Vector spaces, duals and endomorphisms

Vector spaces, duals and endomorphisms A real vector space V is a set equipped with an additive operation which is commutative and associative, has a zero element 0 and has an additive inverse v for any v V (so V is an abelian group under addition). Further there is an operation of multiplication of the reals on the vectors (r, v) rv V, for each real r and each vector v V, called scalar multiplication, which obeys, for any reals r and s and any vectors v V and w V, the relations: 0v = 0, 1v = v, ( v = v, r(sv) = (rs)v, (r + s)v = rs + sv, r(v + w) = rv + rw. The trivial vector space, said to be of zero dimensions, is a vector space consisting of only the zero vector. The basic family of non-trivial examples of a vector space are the spaces R n, n N. Here R n consists of all n-tuples x = (x 1, x 2,..., x n ), with each x i real. The operations of R n, valid for any reals x i, y i, i = 1, 2,..., n and any real s are: x + y = (x 1, x 2,..., x n ) + (y 1, y 2,..., y n ) = (x 1 + y 1, x 2 + y 2,..., x n + y n ), x = (sx 1, sx 2,..., sx n ). The zero element is 0 = (0, 0,..., 0) (often this element is just written 0). The additive inverse of x is x = ( x 1, x 2,..., x n ). Given a pair of vector spaces V and W a map f : V W is said to be linear if f(rx + sy) = rf(x) + sf(y), for any reals r and s and any x V and y V. In particular we have f(0) = 0 and f( v) = f(v), for any v V. Denote the space of linear maps from V to W by Hom(V, W). If f and g are in Hom(V, W), their sum map is defined by (f + g)(v) = f(v) + g(v), for any v V. Also we can multiply f Hom(V, W) by a real scalar r, giving the map rf, such that (rf)(v) = rf(v) for any v V. Then f + g and rf each lie in Hom(V, W) and these operations give Hom(V, W) the natural structure of a vector space. If f Hom(V, W) and g Hom(W, X), for vector spaces V, W and X, then the composition g f : V X is well-defined and linear, so lies in Hom(V, X). Also the composition map (f, g) f g is linear in each argument. 1

A linear map f Hom(V, W) is said to be an epimorphism if f is surjective, a monomorphism if f is injective (if and only if the equation f(v) = 0 has as its only solution the vector v = 0 and an isomorphism if f is bijective, in which case f has an inverse f 1 such that f f 1 = id W and f 1 f = id V are the identity maps (each of the latter is an isomorphism, each its own inverse). All trivial vector spaces are isomorphic. The space Hom(R n, R m ) is isomorphic to the space R mn. An element f Hom(R n, R m ) is given by the formula, for any x R n, f(x) = y where y i = n j=1 f jx i j, i = 1, 2..., m, for an m by n matrix fj. i Then f is surjective if and only if the matrix fj i has rank m and is injective, if and only if there are no solutions to the matrix equation fjx i j = 0, except for the solution x i = 0, i = 1, 2,..., n. Then f is an isomorphism if and only if m = n and the equation f(x) = 0 has as its only solution the vector x = 0, if and only if m = n and f has rank n. A linear map from V to itself is called an endomorphism. Then the space Hom(V, V) of all endomorphisms of V is an algebra, with associative multiplication (distributive over addition) given by composition. The space Hom(R, V) is naturally isomorphic to V itself: simply map f in Hom(R, V) to f( V. The space Hom(V, R) is called the dual vector space of V and is written V. If f Hom(V, W) and α Hom(W, R), then f (α) = α f is an element of Hom(V, R). As α W varies, f (α) depends linearly on α, so f gives a linear map from W to V. The map f Hom(W, V ) is called the adjoint of the map f. Then the map f f is linear and (g f) = f g, for any f Hom(V, W) and g Hom(W, X) and any vector spaces V, W and X. The adjoint of an identity map is itself, so the adjoint of an isomorphism is an isomorphism. 2

