Multilinear Algebra notes [Lecture notes]

10 Pages·2011·0.191 MB·English

by  Eugene Lerman

MULTILINEAR ALGEBRA NOTES
EUGENELERMAN
Contents
1. (Multi)linear algebra 1
1.1. Tensor 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 tensors, tensor algebra and Grassmann (exterior) algebra. Unless noted
otherwise all vector spaces are over the real number 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 product” 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 V ,...,V and U be vector spaces. A map
1 n
nfactors
(cid:122) (cid:125)(cid:124) (cid:123)
f :V ×···×V →U, (v ,...,v )(cid:55)→f(v ,...,v )
1 n 1 n 1 n
is multi-linear if for each fixed index i and a fixed (n−1)-tuple of vectors v ,...,v ,v ,...,v the map
1 i−1 i+1 n
V →U, w (cid:55)→f(v ,...,v ,w,v ,...,v )
i 1 i−1 i+1 n
is linear. When the number of factors is n, as above, we will also say that f is n-linear.
nfactors
(cid:122) (cid:125)(cid:124) (cid:123)
For example, if we identify Rn2 (cid:39) Rn×···×Rn by thinking of an n×n matrix as an n-tuple of column
vectors, then the determinant
nfactors
(cid:122) (cid:125)(cid:124) (cid:123)
det:Rn×···×Rn→R, (v ,...,v )(cid:55)→det(v |...|v )
1 n 1 n
is an n-linear map. Here is an example of a bilinear map. Any inner product on a vector space V:
V ×V (cid:51)(v,w)(cid:55)→v·w ∈R
is bilinear. There is no standard notation for the space of n-linear maps from V ×···×V to U. We will
1 n
denote it by
Mult(V ×···×V ,U)=Mult (V ×···×V ,U)
1 n n 1 n
(n is to indicate that these are n-linear maps). This space, Mult(V ×···×V ,U), is a vector space: any
1 n
linear combination of two n-linear maps is n-linear. We now take a closer look at the space of bilinear maps
Mult (V ×W,U). This case is complicated enough to understand what happens with multi-linear maps in
2
general, but simple enough not to bog down in notation.
typesetFebruary21,2011.
1
Lemma 1.2. Let {v }, {w } and {u } denote the bases of V, W and U respectively and {v∗}, {w∗} and
i j k i j
{u∗} the corresponding dual bases. Then the maps
k
φk :V ×W →U, φk(v,w)=v∗(v)w∗(w)u
ij ij i j k
are bilinear and form a basis of Mult (V ×W,U). Hence
2
dimMult (V ×W,U)=dimV dimW dimU.
2
Proof. Itis easytosee that φk arebilinear. Next, forany b∈Mult (V ×W,U), any w ∈W andany v ∈V,
ij 2
(cid:88) (cid:88)
b(v,w)=b( v∗(v)v , w∗(w)w )
i i j j
(cid:88)
= v∗(v)w∗(w)b(v ,w )
i j i j
i,j
(cid:88)
= v∗(v)w∗(w)u∗(b(v ,w ))u
i j k i j k
i,j,k
(cid:88)
= u∗(b(v ,w ))φk(v,w).
k i j ij
i,j,k
Hencethemapsφk spanMult (V ×W,U). Also, thecollectionofnumbersu∗(b(v ,w ))uniquelydetermine
ij 2 k i j
the bilinear form b. Hence φk’s are linearly independent. (cid:3)
ij
We now turn to the definition of the tensor product 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:
(v +v )⊗w−v ⊗w−v ⊗w =0
1 2 1 2
v⊗(w +w )−v⊗w −v⊗w =0
1 2 1 2
α(v⊗w)−(αv)⊗w =0
α(v⊗w)−v⊗(αw)=0,
forallv,v ,v ∈V,w,w ,w ∈W andα∈R. Theseidentitiessimplysaythatthemap⊗:V×W →V⊗W,
1 2 1 2
(v,w) (cid:55)→ v⊗w, is a bilinear map. The fact that everything in V ⊗W is a linear combination of symbols
v⊗w means that the image 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. AtensorproductoftwofinitedimensionalvectorspacesV andW isavectorspaceV ⊗W
together with a bilinear map ⊗ : V × W → V ⊗ W, (v,w) (cid:55)→ 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 (cid:47)(cid:47)
V ×W (cid:59)(cid:59)U
⊗
(cid:15)(cid:15) ¯b
V ⊗W
commutes. The existence of the map¯b satisfying the above conditions is called the universal property 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 isomorphism
(cid:39)
Hom(V ⊗W,U)−→Mult(V ×W,U).
