Vector Spaces

Vector spaces

We fix a field F. A vector space, V, over the field F, is a set V, equipped with:

• An additive operation + : V × V → V, denoted (v, w) → v + w, for any v and w ∈ V, which makes V into an abelian group with additive identity denoted 0 ∈ V and additive inverse of v ∈ V denoted −v, for each v ∈ V.

• An operation of scalar multiplication F × V → V, denoted (λ, v) → λv = vλ, for each λ ∈ F and each v ∈ V, such that, for any v and w in V and any λ and µ in F, we have the relations: 1v = v, 0v = 0, (−1)v = −v,

λ(v + w) = λv + λw, (λ + µ)v = λv + µv, λ(µv) = (λµ)v = µ(λ)v.

Examples of vector spaces

• The trivial vector space over a ﬁeld F is a set with one element, denoted 0, with the operations 0 + 0 = 0 and λ0 = 0, for each λ ∈ F.

• The ﬁeld F is a vector space over itself, with its usual operations.

• Let S be a set. Then the space FS of all maps from S to F has the natural structure of a vector space, via the formulas, valid for each f : S → F, g : S → F and for each λ and µ in F:

(λf + µg)(s) = λf(s) + µg(s), for each s ∈ S.

FS is called the free vector space over S.

1 Constructions of vector spaces

Subspaces

Given a vector space V, and a subset X of V, if X is invariant under the operations of V, we may restrict these operations to X and then X becomes a vector space in its own right, called a subspace of V.

To verify that X ⊂ V qualiﬁes as a subspace, one need only show that λx + µy ∈ X, for any x and y in X and any λ and µ in F. In general a linear combination of vectors of X is a vector x of the form k x = Σi=1λixi, for some xi ∈ X, i = 1, 2, . . . , k, where λi ∈ F, i = 1, 2, . . . , k, for some positive integer k. Then X is a subspace if and only it contains all linear combinations of vectors of X.

The zero element of any vector space constitutes a subspace, called the trivial subspace.

The intersection of any family of subspaces of a vector space V is a subspace. If a set X is a subset of a vector space V, then the intersection of all subspaces of V containing X is called [X], the subspace spanned by X. Then [[X]] = [X] and X = [X] if and only if X is a subspace. The space [X] consists of all possible linear combinations of the vectors of X. Every vector space is spanned by itself. The dimension of a non-trivial vector space V is the small- est cardinality of a set that spans it. The trivial vector space is assigned dimension zero. A space V is called ﬁnite dimensional if it is spanned by a ﬁnite set: it then has dimension zero if and only if it is trivial and otherwise its dimension is a positive integer.

If X = {x1, x2, . . . xn} ⊂ V, then [X] is written [x1, x2, . . . , xn]. Then [X] has dimension at most n and has dimension n if and only if the vectors xj, j = 1, 2, . . . , n are linearly independent:

n Σj=1λjxj = 0 with λj ∈ F, for each j = 1, 2, . . . , n, if and only if λj = 0, for j = 1, 2, . . . , n. The vectors are linearly dependent if and only if they are not linearly inde- n pendent, if and only if there is a relation 0 = Σj=1λjxj = 0, with all λi ∈ F and at least one λk non-zero, if and only if the dimension of [X] is less than n.

2 If a collection of vectors {x1, x2, . . . , xn} ⊂ V spans a subspace X, then either X = V, or one can enlarge this collection to a list {x1, x2, . . . , xn, xn+1} ⊂ V, such that X ⊂ [x1, x2, . . . , xn+1] 6= X: the new vector xn+1 ∈ V just has to be chosen to lie in the complement of X in V. Then if the {x1, x2, . . . , xn} are linearly independent, so span a space of dimension n, so are {x1, x2, . . . , xn+1} and they span a space of dimension n+1.

The collection of all n-dimensional subspaces of V is called the Grassmannian of all n-planes in V, Gr(n, V). If V has dimension m, then Gr(n, V) is a space of dimension n(m − n).

Each one-dimensional subspace of V is a set [x] of the form [x] = {λx : λ ∈ F} for some 0 6= x ∈ V. We have [x] = [y], for 0 6= y ∈ V if and only if x and y are linearly dependent. The space of all one-dimensional subspaces of V is called P(V) = Gr(1, V), the projective space of V.

Each two-dimensional subspace of V is a set [x, y] of the form [x, y] = {λx + µy : λ ∈ F, µ ∈ F} for some linearly independent vectors x and y in V. a b We have [x, y] = [p, q] if and only if there is an invertible matrix, , c d with coeﬃcients in F, such that p = ax + by, q = cx + dy, if and only if each x and y are linearly independent, as are p and q, but any three of the four vectors x, y, p and q are linearly dependent.

A subset X of V is a subspace if and only if [x, y] ⊂ X, for any x and y in X.

3 The direct sum of vector spaces

Let S and X be sets equipped with a surjective map π : X → S, such that −1 Xs = π ({s}) ⊂ S has the structure of a vector space over F, for each s ∈ S.

A section of π is a map f : S → X, such that π ◦ f = idS. Then the space Γ of all sections of π becomes a vector space by ﬁbrewise addition and scalar multiplication: for any f and g in Γ, any λ ∈ F and each s ∈ S: (f + g)(s) = f(s) + g(s), (λf)(s) = λf(s).

We write Γ = ⊕s∈SXs, called the direct sum of the family of vector spaces Xs. An element f of Γ is written f = ⊕s∈Sf(s). The support of a section f ∈ Γ is the set of all s ∈ S such that f(s) = 0. The collection of all elements of Γ with ﬁnite support is a subspace of Γ, called the restricted direct sum of the family of spaces {Xs : s ∈ S}. These two notions of direct sum coincide when S is ﬁnite.

S In particular if Xs = F for each s ∈ S, then Γ = F .

Homomorphisms

Let V and W be vector spaces. Then a homomorphism or linear map, T from V to W is a set map T from V to W, respecting the vector space structure:

T (λx + µy) = λT (x) + µT (y), for any x and y in V and any λ and µ in F.

Then the image T (X) of a subspace X of V under T is a subspace of W, since if p and q lie in T (X) and if λ and µ are in F, then x and y in X exist, with T (x) = p and T (y) = q, so then z = λx + µy lies in X, since X is a subspace, so T (z) = λT (x) + µT (y) = λp + µq ∈ T (X), so T (X) satisﬁes the subspace criterion. In particular the image of V itself is a subspace. • T is a monomorphism if and only if T is injective. • T is a epimorphism if and only if T is surjective. • T is an isomorphism if and only if T is invertible, if and only if T is both an epimorphism and a monomorphism. In this case the inverse map T −1 lies in Hom(W, V).

4 • T is called an endomorphism in the case that V = W and an automorphism if T is an invertible endomorphism. The automorphisms of V form a group, under composition, called GL(V).

The collection of all homomorphisms, T : V → W itself naturally forms a vector space denoted Hom(V, W).

