Multilinear (tensor) algebra

Transcription

1 Multilinear (tensor) algebra In these notes, V will denote a fixed, finite dimensional vector space over R. Elements of V will be denoted by boldface Roman letters: v, w,.... Bookkeeping: We are going to develop an efficient notation that permits us to keep track of algebraic operations on vectors and tensors. Good notation prevents us from having to do the same computation over and over. If dim(v) = n, then each basis {e 1, e 2,..., e n } of V has precisely n elements, and every element v V can be written uniquely as a linear combination of them: n v = v 1 e 1 + v 2 e v n e n = v a e a. The numbers (v 1, v 2,..., v n ) are called the components of v in the given basis. The reason for the superscripts will become evident shortly. If {ẽ a : 1 a n} is a second basis for V, then the same vector v will have the expression v = n ṽ a ẽ a a=1 in the new basis. The new basis is related to the first one through a non-singular n n matrix P : n ẽ b = Pb a e a. a=1 P is called the change of basis matrix. Now we have two expressions for the vector v, which is a geometric object, so the expressions must be equal: v = ṽ b ẽ b = ( ) ṽ b Pb a e a = ( ) ṽ b Pb a e a = v a e a, so b b a a b a ṽ b Pb a = v a. Solving for the components ṽ b, we find b a=1 ṽ b = a (P 1 ) b av a. Even at this early stage in our deliberations, it s clear that repeated summations and nested parentheses are going to be both distractions and potential sources of silly algebraic errors. We shall adopt Einstein s summation convention: we omit the summation signs and agree that any pair of repeated upper and lower indices is to be summed over. 1

2 For example v a e a means v 1 e v n e n. So v a e a = v c e c. Both a and c are dummy indices, and can be changed to something else whenever the situation seems to call for it. T a a means T T n n. This is the trace of the matrix with entries T a b. An expression of the form which reads T ab x a This works quite nicely, provided that: = 0 stands for a system of n PDEs, the first one of T T + x1 x 2 n1 T + + x = 0. n 1. The sum is taken only over upper and lower indices, as in the examples above. Any expression in which two identical indices appear as upper or lower indices such as x aa, is wrong. If something like this shows up, a mistake has been made. 2. A given index can appear at most twice in a given expression. Something like W a P b av a has no meaning. To solve for ṽ a in the expression ṽ b Pb a = va, we need to multiply both sides of this expression by P 1. Notice that the upstairs index is the row index. We can t use (P 1 ) b a: although this would look fine on the right hand side, it would give us 3 b s on the left. Instead we write ṽ b P a b (P 1 ) c a = v a (P 1 ) c a. Now Pb a(p 1 ) c a is the (c, b) entry of the product matrix P 1 P = I. That is, { 0 if c b Pb a (P 1 ) c a = (P 1 ) c apb a = δb, c where δb c = 1 if c = b, where the quantity on the right is called the Kronecker delta. So he left hand side becomes ṽ b P a b (P 1 ) c a = ṽ b δ c b = ṽ c. Finally, notice that quantities such as Pb a from matrices, and so they commute: are just numbers, even though they may come L a bm b c = L a 1M 1 c + L a nm n c = M 1 c L a M n c L a n = M b c L a b. This is an overcomplicated notation for this particular problem: we could just have written P ṽ = v = ṽ = P 1 v - much simpler. Unfortunately, this standard notation is of little help in computations involving higher order objects like the curvature tensor, while the index notation doesn t get any more difficult than what you ve just seen. You ll get plenty of practice with this notation in the following sections. The positioning of the indices (up or down) is crucial. 2