Bases and finite dimensionality A vector space is said to be finite dimensional if there is an isomorphism f : V R n, for some integer n. Any such isomorphism is called a basis for V. If v V and f is a basis, then f(v) R n is called the co-ordinate vector of v in the basis f. The basis elements of R n are the defined to be the n-vectors, {e i, i = 1, 2,..., n}, such that the j-the entry of e i is δ j i, the Kronecker delta, so is 1 if j = i and zero otherwise. Given a basis f : V R n of a vector space V, the corresponding basis elements of V are the vectors {f i = f 1 (e i ), i = 1, 2,..., n}. Then for each v V, we have f(v) = v = (v 1, v 2,..., v n ) if and only if v = n i=1 vi f i. If f : V R n and g : V R m are bases for V, the map f g 1 : R m R n is an isomorphism (with inverse g f 1 ), so m = n. So an n N, such that a basis f : V R n exists, is unique. It is called the dimension of V. If f : V R n is a basis, then the adjoint f : (R n ) V is an isomorphism, so (f ) 1 : V (R n ) is an isomorphism. Now (R n ) = Hom(R n, R), so is isomorphic to R n itself. This isomorphism maps t (R n ) to the element t = (t(e 1 ), t(e 2 ),..., t(e n )) of R n, where e 1, e 2,..., e n are the standard basis elements for R n. We call this isomorphism T. Then f : V R n is a basis, the map f T = T (f ) 1 : V R n is a basis for V called the dual basis to that of f. Then (f T ) T = f. In particular V and V have the same dimension. If v V a vector space, then v gives an element v of (V ) = Hom(V, R) by the formula: v (α) = α(v), for any α V Then the map V (V ), v v is an injection and is an isomorphism if V is finite dimensional. Let V and W be vector spaces of dimensions n and m respectively. Let s : V R n and t : W R m be bases. Then if f Hom(V, W), define µ(f) = t f s 1. Then µ(f) Hom(R n, R m ) so is represented by a matrix. If {s i, i = 1, 2,... n} are the basis elements of V for the basis s and {t j, j = 1, 2,... m} are the basis elements of W for the basis t, then we have: f(s i ) = m j=1 f j i t j, where f j i is the matrix of µ(f). If v V has s-co-ordinate vector v, then f(v) has t-co-ordinate vector w, where w i = n j=1 f jv i j. 3

Tensor algebras Let V be a real vector space. The tensor algebra of V, denoted T (V), is the associative algebra with identity over the reals, spanned by all the monomials of length k, v 1 v 2... v k, for all integers k, where each v i lies in V, subject to the relations of V: If av + bw + cx = 0, with v, w and x in V and a, b and c real numbers, then aαvβ + bαwβ + cαxβ = 0, for any tensors α and β. If a tensor is a linear combination of monomials all of the same length k, the tensor is said to be of type k. The vector space of tensors of type k is denoted T k (V). We allow k = 0, in which case the tensor is just a real number. The tensors of type one are naturally identified with the vector space V itself. If µ : W V is a homomorphism of vector spaces, then there is a unique algebra homomorphism T (µ) : T (W) T (V), which reduces to µ when acting on W. Then µ maps each monomial w 1 w 2... w k to µ(w 1 )µ(w 2 )... µ(w k ), for any w 1, w 2... w k in W and any integer k. If also λ : X W is also a vector space homomorphism, then we have: T (µ λ) = T (µ) T (λ). Using the standard basis elements {e i, i = 1, 2,..., n} of R n, the tensor algebra T (R n ) has a natural basis given by the monomials e j1 e j2... e jk where 1 j r n, for r = 1, 2,..., k and any k N, together with the number 1, the natural basis for tensors of type zero. Then every tensor of type k in T (R n ) can be written uniquely: T = T i 1i 2...i k e i1... e ik. Here the Einstein summation convention is used: repeated indices are summed over. If λ : V R n is a basis, so an isomorphism, then T (λ) gives an isomorphism of T (V) with T (R n ). In particular, every tensor T of type k in T (V) has a unique expression: T = T i 1i 2...i k f i1... f ik, where f i = λ 1 (e i ), i = 1, 2,..., n. The vector space T k (V) has dimension n k, for each nonnegative integer k. Finally, it is sometimes necessary, for clarity, to use a notation for the tensor product operation: then T U is written T U. 4

