tensor product and dual spaces

tensor product and dual spaces

Let

k

be a field and

V

be a

vector space

over

k

. Recall that

V

*

=

{

f

:

V

→

k

|

f

is linear

}

denotes the

dual space

of

V

(which is also a vector space over

k

).

Proposition

. Let

V

and

W

be vector spaces. Consider the map

ϕ

:

V

*

⊗

W

*

→

(

V

⊗

W

)

*

such that

ϕ

⁢

(

f

⊗

g

)

⁢

(

v

⊗

w

)

=

f

⁢

(

v

)

⁢

g

⁢

(

w

)

.

Then

ϕ

is a

monomorphism

. Moreover if one of the spaces

V

,

W

is finite dimensional, then

ϕ

is an

isomorphism

.

Proof

. One can easily check that

ϕ

is a well defined

linear map

, thus it is

sufficient

to show that

Ker

⁢

(

ϕ

)

=

0

. So assume that

F

∈

V

*

⊗

W

*

is such that

ϕ

⁢

(

F

)

=

0

. It is clear that

F

can be (uniquely) expressed in the form

F

=

∑

i

,

j

α

i

,

j

⁢

f

i

⊗

g

j

,

where

(

f

i

)

is a basis of

V

*

,

(

g

j

)

is a basis of

W

*

and

α

i

,

j

∈

k

. Then for any

v

∈

V

and

w

∈

W

we have:

0

=

ϕ

⁢

(

F

)

⁢

(

v

⊗

w

)

=

ϕ

⁢

(

∑

i

,

j

α

i

,

j

⁢

f

i

⊗

g

j

)

⁢

(

v

⊗

w

)

=

=

∑

i

,

j

α

i

,

j

⁢

ϕ

⁢

(

f

i

⊗

g

j

)

⁢

(

v

⊗

w

)

=

∑

i

,

j

α

i

,

j

⁢

f

i

⁢

(

v

)

⁢

g

j

⁢

(

w

)

.

Since

w

∈

W

is arbitrary then we can write this

equality

in the form:

0

=

∑

i

,

j

α

i

,

j

⁢

f

i

⁢

(

v

)

⁢

g

j

=

∑

j

(

∑

i

α

i

,

j

⁢

f

i

⁢

(

v

)

)

⁢

g

j

and since

(

g

j

)

are

linearly independent

we obtain that

∑

i

α

i

,

j

⁢

f

i

⁢

(

v

)

=

0

for all

j

. Again since

v

∈

V

was arbitrary we obtain that

∑

i

α

i

,

j

⁢

f

i

=

0

for all

j

. Now since

(

f

i

)

are linearly independent we obtain that

α

i

,

j

=

0

for all

i

,

j

. Thus

F

=

0

.

Now assume that

dim

k

⁢

V

=

q

<

+

∞

. Let

(

v

i

)

i

=

1

q

be a basis of

V

and let

(

v

i

*

)

i

=

1

q

be an induced basis of

V

*

. Moreover let

(

w

p

)

p

∈

P

be a basis of

W

. We wish to show that

ϕ

is onto, so let

f

:

V

⊗

W

→

k

be an element of

(

V

⊗

W

)

*

. Define

F

∈

V

*

⊗

W

*

by the

formula

:

F

=

∑

i

=

1

q

v

i

*

⊗

g

i

,

where

g

i

:

W

→

k

is such that

g

i

⁢

(

w

p

)

=

f

⁢

(

v

i

⊗

w

p

)

. Then for any

v

j

from

(

v

i

)

i

=

1

q

and for any

w

p

from

(

w

p

)

p

∈

P

we have:

ϕ

⁢

(

F

)

⁢

(

v

j

⊗

w

p

)

=

ϕ

⁢

(

∑

i

=

1

q

v

i

*

⊗

g

i

)

⁢

(

v

j

⊗

w

p

)

=

∑

i

=

1

q

ϕ

⁢

(

v

i

*

⊗

g

i

)

⁢

(

v

j

⊗

w

p

)

=

=

∑

i

=

1

q

v

i

*

⁢

(

v

j

)

⁢

g

i

⁢

(

w

p

)

=

g

j

⁢

(

w

p

)

=

f

⁢

(

v

j

⊗

w

p

)

and thus

ϕ

⁢

(

F

)

=

f

.

□

Remark

. The map

ϕ

from the previous proposition is very important in studying

algebras

and coalgebras (more precisly it is an essence in defining dual (co)algebras). Unfortunetly

ϕ

does not have to be an isomorphism in general. Nevertheless, the spaces

(

V

⊗

W

)

*

and

V

*

⊗

W

*

are always isomorphic (see this entry (

http://planetmath.org/TensorProductOfDualSpacesIsADualSpaceOfTensorProduct

) for more details).

Title

tensor product

and dual spaces

Canonical name

TensorProductAndDualSpaces

Date of creation

2013-03-22 18:31:51

Last modified on

2013-03-22 18:31:51

Owner

joking (16130)

Last modified by

joking (16130)

Numerical id

6

Author

joking (16130)

Entry type

Theorem

Classification

msc 15A69

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