3 The dual space V Definition: The dual space of V is Examples: Fix any w V and define V = {φ : V R with φ a linear function}. φ w (v) = w t v, v V. Then φ w is linear and thus an element of V. Let f : R n R be differentiable at the point x 0. For any v R n, let φ(v) = df(x 0 + tv) t=0 dt be the derivative of f in the direction v. Writing out this expression in the standard basis, we find φ(v) = f x v1 + + f x n vn = f x a va, (1) all the partial derivatives being evaluated at x 0. Then φ is a linear function on V = R n called the differential of f at x 0, often written as df( x) = f x a x a. The elements of V are called covariant vectors to distinguish them from elements of V which are called contravariant. They are different: getting ahead of ourselves a bit, look at the expression in (1) - it s a scalar and must be independent of the basis. If we change the basis in R n using the matrix P, then we know the components of v get multiplied by P 1. In order for v a f/ x a to remain the same, the components of df, namely the partial derivatives, must get multiplied by P. So the components of df don t behave the same as those of v under a change of basis. Returning to the general subject, any linear function with domain V is completely determined by what it does to a basis {e a } of V. We define n numbers φ a by φ a = φ(e a ). Since φ is linear it follows that for any bv V, φ(v) = φ(v a e a ) = v a φ(e a ) = v a φ a. Here, the linearity of φ has been used in writing φ( v a e a ) = v a φ a. The numbers φ(e a ) = φ a are the component of φ. If these are the components, what s the basis? 3

4 Definition: Given a basis {e a : 1 a n} of V, the dual basis {e 1, e 2,..., e n } of V is defined by e a (e b ) = δ a b, (2) and extending by linearity, where this last means that for any v, we define e a (v) by requiring it to be linear. Thus e a (v) = e a (v b e b ) = (linearity!) v b e a (e b ) = v b δ a b = v a. Under a change of basis for V given by the matrix P, we have What is ẽ b? Well, we must have ẽ b = P a b e a. ẽ b (ẽ a ) = δ b a = ẽ b (P c ae c ) = P c aẽb (e c ). If we write ẽ b (e c ) = Q b c, then this says P c aq b c = δ b a, and therefore, Q = P 1 : Exercise: ẽ b = (P 1 ) b ce c. 1. Show that {e a : 1 a n} is a basis for V and that any φ V can be written uniquely in the form φ a e a. 2. Show that V = V. Hint: φ(v) is linear in φ as well as v. The proof does not need coordinates or bases; the isomorphism of these two vector spaces is natural. Because of this, mathematicians often write something like < φ, v > instead of φ(v) to indicate the bilinearity. Unfortunately this is too easily confused with Dirac s notation (which involves a Hermitian metric) so we won t do it. What s the geometric meaning of a covariant vector? Since it s a function, we can look at the surfaces on which it s constant: φ(v) = c φ a v a = c. This is the equation of a hyperplane in V; it passes through the origin if c = 0. It is not defined using the dot product; we ll come back to this. Example: Let L(x a, ẋ a, t) be the Lagrangian of some physical system. The Euler-Lagrange equations are ( ) d L = L for 1 a n. dt ẋ a xa The conjugate momenta are defined by p a = L ẋ a. 4

5 For a conservative system, with potential energy V (x a ), the equations of motion are just dp a dt = V x a. Both sides of this equation are the components of covariant vectors. Indeed, the Legendre transformation, connecting the Hamiltonian and the Lagrangian, has the form H = p a ẋ a L, in which both H and L are scalars and so, therefore is p a ẋ a ; the ẋ a are the components of the velocities and transform as such. So the conjugate momenta must transform oppositely. Tensors We ll give the general definition here, for completeness, but we re only going to be dealing with tensors of relatively low rank. Definition: A tensor T of rank (r, s) is a multilinear function T : V V V V V V R. There are r copies of V and s copies of V. So this is a function of r + s vector or (r + s)n scalar variables. The word multilinear means that T is linear in each of its arguments. T is said to be contravariant of rank r and covariant of rank s. 5

