{1}+···+r_{k})!}{r_{1}!...r_{k}!}A(R_{1} ⊗ ⋯ ⊗ R_{k}). (i), (ii) and (iii) imply bilinear; (iv) implies associative; (v) implies graded commutative. Proof ▶ e.g. 4.

(R ∧ S) ∧ T = \frac{(r + s + t)!}{(r + s)!t!}A [\frac{(r + s)!}{r!s!}(A(R ⊗ S) ⊗ T ]

= \frac{(r + s + t)!}{r!s!t!}A(R ⊗ S ⊗ T)

=^{sim.} R ∧ (S ∧ T).

5.

S ∧ T = \frac{(s + t)!}{s!t!}A(S ⊗ T) = (−1)^{st} \frac{(t + s)!}{t!s!}A(T ⊗ S) = (−1)^{st}T ∧ S.

6. By induction on k: true for k = 1, assume true for k − 1. Then: (R_{1} ∧ ⋯ ∧ R_{k−1}) ∧ R_{k}

= \frac{(r_{1} + ⋯ + r_{k−1} + r_{k})!}{(r_{1} + ⋯ + r_{k−1})!r_{k}!}A [\frac{r_{1} + ⋯ + r_{k−1})!}{r_{1}! … r_{k−1}!}A((R_{1} ⊗ ⋯ ⊗ R_{k−1}) ⊗ R_{k})]

= \frac{(r_{1} + ⋯ + r_{k})!}{r_{1}! … r_{k}!}A[R_{1} ⊗ ⋯ ⊗ R_{k}].

◀

Note. For each integer r > 0 we write

M^{(r)} for the space of all skew-symmetric tensors of type

M × ⋯ × M_{←r→} → K;

9–6 M_{(r)} for the space of all skew-symmetric tensors of type

M^{∗} × ⋯ × M^{∗}_{←r→} → K;

M^{(0)} = K = M_{(0)}. If S ∈ M^{(s)} or S ∈ M_{(s)}, we say that S has degree s, and we have

S ∧ T = (−1)^{st}T ∧ S if s = deg S, t = deg T.

Thus 1. S ∧ T = T ∧ S if either s or T has even degree;

2. S ∧ T = −T ∧ S if both S and T have odd degree;

3. S ∧ S = 0 if S has odd degree, since S ∧ S = −S ∧ S;

4. T_{1} ∧ T_{2} ∧ ⋯ ∧ S ∧ ⋯ ∧ S ∧ ⋯ ∧ T_{k} = 0 if S has odd degree;

5. If x_{1}, …, x_{r} ∈ M and i_{1}, …, i_{r} are selected from {1, 2, …, r} then

x_{i_{1}} ∧ ⋯ ∧ x_{i_{r}} = ϵ_{i_{1}...i_{r}}x_{1} ∧ x_{2} ∧ ⋯ ∧ x_{r},

where

ϵ_{i_{1}...i_{r}} = \{ \begin{matrix}1 & if i_{1}, …, i_{r} is an even permutation of 1, …, r; \\ −1 & if i_{1}, …, i_{r} is an odd permutation of 1, …, r; \\ 0 & otherwise \end{matrix} \} = ϵ^{i_{1}...i_{r}}

is called a permutation symbol;

6. If x_{j} = α_{j}^{i}y_{i}, (α_{j}^{i}) = A ∈ K^{r×r} then

x_{1} ∧ ⋯ ∧ x_{r} = (α_{1}^{i_{1}}y_{i_{1}}) ∧ ⋯ ∧ (α_{r}^{i_{r}}y_{i_{r}})
= α_{1}^{i_{1}} … α_{r}^{i_{r}} y_{i_{1}} ∧ ⋯ ∧ y_{i_{r}}
= ϵ_{i_{1}...i_{r}}α_{1}^{i_{1}} … α_{r}^{i_{r}} y_{1} ∧ ⋯ ∧ y_{r}
= (det A)y_{1} ∧ ⋯ ∧ y_{r}.

Theorem 9.5. Let M be n-dimensional, with basis u_{1}, …, u_{n}. Then (i) if r > n then M^{(r)} = {0},

9–7 (ii) if 1 ≤ r ≤ n then

dim M^{(r)} = \frac{n!}{r!(n − r)!},

and {u^{i_{1}} ∧ ⋯ ∧ u^{i_{r}}}_{i_{1}<···<i_{r}} is a basis for M^{(r)}.

Proof ▶

(i) If r > n and T ∈ M^{(r)} has components α_{i_{1}...i_{r}} then the indices i_{1}, …, i_{r} cannot be distinct. Therefore T = 0.

(ii) We have

⟨u^{i}, u_{j}⟩ = \{ \begin{matrix}1 & i = j; \\ 0 & i ≠ j \end{matrix} \} = δ_{j}^{i}.

More generally:

u^{i_{1}} ∧ ⋯ ∧ u^{i_{r}}[u_{j_{1}}, …, u_{j_{r}}]

= ∑_{σ∈S_{r}} ϵ^{σ}u^{i_{1}} ⊗ ⋯ ⊗ u^{i_{r}}[u_{j_{σ(1)}}, …, u_{j_{σ(r)}}]

= ∑_{σ∈S_{r}} ϵ^{σ}δ_{j_{σ(1)}}^{i_{1}} … δ_{j_{σ(r)}}^{i_{r}}

= \begin{cases}1 & if i_{1}, …, i_{r} are distinct and an even permutation of j_{1}, …, j_{r}; \\ −1 & if i_{1}, …, i_{r} are distinct and an odd permutation of j_{1}, …, j_{r}; \\ 0 & otherwise \end{cases}

= δ^{i_{1}...i_{r}}_{j_{1}...j_{r}} (general Kronecker delta).

It follows that if 1 ≤ r ≤ n, and if T ∈ M^{(r)} has components α_{i_{1}...i_{r}} then the tensor:

(∗) ∑_{i_{1}<···<i_{r}}α_{i_{1}...i_{r}}u^{i_{1}} ∧ ⋯ ∧ u^{i_{r}}

has components

∑_{i_{1}<···<i_{r}} α_{i_{1}...i_{r}}u^{i_{1}} ∧ ⋯ ∧ u^{i_{r}}[u_{j_{1}}, …, u_{j_{r}}] = ∑_{i_{1}<···<i_{r}} α_{i_{1}...i_{r}}δ^{i_{1}...i_{r}}_{j_{1}...j_{r}}
= α_{j_{1}...j_{r}},

provided j_{1} < ⋯ < j_{r}. Therefore (∗) has the same components as T. Therefore

(∗∗) {u^{i_{1}} ∧ ⋯ ∧ u^{i_{r}}}_{i_{1}<···<i_{r}}

9–8 generate M^{(r)}. Also

(∗) = 0 ⇒ α_{j_{1}...j_{r}} = 0.

Therefore (∗∗) are linearly independent. Therefore (∗∗) form a basis for M^{(r)}. ◀

9–9

Classification of Linear Operators

10.1 Hamilton-Cayley and Primary Decomposition

Let M →^{T} M be a linear operator on a vector space M over a field K. Let K[X] be the ring of polynomials in X with coefficients in K. If p = α_{0} + α_{1}X + α_{2}X^{2} + ⋯ + α_{r}X^{r}, write

p(T) = α_{0}1 + α_{1}T + α_{2}T^{2} + ⋯ + α_{r}T^{r} ∈ L(M).

Theorem 10.1 (Hamilton-Cayley). Let T ∈ L(M) have characteristic polynomial p, and let dim M < ∞. Then p(T) = 0.

Proof ▶ Let T have matrix α_{j}^{i} wrt basis u_{i}. Put P = (p_{j}^{i}), where p_{j}^{i} = α_{j}^{i} − Xδ_{j}^{i}. Then P is an n × n matrix of polynomials, and det P = p is the characteristic polynomial.

Let q_{j}^{i} = (−1)^{i+j} times the determinant of the matrix got from P by removing the i^{th} column and j^{th} row. Then Q = (q_{j}^{i}) is also an n × n matrix of polynomials, and

P Q = (det P)I,

i.e.

p_{k}^{i}q_{j}^{k} = pδ_{j}^{i}

Therefore

p(T)u_{j} = p(T)δ_{j}^{i}u_{i} = p_{k}^{i}(T)q_{j}^{k}(T)u_{i}

= q_{j}^{k}(T)[α_{k}^{i} − T δ_{k}^{i}]u_{i} = q_{j}^{k}(T)[α_{k}^{i}u_{i} − T u_{k}] = 0

10–1 Therefore p(T) = 0, as required. ◀

Example: ( \begin{matrix}α & β \\ γ & δ \end{matrix} ) : ℝ^{2} → ℝ^{2}.

p = | \begin{matrix}α − X & β \\ γ & δ − X \end{matrix} | = X^{2} − (α + δ)X + αδ − βγ.

Therefore

p ( \begin{matrix}α & β \\ γ & δ \end{matrix} ) = ( \begin{matrix}α & β \\ γ & δ \end{matrix} )^{2} − (α + δ) ( \begin{matrix}α & β \\ γ & δ \end{matrix} ) + (αδ − βγ) ( \begin{matrix}1 & 0 \\ 0 & 1 \end{matrix} )

= ( \begin{matrix}0 & 0 \\ 0 & 0 \end{matrix} ) .

Theorem 10.2 (Primary Decomposition Theorem). Let T ∈ L(M), and let

(T − λ_{1}1)^{r_{1}} … (T − λ_{k}1)^{r_{k}} = 0,

where λ_{1}, …, λ_{k} are distinct scalars, and r_{1}, …, r_{k} are positive integers. Then

M = M_{1} ⊕ ⋯ ⊕ M_{k},

where M_{i} = ker(T − λ_{i}1)^{r_{i}} for i = 1, …, k.

Proof ▶ Let

f = (X − λ_{1})^{r_{1}} … (X − λ_{k})^{r_{k}},
g_{i} = (X − λ_{i})^{r_{i}},
f = g_{i}h_{i}

(say), so f(T) = 0 and M_{i} = ker g_{i}(T).

Now h_{1}, …, h_{k} have hcf 1. Therefore there exist

Θ_{1}, …, Θ_{k} ∈ K[X]

such that

Θ_{1}h_{1} + ⋯ + Θ_{k}h_{k} = 1.

Put P_{i} = Θ_{i}(T)h_{i}(T). Then

10–2 (i) P_{1} + ⋯ + P_{k} = 1. Also: (ii) for each x ∈ M,

g_{i}(T)P_{i}x = g_{i}(T)Θ_{i}(T)h_{i}(T)x = f(T)Θ_{i}(T)x = 0.

Therefore P_{i}x ∈ M_{i}, (iii) if x_{i} ∈ M_{i} and j ≠ i then

P_{j}x_{i} = Θ_{j}(T)h_{j}(T)x_{i} = 0,

since g_{i} is a factor of h_{j}, and g_{i}(T)x_{i} = 0. Thus

1. for each x ∈ M we have x = 1x = P_{1}x + ⋯ + P_{k}x, P_{i}x ∈ M_{i};

2. if x = x_{1} + ⋯ + x_{k}, with x_{i} ∈ M_{i}, then (for example, see Figure 10.1)

x_{i} = (P_{1} + ⋯ + P_{k})x_{i} = P_{i}x_{i} = P_{i}(x_{1} + ⋯ + x_{k}) = P_{i}x.

Therefore x_{i} is uniquely determined by x. Therefore M = M_{1} ⊕ ⋯ ⊕ M_{k},

10–3 P_{2}x

P_{1}x

1

Figure 10.1

as required. ◀ Note. Each subspace M_{i} is invariant under T, since

x ∈ M_{1} ⇒ g_{i}(T)x = 0
⇒ T g_{i}(T)x = 0
⇒ g_{i}(T)T x = 0
⇒ T x ∈ M_{i}.

Therefore, if we take bases for M_{1}, …, M_{k}, and put them together to get a basis for M, then wrt this basis T has matrix

( \begin{matrix}A_{1} & & & & & 0 \\ & A_{2} \\ & & . & . & . \\ 0 & & & & & A_{k} \end{matrix} ) ,

where A_{i} is the matrix of T_{M_{i}}, the restriction of T to M_{i}.

10–4 Note also that

(T_{M_{i}} − λ_{i}1)^{r_{i}} = 0.

Example: Let M →^{T} M and T^{2} = T (T a projection operator). Then

T (T − 1) = 0.

Therefore, by primary decomposition,

M = ker(T − 1) ⊕ ker T

= 1-eigenspace ⊕ 0-eigenspace.

