tensor product of dual spaces is a dual space of tensor product
Proof. If or is finite dimensional, then there is an explicit isomorphism between and (see this entry (http://planetmath.org/TensorProductAndDualSpaces) for more details). So assume that both and are infinite dimensional.
First of all, note that if is a vector space, then denotes its dimension, that is is a cardinality of any basis of . Thus we can compare dimensions of spaces, so if and only if there is an injection from a basis of to a basis of (note that this relation is well defined, i.e. it does not depend on the choice of bases). One can easily show that if and only if there is an injective linear map from to and this is if and only if there is a surjective linear map from to . Therefore is isomorphic to if and only if (which here means that and ).
Without loss of generality, we may assume that (we can always compare any two sets). Note that the basis of is the product of bases of and (namely if is a basis of and a basis of , then is a basis of ). If are sets such that is infinite and there is an injection , then it is well known that there is a bijection from to . Thus (since is infinite dimensional) we have:
Therefore is isomorphic to , so is isomorphic to .
Now the inequality implies that . Indeed, assume that there is an injective linear map . Then there is a (surjective) linear map such that (see this entry (http://planetmath.org/SomeFactsAboutInjectiveAndSurjectiveLinearMaps) for more details). Therefore (since is a contravariant functor) we have that
and this implies that is a surjective linear map (on the other hand is an injective linear map), so .
Now (due to previous arguments) we have
so and are isomorphic.
All in all, we have that is isomorphic to , which is isomorphic to . This completes the proof.
Remarks. We know that there is an isomorphism between and , 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|
|Date of creation||2013-03-22 18:32:19|
|Last modified on||2013-03-22 18:32:19|
|Last modified by||joking (16130)|