1Buttheimageof⊗isnotall ofV ⊗W. Theelementsintheimagearecalleddecomposabletensors.
2Thesymbolv⊗w standsforthevalueofthemap⊗onthepair(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)(cid:51)A(cid:55)→A◦⊗∈Mult(V ×W,U) and Mult(V ×W,U)(cid:51)b(cid:55)→¯b∈
Hom(V ⊗W,U) are inverses of each other. (cid:3)
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
spacesV⊗ W andV⊗ W withcorrespondingbilinearmaps⊗ :V×W →V⊗ W and⊗ :V×W →V⊗ W
1 2 1 1 2 2
whichsatisfytheconditionsoftheDefinition1.3. Wewillarguethatthesevectorspacesareisomorphic. By
the universal property there exist a unique linear map ⊗ :V ⊗ W →V ⊗ W so that the diagram
1 2 1
V ×W ⊗1 (cid:47)(cid:47)V(cid:56)(cid:56) ⊗1W
⊗2
(cid:15)(cid:15) ⊗1
V ⊗ W
2
commutes. By the same argument, switching the roles of ⊗ and ⊗ , there is a unique linear map ⊗ :
1 2 2
V ⊗ W →V ⊗ W making the diagram
1 2
V ×W ⊗2 (cid:47)(cid:47)V(cid:56)(cid:56) ⊗2W
⊗1
(cid:15)(cid:15) ⊗2
V ⊗ W
1
commute. Define
T = ⊗ ◦⊗ :V ⊗ W →V ⊗ W
1 1 2 1 1
T = ⊗ ◦⊗ :V ⊗ W →V ⊗ W.
2 2 1 2 2
These are linear maps making the diagrams
V ×W ⊗1 (cid:47)(cid:47)V(cid:56)(cid:56) ⊗1W and V ×W ⊗2 (cid:47)(cid:47)V(cid:56)(cid:56) ⊗2W
⊗1 ⊗2
(cid:15)(cid:15) T1 (cid:15)(cid:15) T2
V ⊗ W V ⊗ W
1 2
commute. But the identity maps id : V ⊗ W → V ⊗ W, i =1,2, are linear and also make the respective
i i i
diagrams commute. By uniqueness T = id . Hence ⊗ and ⊗ are inverses of each other and provide the
i i 1 2
desired isomorphisms. (cid:3)
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 set {(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)(cid:54)=0 for only finitely many pairs (v,w)}.
3
Thissetoffunctionsisaninfinitedimensionalvectorspace. Itsbasisconsistsoffunctionsthattake
value1onagivenpair(v ,w )and0onallotherpairs. It’stemptingtocallthisfunction(v ,w ).
0 0 0 0
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
(v +v ,w)(cid:54)=(v ,w)+(v ,w) in F(V,W).
1 2 1 2
Consider the smallest subspace K of F(V,W) containing the following collection of vectors:
 (cid:12) 
S = ((vv,1w+1v+c2c(,(wvwv,2,)w)w−−))−−((vv(1(,v,cw,wv1c,))ww−−)),((vv2,,ww2)) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)v,v1,v2 ∈V, w,w1,w2 ∈W and c∈R
Inotherwords,considerthesubspaceK ofF(V ×W)spannedbythesetS. DefineV ⊗W tobethequotient
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 (cid:44)→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. Wewritev⊗wforthevalueof⊗onthepair(v,w). Byconstructiontheset{v⊗w |(v,w)∈V×W}
spans V ⊗W [but it’s much too big to be a basis].
Wecheckthatthemap⊗:V ×W →V ⊗W hastherequireduniversalproperty. Supposeb: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. (cid:3)
Lemma 1.7. For any vector spaces V and W
dim(V ⊗W)=dimV ·dimW.
Proof.
dimV ⊗W =dim(V ⊗W)∗ =dimHom(V ⊗W,R)
=dimMult(V ×W,R) by Lemma 1.4
=dimV ·dimW ·dimR.
(cid:3)
We are now in position to quickly prove a number of results about tensor products.