If V, W and X are vector spaces, then the composition map Hom(V, W) × Hom(W, X), (f, g) → g ◦f, for any f ∈ Hom(V, W) and any g ∈ Hom(W, X), is bilinear (i.e. linear in each argument).

In the particular case that W = F, we put V∗ = Hom(V, F). Then V∗ is called the dual space of V. There is a natural homomorphism J from V → (V∗)∗ given by the formula, valid for any v ∈ V and any α ∈ V∗: (Jv)(α) = α(v).

If V is ﬁnite dimensional, then J is an isomorphism.

A vector space V over F is ﬁnite dimensional if and only if any one of the following three conditions is met:

• There is an epimorphism from FP to V, for some ﬁnite set P

• There is a monomorphism from V to FQ, for some ﬁnite set Q.

• There is an isomorphism from FS to V for some ﬁnite set S.

5 Matrix representations for linear transformations

If S is a ﬁnite set put δs(t) = 0, if S 3 t 6= s and δs(s) = 1. S S Then δs ∈ F and for any f ∈ F , we have the unique expression f = P f(s)δ , so δ = {δ : s ∈ } forms a basis for S called the canonical s∈S s S s S F basis. T If T is also a ﬁnite set and the canonical basis for F is T = {t : t ∈ T}, then a homomorphism M from FS to FT is given by the formulas: X M(δs) = tMts, for each s ∈ S, t∈T

X X S M(f) = tMtsf(s), for any f ∈ F , t∈T s∈S X S M(f)(t) = Mtsf(s), for any f ∈ F and any t ∈ T. s∈S Here the matrix entry Mts is in F, for each s ∈ S and t ∈ T. Next, if U is another ﬁnite set and N a homomorphism from FT to FU, where U the canonical basis of F is ζU = {ζu : u ∈ U}, then we have: X N(t) = ζuNut, u∈U X (N ◦ M)(δs) = ζu(NM)us, u∈U X (NM)us = NutMts. t∈T This formula gives by deﬁnition the matrix product of the matrices repre- senting M and N.

If e is an isomorphism of FS and a vector space V and f is an isomorphism T of F and a vector space W, then the set of vectors E = {es = e(δs): s ∈ S} forms a basis for V, whereas the set of vectors F = {ft = f(t): t ∈ T} forms a basis of W. Then the map M : Hom(V, W) → Hom(FS, FT) given by the composition: M(T ) = f −1 ◦ T ◦ e, for any T ∈ Hom(V, W) is an isomorphism of vector spaces.

6 For T ∈ Hom(V, W) its image M(T ) is called the matrix of T with respect to the the bases e and f. We have the formula: X T (es) = ft(M(T ))ts, for any s ∈ S and any T ∈ Hom(V, W). t∈T

If e0 and f 0 are also isomorphisms of FS and FT with V and W, respectively, then we have:

M 0(T ) = (f 0)−1 ◦ T ◦ e0 = ((f 0)−1 ◦ f) ◦ f −1 ◦ T ◦ e(◦e−1 ◦ e0) = FM(T )E−1.

Here F = (f 0)−1 ◦ f is an automorphism of FT and E = (e0)−1 ◦ e, whose matrix representatives are invertible.

7 Tensor products

Let X be a vector space over a ﬁeld F. The tensor algebra of X, denoted T (X), is the associative algebra over F, generated by X and F, such that the multiplication operations of the algebra are bilinear.

More explicitly, we may define a word w in X of length k, a nonnegative integer, to be an ordered (k + 1)-tuple w = {λ, w1, w2 . . . , wk}, where λ ∈ F and each wi, i = 1, 2, . . . , k lies in X. If also x = {µ, x1, x2 . . . , xm} is a word of length m, the product z = wx is the word of length k + m, given by z = {ν, z1, z2, . . . , zm+k}, where ν = λµ ∈ F, zi = wi ∈ X, for 1 ≤ i ≤ k and zi = xi−k ∈ X, for k + 1 ≤ i ≤ k + m. Then we have (xy)z = x(yz), for all words x, y and z in X. The tensor algebra T (X) is then the span of all the words, subject to the conditions than the multiplication of words is bilinear in its arguments and to the requirement that if w = {λ, w1, w2 . . . , wk} is a 0 0 0 0 0 word and w = {λ , w1, w2 . . . , wk} is another word of the same length, then 0 0 0 k 0 w = w if wi = λiwi, for all i and some λi ∈ F, where λ Πi=1λi = λ. Note that the word w = {λ, w1, w2 . . . , wk} is then the (ordered) product of the word {λ} of length zero and the words {1, wi} of length one. The words of length zero may be identified with the field F and the words of length one with the vector space X. Also note that any word with one or more entries zero gives the zero element of the tensor algebra. We abbrevi- ate the words w, x and z = wx as w = λw1w2 . . . wk, x = µx1x2 . . . xm and z = wx = λµw1w2 . . . wkx1x2 . . . xm.

By deﬁnition, for each non-negative integer k, T k(X) is the subspace of T (X) spanned by the words of length k. Then the algebra product maps T k(X)×T m(X) into T k+m(X), for any non-negative integers k and m (where T 0(X) = F and T 1(X) = X.

The full tensor algebra of X, is the quotient of T (X ⊕ X∗) by the relations αv = vα, for any v ∈ T (X) and any α ∈ T (X∗). In this algebra, the subspace spanned by the words αw, with α ∈ T q(X∗) and with w ∈ T p(X) is denoted p p r p+r Tq (X) and we have that the product maps Tq (X) × Ts (X) into Tq+s (X), for p any non-negative integers p, q, r and s. When X is ﬁnite dimensional, Tq (X) q ∗ and Tp (X ) are naturally isomorphic, for any nonnegative integers p and q. p The words spanning Tq (X) may be written λw1w2 . . . wpα1α2 . . . αq, where ∗ λ ∈ F, w1, w2, . . . , wp lie in X and α1, α2 . . . , αq lie in X .

8 If the vector space X has dimension n and if {ei, i = 1, 2, . . . , n} is a set of vectors of X that spans X, so forms a basis, then any tensor τ of type (p, 0) can be written uniquely:

n n n X X X i1i2...ip τ = ··· τ ei1 ei2 . . . eip .

i1=1 i2=1 ip=1

j ∗ j j If also {e , i = 1, 2, . . . , n} is the dual basis of X , so we have e (ei) = δi , for any 1 ≤ i ≤ n and 1 ≤ j ≤ n, then any tensor υ of type (0, q) can be written uniquely: n n n X X X j1 j2 jq υ = ··· υj1j2...jq e e . . . e .

j1=1 j2=1 jq=1 Also any tensor φ of type (p, q) can be written uniquely:

n n n n n n X X X X X X i i ...i φ = ··· ··· φ 1 2 p e e . . . e ej1 ej2 . . . ejq . j1j2...jq i1 i2 ip i1=1 i2=1 ip=1 j1=1 j2=1 jq=1

i i ...i Here each of the coeﬃcients τ i1i2...ip , υ and φ 1 2 p lies in and each j1j2...jq j1j2...jq F coeﬃcient may be chosen arbitrarily. In particular, it follows that the vector space of tensors of type (p, q) is isomorphic to Fr, where r = np+q.

