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Home , Exterior algebra, Linear map, Multilinear algebra, Tensor product

MULTILINEAR NOTES

EUGENE LERMAN

Contents 1. (Multi) 1 1.1. products 1 1.2. The Grassmann (exterior) algebra and alternating maps 7 1.3. Pairings 9

1. (Multi)linear algebra The goal of this note is to define , and Grassmann (exterior) algebra. Unless noted otherwise all vector spaces are over the and are finite dimensional. There are two ways to think about tensors: (1) tensors are multi-linear maps; (2) tensors are elements of a “tensor ” of two or more vector spaces. The first way is more concrete. The second is more abstract but also more powerful. 1.1. Tensor products. We start by reviewing multi-linear maps.

Definition 1.1. Let V1,...,Vn and U be vector spaces. A n factors z }| { f :V1 × · · · × Vn→ U, (v1, . . . , vn) 7→ f(v1, . . . , vn)

is multi-linear if for each fixed index i and a fixed (n − 1)-tuple of vectors v1, . . . , vi−1, vi+1, . . . , vn the map

Vi → U, w 7→ f(v1, . . . , vi−1, w, vi+1, . . . , vn) is linear. When the number of factors is n, as above, we will also say that f is n-linear.

n factors 2 z n }| n{ For example, if we identify Rn ' R × · · · × R by thinking of an n × n as an n-tuple of column vectors, then the n factors z n }| n{ det :R × · · · × R → R, (v1, . . . , vn) 7→ det(v1| ... |vn) is an n-. Here is an example of a . Any inner product on a V : V × V 3 (v, w) 7→ v · w ∈ R

is bilinear. There is no standard notation for the space of n-linear maps from V1 × · · · × Vn to U. We will denote it by Mult(V1 × · · · × Vn,U) = Multn(V1 × · · · × Vn,U)

(n is to indicate that these are n-linear maps). This space, Mult(V1 × · · · × Vn,U), is a vector space: any of two n-linear maps is n-linear. We now take a closer look at the space of bilinear maps Mult2(V × W, U). This case is complicated enough to understand what happens with multi-linear maps in general, but simple enough not to bog down in notation.

typeset February 21, 2011. 1 ∗ ∗ Lemma 1.2. Let {vi}, {wj} and {uk} denote the bases of V , W and U respectively and {vi }, {wj } and ∗ {uk} the corresponding dual bases. Then the maps k k ∗ ∗ φij : V × W → U, φij(v, w) = vi (v)wj (w)uk

are bilinear and form a of Mult2(V × W, U). Hence

dim Mult2(V × W, U) = dim V dim W dim U. k Proof. It is easy to see that φij are bilinear. Next, for any b ∈ Mult2(V × W, U), any w ∈ W and any v ∈ V , X ∗ X ∗ b(v, w) = b( vi (v)vi, wj (w)wj) X ∗ ∗ = vi (v)wj (w)b(vi, wj) i,j X ∗ ∗ ∗ = vi (v)wj (w)uk(b(vi, wj))uk i,j,k X ∗ k = uk(b(vi, wj))φij(v, w). i,j,k k ∗ Hence the maps φij span Mult2(V × W, U). Also, the collection of numbers uk(b(vi, wj)) uniquely determine k the b. Hence φij’s are linearly independent.  We now turn to the definition of the V ⊗ W [pronounced “V tensor W ”] of two vector spaces V and W . Informally it consists of finite linear combinations of symbols v ⊗ w, where v ∈ V and w ∈ W . Additionally, these symbols are subject to the following identities:

(v1 + v2) ⊗ w − v1 ⊗ w − v2 ⊗ w = 0

v ⊗ (w1 + w2) − v ⊗ w1 − v ⊗ w2 = 0 α (v ⊗ w) − (αv) ⊗ w = 0 α (v ⊗ w) − v ⊗ (αw) = 0,

for all v, v1, v2 ∈ V , w, w1, w2 ∈ W and α ∈ R. These identities simply say that the map ⊗ : V ×W → V ⊗W , (v, w) 7→ v ⊗ w, is a bilinear map. The fact that everything in V ⊗ W is a linear combination of symbols v ⊗ w means that the of the map ⊗ : V × W → V ⊗ W spans V ⊗ W .1 Here is the formal definition of the tensor product of two vector spaces. Definition 1.3. A tensor product of two finite dimensional vector spaces V and W is a vector space V ⊗ W together with a bilinear map ⊗ : V × W → V ⊗ W ,(v, w) 7→ v ⊗ w2 such that for any bilinear map b : V × W → U there is a unique linear map ¯b : V ⊗ W → U with ¯b(v ⊗ w) = b(v, w). That is, the diagram

