and Exterior Algebras
23.1
Tensors Products
We begin by defining tensor products of vector spaces over a field and then we investigate
some basic properties of these tensors, in particular the existence of bases and duality. After
this, we investigate special kinds of tensors, namely, symmetric tensors and skew-symmetric
tensors. Tensor products of modules over a commutative ring with identity will be discussed
in Chapter 24.
Given a linear map, f : E → F , we know that if we have a basis, (ui)i∈I, for E, then f
is completely determined by its values, f (ui), on the basis vectors. For a multilinear map,
f : En → F , we don’t know if there is such a nice property but it would certainly be very
useful.
In many respects, tensor products allow us to define multilinear maps in terms of their
action on a suitable basis. The crucial idea is to linearize, that is, to create a new vector space,
E⊗n, such that the multilinear map, f : En → F , is turned into a linear map, f⊗ : E⊗n → F ,
which is equivalent to f in a strong sense. If in addition, f is symmetric, then we can define
a symmetric tensor power, Symn(E), and every symmetric multilinear map, f : En → F , is
turned into a linear map, f : Symn(E) → F , which is equivalent to f in a strong sense.
Similarly, if f is alternating, then we can define a skew-symmetric tensor power,
n(E), and
every alternating multilinear map is turned into a linear map, f∧ :
n(E) → F, which is
equivalent to f in a strong sense.
Tensor products can be defined in various ways, some more abstract than others. We
tried to stay down to earth, without excess!
Let K be a given field, and let E1, . . . , En be n ≥ 2 given vector spaces. For any vector
space, F , recall that a map, f : E1 × · · · × En → F , is multilinear iff it is linear in each of
605
606
CHAPTER 23. TENSOR ALGEBRAS
its argument, that is,
f (u1, . . . ui , v + w, u
, v, u
1
i+1, . . . , un)
= f (u1, . . . ui1
i+1, . . . , un)
+ f (u1, . . . ui , w, u
1
i+1, . . . , un)
f (u1, . . . ui , λv, u
, v, u
1
i+1, . . . , un)
= λf (u1, . . . ui1
i+1, . . . , un),
for all uj ∈ Ej (j = i), all v, w ∈ Ei and all λ ∈ K, for i = 1 . . . , n.
The set of multilinear maps as above forms a vector space denoted L(E1, . . . , En; F ) or
Hom(E1, . . . , En; F ). When n = 1, we have the vector space of linear maps, L(E, F ) or
Hom(E, F ). (To be very precise, we write HomK(E1, . . . , En; F ) and HomK(E, F ).) As
usual, the dual space, E∗, of E is defined by E∗ = Hom(E, K).
Before proceeding any further, we recall a basic fact about pairings. We will use this fact
to deal with dual spaces of tensors.
Definition 23.1. Given two vector spaces, E and F , a map, (−, −) : E × F → K, is a
nondegenerate pairing iff it is bilinear and iff (u, v) = 0 for all v ∈ F implies u = 0 and
(u, v) = 0 for all u ∈ E implies v = 0. A nondegenerate pairing induces two linear maps,
ϕ : E → F ∗ and ψ : F → E∗, defined by
ϕ(u)(y) = (u, y)
ψ(v)(x) = (x, v),
for all u, x ∈ E and all v, y ∈ F .
Proposition 23.1. For every nondegenerate pairing, (−, −) : E ×F → K, the induced maps
ϕ : E → F ∗ and ψ : F → E∗ are linear and injective. Furthermore, if E and F are finite
dimensional, then ϕ : E → F ∗ and ψ : F → E∗ are bijective.
Proof. The maps ϕ : E → F ∗ and ψ : F → E∗ are linear because u, v → (u, v) is bilinear.
Assume that ϕ(u) = 0. This means that ϕ(u)(y) = (u, y) = 0 for all y ∈ F and as our
pairing is nondegenerate, we must have u = 0. Similarly, ψ is injective. If E and F are finite
dimensional, then dim(E) = dim(E∗) and dim(F ) = dim(F ∗). However, the injectivity of ϕ
and ψ implies that that dim(E) ≤ dim(F ∗) and dim(F ) ≤ dim(E∗). Consequently dim(E) ≤
dim(F ) and dim(F ) ≤ dim(E), so dim(E) = dim(F ). Therefore, dim(E) = dim(F ∗) and ϕ
is bijective (and similarly dim(F ) = dim(E∗) and ψ is bijective).