If T has rank r, and u_{1}, …, u_{r} is basis of 1-eigenspace; u_{r+1}, …, u_{n} is basis of 0-eigenspace then, wrt u_{1}, …, u_{n} T has matrix

( \begin{matrix}\begin{matrix}1 & … & 0 \\ ⋮ & ... & ⋮ \\ 0 & … & 1 \end{matrix} & 0 \\ 0 & \begin{matrix}0 & … & 0 \\ ⋮ & ... & ⋮ \\ 0 & … & 0 \end{matrix} \end{matrix} ) = ( \begin{matrix}I & 0 \\ 0 & 0 \end{matrix} ) .

10.2 Diagonalisable Operators

Let M →^{T} M; dim M < ∞. Then:

(T − λ_{1}1) … (T − λ_{k}1) = 0 (λ_{1}, …, λ_{k} distinct) ⇒ M = ker(T − λ_{1}1) ⊕ ⋯ ⊕ (T − λ_{k}1) by Primary Decomposition ⇒ M = (λ_{1}-eigenspace) ⊕ ⋯ ⊕ (λ_{k}-eigenspace) ⇒ T has a diagonal matrix wrt some basis of M ⇒ M has basis consisting of eigenvectors of T; and (T − λ_{1}1) … (T − λ_{k}1)u = 0

for each eigenvector u, where λ_{1}, …, λ_{k} are the distinct eigenvalues of T ⇒ (T − λ_{1}1) … (T − λ_{k}1) = 0 (λ_{1}, …, λ_{k} distinct).

Definition. A linear operator T with any one (and hence all) of the above properties is called diagonalisable.

10–5 Example: The operator

( \begin{matrix}1 & 0 \\ 1 & 1 \end{matrix} ) : ℝ^{2} → ℝ^{2}

is not diagonalisable.

Proof of This ▷ The characteristic polynomial is

| \begin{matrix}1 − X & 0 \\ 1 & 1 − X \end{matrix} | = (1 − X)^{2}.

Therefore 1 is the only eigenvalue.

Also

(α, β) ∈ 1-eigenspace ⇔ ( \begin{matrix}1 & 0 \\ 1 & 1 \end{matrix} ) ( \begin{matrix}α \\ β \end{matrix} ) = ( \begin{matrix}α \\ β \end{matrix} )

⇔ ( \begin{matrix}α \\ α + β \end{matrix} ) = ( \begin{matrix}α \\ β \end{matrix} )

⇔ α = 0.

Therefore 1-eigenspace = {β(0, 1) : β ∈ ℝ} is 1-dimensional. Therefore ℝ^{2}

does not have a basis of eigenvectors of ( \begin{matrix}1 & 0 \\ 1 & 1 \end{matrix} ). ◁

Theorem 10.3. Let S, T ∈ L(M) be diagonalisable (dim M < ∞). Then there exists a basis wrt which both S and T have diagonal matrices (S, T simultaneously diagonalisable) iff ST = T S (S, T commute).

Proof ▶

(i) Let M have a basis wrt which S has diagonal matrix

A = ( \begin{matrix}λ_{1} \\ & . & . & . \\ & & & & λ_{n} \end{matrix} )

and T has diagonal matrix

B = ( \begin{matrix}μ_{1} \\ & . & . & . \\ & & & & μ_{n} \end{matrix} )

Then AB = BA. Therefore ST = T S.

10–6 (ii) Let ST = T S. Since S is diagonalisable we have:

M = M_{1} ⊕ ⋯ ⊕ M_{i} ⊕ ⋯ ⊕ M_{k},

distinct sum of eigenspaces of S. Since S and T commute, T leaves each M_{i} invariant:

T_{M_{i}} : M_{i} → M_{i}.

Since T is diagonalisable we have:

(T − μ_{1}1) … (T − μ_{l}1) = 0,

distinct μ_{1}, …, μ_{l}. Therefore

(T_{M_{i}} − μ_{1}1_{M_{i}}) … (T_{M_{i}} − μ_{l}1_{M_{i}}) = 0.

Therefore T_{M_{i}} is diagonalisable. Therefore M_{i} has a basis of eigenvectors of T. Therefore M has a basis of eigenvectors of S, and of T.

◀

10.3 Conjugacy Classes

Problem: Given two linear operators

S, T : M → M,

to determine whether they are equivalent up to an isomorphism of M, i.e. is there a linear isomorphism M →^{R} M so that the diagram

M ⟶^{S} M
R↓  ↓^{R}
M ⟶_{T} M

commutes, i.e.

RS = T R,

i.e.

RSR^{−1} = T?

10–7 Definition. S is conjugate to T if there exists a linear isomorphism R such that

RSR^{−1} = T.

Conjugacy is an equivalence relation on L(M); the equivalence classes are called conjugacy classes.

If T ∈ L(M) has matrix A ∈ K^{n×n} wrt some basis of M then the set of all matrices which can represent T is:

{P AP^{−1} : P ∈ K^{n×n} is invertible},

which is a conjugacy class in K^{n×n}.

Conversely, the set of all linear operators on M which can be represented by A is:

{RT R^{−1} : R is a linear isomorphism of M},

which is a conjugacy class in L(M).

Hence we have a bijective map from the set of conjugacy classes in L(M) to the set of conjugacy classes in K^{n×n} (see Figure 10.2).

−1
−1

Figure 10.2

10–8

The problem of determining which conjugacy class T belongs to is thus equivalent to determining which conjugacy class A belongs to.

A simple way of distinguishing conjugacy classes is to use properties such as: rank, trace, determinant, eigenvalues, characteristic polynomial, which are the same for all elements of a conjugacy class.

Examples:

1. Let

J = ( \begin{matrix}λ & 0 & 0 & 0 \\ 1 & λ & 0 & 0 \\ 0 & 1 & λ & 0 \\ 0 & 0 & 1 & λ \end{matrix} ) , a Jordan λ-block of size 4.

(4 × 4, λ on diagonal, 1 just below diagonal, zero elsewhere).

J − λI = ( \begin{matrix}0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{matrix} ) .

Therefore

(J − λI)e_{1} = e_{2},
(J − λI)e_{2} = e_{3},
(J − λI)e_{3} = e_{4},
(J − λI)e_{4} = 0.

Thus

K^{4}  e_{1} e_{2} e_{3} e_{4}
↓  ↓ ↓ ↓ ↓
im(J − λI) e_{2} e_{3} e_{4} 0
↓  ↓ ↓ ↓
im(J_{λ}I)^{2}  e_{3} e_{4} 0
↓  ↓ ↓
im(j − λI)^{3} e_{4} 0
↓  ↓
{0}  0

Thus

im(J − λI) has basis e_{2}, e_{3}, e_{4}, rank(J − λI) = 3,

10–9 im(J − λI)^{2} has basis e_{3}, e_{4}, rank(J − λI)^{2} = 2,

im(J − λI)^{3} has basis e_{4}, rank(J − λI)^{3} = 1,

(J − λI)^{4} = 0.

char J = | \begin{matrix}λ − X & 0 & 0 & 0 \\ 1 & λ − X & 0 & 0 \\ 0 & 1 & λ − X & 0 \\ 0 & 0 & 1 & λ − X \end{matrix} | = (λ − X)^{4}.

Therefore λ is the only eigenvalue of J, and the λ-eigenspace = ker(J − λI) has basis e_{4}.

2. Let

J = ( \begin{matrix}\begin{matrix}λ & 0 & 0 \\ 1 & λ & 0 \\ 0 & 1 & λ \end{matrix} & 0 & 0 \\ 0 & \begin{matrix}λ & 0 \\ 1 & λ \end{matrix} & 0 \\ 0 & 0 & \begin{matrix}λ & 0 \\ 1 & λ \end{matrix} \end{matrix} )

(Jordan λ-blocks on diagonal: 3 × 3, 2 × 2, 2 × 2).

K^{7}  \underbrace{e_{1}}_{s_{1}} \underbrace{e_{2}e_{4}e_{6}}_{s_{2}} \underbrace{e_{3}e_{5}e_{7}}_{s_{3}}
↓
im(J − λI)  \underbrace{e_{2}}_{s_{1}} \underbrace{e_{3}e_{5}e_{7}}_{s_{2}}
↓
im(J − λI)^{2}  \underbrace{e_{3}}_{s_{1}}
↓
{0}

where s_{1}, s_{2} and s_{3} are the dimensions of the kernel of (J−λI) restricted to im(J − λI)^{2}, im(J − λI) and K^{7} respectively.

10–10 char J = (λ − X)^{7}. λ is the only eigenvector; dim λ-eigenspace = 3 = number of Jordan blocks.

(J − λ)e_{1} = e_{2}
(J − λ)e_{2} = e_{3}
(J − λ)e_{3} = 0 eigenvector
(J − λ)e_{4} = e_{5}
(J − λ)e_{5} = 0 eigenvector
(J − λ)e_{6} = e_{7}
(J − λ)e_{7} = 0 eigenvector.

10.4 Jordan Forms

Definition. A square matrix J ∈ K^{n×n} is called a Jordan matrix if it is of the form

J = ( \begin{matrix}J_{1} \\ & J_{2} \\ & & . & . & . \\ & & & & & J_{l} \end{matrix} ) ,

where each J_{i} is a Jordan block.

Example:

J = ( \begin{matrix}λ & 0 & 0 \\ 1 & λ & 0 \\ 0 & 1 & λ \end{matrix} ; \begin{matrix}λ & 0 \\ 1 & λ \end{matrix} ; \begin{matrix}μ & 0 \\ 1 & μ \end{matrix} ; \begin{matrix}μ & 0 \\ 1 & μ \end{matrix} ) ,

(where λ ≠ μ (say)) is a 9 × 9 Jordan matrix.

Note that

10–11 (i) char J = (λ − X)^{5}(μ − X)^{4}; eigenvalue λ, with algebraic multiplicity 5;

eigenvalue μ, with algebraic multiplicity 4.

(ii)

dimension of λ-eigenspace = number of λ-blocks

= geometric multiplicity of eigenvalue λ = 2;

geometric multiplicity of eigenvalue μ = 2.

10–12

J − λI = ( 0 0 0 1 0 0 0 1 0 0 0 1 0 \begin{matrix}μ − λ & 0 \\ 1 & μ − λ \end{matrix}  ← non-sing. non-sing. →  \begin{matrix}μ − λ & 0 \\ 1 & μ − λ \end{matrix} ) ,

(J − λI)^{2} = ( 0 0 0 0 0 0 1 0 0 0 0 0 0 \begin{matrix}(μ − λ)^{2} & 0 \\ 1 & (μ − λ)^{2} \end{matrix}  ← non-sing. non-sing. →  \begin{matrix}(μ − λ)^{2} & 0 \\ 1 & (μ − λ)^{2} \end{matrix} ) ,

(J − λ)^{3} = ( 0 0 0 0 0 0 0 0 0 0 0 0 0 N.S. N.S. ) .

Therefore

rank(J − λI) = 2 + 1 + 4,
rank(J − λI)^{2} = 1 + 0 + 4,
rank(J − λI)^{3} = 0 + 0 + 4.

10–13 More generally, if J is a Jordan n × n matrix, with

b_{1} λ-blocks of size 1,
b_{2} λ-blocks of size 2,
⋮
b_{k} λ-blocks of size k,

and if λ has algebraic multiplicity m then

b_{1} + 2b_{2} + 3b_{3} + ⋯ = m,

rank(J − λI) has rank:

0b_{1} + 1b_{2} + 2b_{3} + 3b_{4} + ⋯ + (n − m),

rank(J − λI)^{2} has rank:

0b_{1} + 0b_{2} + 1b_{3} + 2b_{4} + ⋯ + (n − m),

rank(J − λI)^{3} has rank:

0b_{1} + 0b_{2} + 0b_{3} + 1b_{4} + 2b_{5} + ⋯ + (n − m),

and so on. Hence the number b_{k} of λ-blocks of size k in J is uniquely determined by the conjugacy class of J.

Theorem 10.4. Let T ∈ L(M) be a linear operator on a finite dimensional vector space over a field K which is algebraically closed. Then T can be represented by a Jordan matrix J. The matrix J, which by the preceding is uniquely determined, apart from the arrangement of the blocks on the diagonal, is called the Jordan form of T.

Proof ▶ Since K is algebraically closed, the characteristic polynomial is a product of linear factors; so, by Hamilton-Cayley we have

(T − λ_{1}1)^{r_{1}} … (T − λ_{k}1)^{r_{k}} = 0

(say), with λ_{1}, …, λ_{k} distinct factors.

By primary decomposition:

M = M_{1} ⊕ ⋯ ⊕ M_{k},

10–14 where M_{i} = ker(T − λ_{i}1)^{r_{i}}. We will show that M_{i} has a basis wrt which T_{M_{i}} has a Jordan matrix with λ_{i} on the diagonal.