b V × W / U ; ⊗ ¯b  V ⊗ W commutes. The existence of the map ¯b satisfying the above conditions is called the of the tensor product. This definition is quite abstract. It is not clear that such objects exist and, if they exist, that they are unique. Setting the question of existence and uniqueness of tensor products aside, let’s us sort out the relationship between V ⊗ W and bilinear maps Mult(V × W, U). Recall that Hom(X,Y ) is the space of all linear maps from a vector space X to a vector space Y and is itself a vector space. Lemma 1.4. Assume that V ⊗ W exists. Then there is a canonical Hom(V ⊗ W, U) −→' Mult(V × W, U).

1But the image of ⊗ is not all of V ⊗ W . The elements in the image are called decomposable tensors. 2The symbol v ⊗ w stands for the value of the map ⊗ on the pair (v, w) 2 Proof. The isomorphism in question is built into the definition of the tensor product. Given a linear map A : V ⊗ W → U the composition A ◦ ⊗ : V × W → U is bilinear. And conversely, given a bilinear map b ∈ Mult(V × W, U) there is a unique linear map ¯b : V ⊗ W → U so that (¯b ◦ ⊗)(v, w) = b(v, w) for all (v, w) ∈ V × W . In other words the maps Hom(V ⊗ W, U) 3 A 7→ A ◦ ⊗ ∈ Mult(V × W, U) and Mult(V × W, U) 3 b 7→ ¯b ∈ Hom(V ⊗ W, U) are inverses of each other.  Next we observed that the uniqueness of the tensor product is also built into the definition of the tensor product. Proposition 1.5. If tensor products exist, they are unique up to isomorphism. Proof. The proof is quite formal and uses nothing but the universal property. Suppose there are two vector spaces V ⊗1W and V ⊗2W with corresponding bilinear maps ⊗1 : V ×W → V ⊗1W and ⊗2 : V ×W → V ⊗2W which satisfy the conditions of the Definition 1.3. We will argue that these vector spaces are isomorphic. By the universal property there exist a unique linear map ⊗1 : V ⊗2 W → V ⊗1 W so that the diagram

⊗1 V × W / V ⊗1 W 8

⊗2 ⊗  1 V ⊗2 W

commutes. By the same argument, switching the roles of ⊗1 and ⊗2, there is a unique linear map ⊗2 : V ⊗1 W → V ⊗2 W making the diagram

⊗2 V × W / V ⊗2 W 8

⊗1 ⊗  2 V ⊗1 W commute. Define

T1 = ⊗1 ◦ ⊗2 : V ⊗1 W → V ⊗1 W

T2 = ⊗2 ◦ ⊗1 : V ⊗2 W → V ⊗2 W. These are linear maps making the diagrams

⊗1 ⊗2 V × W / V ⊗1 W and V × W / V ⊗2 W 8 8

⊗1 ⊗2 T1 T2   V ⊗1 W V ⊗2 W

commute. But the identity maps idi : V ⊗i W → V ⊗i W , i = 1, 2, are linear and also make the respective diagrams commute. By uniqueness Ti = idi. Hence ⊗1 and ⊗2 are inverses of each other and provide the desired .  Now we construct the tensor product as a quotient of an infinite dimensional vector space by an infinite dimensional subspace thereby proving its existence. Proposition 1.6. Tensor products exist. Proof. Let V and W be two finite dimensional vector spaces. We want to construct a new vector space V ⊗ W and a bilinear map ⊗ : V × W → V ⊗ W satisfying the conditions of Definition 1.3. We start with a vector space F (V × W ) made of formal finite linear combinations of ordered pairs (v, w), v ∈ V , w ∈ W . Its basis is the {(v, w) | v ∈ V, w ∈ W } = V × W . If you prefer you can think of F (V × W ) as the set of functions