Covariant and contravariant tensors Let V be a vector space of dimension n with dual space V. The full tensor algebra of V is the sub-algebra of the tensor algebra T (V V ) generated by monomials v i1 v i2... v ik such that each v i belongs either to V or to V. If a monomial is a product of p elements of V with q elements of V, then the tensor is said to be contravariant of type p and covariant of type q and of type ( p. It is traditional to quotient out this tensor algebra by the relations: T vαu = T αvu, for any tensors T and U and and v V and any α V. So the relative ordering of elements of V vis à vis elements of V in any tensor monomial is immaterial. If λ : V W is a homomorphism of vector spaces, then the adjoint of λ, denoted λ is the map λ : W V given by the formula: λ (β)(v) = β(λ(v)), for any v V. If λ is an isomorphism, then so is λ and then the sum λ (λ ) 1 is an isomorphism, so induces an isomorphism T (λ (λ ) 1 ) of T (V V ) with T (W W ). In particular, if λ : V R n is a basis, so an isomorphism, every tensor T of T (V V ) of type ( p has a unique expression: T = T i 1i 2...i p f i1... f ip f j 1... f jq. Here the vectors f i V, i = 1, 2,..., n are determined by the formula, λ(f i ) = e i, i = 1, 2,... n and the dual vectors f j V, j = 1, 2,..., n are determined by the relations f j (v) = λ(v) j, for any v V, or, equivalently, by the duality relations f j (f i ) = δ j i. The quantities T i 1i 2...i p are called the components of T. If U is another tensor, of type ( ) r s, with components U k 1 k 2...k r m 1 m 2...m s, then the tensor T U has components: (T U) i 1i 2...i p+r +s = T i 1i 2...i p U i p+1i p+2...i p+r j q+1 j j+2...j q+s. 5

Tensor contraction and multilinear maps If τ = v 1 v 2... v p α 1 α 2... α q is a tensor monomial in T (V V ) of type ( p, with v i, i = 1, 2,..., p in V and α j, j = 1, 2,... q in V, then the ( k l) contraction of τ, denoted C ( k (τ), is the tensor: l) C ( k l) (τ) = α l(v k )v 1 v 2... v k 1 v k+1... v p α 1 α 2... α l 1 α l+1... α q. Then C k l) extends to give a linear map from tensors of type ( p to tensors of type ( ( p 1 q. This map is called the trace map over the index pair k l). If T has components T i 1i 2...i p the trace C ( k l) (T ) has components: T i 1i 2...i k 1 mi k+1...i p It follows that if S is a tensor of type ( ) p+q i p+q, with components S 1 i 2...i p+q j 1 j 2...j p+q, then the tensor C ( 1 q+ C ( q+... C 1 ( q+ C 1 ( (S) (with p contractions) has components: S m 1m 2...m pi 1...i q q+ 1 j 1 j 2 j qm 1 m 2...m p. Then the tensor: C ( 1 C ( 1... C ( 1 C ( q+... C 1 ( q+ C 1 ( (S), with q contractions of the kind C ( q+ 1 1 and p contractions of the kind C (, has components: q+ 1 S m 1m 2...m pn 1...n q n 1 n 2 n qm 1 m 2...m p. In particular if T is of type ( ) p i q, with components T 1 i 2...i p and U is a tensor of type ( ) q k p, with components U 1 k 2...k q l 1 l 2...l p, the complete contraction of T and U is: T.U = C ( 1 C ( 1... C ( 1 C ( q+... C 1 ( q+ C 1 ( 1 q+ (T U) = T m 1m 2...m p n 1 n 2...n q j 1 j 2 j l 1 mj l+1...j q. U n 1n 2...n q m 1 m 2...m q. This complete contraction renders the vector spaces of tensors of types ( p q and ( q p) dual to each other. In particular, consider the special case that U is a monomial: U = v 1 v 2... v q α 1 α 2... α p, for some vectors v j V, j = 1, 2,... q and some co-vectors, α k V, k = 1, 2,..., p. Then put: T (α 1, α 2... α p, v 1, v 2..., v q ) = T.(v 1 v 2... v q α 1 α 2... α p ). Then this gives a well-defined map from (V ) p V q to the reals, which is linear in each variable, called a multilinear map. This gives a natural isomorphism of the space of tensors of type ( p with the space of such multilinear maps. ) 6