Put S = T_{M_{i}} − λ_{i}1_{M_{i}}. Then

M_{i} →^{S} M_{i}

and S^{r_{i}} = 0, i.e. S is a nilpotent operator. Suppose S^{r} = 0 but S^{r−1} ≠ 0, and consider:

M_{i} a_{1} … a_{s_{1}}b_{1} … b_{s_{2}} … y_{1} … y_{s_{r+1}}z_{1} … z_{s_{r}} ↓ im S ↓ im S^{2} ↓ ⋮ ↓ im S^{r−3} x_{1} … x_{s_{1}}y_{1} … y_{s_{2}}z_{1} … z_{s_{1}} … z_{s_{2}} … z_{s_{3}} ↓ im S^{r−2} y_{1} … y_{s_{1}}z_{1} … z_{s_{1}} … z_{s_{2}} ↓ im S^{r−1} z_{1} … z_{s_{1}} ↓ {0} (∗)

Choose a basis z_{1}, …, z_{s_{1}} for im S^{r−1}. Choose y_{1}, …, y_{s_{1}} ∈ im S^{r−2} such that Sy_{j} = z_{j}. Extend to a basis z_{1}, …, z_{s_{1}}, …, z_{s_{2}} for the kernel of im S^{r−2} → im S^{r−1}. Thus y_{1}, …, y_{s_{1}}, z_{1}, …, z_{s_{2}} is a basis for im S^{r−2}.

Now repeat the construction: choose x_{1}, …, x_{s_{1}}, y_{1}, …, y_{s_{2}} ∈ im S^{r−3} such that Sx_{j} = y_{j}, Sy_{i} = z_{i}. Extend to a basis z_{1}, …, z_{s_{2}}, …, z_{s_{3}} for the kernel of im S^{r−3} → im S^{r−2}. Thus x_{1}, …, x_{s_{1}}, y_{1}, …, y_{s_{2}}, z_{1}, …, z_{s_{3}} is a basis for im S^{r−3}.

Continue in this way until we get a basis

a_{1}, …, a_{s_{1}}, b_{1}, …, b_{s_{2}}, …, y_{1}, …, y_{s_{r−1}}, z_{1}, …, z_{s_{r}}

(say), for M_{i}, with

Sa_{j} = b_{j}, Sb_{j} = c_{j}, …, Sy_{j} = z_{j}, Sz_{j} = 0.

Now write the basis elements in order, by going down each column of (∗) in turn, starting at the left most column (and leaving out any column whose elements have already been written down).

10–15

Relative to this basis for M_{i} the matrix of S is a Jordan matrix with zeros on the diagonal. Therefore the matrix of T_{M_{i}} = S + λ_{i}1_{M_{i}} is a Jordan matrix with λ_{i} on the diagonal.

Putting together these bases for M_{1}, …, M_{k} we get a basis for M wrt which T has a Jordan matrix, as required. ◀

Example: To find a Jordan matrix J conjugate to the matrix

A = ( \begin{matrix}5 & 4 & 3 \\ −1 & 0 & −3 \\ 1 & −2 & 1 \end{matrix} ) .

The characteristic polynomial is

p = | \begin{matrix}5 − X & 4 & 3 \\ −1 & −X & −3 \\ 1 & −2 & 1 − X \end{matrix} |
= (5 − X)[−X(1 − X) − 6] − 4[−(1 − X) + 3] + 3[2 + X]
= (5 − X)[X^{2} − X − 6] − 4[X + 2] + 3[2 + X]
= 5X^{2} − 5X − 30 − X^{3} + X^{2} + 6X − 4X − 8 + 3X + 6
= −X^{3} + 6X^{2} − 32
= (X + 2)(−X^{2} + 8X − 16)
= −(X + 2)(X − 4)^{2}.

Therefore, by Hamilton-Cayley the operator ℝ^{3} →^{A} ℝ^{3} satisfies:

(A + 2I)(A − 4I)^{2} = 0.

Therefore, by primary decomposition:

ℝ^{3} = ker(A + 2I) ⊕ ker(A − 4I)^{2}.

Now

ker(A + 2I) = ker ( \begin{matrix}7 & 4 & 3 \\ −1 & 2 & −3 \\ 1 & −2 & 3 \end{matrix} )

= ker ( \begin{matrix}7 & 4 & 3 \\ 0 & 18 & −18 \\ 0 & 0 & 0 \end{matrix} ) row 3 + row 2 7 row 2 + row 1

10–16

(α, β, γ) ∈ ker(A + 2I) ⇔ 7α + 4β + 3γ = 0
β − γ = 0
⇔ 7α + 7γ = 0
β − γ = 0
⇔ (α, β, γ) = (α, −α, −α) = α(1, −1, −1).

Therefore ker(A + 2I) has basis u_{1} = (1, −1, −1).

ker(A − 4I)^{2} = ker ( \begin{matrix}1 & 4 & 3 \\ −1 & −4 & −3 \\ 1 & −2 & −3 \end{matrix} )^{2}

= ker ( \begin{matrix}0 & −18 & −18 \\ 0 & 18 & 18 \\ 0 & 18 & 18 \end{matrix} )

= ker ( \begin{matrix}0 & 1 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} )

Therefore

(α, β, γ) ∈ ker(A − 4I)^{2} ⇔ β + γ = 0
⇔ (α, β, γ) = (α, β, −β) = α(1, 0, 0) + β(0, 1, −1).

Therefore (1, 0, 0), (0, 1, −1) is a basis for ker(A − 4I)^{2}. Put

u_{2} = (1, 0, 0), u_{3} = (A − 4I)u_{2} = (1, −1, 1).

So u_{1}, u_{2}, u_{3} is a basis for ℝ^{3} such that

(A + 2I)u_{1} = 0,
(A − 4I)u_{2} = u_{3},
(A − 4I)u_{3} = 0,

i.e.

Au_{1} = −2u_{1},
Au_{2} = 4u_{2} + u_{3},
Au_{3} = 4u_{3}.

10–17 Therefore wrt basis

u_{1} = (1, −1, −1) = e_{1} − e_{2} − e_{3},
u_{2} = (1, 0, 0) = e_{1},
u_{3} = (1, −1, 1) = e_{1} − e_{2} + e_{3}

the operator A has matrix

J = ( \begin{matrix}−2 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 1 & 4 \end{matrix} )

Let P be the transition matrix from (e_{1}, e_{2}, e_{3}) to (u_{1}, u_{2}, u_{3}).

P = ( \begin{matrix}1 & 1 & 1 \\ −1 & 0 & −1 \\ −1 & 0 & 1 \end{matrix} )^{−1} = \frac{1}{2} ( \begin{matrix}0 & −1 & −1 \\ 2 & 2 & 0 \\ 0 & −1 & 1 \end{matrix} ) = ( \begin{matrix}0 & −\frac{1}{2} & −\frac{1}{2} \\ 1 & 1 & 0 \\ 0 & −\frac{1}{2} & −\frac{1}{2} \end{matrix} ) .

Therefore

P AP^{−1} = ( \begin{matrix}0 & −\frac{1}{2} & −\frac{1}{2} \\ 1 & 1 & 0 \\ 0 & −\frac{1}{2} & \frac{1}{2} \end{matrix} ) ( \begin{matrix}5 & 4 & 3 \\ −1 & 0 & −3 \\ 1 & −2 & 1 \end{matrix} ) ( \begin{matrix}1 & 1 & 1 \\ −1 & 0 & −1 \\ −1 & 0 & 1 \end{matrix} )

= ( \begin{matrix}0 & 1 & 1 \\ 4 & 4 & 0 \\ 1 & −1 & 2 \end{matrix} ) ( \begin{matrix}1 & 1 & 1 \\ −1 & 0 & −1 \\ −1 & 0 & 1 \end{matrix} )

= ( \begin{matrix}−2 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 1 & 4 \end{matrix} ) = J,

as required.

Note. (i)
\begin{matrix}u_{1} = e_{1} − e_{2} − e_{3} \\ u_{2} = e_{1} \\ u_{3} = e_{1} − e_{2} + e_{3} \end{matrix} \} ⇒ \begin{matrix}e_{1} = u_{2} \\ u_{1} + u_{3} = 2e_{1} − 2e_{2} \\ u_{1} − u_{3} = −2e_{3} \end{matrix} \}

⇒ \begin{matrix}e_{1} = u_{2} \\ e_{2} = −\frac{1}{2}u_{1} + u_{2} − \frac{1}{2}u_{3} \\ e_{3} = −\frac{1}{2}u_{1} + \frac{1}{2}u_{3} \end{matrix} \}

⇒ P = ( \begin{matrix}0 & −\frac{1}{2} & −\frac{1}{2} \\ 1 & 1 & 0 \\ 0 & −\frac{1}{2} & \frac{1}{2} \end{matrix} ) .

10–18

( \begin{matrix}1 & 1 & 1 & 1 & 0 & 0 \\ −1 & 0 & −1 & 0 & 1 & 0 \\ −1 & 0 & 1 & 0 & 0 & 1 \end{matrix} ) ∼ ( \begin{matrix}1 & 1 & 1 & 1 & 0 & 0 \\ 0 & 1 & 0 & 1 & 1 & 0 \\ 0 & 1 & 2 & 1 & 0 & 1 \end{matrix} ) ∼ ( \begin{matrix}1 & 1 & 1 & 1 & 0 & 0 \\ 0 & 1 & 0 & 1 & 1 & 0 \\ 0 & 0 & 2 & 0 & −1 & 1 \end{matrix} )

∼ ( \begin{matrix}2 & 2 & 0 & 2 & 1 & −1 \\ 0 & 1 & 0 & 1 & 1 & 0 \\ 0 & 0 & 2 & 0 & −1 & 1 \end{matrix} ) ∼ ( \begin{matrix}2 & 0 & 0 & 0 & −1 & −1 \\ 0 & 1 & 0 & 1 & 1 & 0 \\ 0 & 0 & 2 & 0 & −1 & 1 \end{matrix} )

Example: To find the Jordan form of

A = ( \begin{matrix}1 & 1 & 3 \\ 5 & 2 & 6 \\ −2 & −1 & −3 \end{matrix} ) .

char A = | \begin{matrix}1 − X & 1 & 3 \\ 5 & 2 − X & 6 \\ −2 & −1 & −3 − X \end{matrix} |

= (1 − X)[(2 − X)(−3 − X) + 6] − [5(−3 − X) + 12] + 3[−5 + 2(2 − X)] = (1 − X)[X^{2} + X] − [−5X − 3] + 3[−1 − 2X] = X^{2} + X − X^{3} − X^{2} + 5X + 3 − 6X − 3

= −X^{3}.

Therefore operator ℝ^{3} →^{A} ℝ^{3} satisfies

A^{3} = 0.

Now

A^{2} = ( \begin{matrix}1 & 1 & 3 \\ 5 & 2 & 6 \\ −2 & −1 & −3 \end{matrix} ) ( \begin{matrix}1 & 1 & 3 \\ 5 & 2 & 6 \\ −2 & −1 & −3 \end{matrix} ) = ( \begin{matrix}0 & 0 & 0 \\ 3 & 3 & 9 \\ −1 & −1 & −3 \end{matrix} ) .

So put

u_{1} = e_{1} = (1, 0, 0) = e_{1},
u_{2} = Ae_{1} = (1, 5, −2) = e_{1} + 5e_{2} − 2e_{3},
u_{3} = A^{2}e_{1} = (0, 3, −1) = 3e_{2} − e_{3}.

So wrt new basis u_{1}, u_{2}, u_{3} operator A has matrix

J = ( \begin{matrix}0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{matrix} ) .

10–19 Also

u_{2} − 2u_{3} = e_{1} − e_{2}.

So

e_{2} = u_{1} − u_{2} + 2u_{3},
e_{3} = 3e_{2} − u_{3} = 3u_{1} − 3u_{2} + 5u_{3}.

Thus

e_{1} = u_{1},
e_{2} = u_{1} − u_{2} + 2u_{3},
e_{3} = 3u_{1} − 3u_{2} + 5u_{3}.

Therefore P AP^{−1} = J, where

P = ( \begin{matrix}1 & 1 & 3 \\ 0 & −1 & −3 \\ 0 & 2 & 5 \end{matrix} ) .

Check:

P A = JP ⇔ ( \begin{matrix}1 & 1 & 3 \\ 0 & −1 & −3 \\ 0 & 2 & 5 \end{matrix} ) ( \begin{matrix}1 & 1 & 3 \\ 5 & 2 & 6 \\ −2 & −1 & −3 \end{matrix} ) = ( \begin{matrix}0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{matrix} ) ( \begin{matrix}1 & 1 & 3 \\ 0 & −1 & −3 \\ 0 & 2 & 5 \end{matrix} )

⇔ ( \begin{matrix}0 & 0 & 0 \\ 1 & 1 & 3 \\ 0 & −1 & −3 \end{matrix} ) = ( \begin{matrix}0 & 0 & 0 \\ 1 & 1 & 3 \\ 0 & −1 & −3 \end{matrix} ) .