{f : V × W → R | f(v, w) 6= 0 for only finitely many pairs (v, w)}. 3 This set of functions is an infinite dimensional vector space. Its basis consists of functions that take value 1 on a given pair (v0, w0) and 0 on all other pairs. It’s tempting to call this (v0, w0). The vector space F (V × W ) is called the free vector space generated by the set V × W . Note that we have an inclusion map ι : V × W → F (V × W ), ι(v, w) = (v, w). It is not bilinear since (v1 + v2, w) 6= (v1, w) + (v2, w) in F (V,W ). Consider the smallest subspace K of F (V,W ) containing the following collection of vectors:   (v1 + v2, w) − (v1, w) − (v2, w)    (v, w1 + w2) − (v, w1) − (v, w2)  S = v, v1, v2 ∈ V, w, w1, w2 ∈ W and c ∈ R c(v, w) − (cv, w)    c(v, w) − (v, cw),  In other words, consider the subspace K of F (V ×W ) spanned by the set S. Define V ⊗W to be the quotient of F (V × W ) by K: V ⊗ W := F (V × W )/K. Define the map ⊗ : V × W → V ⊗ W to be the composite of the inclusion ι : V × W,→ F (V × W ) and the quotient map F (V × W ) → F (V × W )/K. The definition of K is rigged precisely so that this composite is bilinear. We write v⊗w for the value of ⊗ on the pair (v, w). By construction the set {v⊗w | (v, w) ∈ V ×W } spans V ⊗ W [but it’s much too big to be a basis]. We check that the map ⊗ : V ×W → V ⊗W has the required universal property. Suppose b : V ×W → U is bilinear. Since V × W is a basis for F (V × W ), b defines a unique linear map ˜b : F (V × W ) → U given on the basis by ˜b((v, w)) = b(v, w). As b is bilinear, ˜b is 0 on K by the definition of K. Thus we obtain a linear map b : F (V × W )/K = V ⊗ W → U with b(v ⊗ w) = ˜b((v, w)) = b(v, w). Since the vectors of the form v ⊗ w span V ⊗ W , b is unique. This verifies the universal property and thereby proves the existence of the tensor product.  Lemma 1.7. For any vector spaces V and W dim(V ⊗ W ) = dim V · dim W. Proof. ∗ dim V ⊗ W = dim(V ⊗ W ) = dim Hom(V ⊗ W, R) = dim Mult(V × W, R) by Lemma 1.4 = dim V · dim W · dim R.  We are now in position to quickly prove a number of results about tensor products.

Corollary 1.7.1. If {vi} and {wj} are a basis of V and W respectively, then {vi ⊗ wj} is a basis of V ⊗ W .

Proof. Since the vectors of the form v ⊗ w, v ∈ V , w ∈ W , span V ⊗ W , the much smaller set {vi ⊗ wj} also 3 spans V ⊗ W . But dim(V ⊗ W ) = dim V · dim W is precisely the number of elements in the set {vi ⊗ wj}. Hence the set {vi ⊗ wj} is a basis.  Lemma 1.8. V ⊗ W is isomorphic to W ⊗ V . Proof. Consider the map b : W × V → V ⊗ W defined by b(w, v) = v ⊗ w. Since b is bilinear, there is a unique linear map b : W ⊗ V → V ⊗ W with b(w ⊗ v) = v ⊗ w. Since the set {v ⊗ w | v ∈ V, w ∈ W } generates V ⊗ W , the map b is surjective. It is an isomorphism by count.  Lemma 1.9. V ∗ ⊗ W is isomorphic to Hom(V,W ).

3 We are using here the fact that for any (v, w) ∈ V × W , the tensor v ⊗ w is a linear combination of vi ⊗ wj ’s. 4 Proof. Consider b : V ∗ × W → Hom(V,W ) defined by (b(v∗, w))(v) = v∗(v)w for all v∗ ∈ V ∗, v ∈ V, w ∈ W. Since b is bilinear, it induces a linear map b : V ∗ ⊗ W → Hom(V,W ) with (b(v∗ ⊗ w))(v) = v∗(v)w for all v∗ ∈ V ∗, v ∈ V, w ∈ W. Observe that linear maps of the form v 7→ v∗(v)w span Hom(V,W ) (The proof of this fact is very similar to the proof of Lemma 1.2 and is left as an exercise). Hence b is an isomorphism by dimension count. 