Proposition 23.1 shows that when E and F are finite dimensional, a nondegenerate pairing
induces canonical isomorphims ϕ : E → F ∗ and ψ : F → E∗, that is, isomorphisms that do
not depend on the choice of bases. An important special case is the case where E = F and
we have an inner product (a symmetric, positive definite bilinear form) on E.
Remark: When we use the term “canonical isomorphism” we mean that such an isomor-
phism is defined independently of any choice of bases. For example, if E is a finite dimensional
23.1. TENSORS PRODUCTS
607
vector space and (e1, . . . , en) is any basis of E, we have the dual basis, (e∗1, . . . , e∗n), of E∗
(where, e∗i(ej) = δi j) and thus, the map ei → e∗i is an isomorphism between E and E∗. This
isomorphism is not canonical.
On the other hand, if −, − is an inner product on E, then Proposition 23.1 shows that
the nondegenerate pairing, −, − , induces a canonical isomorphism between E and E∗. This
isomorphism is often denoted : E → E∗ and we usually write u for (u), with u ∈ E.
Given any basis, (e1, . . . , en), of E (not necessarily orthonormal), if we let gij = (ei, ej),
then for every u =
n
u
j=1
j ej , since u (v) = u, v
for all v ∈ V , we have
n
n
n
u (ei) = (u, ei) =
ujej, ei =
uj(ej, ei) =
gijuj,
j=1
j=1
j=1
so we get
n
n
u =
ωie∗i, with ωi =
gijuj.
i=1
j=1
If we use the convention that coordinates of vectors are written using superscripts
(u =
n
uie
i=1
i) and coordinates of one-forms (covectors) are written using subscripts
(ω =
n
ω
i=1
ie∗
i ), then the map,
, has the effect of lowering (flattening!) indices. The
inverse of
is denoted : E∗ → E. If we write ω ∈ E∗ as ω =
n
ω
i=1
ie∗
i and ω
∈ E as
ω =
n
(ω )je
j=1
j , since
n
ωi = ω(ei) = ω , ei =
(ω )jgij,
1 ≤ i ≤ n,
j=1
we get
n
(ω )i =
gijωj,
j=1
where (gij) is the inverse of the matrix (gij).
The inner product, (−, −), on E induces an inner product on E∗ also denoted (−, −)
and given by
(ω1, ω2) = (ω1, ω2), for all ω1, ω2 ∈ E∗.
Then, it is obvious that
(u, v) = (u , v ),
for all u, v ∈ E.
If (e1, . . . , en) is a basis of E and gij = (ei, ej), as
n
(e∗i) =
gikek,
k=1
608
CHAPTER 23. TENSOR ALGEBRAS
an easy computation shows that
(e∗i, e∗j) = ((e∗i) , (e∗j) ) = gij,
that is, in the basis (e∗1, . . . , e∗n), the inner product on E∗ is represented by the matrix (gij),
the inverse of the matrix (gij).
The inner product on a finite vector space also yields a natural isomorphism between
the space, Hom(E, E; K), of bilinear forms on E and the space, Hom(E, E), of linear maps
from E to itself. Using this isomorphism, we can define the trace of a bilinear form in an
intrinsic manner. This technique is used in differential geometry, for example, to define the
divergence of a differential one-form.
Proposition 23.2. If −, − is an inner product on a finite vector space, E, (over a field,
K), then for every bilinear form, f : E × E → K, there is a unique linear map, f : E → E,
such that
f (u, v) = f (u), v ,
for all u, v ∈ E.
The map, f → f , is a linear isomorphism between Hom(E, E; K) and Hom(E, E).