Finally, if {As : s ∈ S} is a set of subspaces of X labelled by a totally ordered ﬁnite set S, then the tensor product ⊗s∈SA is the subspace of the tensor algebra of X spanned by the words formed from the ordered product of the elements of the image of maps x : S → X, such that x(s) ∈ As, for all s ∈ S.

9 The Kronecker delta and the trace

Let V and W be vector spaces over F. The vector space V∗ ⊗W acts naturally on V, such that for any α ∈ V∗, any v ∈ V and w ∈ W, we have: (α ⊗ w)(v) = α(v)w.

This gives an element of Hom(V, W). This induces a natural map, easily seen to be an endomorphism, from V∗ ⊗W to Hom(V, W). This map is an isomorphism if V is ﬁnite dimensional.

When V and W are both ﬁnite dimensional, the following spaces are all naturally isomorphic: • Hom(V, W) • V∗ ⊗ W • W ⊗ V∗ • Hom(W∗, V∗) . Also the dual space of this quartet of spaces has four corresponding versions: • Hom(W, V) • W∗ ⊗ V • V ⊗ W∗ • Hom(V∗, W∗) In particular, if V = W, we see that Hom(V, V) is its own dual. This entails that there is a natural map E : Hom(V, V) × Hom(V, V) → F, which is bilinear in its arguments. Speciﬁcally we have the formulas, valid for any α and β in V∗, v and w in V and any T in Hom(V, V): E(α ⊗ v, β ⊗ w) = β(v)α(w), E(α ⊗ v, T ) = α(T (v)). Then E is a non-degenerate symmetric bilinear form on Hom(V, V): for any S and T in Hom(V, V), we have E(S,T ) = E(T,S) and if S ∈ Hom(V, V), then E(S,W ) = 0, for all W ∈ Hom(V, V), if and only if S = 0.

10 There is a distinguished element of Hom(V, V), the identity operator δV, deﬁned by the formula, valid for any v ∈ V.

δV(v) = v.

Then δV is called the Kronecker delta tensor for V. Then δV gives a natural element of Hom(V, V), called trV, the trace, given by the formula, valid for any T ∈ Hom(V, V):

trV(T ) = E(δV,T ).

In particular we have, for any α ∈ V∗ and any v ∈ V the formula:

trV(α ⊗ v) = α(v).

Also we have, for any S and T in Hom(V, V):

trV(S ◦ T ) = trV(S ◦ T ). Finally we have a key formula:

trV(δV) = n.

Here n is the dimension of V, but here regarded as an element of F, not as a positive integer.

If X is a vector space with dual X∗, the full tensor algebra of X is the tensor algebra of X ⊕ X∗. It is usual to quotient this algebra by the relations αw = wα, for any word w in X and any word α in X∗. This gives the algebra of mixed tensors of X. If a word of this algebra is the concatenation of p letters form X and q from X∗, the word is said to be of type (p, q). The words of type (p, 0) are said to be contravariant, the words of type (0, q) co- p variant. Then Tq (X) is the subspace of the tensor algebra spanned by words p r p+r of type (p, q). Under the tensor product, we have Tq (X)⊗Ts (X) ⊂ Tq+s (X), 0 for any nonnegative integers p, q, r and s (Here T0 (X) = F). Note that p p 0 p q ∗ p p Tq (X) = T0 (X) ⊗ Tq (X) = T (X) ⊗ T (X ) and T0 (X) = T (X), whereas 0 q ∗ Tq (X) = T (X ), for all non-negative integers p and q.

11 For each positive integer s, denote by Ns the (totally ordered) set Ns = p {t ∈ N : 1 ≤ t ≤ s}. Now let τ be a word in Tq (X), so τ has the form

τ = x ⊗ α, where x = ⊗k∈Np xk and α = ⊗j∈Nq αj and each xk, k = 1, 2, . . . , p ∗ is in X, whereas each αj, j = 1, 2, . . . , q is in X . For some integer r with 1 ≤ r ≤ min(p, q), let there be given a bijection µ from a subset Ar of Np with r elements to a subset Br of Nq, also with r elements. Then deﬁne p−r trµ(τ) ∈ Tq−r (X) by the formula:

0 0 trµ(τ) = λx ⊗ α ,

0 p−r x = ⊗k∈Np−Ar xk ∈ T0 (X), 0 0 α = ⊗j∈Nq−Br αj ∈ Tq−r(X), r λ = Πi=1αµ(r)(xr) ∈ F. −1 Note that the maps µ and µ are not, in general, order preserving. Then trµ p p−r extends naturally to a linear map, still called trµ, from Tq (X) to Tq−r (X): it p is called the µ trace applied to Tq (X). In the special case that p = q = r = 1, the map µ is the identity and trµ equals the ordinary trace trX, deﬁned above.

Tensors in the one-dimensional case

Let L be a one-dimensional space over a ﬁeld F. Then the tensor algebra of L is abelian. For any positive integer r, denote by Lr the r-fold tensor product of L with itself, with, in particular, L1 = L. Then the dual space of L may be denoted L−1 and for any positive integer s, its s-fold tensor product with itself may be denoted L−s. Finally we put L0 = F. Then there is a natural isomorphism of Lr ⊗ Ls with Lr+s, for any integers r and s. If {e} is a basis for L, the induced basis for Lr is denoted {er}, for any integer r. In particular e0 = 1 and e−1 is dual to e, e−1e = 1 and {e−1} is a basis for L−1. Then a general tensor T can be written uniquely in the form:

r T = Σr∈Ztre .

Here the map Z → F, r → tr is required to be non-zero for only a ﬁnite number of integers r. The (associative, commutative) tensor multiplication of T with the tensor s U = Σr∈Zuse is the tensor:

n TU = UT = Σn∈Nvne , vn = Σr+s=ntrus.

12 The rank theorem for homomorphisms

Let V and W be ﬁnite dimensional vector spaces over F, with dimensions m and n, respectively and let T : V → W be a linear transformation. Associated to T is its image, T (V), which we have proved above to be a subspace of W. Also associated to T is its kernel ker(T ) = {v ∈ V : T (v) = 0}. If v and w are in ker(T ), so T (v) = T (w) = 0, then for any λ and µ in F, we have: T (λv + µw) = λT (v) + µT (w) = 0. So λv + µw ∈ ker(T ). So ker(T ) is a subspace of V. T is a monomorphism if and only if ker(T ) = 0. Let E2 = {eα, α ∈ S} form a basis of ker(T ) (where E2 is empty if ker(T ) = {0}. This has at most m elements, since any m + 1 vectors of V are linearly dependent. Since any n + 1 vectors of W are linearly dependent, the same is true of any subspace of W, so T (V) has ﬁnite dimension k, for some non-negative integer k ≤ n.