10.5 Determinants

Let M be a vector space of finite dimension n, and M →^{T} N be a linear operator. The pull-back (or transpose) of T is the operator

M^{∗} ←^{T^{∗}} N^{∗}

defined by

⟨T^{∗}f, x⟩ = ⟨f, T x⟩.

10–20 The transpose of T^{∗} is T itself (by duality) written T_{∗}. So:

M →^{T_{∗}} N (T_{∗} = T),
M^{∗} ←^{T^{∗}} N^{∗},
⟨T^{∗}f, x⟩ = ⟨f, T_{∗}x⟩.

If M = N and T has matrix A = (α_{j}^{i}) wrt basis u_{i} then

⟨T^{∗}u^{j}, u_{i}⟩ = ⟨u^{j}, T_{∗}u_{i}⟩ = ⟨u^{j}, α_{i}^{k}u_{k}⟩ = α_{i}^{j}.

Therefore T^{∗} has matrix A^{t} wrt basis u^{i}.

More generally we have the pull-back

M^{(r)} ←^{T^{∗}} N^{(r)},

and push-forward

M_{(r)} →^{T_{∗}} N_{(r)}

defined by

(T^{∗}S)(x_{1}, …, x_{r}) = S(T_{∗}x_{1}, …, T_{∗}x_{r}),
(T_{∗}S)(f^{1}, …, f^{r}) = S(T^{∗}f^{1}, …, T^{∗}f^{r}).

These maps T^{∗}, T_{∗} are linear, and preserve the wedge-product. In particular the spaces M^{(n)} and M_{(n)} are 1-dimensional. Therefore for M →^{T} M the push-forward:

M_{(n)} →^{T_{∗}} M_{(n)},

and the pull-back

M^{(n)} ←^{T^{∗}} M^{(n)}

must each be multiplication by a scalar (called det T, det T^{∗} respectively).

To see what these scalars are let T have matrix A = (α_{j}^{i}) wrt basis u_{i}. Then

T_{∗}(u_{1} ∧ ⋯ ∧ u_{n}) = T u_{1} ∧ ⋯ ∧ T u_{n}

= (α_{1}^{i_{1}}u_{i_{1}}) ∧ ⋯ ∧ (α_{n}^{i_{n}}u_{i_{n}}) = det A u_{1} ∧ ⋯ ∧ u_{n}.

10–21 Therefore M_{(n)} →^{T_{∗}} M_{(n)} is multiplication by det A. Similarly M^{(n)} ←^{T^{∗}} M^{(n)} is multiplication by det A^{t} = det A. Therefore

det T = det T^{∗} = det A,

independent of choice of basis.

Example: Let dim M = 4, and M →^{T} M have matrix A = (α_{j}^{i}) wrt basis u_{i}. Then

dim M_{(2)} = \frac{4!}{2!2!} = 6,

and wrt basis u_{1} ∧ u_{2}, u_{1} ∧ u_{3}, u_{1} ∧ u_{4}, u_{2} ∧ u_{3}, u_{2} ∧ u_{4}, u_{3} ∧ u_{4}

T_{∗} : M_{(2)} → M_{(2)}

satisfies

T_{∗}(u_{1} ∧ u_{2}) = T u_{1} ∧ T u_{2}
= (α_{1}^{1}u_{1} + α_{1}^{2}u_{2} + α_{1}^{3}u_{3} + α_{1}^{4}u_{4}) ∧ (α_{2}^{1}u_{1} + α_{2}^{2}u_{2} + α_{2}^{3}u_{3} + α_{2}^{4}u_{4})
= (α_{1}^{1}α_{2}^{2} − α_{1}^{2}α_{2}^{1})u_{1} ∧ u_{2} + ….

Therefore matrix of T_{∗} is a 6 × 6 matrix whose entries are 2 × 2 subdeterminants of A.

Theorem 10.5. If M →^{T} M has rank r then

(i) M_{(r)} →^{T_{∗}} M_{(r)} is non-zero,

(ii) M_{(r+1)} →^{T_{∗}} M_{(r+1)} is zero.

Proof ▶

(i) Let y_{1}, …, y_{r} be a basis for im T, and let y_{i} = T x_{i}. Then

T_{∗} x_{1} ∧ ⋯ ∧ x_{r} = T x_{1} ∧ ⋯ ∧ T x_{r} = y_{1} ∧ ⋯ ∧ y_{r} ≠ 0.

(ii) Let u_{i} be a basis for M. Then

T_{∗} u_{i_{1}} ∧ ⋯ ∧ u_{i_{r+1}} = T u_{i_{1}} ∧ ⋯ ∧ T u_{i_{r+1}} = 0,

since T u_{i_{1}}, …, T u_{i_{r+1}} ∈ im T, which has dimension r. Therefore linearly independent. ◀

Corollary 10.1. If T has matrix A then rank T = r ⇔ all (r + 1) × (r + 1) subdeterminants are zero, and there exists at least one non-zero r × r subdeterminant.

10–22

Orientation

11.1 Orientation of Vector Spaces

Let M be a finite dimensional real vector space. Let P = (p_{j}^{i}) be the transition matrix from basis u_{1}, …, u_{n} to basis w_{1}, …, w_{n}:

u_{j} = p_{j}^{i}w_{i}.

Then

u_{1} ∧ ⋯ ∧ u_{n} = det P w_{1} ∧ ⋯ ∧ w_{n}.

Definition. u_{1}, …, u_{n} has same orientation as w_{1}, …, w_{n} if u_{1} ∧ ⋯ ∧ u_{n} is a positive multiple of w_{1} ∧ ⋯ ∧ w_{n}, i.e. det P > 0. Otherwise opposite orientation as, i.e. det P < 0.

‘Same orientation as’ is an equivalence relation on the set of all bases for M. There are just two equivalence classes. We call M an oriented vector space if one of these classes has been designated as positively oriented bases and the other as negatively oriented bases. We call this choosing an orientation for M.

For ℝ^{n} we may designate the equivalence class of the usual basis e_{1}, …, e_{n} as being positively oriented bases. This is called the usual orientation of ℝ^{n}.

Example: In ℝ^{3}, with usual orientation. e_{1}, e_{2}, e_{3} (see Figure 11.1) is positively oriented (by definition).

e_{2} ∧ e_{1} ∧ e_{3} = −e_{1} ∧ e_{2} ∧ e_{3},
e_{2} ∧ e_{3} ∧ e_{1} = e_{1} ∧ e_{2} ∧ e_{3}.

11–1

e_{3}

e_{2}

e_{1}

Figure 11.1

Therefore e_{2}, e_{1}, e_{3} is negatively oriented and e_{2}, e_{3}, e_{1} is positively oriented.

Definition. Let M be a real vector space of finite dimension n with a nonsingular symmetric scalar product (·|·). We call u_{1}, …, u_{n} a standard basis if

(u_{i}|u_{j}) = ( \begin{matrix}±1 \\ & . & . & . \\ & & & & ±1 \end{matrix} ) .

Recall that such bases for M exist, and the numbers of − signs is uniquely determined.

Theorem 11.1. Let M be oriented. Then the n-form

vol = u^{1} ∧ u^{2} ∧ ⋯ ∧ u^{n}

is independent of the choice of positively oriented standard basis for M. It is called the volume form on M.

If v_{1}, …, v_{n} is any positively oriented basis for M then

vol = \sqrt{(−1)^{s} det(v_{i}|v_{j})} v^{1} ∧ ⋯ ∧ v^{n}.

Proof ▶

11–2 1. Let w_{1}, …, w_{n} be another positively oriented standard basis for M:

w^{i} = p_{j}^{i}u^{j} (say). Therefore

w^{1} ∧ ⋯ ∧ w^{n} = det P u^{1} ∧ ⋯ ∧ u^{n}.

But det P > 0 and

P^{t} ( \begin{matrix}±1 \\ & . & . & . \\ & & & & ±1 \end{matrix} ) P = ( \begin{matrix}±1 \\ & . & . & . \\ & & & & ±1 \end{matrix} ) .

Therefore

(−1)^{s}(det P)^{2} = (−1)^{s}.

Therefore

(det P)^{2} = 1.

Therefore det P = 1. Therefore

w^{1} ∧ ⋯ ∧ w^{n} = u^{1} ∧ ⋯ ∧ u^{n} = vol,

as required.

2. Let u^{i} = p_{j}^{i}v^{j} (say), det P > 0. Then

P^{t} ( \begin{matrix}±1 \\ & . & . & . \\ & & & & ±1 \end{matrix} ) P = G,

where g_{ij} = (v_{i}|v_{j}). Therefore

(−1)^{s}(det P)^{2} = det G.

Therefore

det P = \sqrt{(−1)^{s} det G}.

Therefore vol = u^{1} ∧ ⋯ ∧ u^{n} = det P v^{1} ∧ ⋯ ∧ v^{n} = \sqrt{(−1)^{s} det G} v^{1} ∧ ⋯ ∧ v^{n}, as required. ◀

11–3 Corollary 11.1. vol has components \sqrt{| det g_{ij}|} ϵ_{i_{1}...i_{n}} wrt any positively oriented basis.

Example: Take ℝ^{n} with usual orientation and dot product. Let

D = {t^{1}v_{1} + ⋯ + t^{n}v_{n} : 0 ≤ t^{i} ≤ 1}

be the parallelogram spanned by vectors v_{1}, …, v_{n}. Let A be the matrix having v_{1}, …, v_{n} as columns.

ℝ^{n} →^{A} ℝ^{n}, v_{i} = Ae_{i}.

e_{3}

v_{3} v_{2} e_{2}

v_{1} e_{1}

Figure 11.2

(for example, see Figure 11.2).

vol(v_{1}, …, v_{n}) = vol(Ae_{1}, …, Ae_{n})

= det A vol(e_{1}, …, e_{n})

= det A = ±| det A| = ±Lebesgue measure of D,

and Lebesgue measure of D = \sqrt{| det(v_{i}|v_{j})|}.

We continue to consider a real oriented vector space M of finite dimension n with a non-singular symmetric scalar product (·|·) with s signs.

M^{(r)} denotes the vector space of skew-symmetric tensors of type

M × ⋯ × M_{←r→} → K.

11–4 Theorem 11.2. There exists a unique linear operator

M^{(r)} →^{∗} M^{(n−r)},

called the Hodge star operator, with the property that for each positively oriented standard basis u_{1}, …, u_{n} we have

∗(u^{1} ∧ ⋯ ∧ u^{r}) = s_{r+1} … s_{n}u^{r+1} ∧ ⋯ ∧ u^{n} (no summation here),

where

g_{ij} = ( \begin{matrix}s_{1} & & & 0 \\ & s_{2} \\ & & . & . & . \\ & 0 & & & & s_{n} \end{matrix} ) = g^{ij}, s_{i} = ±1.

Example: If M is 3-dimensional oriented Euclidean, and u_{1}, u_{2}, u_{3} is any positively oriented orthonormal then

M^{(1)} →^{∗} M^{(2)}, M^{(2)} → M^{(1)},

with

∗(α_{1}u^{1} + α_{2}u^{2} + α_{3}u^{3}) = α_{1} u^{2} ∧ u^{3} + α_{2} u^{3} ∧ u^{1} + α_{3} u^{1} ∧ u^{2},
∗(α_{1}u^{2} ∧ u^{3} + α_{2}u^{3} ∧ u^{1} + α_{3}u^{1} ∧ u^{2}) = α_{1}u^{1} + α_{2}u^{2} + α_{3}u^{3}.

Thus, if v has components α_{i} wrt u^{i}, and w has components β_{i} wrt u^{i} then ∗(v ∧w) has components ϵ^{ijk}α_{j}β_{k} wrt u^{i} for any positively oriented orthonormal basis u_{i}.

We write v × w = ∗(v ∧ w), and call it the vector product of v and w because:

v × w = ∗(v ∧ w)
= ∗[(α_{1}u^{1} + α_{2}u^{2} + α_{3}u^{3}) ∧ (β_{1}u^{1} + β_{2}u^{2} + β_{3}u^{3})]
= ∗[(α_{2}β_{3} − α_{3}β_{2})u^{2} ∧ u^{3} + …]
= (α_{2}β_{3} − α_{3}β_{2})u^{1} + …,

as required. Proof ▶(of theorem) If ∗ exists then it must be unique, from the definition. Thus it is sufficient to define one such operator ∗. For any positively oriented basis we define ∗ω by contraction as:

(∗ω)_{i_{r+1}...i_{n}} = \frac{(−1)^{s}}{r!} g^{i_{1}j_{1}} … g^{i_{r}j_{r}} \underbrace{ω_{j_{1}...j_{r}}}_{ω} \underbrace{\sqrt{| det g_{ij}|} ϵ_{i_{1}...i_{r}i_{r+1}...i_{n}}}_{vol} .