6 Examples: 1. A vector is a tensor of rank (1, 0); a dual vector is a tensor of rank (0, 1). 2. A covariant tensor g of rank 2 is a tensor of rank (0, 2); it s a bilinear function of vectors. So for any pair of vectors, g(v, w) is a real number, and for all scalars c 1, c 2, g(c 1 v 1 + c 2 v 2, w) = c 1 g(v 1, w) + c 2 g(v 2, w) and g(v, c 1 w 1 + c 2 w 2 ) = c 1 g(v, w 1 ) + c 2 g(v, w 2 ). Specific examples of rank 2 covariant tensors include the metric tensors of Euclidean and Minkowski space, the electromagnetic field tensor and the energy-momentum tensor; the latter two also have contravariant forms - see below. 3. The curvature tensor R has rank (1, 3), so it s a multilinear function of 4 arguements: R(u, v, w, φ). 4. A linear transformation L : V V is (naturally identified with) a tensor ˆL of rank (1, 1): let φ V, v V, and define ˆL(φ, v) = φ(l(v)). 5. In conjunction with the above, the identity map I(v) = v is identified with a tensor of rank (1, 1) called the Kronecker delta. We ll talk a bit more about these in a minute. Definition: The set of all tensors of rank (r, s) is a vector space over R with the usual pointwise definitions of addition and scalar multiplication: (T + S)(φ,..., v) = T (φ,..., v) + S(φ,..., v) (ct )(φ,..., v) = ct (φ,..., v) This vector space is denoted V V V V V V, where there are r copies of V and s of V. Example: Suppose v, w V. Then we define an element of V V, denoted v w by the requirement (v w)(φ, µ) = φ(v)µ(w). This is evidently bilinear in φ and µ and so it s a tensor of rank (2, 0) called the tensor product of v and w. In a similar fashion, we can define things like φ v V V. A tensor which can be written in the form v w φ is said to be decomposable. Components of tensors: We give an example; generalization should be easy. Suppose T has rank (2, 1); then we define T ab c = T (e a, e b, e c ). 6

7 (There are n 3 numbers here!), and we write T = T ab ce a e b e c, where, by definition, e a e b e c is the tensor of rank (2, 1) such that (e a e b e c )(φ, µ, v) = φ a µ b v c. Equivalently, Exercise: (e a e b e c )(e e, e f, e g ) = δ e aδ f b δc g. 1. Show that {e a e b e c : 1 a, b, c n} is a basis for V V V. 2. How do the components T ab c transform under a change of basis? (We always assume that the new basis in V is the dual basis of the new one in V.) 3. Show that the identity map on V can be written as I = e a e a. What are its components? Is this true in every basis? Further examples: We can fool around algebraically with tensors, getting maps into various vector spaces other than just R: Suppose, as above, that T has rank (1.2). For any v V, let Thus T is a map from V V V. For φ V, we can define T (v) = T a bce a e b e c (v) = T a bcv c e a e b. T (φ) = T a bce a (φ)e b e c = φ a T a bce b e c V V. If L = L a b e a e b has rank (1, 1), then as in the first example, we can define L(v) = L a be a e b (v) = L a bv b e a V, as mentioned above in a slightly different form. Metric tensors Notation: From now on, we drop the boldface fonts for simplicity. A vector will be written simply as V (upper case) and its components in a given basis as V a. We ll retain the boldface font for basis vectors and the zero vector. Suppose g is a rank (0, 2) tensor (also called a bilinear form) which is 7

8 1. Symmetric: g(u, V ) = g(v, U), U, V V. 2. Non-degenerate: g(u, V ) = 0, U = V = 0 Symmetry implies that, in any basis, g(e a, e b ) = g ab = g ba = g(e b, e a ), so the components of g form a symmetric n n matrix. As we know (notes on SR), there exists a basis in which the matrix of components can be brought to canonical form g ab = g cd P c ap d b = ±δ ab. If all the signs preceding δ ab are positive, then g is said to be positive-definite, meaning that, in this coordinate system g(v, V ) = (V 1 ) 2 + (V 2 ) (V n ) 2 > 0 The particular expression for g(v, V ) as the sum of squares is only valid in a coordinate system in which g ab = δ ab, but the value of g(v, V ) and the fact that it s positive, unless V = 0 is independent of the basis. A bilinear form with all these properties is called a (positive-definite) metric tensor. If all properties except that of positive-definiteness hold, then g is called indefinite. If the canonical form of g is Diag{1, 1, 1,..., 1}, then g is called a Lorentz metric. If g is a metric, then the number g(u, V ) is called the scalar product of U and V. Definition: The pair (R n, g) where g is positive-definite, is called n-dimensional Euclidean space, and denoted E n.. Definition: The pair (R n, g), where g is a Lorentz metric, is called n-dimensional Minkowski space and denoted M n. The metric g defines a vector space isomorphism between V and its dual, known in the trade as lowering indices: Fix a V V, and use it to define the linear function Since φ V V, its components are given by φ V (U) = g(v, U). (φ V ) a = φ V (e a ) = g(v, e a ) = g(v b e b, e a ) = V b g(e b, e a ) = V b g ab. The matrix of this transformation, in the given bases, is just ((g ab )). It is non-singular, so the map is an isomorphism between V and its dual. From the definition, it s clearly well-defined, and therefore independent of the basis. Definition: We write the component (φ V ) a as V a ; that is, V a = V b g ab, and say that the covariant vector V b e b has been obtained from V by lowering an index. We can raise indices as well: since g gives an isomorphism from V V, it has an inverse g 1 : V V which is called raising an index. The components of g 1 are, somewhat 8