Let {f1, f2, . . . , fk} ∈ T (V) be a minimal spanning set for T (V) (regarded as the empty set if T = 0). Then by repeatedly adding new vectors fk+1, fk+2,... , whilst maintaining linear independence, we can extend to a list F = {f1, f2, . . . , fn}, yielding a basis for W. Let ei ∈ V obey the relations T (ei) = fi, for each i = 1, 2, . . . k. Pk Then if λi ∈ F obey the relation i=1 λiei = 0, applying T to this equation, we get: k ! k k X X X 0 = T (0) = T λiei = λiT (ei) = λifi. i=1 i=1 i=1

But the {fi; i ∈ Nk} are linearly independent in W, so all the λi must vanish, for i ∈ Nk. So the set E1 = {ei; i ∈ Nk} is a linearly independent set of vectors in V.

13 Now consider the union E of the sets E1 and E2. If Pk λ e + P µ e = 0, for some {λ : i ∈ } and {µ : α ∈ }, then i=1 i i α∈S α α i Nk α S applying T , we get the formula:

k ! X X 0 = T (0) = T λiei + µαeα i=1 α∈S

k X X = λiT (ei) + µαT (eα) i=1 α∈S k k X X X = λifi + µα(0) = λifi. i=1 α∈S i=1 But the {fi : i ∈ Nk} is linearly independent. So the λi, i = 1, 2, . . . k are all zero. Back substituting, we get: 0 = P µ e . α∈S α α But the {eα : α ∈ S} are linearly independent. So all the µα vanish, for each α ∈ S. So the set E is a linearly independent set. Next let v ∈ V. Then T (v) ∈ T (V), so there is an expression: k X T (v) = vifi, for some vi ∈ F, i = 1, 2, . . . , k. i=1 Pk Put w = v − i=1 viei ∈ V. Then we have: k ! k X X T w = T v − viei = T v − viT (ei) i=1 i=1

k X = T v − vifi = 0. i=1 So w ∈ ker(T ), so we have an expression w = P w e , for some {w ∈ α∈S α α α F : α ∈ S}. Then we have the formula: k k X X X v = w + viei = wαeα + viei. i=1 α∈S i=1

14 So the set E is linearly independent and spans V, so is a basis for V. In particular the cardinality of S is m − k and we have proved the formula:

ν(T ) + ρ(T ) = dim(V). Here ρ(T ), called the rank of T is the dimension of the image of T and ν(T ), called the nullity of T is the dimension of the kernel of T .

Now by relabeling the elements of E2, we may write E = {ei : i ∈ Nm}. Now we have the Rank Theorem in two versions:

• With respect to the bases E of V and F of W, the matrix M of T is:

n X T (ei) = fjMji, for any i ∈ Nm. j=1

Here we have T (ei) = 0, for i > k and T (ei) = fi, for 1 ≤ i ≤ k, so the matrix Mji has the block form:

I 0 k . 0 0

Here Ik is the k × k identity matrix.

• More geometrically, we can decompose V = V1 ⊕V2 and W = W1 ⊕W2, where T gives an isomorphism of V1 with W1 = T (V), both spaces of dimension k and T is zero on V2 = ker(T ). A corollary is that given any n × m-matrix M, there is an invertible n × n- I 0 matrix E and an invertible m × m matrix F , such that EMF −1 = k , 0 0 for some non-negative integer k ≤ min(m, n) and the number k, called the rank of the matrix M, is independent of the choice of the matrices E and F .

15 Monomorphisms, epimorphisms and morphisms We ﬁrst describe monomorphisms and epimorphisms, in term of each other.

• If V and W are vector spaces, a map L : V → W is a monomorphism if and only if there exists a vector space X and a linear map M : W → X, such that ML is an isomorphism.

• If W and X are vector spaces, a linear map M : W → X is an epimorphism if and only if there exists a vector space V and a linear map L : V → W, such that ML is an isomorphism.

Given vector spaces V, W and X, L ∈ Hom(V, W) and M ∈ Hom(W, X), we say that the pair (L, M) forms a dual pair if and only if ML is an isomorphism. If (L, M) is a dual pair, then L is necessarily a monomorphism and M is necessarily an epimorphism. So given L ∈ Hom(V, W), L is a monomorphism if and only if X and M ∈ Hom(W, X) exist such that (L, M) is a dual pair. Conversely, given M ∈ Hom(W, X), M is an epimorphism if and only if V and L ∈ Hom(V, W) exist such that (L, M) is a dual pair.

Note that if L is a monomorphism from V to W, then there is a dual pair (L, M) with M : W → X and ML = idV. Also if M is an epimorphism from W to V, then there is a dual pair (L, M) with L : V → W and ML = idV.

Now let T : X → Z be any homomorphism from the vector space X to the vector space Z. Then there exists a factorization T = LM, with M an epimorphism from X to a vector space Y and L a monomorphism from Y to Z. Also there exist a monomorphism S from Y to X and an epimorphism U from Z to Y, such that UTS is an isomorphism.

If T = LM = L0M 0 with M : X → Y, M 0 : X → Y0, L : Y → Z and L0 : Y0 → Z and with L and L0 monomorphisms, whereas M and M 0 are epimorphisms, then there is an isomorphism N : Y → Y0, such that L0N = L and M 0 = NM. In particular the dimension of Y is independent of all choices. It is called the rank of T and is bounded by the min(dim(X), dim(Z)), since if dim(Y) > dim(Z), there is no monomorphism from Y to Z and if dim(Y) > dim(X), there is no epimorphism from X to Y. The rank of T is equal to dim(X) if and only if T is a monomorphism and the rank of T is equal to dim(Z) if and only if T is an epimorphism.

16 Pictures We represent vectors and tensors by pictures:

This represents a vector of a vector space V. The line represents that the vector space in question is V. For another vector in the same space, we can use a diﬀerent shape:

Addition and scalar multiplication can be represented naturally:

2 + 3

Depending on the context it might be more appropriate to bring the lines together, to emphasize that they represent only a single vector:

2 + 3

Also, depending on the context it might be more appropriate to tie the scalars to the vector that they act on, in which case, if the lines are brought together, the plus sign becomes superﬂuous:

2 3

17 Alternatively, we can tie the scalar to the vector:

2 3

If we have a vector of another vector space W, we use a diﬀerent colour:

If we have an element of the dual vector space V∗, it is represented by a downward oriented line:

Objects without lines emerging from them represent scalars, or elements of the ﬁeld. So for example the dual pairing of a co-vector and a vector is represented by joining the two:

Tensors based on V and V∗ are represented by multiple lines: • A tensor of type (3, 0):

• A tensor of type (0, 3):

18 • A tensor T of mixed type (3, 3):

Tensors act on bunches of vectors and dual vectors, as appropriate to get scalars. So, for example, when the (3, 3) tensor T acts on three vectors v, w and x in V and three dual vectors α, β and γ in V∗, we get the scalar:

α β γ

v w x

This scalar, usually written T (v, w, x, α, β, γ) is linear in all seven arguments.