11–5 ∗ω is then well-defined independent of choice of basis, since contraction is independent of a choice of basis. Thus wrt a positively oriented standard basis u_{1}, …, u_{n},

g^{ij} = \{ \begin{matrix}s_{i} & i = j \\ 0 & i ≠ j \end{matrix} \} and ω = u^{1} ∧ ⋯ ∧ u^{r}.

ω_{i...r} = 1, other components of ω by skew-symmetry, otherwise zero. Therefore

(∗ω)_{r+1,...,n} = \frac{(−1)^{s}}{r!} s_{1}s_{2} … s_{r}ω_{1...r}r!1.1 = s_{r+1} … s_{n},

as required as other components of ∗ω by skew-symmetry are zero.. ◀

Theorem 11.3. There is a unique scalar product (·|·) on each M^{(r)} such that

ω ∧ η = (∗ω|η) vol

for each ω ∈ M^{(r)}, η ∈ M^{(n−r)}. The scalar product is non-singular and symmetric for a standard basis

(u^{1} ∧ ⋯ ∧ u^{r}|u^{1} ∧ ⋯ ∧ u^{r}) = (u^{1}|u^{1}) … (u^{r}|u^{r}) = ±1,

and u^{1}∧… u^{r} is orthogonal to the other basis element of {u^{i_{1}}∧⋯∧u^{i_{r}}}_{i_{1}<···<i_{r}}.

Proof ▶ Define (·|·) by ω ∧ η = (∗ω|η) vol. Then (·|·) is a bilinear form on M^{(n−r)}.

If u_{1}, …, u_{n} is a positively oriented standard basis for M then

vol = \underbrace{u^{1} ∧ ⋯ ∧ u^{r}}_{ω} ∧ \underbrace{u^{r+1} ∧ ⋯ ∧ u^{n}}_{η}

= (\underbrace{s_{r+1} … s_{n}u^{r+1} ∧ ⋯ ∧ u^{n}}_{∗ω} | \underbrace{u^{r+1} ∧ ⋯ ∧ u^{n}}_{η}) vol .

Therefore

(u^{r+1} ∧ ⋯ ∧ u^{n}|u^{r+1} ∧ ⋯ ∧ u^{n}) = s_{r+1} … s_{n} = (u^{r+1}|u^{r+1}) … (u^{n}|u^{n}),

as required. Similarly other scalar products give zero. ◀

Similarly other scalar products give zero.

Example: If u_{1}, u_{2}, u_{3} is an orthonormal basis for M then u^{2}∧u^{3}, u^{3}∧u^{1}, u^{1}∧u^{3} is an orthonormal basis for M^{(2)}, since

(u^{2} ∧ u^{3}|u^{2} ∧ u^{3}) = (u^{2}|u^{2})(u^{3}|u^{3}) = 1.1 = 1,
(u^{2} ∧ u^{3}|u^{3} ∧ u^{1}) = (u^{2}|u^{3})(u^{3}|u^{1}) = 0.0 = 0.

11–6

11.2 Orientation of Coordinate Systems

Definition. Let X be an n-dimensional manifold. Two coordinate systems on X: y^{1}, …, y^{n} with domain V, and z^{1}, …, z^{n} with domain W have the same orientation if

\frac{∂(y^{1}, …, y^{n})}{∂(z^{1}, …, z^{n})} > 0

on V ∩ W. We call X oriented if a family of coordinate systems is given on X whose domains cover X, and such that any two have the same orientation. We then call these coordinate systems positively oriented.

Note. On V ∩ W,

dy^{i} = \frac{∂y^{i}}{∂z^{j}} dz^{j}.

Therefore

dy^{1} ∧ ⋯ ∧ dy^{n} = det (\frac{∂y^{i}}{∂z^{j}} ) dz^{1} ∧ ⋯ ∧ dz^{n}

= \frac{∂(y^{1}, …, y^{n})}{∂(z^{1}, …, z^{n})}dz^{1} ∧ ⋯ ∧ dz^{n}.

Therefore for each a ∈ V ∩W, \frac{∂}{∂y_{a}^{1}} , …, \frac{∂}{∂y_{a}^{n}} has same orientation as \frac{∂}{∂z_{a}^{1}} , …, \frac{∂}{∂z_{a}^{n}}. Thus each tangent space T_{a}X is an n-dimensional oriented vector space.

If X has a metric tensor (·|·) then T_{a}X has a non-singular symmetric scalar product (·|·)_{a} for each a ∈ X. Therefore we can define a differential n-form vol on X, called the volume form on X by:

(vol)_{a} = the volume form on T_{a}X.

Also, if ω is a differential r-form on X then we can define a differential (n − r)-form on X, ∗ω, called the Hodge star of ω by

(∗ω)_{a} = ∗(ω_{a}) for each a ∈ X.

If u_{1}, …, u_{n} are positively oriented vector fields on X with domain V, and

ds^{2} = ±(u^{1})^{2} ± ⋯ ± (u^{n})^{2}

then

vol = u^{1} ∧ ⋯ ∧ u^{n}

11–7 on V.

If y^{1}, …, y^{n} is a positively oriented coordinate system on X with domain V then

vol = \sqrt{| det g_{ij}|} dy^{1} ∧ ⋯ ∧ dy^{n}

on V, where

g_{ij} = (\frac{∂}{∂y^{i}} |\frac{∂}{∂y^{j}} ) .

Examples:

1. ℝ^{2}, with usual coordinates: x, y, and polar coordinates: r, θ (see Figure 11.3). Take x, y as positively oriented:

x = r cos θ, y = r sin θ.

Therefore

dx ∧ dy = (cos θ dr − r sin θ dθ) ∧ (sin θ dr + r cos θ dθ) = r dr ∧ dθ.

\frac{∂}{∂y} \frac{∂}{∂r}

\frac{∂}{∂θ}

\frac{∂}{∂x}

Figure 11.3

r > 0. Therefore r, θ is positively oriented.

area element = dx ∧ dy = r dr ∧ dθ,
ds^{2} = (dx)^{2} + (dy)^{2} = (dr)^{2} + (r dθ)^{2}.

Therefore

∗dx = dy, ∗ dy = −dx,
∗dr = r dθ, ∗ (r dθ) = −dr.

11–8 2. Unit sphere S^{2} in ℝ^{3} : x^{2} + y^{2} + z^{2} = 1. On S^{2} we have:

x dx + y dy + z dz = 0.

Therefore (wedge with dx):

y dx ∧ dy + z dx ∧ dz = 0.

Therefore

dx ∧ dy = \frac{z}{y} dz ∧ dx.

Therefore the coordinate system x, y on z > 0 has the same orientation as the coordinate system z, x on y > 0.

We orient S^{2} so that these coordinates are positively oriented. Now

ds^{2} = (dx)^{2} + (dy)^{2} + (dz)^{2}

= (dx)^{2} + (dy)^{2} + (−\frac{x}{z}dx − \frac{y}{z} dy)^{2} (on z > 0)

= (1 + \frac{x^{2}}{z^{2}} ) (dx)^{2} + 2\frac{xy}{z^{2}} dx dy + (1 + \frac{y^{2}}{z^{2}} ) (dy)^{2}.

Therefore wrt coordinates x, y,

g_{ij} = ( \begin{matrix}1 + \frac{x^{2}}{z^{2}} & \frac{xy}{z^{2}} \\ \frac{xy}{z^{2}} & 1 + \frac{y^{2}}{z^{2}} \end{matrix} ) .

Therefore

det g_{ij} = 1 + \frac{x^{2}}{z^{2}} + \frac{y^{2}}{z^{2}} = \frac{x^{2} + y^{2} + z^{2}}{z^{2}} = \frac{1}{z^{2}} .

Therefore

area element = \frac{1}{|z|}dx ∧ dy.

11–9

Manifolds and (n)-dimensional Vector Analysis

This chapter could be considered a continuation of Chapter 11.

12.1 Gradient

Definition. If f is a scalar field on a manifold X, with non-singular metric tensor (·|·), then we define the gradient of f to be the vector field grad f such that

(grad f|v) =< df|v >= vf = rate of change of f along v

for each vector field v.

Thus grad f is raising the index of df.

df = \frac{∂f}{∂y^{i}} dy^{i}

has components

\frac{∂f}{∂y^{i}} ,

and

grad f = g^{ij} \frac{∂f}{∂y^{j}} \frac{∂}{∂y^{i}}

has components

g^{ij} \frac{∂f}{∂y^{j}} .

12–1 Theorem 12.1. If the metric tensor is positive definite then

(i) grad f is in the direction of fastest increase of f; (ii) the rate of change of f in the direction of fastest increase is ∥ grad f∥; (iii) grad f is orthogonal to the level surfaces of f (see Figure 12.1).

Figure 12.1

Proof ▶

(i) The rate of change of f along any unit vector field v has absolute value

|vf| = |(grad f|v)| ≤ ∥ grad f∥∥v∥ = ∥ grad f∥,

by Cauchy-Schwarz. (ii) For

v = \frac{grad f}{∥ grad f∥}

the maximum is attained:

|vf| = |(grad f|\frac{grad f}{∥ grad f∥})| = ∥ grad f∥.

(iii) If v is tangented to the level surface f = c then

(grad f|v) = vf = 0.

◀

12–2 Definition. If f is a scalar field on a 2n-dimensional manifold X, with coordinates

x^{i} = (q^{1} … q^{n} p_{1} … p_{n})

and skew-symmetric tensor

(·|·) = ∑_{i=1}^{n} dp_{i} ∧ dq^{i}

then along a curve α whose velocity vector is grad f we have:

\frac{dx^{i}}{dt} = g^{ij} \frac{∂}{∂x^{j}} ,

i.e.

( \frac{dq^{1}}{dt} ; ⋮ ; \frac{dq^{n}}{dt} ; \frac{dp_{1}}{dt} ; ⋮ ; \frac{dp_{n}}{dt} ) = ( \begin{matrix}0 & 1 ... 1 \\ \begin{aligned}−1 \\ & . & . & . \\ & & & & −1 \end{aligned} & 0 \end{matrix} ) ( \frac{∂f}{∂q^{1}} ; ⋮ ; \frac{∂f}{∂q^{n}} ; \frac{∂f}{∂p_{1}} ; ⋮ ; \frac{∂f}{∂p_{n}} ) ,

i.e.

\frac{dq^{i}}{dt} = \frac{∂f}{∂p_{i}} ,
\frac{dp_{i}}{dt} = −\frac{∂f}{∂q^{i}}

(Hamiltonian Equations of Motion).

Note.

\frac{d}{dt}f(α(t)) = \frac{∂f}{∂x^{i}} (α(t))\frac{dx^{i}}{dt} (α(t)) = g^{ij}(α(t))\frac{∂f}{∂x^{j}} (α(t))\frac{∂f}{∂x^{i}} (α(t)) = 0,

since g^{ij} is skew-symmetric.

i.e.

rate of change of f along grad f = ⟨df, grad f⟩ = (grad f| grad f) = 0,

since (·|·) is skew-symmetric.

12–3

12.2 3-dimensional Vector Analysis

Given X a 3-dimensional manifold, x, y, z positively oriented orthonormal coordinates, i.e. metric tensor with line element (dx)^{2} + (dy)^{2} + (dz)^{2}. Write

∇ = (\frac{∂}{∂x} \frac{∂}{∂y} \frac{∂}{∂z} ) ,  dr = (dx dy dz),
dS = (dy ∧ dz, dz ∧ dx, dx ∧ dy), dV = dx ∧ dy ∧ dz,
F = (F^{1} F^{2} F^{3}) = (F_{1} F_{2} F_{3}),  ∇f = (\frac{∂f}{∂x} \frac{∂f}{∂y} \frac{∂f}{∂z} ) for a scalar field f,
∇ × F = (\frac{∂F^{3}}{∂y} − \frac{∂F^{2}}{∂z} , , ) ,  ∇.F = \frac{∂F^{1}}{∂x} + \frac{∂F^{2}}{∂y} + \frac{∂F^{3}}{∂z} .

The vector field

F^{→} = F.∇ = F^{1} \frac{∂}{∂x} + F^{2} \frac{∂}{∂y} + F^{3} \frac{∂}{∂z}

components F, corresponds, lowering the index, to the 1-form (components F)

F.dr = F_{1}dx + F_{2}dy + F_{3}dz,
df = (∇f).dr,
d[F.dr] = (∇ × F).dS,
d[F.dS] = (∇.F)dV,
∗1 = dV,
∗dr = dS,
∗dS = dr,
∗V = 1.