Proof. For every g ∈ Hom(E, E), the map given by
f (u, v) = g(u), v ,
u, v ∈ E,
is clearly bilinear. It is also clear that the above defines a linear map from Hom(E, E) to
Hom(E, E; K). This map is injective because if f (u, v) = 0 for all u, v ∈ E, as −, − is
an inner product, we get g(u) = 0 for all u ∈ E. Furthermore, both spaces Hom(E, E) and
Hom(E, E; K) have the same dimension, so our linear map is an isomorphism.
If (e1, . . . , en) is an orthonormal basis of E, then we check immediately that the trace of
a linear map, g, (which is independent of the choice of a basis) is given by
n
tr(g) =
g(ei), ei ,
i=1
where n = dim(E). We define the trace of the bilinear form, f , by
tr(f ) = tr(f ).
From Proposition 23.2, tr(f ) is given by
n
tr(f ) =
f (ei, ei),
i=1
for any orthonormal basis, (e1, . . . , en), of E. We can also check directly that the above
expression is independent of the choice of an orthonormal basis.
We will also need the following Proposition to show that various families are linearly
independent.
23.1. TENSORS PRODUCTS
609
Proposition 23.3. Let E and F be two nontrivial vector spaces and let (ui)i∈I be any family
of vectors ui ∈ E. The family, (ui)i∈I, is linearly independent iff for every family, (vi)i∈I, of
vectors vi ∈ F , there is some linear map, f : E → F , so that f(ui) = vi, for all i ∈ I.
Proof. Left as an exercise.
First, we define tensor products, and then we prove their existence and uniqueness up to
isomorphism.
Definition 23.2. A tensor product of n ≥ 2 vector spaces E1, . . . , En, is a vector space T ,
together with a multilinear map ϕ : E1 × · · · × En → T , such that, for every vector space F
and for every multilinear map f : E1×· · ·×En → F , there is a unique linear map f⊗ : T → F ,
with
f (u1, . . . , un) = f⊗(ϕ(u1, . . . , un)),
for all u1 ∈ E1, . . . , un ∈ En, or for short
f = f⊗ ◦ ϕ.
Equivalently, there is a unique linear map f⊗ such that the following diagram commutes:
ϕ
E
/
1 × · · · × En
T
f⊗
f
&◆
◆
◆
◆
◆
◆
◆
◆
◆
◆
◆
F
First, we show that any two tensor products (T1, ϕ1) and (T2, ϕ2) for E1, . . . , En, are
isomorphic.
Proposition 23.4. Given any two tensor products (T1, ϕ1) and (T2, ϕ2) for E1, . . . , En, there
is an isomorphism h : T1 → T2 such that
ϕ2 = h ◦ ϕ1.
Proof. Focusing on (T1, ϕ1), we have a multilinear map ϕ2 : E1 × · · · × En → T2, and thus,
there is a unique linear map (ϕ2)⊗ : T1 → T2, with
ϕ2 = (ϕ2)⊗ ◦ ϕ1.
Similarly, focusing now on on (T2, ϕ2), we have a multilinear map ϕ1 : E1 × · · · × En → T1,
and thus, there is a unique linear map (ϕ1)⊗ : T2 → T1, with
ϕ1 = (ϕ1)⊗ ◦ ϕ2.
610
CHAPTER 23. TENSOR ALGEBRAS
But then, we get
ϕ1 = (ϕ1)⊗ ◦ (ϕ2)⊗ ◦ ϕ1,
and
ϕ2 = (ϕ2)⊗ ◦ (ϕ1)⊗ ◦ ϕ2.
On the other hand, focusing on (T1, ϕ1), we have a multilinear map ϕ1 : E1 × · · · × En → T1,
but the unique linear map h : T1 → T1, with
ϕ1 = h ◦ ϕ1
is h = id, and since (ϕ1)⊗ ◦ (ϕ2)⊗ is linear, as a composition of linear maps, we must have
(ϕ1)⊗ ◦ (ϕ2)⊗ = id.
Similarly, we must have
(ϕ2)⊗ ◦ (ϕ1)⊗ = id.
This shows that (ϕ1)⊗ and (ϕ2)⊗ are inverse linear maps, and thus, (ϕ2)⊗ : T1 → T2 is an
isomorphism between T1 and T2.