Corollary 1.7.1. If {v } and {w } are a basis of V and W respectively, then {v ⊗w } is a basis of V ⊗W.
i j i j
Proof. Since the vectors of the form v⊗w, v ∈V, w ∈W, span V ⊗W, the much smaller set {v ⊗w } also
i j
spans V ⊗W3. But dim(V ⊗W)=dimV ·dimW is precisely the number of elements in the set {v ⊗w }.
i j
Hence the set {v ⊗w } is a basis. (cid:3)
i j
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 dimension
count. (cid:3)
Lemma 1.9. V∗⊗W is isomorphic to Hom(V,W).
3Weareusingherethefactthatforany(v,w)∈V ×W,thetensorv⊗w isalinearcombinationofvi⊗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 (cid:55)→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. (cid:3)
Exercise 1.1. Show that if {v } is a basis of a vector space V, {v∗} the dual basis and {w } the basis of a
i i j
vector space W, then {v∗(·)w } is a basis of Hom(V,W).
i j
Lemma 1.10. If A : V → W and B : V(cid:48) → W(cid:48) are two linear maps, then there is a unique linear map
A⊗B :V ⊗V(cid:48) →W ⊗W(cid:48) such that (A⊗B)(v⊗w)=A(v)⊗B(w) for all (v,w)∈V ×V(cid:48).
Proof. Consider b:V ×W →V(cid:48)⊗W(cid:48) 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(cid:48) ⊗W(cid:48)
with
¯b(v⊗w)=Av⊗Bw
for all (v,w)∈V ×W. (cid:3)
Exercise 1.2. Show that if A : V → W is represented by a matrix (a ) with respect to some bases of V
ij
and W and B : V(cid:48) → W(cid:48) is represented by a matrix (b ) with respect to bases of V(cid:48) and W(cid:48), then A⊗B
kl
is represented by the matrix (a b ) with respect to the appropriate bases.
ij kl
Exercise 1.3. Show that there is a natural isomorphism φ:V∗⊗W∗ →(cid:39) 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)(cid:55)→av gives rise to an isomorphism R⊗V →(cid:39) 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)(cid:39)(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
V⊗0 :=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
(cid:122) (cid:125)(cid:124) (cid:123)
f :V ×···×V→U,
then there exists a unique linear map f :V⊗n →U with
f(v ⊗···⊗v )=f(v ,...,v ).
1 n 1 n
for all (v ,...,v )∈V ×···×V.
1 n
Moreover, given a∈V⊗n and b∈V⊗m, a⊗b is in V⊗n⊗V⊗m (cid:39)V⊗(n+m). This gives us an R-bilinear
map,
V⊗n×V⊗m →V⊗(n+m), (a,b)(cid:55)→a⊗b.
Note that if n=0 the map above is simply
R×V⊗m →V⊗m, (a,t)(cid:55)→at.
(cf. Exercise 1.4).
Definition1.11. Analgebra overRisavectorspaceAtogetherwithabilinearmapA×A→A,(a,a(cid:48))(cid:55)→aa(cid:48)
(“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(cid:55)→x1. We will always identify
R with its image in A.
Example 1.14. A Lie algebra is an algebra. It is not associative and does not have 1 (why not?).
Example 1.15. The space M (R) of n×n matrices forms an algebra under matrix multiplication. It is an
n
algebra with unity: the identity matrix I is the unity.
Definition 1.16. An algebra A is graded if
∞
(cid:88)
A= A direct sum
i
i=0
and if for any a ∈ A and b ∈ A we have a·b ∈ A . We will refer to the elements of A as elements of
i j i+j k
degree k.
GivenavectorspaceV weconstructthecorrespondingtensor algebra T(V)asfollows. Asavectorspace
T(V) is the direct sum:
∞
(cid:88)
T(V)=R⊕V ⊕V⊗2⊕···⊕V⊗n⊕···= V⊗i.
i=0
Thus the elements of T(V) are finite sums ai1+ai2+···aik, aij ∈V⊗ij. We define the multiplication on T
by extending the multiplication
V⊗n×V⊗m →V⊗(n+m) (a,b)(cid:55)→a⊗b.
bilinearly to all of T(V). The tensor algebra T(V) of a vector space V is a graded associative algebra 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) algebras 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(v ,...,v ,v ,...,v )=−f(v ,...,v ,v ,...,v )
1 i i+1 n 1 i+1 i n
for all (v ,...,v )∈V ×···×V and any index i.
1 n
Example 1.17. The determinant
nfactors
(cid:122) (cid:125)(cid:124) (cid:123)
det:Rn×···×Rn→R, (v ,...,v )→det(v |...|v )
1 n 1 n
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)(v ,v ):=a(v )b(v )−a(v )b(v ), v ,v ∈V.