∗ Exercise 1.1. Show that if {vi} is a basis of a vector space V , {vi } the and {wj} the basis of a ∗ vector space W , then {vi (·)wj} is a basis of Hom(V,W ). Lemma 1.10. If A : V → W and B : V 0 → W 0 are two linear maps, then there is a unique linear map A ⊗ B : V ⊗ V 0 → W ⊗ W 0 such that (A ⊗ B)(v ⊗ w) = A(v) ⊗ B(w) for all (v, w) ∈ V × V 0. Proof. Consider b : V × W → V 0 ⊗ W 0 given by b(v, w) = Av ⊗ Bw. The map b is bilinear, whence the universal property gives us a unique linear map ¯b : V ⊗ W → V 0 ⊗ W 0 with ¯b(v ⊗ w) = Av ⊗ Bw

for all (v, w) ∈ V × W . 

Exercise 1.2. Show that if A : V → W is represented by a matrix (aij) with respect to some bases of V 0 0 0 0 and W and B : V → W is represented by a matrix (bkl) with respect to bases of V and W , then A ⊗ B is represented by the matrix (aijbkl) with respect to the appropriate bases.

' Exercise 1.3. Show that there is a natural isomorphism φ : V ∗ ⊗ W ∗ → Mult(V × W, R) with φ(v∗ ⊗ w∗)(v, w) = v∗(v)w∗(w) for all v∗, w∗, v, w. Show that there is a natural isomorphism ψ : V ∗ ⊗ W ∗ → (V ⊗ W )∗ with ψ(v∗ ⊗ w∗)(v ⊗ w) = v∗(v)w∗(w) for all v∗, w∗, v, w.

' Exercise 1.4. Show that the map R × V → V ,(a, v) 7→ av gives rise to an isomorphism R ⊗ V → V which sends a ⊗ v to av for all a ∈ R and v ∈ V . Exercise 1.5. Show that taking tensor product is associative: V ⊗ (U ⊗ W ) ' (V ⊗ U) ⊗ W for any three vector spaces V,U and W . From now on we write V ⊗ U ⊗ W for V ⊗ (U ⊗ W ) since the order of taking tensor products doesn’t matter. Exercise 1.5 above also allows us to define recursively tensor powers of a vector space V . We define

⊗0 V := R, V ⊗1 := V and V ⊗n := V ⊗(n−1) ⊗ V for n > 1. It is not hard to generalize the relationship between bilinear maps and tensor products to the relationship between n-linear maps and n-fold tensor products. For example: 5 Exercise 1.6. Prove that given a n-linear map

n z }| { f :V × · · · × V → U, then there exists a unique linear map f : V ⊗n → U with

f(v1 ⊗ · · · ⊗ vn) = f(v1, . . . , vn). for all (v1, . . . , vn) ∈ V × · · · × V .

Moreover, given a ∈ V ⊗n and b ∈ V ⊗m, a ⊗ b is in V ⊗n ⊗ V ⊗m ' V ⊗(n+m). This gives us an R-bilinear map, V ⊗n × V ⊗m → V ⊗(n+m), (a, b) 7→ a ⊗ b. Note that if n = 0 the map above is simply

⊗m ⊗m R × V → V , (a, t) 7→ at. (cf. Exercise 1.4).

Definition 1.11. An algebra over R is a vector space A together with a bilinear map A×A → A,(a, a0) 7→ aa0 (“multiplication”). An algebra A is said to be an algebra with unity if there is an element 1 ∈ A such that 1 · a = a for all a ∈ A. An algebra A is associative if the multiplication is associative.

Remark 1.12. Note that in any algebra A, 0a = a0 = 0 for all a ∈ A (this is because multiplication is required to be bilinear).

Remark 1.13. If A is an algebra with 1 then there is an injection R → A, x 7→ x1. We will always identify R with its image in A. Example 1.14. A is an algebra. It is not associative and does not have 1 (why not?).

Example 1.15. The space Mn(R) of n × n matrices forms an algebra under . It is an algebra with unity: the I is the unity.

Definition 1.16. An algebra A is graded if

∞ X A = Ai i=0 and if for any a ∈ Ai and b ∈ Aj we have a · b ∈ Ai+j. We will refer to the elements of Ak as elements of degree k.