Now

1-forms  2-forms  3-froms
Ω(X) →  Ω^{1}(X)  →  Ω^{2}(X)  → Ω^{3}(X)
F.dr  ↦^{d} (∇ × F).dS

↘ ∗

f ↦ ∇F.dr F.dS ↦^{d} (∇.F)dV
(∇ × F).dr,
F⃗,
grad f,
curl F⃗

12–4

12.3 Results

Field Components Type
grad f  ∇f  vector
curl F⃗  ∇ × F  vector
div F⃗  ∇.F  scalar

Theorem 12.2. Let v be a vector field on a manifold, with non-singular symmetric metric tensor. Let ω be the 1-form given by lowering the index of v. Then the scalar field divv defined by:

d ∗ ω = (div v) vol

is called the divergence of v. If v has components v^{i} wrt coordinates y^{i} then

div v = \frac{1}{\sqrt{g}} \frac{∂}{∂y^{i}} (\sqrt{g} v^{i}),

where g = | det g_{ij}|.

Proof ▶

(∗ω)_{j_{1}...j_{n−1}} = g^{ij}v_{j}\sqrt{g} ϵ_{ij_{1}...j_{n−1}} = \sqrt{g} v^{i}ϵ_{ij_{1}...j_{n−1}}.

Therefore

∗ω = \sqrt{g} v^{1}dy^{2} ∧ dy^{3} ∧ ⋯ ∧ dy^{n} − \sqrt{g} v^{2}dy^{1} ∧ dy^{3} ∧ ⋯ ∧ dy^{n} + ….

Therefore

d ∗ ω = \frac{∂}{∂y^{i}} (\sqrt{g} v^{i})dy^{1} ∧ dy^{2} ∧ ⋯ ∧ dy^{n} = \frac{1}{\sqrt{g}} \frac{∂}{∂y^{i}} (\sqrt{g} v^{i}) vol,

as required. ◀

Theorem 12.3. Let f be a scalar field on a manifold, with non-singular symmetric metric tensor. Then the scalar field

Δf = div grad f = ∗d ∗ df

is called the Laplacian of f. Wrt coordinates y^{i} we have

Δf = \frac{1}{\sqrt{g}} \frac{∂}{∂y^{i}} (\sqrt{g} g^{ij} \frac{∂f}{∂y^{j}} ) .

Thus

Δf = \frac{1}{\sqrt{g}} ( \frac{∂}{∂y^{1}} … \frac{∂}{∂y^{n}} ) \sqrt{g}( \begin{matrix}g^{11} & … & g^{1n} \\ ⋮ & & ⋮ \\ g^{n1} & … & g^{nn} \end{matrix} )( ⋮_{\frac{∂f}{∂y^{n}}}^{\frac{∂f}{∂y^{1}}} ).

12–5 Examples:

(i) ℝ^{n}, with (ds)^{2} = (dx^{1})^{2} + ⋯ + (dx^{n})^{2}, (g^{ij}) = I, g = 1. Therefore

Δf = ( \frac{∂}{∂x^{1}} … \frac{∂}{∂x^{n}} ) ( ⋮_{\frac{∂}{∂x^{n}}}^{\frac{∂f}{∂x^{1}}} ) = \frac{∂^{2}f}{∂x^{1}^{2}} + ⋯ + \frac{∂^{2}f}{∂x^{n}^{2}} .

Therefore

Δ = \frac{∂^{2}}{∂x^{1}^{2}} + ⋯ + \frac{∂^{2}}{∂x^{n}^{2}} usual Laplacian.

(ii) ℝ^{4}, with (ds)^{2} = (dx)^{2} + (dy)^{2} + (dz)^{2} − (dt)^{2}: Minkowski.

Δf = ( \frac{∂}{∂x} \frac{∂}{∂y} \frac{∂}{∂z} \frac{∂}{∂t} ) ( \begin{matrix}1 \\ & 1 \\ & & 1 \\ & & & −1 \end{matrix} )( \frac{∂f}{∂x} ; \frac{∂f}{∂y} ; \frac{∂f}{∂z} ; \frac{∂f}{∂t} )

= \frac{∂^{2}f}{∂x^{2}} + \frac{∂^{2}f}{∂y^{2}} + \frac{∂^{2}f}{∂z^{2}} − \frac{∂^{2}f}{∂t^{2}} .

Therefore

Δ = \frac{∂^{2}}{∂x^{2}} + \frac{∂^{2}}{∂y^{2}} + \frac{∂^{2}}{∂z^{2}} − \frac{∂^{2}}{∂t^{2}} wave operator.

(iii) S^{2}, with (ds)^{2} = (dθ)^{2} + (sin θ dφ)^{2}, g_{ij} = ( \begin{matrix}1 & 0 \\ 0 & sin^{2} θ \end{matrix} ) , g = sin^{2} θ.

Δf = \frac{1}{sin θ} (\frac{∂}{∂θ} \frac{∂}{∂φ}) sin θ ( \begin{matrix}1 & 0 \\ 0 & \frac{1}{sin^{2} θ} \end{matrix} ) ( \frac{∂f}{∂θ} ; \frac{∂f}{∂φ} )

= \frac{1}{sin θ} (\frac{∂}{∂θ} \frac{∂}{∂φ}) ( sin θ\frac{∂f}{∂θ}_{\frac{1}{sin θ} \frac{∂f}{∂φ}} )

= \frac{1}{sin θ} [\frac{∂}{∂θ} (sin θ\frac{∂f}{∂θ} ) + \frac{∂}{∂φ} (\frac{1}{sin θ} \frac{∂f}{∂φ})] .

Definition. Let X be a 3-dimensional oriented manifold with non-singular symmetric metric tensor. Let v be a vector field corresponding to the 1-form ω. Then curl v is the vector field corresponding to the 1-form ∗dω. Wrt positively oriented coordinates y^{i}, curl v has components:

ϵ^{ijk} \frac{1}{\sqrt{g}} (\frac{∂v_{k}}{∂y^{j}} − \frac{∂v_{j}}{∂y^{k}} ) ,

where g = | det g_{ij}|.

12–6

12.4 Closed and Exact Forms

Definition. A differential form ω ∈ Ω^{r}(X) is called

(i) closed if dω = 0; (ii) exact if ω = dη for some η ∈ Ω^{r−1}(X).

We note that

1. ω exact ⇒ ω closed (d dη = 0),

2. ω an exact 1-form ⇒ ω = df (say)

⇒ ∫_{α} ω = ∫_{α} df = 0

for each closed curve α. Examples:

1. If ω = P dx + Q dy is a 1-form on an open V ⊂ ℝ^{2} then

ω = dP ∧ dx + dQ ∧ dy

= (\frac{∂P}{∂x} dx + \frac{∂P}{∂y} dy) ∧ dx + (\frac{∂Q}{∂x}dx + \frac{∂Q}{∂y}dy) ∧ dy

= (\frac{∂Q}{∂x} − \frac{∂P}{∂y} ) dx ∧ dy.

Therefore

ω is closed ⇔ \frac{∂P}{∂y} = \frac{∂Q}{∂x} on V ,
ω is exact ⇔ P = \frac{∂f}{∂x}, Q = \frac{∂f}{∂y}

for some scalar field f on V.

2. The 1-form

ω = \frac{x dy − y dx}{x^{2} + y^{2}}

on ℝ^{2} − {0} is called the angle-form about 0. We have:

\frac{∂}{∂x} (\frac{x}{x^{2} + y^{2}} ) = +\frac{1}{x^{2} + y^{2}} − \frac{2x^{2}}{(x^{2} + y^{2})^{2}} = \frac{y^{2} − x^{2}}{(x^{2} + y^{2})^{2}} ,
\frac{∂}{∂y} (\frac{−y}{x^{2} + y^{2}} ) = −\frac{1}{x^{2} + y^{2}} + \frac{2y^{2}}{(x^{2} + y^{2})^{2}} = \frac{y^{2} − x^{2}}{(x^{2} + y^{2})^{2}} .

12–7 Therefore ω is closed. ω is not exact because if α(t) = (cos t, sin t) is the unit circle about 0 (0 ≤ t ≤ 2π) then

∫_{α} ω = ∫_{0}^{2π} \frac{cos t. sin t − sin t.(− sin t)}{cos^{2} t + sin^{2} t}dt = ∫_{0}^{2π} 1 dt = 2π ≠ 0.

However, on ℝ^{2} − {negative or zero x-axis} we have

x = r cos θ, y = r sin θ,

where θ is a scalar field, with −π ≤ θ ≤ π and

ω = \frac{r cos θ.(−r cos θ) − r sin θ.(−r sin θ)}{r^{2} cos^{2} θ + r^{2} sin^{2} θ}dθ = dθ.

Figure 12.2

Therefore, if α is a curve from a to b (see Figure 12.2),

∫_{α} ω = ∫_{α} dθ = θ(b) − θ(a) = change in angle along α.

Note.

\frac{dz}{z} = \frac{\overline{z} dz}{\overline{z}z} = \frac{(x − iy)(dx + i dy)}{x^{2} + y^{2}} = \frac{x dx + y dy}{x^{2} + y^{2}} + i\frac{x dy − y dx}{x^{2} + y^{2}}.

Therefore ω = im \frac{∂z}{∂}.

12–8 3. If

F⃗ = F^{1} \frac{∂}{∂x} + F^{2} \frac{∂}{∂y} + F^{3} \frac{∂}{∂z}

is a force field in ℝ^{3} then

F.dr = F_{1}dx + F_{2}dy + F_{3}dz, F_{i} = F^{i}

is called the work element.

∫_{α} F.dr = work done by the force F⃗ along the curve α. F⃗ is conservative if F.dr is exact, i.e.

F.dr = dV,

where V is a scalar field. V is called a potential function for F⃗. Work done by F⃗ along α from a to b (see Figure 12.3) is

∫_{α} F.dr = ∫_{α} dV = V (b) − V (a) = potential difference.

α

Figure 12.3

A necessary condition that F⃗ be conservative is that F.dr be closed, i.e.

∇ × F = 0.

12–9

12.5 Contraction and Results

Definition. An open set V ⊂ ℝ^{n} is called contractible to a ∈ V if there exists a C^{∞} map

V × [0, 1] →^{φ} V

such that

φ(x, 1) = x,
φ(x, 0) = a

for all x ∈ V.

Example: V star-shaped ⇒ V contractible. φ(x, t) = tx+(1−t)a (see Figure 12.4).

ϕ(x, t)

T

Figure 12.4

Theorem 12.4 (Poincaré Lemma). Let ω ∈ Ω^{r}(V), where V is a contractible open subset of ℝ^{n}, and r ≥ 1. Then ω is exact iff ω is closed.

Proof ▶ Let I = [0, 1] be the unit interval 0 ≤ t ≤ 1, and define a linear operator (‘homotopy’)

Ω^{r}(V × I) →^{H} Ω^{r−1}(V)

for each r ≥ 1 by

H[f dt ∧ dx^{J}] = (∫_{0}^{1} f dt) dx^{J},
H[f dx^{J}] = 0.

Now calculate the operator dH + Hd:

12–10 (i) if η = f dt ∧ dx^{J} then

dHη + H dη = d [(∫_{0}^{1} f dt) dx^{J}] + H [−\frac{∂f}{∂x^{i}} dt ∧ dx^{i} ∧ dx^{J}]

= (∫_{0}^{1} \frac{∂f}{∂x^{i}} dt) dx^{i} ∧ dx^{J} − (∫_{0}^{1} \frac{∂f}{∂x^{i}} dt) dx^{i} ∧ dx^{J}

= 0;

(ii) if η = f dx^{J} then

dHη + H dη = 0 + H [\frac{∂f}{∂t} dt ∧ dx^{J} + \frac{∂f}{∂x^{i}} dx^{i} ∧ dx^{J}]

= (∫_{0}^{1} \frac{∂f}{∂t} dt) dx^{J} + 0

= [f(x, t)]_{t=0}^{t=1}dx^{J}.

(iii) Now let V be contractible, with

V × I →^{φ} V

ϕ(x, t)

V

Figure 12.5

a C^{∞} map such that (see Figure 12.5)

φ(x, 1) = x,
φ(x, 0) = a.

So

φ^{i}(x, 1) = x^{i},
φ^{i}(x, 0) = a^{i}.

12–11 Therefore

\frac{∂φ^{i}}{∂x^{i}} = \begin{cases}δ_{j}^{i} & at t = 1; \\ 0 & at t = 0. \end{cases}

Let ω ∈ Ω^{r}(V), say ω = g(x)dx^{i_{1}} ∧ ⋯ ∧ dx^{i_{r}}. Apply φ^{∗}:

φ^{∗}ω = g(φ(x, t))dφ^{i_{1}} ∧ ⋯ ∧ dφ^{i_{r}}

= g(φ(x, t)) [\frac{∂φ^{i_{1}}}{∂x^{j_{1}}} dx^{j_{1}} + \frac{∂φ^{i_{1}}}{∂t} dt] ∧ ⋯ ∧ [\frac{∂φ^{i_{r}}}{∂x^{j_{r}}} dx^{j_{r}} + \frac{∂φ^{i_{r}}}{∂t} dt]

= g(φ(x, t)) [\frac{∂φ^{i_{1}}}{∂x^{j_{1}}} … \frac{∂φ^{i_{r}}}{∂x^{j_{r}}} dx^{j_{1}} ∧ ⋯ ∧ dx^{j_{r}} + (\frac{∂φ^{i_{1}}}{∂t} … \frac{∂φ^{i_{r}}}{∂t} dt ∧ ⋯ ∧ dt) ∧ dt] .