Sometimes we need to permute the arguments of tensors, so for example if we permute the ﬁrst and third vector arguments and cyclically permute the dual vector arguments of T , we can create the following tensor U:

U = T

In general the tensor U is not the same tensor as T , since its action on the same vector and dual vector arguments is diﬀerent.

The (associative) tensor product is just denoted by juxtaposition: • A tensor product T ⊗ U:

T U

19 Here the tensor product of the (3, 0) tensor T and the (2, 0) tensor U is the (5, 0) tensor T ⊗ U.

• A mixed tensor product T ⊗ W :

T W

This tensor product produces a tensor T ⊗ W of mixed type (3, 2).

• It is usual when considering mixed tensors, to assume that tensors based on V commute with those based on V∗, so we can also write T ⊗ W , without ambiguity as, for example:

• The contraction of tensors is represented by joining up the lines. So, for example, if we contract the third argument of T with the ﬁrst argument of W and the second argument of T with the second argument of W , we get a dual vector:

Notice in particular that the action of the (3, 3)-tensor T on the vectors v, w and x and the dual vectors α, β and γ given above is a complete contraction (meaning that the result is a scalar) of the tensor product T ⊗ α ⊗ β ⊗ γ ⊗ v ⊗ w ⊗ x.

20 An endomorphism L of a vector space is represented as a (1, 1) tensor:

This acts in many diﬀerent ways, every one linear in each argument, for example:

• Mapping a vector v to a vector L(v):

→ L v v

• Mapping a dual vector α to a dual vector α(L):

α α → L

• Mapping a vector v and a dual vector α to a scalar α(L(v)):

α ! α , v → L

21 • Mapping an endomorphism M to to a scalar tr(LM):

L M → M

The Kronecker delta tensor, the identity endomorphism, is just represented by a line, so we have the trace of L as:

L →

The trace of the Kronecker delta is n the dimension of the vector space, regarded as an element of the ﬁeld of scalars F, not as an integer:

n =

22 The elements of the direct sum of two (or more) vector spaces are represented by the merging of their colors to form a new color:

Here the blue/green and the yellow/green endomorphism are injections of the blue and yellow vector spaces into the green one, respectively.

An (L, M) pair, with L : V → W an injection and M : W → V a sur- jection, such that ML = idV is written as follows:

L = , M = ,

23 = .

A V basis for a vector space W is represented by a pair of isomorphisms, the basis and the dual basis:

, .

24 These are required to compose both ways to give the Kronecker delta:

= , = .

25 If the vector space V is Fn, the vectors of V and V∗ are ordered lists of n elements of F:

−5, 20, −10, −35, 10 −5 8, −32, 16, 56, −16 8 4 1, −4, 2, 7, −2 = 4, −16, 8, 28, −8 . −3 −3, 12, −6, −21, 6 6 6, −24, 12, 42, −12

Here the 5 by 5 matrix is called the Kronecker product of the constituent vectors. The dual pairing (also the trace of the matrix) is then:

1, −4, 2, 7, −2

−5 = 1(−5) + (−4)(8) + 2(4) + 7(−3) + (−2)(6) = −62. 8 4 −3 6

26 The sign of a permutation

For n ∈ N, the permutation group Sn is the group of all bijections from Nn to itself, with the group multiplication given by composition. Then Sn has n! elements. Note that S1 has just one element, the identity element.

For n ≥ 2, put D(x) = Πipolynomial D(x), so by 2 the pigeonhole principle, every factor of D(x) occurs exactly once in s(D)(x), so we arrive at the formula:

s(D)(x)) = (s)D(x), (s) = ±1.

If also t ∈ Sn and we put yi = xt(i), for 1 ≤ i ≤ n, then we get: s(D)(y) = (s)D(y) = (s)t(D)(x) = (t)(s)D(x)

= Πi

This shows that the map : Sn → Z2, s → (s) is a group homomorphism, where Z2 = {1, −1} is the multiplicative group with two elements, 1, the identity and −1, with (−1)2 = 1. Then (s) is called the sign of the permutation s. The permutation s is called even if (s) = 1 and odd if (s) = −1.

27 Next deﬁne the auxiliary polynomials, for any i and j with 1 ≤ i 6= j ≤ n:

i+j+1 Eij(x) = (−1) Π1≤k≤n,k6=i,k6=j((xi − xk)(xj − xk)),

Fij(x) = Π1≤k

Note that, by inspection, Eij(x) and Fij(x) are symmetric under the inter- change of i and j. Also Fij(x) is independent of the variables xi and xj. Then we have the formula, highlighting the dependence of the polynomial D(x) on the variables xi and xj, with i < j:

D(x) = (xi − xj)Eij(x)Fij(x).

If now s ∈ Sn satisﬁes s(i) = j, s(j) = i and s(k) = k, for all k with 1 ≤ k ≤ n, k 6= i and k 6= j, then we have:

s(D)(x) = (xs(i) − xs(j))Hij(x)Mij(x),

i+j+1 Hij(x) = (−1) Π1≤k≤n,k6=i,k6=j((xs(i) − xs(k))(xs(j) − xs(k))) i+j+1 = (−1) Π1≤k≤n,k6=i,k6=j((xj − xk)(xi − xk)) = Eij(x),

Mij(x) = Π1≤k

s(D)(x) = (xs(i) − xs(j))Hij(x)Mij(x) = (xj − xi)Eij(x)Fij(x) = −D(x).

This proves that (s) = −1. The mapping s is called the (i, j) transposition and is denoted (ij). So the map : Sn → Z2 is surjective, when n ≥ 2 (and not when n = 1, since then S1 consists only of the identity element, which is deﬁned to be even (1) = 1). For n ≥ 2, the kernel of is called the alternating group An, the normal subgroup of Sn of all even permutations, a subgroup of index two in Sn, so with 2−1n! elements. Every element of the symmetric group is a product of transpositions, so the sign of the element is therefore positive or negative according to the evenness or oddness of the number of transpositions in the product, so the parity of the number of such transpositions is an invariant.

28 Acting on tensors, we can use permutations to rearrange the arguments of the tensor. This is achieved by acting on the tensor with suitable products of the Kronecker delta tensor. For example in the case of S3, its six elements are represented as:

(1) = , (231) = , (312) = ,

(23) = , (13) = , (12) = .

Here (231) denotes the cyclic permutation 1 → 2 → 3 → 1 and (312) denotes the cyclic permutation 1 → 3 → 2 → 1.

p In general, given a word w = fw1w2 . . . wp in T (X), for some ﬁxed non- negative integer p and given a permutation σ ∈ Sp, we deﬁne:

σ(w) = fwσ(1)wσ(2) . . . wσ(p).

If now ρ is another permutation in Sp, we have:

ρ(σ(w)) = fwρ(σ(1))wρ(σ(2)) . . . wρ(σ(p))

= fw(ρ◦σ)(1))w(ρ◦σ)(2)) . . . wρ◦σ)(p)) = (ρ ◦ σ)(w). Then the assignment w → σ(w) for each word extends naturally to give an endomorphism of T p(X), still denoted σ, such that we have ρ(σ(t)) = p (ρ ◦ σ)(t), for any permutations ρ and σ in Sp and any t ∈ T (X). So for p each p, we get a linear representation of the group Sp on the n dimensional vector space T p(X).