Given a vector space V we construct the corresponding tensor algebra T (V ) as follows. As a vector space T (V ) is the direct sum:

∞ ⊗2 ⊗n X ⊗i T (V ) = R ⊕ V ⊕ V ⊕ · · · ⊕ V ⊕ · · · = V . i=0

⊗ij Thus the elements of T (V ) are finite sums ai1 + ai2 + ··· aik , aij ∈ V . We define the multiplication on T by extending the multiplication

V ⊗n × V ⊗m → V ⊗(n+m) (a, b) 7→ a ⊗ b. bilinearly to all of T (V ). The tensor algebra T (V ) of a vector space V is a graded with 1. Note that by construction the elements of T (V ) are sums of products of elements of V , that is, T (V ) is generated by V . 6 1.2. The Grassmann (exterior) algebra and alternating maps. We have seen that tensor products are intimately related to multi-linear maps. Exterior (Grassmann) are just as intimately related to alternating multi-linear maps. Recall that an n-linear map f : V × · · · × V → U is alternating if it changes sign whenever we switch to adjacent entries:

f(v1, . . . , vi, vi+1, . . . , vn) = −f(v1, . . . , vi+1, vi, . . . , vn)

for all (v1, . . . , vn) ∈ V × · · · × V and any index i. Example 1.17. The determinant n factors z n }| n{ det :R × · · · × R → R, (v1, . . . , vn) → det(v1| ... |vn) is an alternating map. Example 1.18. Consider a vector space V and a, b ∈ V ∗. Define the bilinear map a ∧ b by

(a ∧ b)(v1, v2) := a(v1)b(v2) − a(v2)b(v1), v1, v2 ∈ V. The map a ∧ b (“a wedge b”) is alternating.

Definition 1.19 (Grassmann (exterior) algebra). Let V be a finite dimensional vector space over R. The Grassmann (exterior) algebra Λ∗(V ) is an algebra over R with unity together with an injective linear map i : V → Λ∗(V ) called the structure map which has the following universal property: If A is an algebra over R with unity and j : V → A is a linear map such that j(v) · j(v) = 0 for all v ∈ V , then there is a unique algebra map 4  :Λ∗(V ) → A such that the following diagram commutes: V OOO OOO j i OOO OOO   OOO Λ∗(V ) /' A. Proposition 1.20. If the Λ∗(V ) exists, it is unique (up to isomorphism).

Proof. This is a formal exercise and is left to the reader.  Proposition 1.21. For every vector space V the exterior algebra Λ∗(V ) exists. Proof. Let I be the two-sided in the tensor algebra T (V ) generated by the set {v ⊗ v : v ∈ V }. Note that R ∩ I = 0 and V ∩ I = 0 for degree reasons. Define Λ∗(V ) := T (V )/I, the quotient of the tensor algebra by the ideal I. Then Λ∗(V ) is an algebra — it inherits the multiplication from T (V ). The induced multiplication in Λ∗(V ) is denoted by ∧ (“wedge”). Since the tensor algebra is graded, so is I, and I = (I ∩ V ⊗2) ⊕ (I ∩ V ⊗3) ⊕ · · · Since V ∩ I = 0, the composite i : V → T (V ) → T (V )/I = Λ∗(V ) is an injection. Note that any element of Λ∗(V ) is a finite linear combination of products of elements of V . Now that we have constructed the exterior Λ∗(V ), let us prove the universal property. Suppose that A is an algebra and that we are given a linear map j : V → A with j(v) · j(v) = 0 for all v ∈ V . Consider the map b : V × V → A given by b(v, w) = j(v) · j(w). Since the map b is bilinear, there is a unique linear map j(2) : V ⊗ V → A with j(2)(v ⊗ w) = j(v) · j(w). Similarly, for all positive integers k, we have k-linear maps j(k) : V ⊗k → A with (k) j (v1 ⊗ · · · ⊗ vk) = j(v1) ··· j(vk). (0) In , we define j (a) = a · 1A, for all a ∈ R. In this way, we obtain an algebra map ˜ : T (V ) → A. By assumption, ˜(v ⊗ v) = 0 for all v ∈ V . Therefore ˜ vanishes on the ideal I. This implies that ˜ descends to an algebra map  :Λ∗(V ) = T /I → A with (v) = j(v) for all v ∈ V . Since an algebra map is uniquely ∗ determined on generators, and since V generated Λ (V ), the map  is unique. 