1 2 1 2 2 1 1 2
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(cid:15)(cid:15)i(cid:79)(cid:79)(cid:79)(cid:79)(cid:79)(cid:79)(cid:79)(cid:79)j(cid:79)(cid:79)(cid:79)(cid:79)(cid:79)(cid:79)(cid:79)(cid:39)(cid:39)
 (cid:47)(cid:47)
Λ∗(V) A.
Proposition 1.20. If the exterior algebra Λ∗(V) exists, it is unique (up to isomorphism).
Proof. This is a formal exercise and is left to the reader. (cid:3)
Proposition 1.21. For every vector space V the exterior algebra Λ∗(V) exists.
Proof. Let I be the two-sided ideal 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
j(k)(v ⊗···⊗v )=j(v )···j(v ).
1 k 1 k
In addition, we define j(0)(a) = a·1 , for all a ∈ R. In this way, we obtain an algebra map ˜: T(V) → A.
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. (cid:3)
4Amapf :A→Bbetweentwoalgebrasisanalgebramapiff islinearandpreservesmultiplication: 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=(v +v )∧(v +v )=v ∧v +v ∧v +v ∧v +v ∧v
1 2 1 2 1 1 1 2 2 1 2 2
gives that
v ∧v =−v ∧v ;
1 2 2 1
That is, the wedge product is skew-commutative.
Remark 1.23. Let Λk(V)=Tk(V)/(Tk(V)∩I). The vector space Λk(V) is called the kth exterior power
of V. Then
∞
(cid:88)
Λ∗(V)= Λk(V),
k=0
where
Λ0(V)=R and Λ1(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 {v ,...,v } is a basis for V, then {v ⊗v } is a basis for V ⊗V. By
1 n i j
induction, {v ⊗···⊗v } is a basis for V⊗k. Thus, {v ∧···∧v } generates Λk(V) = V⊗k/(I ∩V⊗k).
i1 ik i1 ik
Since ∧ is skew-commutative, however, we can reduce this generating set to a smaller one:
(1.1) {v ∧···∧v |i <···<i },
i1 ik 1 k
This implies that
Λl(V)=0 whenever l>dimV.
We will see below that the set (1.1) is a basis of Λk(V).
We now investigate the connection 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
(cid:122) (cid:125)(cid:124) (cid:123)
spaces. If f :V ×···×V→U is alternating then there is a unique linear map f :Λk(V)→U with
f(v ∧···∧v )=f(v ,...,v ).
1 k 1 k
Proof. BytheuniversalpropertyofV⊗k,thereisauniquelinearmapf˜:V⊗k →U suchthatf˜(v ⊗···⊗v )=
1 k
(cid:12)
f(v ,...,v ). Since f is alternating, f(cid:12) =0, where I is the ideal defined in the construction of Λ∗(V).
1 k (cid:12)
I∩V⊗k
This gives us the linear map f :Λk(V)=V⊗k/(I∩V⊗k)→U with the desired property. (cid:3)
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.
Proof. We may assume that V =Rn. Let e ,...,e be the standard basis. Then e ∧···∧e spans Λn(V).
1 n 1 n
Weneedtoshowthate ∧···∧e (cid:54)=0. Thedeterminantdet:Rn×···×Rn →Ris1ontheidentitymatrix
1 n
I = (e |...|e ): det(e |...|e ) = 1. Hence the induced linear map det : Λn(Rn) → R is 1 on e ∧···∧e .
1 n 1 n 1 n
Therefore e ∧···∧e (cid:54)=0. (cid:3)
1 n
Corollary 1.26.1. If{f ,...,f }isabasisforavectorspaceV, then{f ∧···∧f :1≤i <···<i ≤n}
1 n i1 ik 1 k
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
(cid:88)
0= a f ∧···∧f for some a ∈R
i1,...,ik i1 ik i1,...,ik
i1<···<ik
Pick a sequence j <j <···<j . Let j <···<j be the remaining indices. Then
1 2 k k+1 n
(cid:88)
( a f ∧···∧f )∧f ∧···∧f
i1,...,ik i1 ik jk+1 jn
= a f ∧···f ∧f ∧···∧f ,
j1,...,jk j1 jk jk+1 jn
8
since a f ∧···f ∧f ∧···∧f =0 whenever i =j for some s,r. This gives a =0. Also,
i1,...,ik i1 ik jk+1 jn s r j1,...,jk
f ∧···∧f ∧f ∧···∧f =±f ∧···∧f (cid:54)=0. Hence f ∧···∧f (cid:54)=0. (cid:3)
j1 jk jk+1 jn 1 n j1 jk
Corollary 1.26.2. For any finite dimensional vector space V
(cid:18) (cid:19)
dimV (dimV)!
dimΛk(V)= = .
k k!(dimV −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
(Λk(A))(v ∧···∧v )=Av ∧···∧Av
1 k 1 k
for all v ,...,v ∈V.