Now that we have shown that tensor products are unique up to isomorphism, we give a
construction that produces one.
Theorem 23.5. Given n ≥ 2 vector spaces E1, . . . , En, a tensor product (E1 ⊗ · · · ⊗ En, ϕ)
for E1, . . . , En can be constructed. Furthermore, denoting ϕ(u1, . . . , un) as u1 ⊗ · · · ⊗ un,
the tensor product E1 ⊗ · · · ⊗ En is generated by the vectors u1 ⊗ · · · ⊗ un, where u1 ∈
E1, . . . , un ∈ En, and for every multilinear map f : E1 × · · · × En → F , the unique linear
map f⊗ : E1 ⊗ · · · ⊗ En → F such that f = f⊗ ◦ ϕ, is defined by
f⊗(u1 ⊗ · · · ⊗ un) = f(u1, . . . , un),
on the generators u1 ⊗ · · · ⊗ un of E1 ⊗ · · · ⊗ En.
Proof. Given any set, I, viewed as an index set, let K(I) be the set of all functions, f : I → K,
such that f (i) = 0 only for finitely many i ∈ I. As usual, denote such a function by (fi)i∈I,
it is a family of finite support. We make K(I) into a vector space by defining addition and
scalar multiplication by
(fi) + (gi) = (fi + gi)
λ(fi) = (λfi).
The family, (ei)i∈I, is defined such that (ei)j = 0 if j = i and (ei)i = 1. It is a basis of
the vector space K(I), so that every w ∈ K(I) can be uniquely written as a finite linear
combination of the ei. There is also an injection, ι : I → K(I), such that ι(i) = ei for every
23.1. TENSORS PRODUCTS
611
i ∈ I. Furthermore, it is easy to show that for any vector space, F , and for any function,
f : I → F , there is a unique linear map, f : K(I) → F , such that f = f ◦ ι, as in the following
diagram:
I
ι
/ K(I)
f
f
!❈
❈
❈
❈
❈
❈
❈
❈
❈
F
This shows that K(I) is the free vector space generated by I. Now, apply this construction
to the cartesian product, I = E1 × · · · × En, obtaining the free vector space M = K(I) on
I = E1 ×· · ·×En. Since every, ei, is uniquely associated with some n-tuple i = (u1, . . . , un) ∈
E1 × · · · × En, we will denote ei by (u1, . . . , un).
Next, let N be the subspace of M generated by the vectors of the following type:
(u1, . . . , ui + vi, . . . , un) − (u1, . . . , ui, . . . , un) − (u1, . . . , vi, . . . , un),
(u1, . . . , λui, . . . , un) − λ(u1, . . . , ui, . . . , un).
We let E1 ⊗ · · · ⊗ En be the quotient M/N of the free vector space M by N, π : M → M/N
be the quotient map and set
ϕ = π ◦ ι.
By construction, ϕ is multilinear, and since π is surjective and the ι(i) = ei generate M,
since i is of the form i = (u1, . . . , un) ∈ E1 ×· · ·×En, the ϕ(u1, . . . , un) generate M/N. Thus,
if we denote ϕ(u1, . . . , un) as u1 ⊗ · · · ⊗ un, the tensor product E1 ⊗ · · · ⊗ En is generated by
the vectors u1 ⊗ · · · ⊗ un, where u1 ∈ E1, . . . , un ∈ En.
For every multilinear map f : E1 × · · · × En → F , if a linear map f⊗ : E1 ⊗ · · · ⊗ En → F
exists such that f = f⊗ ◦ ϕ, since the vectors u1 ⊗ · · · ⊗ un generate E1 ⊗ · · · ⊗ En, the map
f⊗ is uniquely defined by
f⊗(u1 ⊗ · · · ⊗ un) = f(u1, . . . , un).