4 A map f : A → B between two algebras is an algebra map if f is linear and preserves multiplication: f(a1a2) = f(a1)f(a2) 7 Remark 1.22. For any v ∈ V , we have v ∧ v = 0 in the exterior algebra Λ∗(V ). Also,

0 = (v1 + v2) ∧ (v1 + v2) = v1 ∧ v1 + v1 ∧ v2 + v2 ∧ v1 + v2 ∧ v2 gives that v1 ∧ v2 = −v2 ∧ v1; That is, the wedge product is skew-commutative. Remark 1.23. Let Λk(V ) = T k(V )/(T k(V ) ∩ I). The vector space Λk(V ) is called the kth exterior power of V . Then ∞ X Λ∗(V ) = Λk(V ), k=0 where 0 1 Λ (V ) = R and Λ (V ) = V. Also, if α ∈ Λk(V ) and β ∈ Λl(V ), then α ∧ β ∈ Λk+1(V ). Thus, Λ∗(V ) is a graded algebra with 1.

Remark 1.24. We know that if {v1, . . . , vn} is a basis for V , then {vi ⊗ vj} is a basis for V ⊗ V . By ⊗k k ⊗k ⊗k induction, {vi1 ⊗ · · · ⊗ vik } is a basis for V . Thus, {vi1 ∧ · · · ∧ vik } generates Λ (V ) = V /(I ∩ V ). Since ∧ is skew-commutative, however, we can reduce this generating set to a smaller one:

(1.1) {vi1 ∧ · · · ∧ vik | i1 < ··· < ik}, This implies that Λl(V ) = 0 whenever l > dim V. We will see below that the set (1.1) is a basis of Λk(V ). We now investigate the between the k-th exterior power Λk(V ) of a vector space V and alternating maps. Proposition 1.25 (Universal property of k-th exterior power of a vector space). Let U and V be vector k z }| { spaces. If f :V × · · · × V → U is alternating then there is a unique linear map f :Λk(V ) → U with

f(v1 ∧ · · · ∧ vk) = f(v1, . . . , vk). ⊗k ˜ ⊗k ˜ Proof. By the universal property of V , there is a unique linear map f : V → U such that f(v1⊗· · ·⊗vk) = ∗ f(v1, . . . , vk). Since f is alternating, f = 0, where I is the ideal defined in the construction of Λ (V ). I∩V ⊗k k ⊗k ⊗k This gives us the linear map f :Λ (V ) = V /(I ∩ V ) → U with the desired property.  Corollary 1.25.1. The space of k-linear alternating maps {f : V × · · · × V → U | f is alternating} is isomorphic to the space Hom(Λk(V ),U). Lemma 1.26. Let V be an n-dimensional vector space. Then Λn(V ) is 1-dimensional. n n Proof. We may assume that V = R . Let e1, . . . , en be the . Then e1 ∧ · · · ∧ en spans Λ (V ). n n We need to show that e1 ∧ · · · ∧ en 6= 0. The determinant det : R × · · · × R → R is 1 on the identity matrix n n I = (e1| ... |en): det(e1| ... |en) = 1. Hence the induced linear map det : Λ (R ) → R is 1 on e1 ∧ · · · ∧ en. Therefore e1 ∧ · · · ∧ en 6= 0. 

Corollary 1.26.1. If {f1, . . . , fn} is a basis for a vector space V , then {fi1 ∧· · ·∧fik : 1 ≤ i1 < ··· < ik ≤ n} is a basis for its k-th exterior power Λk(V ). Proof. By Remark 1.24 the above set generates Λk(V ). So we only need to check independence. Suppose X 0 = ai1,...,ik fi1 ∧ · · · ∧ fik for some ai1,...,ik ∈ R i1<···

Pick a j1 < j2 < ··· < jk. Let jk+1 < ··· < jn be the remaining indices. Then X ( ai1,...,ik fi1 ∧ · · · ∧ fik ) ∧ fjk+1 ∧ · · · ∧ fjn

