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If
and
are finite dimensional vector spaces then
the Cartesian product
is naturally a
vector space called the direct sum of
and
and
denoted . The tensor product is a more complicated
object. To define it we start by defining for any *set*
the free vector space over , . This is the
set of all maps from
to
which are zero except at
a finite number of points. We define
the vector space structure by adding and scalar multiplying
maps. Each
gives rise to a function
which is one at
and zero elsewhere.
We therefore have a map
.
By construction the span of the image of
is all of
.
The special property of the free vector space over
is the following.

**Proposition D.1**
Let

be any map from

into a vector space

then there is a unique
linear map

such that

.

*Proof*.
The general element of

is

for

. We define

Given two vector spaces
and
we can define
. This is an infinite dimensional
vector space. We shall denote
by
.
Consider the subspace
defined
as the span of all elements of the form

and
for any real numbers
and
and vectors
and
.
Let us denote
and define a map
by
We have

**Proposition D.2**
The map

is bilinear.

*Proof*.
We check the first factor only

From Proposition D.1 we know that any map
, where
is a vector
space extends to a map
.
Standard linear algebra tells us that we can take
the quotient to get a map
if
. The map is defined by
.
For example if
and
then
defines a linear map from
.

Let
be a basis of
and
be a basis of .
Consider the set of
vectors
in
. We wish to show that they form a
basis. First we check that they span the space
. As the elements of
are
finite linear combinations of elements of the form
it suffices to show that these are all in the span of the
vectors
. But this follows from
the bilinearity. If
and
then

To show that they are linearly independent assume that
Let
and
be the dual bases of and .
That is
and
.
Then apply the map
defined by
and
to this equation to obtain
.
So we have proved.

**Proposition D.3**
If

and

are finite dimensional
vector spaces then

We can iterate tensor products. If
and
and
are vector spaces we can form
and
. These different vector
spaces are in fact isomorphic via the map

We use this map to identify these two spaces and ignore the
brackets. We write
for
the triple tensor product. More generally we can form
finitely many tensor products.
We also need to know about tensor products of maps. If
is linear and
is
linear then we can define a map

by
. This is a bilinear
map so factors to a map
which we denote by
. It is defined by
We have seen that any bilinear map
gives rise to a linear map
. It
is easy to show that this is an isomorphism. More
generally if for any collection of vector spaces
we denote by
Mult
the
space of all multilinear maps from
we have

**Proposition D.4**
If

are vectors spaces then
there is a natural isomorphism

defined by

** Next:** About this document ...
** Up:** Differential Geometry. Honours 1996
** Previous:** Vector fields and derivations.
** Contents**
*Michael Murray*

*1998-09-16*