Tensors of type ( 1 and endomorphisms Let M be a tensor of type ( 1, so an element of V V. We have M = M i jf i f j, in terms of basis elements {f i, i = 1, 2,..., n} of V and dual basis elements {f j, j = 1, 2,..., n} of V. We call M i j the matrix of M. Then M has many interpretations: As a linear operator on tensors N of type ( 1 : N M.N Here M.N is the complete contraction. In terms of components, M.N = MjN i j i matrix for M times the matrix for N. = tr(mn), the trace of the As a bilinear map from V V to the reals. This is the special case when N is a monomial vα, with v V and α V : M(v, α) = M.(vα). In terms of components this is α i M i jv j. This is just the matrix product αmv, where α is regarded as a row matrix and v as a column matrix. Here v = v i f i and α = α j f j. As an endomorphism of V. This is the ( 2 contraction of the tensor product Mv. Equivalently it is the map: v M(v, ). In terms of components this is v i Mjv i j. This is just the matrix product Mv, where v = v i f i is regarded as a column vector As an endomorphism of V. This is the ( 1 contraction of the tensor product αm. Equivalently it is the map: α M(, α). In terms of components this is α j α i Mj. i This is just the matrix product αmv, where α = α j f j is regarded as a row vector. Finally the trace of M is the number M.I = Mi i, where I is the identity endomorphism of V. The trace itself is a linear operator on ( 1 tensors, so as such is represented by a ( 1 tensor, called the Kronecker delta tensor, the identity endomorphism, whose trace I.I = n is just the dimension of the vector space V. 7

Tensors and non-commutative polynomials A polynomial f(x) in n-variables x i, i = 1, 2,..., n is a linear combination of monomials x i1 x i1... x ik where 1 i j n, for each j = 1, 2,..., k and for any k, with real coefficients. Usually one assumes that these variables commute, x i x j = x j x i, for any i and j. Then polynomials form a commutative associative algebra under multiplication. However it is also possible to consider non-commutative polynomials where we make no assumption that the variable commute. This is natural, for example, when we consider polynomials in matrices. If now V is a vector space, we can associate to V a variable x which takes values in the dual vector space, V, such that to each vector v we have the associated polynomial x(v). If v = v i f i, with {f i, i = 1, 2,..., n} a basis for V, we have x(v) = v i x i, where x i = x(f i ), i = 1, 2,..., n is a list of n-variables. Then by addition and associative multiplication, every tensor of T (V) may be thought of as a polynomial T (x) in the variable x, such that if it has type p and its components are T i 1i 2...i p, then the corresponding polynomial is T (x) = T i 1i 2...i p x i1 x i2... x ip. If T is thought of as a multilinear map, T (α 1, α 2,..., α p ), with α i in V, the non-commutative polynomial of T is just: T (x) = T (x, x,..., x). Then the full tensor algebra of contravariant and covariant tensors based on the vector space V may be regarded as all polynomials T (x, y), where x takes values in V, y takes values in V and x commutes with y (but the x-variables do not commute with each other, nor do the y-variables commute with each other. If T (x, y) is of type ( p so has p x-variables and q y-variables, so can be written T (x, y) = T (x, x,..., x, y, y,... y) with p x-variables and q y-variables, denote by λ k the operation of removing the k-th x-variable, so we have: λ k (T )(x, y) = T (x, x,..., x,, x,..., x, y, y,..., y), which gives a vector-valued polynomial. Similarly, denote by λ l the operation of removing the l-th y- variable, so we have: λ l (T )(x, y) = T (x, x,... x,..., x, y,..., y,, y,..., y), which gives a co-vector-valued polynomial. Then λ l k = λl λ k = λ k λ l removes the k-th x-variable and the l-th y-variable. This gives a polynomial, denoted λ l k (T ), with values in V V, the tensors of type ( 1, given by the formula: λ l k(t )(x, x,..., x, y, y,..., y) = T (x, x,... x,, x,..., x, y,..., y,, y,..., y). 8