= aj1,...,jk fj1 ∧ · · · fjk ∧ fjk+1 ∧ · · · ∧ fjn , 8 since ai1,...,ik fi1 ∧ · · · fik ∧ fjk+1 ∧ · · · ∧ fjn = 0 whenever is = jr for some s, r. This gives aj1,...,jk = 0. Also,

fj1 ∧ · · · ∧ fjk ∧ fjk+1 ∧ · · · ∧ fjn = ±f1 ∧ · · · ∧ fn 6= 0. Hence fj1 ∧ · · · ∧ fjk 6= 0.  Corollary 1.26.2. For any finite dimensional vector space V  dim V  (dim V )! dim Λk(V ) = = . k k!(dim V − k)! Lemma 1.27. Let A : V → W be a linear map. Then there is a unique linear map Λk(A):Λk(V ) → Λk(W ) such that k (Λ (A))(v1 ∧ · · · ∧ vk) = Av1 ∧ · · · ∧ Avk for all v1, . . . , vk ∈ V . Proof. Consider the map b : V × · · · × V → Λk(W ) given by

b(v1, . . . , vk) = Av1 ∧ · · · ∧ Avk. Since ∧ is skew-commutative, b is an alternating map. By Proposition 1.25 there exists a unique linear map k k k Λ (A):Λ (V ) → Λ (W ) with the required properties.  Exercise 1.7. Let A : V → W be a linear map as above. Choose bases of V and W and the corresponding bases of Λk(V ) and of Λk(W ). Show that the entries of the matrix representing Λk(A) are in the entries of the matrix representing A. 1.3. Pairings.

Definition 1.28. Let V and W be two vector spaces. A pairing is a bilinear map h·, ·i : V × W → R. Example 1.29. Let V be a vector space and V ∗ be its dual. The evaluation map ∗ V × V → R h`, vi = `(v) is a pairing.

Definition 1.30. A pairing h·, ·i : V × W → R is non-degenerate if

hv0, wi = 0 ∀w ∈ W ⇒ v0 = 0

hv, w0i = 0 ∀v ∈ V ⇒ w0 = 0. Example 1.31. The evaluation map ∗ V × V → R h`, vi = `(v) is a non-degenerate pairing. In a sense it is the only nondegenerate pairing:

Proposition 1.32. If b : V × W → R is a nondegenerate pairing, then V ' W ∗ and W ' V ∗. # ∗ Proof. Consider b1 : V → W given by # (b1 (v))(w) = b(v, w). # The map b1 is linear, and # # ker b1 = {v0 ∈ V : b1 (v0) = 0} = {v0 ∈ V : b(v0, w) = 0 ∀w} = {0}. Thus dim V ≤ dim W ∗ = dim W . By the same argument, we have dim W ≤ dim V ∗ = dim V . Therefore # dim V = dim W . Hence b1 is an isomorphism. # ∗ By the same argument, b2 : W → V given by w 7→ b(·, w) is an isomorphism as well.  Proposition 1.33. There is a nondegenerate pairing k ∗ k h·, ·i :Λ (V ) × Λ (V ) → R with ∗ ∗  ∗  hv1 ∧ · · · ∧ vk, v1 ∧ · · · ∧ vki = det vi (vj) . Hence Λk(V ∗) ' (Λk(V ))∗. 9 k k z ∗ }| ∗{ z }| { Proof. Consider b :V × · · · × V × V × · · · × V → R given by   b(l1, . . . , lk, v1, . . . , vk) = det li(vj) .

∗ ∗ ∗ ∗ k For a fixed (l1, . . . , lk) ∈ V ×· · ·×V , b is alternating in the v’s. So there is a map b :(V ×· · ·×V )×Λ (V ) → with R   (l1, . . . , lk, v1 ∧ · · · ∧ vk) 7→ det li(vj) . k Similarly, for a fixed v1 ∧ · · · ∧ vk ∈ Λ (V ), b is alternating in the l’s, which means that there is a map ˜b :Λk(V ∗) × Λk(V ) → R with the desired property. To check non- evaluate the pairing on the respective bases.  Combining the proposition above with Corollary 1.25.1 we get:

Corollary 1.33.1. The space of k-linear alternating maps {f : V × · · · × V → R | f is alternating } is isomorphic to the k-th exterior power Λk(V ∗). k ∗ Remark 1.34. Explicitly `1 ∧ · · · ∧ `k ∈ Λ (V ) defines a k-linear alternating map by

`1 ∧ · · · ∧ `k (v1, ··· , vk) = det(`i(vj))

for all v1, . . . , vk ∈ V . In particular

`1 ∧ `2 (v1, v2) = `1(v1)`2(v2) − `1(v2)`2(v1) Exercise 1.8. Suppose that V is an n-dimensional vector space. Given a linear map A : V → V , we get a map Λn(A):Λn(V ) → Λn(V ), and since dim Λn(V ) = 1, the map Λn(A) is multiplication by a . Show that this scalar is det A.

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