9 surprisingly, denoted by g ab (no inverse). It is (almost) impossible to make a mistake since the metric has its indices downstairs and the inverse upstairs. Since the two matrices are inverses, in any coordinate system, we must have g ab g bc = δ a c. (3) The tensor of rank (2, 0) given by g ab e a e b is called the contravariant metric tensor. Explicitly, the isomorphism defined by the contravariant metric is given as follows: For any fixed φ V, the vector V φ is the linear function µ(v φ ) = g(µ, φ) = g ab µ a φ b. As above, we will simply write φ a = g ab φ b, and we ll say that we ve raised the index on φ. Exercise: Show that for any vectors V and W, V a W a = V a W a. Is V a W b = V a W b? Why or why not? The scalar product of U and V is often written without explicitly showing the components of the metric tensor: g(u, V ) = g ab U a V b = U b V b = U a V a. The same process allows us to raise and lower indices on tensors of any rank: T ab = g ac g bd T cd, or T a b = T ac g cb. Note: It is important to keep track of the relative positions of indices. It is not generally true that T a b = T a b, although it might happen in some particular case. In E n, in a Cartesian (orthonormal) basis, the metric has the components Diag{1, 1,..., 1}, and the covariant and contravariant component of tensor are numerically identical in this basis, since, for example T 1b = g 1c T c b = δ 1c T c b = δ 11 T 1 b = T 1 b. This leads to some confusion in scientific texts. A simple example is given by the usual calculus book definition of the derivative of f in the direction V as f V. In fact, the directional derivative is given, as shown above, by the expression V a f/ x a, and doesn t involve the dot product at all. The numbers f/ x a are the components of the covariant vector df. To get a (contravariant) vector involves raising an index. The correct definition of the gradient is ( f) a ab f = g x often written as b a f. 9

10 In Cartesian coordinates, these are numerically equal, but in spherical polar coordinates, the components of the gradient vector are not ( f/ r, f/ θ, f/ φ). These are the components of df in the appropriate basis. The components of f, on the other hand, must be found by raising indices with the metric tensor, whose components are not constants in this coordinate system. In fact, as we ll see when we look at tensor analysis, the components of the gradient are ( f r, 1 r 2 f θ, 1 r 2 sin 2 θ f φ ). In M 4, the preferred coordinates are the inertial frames, in which the Lorentz metric takes the form g ab = Diag{1, 1, 1, 1} = g ab. So raising and lowering indices is simple but not trivial: writing the components of V as (V 0, V 1, V 2, V 3 ), we have and Or, as we ve written it elsewhere, V 0 = V 0, V 1 = V 1, V 2 = V 2, V 3 = V 3, so that g(u, V ) = U a V a = U 0 V 0 U 1 V 1 U 2 V 2 U 3 V 3, g(v, V ) = V a V a = (V 0 ) 2 (V 1 ) 2 (V 2 ) 2 (V 3 ) 2. τ 2 = t 2 dx dx. Odds and ends: Contraction Definition: rs new tensors of rank (r 1, s 1) can be produced from a tensor T of rank (r, s) by picking one upper and one lower index and summing over them to obtain the new tensor with components in a process called contraction. T a 1...a i 1 aa i+1...a r b 1...b j 1 ab j+1...b s, (Coordinate-free definition: To contract on the i th contravariant and j th covariant arguments, take T (φ 1,..., φ r 1, V 1,..., V s 1 = tr(t (φ 1,..., φ i 1,, φ i,..., φ r 1, V 1,..., V j 1,, V j,..., V s 1 )), where the circles denote the slots of the omitted arguments. The quantity in the outer parenthesis is of rank (1, 1) and tr denotes the trace. This is just written down so you know it s possible to do it without indices, but we ll never use this form.) Examples: 1. Contracting the tensor T a be a e b gives the scalar T a a. This is called the trace of T. 10