Apply dH + Hd:

(dH + Hd)φ^{∗}ω = [g(φ(x, t))\frac{∂φ^{i_{1}}}{∂x^{j_{1}}} … \frac{∂φ^{i_{r}}}{∂x^{j_{r}}} ]_{t=0}^{t=1} dx^{j_{1}} ∧ ⋯ ∧ dx^{j_{r}} + 0
= g(x)δ_{j_{1}}^{i_{1}} … δ_{j_{r}}^{i_{r}}dx^{j_{1}} ∧ ⋯ ∧ dx^{j_{r}}
= g(x)dx^{i_{1}} ∧ ⋯ ∧ dx^{i_{r}}
= ω.

Hence:

(dH + Hd)φ^{∗}ω = ω

for all ω ∈ Ω^{r}(V). (iv) Let ω be closed. Then

dφ^{∗}ω = φ^{∗}dω = 0.

Therefore

dHφ^{∗}ω = ω.

Therefore ω is exact. ◀

12–12 Theorem 12.5. Let ω be a closed r-form, with domain V open in manifold X. Let a ∈ V. Then there exists an open neighbourhood W of a such that ω is exact on W.

Proof ▶ Let y be a coordinate system on X at a, domain U ⊂ V, say. Let W ⊂ U be an open neighbourhood of a such that y(W) is an open ball (see Figure 12.8).

Figure 12.6

Consider

W →^{y} y(W), W ←_{φ} y(W)

(open ball), where φ is the inverse map.

dφ^{∗}ω = φ^{∗}dω = 0, since ω is closed. Therefore φ^{∗}ω is closed. Therefore φ^{∗}ω = dη on y(W), by Poincaré. Therefore

dy^{∗}η = y^{∗}dη = y^{∗}φ^{∗}ω = ω

on W. Therefore ω is exact on W. ◀

Theorem 12.6. Let X be a 2-dimensional oriented manifold with a positive definite metric tensor. Let u_{1}, u_{2} be positively oriented orthonormal vector fields on X, with domain V (moving frame). Then there exists a unique 1-form ω on V such that

(∗) ( \begin{matrix}du^{1} \\ du^{2} \end{matrix} ) = ( \begin{matrix}0 & −ω \\ ω & 0 \end{matrix} ) ∧ ( \begin{matrix}u^{1} \\ u^{2} \end{matrix} )

on V. ω is called the connection form (gauge field) wrt moving frame u_{1}, u_{2}.

12–13 Proof ▶ Any 1-form ω on V can be written uniquely as:

ω = αu^{1} + βu^{2}, α, β scalar fields.

ω satisfies (∗) ⇔ du^{1} = −ω ∧ u^{2},
du^{2} = ω ∧ u^{1}
⇔ du^{1} = −(αu^{1} + βu^{2}) ∧ u^{2} = −αu^{1} ∧ u^{2},
du^{2} = (αu^{1} + βu^{2}) ∧ u^{1} = −βu^{1} ∧ u^{2}.

Thus α, β are uniquely determined. ◀

Theorem 12.7. Let X be a 2-dimensional oriented manifold with positive definite metric tensor. Let u_{1}, u_{2} be a moving frame with domain V, with connection form ω and

dω = Ku^{1} ∧ u^{2} = K.area element.

Then the scalar field K is independent of the choice of moving frame, and is called the Gaussian curvature of X.

Proof ▶ Let w_{1}, w_{2} be another moving frame with domain V:

( \begin{matrix}w^{1} \\ w^{2} \end{matrix} ) = ( \begin{matrix}cos θ & − sin θ \\ sin θ & cos θ \end{matrix} ) ( \begin{matrix}u^{1} \\ u^{2} \end{matrix} )

(say). Write this in matrix form as:

w = P u.

Also

( \begin{matrix}du^{1} \\ du^{2} \end{matrix} ) = ( \begin{matrix}0 & −ω \\ ω & 0 \end{matrix} ) ∧ ( \begin{matrix}u^{1} \\ u^{2} \end{matrix} ) .

Write this in matrix form as:

du = Ω ∧ u.

12–14

dw = (dP) ∧ u + P du

= (dP) ∧ u + P Ω ∧ u = (dP + P Ω) ∧ u = (dP + P Ω) ∧ P^{−1}w = [(dP)P^{−1} + P ΩP^{−1}] ∧ w

= [( \begin{matrix}− sin θ dθ & − cos θ dθ \\ cos θ dθ & − sin θ dθ \end{matrix} ) ( \begin{matrix}cos θ & sin θ \\ − sin θ & cos θ \end{matrix} )

+ ( \begin{matrix}cos θ & − sin θ \\ sin θ & cos θ \end{matrix} ) ( \begin{matrix}0 & −ω \\ ω & 0 \end{matrix} ) ( \begin{matrix}cos θ & sin θ \\ − sin θ & cos θ \end{matrix} )] ∧ w

= [( \begin{matrix}0 & −dθ \\ dθ & 0 \end{matrix} ) + ( \begin{matrix}0 & −ω \\ ω & 0 \end{matrix} )] ∧ w

= ( \begin{matrix}0 & −(ω + dθ) \\ ω + dθ & 0 \end{matrix} ) ∧ w.

Therefore ω + dθ is the connection form wrt moving frame w_{1}, w_{2} and

d[ω + dθ] = dω + d dθ = dω,

as required. ◀

Example: On S^{2}, with angle coordinates θ, φ,

ds^{2} = (dθ)^{2} + (sin θ dφ)^{2}, u^{1} = dθ, u^{2} = sin θ dφ.

Recall

ω = − cos θ dφ,
dω = sin θ dθ ∧ dφ = u^{1} ∧ u^{2}.

Therefore Gaussian curvature is constant function 1.

Theorem 12.8. Let X be an oriented 2-dimensional manifold with positive definite metric tensor. Then the Gaussian curvature of X is zero iff for each a ∈ X there exists local coordinates x, y such that

ds^{2} = (dx)^{2} + (dy)^{2},

i.e.

g_{ij} = ( \begin{matrix}1 & 0 \\ 0 & 1 \end{matrix} ) .

12–15 Proof ▶ Let u_{1}, u_{2} be a moving frame with connection form ω on an open neighbourhood V of a on which the Poincaré lemma holds. Then, on V:

Gaussian curvature is zero ⇔ dω = 0

⇔ ω = −dθ (say), by Poincaré

⇔ ω + dθ = 0

⇔ u_{1}, u_{2} can be rotated to a new frame w_{1}, w_{2}

having connection form 0 ⇔ dw^{1} = 0, dw^{2} = 0

⇔ w^{1} = dx, w^{2} = dy (say), by Poincaré
⇔ ds^{2} = (dx)^{2} + (dy)^{2}.

N⃗ ◀

12.6 Surface in ℝ^{3}

Let X be a 2-dimensional submanifold of ℝ^{3}. Denote all vectors by their usual components in ℝ^{3}. Let N⃗ be a field of unit vectors normal to X, t^{1}, t^{2} be coordinates on X, and let r⃗ = (x, y, z). Let \frac{∂r⃗}{∂t^{1}} , \frac{∂r⃗}{∂t^{2}} be a basis for vectors tangent to X (see Figure 12.9).

\frac{∂r⃗}{∂t^{i}} .N⃗ = 0.

Therefore

\frac{∂^{2}r⃗}{∂t^{i}∂t^{j}} .N⃗ + \frac{∂r⃗}{∂t^{i}} .\frac{∂N⃗}{∂t^{j}} = 0.

N

\frac{∂r⃗}{∂t^{2}}

\frac{∂r⃗}{∂t^{1}}

Figure 12.7

12–16 If u⃗ is a vector field on X tangent to X then

N⃗.N⃗ = 1

along u⃗. Therefore

(∇u⃗N⃗).N⃗ + N⃗.∇u⃗N⃗ = 0.

Therefore ∇u⃗N⃗ is tangential to X.

So define tensor field S on X

Su⃗ = −∇u⃗N⃗.

S is called the shape operator, and measures the amount of curvature of X in ℝ^{3}. Wrt coordinates t^{1}, t^{2} S has covariant components

S_{ij} = (\frac{∂}{∂t^{i}} |S \frac{∂}{∂t^{j}} ) = \frac{∂r⃗}{∂t^{i}} . − \frac{∂N⃗}{∂t^{j}} = \frac{∂^{2}x}{∂t^{i}∂t^{j}} .N⃗ (symmetric).

Therefore S_{a} is a self-adjoint operator on T_{a}X for each a. Therefore it has real eigenvalues K_{1}, K_{2} and orthonormal eigenvectors u_{1}, u_{2} (see Figure 12.8) exist, (say) K_{1} ≥ K_{2}.

N

u

u_{2}
u_{1}

Figure 12.8

If we intersect X by a plane normal to X containing the vector

cos θ u_{1} + sin θ u_{2}

at a, we have a curve α of intersection along which the unit tangent vector \underline{t} satisfies:

t⃗.N⃗ = 0.

Therefore

(∇t⃗t⃗).N⃗ + t⃗.(∇t⃗N⃗) = 0.

12–17

κN⃗.N⃗ − t⃗.St⃗ = 0

at a, where κ is the curvature of α at a. Therefore

κ = t⃗.St⃗ = (cos θ u_{1} + sin θ u_{2}).S(cos θ u_{1} + sin θ u_{2}) (at a)
= κ_{1} cos^{2} θ + κ_{2} sin^{2} θ.

Therefore u_{1} is the direction of maximum curvature K_{1}, and u_{2} is the direction of minimum curvature K_{2}.

Put N⃗ = u_{3}, with u_{1}, u_{2}, u_{3} a moving frame.

∇u_{3} = −ω_{3}^{1} ⊗ u_{1} − ω_{3}^{2} ⊗ u_{2} − ω_{3}^{3} ⊗ u_{3}.

Therefore

Su_{1} = −∇_{u_{1}}u_{3} = ⟨ω_{3}^{1}, u_{1}⟩u_{1} + ⟨ω_{3}^{2}, u_{1}⟩u_{2},
Su_{2} = −∇_{u_{2}}u_{3} = ⟨ω_{3}^{1}, u_{2}⟩u_{1} + ⟨ω_{3}^{2}, u_{2}⟩u_{2}.

Therefore

κ_{1}κ_{2} = det S
= ⟨ω_{3}^{1}, u_{1}⟩⟨ω_{3}^{2}, u_{2}⟩ − ⟨ω_{3}^{1}, u_{2}⟩⟨ω_{3}^{2}, u_{1}⟩
= ω_{3}^{1} ∧ ω_{3}^{2}(u_{1}, u_{2})
= dω_{2}^{1}(u_{1}, u_{2})
= κ u^{1} ∧ u^{2}(u_{1}, u_{2})
= κ

(since dΩ = −Ω ∧ Ω, so dω_{2}^{1} = −ω_{k}^{1} ∧ ω_{2}^{k} = −ω_{3}^{1} ∧ ω_{2}^{3} = ω_{3}^{1} ∧ ω_{3}^{2}).

12.7 Integration on a Manifold (Sketch)

Let ω be a differential n-form on an oriented n-dimensional manifold X. We want to define

∫_{X} ω,

the integral of ω over X.

To justify in detail the construction which follows X must satisfy some conditions. It is sufficient, for instance, that X be a submanifold of some ℝ^{N}.

12–18 (i) Suppose

ω = f(y^{1}, …, y^{n})dy^{1} ∧ ⋯ ∧ dy^{n}
= y^{∗}[f(x^{1}, …, x^{n})dx^{1} ∧ ⋯ ∧ dx^{n}]
= y^{∗}ω_{1}

on the domain V of a positively oriented coordinate system y^{i} (see Figure 12.9), and that ω is zero outside V, and that

supp f = closure of {x ∈ y(V) : f(x) ≠ 0}

is a bounded set contained in y(V). Then we define

∫_{X} ω = ∫_{y(V)} ω_{1},

i.e.