29 The symmetric tensor algebra

The symmetric tensor algebra S(X) of a vector space X is the algebra obtained from the tensor algebra T (X), by requiring that all the letters of all words commute, so the multiplication in this algebra is commutative and indeed is isomorphic to a standard commutative algebra of polynomials in n-variables over the ﬁeld F, when X has dimension n. So there is a surjective algebra homomorphism from T (X) to S(X), which maps each word to itself. For clarity the symmetric tensor product is often represented by the p p notation . The images of the n basis words ei1 ei2 . . . eip for T (X), where ei, i = 1, 2, . . . , n is a basis for X, gives the following basis words for the image p p S (X) of T (X) in S(X): ei1 ei2 · · · eip , for 1 ≤ i1 ≤ i2 · · · ≤ ip ≤ n. n + p − 1 This basis has elements. p There is a natural subspace of T (X) that maps isomorphically to the symmetric algebra: it is the subspace spanned by all words of the form fwp, for w ∈ X and f ∈ F. Explicitly there is an F-linear idempotent symmetrization map S : T (X) → T (X), which preserves tensor type and which for a given positive integer p, we describe in terms of the action of the symmetric group p p Sp on T (X) given above: For each tensor t ∈ T (X), we deﬁne: 1 X S(t) = σ(t). p! σ∈Sp

This gives an endomorphism of T p(X). We prove that it is idempotent:   1 X 1 X X 1 X X S2(t) = S σ(t) = ρ(σ(t)) = (ρ ◦ σ)(t) p!  p! p! σ∈Sp ρ∈Sp σ∈Sp ρ∈Sp σ∈Sp

1 X X 1 X X X = (ρ ◦ (ρ−1 ◦ σ))(t) = σ(t) = σ(t) = S(t). p! p! ρ∈Sp σ∈Sp ρ∈Sp σ∈Sp σ∈Sp

Here we used the fact that as σ runs once through all permutations in Sp, −1 so does ρ ◦ σ, for each ﬁxed permutation ρ in Sp. The image of S is then called the subspace of symmetric tensors. On this space we can deﬁne the (associative) symmetric tensor product by the formula A B = S(AB), for any symmetric tensors A and B. This product makes the space of symmetric tensors into an algebra naturally isomorphic to the algebra S(X).

30 For example, given a tensor T in T 3(X), we can represent its symmetrization S(T ) diagrammatically as follows:

= + + 6S(T ) T T T

+ + + . T T T

Just like any algebra of polynomials, the algebra S(X) carries a natural collection of derivations (or ﬁrst order derivative operators). A derivation D of S(X) is an endomorphism of S(X), obeying the rules:

• Dc = 0, for any c ∈ F = S0(X),

• D(f + g) = Df + Dg, for any f and g in S(X),

• D(fg) = (Df)g + (Dg)f, for any f and g in S(X).

Then D is determined by its action on S1(X) = X, which is an arbitrary linear map from X to S(X), so D is determined by an element of S(X) ⊗ X∗. If D and E are derivations, so is their bracket [D,E] = DE − ED and this bracket obeys all the usual rules of a Lie bracket (see the section on the Lie derivative along a vector ﬁeld). Indeed we have a natural linear derivative operator denoted ∂, which takes S(X) to X⊗S(X) and is determined uniquely by the derivative formula, valid for any t ∈ X and any positive integer p: tp tp−1 ∂ ⊗ = t ⊗ . p! (p − 1)!

Alternatively the operator ∂ is determined by the formal power series formula (obtained by summing the previous formula over p):

∂ ⊗ et = t ⊗ et.

31 Equivalently, we have, acting on the word w = fw1w2 . . . wp, with f ∈ F and each wi ∈ X, the derivative formula:

∂⊗(fw1w2w3 . . . wp) = fw1⊗w2w3 . . . wp+fw2⊗w1w3 . . . wp+fw3⊗w1w2 . . . wp+···+fwp⊗w1w2w3 . . . wp−1.

Then the general derivation is given by D.∂, where D lies in S(X) ⊗ X∗. Finally, we can compute multiple derivatives, which (since they commute with each other) give natural maps from S(X) to S(X)⊗S(X); so for example, we have (when p ≥ k) and for any t ∈ X: tp tp−k ∂k ⊗ = tk ⊗ , p! (p − k)!

∂k ⊗ et = tk ⊗ et, e∂ ⊗ et = et ⊗ et.

The skew-symmetric tensor algebra: the Grassmann or exterior algebra

The skew-symmetric tensor algebra Ω(X) of a vector space X, also called its Grassmann algebra, or its exterior algebra, is the algebra obtained from the tensor algebra, T (X), by requiring that all the individual letters of all words anti-commute, so the multiplication in this algebra is graded commutative: an element is even if it is a sum of words each with an even number of letters and is odd if it is a sum of words each with an odd number of letters. Then even elements commute with everything, whereas odd elements commute with even and anti-commute with each other. In particular, the square of any odd element is zero. There is then a surjective algebra homomorphism from T (X) to Ω(X), which maps each word to itself. For clarity, the skew-symmetric product is often represented by the notation ∧. The image p p of the n basis words ei1 ei2 . . . eip for T (X), where {ei, i = 1, 2, . . . , n} is a basis for X, gives the following basis words for the image Ωp(X) of T p(X) in n Ω( ): e ∧ e ∧ · · · ∧ e , for 1 ≤ i < i ··· < i ≤ n. This basis has X i1 i2 ip 1 2 p p elements. In particular the Grassmann algebra of X is ﬁnite dimensional, of n X n total dimension = 2n. p p=0

32 There is a natural subspace of T (X) that maps isomorphically to the Grass- mann algebra. Explicitly there is an F-linear idempotent skew-symmetrization map Ω : T (X) → T (X), which preserves tensor type and which for a given positive integer p, we describe in terms of the action of the symmetric group p p Sp on T (X) given above. For each tensor t ∈ T (X), we deﬁne: 1 X Ω(t) = (σ)σ(t). p! σ∈Sp

Here (σ) is the sign of the permutation σ. This gives an endomorphism of T p(X). We prove that the endomorphism Ω idempotent: for any ﬁxed positive integer p and any t ∈ T p(X), we have:   1 X 1 X X Ω2(t) = Ω (σ)σ(t) = (ρ) (σ)ρ(σ(t)) p!  p! σ∈Sp ρ∈Sp σ∈Sp

1 X X 1 X X = (ρ ◦ σ)(ρ ◦ σ)(t) = (ρ ◦ (ρ−1 ◦ σ))(ρ ◦ (ρ−1 ◦ σ))(t) p! p! ρ∈Sp σ∈Sp ρ∈Sp σ∈Sp 1 X X X = (σ)σ(t) = σ(t) = Ω(t). p! ρ∈Sp σ∈Sp σ∈Sp Here we used the fact is a group homomorphism and that as σ runs once −1 through all permutations in Sp, so does ρ ◦ σ, for each ﬁxed permutation ρ in Sp. The image of Ω is then called the subspace of skew-symmetric tensors. On this space we can deﬁne the skew-symmetric tensor product by the formula A ∧ B = Ω(AB), for any skew-symmetric tensors A and B. This product makes the space of skew-symmetric tensors into an algebra naturally isomorphic to the algebra Ω(X).