11 2. If U = U ab e a e b, then we can t sum on the two indices directly, but we can first lower an index and then contract to get g ab U ab. 3. The 2-covariant tensor obtained from the curvature tensor via is called the Ricci tensor. R ab = R acb c 4. For those not satisfied with the level of mathematical rigor, we could write, in any basis, R ab = Ric(e a, e b ) = tr(r(e a,, e b, )). 5. If we use the shorthand notation a = x a, then a number of differential operators can be written as contractions: φ = a a φ (the Laplace operator in E n ) φ = a a φ (the d Alembertian operator in M 4 ) In the first equation, we re using the Cartesian components of the Euclidean metric to raise the index prior to contraction; in the second, the Lorentz metric is used. Symmetric and skew-symmetric tensors Definition: A covariant tensor T is said to be symmetric in two of its arguments if. In terms of components, this reads T (..., U,..., V,...) = T (..., V,..., U,...) T...a...b... = T...b...a.... It is symmetric if this holds for all pairs of arguments that is, the value remains unchanged under any transposition of two arguments, and hence under any permutation of the arguments. Similarly for contravariant tensors. Mixed tensors can be symmetric under the interchange of two contravariant or two covariant vectors, but there s no notion of a totally symmetric tensor of mixed rank. Definition: The covariant tensor is said to be skew-symmetric if it changes sign under a transposition of any two of its arguments: F (..., U,..., V,...) = F (..., V,..., U,...). Under an arbitrary permutation of its arguments, F changes sign under odd permutations, but not under even ones. Examples: 11

12 1. Any 2-covariant tensor can be decomposed into the sum of a symmetric and a skew-symmetric tensor: T (U, V ) = S(U, V ) + A(U, V ), where S(U, V ) = 1 (T (U, V ) + T (V, U)), 2 A(U, V ) = 1 (T (U, V ) T (V, U)) 2 2. Higher rank tensors have totally symmetric and skew-symmetric parts in the following sense: if T is of rank r, we can define S(T )(U 1, U 2,..., U r ) = 1 T (U σ1, U σ2,..., U σr ) r! σ A(T )(U 1, U 2,..., U r ) = sign(σ) 1 T (U σ1, U σ2,..., U σr ) r! where the sum is over all permutations of order r, and, in the second case, sign(σ) = ±1 depending on whether σ is even or odd. (Remark: S and A are homomorphisms from the tensor algebra r V onto subalgebras of r V which are called, not surprisingly, the symmetric and skewsymmetric subalgebras.) 3. In terms of components, a rank 2 tensor is symmetric if T ab = T ba, and skewsymmetric if T ab = T ba, for all pairs of indices. Similarly for higher rank tensors. σ 12

13 Exercises 1. Show that {e a e b e c : 1 a, b, c n } forms a basis for the vector space V V V. 2. Show that, under the change of basis ẽ a = Pae b b, the components Tbc a rank (1, 2) undergo the transformation of a tensor of T a bc = T d ef(p 1 ) a dp e b P f c. 3. Is the bilinear form defined by g(x, Y ) = X t AY, where A = positive definite? What s the canonical form of g? 4. For φ, µ V, define their wedge product by Show that (a) φ µ = µ φ; φ φ = 0. (b) If µ = µ a e a, φ = φ b e b, then φ µ = 1 (φ µ µ φ). 2 µ φ = µ a φ b e a e b ( ) = (µ a φ b φ a µ b )e a e b a<b = µ [a φ b] e a e b, where, in the last line, µ [a φ b] = 1 2 (µ aφ b φ a µ b ). 5. Show that if T V V is skew-symmetric (i.e., T (V, W ) = T (W, V ) V, W V), then T = T ab e a be b = T [ab] e a e b = T ab e a e b. 13