∫_{X} f(y^{1}, …, y^{n})dy^{1} ∧ ⋯ ∧ dy^{n} = ∫_{y(V)} f(x_{1}, …, x_{n})dx_{1} … dx_{n}

X  ω  ω_{1}

Figure 12.9

(Lebesgue integral). The definition of ∫_{X} ω does not depend on the choice of coordinates, since if z^{i} with domain W is another such coordinate system,

ω = z^{∗}ω_{2}

(say), then

φ^{∗}ω_{2} = ω_{1},

12–19 where φ = z ∘ y^{−1} (see Figure 12.10). Therefore

∫_{y(V)} ω_{1} = ∫_{y(V ∩W)} ω_{1} = ∫_{y(V ∩W)} φ^{∗}ω_{2} = ∫_{z(V ∩W)} ω_{2} = ∫_{z(W)} ω_{2}.

ω

ω_{1} ω_{2}

ϕ

Figure 12.10

(ii) For a general n-form ω on X we write

ω = ω_{1} + ⋯ + ω_{r},

where each ω_{i} is an n-form satisfying the conditions of (i), and define

∫_{X} ω = ∫_{X} ω_{1} + ⋯ + ∫_{X} ω_{r},

and check that the result is independent of the choice of ω_{1}, …, ω_{r}. Definition. If X has a metric tensor then

volume of X = ∫_{X} volume form.

12–20 Example: If

v⃗ = v^{1} \frac{∂}{∂x} + v^{2} \frac{∂}{∂y} + v^{3} \frac{∂}{∂z} = v⃗.∇

is a vector field in ℝ^{3},

v = v_{1} dx + v_{2} dy + v_{3} dz = v⃗.dr⃗

is the corresponding 1-form (v_{i} = v^{i}), and

∗v = v_{1} dy ∧ dz + v_{2} dz ∧ dx + v_{3} dx ∧ dy = v⃗.ds

is the corresponding 2-form then if u_{i} is a moving frame, with u_{3} = N⃗ normal to surface S (see Figure 12.11), then

v⃗ = α^{i}u_{i},
v = α_{i}u^{i},
∗v = α_{1} u^{2} ∧ u^{3} + α_{2} u^{3} ∧ u^{1} + α_{3} u^{1} ∧ u^{2},

where α_{i} = α^{i}. Therefore pull-back of ∗v to X is

α_{3}u^{1} ∧ u^{2} = (v⃗.N⃗)dS,

where dS = u^{1} ∧ u^{2} (area form). Therefore

∫ \underline{v}.\underline{dS} = ∫ (v⃗.N⃗)dS = flux of v⃗ across S
∫ N⃗.\underline{dS} = ∫ dS = area of S.

Note. v⃗.dr is work element, v⃗.ds is flux element, N⃗.ds is area element of vector field v⃗.

12–21

u_{3} = N⃗

u_{1}
u_{2}

Figure 12.11

12.8 Stokes Theorem and Applications

Theorem 12.9 (Stokes). (George Gabriel Stokes 1819 - 1903, Skreen Co. Sligo) Let ω be an (n − 1)-form on an n-dimensional manifold X with an (n − 1)-dimensional boundary ∂X (see Figure 12.12). Then

∫_{X} dω = ∫_{∂X} i^{∗}ω,

where ∂X →^{i} X is the inclusion map.

Proof ▶(Sketch) We write

ω = ω_{1} + ⋯ + ω_{r},

where each ω_{i} satisfies the conditions of either (i), (ii) or (iii) below, and we prove it for each ω_{i}. It then follows for ω.

12–22

∂X

Figure 12.12

(i) Let

ω = f(y^{1}, …, y^{n})dy^{2} ∧ ⋯ ∧ dy^{n}

on the domain V of a positively oriented coordinate system y^{i} such that

y(V) = (−1, 1) × ⋯ × (−1, 1) (cube),

ω zero outside V, and supp f a closed bounded subset of y(V) (see Figure 12.13).

∫_{∂X} i^{∗}ω = 0,

since ω is zero on ∂X.

dω = \frac{∂f}{∂y^{1}} dy^{1} ∧ dy^{2} ∧ ⋯ ∧ dy^{n}.

Therefore

∫_{X} dω = ∫_{y(V)} \frac{∂f}{∂x^{1}} dx_{1}dx_{2} … dx_{n}

= ∫_{−1}^{1}⋯ ∫_{−1}^{1} [∫_{−1}^{1} \frac{∂f}{∂x^{1}} dx_{1}] dx_{2} … dx_{n}

= ∫_{−1}^{1}⋯ ∫_{−1}^{1}[f(1, x_{2}, …, x_{n}) − f(−1, x_{2}, …, x_{n})]dx_{2} … dx_{n} = 0.

12–23

∫_{X} dω = 0 = ∫_{∂X} i^{∗}ω.

For (ii) and (iii), let ω be zero outside the domain W of functions y^{1}, y^{2}, …, y^{n}, with y a homeomorphism:

y(W) = (−1, 0] × (−1, 1) × ⋯ × (−1, 1)

Figure 12.13

12–24 (see Figure 12.14), where y^{1}, y^{2}, …, y^{n} are positively oriented coordinates on X with domain W − (W ∩ ∂X), y^{1} = 0 on W ∩ ∂X, and y^{2}, …, y^{n} are positively oriented coordinates on ∂X with domain W ∩ ∂X. Then

(ii) if

ω = f(y^{1}, y^{2}, …, y^{n})dy^{2} ∧ ⋯ ∧ dy^{n} (say),

(supp f closed bounded subset of y(W)) then

i^{∗}ω = f(0, y^{2}, …, y^{n})dy^{2} ∧ ⋯ ∧ dy^{n},
dω = \frac{∂f}{∂y^{1}} dy^{1} ∧ dy^{2} ∧ ⋯ ∧ dy^{n}.

12–25 Therefore

∫_{X} dω = ∫_{y(W)} \frac{∂f}{∂x^{1}} dx_{1}dx_{2} … dx_{n}
= ∫_{y(W ∧∂X)} f(0, x_{2}, …, x_{n})dx_{2} … dx_{n}
= ∫_{∂X} i^{∗}ω;

(iii) if

ω = f(y^{1}, …, y^{n})dy^{1} ∧ dy^{3} ∧ ⋯ ∧ dy^{n}

(say), (supp f closed bounded subset of y(W)) then

i^{∗}ω = 0,

since y^{1} = 0 on W ∧ ∂X. Also

dω = −\frac{∂f}{∂y^{2}} dy^{1} ∧ dy^{2} ∧ ⋯ ∧ dy^{n}.

Therefore

∫_{X} dω = − ∫_{−1}^{1}⋯ ∫_{−1}^{0} \underbrace{[∫_{−1}^{1} \frac{∂f}{∂x^{2}} dx_{2}]}_{0} dx_{1} … dx_{n} = 0.

Therefore

∫_{X} dω = 0 = ∫_{∂X} i^{∗}ω.

◀

12–26 Applications of Stokes Theorem:

1. In ℝ^{2}: ω 1-form (see Figure 12.17). ∫_{∂D} i^{∗}ω = ∫_{ω} dω,

∫_{∂D}(P dx + Q dy) = ∫_{D} (\frac{∂Q}{∂x} − \frac{∂P}{∂y} ) dx ∧ dy Green’s Theorem.

In particular:

(a)

∫_{∂D} x dy = − ∫_{∂D} y dx = \frac{1}{2i} ∫_{∂D} \overline{z} dz = ∫_{D} dx ∧ dy = area of D.

(b) if f = u + iv, ω = f dz then

dω = [−\frac{∂u}{∂y} − \frac{∂v}{∂x} + i (\frac{∂u}{∂x} − \frac{∂v}{∂y} )] dx ∧ dy.

Now

dω = 0 ⇔ f holomorphic ,

by Cauchy-Riemann. Therefore ∫_{∂D} f(z)dz = 0 if f holomorphic (Cauchy).

2. In ℝ^{3}: X surface (see Figure 12.18).

∫_{∂X} \underline{F} .\underline{dr} = ∫_{X}(∇ × \underline{F}).\underline{dS} = ∫_{X}(∇ × \underline{F}).N⃗ dS,

i.e.

work of F around loop ∂X = flux of ∇ × F across surface X

= flux of ∇ × F through loop ∂X.

3. In ℝ^{3}:

∫_{∂D} \underline{F} .N⃗ dS = ∫_{∂D} \underline{F} .\underline{dS} = ∫_{D}(∇.\underline{F})dV,

i.e. (see Figure 12.19) flux of F out of region D = integral of ∇.F over interior of D.

12–27 If ∇.\underline{F} > 0 at a then a is a source for \underline{F}, if ∇.\underline{F} < 0 at a then a is a sink for \underline{F}, if ∇.\underline{F} = 0 then \underline{F} is source-free (see Figure 12.20).

4. If X is n-dimensional and ω ∧ η an (n − 1)-form then

∫_{∂X} ω ∧ η = ∫_{X} d(ω ∧ η) (by Stokes)
= ∫_{X}(dω) ∧ η + (−1)^{r}∫_{X} ω ∧ dη,

ω an r-form, by Leibsing. Therefore

∫_{X}(dω) ∧ η = \underbrace{∫_{∂X} ω ∧ η}_{boundary term} +(−1)^{r+1}∫_{X} ω ∧ dη

(integration by parts) (see Figure 12.21). Example: X connected, 3-dimensional in ℝ^{3}.

∫_{∂X} f ∇f.\underline{n} dS = ∫_{∂X} f(∇f.\underline{dS})
= ∫_{X}[∇f.∇f + f∇.(∇f)]
= ∫_{X}[∥∇f∥^{2} + f∇^{2}f]dV.

Therefore

∇^{2}f = 0; f or ∇f.\underline{n} = 0 on ∂X ⇒ ∇f = 0 on X ⇒ f = 0 on X (Dirichlet Neumann). If X has a metric tensor and no boundary then

∫_{X}(∗dω|η) vol = (−1)^{r+1}∫_{X}(∗ω|dη) vol .

If (say) the metric is positive definite and n odd then ∗∗ = 1, so putting ∗ω in place of ω:

∫_{X} ((−1)^{r} ∗ d ∗ ω|η) vol = ∫_{X}(ω|dη) vol .

12–28 Therefore (−1)^{r} ∗ d∗ is the adjoint of the operator d. Hence δ = ± ∗ d∗ is the operator adjoint to d.

Δ = dδ + δd

is self-adjoint, and is called the Laplacian on f.

5. If X is the unit ball in ℝ^{n} (see Figure 12.22), with n ≥ 2:

X = {x ∈ ℝ^{n} : ∑ x_{i}^{2} ≤ 1}

then ∂X is the (n − 1)-dimensional sphere

∂X = {x ∈ ℝ^{n} : ∑ x_{i}^{2} = 1}.

Theorem 12.10. There is no C^{∞} map

X →^{φ} ∂X

which leaves each point of the sphere ∂X fixed.

Proof ▶ Suppose φ exists. Then we have a commutative diagram:

ω X →^{φ} ∂X
i ↑ ↗_{1}
α ∂X

where i is the inclusion map and 1 the identity map. Let

ω = x^{1} dx^{2} ∧ ⋯ ∧ dx^{n}, α = i^{∗}ω.

So

dω = dx^{1} ∧ dx^{2} ∧ ⋯ ∧ dx^{n} (volume form on X),
dα = 0,

since dα is an n-form on (n − 1)-dimensional ∂X. Therefore

i^{∗}ω = α = i^{∗}φ^{∗}α.

Therefore

∫_{∂X} i^{∗}ω = ∫_{∂X} i^{∗}φ^{∗}α.

Therefore volume of X is

∫_{X} dω = ∫_{X} dφ^{∗}α = ∫_{X} φ^{∗}dα ∫_{X} φ^{∗}0 = 0.

This is a contradiction so the result follows. ◀

12–29 6. In ℝ^{4}, Minkowski: the electromagnetic field is a 2-form:

F = (\underline{E}.\underline{dr}) ∧ dt + \underline{B}.\underline{dS},

where \underline{E} is components of electric field, and \underline{B} are components of magnetic field.

One of Maxwell’s equations is:

dF = 0

(the other is d ∗ F = J charge-current), i.e.

d[(\underline{E}.\underline{dr}) ∧ dt + \underline{B}.\underline{dS}] = 0,

i.e.

(∇ × \underline{E}).\underline{dS} ∧ dt + (∇.\underline{B})dV + \frac{∂\underline{B}}{∂t} .\underline{dS} ∧ dt = 0,

i.e.

∇ × E = −\frac{∂B}{∂t} ,
∇.B = 0,

i.e. magnetic field is source free.

Therefore electromotive force (EMF) around loop ∂X (see Figure 12.23) is

∫_{∂X} \underline{E}.\underline{dr} =^{Stokes} ∫_{X}(∇ × \underline{E}).\underline{dS}

=^{Maxwell} −\frac{d}{dt} ∫_{X} \underline{B}.\underline{dS}

= rate of decrease of magnetic flux through loop.

12–30