On the other hand, because M = K(E1×···×En) is free on I = E1 × · · · × En, there is a unique
linear map f : K(E1×···×En) → F , such that
f = f ◦ ι,
as in the diagram below:
E
ι
/
1 × · · · × En
K(E1×···×En)
f
f
(❘
❘
❘
❘
❘
❘
❘
❘
❘
❘
❘
❘
❘
❘
❘
F
Because f is multilinear, note that we must have f (w) = 0, for every w ∈ N. But then,
f : M → F induces a linear map h: M/N → F , such that
f = h ◦ π ◦ ι,
612
CHAPTER 23. TENSOR ALGEBRAS
by defining h([z]) = f (z), for every z ∈ M, where [z] denotes the equivalence class in M/N
of z ∈ M:
E
π◦ι /
1 × · · · × En
K(E1×···×En)/N
h
f
)❙
❙
❙
❙
❙
❙
❙
❙
❙
❙
❙
❙
❙
❙
❙
❙
❙
F
Indeed, the fact that f vanishes on N insures that h is well defined on M/N , and it is clearly
linear by definition. However, we showed that such a linear map h is unique, and thus it
agrees with the linear map f⊗ defined by
f⊗(u1 ⊗ · · · ⊗ un) = f(u1, . . . , un)
on the generators of E1 ⊗ · · · ⊗ En.
What is important about Theorem 23.5 is not so much the construction itself but the
fact that it produces a tensor product with the universal mapping property with respect to
multilinear maps. Indeed, Theorem 23.5 yields a canonical isomorphism,
L(E1 ⊗ · · · ⊗ En, F ) ∼
= L(E1, . . . , En; F ),
between the vector space of linear maps, L(E1 ⊗ · · · ⊗ En, F ), and the vector space of
multilinear maps, L(E1, . . . , En; F ), via the linear map − ◦ ϕ defined by
h → h ◦ ϕ,
where h ∈ L(E1 ⊗ · · · ⊗ En, F ). Indeed, h ◦ ϕ is clearly multilinear, and since by Theorem
23.5, for every multilinear map, f ∈ L(E1, . . . , En; F ), there is a unique linear map f⊗ ∈
L(E1 ⊗ · · · ⊗ En, F ) such that f = f⊗ ◦ ϕ, the map − ◦ ϕ is bijective. As a matter of fact,
its inverse is the map
f → f⊗.
Using the “Hom” notation, the above canonical isomorphism is written
Hom(E1 ⊗ · · · ⊗ En, F ) ∼
= Hom(E1, . . . , En; F ).
Remarks:
(1) To be very precise, since the tensor product depends on the field, K, we should subscript
the symbol ⊗ with K and write
E1 ⊗K · · · ⊗K En.
However, we often omit the subscript K unless confusion may arise.
23.2. BASES OF TENSOR PRODUCTS
613
(2) For F = K, the base field, we obtain a canonical isomorphism between the vector
space L(E1 ⊗ · · · ⊗ En, K), and the vector space of multilinear forms L(E1, . . . , En; K).
However, L(E1 ⊗ · · · ⊗ En, K) is the dual space, (E1 ⊗ · · · ⊗ En)∗, and thus, the vector
space of multilinear forms L(E1, . . . , En; K) is canonically isomorphic to (E1 ⊗ · · · ⊗
En)∗. We write
L(E1, . . . , En; K) ∼
= (E1 ⊗ · · · ⊗ En)∗.
The fact that the map ϕ : E1 × · · · × En → E1 ⊗ · · · ⊗ En is multilinear, can also be
expressed as follows:
u1 ⊗ · · · ⊗ (vi + wi) ⊗ · · · ⊗ un = (u1 ⊗ · · · ⊗ vi ⊗ · · · ⊗ un) + (u1 ⊗ · · · ⊗ wi ⊗ · · · ⊗ un),
u1 ⊗ · · · ⊗ (λui) ⊗ · · · ⊗ un = λ(u1 ⊗ · · · ⊗ ui ⊗ · · · ⊗ un).
Of course, this is just what we wanted! Tensors in E1 ⊗ · · · ⊗En are also called n-tensors,
and tensors of the form u1 ⊗ · · · ⊗ un, where ui ∈ Ei, are called simple (or indecomposable)
n-tensors. Those n-tensors that are not simple are often called compound n-tensors.