In terms of components we have: T (x, y) = T i 1i 2...i p x i1 x i2... x ip y j 1 y j 2... y jq, λ k (T )(x, y) = T i 1i 2...i p x i1 x i2 x ik 1 x ik+1... x ip y j 1 y j 2... y jq, λ l (T )(x, y) = T i 1i 2...i p x i1 x i2... x ip y j 1 y j 2... y jl 1 y jl+1... y jq, λ l k(t )(x, y) = T i 1i 2...i p x i1 x i2... x ik 1 x ik+1... x ip y j 1 y j 2... y jl 1 y jl+1... y jq. Abstractly, since elements of V V have a trace, we can take the trace of λ l k (T )(x, y). In terms of variables this is: tr(λ l k(t ))(x, y) = T i 1i 2...i k 1 mi k+1...i p j 1 j 2...j l 1 mj l+1...j q x i1 x i2... x ik 1 x ik+1... x ip y j 1 y j 2... y jl 1 y jl+1... y jq. We see that this operation is the non-commutative polynomial version of the tensor trace operation, C ( k l). 9

From tensors to the symmetric and Grassmann algebras Let T be a tensor algebra based on a real vector space V of dimension n. Let V have basis {f i, i = 1, 2,... n}, with dual basis {f j, j = 1, 2,..., n}. Then the elements of T may be regarded as polynomials in a non-commuting variable x which takes values in V. Contracting with the basis elements, the variable x gives rise to the variables {x i = x(f i ), i = 1, 2,..., n}, so that we have x = x i f i. If T is a tensor of type ( k 0) the corresponding polynomial is: T (x, x,..., x) = T i 1i 2...i k x i1 x i2... x ik. Here T i 1i 2...i k = T (f i 1, f i 2,... f i k ) are the components of T in the basis {f i }. The multiplication of tensors is associative but not commutative. We can force the multiplication to be commutative, if we like, by simply requiring that x i x j = x j x i, for all i and j. This gives the symmetric tensor algebra ST, which is just an ordinary associative and commutative algebra of polynomials in n variables. Equivalently, we can take any tensor and symmetrize it: for s a permutation of the numbers 1, 2,..., p, and T is a tensor of type p, so has p co-vector arguments, denote the tensor with its arguments switched according to the permutation s by s(t ). There are p! such permutations, forming the symmetric group S p. Then we define: S(T ) = (p!) 1 s S p s(t ), S(T ) S(U) = S(T U). This converts the tensor algebra into the symmetric tensor algebra. Similarly, we can force the multiplication to be skew-commutative, if we like, by simply requiring that x i x j = x j x i, for all i and j. This gives the exterior or Grassmann algebra ΩT of T, which is associative. If T (x) and U(x) are of types ( p 0) and ( q 0), we have the graded commutation rule: T (x)u(x) = ( pq U(x)T (x). Equivalently we can take any tensor T of type p and skew-symmetrize it: for s a permutation of the numbers 1, 2,..., p, denote by sgn(s) its sign. Then we define: Ω(T ) = (p!) 1 s S p sgn(s)s(t ), Ω(T ) Ω(U) = Ω(T U). This converts the tensor algebra into the Grassmann tensor algebra. 10