tensor product of dual spaces is a dual space of tensor product

tensor product of dual spaces is a dual space of tensor product

Proposition

. Let

k

be a field and

V

,

W

be

vector spaces

over

k

. Then

(

V

⊗

W

)

*

is

isomorphic

to

V

*

⊗

W

*

.

Proof

. If

V

or

W

is finite dimensional, then there is an explicit

isomorphism

between

(

V

⊗

W

)

*

and

V

*

⊗

W

*

(see this entry (

http://planetmath.org/TensorProductAndDualSpaces

) for more details). So assume that both

V

and

W

are infinite dimensional.

First of all, note that if

V

is a vector space, then

dim

k

⁢

(

V

)

denotes its

dimension

, that is

dim

k

⁢

(

V

)

is a

cardinality

of any basis of

V

. Thus we can compare dimensions of spaces, so

dim

k

⁢

(

V

)

≤

dim

k

⁢

(

W

)

if and only if there is an

injection

from a basis of

V

to a basis of

W

(note that this

relation

is well defined, i.e. it does not depend on the choice of bases). One can easily show that

dim

k

⁢

(

V

)

≤

dim

k

⁢

(

W

)

if and only if there is an injective

linear map

from

V

to

W

and this is if and only if there is a

surjective

linear map from

W

to

V

. Therefore

V

is isomorphic to

W

if and only if

dim

k

⁢

(

V

)

=

dim

k

⁢

(

W

)

(which here means that

dim

k

⁢

(

V

)

≤

dim

k

⁢

(

W

)

and

dim

k

⁢

(

W

)

≤

dim

k

⁢

(

V

)

).

Without loss of generality, we may assume that

dim

k

⁢

(

V

)

≤

dim

k

⁢

(

W

)

(we can always compare any two sets). Note that the basis of

V

⊗

W

is the

product

of bases of

V

and

W

(namely if

(

e

i

)

i

∈

I

is a basis of

V

and

(

e

j

′

)

j

∈

J

a basis of

W

, then

(

e

i

⊗

e

j

′

)

i

∈

I

,

j

∈

J

is a basis of

V

⊗

W

). If

X

,

Y

are sets such that

Y

is

infinite

and there is an injection

X

→

Y

, then it is well known that there is a

bijection

from

Y

to

X

×

Y

. Thus (since

W

is infinite dimensional) we have:

dim

k

⁢

(

V

⊗

W

)

=

dim

k

⁢

(

W

)

.

Therefore

V

⊗

W

is isomorphic to

W

, so

(

V

⊗

W

)

*

is isomorphic to

W

*

.

Now the inequality

dim

k

⁢

(

V

)

≤

dim

k

⁢

(

W

)

implies that

dim

k

⁢

(

V

*

)

≤

dim

k

⁢

(

W

*

)

. Indeed, assume that there is an injective linear map

f

:

V

→

W

. Then there is a (surjective) linear map

g

:

W

→

V

such that

g

∘

f

=

id

V

(see this entry (

http://planetmath.org/SomeFactsAboutInjectiveAndSurjectiveLinearMaps

) for more details). Therefore (since

(

⋅

)

*

is a

contravariant functor

) we have that

id

V

*

=

(

id

V

)

*

=

(

g

∘

f

)

*

=

f

*

∘

g

*

and this implies that

f

*

:

W

*

→

V

*

is a surjective linear map (on the other hand

g

*

:

V

*

→

W

*

is an injective linear map), so

dim

k

⁢

(

V

*

)

≤

dim

k

⁢

(

W

*

)

.

Now (due to previous

arguments

) we have

dim

k

⁢

(

V

*

⊗

W

*

)

=

dim

k

⁢

(

W

*

)

,

so

V

*

⊗

W

*

and

W

*

are isomorphic.

All in all, we have that

(

V

⊗

W

)

*

is isomorphic to

W

*

, which is isomorphic to

V

*

⊗

W

*

. This

completes

the proof.

□

Remarks

. We know that there is an isomorphism between

(

V

⊗

W

)

*

and

V

*

⊗

W

*

, but generally we know nothing about it, about its behaviour. Thus it is hard to find imprortant applications for this proposition. Also note, that this proposition is true for

free modules

over any

unital ring

.

Title

tensor product

of

dual spaces

is a dual space of tensor product

Canonical

name

TensorProductOfDualSpacesIsADualSpaceOfTensorProduct

Date of creation

2013-03-22 18:32:19

Last modified on

2013-03-22 18:32:19

Owner

joking (16130)

Last modified by

joking (16130)

Numerical id

18

Author

joking (16130)

Entry type

Theorem

Classification

msc 15A69

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