Not only do tensor products act on spaces, but they also act on linear maps (they are
functors). Given two linear maps f : E → E and g : F → F , we can define h: E × F →
E ⊗ F by
h(u, v) = f (u) ⊗ g(v).
It is immediately verified that h is bilinear, and thus, it induces a unique linear map
f ⊗ g : E ⊗ F → E ⊗ F ,
such that
(f ⊗ g)(u ⊗ v) = f(u) ⊗ g(u).
If we also have linear maps f : E → E and g : F → F , we can easily verify that
the linear maps (f ◦ f) ⊗ (g ◦ g) and (f ⊗ g ) ◦ (f ⊗ g) agree on all vectors of the form
u ⊗ v ∈ E ⊗ F . Since these vectors generate E ⊗ F , we conclude that
(f ◦ f) ⊗ (g ◦ g) = (f ⊗ g ) ◦ (f ⊗ g).
The generalization to the tensor product f1 ⊗ · · · ⊗ fn of n ≥ 3 linear maps fi : Ei → Fi
is immediate, and left to the reader.
23.2
Bases of Tensor Products
We showed that E1 ⊗ · · · ⊗ En is generated by the vectors of the form u1 ⊗ · · · ⊗ un. However,
there vectors are not linearly independent. This situation can be fixed when considering
bases, which is the object of the next proposition.
614
CHAPTER 23. TENSOR ALGEBRAS
Proposition 23.6. Given n ≥ 2 vector spaces E1, . . . , En, if (uki)i∈I is a basis for E
k
k,
1 ≤ k ≤ n, then the family of vectors
(u1i ⊗ · · · ⊗ un )(i
1
in
1,...,in)∈I1×...×In
is a basis of the tensor product E1 ⊗ · · · ⊗ En.
Proof. For each k, 1 ≤ k ≤ n, every vk ∈ Ek can be written uniquely as
vk =
vkjukj,
j∈Ik
for some family of scalars (vkj)j∈I . Let F be any nontrivial vector space. We show that for
k
every family
(wi
)
,
1,...,in
(i1,...,in)∈I1×...×In
of vectors in F , there is some linear map h : E1 ⊗ · · · ⊗ En → F , such that
h(u1i ⊗ · · · ⊗ un ) = wi
.
1
in
1,...,in
Then, by Proposition 23.3, it follows that
(u1i ⊗ · · · ⊗ un )(i
1
in
1,...,in)∈I1×...×In
is linearly independent. However, since (uki)i∈I is a basis for E
⊗ · · · ⊗ un also
k
k, the u1
i1
in
generate E1 ⊗ · · · ⊗ En, and thus, they form a basis of E1 ⊗ · · · ⊗ En.
We define the function f : E1 × · · · × En → F as follows:
f (
v1j u1 , . . . ,
vn un ) =
v1 · · · vn wj
.
1
j1
jn jn
j1
jn
1,...,jn
j1∈I1
jn∈In
j1∈I1,...,jn∈In
It is immediately verified that f is multilinear. By the universal mapping property of the
tensor product, the linear map f⊗ : E1 ⊗ · · · ⊗ En → F such that f = f⊗ ◦ ϕ, is the desired
map h.
In particular, when each Ik is finite and of size mk = dim(Ek), we see that the dimension
of the tensor product E1 ⊗· · ·⊗En is m1 · · · mn. As a corollary of Proposition 23.6, if (uki)i∈Ik
is a basis for Ek, 1 ≤ k ≤ n, then every tensor z ∈ E1 ⊗ · · · ⊗ En can be written in a unique
way as
z =
λi
u1 ⊗ · · · ⊗ un ,
1,...,in
i1
in
(i1,...,in) ∈ I1×...×In
for some unique family of scalars λi
∈ K, all zero except for a finite number.
1,...,in
23.3. SOME USEFUL ISOMORPHISMS FOR TENSOR PRODUCTS
615
23.3
Some Useful Isomorphisms for Tensor Products
Proposition 23.7. Given three vector spaces E, F, G, there exists unique canonical isomor-
phisms