33 For example, given a tensor T in T 3(X), we can represent its skew-symmetrization Ω(T ) diagrammatically as follows:

= + + 6Ω(T ) T T T

− − − . T T T

Just like the symmetric algebra, the Grassmann algebra Ω(X) carries a natural derivative operator. A derivation D of Ω(X) is an endomorphism of Ω(X), obeying the rules: • Dc = 0, for any c ∈ F = Ω0(X), • D(f + g) = Df + Dg, for any f and g in Ω(X), • D(fg) = (Df)g + (−1)pq(Dg)f, for any f ∈ Ωp(X) and g ∈ Ωq(X) and any non-negative integers p and q. Then D is determined by its action on Ω1(X) = X, which is an arbitrary linear map from X to Ω(X), so D is determined by an element of Ω ⊗ X∗.A derivation D is said to have integral degree k if it maps Ωp(X) to Ωp+k(X), for each non-negative integer p. If D is a derivation of degree p and E a derivation of degree q, so is their bracket [D,E] = DE − (−1)pqED and this bracket obeys all the usual rules of a graded Lie bracket (see the section on diﬀerential forms). Indeed we have a natural linear derivation of degree minus one, denoted δ, which takes Ω(X) to X ⊗ Ω(X) and is determined uniquely by the derivative formula acting on any word w = fw1w2 . . . wk, with f ∈ F and each wi ∈ X, the derivative formula: k−1 δ⊗(fw1w2w3 . . . wk) = fw1⊗w2w3 . . . wk−fw2⊗w1w3 . . . wk+fw3⊗w1w2 . . . wk+···+(−1) fwk⊗w1w2w3 . . . wk−1. Then the general derivation is given by D.δ, where D lies in Ω(X) ⊗ X∗. Finally, we can compute multiple derivatives, which (since they anti-commute with each other) give natural maps from Ω(X) to Ω(X) ⊗ Ω(X). In particular if X has dimension n all derivatives of order n + 1 or more vanish identically.

34 Determinants

Consider a vector space X with dual space X∗, defined over a field F, with dimension n, a positive integer. Then X acts on Ω(X∗) the exterior algebra ∗ of X by the derivation δv, defined for each v ∈ X. The operator δv is the derivation of degree minus one, which kills Ω0(X∗) = F and acting on Ω1(X∗) = X∗ obeys the relation, valid for any v ∈ X and α ∈ X∗:

δv(α) = α(v). Consider now the vector space L = Ωn(X∗), a vector space over F of dimension 1, so isomorphic to F. Given n-vectors, v1, v2, . . . , vn in X and ∈ L, the following quantity is a well-deﬁned element of F:

D(v1, v2, . . . , vn)() = δvn δvn−1 . . . δv2 δv1 .

This quantity is linear in all (n+1)-variables (v1, v2, . . . , vn, ). In particular, −1 the quantity D(v1, v2, . . . , vn) gives an element of L the dual space of L (so L−1 is also a vector space of dimension one). Then, from its deﬁnition, D(v1, v2, . . . , vn) is linear in each argument and totally skew, since the derivations δv, for v ∈ X pairwise anti-commute and each is linear in v.

Let {e = e1} be a basis of L (so e is any non-zero element of L), with dual −1 −1 −1 basis e of L , so we have e (e) = 1. Then there is a basis {e1, e2, . . . , en} for X, with dual basis {e1, e2, . . . , en}, a basis of X∗, such that e = e1e2 . . . en. Then we have: 1 2 n −1 D(e1, e2, . . . , en)(e) = δen δen−1 . . . δe2 δe1 (e e . . . e ) = 1 = e (e). −1 So we have the formula D(e1, e2, . . . en) = e . This, in turn, yields the general formula: −1 D(v1, v2, . . . , vn) = dete(v1, v2, . . . , vn)e = dete(v1, v2, . . . , vn)D(e1, e2, . . . en).

Here the quantity dete(v1, v2, . . . , vn) is an element of F, and is linear in each vector argument, v1, v2, . . . , vn and totally skew. Then dete is called the determinant function, relative to the basis {e} of L. Under the replacement e → f, where {f} is also a basis for L we have detf = r dete, where f = re. The argument above shows that when the vectors {vi, i = 1, 2, . . . , n} are linearly independent, so constitute a basis, then D(v1, v2, . . . , vn) is non-zero. Conversely, it is easy to see that if the vectors vi, i = 1, 2, . . . , n are linearly dependent, then D(v1, v2, . . . , vn) is zero. So we have that dete(v1, v2, . . . , vn) vanishes if and only if the vectors {vi, i = 1, 2, . . . , n} are linearly dependent.

35 ∗ Consider now n-elements α1, α2, . . . , αn of the space X . Then we have:

α1α2 . . . αn = E(α1, α2, . . . , αn) ∈ L.

Here the quantity E(α1, α2, . . . , αn) is linear in each argument and totally skew. The arguments given above, show that E(α1, α2, . . . , αn) vanishes if and only if the {αi, i = 1, 2, . . . , n} are linearly dependent. Further we may write E(α1, α2, . . . , αn) = dete−1 (α1, α2, . . . , αn), where dete−1 (α1, α2, . . . , αn) ∈ F is linear in each argument and totally skew.

Now consider the element of F:

D(v1, v2, . . . , vn)E(α1, α2, . . . , αn) = dete(v1, v2, . . . , vn)dete−1 (α1, α2, . . . , αn)

n = δvn δvn−1 . . . δv2 δv1 (α1α2 . . . αn) = Σσ∈Sn (σ) Πi=1(ασ(i))(vi) . Note that this relation is symmetrical between the vector space X and its dual X∗:

D(v1, v2, . . . , vn)E(α1, α2, . . . , αn) = δvn δvn−1 . . . δv2 δv1 (α1α2 . . . αn) = δαn δαn−1 . . . δα2 δα1 (v1v2 . . . vn)

n n = Σσ∈Sn (σ) Πi=1vi(ασ(i)) = Στ∈Sn (τ) Πi=1αj(vτ(j)) = det(αi(vj)). Also this relation embodies the fact that the one-dimensional vector spaces Ωn(X) and Ωn(X∗) are naturally dual.