1 k
Proof. Consider the map b:V ×···×V →Λk(W) given by
b(v ,...,v )=Av ∧···∧Av .
1 k 1 k
Since ∧ is skew-commutative, b is an alternating map. By Proposition 1.25 there exists a unique linear map
Λk(A):Λk(V)→Λk(W) with the required properties. (cid:3)
Exercise 1.7. Let A:V →W be a linear map as above. Choose bases of V and W and the corresponding
basesofΛk(V)andofΛk(W). ShowthattheentriesofthematrixrepresentingΛk(A)arepolynomialinthe
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 (cid:104)·,·(cid:105):V ×W →R.
Example 1.29. Let V be a vector space and V∗ be its dual. The evaluation map
V∗×V →R (cid:104)(cid:96),v(cid:105)=(cid:96)(v)
is a pairing.
Definition 1.30. A pairing (cid:104)·,·(cid:105):V ×W →R is non-degenerate if
(cid:104)v ,w(cid:105)=0 ∀w ∈W ⇒v =0
0 0
(cid:104)v,w (cid:105)=0 ∀v ∈V ⇒w =0.
0 0
Example 1.31. The evaluation map
V∗×V →R (cid:104)(cid:96),v(cid:105)=(cid:96)(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 (cid:39)W∗ and W (cid:39)V∗.
Proof. Consider b# :V →W∗ given by
1
(b#(v))(w)=b(v,w).
1
The map b# is linear, and
1
kerb# ={v ∈V :b#(v )=0}={v ∈V :b(v ,w)=0∀w}={0}.
1 0 1 0 0 0
Thus dimV ≤ dimW∗ = dimW. By the same argument, we have dimW ≤ dimV∗ = dimV. Therefore
dimV =dimW. Hence b# is an isomorphism.
1
By the same argument, b# :W →V∗ given by w (cid:55)→b(·,w) is an isomorphism as well. (cid:3)
2
Proposition 1.33. There is a nondegenerate pairing
(cid:104)·,·(cid:105):Λk(V∗)×Λk(V)→R
with
(cid:16) (cid:17)
(cid:104)v∗∧···∧v∗,v ∧···∧v (cid:105)=det v∗(v ) .
1 k 1 k i j
Hence
Λk(V∗)(cid:39)(Λk(V))∗.
9
k k
(cid:122) (cid:125)(cid:124) (cid:123) (cid:122) (cid:125)(cid:124) (cid:123)
Proof. Consider b:V∗×···×V∗ ×V ×···×V→R given by
(cid:16) (cid:17)
b(l ,...,l ,v ,...,v )=det l (v ) .
1 k 1 k i j
Forafixed(l ,...,l )∈V∗×···×V∗,bisalternatinginthev’s. Sothereisamapb:(V∗×···×V∗)×Λk(V)→
1 k
R with
(cid:16) (cid:17)
(l ,...,l ,v ∧···∧v )(cid:55)→det l (v ) .
1 k 1 k i j
Similarly, for a fixed v ∧···∧v ∈ Λk(V), b is alternating in the l’s, which means that there is a map
1 k
˜b:Λk(V∗)×Λk(V)→R with the desired property.
To check non-degeneracy evaluate the pairing on the respective bases. (cid:3)
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∗).
Remark 1.34. Explicitly (cid:96) ∧···∧(cid:96) ∈Λk(V∗) defines a k-linear alternating map by
1 k
(cid:96) ∧···∧(cid:96) (v ,··· ,v )=det((cid:96) (v ))
1 k 1 k i j
for all v ,...,v ∈V. In particular
1 k
(cid:96) ∧(cid:96) (v ,v )=(cid:96) (v )(cid:96) (v )−(cid:96) (v )(cid:96) (v )
1 2 1 2 1 1 2 2 1 2 2 1
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 dimΛn(V)=1, themap Λn(A) is multiplicationbya scalar. Show
that this scalar is detA.
10