Let V and W be vector spaces (say over the reals). There is a linear injection V∗⊗W∗→(V⊗W)∗ which sends ∑ifi⊗gi∈V∗⊗W∗ to the unique functional in (V⊗W)∗ sending v⊗w↦∑ifi(v)⋅gi(w) for all (v,w)∈V×W. In the finite dimensional case it is easy to see this is an isomorphism by comparing dimensions on both sides. I’m aware that this is not an isomorphism in the infinite-dimensional case (see Relation with the dual space in the wikipedia article on tensor products) but I must admit that I’m not sure why. Cardinal arithmetic does not seem help here and, even if it did, I would be much happier to see an explicit example of a functional in (V⊗W)∗ outside the range of this map. I’m having trouble cooking one up myself.

Added:Let me elaborate on my comment about cardinal arithmetic. Suppose that X and Y areinfinite dimensionalvector spaces over a field k. I’m confident that dim(X⊗Y)=dimX⋅dimY holds. It is clear that |X∗|=|k|dimX since we may identify a functional on X with a function from a basis for X to k. Also I think I’ve convinced myself that if dimX≥|k| then in fact |X|=dimX ie. the cardinality and dimension of a vector space agree when the dimension is larger than the cardinality of the ground field. Consequently, assuming say that |k|≤dimX≤dimY it seems we have

dim(X⊗Y)∗=|k|dimX⋅dimY=|k|dimY=dimY∗=dimX∗⋅dimY∗=dim(X∗⊗Y∗)which implies that in fact (X⊗Y)∗ and X∗⊗Y∗

areisomorphic but, rather frustratingly, the obvious map does not do the job. Did I make a mistake here? If not, is it generally true (ie without making assumptions about |k|) that there exists some isomorphism (X⊗Y)∗→X∗⊗Y∗?

**Answer**

The map is not an isomorphism because an element in X∗⊗Y∗ is a finite

sum of functionals of the form x∗⊗y∗, where x∗∈X∗ and y∗∈Y∗. However, when X and Y are infinite dimensional, not every functional on X⊗Y will be of this form.

One case to consider, which will make this clear, is the case when Y=X∗.

There is one obvious element of (X⊗Y)∗, namely the evaluation map

which takes a tensor x⊗y to the value of the functional y on the vector x. Now one can check that this element is *not* in the image

of the map from X∗⊗Y∗.

I am going to take a little time to rewrite the preceding example in a different language, because I think that it helps illustrate what is going on.

First note that if Y=X∗, then

X⊗Y embeds into End(X)

(the space of linear operators from X to itself) as the space of finite rank linear operators (i.e. those whose image is finite dimensional). Denote

this image by FREnd(X)⊂End(X). Note that any element of FREnd(X)

has a well-defined trace (because even though the domain is infinite dimensional, the range is finite dimensional); in the tensor product

description, this is just the natural map from X⊗Y to the ground field

given by evaluation of the functionals in Y on the vectors in X.

(This is precisely the functional on X⊗Y that we considered above,

reexpressed in the language of operators.)

Replacing X by X∗ in the preceding paragraph, we see that

X∗⊗Y∗=FREnd(X∗). Note that there is a natural map

FREnd(X)→FREnd(X∗) given by mapping an endomorphism ϕ to its

transpose ϕt. (In terms of tensor products, this is the natural

map X⊗Y=X⊗X∗→X∗∗⊗X∗≅X∗⊗X∗∗=X∗⊗Y∗, the isomorphism being the canonical one which switches the

two factors.)

The map X∗⊗Y∗→(X⊗Y)∗

can then be reintrepreted as the pairing between

FREnd(X∗) and FREnd(X)∗

defined as follows:

for ϕ∈FREnd(X∗) and ψ∈FREnd(X),

⟨ϕ,ψ⟩:=trace(ϕ∘ψt).

And now we see why this map is not surjective: for example, if

ϕ is *any* endomorphism of X∗, i.e. any element of End(X∗),

then the composite ϕ∘ψt has finite rank (since ψt does),

and so trace(ϕ∘ψt) is defined.

Thus we in fact have an embedding of all of End(X∗) into FREnd(X)∗. With a little more work you can check that this latter embedding is an isomorphism. The conclusion in this case is that the embedding X∗⊗Y∗→(X⊗Y)∗ can be reinterpreted as the embedding

FREnd(X∗)→End(X∗),

which is not surjective when X∗ (or equivalently, X) is infinite dimensional, since it does not contain the identity map (for example).

Note that under the identification of End(X∗) with FREnd(X)∗,

the identity map is identified precisely with the trace on FREnd(X),

and so we get a reinterpretation of our original example, and see more clearly

what is going on: the point is that the identity endomorphism of an infinite dimensional vector space does not have finite rank.

**Attribution***Source : Link , Question Author : Mike F , Answer Author : jimmij*