36 Volume In a Euclidean space, of finite dimension, denote the inner product of vectors v and w by v.w. If we represent the elements of the space by column vectors, we can also T T represent this inner product by w.v = w v = v√w = v.w√. T Then the length Lv of a vector v is Lv = |v| = v.v = v v. Here T applied to a matrix denotes its transpose. Then Lv is well-defined and is zero if and only if v = 0. The Cauchy-Schwarz inequality is the inequality |v.w| ≤ |v||w|, with equality if and only if the vectors v and w are linearly dependent. The real angle θ, with 0 ≤ θ ≤ π, between non-zero vectors v and w is given by the formula: |v||w| cos(θ) = v.w. The Cauchy-Schwarz inequality implies that θ is well-defined. We say that (possibly zero) vectors v and w are orthogonal or perpendicular, if v.w = 0. The orthogonal group is the group of linear transformations v → L(v), preserving length, so for any vector v we have: |L(v)| = |v|, vT v = vT LT Lv. This gives the defining relation for elements of the orthogonal group: LLT = LT L = I. Here I denotes the identity transformation. A reflection is a linear transformation of the form:

v → Tn(v) = v − 2(v.n)n, |n| = 1,

T T Tn = I − 2n ⊗ n , n n = 1. If v is any vector we can write v uniquely in the form v = a + b, where a is perpendicular to the unit vector n and b is parallel to n. Explicitly we have b = (v.n)n and a = v − (v.n)n. Then the reflection Tn acts so that Tn(v) = a − b. There are then two basic facts about the orthogonal group: • The group is generated by reflections. • The parity (even/odd) of the number of reflections whose product is a given element of the orthogonal group is independent of the choice of the reflections producing that element.

37 More precisely, from the relation AAT = I, we infer that det(A)2 = 1, so det(A) = ±1. Then every reflection Tn has det(Tn) = −1 and an orthogonal transformation A is the product of an even number of reflections if and only if det(A) = 1 and is otherwise the product of an odd number of reflections. The orthogonal transformations of unit determinant form a subgroup of the orthogonal group, called the rotation group.

If now v and w are vectors, they deﬁne a parallelogram in the Euclidean space. It has area Av,w given by the formula:

p 2 Av,w = (v.v)(w.w) − (v.w) .

The Cauchy-Schwarz inequality gives that Av,w is well-deﬁned and Av,w = 0 if and only if v and w are linearly dependent, precisely the case where the parallelogram collapses onto a line. More generally, if P = {v1, v2, . . . vk} is a collection of vectors of the space, then the volume VP of the parallelepiped determined by them is: q VP = det(vi.vj).

One can prove that VP is well-deﬁned and zero if and only if the vectors of P are linearly dependent.

We can remove the dependence on the standard basis of Euclidean space, by passing to a real vector space V with a symmetric bilinear form g : V×V → R, that is positive deﬁnite: for any 0 6= v ∈ V, we have g(v, v) > 0. If now P = {v1, v2, . . . vk} is a collection of vectors of the space V, then the volume VP of the parallelepiped determined by them is: q VP = det(g(vi, vj)).

Again VP is well-deﬁned and vanishes if and only if the elements of P are linearly dependent.

38 The metric of V induces a metric still called g on the tensor algebra of V, such that if v1v2 . . . vk and w1w2 . . . wm are words of the algebra, then their inner product is zero, unless k = m and when k = m we have:

k g(v1v2 . . . vk, w1w2 . . . wk) = Πj=1g(vi, wi). This in turn induces inner products on the symmetric tensor algebra, such that in the case of the symmetric algebra we have:

g(ev, ew) = eg(v,w).

Equivalently, we have, for each nonnegative integer k:

g(vk, wk) = k!g(v, w)k.

Then for example, we have: 1 g(v v , w w ) = g((v + v )2 − (v − v )2, (w + w )2 − (w − w )2) 1 2 1 2 16 1 2 1 2 1 2 1 2 1 = g(v + v , w + w )2 − g(v + v , w − w )2 − g(v − v , w + w )2 + g(v − v , w − w )2 8 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 = (g(v +v , w +w )−g(v +v , w −w ))(g(v +v , w +w )+g(v +v , w −w )) 8 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 − ((g(v −v , w +w )−g(v −v , w −w ))(g(v −v , w +w )+g(v −v , w −w )) 8 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 = (g(v + v , w )g(v + v , w ) − g(v − v , w )g(v − v , w )) 2 1 2 1 1 2 2 1 2 1 1 2 2

= g(v1, w1)g(v2, w2) + g(v2, w1)g(v1, w2). The right-hand-side is called a permanent, a variation of the formula for the determinant, where all signs in the formula are taken to be positive. In general, we could use the Laplace or Fourier transform to compute the inner products and hence permanents.

In the case of the exterior algebra, we have:

g(v1v2 . . . vk, w1w2 . . . wk) = det(g(vi, wj)).

2 Note in particular that g(v1v2 . . . vk, v1v2 . . . vk) = det(g(vi, vj)) = VP . Here VP is the volume of the parallelepiped determined by the vectors v1, v2, . . . , vk.

39 Determinants and linear transformations

Let X and Y be vector spaces (over the same ﬁeld F) of the same dimension n n n. Put ωX = Ω (X) and ωY = Ω (Y). These are each one-dimensional vector spaces. Let L : X → Y be a linear transformation. Then L induces a natural linear mapping called T (L) mapping the tensor algebra of X to that of Y, such that if x = fx1x2 . . . xk is a word of T(X), then we have T (L)(x) = fL(x1)L(x2) ...L(xk). This mapping is compatible with the passage to the exterior algebra, so we induce a natural map called Ω(L) mapping Ω(X) to Ω(Y). This mapping preserves degree, so, in particular, in the highest dimension, it induces a natural map, denoted det(L) from the line ωX to the line ωY. Clearly this map is functorial: if M : Y → Z is also a linear map with Z a vector space of dimension n over F, then we get det(M ◦ L) = det(M) ◦ det(L). Also we have that det(cL) = cn det(L), for any c ∈ F and det(L) 6= 0, if and only if L is an isomorphism. If K and L are linear maps from X to Y, their relative characteristic polynomial is the polynomial det(sK +tL) and is a homogeneous polynomial of total degree n, of the form sn det(K) + ··· + tn det(L). Of interest also is the restriction of Ω(L) to the n-dimensional space Ωn−1(X). This is called the adjoint mapping of L, and is denoted adj(L):Ωn−1(X) → Ωn−1(Y). Note that we have the formulas: n−1 n X ∧ Ω (X) = Ω (X), n−1 L(w) ∧ adj(L)(α) = det(L)(w ∧ α), for any w ∈ X and any α ∈ Ω (X). s

In the special case that Y = X, so L is an endomorphism of X, then det(L) is a multiple of the identity isomorphism of ωX, by an element of F, so may be regarded canonically as an element of F. Then det gives a homomorphism from the space of endomorphisms of X under composition to the ﬁeld F under multiplication, whose restriction to the group of isomorphisms of X is a group homomorphism to the group F∗ of non-zero elements of F under multiplication. In particular we have det(idX) = 1. Also the characteristic polynomial det(s idX + tL) is a homogeneous polynomial in the variables s and t of total degree n, of the form sn + tr(L)sn−1t + ··· + tn det(L).